(Baøi Giaûng Toùm Taét)€¦ · TRÖÔØNG ÑAÏI HOÏC ÑAØ LAÏT KHOA TOAÙN - TIN HOÏC Y Z...

TRÖÔØNG ÑAÏI HOÏC ÑAØ LAÏT KHOA TOAÙN - TIN HOÏC

PHAÏM TIEÁN SÔN

TOAÙN RÔØI RAÏC 1 (Baøi Giaûng Toùm Taét)

-- Löu haønh noäi boä -- Ñaø Laït 2008

Mu. c lu. c

MO.’ D- AU iv

1 TA. P HO..P VA ANH XA. 1

1.1 Ta.p ho..p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Khai nie.m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.2 Cac phep toan tren ta.p ho..p . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.3 Tıch Descartes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 Anh xa. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2.1 D- i.nh nghıa va tınh chat . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2.2 Anh xa. ha.n che . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.2.3 Ho..p cu’a cac anh xa. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.2.4 Anh xa. ngu.o..c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.2.5 Lu..c lu.o.

.ng cu’a mo.t ta.p ho..p . . . . . . . . . . . . . . . . . . . . . . . 12

2 LOGIC VA CAC PHU.O.NG PHAP CHU

.NG MINH 17

2.1 Me.nh de . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.2 Me.nh de co dieu kie.n va cac me.nh de tu.o.ng du.o.ng . . . . . . . . . . . . . . 20

2.3 Lu.o..ng hoa . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.4 Phu.o.ng phap chu.ng minh . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

i

2.5 Quy na.p toan ho.c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3 THUA. T TOAN 33

3.1 Mo.’ dau . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.1.1 Tım so lo.n nhat trong ba so . . . . . . . . . . . . . . . . . . . . . . . 33

3.1.2 Tım so lo.n nhat trong day hu.u ha.n cac so thu..c . . . . . . . . . . . . 33

3.2 Thua. t toan Euclid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.2.1 Thua. t toan Euclid . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.3 Thua. t toan de. quy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.3.1 Tınh n giai thu.a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.3.2 Tım u.o.c so chung lo.n nhat . . . . . . . . . . . . . . . . . . . . . . . 40

3.3.3 Thua. t toan xac di.nh day Fibonacci . . . . . . . . . . . . . . . . . . . 41

3.4 D- o. phu.c ta.p cu’a thua.t toan . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

3.5 Phan tıch thua.t toan Euclid . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

4 PHEP D- EM 51

4.1 Cac nguyen ly co. ba’n cu’a phep dem . . . . . . . . . . . . . . . . . . . . . . 51

4.1.1 Nguyen ly to’ng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.1.2 Nguyen ly tıch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4.1.3 Nguyen ly bao ham-loa. i tru. . . . . . . . . . . . . . . . . . . . . . . . 54

4.2 Hoan vi. va to’ ho..p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4.3 Cac thua.t toan sinh ra hoan vi. va to’ ho..p . . . . . . . . . . . . . . . . . . . . 62

4.4 Hoan vi. va to’ ho..p suy ro.ng . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

4.5 He. so cu’a nhi. thu.c va cac dong nhat thu.c . . . . . . . . . . . . . . . . . . . 73

4.6 Nguyen ly chuong chim bo cau . . . . . . . . . . . . . . . . . . . . . . . . . . 77

4.6.1 Nguyen ly chuong chim bo cau (da.ng thu. nhat) . . . . . . . . . . . . 77

ii

4.6.2 Nguyen ly chuong chim bo cau (da.ng thu. hai) . . . . . . . . . . . . . 78

4.6.3 Nguyen ly chuong chim bo cau (da.ng thu. ba) . . . . . . . . . . . . . 80

5 QUAN HE. 85

5.1 Quan he. hai ngoi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

5.2 Quan he. va ma tra.n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

5.3 Quan he. thu. tu.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

5.4 Quan he. tu.o.ng du.o.ng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

5.5 Bao dong cu’a quan he. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

5.6 Lattice cu’a cac phan hoa.ch . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

5.6.1 Thua.t toan xac di.nh ho. i cu’a hai phan hoa.ch . . . . . . . . . . . . . 118

5.6.2 Thua. t toan xac di.nh tuye’n cu’a hai phan hoa.ch . . . . . . . . . . . . 119

6 D- A. I SO BOOLE 123

6.1 Lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

6.2 Lattice phan bo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

6.3 D- a. i so Boole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

6.4 Ham Boole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

6.5 Bie’u dien cac ham Boole qua he. tuye’n, ho. i va phu’ di.nh . . . . . . . . . . . 149

6.6 Bie’u dien toi thie’u cu’a ham Boole . . . . . . . . . . . . . . . . . . . . . . . 152

6.6.1 Khai nie.m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

6.6.2 Phu.o.ng phap ba’n do Karnaugh . . . . . . . . . . . . . . . . . . . . . 153

7 MA TUYEN TINH 159

7.1 Mo.’ dau . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

7.1.1 Khai nie.m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

iii

7.1.2 Ma phat hie.n loi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

7.1.3 Ma su.’ a sai . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

7.2 Cac khai nie.m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

7.3 Khoa’ng cach Hamming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

7.4 Ho. i chu.ng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

7.4.1 Gia’i ma dung ba’ng chua’n . . . . . . . . . . . . . . . . . . . . . . . . 179

7.5 Ma hoan ha’o . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182

7.6 Ma Hamming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

Tai lie.u tham kha’o 189

iv

MO.’ D- AU

Toan ho.c ro.i ra.c la mo.t bo. pha.n cu’a Toan ho.c nham nghien cu.u cac doi tu.o.. ng ro.i ra. c:nghien cu.u cac cau truc ro.i ra. c khac nhau va cac phu.o.ng phap gia’i cac van de co lien quanden cac cau truc nay.

Thong tin lu.u tru. va va.n hanh trong may tınh du.o.i da.ng cac tın hie.u ro.i ra. c (cac maytınh lien tu.c chı’ la cac may tınh tu.o.ng tu.. , chuyen du.ng). Vı va.y cong cu. dung de’ bie’u dienthong tin trong may va xu.’ ly cac thong tin nay la Toan ho.c ro.i ra.c.

Ngoai ra, cac phu.o.ng phap va ket qua’ cu’a Toan ho.c ro.i ra. c co the’ dung de’ gia’i quyet tru.. ctiep nhieu van de da. t ra cu’a Tin ho.c nhu. logic, ham da. i so logic, to’ ho.. p tren tu.... Toanho.c ro.i ra.c chua’n bi. san va cung cap cac cong cu. , phu.o.ng phap lua.n de’ gia’i quyet nhieuvan de cu’a Tin ho.c. Co the’ noi Toan ho.c ro.i ra.c la nganh Toan ho.c co. so.’ cho Tin ho.c.

Mu.c dıch cu’a giao trınh nham cung cap mo.t so cong cu. Toan ho.c de’ bu.o.c dau di vaoTin ho.c. Giao trınh du.o.. c trınh bay mo.t cach dan tra’i ho.n la di sau vao mo.t van de cu. the’.Cuoi moi phan co cac bai ta.p nham cu’ng co nhu.ng kien thu.c da ho.c. Hy vo.ng rang giaotrınh nay dap u.ng du.o.. c phan nao yeu cau ho.c ta.p cu’a cac ba.n sinh vien.

D- a La. t, ngay 11 thang 2 nam 2008Pha.m Tien So.n

v

Chu.o.ng 1

TA. P HO.

. P VA ANH XA.

1.1 Ta.p ho.. p

1.1.1 Khai nie.m

Mo.t khai nie.m co. ba’n cu’a toan ho.c hie.n da. i la khai nie.m ta. p ho.. p.

Cung giong nhu. die’m, doa.n tha’ng, ma.t pha’ng, ... trong hınh ho.c Euclid, khai nie.m ta.pho.. p khong du.o.. c di.nh nghıa ma chı’ du.o.. c mo ta’ bang nhu.ng vı du. . Cha’ng ha.n, ta.p ho.. p cacsach trong thu. vie.n, ta.p ho.. p cac so thu.. c, ta.p ho.. p cac da thu.c ba. c hai, v.v...

Cac va. t ta.o nen mo.t ta.p ho.. p go. i la cac phan tu.’ cu’a ta.p ho.. p ay. Co hai cach xac di.nhmo.t ta.p ho.. p:

(a) Lie. t ke danh sach cac phan tu.’ cu’a no. Cha’ng ha.n, ta.p ho.. p gom cac phan tu.’ a, b, c, dthu.o.ng du.o.. c viet

{a, b, c, d}.

(b) Neu len tınh chat da. c tru.ng cu’a cac phan tu.’ cu’a ta. p ho.. p. Cha’ng ha.n, ta.p ho.. p {1, 3}co the’ mo ta’ la ta.p ho.. p hai so tu.. nhien le’ nho’ nhat hay ta.p ho.. p cac nghie.m cu’aphu.o.ng trınh ba.c hai x2 − 4x + 3 = 0.

Ky hie.u x ∈ A (va do.c la x thuo.c A) co nghıa x la phan tu.’ cu’a ta.p ho.. p A. Khi x khongpha’i la phan tu.’ cu’a ta.p ho.. p A ta viet x 6∈ A (va do. c la x khong thuo.c A). Cha’ng ha.n, neugo. i N la ta.p cac so tu.. nhien thı 7 ∈ N nhu.ng 12

56∈ N.

Chu y 1. (a) D- e’ do.n gia’n, doi khi ta chı’ dung tu. “ta.p” thay cho cu.m tu. “ta.p ho.. p”.

(b) Ky hie.u := thu.o.ng dung de’ du.a vao di.nh nghıa, no thay cho cu.m tu. “di.nh nghıa bo.’ i”.Cha’ng ha.n, N := {0, 1, 2, . . .}.

1

(c) Ta thu.o.ng dung ky hie.u | de’ dien da. t y “sao cho” (hoa. c “trong do”). Cha’ng ha.n, ta.pho.. p tat ca’ cac so tu.. nhien chan co the’ mo ta’ nhu. sau:

{n ∈ N | n chia het cho 2}.

Vı du. 1.1.1. Mo.t vai ta.p ho.. p so thu.o.ng ga.p:

(a) Ta.p ho.. p cac so tu.. nhien N := {0, 1, 2, . . .}.

(b) Ta.p ho.. p cac so nguyen du.o.ng P := {1, 2, . . .}.

(c) Ta.p ho.. p cac so nguyen Z := {0, 1,−1, 2,−2, . . .}.

(d) Ta.p ho.. p cac so hu.u tı’Q := {pq| p, q ∈ Z, q 6= 0}.

(e) Ta.p ho.. p cac so thu.. c R.

(f) Ta.p ho.. p cac so phu.c C := {a +√−1b | a, b ∈ R}.

Mo.t ta.p ho.. p khong co phan tu.’ nao ca’ go. i la ta.p ho.. p trong (hay rong) va du.o.. c ky hie.u la∅. Cha’ng ha.n ta.p ho.. p gom cac nghie.m so thu.. c cu’a phu.o.ng trınh ba. c hai x2 + 1 = 0 la mo.tta.p ho.. p trong.

Ta.p ho.. p B go. i la ta. p ho.. p con cu’a ta.p ho.. p A neu mo.i phan tu.’ cu’a ta.p ho.. p B deu laphan tu.’ cu’a ta.p ho.. p A; trong tru.o.ng ho.. p nay ta ky hie.u B ⊆ A hay A ⊇ B. Hie’n nhienA ⊆ A. Ho.n nu.a, de’ thua.n tie.n, ta thu.o.ng coi ta.p ho.. p trong la mo.t ta.p ho.. p con cu’a ta.pbat ky, tu.c la ∅ ⊆ A vo.i mo.i ta.p ho.. p A. Hai ta.p ho.. p A va B go. i la bang nhau neu B ⊆ Ava A ⊆ B; khi do ta viet A = B. Neu B ⊆ A nhu.ng A 6= B ta noi B la ta. p ho.. p con thu.. csu.. cu’a ta.p ho.. p A va viet B A.

Vı du. 1.1.2. (a) Neu

A := {x ∈ R | x2 + x − 6 = 0}, B := {2,−3}

thı A = B.

(b) Ta co cac bao ham thu.c thu.. c su.. sau

P N Z Q R.

Mo.t ta.p ho.. p ma phan tu.’ cu’a no la nhu.ng ta.p ho.. p thu.o.ng du.o.. c go. i la mo.t ho. cac ta.pho.. p, hoa. c mo.t he. cac ta.p ho.. p. Noi cach khac, “ta.p ho.. p”, “ho.”, “he.” la nhu.ng thua.t ngu.

dong nghıa.

D- e’ neu len danh sach cac ta.p ho.. p cu’a mo.t ho. ta.p ho.. p A, ta hay go. i moi ta.p ho.. p cu’a Ala Ai; ky hie.u i du.o.. c go. i la chı’ so de’ danh dau ta.p ho.. p ay, hai ta.p ho.. p khac nhau cu’a ho.

2

A du.o.. c danh dau bo.’ i hai chı’ so khac nhau. Neu I la ta.p ho.. p tat ca’ cac chı’ so da dung de’

danh dau cac ta.p ho.. p cu’a ho. A thı ta co the’ viet

A := {Ai | i ∈ I},

hay

A := {Ai}i∈I .

Cung co the’ su.’ du.ng phu.o.ng phap nay de’ danh dau tat ca’ cac phan tu.’ cu’a mo.t ta.p ho.. pA tuy y.

1.1.2 Cac phep toan tren ta.p ho.. p

Cho tru.o.c cac ta.p A va B ta co the’ thanh la.p cac ta.p mo.i bang cac phep toan sau:

D- i.nh nghıa 1.1.1. Ho.. p cu’a hai ta.p A va B la mo.t ta.p ho.. p, ky hie.u A∪B, gom tat ca’ cacphan tu.’ hoa. c thuo.c A hoa. c thuo.c B (hoa.c thuo.c ca’ hai).

Giao cu’a hai ta.p A va B la mo.t ta.p ho.. p, ky hie.u A ∩ B, gom tat ca’ cac phan tu.’ vu.athuo.c A vu.a thuo. c B.

Hie.u cu’a ta.p ho.. p A vo.i ta.p ho.. p B la mo.t ta.p ho.. p, ky hie.u A \ B, gom tat ca’ cac phantu.’ thuo.c A nhu.ng khong thuo.c B.

Hie.u doi xu.ng cu’a hai ta.p ho.. p A va B la ta.p ho.. p

A ∆ B := (A \ B) ∪ (B \ A).

Nha.n xet 1. (a) Mo. t cach tu.o.ng tu.. , co the’ di.nh nghıa ho.. p ∪i∈IAi va giao ∩i∈IAi cu’amo.t ho. ta.p ho.. p A := {Ai | i ∈ I}.

(b) Ta luon co A ∆ B = B ∆ A. Nhu.ng nhu. vı du. du.o.i day chı’ ra, noi chung A\B 6= B\A.

Vı du. 1.1.3. Gia’ su.’ A := {a, b, c, d} va B := {c, d, e}. Khi do

A ∪ B = {a, b, c, d, e},A ∩ B = {c, d},A \ B = {a, b},B \ A = {e},

A ∆ B = {a, b, e}.

Vı du. 1.1.4. Gia’ su.’ A (tu.o.ng u.ng, B) la ta.p nghie.m cu’a phu.o.ng trınh x2 − 3x + 2 = 0

3

(tu.o.ng u.ng, x2 − 4x + 3 = 0). Ta co A = {1, 2}, B = {1, 3} va

A ∪ B = {1, 2, 3},A ∩ B = {1},A \ B = {2},B \ A = {3},

A ∆ B = {2, 3}.

Ta.p nghie.m cu’a phu.o.ng trınh

(x2 − 3x + 2)(x2 − 4x + 3) = 0

la A ∪ B = {1, 2, 3}. Ta.p nghie.m cu’a he. hai phu.o.ng trınh

x2 − 3x + 2 = 0,

x2 − 4x + 3 = 0,

la A ∩ B = {1}.

Vı du. 1.1.5. Gia’ su.’

Ai := {i, i + 1, . . .}, i ∈ N.

Khi do ⋃

i∈N

Ai = N va⋂

i∈N

Ai = ∅.

Cac phep toan ho.. p va giao tren cac ta.p ho.. p co nhu.ng tınh chat sau:

Tınh chat 1.1.2. Tınh giao hoan

A ∪ B = B ∪ A,

A ∩ B = B ∩ A.

Tınh ket ho.. p

(A ∪ B) ∪ C = A ∪ (B ∪ C),

(A ∩ B) ∩ C = A ∩ (B ∩ C).

Tınh phan phoi

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C),

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).

Chu.ng minh. Bai ta.p. 2

4

Neu cac ta.p ho.. p A va B co giao bang trong, tu.c la neu A ∩ B = ∅, thı cac ta.p ho.. p naygo. i la khong co phan tu.’ chung, hoa. c la ro.i nhau.

Thu.o.ng cac ta.p ho.. p du.o.. c xet to.i trong cung mo.t van de deu la cac bo. pha.n cu’a mo.t ta.pho.. p X co di.nh nao do. Khi ay, ta.p ho.. p X nay go. i la “khong gian”. Hie.u X \A go. i la phanbu cu’a ta.p A va ky hie.u la Ac. Hie’n nhien A va Ac la ro.i nhau, A \ B = A ∩ Bc. Ho.n nu.a

Tınh chat 1.1.3. (Cong thu.c De Morgan) Gia’ su.’ {Ai}i∈I la ho. cac ta. p ho.. p con cu’a khonggian X. Khi do

(⋃

i∈I

Ai

)c

=⋂

i∈I

(Ai)c,

(⋂

i∈I

Ai

)c

=⋃

i∈I

(Ai)c.


D- i.nh nghıa 1.1.4. Ho. cac ta.p ho.. p A := {Ai | i ∈ I} go. i la phu’ cu’a ta.p X neu X = ∪i∈IAi.Neu ngoai ra Ai 6= ∅ vo.i mo.i i ∈ I va Ai ∩Aj = ∅ vo.i mo.i i, j ∈ I, i 6= j, thı ta noi A la mo.tphan hoa. ch cu’a ta.p X.

Vı du. 1.1.6. D- a.t A1 (tu.o.ng u.ng, A2) la ta.p cac so nguyen chan (tu.o.ng u.ng, le’). Khi do{A1, A2} la mo.t phan hoa.ch cu’a ta.p cac so nguyen Z.

1.1.3 Tıch Descartes

Tıch Descartes, hay van tat tıch, cu’a cac ta.p ho.. p Ai, i ∈ I, la mo.t ta.p ho.. p, ky hie.u la

∏

i∈I

Ai,

du.o.. c xac di.nh nhu. sau: tat ca’ cac phan tu.’ cu’a no co da.ng x := (xi)i∈I vo.i xi ∈ Ai. Khi do,xi go. i la thanh phan (hay to. a do. ) thu. i cu’a x.

Tıch cu’a mo.t so hu.u ha.n cac ta.p ho.. p Ai, i = 1, 2, . . . , n, thu.o.ng du.o.. c ky hie.u la

n∏

i=1

Ai hoa. c A1 × A2 × · · · × An.

Moi phan tu.’ cu’a tıch nay la mo.t vector (x1, x2, . . . , xn) vo.i xi ∈ Ai, i = 1, 2, . . . , n. Noi cachkhac

n∏

i=1

Ai = {(x1, x2, . . . , xn) | xi ∈ Ai, i = 1, 2, . . . , n}.

5

Neu A1 = A2 = · · · = An = A thı tıch A×A× · · · ×A (A co ma.t n lan) thu.o.ng du.o.. c kyhie.u la An.

Chu y rang, noi chung, A ×B 6= B × A. Dı nhien A × ∅ = ∅.

Vı du. 1.1.7. Gia’ su.’ A := {1, 2}, B = {a, b, c}. Khi do

A× B = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)},B × A = {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)},A× A = {(1, 1), (1, 2), (2, 1), (2, 2)}.

Bai ta.p

1. Gia’ su.’ X := {1, 2, . . . , 10}. D- a.t A := {1, 4, 7, 10}, B := {1, 2, 3, 4, 5} va C := {2, 4, 6, 8}.Lie.t ke cac phan tu.’ cu’a moi ta.p ho.. p sau:

(a) A ∪ B.

(b) B ∪ C.

(c) A ∩ B.

(d) B ∩ C.

(e) A \ B.

(f) Ac.

(g) (Bc ∩ (C \ A)).

(h) (A ∩ B)c ∪ C.

(i) B \ A.

(j) A ∩ (B ∪ C).

(k) ((A ∩ B) \ C).

(l) (A ∩ B) \ (C \ B).

2. Gia’ su.’ X := {1, 2, 3} va Y := {x, y}. Lie.t ke cac phan tu.’ cu’a moi ta.p ho.. p sau:

(a) X2.

(b) X × Y.

(c) Y × X.

(d) Y 3.

3. Lie.t ke tat ca’ cac phan hoa.ch cu’a cac ta.p ho.. p sau:

(a) {1}.(b) {1, 2}.

6

(c) {a, b, c}.(d) {a, b, c, d}.

4. Xac di.nh moi quan he. giu.a cac ca.p ta.p ho.. p sau:

(a) {1, 2, 3} va {1, 3, 2}.(b) {1, 2, 2, 3} va {1, 2, 3}.(c) {1, 1, 3} va {3, 3, 1}.(d) {x ∈ R | x2 + x = 2} va {1,−2}.(e) {x ∈ R | 0 < x ≤ 2} va {1, 2}.

5. Ky hie.u P(X) la ta.p ho.. p ma cac phan tu.’ cu’a no la cac ta.p con cu’a X. Lie.t ke tat ca’

cac phan tu.’ cu’a P({a, b}) va P({a, b, c}).

6. Gia’ su.’ X co 10 phan tu.’ . Co bao nhieu ta.p ho.. p con thu.. c su.. cu’a ta.p ho.. p X? To’ngquat?

7. Gia’ su.’ X va Y la cac ta.p ho.. p khac trong sao cho X × Y = Y ×X. Cac ta.p ho.. p X vaY pha’i tho’a nhu.ng dieu kie.n gı?

8. Chu.ng minh hoa. c cho pha’n vı du. cac quan he. (A,B,C la nhu.ng ta.p ho.. p con cu’a ta.pho.. p X) sau:

(a) A ∩ (B \ C) = (A ∩ B) \ (A ∩ C).

(b) (A \ B) ∩ (B \ A) = ∅.(c) A \ (B ∪ C) = (A \ B) ∪ C.

(d) (A \ B)c = (B \ A)c.

(e) (A ∩ B)c ⊆ A.

(f) (A ∩ B) ∪ (B \ A) = A.

(g) A × (B ∪ C) = (A × B) ∪ (A× C).

(h) (A× B)c = Ac ×Bc.

9. D- a’ng thu.c nao du.o.i day la dung?

(a) A ∩ B = A.

(b) A ∪ B = A.

(c) (A ∩ B)c = Bc.

10. Tım hie.u doi xu.ng cu’a hai ta.p ho.. p A := {1, 2, 3} va B := {2, 3, 4, 5}.

11. Gia’ su.’ C la mo.t du.o.ng tron va A la ta.p tat ca’ cac du.o.ng kınh cu’a du.o.ng tron C.Xac di.nh ∩A∈AA.

7

12. Ky hie.u P la ta.p ho.. p tat ca’ cac so nguyen lo.n ho.n 1. Vo.i moi so tu.. nhien i ≥ 2, da. t

Ai := {ik | k ≥ 2, k ∈ P}.

Mo ta’ ta.p ho.. p P \⋃∞

i=2 Ai.

13. Chu.ng minh cac da’ng thu.c sau (gia’ su.’ cac ta.p du o.. c xet deu la ta.p con cu’a ta.p Xnao do):

A ∩ (A1 ∪ A2 ∪ · · · ∪ An) = (A ∩ A1) ∪ (A ∩ A2) ∪ · · · ∪ (A ∩ An).

(A1 ∩ A2 ∩ · · · ∩ An)c = Ac1 ∪ Ac

2 ∪ · · · ∪ Acn.

1.2 Anh xa.

1.2.1 D- i.nh nghıa va tınh chat

Mo.t khai nie.m co. ba’n khac cu’a toan ho.c hie.n da. i la khai nie.m anh xa. , mo.’ ro.ng khai nie.mham so.

D- i.nh nghıa 1.2.1. Cho X va Y la hai ta.p ho.. p bat ky. Mo.t anh xa. (hay ham so) tu. ta.pho.. p X vao ta.p ho.. p Y la mo.t tu.o.ng u.ng moi phan tu.’ cu’a X mo.t phan tu.’ xac di.nh cu’a Y.

Gia’ su.’ f la mo.t anh xa. tu. ta.p ho.. p X vao ta.p ho.. p Y. Khi do ta viet f : X → Y ; neux ∈ X thı f(x) chı’ phan tu.’ cu’a Y tu.o.ng u.ng vo.i phan tu.’ x do va ta viet x 7→ f(x); phantu.’ f(x) go. i la a’nh cu’a phan tu.’ x qua anh xa. f, hay la gia tri. cu’a ham f ta. i x. Ta.p ho.. p

{(x, y) ∈ X × Y | y = f(x)}

go. i la do thi. cu’a anh xa. f va ky hie.u la graph(f).

Vı du. 1.2.1. Tu.o.ng u.ng moi so thu.. c x vo.i mo.t so thu.. c x3 cho ta mo.t anh xa. f : R →R, x 7→ x3.

Cho tru.o.c mo.t ta.p ho.. p A ⊆ X thı ta.p ho.. p

f(A) := {f(x) | x ∈ A}

go. i la a’nh cu’a ta.p ho.. p A qua anh xa. f. D- a.c bie.t, ta.p ho.. p f(X) go. i la mien gia tri. cu’a f.

De dang chu.ng minh rang:

Tınh chat 1.2.2. Gia’ su.’ f : X → Y la mo. t anh xa. tu. ta. p ho.. p X vao ta. p ho.. p Y. Khi do

(a) Neu A ⊂ B ⊂ X thı f(A) ⊂ f(B).

8

(b) Neu Ai, i ∈ I, la mo. t ho. cac ta. p ho.. p con cu’a ta. p ho.. p X thı

f

(⋃

i∈I

Ai

)=

⋃

i∈I

f (Ai) ,

f

(⋂

i∈I

Ai

)⊂

⋂

i∈I

f (Ai) .

D- e’ y rang, noi chung da’ng thu.c sau

f

(⋂

i∈I

Ai

)=⋂

i∈I

f (Ai)

khong dung.

D- i.nh nghıa 1.2.3. Gia’ su.’ f : X → Y la mo.t anh xa. tu. ta.p ho.. p X vao ta.p ho.. p Y.

(a) Anh xa. f go. i la mo. t-mo. t (hoa. c do.n anh) neu vo.i mo.i x, x′ ∈ X ma x 6= x′ thıf(x) 6= f(x′).

(b) f go. i la anh xa. len (hoa. c toan anh) neu f(X) = Y.

(c) f go. i la mo. t-mo. t len (hoa. c song anh) neu f dong tho.i la mo.t-mo. t va la len; noi cachkhac, vo.i moi phan tu.’ y ∈ Y co duy nhat mo.t phan tu.’ x ∈ X sao cho f(x) = y.

Vı du. 1.2.2. (a) Anh xa.f : R→ R, x 7→ sin x,

la mo.t-mo.t nhu.ng khong la anh xa. len.

(b) Anh xa.1

g : R→ N, x 7→ [x],

la len nhu.ng khong la anh xa. mo.t-mo.t.

(c) Anh xa.h : R→ R, x 7→ x3,

la mo.t-mo.t va len.

Vo.i mo.t anh xa. tuy y f : X → Y va vo.i mo.t ta.p ho.. p B ⊆ Y, ta.p ho.. p

{x ∈ X | f(x) ∈ B}

go. i la nghi.ch a’nh cu’a ta.p ho.. p B qua anh xa. f va du.o.. c ky hie.u la f−1(B). Ro rang f−1(Y ) =X va f−1(∅) = ∅, nhu.ng co the’ xa’y ra rang ∅ 6= B ⊂ Y va f−1(B) = ∅.

1Phan nguyen cu’a so thu.. c x, ky hie.u [x], la so nguyen lo.n nhat khong vu.o.. t qua x.

9

Neu ta.p ho.. p B ⊂ Y chı’ gom co mo.t phan tu.’ y, tu.c la B = {y}, thı thay cho ky hie.uf−1({y}) ta thu.o.ng ky hie.u van tat la f−1(y).

De dang chu.ng minh rang:

Tınh chat 1.2.4. Gia’ su.’ f : X → Y la mo. t anh xa. tu. ta. p ho.. p X vao ta. p ho.. p Y. Khi do

(a) Neu B ⊂ C ⊂ Y thı f−1(B) ⊂ f−1(C).

(b) Neu Bi, i ∈ I, la mo. t ho. cac ta. p ho.. p con cu’a ta. p ho.. p Y thı

f−1

(⋃

i∈I

Bi

)=

⋃

i∈I

f−1 (Bi) ,

f−1

(⋂

i∈I

Bi

)=

⋂

i∈I

f−1 (Bi) .

(c) Neu B,C la hai ta. p ho.. p con cu’a ta. p ho.. p Y thı

f−1 (B \ C) = f−1 (B) \ f−1(C).

D- a. c bie. tf−1 (Y \ B) = X \ f−1(B).

(d) Vo.i mo. i ta. p ho.. p con B ⊂ Y ta deu co

f [f−1(B)] ⊆ B.

(e) Vo.i mo. i ta. p ho.. p con A ⊂ X ta deu co

f−1[f(A)] ⊇ A.

D- e’ y rang cac da’ng thu.c

f−1[f(A)] = A va f [f−1(B)] = B

noi chung khong dung.

1.2.2 Anh xa. ha.n che

Gia’ su.’ f : X → Y la mo.t anh xa. tu. ta.p ho.. p X vao ta.p ho.. p Y va gia’ su.’ Z la mo.t ta.p ho.. pcon cu’a X. Anh xa.

f |Z : Z → Y

xac di.nh bo.’ if |Z(x) = f(x), x ∈ Z,

du.o.. c go. i la ha. n che cu’a f len Z, con anh xa. f du.o.. c go. i la thac trie’n cu’a f |Z len X. Hie’nnhien

10

(a) Neu f la mo.t-mo.t thı f |Z cung la mo.t-mo.t .

(b) Vo.i mo.i ta.p ho.. p con B cu’a Y ta deu co

(f |Z)−1(B) = f−1(B) ∩ Z.

1.2.3 Ho.. p cu’a cac anh xa.

Gia’ su.’ X,Y va Z la ba ta.p ho.. p va ta co cac anh xa.

f : X → Y, g : Y → Z.

Khi do co the’ thiet la.p anh xa.

g ◦ f : X → Z, x 7→ g[f(x)].

Anh xa. g ◦ f du.o.. c go. i la ho.. p cu’a cac anh xa. f va g.

Vı du. 1.2.3. Cho hai anh xa.

f : R→ R, x 7→ x2,

g : R→ R, y 7→ y − 1.

Ta co anh xa. ho.. p

g ◦ f : R→ R, x 7→ x2 − 1.

Tu. di.nh nghıa de dang suy ra

Tınh chat 1.2.5. Cho hai anh xa.

f : X → Y, g : Y → Z.

(a) Neu f va g la mo. t-mo. t (tu.o.ng u.ng, len, mo. t-mo. t len) thı anh xa. ho.. p g ◦ f cung lamo. t-mo. t (tu.o.ng u.ng, len, mo. t-mo. t len).

(b) Vo.i mo. i ta. p ho.. p con A cu’a X ta deu co

(g ◦ f)(A) = g[f(A)].

(c) Vo.i mo. i ta. p ho.. p con C cu’a Z ta deu co

(g ◦ f)−1(C) = f−1[g−1(C)].

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1.2.4 Anh xa. ngu.o.. c

Gia’ su.’

f : X → Y

la anh xa. mo.t-mo.t len. Khi do vo.i moi phan tu.’ y ∈ Y ton ta. i duy nhat mo. t phan tu.’ x ∈ Xsao cho f(x) = y, va bo.’ i va.y f−1(y) = {x}. Do do ta co the’ thiet la.p mo.t anh xa.

g : Y → X

xac di.nh bo.’ i cong thu.c: vo.i mo.i y ∈ Y,

g(y) = x neu f(x) = y.

Anh xa. g go. i la anh xa. ngu.o.. c cu’a f va ky hie.u la f−1. Hie’n nhien f−1 : Y → X la anh xa.mo.t-mo.t len va (f−1)−1 = f.

Vı du. 1.2.4. (a) Anh xa. dong nhat

idX : X → X, x 7→ x,

la anh xa. mo.t-mo. t len va (idX)−1 = idX .

(b) Anh xa. mo.t-mo.t va lenf : R→ R, x 7→ x3,

co anh xa. ngu.o.. c la

f−1 : R→ R, y 7→ y13 .

Tu. di.nh nghıa de dang suy ra

Tınh chat 1.2.6. (a) Gia’ su.’ f : X → Y la anh xa. mo. t-mo. t len. Khi do

f−1 ◦ f = idX , f ◦ f−1 = idY .

(b) Neu f : X → Y, g : Y → Z la nhu.ng anh xa. mo. t-mo. t len, thı anh xa. ho.. p (g◦f) : X → Zcung mo. t-mo. t len va

(g ◦ f)−1 = f−1 ◦ g−1.

1.2.5 Lu.. c lu.o.. ng cu’a mo. t ta.p ho.. p

D- i.nh nghıa 1.2.7. (a) Hai ta.p ho.. p A va B go. i la co cung lu.. c lu.o.. ng neu ton ta. i anh xa.mo.t-mo.t va len f : A → B.

(b) Ta.p ho.. p trong, ta.p ho.. p {x1, x2, . . . , xn} va cac ta.p ho.. p cung lu.. c lu.o.. ng vo.i no go. i lata. p ho.. p hu.u ha. n.

12

(c) Ta.p ho.. p cac so tu.. nhien N va cac ta.p ho.. p cung lu.. c lu.o.. ng vo.i no go. i la ta. p ho.. p demdu.o.. c.

(d) Ta.p ho.. p cac so thu.. c R va cac ta.p ho.. p cung lu.. c lu.o.. ng vo.i no go. i la ta. p ho.. p khong demdu.o.. c.

(e) Ta.p ho.. p A go. i la khong qua dem du.o.. c, neu A la mo.t ta.p ho.. p hu.u ha.n (va co the’ latrong), hoa. c neu A la mo.t ta.p ho.. p dem du.o.. c.

Gia’ su.’ A := {x1, x2, . . . , xn} la mo.t ta.p ho.. p hu.u ha.n khac trong sao cho xi 6= xj vo.i mo.ii 6= j. Khi do ta noi ta.p ho.. p A co n phan tu.’ va ky hie.u #A := n. Ta.p ho.. p trong ∅ khongco phan tu.’ nao ca’, vı va.y da. t #∅ := 0. Neu A la ta.p ho.. p khac trong va khong pha’i ta.p ho.. phu.u ha.n, da. t #A := +∞.

Bai ta.p

1. Gia’ su.’ X := {1, 2, 3}, Y := {a, b, c, d}, Z := {w, x, y, z}. Xet cac anh xa. f : X → Y vag : Y → Z cho bo.’ i

f(1) = b, f(2) = c, f(3) = a,

g(a) = x, g(b) = x, g(c) = z, g(d) = w.

Xac di.nh anh xa. ho.. p f ◦ g.

2. Gia’ su.’ f : X → N, x 7→ x2, vo.i X := {−5,−4, . . . , 4, 5}. f la anh xa. mo.t-mo. t? f laanh xa. len?

3. Co bao nhieu anh xa. tu. ta.p {a, b} vao ta.p {1, 2}. Nhu.ng anh xa. nao la mo.t-mo.t?Nhu.ng anh xa. nao la len?

4. Gia’ su.’ X := {a, b, c} va f : X → X cho bo.’ i

f(a) = b, f(b) = a, f(c) = b.

D- i.nh nghıa day cac anh xa. fn : X → X,n = 1, 2, . . . , bo.’ i f1 := f va fn := fn−1 ◦ fvo.i mo.i n ≥ 2. Hay xac di.nh cac anh xa. f2, f3, f9, f789.

5. Gia’ su.’ X := {0, 1, 2, 3, 4} va anh xa. f : X → X xac di.nh bo.’ i

f(x) := 4x mod 5.

f la anh xa. mo.t-mo.t? f la anh xa. len?

6. Gia’ su.’ m,n la cac so nguyen du.o.ng. Gia’ su.’ X := {0, 1, 2, . . . ,m − 1}. Xet anh xa.f : X → X cho bo.’ i

f(x) := nx mod m.

Tım nhu.ng dieu kie.n cu’a m va n de’ f la anh xa. mo.t-mo. t va len?

13

7. Cho cac anh xa. f : X → Y va g : Y → Z. Chu.ng minh hoa. c cho pha’n vı du. cac phatbie’u sau:

(a) Neu g la mo.t-mo.t thı g ◦ f la mo.t-mo.t.

(b) Neu f va g la len thı g ◦ f la len.

(c) Neu f va g la mo.t-mo.t va len thı g ◦ f la mo.t-mo. t va len.

(d) Neu g ◦ f la mo.t-mo.t thı f la mo.t-mo.t.

(e) Neu g ◦ f la mo.t-mo.t thı g la mo. t-mo.t.

(f) Neu g ◦ f la len thı f la len.

(g) Neu g ◦ f la len thı g la len.

8. Gia’ su.’ X := {1, 2, 3} va Y := {a, b, c, d}. Xet anh xa. f : X → Y cho bo.’ i

f(1) = a, f(2) = c, f(3) = c.

Xac di.nh cac ta.p ho.. p sau: f({1}), f({1, 3}), f−1({a}) va f−1({a, c}).

9. Cho anh xa. f : X → Y. Chu.ng minh f la mo. t-mo.t neu va chı’ neu

f(A ∩ B) = f(A) ∩ f(B)

vo.i mo.i ta.p con A va B cu’a X.

10. Cho anh xa. f : X → Y. Chu.ng minh rang ho. cac ta.p ho.. p

A := {f−1({y}) | y ∈ Y }

la mo.t phan hoa.ch cu’a ta.p ho.. p X.

11. Cho anh xa. g : X → Y. Chu.ng minh rang g la mo.t-mo. t neu va chı’ neu vo.i mo.i anh xa.mo.t-mo.t f : A → X (A la ta.p ho.. p bat ky) thı anh xa. ho.. p g ◦ f : A → Y la mo.t-mo.t.

12. Cho anh xa. f : X → Y. Chu.ng minh rang f la len neu va chı’ neu vo.i mo.i anh xa. leng : Y → Z (Z la ta.p ho.. p bat ky) thı anh xa. ho.. p g ◦ f : X → Z la len.

13. A la ta.p ho.. p con cu’a ta.p ho.. p X. D- i.nh nghıa ham da. c tru.ng cu’a ta.p ho.. p A (trong X)nhu. sau:

χA(x) :=

{1 neu x ∈ A,

0 neu x 6∈ A.

(a) Chu.ng minh vo.i mo.i x ∈ X ta co cac quan he. sau

χA∩B(x) = χA(x)χB(x),

χA∪B(x) = χA(x) + χB(x)− χA∩B(x),

χAc(x) = 1 − χA(x),

χA\B(x) = χA(x)[1 − χB(x)].

14

(b) Chu.ng minh neu A ⊆ B thı χA(x) ≤ χB(x) vo.i mo.i x ∈ X.

(c) Chu.ng minh χA∪B(x) = χA(x)+ χB(x) vo.i mo.i x ∈ X neu va chı’ neu A∩B = ∅.(d) Tım cong thu.c lien quan den anh xa. χA ∆ B.

14. Xet anh xa. f tu. P(X) vao ta.p ho.. p cac ham da.c tru.ng trong X di.nh nghıa bo.’ i

f(A) := χA.

Chu.ng minh f la mo.t-mo.t va len.

15. Chu.ng minh ta.p ho.. p cac so tu.. nhien N va ta.p cac so tu.. nhien chan 2N la cung lu.. clu.o.. ng.

16. Chu.ng minh ta.p ho.. p khac trong X khong cung lu.. c lu.o.. ng vo.i P(X).

17. Gia’ su.’ X := {0, 1}. Lie.t ke tat ca’ cac chuoi do. dai 2 tren X. Lie.t ke tat ca’ cac chuoido. dai ≤ 2 tren X.

18. Chuoi s go. i la chuoi con cu’a chuoi t neu ton ta. i cac chuoi u, v sao cho t = usv. Lie.tke tat ca’ cac chuoi con cu’a chuoi babc.

19. Chu.ng minh hoa. c cho pha’n vı du. cac phat bie’u sau doi vo.i tat ca’ cac so thu.. c2:

(a) dx + 7e = dxe + 7.

(b) dx + ye = dxe + dye.(c) bx + yc = bxc + dye.

20. Gia’ su.’ n la so nguyen le’. Chu.ng minh cac da’ng thu.c sau

⌊n2

4

⌋=

(n − 1

2

)(n + 1

2

),

⌈n2

4

⌉=

n2 + 3

4.

21. Chu.ng minh rang#(A ∪ B) = #A + #B − #(A ∩ B).

2dxe la so nguyen nho’ nhat lo.n ho.n hoa.c bang x; bxc la so nguyen lo.n nhat nho’ ho.n hoa.c bang x.

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Chu.o.ng 2

LOGIC VA CAC PHU.O.NG PHAP

CHU.NG MINH

2.1 Me.nh de

Mo.t me.nh de toan ho.c co the’ xem la mo.t kha’ng di.nh toan ho. c chı’ co the’ dung hoa. c sai,khong the’ nha.p nhang, nghıa la khong the’ vu.a dung vu.a sai, cung khong the’ vu.a khongdung vu.a khong sai.

Vı du. 2.1.1. Cac phat bie’u sau la cac me.nh de:

(a) Trai dat co da.ng hınh cau.

(b) Vie.t Nam la nu.o.c co so dan dong nhat the gio.i.

(c) 2 + 2 = 4.

(d) 4 la mo.t so du.o.ng va 3 la mo.t so am.

Vı du. 2.1.2. Cac phat bie’u sau khong pha’i la me.nh de:

(a) Hom nay tro.i mu.a.

(b) Xin hay giup do. toi.

(c) x − y = y − x.

(d) x− 3 = 5.

17

Ta thu.o.ng dung cac ky tu.. in thu.o.ng, cha’ng ha.n p, q va r de’ bie’u dien mo.t me.nh de. D- e’

do.n gia’n, chung ta cung ky hie.up : 1 + 1 = 3

de’ di.nh nghıa p la me.nh de 1 + 1 = 3.

D- i.nh nghıa 2.1.1. Gia’ su.’ p va q la cac me.nh de. Ho. i cu’a p va q, ky hie.u la p∧ q, la me.nhde

p va q.

Tuye’n cu’a p va q, ky hie.u la p ∨ q, la me.nh de

p hoa. c q.

Vı du. 2.1.3. Gia’ su.’

p : 1 + 1 = 3,

q : mo.t tha.p ky’ la 10 nam.

Khi do ho. i cu’a p va q la me.nh de

p ∧ q : 1 + 1 = 3 va mo.t tha.p ky’ la 10 nam,

va tuye’n cu’a p va q la me.nh de

p ∨ q : 1 + 1 = 3 hoa. c mo.t tha.p ky’ la 10 nam.

D- i.nh nghıa 2.1.2. Gia tri. cu’a me.nh de p ∧ q du.o.. c cho bo.’ i ba’ng chan tri.

p q p ∧ qT T TT F FF T FF F F

trong do ky hie.u T la dung va F la sai.

Vı du. 2.1.4. Gia’ su.’

p : 1 + 1 = 3,

q : Mo.t tha.p ky’ la 10 nam.

Ta co p la sai va q la dung. Vı va.y ho. i cu’a p va q la me.nh de sai.

D- i.nh nghıa 2.1.3. Gia tri. cu’a me.nh de p ∨ q du.o.. c cho bo.’ i ba’ng chan tri.

p q p ∨ qT T TT F TF T TF F F

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Vı du. 2.1.5. Gia’ su.’

p : 1 + 1 = 3,

q : Mo.t tha.p ky’ la 10 nam.

Ta co p la sai va q la dung. Vı va.y tuye’n cu’a p va q la me.nh de dung.

D- i.nh nghıa 2.1.4. Phu’ di.nh cu’a me.nh de p, ky hie.u p hay p′, la me.nh de

khong pha’i p.

Gia tri. cu’a me.nh de p du.o.. c cho bo.’ i ba’ng chan tri.

p pT FF T

Vı du. 2.1.6. Gia’ su.’

p : π la so hu.u tı’.

Ta co me.nh de p la sai va do va.y phu’ di.nh cu’a no p la dung.

Bai ta.p

1. Gia tri. cu’a cac me.nh de p, q va R tuo.ng u.ng la F, T va F. Xac di.nh gia tri. cu’a cacme.nh de sau:

(a) (p ∨ q) ∨ p.

(b) (p ∨ q) ∧ p.

(c) (p ∧ q) ∧ p.

(d) (p ∧ q) ∨ (p ∨ q).

(e) (p ∧ q) ∨ (r ∧ p).

(f) (p ∨ q) ∧ (p ∨ q) ∧ (p ∨ q) ∧ (p ∨ q).

(g) (p ∨ q) ∨ (q ∨ r).

2. Cho cac me.nh de saup : 5 < 9, q : 9 < 7 va 5 < 7.

Xac di.nh tınh dung sai cu’a cac me.nh de sau

(a) 5 < 9 va 9 < 7.

(b) Phu’ di.nh cu’a me.nh de (5 < 9 va 9 < 7).

(c) 5 < 9 hoa. c phu’ di.nh cu’a me.nh de (9 < 7 va 5 < 7).

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2.2 Me.nh de co dieu kie.n va cac me.nh de tu.o.ng du.o.ng

D- i.nh nghıa 2.2.1. Gia’ su.’ p va q la hai me.nh de. Khi do phat bie’u

neu p thı q

go. i la me.nh de co dieu kie. n va ky hie.u la

p → q.

Me.nh de p go. i la gia’ thiet va me.nh de q go. i la ket lua. n (hay he. qua’).

D- i.nh nghıa 2.2.2. Ba’ng gia tri. cu’a me.nh de co dieu kie.n p → q di.nh nghıa nhu. sau:

p q p → qT T TT F FF T TF F T

Vı du. 2.2.1. Gia’ su.’

p : 1 > 2,

q : 3 < 7.

Ta co p la sai va q la dung. Do do p → q la dung va q → p la sai.

D- i.nh nghıa 2.2.3. Gia’ su.’ p va q la hai me.nh de. Khi do phat bie’u

p neu va chı’ neu q

go. i la me.nh de neu va chı’ neu va du.o.. c ky hie.u la

p ↔ q.

Ba’ng gia tri. cu’a me.nh de p ↔ q du.o.. c di.nh nghıa nhu. sau:

p q p ↔ qT T TT F FF T FF F T

Me.nh de “p neu va chı’ neu q” con du.o.. c dien da. t da.ng “dieu kie.n can va du’ de’ p la q”.

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Vı du. 2.2.2. Cau

1 < 5 neu va chı’ neu 2 < 8

co the’ viet du.o.i da.ng

p ↔ q,

trong do

p : 1 < 5, q : 2 < 8.

Ta co p va q la dung. Do do p ↔ q la dung.

D- i.nh nghıa 2.2.4. Gia’ su.’ P va Q la hai me.nh de du.o.. c xay du.. ng tu. cac me.nh dep1, p2, . . . , pn. Ta noi P tu.o.ng du.o.ng Q va viet

P ≡ Q

neu vo.i mo.i gia tri. cu’a p1, p2, . . . , pn ta co P va Q hoa. c dong tho.i dung, hoa.c dong tho.i sai.

Vı du. 2.2.3. Ta co cong thu.c De Morgan:

p ∨ q ≡ p ∧ q, p ∧ q ≡ p ∨ q.

Vı du. 2.2.4. Ta co

p → q ≡ p ∧ q,

p ↔ q ≡ (p → q) ∧ (q → p).

D- i.nh nghıa 2.2.5. Me.nh de q → p go. i la pha’n da’o cu’a me.nh de p → q.

Vı du. 2.2.5. Gia’ su.’

p : 1 < 4, q : 5 > 8.

Khi do

p → q : neu 1 < 4 thı 5 > 8,

q → p : neu 5 > 8 thı 1 < 4,

q → p : neu 5 khong lo.n ho.n 8 thı 1 khong lo.n ho.n 4.

Ta co p → q la sai. Nen q → p la dung va q → p la sai.

D- i.nh ly 2.2.6. Me.nh de p → q tu.o.ng du.o.ng vo.i me.nh de pha’n da’o cu’a no. Tu.c la

p → q ≡ q → p

Chu.ng minh. Chu.ng minh suy tru.. c tiep tu. ba’ng chan tri. cu’a cac me.nh de p → q va q → p.2

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Bai ta.p

1. Xac di.nh gia tri. cu’a cac me.nh de sau neu gia tri. cu’a cac me.nh de p, q, r, s tu.o.ng u.ngla F, T, F, T :

(a) p → q.

(b) p → q.

(c) p → q.

(d) (p → q) ∧ (q → r).

(e) (p → q) → r.

(f) p → (q → r).

(g) (s → (p ∧ r)) ∧ ((p → (r ∨ q)) ∧ s).

(h) ((p ∧ q) → (q ∧ r)) → (s ∨ q).

2. Cho cac me.nh dep : 4 < 2, q : 7 < 10, r : 6 < 6.

Viet cac phat bie’u du.o.i day da.ng ky hie.u

(a) Neu 4 < 2 thı 7 < 10.

(b) Neu (4 < 2 va 6 < 6) thı 7 < 10.

(c) Neu (6 < 6 va 7 khong nho’ ho.n 10) khong dung thı 6 < 6.

(d) 7 < 10 neu va chı’ neu (4 < 2 va 6 khong nho’ ho.n 6).

3. Vo.i cac phat bie’u du.o.i day, hay viet moi me.nh de va phu’ di.nh cu’a no da.ng ky hie.u.Tım gia tri. cu’a moi me.nh de.

(a) Neu 4 < 6 thı 9 > 12.

(b) Neu 4 > 6 thı 9 > 12.

(c) |1| < 3 neu −3 < 1 < 3.

(d) |4| < 3 neu −3 < 4 < 3.

4. P ≡ Q la dung hay sai neu

(a) P = p,Q = p ∨ q.

(b) P = p ∧ q,Q = p ∨ q.

(c) P = p → q,Q = p ∨ q.

(d) P = p ∧ (q ∨ r), Q = p ∨ (q ∧ r).

(e) P = p ∧ (q ∨ r), Q = (p ∨ q) ∧ (p ∨ r).

(f) P = p → q,Q = q → p.

(g) P = p → q,Q = p ↔ q.

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(h) P = (p → q) ∧ (q → r), Q = p → r.

(i) P = (p → q) → r,Q = p → (q → r).

(k) P = (s → (p ∧ r)) ∧ ((p → (r ∨ q)) ∧ s), Q = p ∨ t.

5. Xet me.nh de p ⊕1 q cho bo.’ i ba’ng gia tri.

p q p ⊕1 qT T TT F FF T FF F T

Chu.ng minh rang

p ⊕1 q = q ⊕1 p.

6. Xet me.nh de p ⊕2 q cho bo.’ i ba’ng gia tri.

p q p ⊕2 qT T TT F FF T TF F F

(a) Chu.ng minh rang

(p ⊕2 q) ∧ (q ⊕2 p) 6≡ p ↔ q.

(b) Chu.ng minh

(p ⊕2 q) ∧ (q ⊕2 p) ≡ p ↔ q

neu ta thay ⊕2 sao cho neu p la sai va q la dung thı p ⊕2 q la sai.

7. Chu.ng minh rang

(p → q) ≡ (p ∨ q).

2.3 Lu.o.. ng hoa

Logic nghien cu.u cac me.nh de trong nhu.ng tiet tru.o.c khong du’ de’ dien ta’ hau het cacme.nh de trong toan ho.c cung nhu. khoa ho.c may tınh. Cha’ng ha.n, xet:

p : n la mo.t so nguyen le’.

D- i.nh nghıa 2.3.1. Cho X la mo.t ta.p ho.. p. P (x) la mo.t phat bie’u lien quan den bienx ∈ X. Ta noi P la ham me.nh de neu vo.i moi x ∈ X thı P (x) la mo.t me.nh de.

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Vı du. 2.3.1. Gia’ su.’ P la ta.p cac so nguyen du.o.ng va vo.i moi n ∈ P da. t

P (n) : n la mo.t so nguyen le’.

Khi do P la ham me.nh de tren P.

D- i.nh nghıa 2.3.2. Gia’ su.’ P la ham me.nh de tren ta.p X. Phat bie’u

vo.i mo.i x, P (x)

go. i la lu.o.. ng hoa pho’ ca. p. Ky hie.u ∀ nghıa la “vo.i mo.i”. Vı va.y phat bie’u

vo.i mo.i x, P (x)

co the’ viet la. i

∀x, P (x).

Ky hie.u ∀ go. i la lu.o.. ng hoa pho’ ca. p.

Phat bie’u

vo.i mo.i x, P (x)

la dung neu P (x) dung vo.i mo.i x ∈ X. Phat bie’u nay la sai neu co ıt nhat mo.t x ∈ X saocho P (x) sai.

Phat bie’u

ton ta. i x, P (x)

go. i la lu.o.. ng hoa ton ta. i. Ky hie.u ∃ nghıa la “ton ta. i”. Vı va.y phat bie’u

ton ta. i x, P (x)

co the’ viet la. i

∃x, P (x).

Ky hie.u ∃ go. i la lu.o.. ng hoa ton ta. i.

Phat bie’u

ton ta. i x, P (x)

la dung neu P (x) dung vo.i ıt nhat mo.t x ∈ X. Phat bie’u nay la sai neu vo.i mo.i x ∈ X deuco P (x) sai.

Vı du. 2.3.2. Phat bie’u

vo.i mo.i so thu.. c x thı x2 ≥ 0

la lu.o.. ng hoa pho’ ca.p va la mo.t kha’ng di.nh dung.

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Vı du. 2.3.3. Phat bie’u lu.o.. ng hoa pho’ ca.p

vo.i mo.i so thu.. c x thı x2 − 1 > 0

la sai vı vo.i x = 1 ta co

12 − 1 > 0

la me.nh de sai.

Vı du. 2.3.4. Phat bie’u lu.o.. ng hoa ton ta. i

ton ta. i so nguyen x de’ x2 − 4 = 0

la dung vı ta co the’ tım du.o.. c ıt nhat mo.t so nguyen x sao cho

x2 − 4 = 0.

Cha’ng ha.n, vo.i x = 2 ta co me.nh de dung:

22 − 4 = 0.

Vı du. 2.3.5. De dang chu.ng minh phat bie’u lu.o.. ng hoa ton ta. i sau la sai:

ton ta. i so thu.. c x de’1

x2 + 1> 1.

Nha.n xet 2. Giu.a cac lu.o..ng hoa pho’ ca.p va ton ta. i co lien he. sau day:

(a) Khong (∃x) P (x) ⇔ (∀x) khong P (x). Tu.c la phu’ di.nh cu’a me.nh de “co ton ta. i mo.tx sao cho P (x)” la “vo.i mo.i x deu khong co P (x)”.

(b) Khong (∀x) P (x) ⇔ (∃x) khong P (x). Tu.c la phu’ di.nh cu’a me.nh de “vo.i mo.i x deuco P (x)” la “co ton ta. i mo.t x sao cho khong co P (x)”.

D- i.nh ly 2.3.3. Gia’ su.’ P la ham me.nh de. Khi do ca. p cac me.nh de (a) va (b) sau hoa. cdong tho.i dung, hoa. c dong tho.i sai:

(a) ∀x, P (x); ∃x, P (x).

(b) ∃x, P (x); ∀x, P (x).


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Bai ta.p

1. Gia’ su.’ P (n) la ham me.nh de “n la u.o.c so cu’a 77”. Kie’m tra tınh dung sai cu’a

(a) P (11).

(b) P (1).

(c) P (3).

(d) P (n) vo.i mo.i so tu.. nhien n.

(e) Ton ta. i so tu.. nhien n sao cho P (n).

2. Xac di.nh gia tri. cu’a cac phat bie’u du.o.i day (xet tren ta.p ho.. p cac so thu.. c R):

(a) x2 > x vo.i mo.i x.

(b) Ton ta. i x sao cho x2 > x.

(c) Vo.i mo.i x vo.i x > 1 thı x2 > x.

(d) Ton ta. i x vo.i x > 1 sao cho x2 > x.

(e) Vo.i mo.i x vo.i x > 1 thı xx2+1

< 13.

(f) Ton ta. i x vo.i x > 1 thı xx2+1

< 13.

(g) Vo.i mo.i x va vo.i mo.i y ma x < y ta co x2 < y2.

(h) Vo.i mo.i x, ton ta. i y vo.i x < y ta co x2 < y2.

(i) Ton ta. i x sao cho vo.i mo.i y ma x < y thı x2 < y2.

(j) Ton ta. i x, ton ta. i y vo.i x < y sao cho x2 < y2.

3. Viet phu’ di.nh cu’a cac phat bie’u trong bai ta.p tren.

2.4 Phu.o.ng phap chu.ng minh

Mo.t he. thong toan ho. c gom cac tien de, di.nh nghıa, va cac thanh phan khong xac di.nh.

• Tien de du.o.. c gia’ thiet la dung.

• D- i.nh nghıa du.o.. c su.’ du.ng de’ xay du.. ng cac khai nie.m mo.i tu. cac khai nie.m da co.

• Mo.t so thanh phan khong du.o.. c di.nh nghıa mo.t cach tu.o.ng minh ma du.o.. c xac di.nhbo.’ i cac tien de.

Tu. he. thong toan ho.c ta co the’ dan den:

• D- i.nh ly la mo.t me.nh de da du.o.. c chu.ng minh la dung.

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• Bo’ de la mo.t di.nh ly khong quan tro.ng lam va du.o.. c su.’ du.ng de’ chu.ng minh mo.tdi.nh ly khac.

• He. qua’ la mo.t di.nh ly du.o.. c suy ra de dang tu. mo.t di.nh ly khac.

• Chu.ng minh la mo.t ly lua.n chı’ ra tınh dung cu’a mo.t di.nh ly.

• Logic la mo.t cong cu. de’ phan tıch cac chu.ng minh.

Vı du. 2.4.1. Hınh ho.c Euclid la mo.t he. toan ho.c. Mo.t so tien de:

• Ton ta. i mo.t va chı’ mo.t du.o.ng tha’ng di qua hai die’m phan bie.t cho tru.o.c.

• Ton ta. i mo.t va chı’ mo.t du.o.ng tha’ng di qua mo.t die’m va song song vo.i mo.t du.o.ngtha’ng (khong chu.a die’m) cho tru.o.c.

D- ie’m va du.o.ng tha’ng la cac thanh phan khong xac di.nh va du.o.. c di.nh nghıa a’n trongcac tien de.

Mo.t so di.nh nghıa:

• Hai tam giac la bang nhau neu co the’ sap xep cac dı’nh thanh nhu.ng ca.p sao cho cacca.nh va cac goc tu.o.ng u.ng la bang nhau.

• Hai goc la bu nhau neu to’ng cu’a chung bang 1800.

Mo.t so di.nh ly:

• Neu hai ca.nh cu’a mo.t tam giac bang nhau thı cac goc doi die.n bang nhau.

• Neu hai du.o.ng cheo cu’a tu. giac cat nhau ta. i cac trung die’m cu’a chung thı tu. giac lahınh bınh hanh.

Tu. di.nh ly thu. nhat suy ra he. qua’ sau:

• Tam giac co ba ca.nh bang nhau thı co cac goc bang nhau.

Vı du. 2.4.2. Ta.p cac so thu.. c R la mo.t he. toan ho.c. Mo.t so tien de:

• Vo.i mo.i x, y ∈ R ta co xy = yx.

• Ton ta. i mo.t ta.p con P ⊂ R sao cho

(a) Neu x, y thuo.c P thı x + y va xy thuo.c P.

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(b) Vo.i mo.i x ∈ R thı mo.t va chı’ mo.t trong cac dieu sau dung:

x ∈ P, x = 0, −x ∈ P.

Phep toan nhan du.o.. c di.nh nghıa a’n trong tien de thu. nhat.

Mo.t so di.nh nghıa:

• Cac phan tu.’ thuo. c P go. i la cac so thu.. c du.o.ng.

• Gia tri. tuye. t doi |x| cu’a so thu.. c x du.o.. c di.nh nghıa la x neu x du.o.ng hoa. c bang 0 vabang −x neu ngu.o.. c la. i.

Mo.t so di.nh ly:

• x · 0 = 0 vo.i mo.i x ∈ R.

• vo.i mo.i x, y, z ∈ R neu x ≤ y va y ≤ z thı x ≤ z.

Mo.t vı du. ve bo’ de:

• neu n la so nguyen du.o.ng thı hoa. c n − 1 la so nguyen du.o.ng hoa. c n − 1 = 0.

D- i.nh ly thu.o.ng co da.ng:

Vo.i mo.i x1, x2, . . . , xn,neu p(x1, x2, . . . , xn) thı q(x1, x2, . . . , xn). (2.1)

D- i.nh nghıa 2.4.1. Chu.ng minh tru.. c tiep cu’a di.nh ly (2.1) co da.ng: Gia’ su.’ p(x1, x2, . . . , xn)dung; su.’ du.ng p(x1, x2, . . . , xn) cung nhu. cac tien de, cac di.nh nghıa, cac di.nh ly da co de’

suy ra q(x1, x2, . . . , xn) la dung.

Vı du. 2.4.3. Chu.ng minh tru.. c tiep kha’ng di.nh sau: vo.i mo. i so thu.. c d, d1, d2 va x ta co:

neu d = min{d1, d2} va x ≤ d thı x ≤ d1 va x ≤ d2.

D- i.nh nghıa 2.4.2. Chu.ng minh pha’n chu.ng (hay chu.ng minh gian tiep) cu’a di.nh ly (2.1)co da.ng: Gia’ su.’ p(x1, x2, . . . , xn) dung va q(x1, x2, . . . , xn) sai; su.’ du.ng p, q cung nhu. cactien de, cac di.nh nghıa, cac di.nh ly da co de’ suy ra mo.t mau thuan. Mo.t mau thuan lame.nh de co da.ng r ∧ r (r la me.nh de nao do).

Tınh dung cu’a chu.ng minh pha’n chu.ng suy tru.. c tiep tu. su.. kie.n sau (ta. i sao):

p → q ≡ p ∧ q → r ∧ r.

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Vı du. 2.4.4. Chu.ng minh bang pha’n chu.ng kha’ng di.nh sau: vo.i mo. i so thu.. c x va y, neux + y ≥ 2 thı hoa. c x ≥ 1 hoa. c y ≥ 1.

D- i.nh nghıa 2.4.3. Day cac me.nh de du.o.. c viet da.ng

p1

p2...pn

∴ q

hay p1, p2, . . . , pn /∴ q go. i la mo.t ly lua. n. Cac me.nh de p1, p2, . . . , pn go. i la cac gia’ thiet vame.nh de q go. i la ket lua. n. Ly lua.n la ho.. p le. neu p1 va p2 va · · · va pn dong tho.i dung thıq cung dung; ngu.o.. c la. i ly lua.n go. i la khong ho.. p le. (hay sai).

Vı du. 2.4.5. Chu.ng minh ly lua.n sau la ho.. p le.:

p → qp

∴ q

Vı du. 2.4.6. Ly lua.n sau khong ho.. p le.:

Neu 2 = 3 thı toi an cai mu nay.Toi an cai mu nay.

∴ 2 = 3

Bai ta.p

1. Cho mo.t vı du. ve tien de, di.nh nghıa va di.nh ly cu’a hınh ho.c Euclid.

2. Cho mo.t vı du. ve tien de, di.nh nghıa va di.nh ly cu’a he. cac so thu.. c.

3. Gia’ su.’ ta da co cac di.nh ly sau: vo.i mo.i a, b, c ∈ R thı b+0 = b; a(b+ c) = ab+ ac; vaneu a + b = a + c thı b = c. Hay kie’m tra cac bu.o.c chu.ng minh tru.. c tiep cu’a kha’ngdi.nh: “x · 0 = 0 vo.i mo.i so thu.. c x.”

Chu.ng minh. (x · 0) + 0 = x · 0 = x · (0 + 0) = x · 0 + x · 0; do do x · 0 = 0. 2

4. Gia’ su.’ da co di.nh ly sau: vo.i mo.i a, b, c ∈ R, neu ab = ac va a 6= 0 thı b = c. Haykie’m tra cac bu.o.c chu.ng minh bang pha’n chu.ng cu’a kha’ng di.nh: “neu x · y = 0 thıhoa. c x = 0 hoa. c y = 0.”

Chu.ng minh. Gia’ su.’ x · 0 = 0 va x 6= 0, y 6= 0. Tu. xy = 0 = x · 0 va n 6= 0 ta co y = 0ma la mo.t mau thuan. 2

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5. Chu.ng minh bang pha’n chu.ng kha’ng di.nh sau: “neu da. t 100 qua’ bong vao trong 9ho.p thı co ıt nhat mo.t ho.p chu.a ıt nhat 12 qua’ bong”.

6. Viet cac ly lua.n sau du.o.i da.ng ky hie.u va xac di.nh tınh dung-sai:

(a)

Neu toi ho.c ta.p cham chı’ thı toi se da.t die’m tot.Toi ho.c ta.p cham chı’.∴ Toi se da. t die’m tot.

(b)

Neu toi ho.c ta.p cham chı’ thı toi se da.t die’m tot.Neu toi khong giau co thı toi se khong da.t die’m tot.∴ Toi giau co

(c)Toi ho.c ta.p cham chı’ neu va chı’ neu toi giau co.Toi giau co.∴ Toi ho.c ta.p cham chı’.

(d)

Neu toi ho.c ta.p cham chı’ hoa. c toi giau co thı toi se da.t die’m tot.Toi se da.t die’m tot.∴ Neu toi khong ho.c ta.p cham chı’ thı toi se giau co.

(e)

Neu toi ho.c ta.p cham chı’ thı hoa. c toi giau co hoa. c toi se da. t die’m tot.Toi khong da.t die’m tot va toi khong giau co.∴ Toi khong ho.c ta.p cham chı’.

7. Gia’ su.’p : Co 64K bo. nho. thı tot ho.n khong co bo. nho..q : Toi se mua bo. nho. mo.i.r : Toi se mua mo.t may tınh mo.i.

Hay viet cac ly lua.n du.o.i day da.ng cau va xac di.nh tınh dung-sai cu’a cac ly lua.n.

(a)

p → rp → q

∴ p → (r ∧ q)

8. Chu.ng minh rang neup1, p2/ ∴ p

vap, p3, . . . , pn/ ∴ c

la nhu.ng ly lua.n ho.. p le. thı ly lua.n sau cung ho.. p le.:

p1, p2, . . . , pn/ ∴ c

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9. Bınh lua.n ve ly lua.n sau

Co dıa mem thı tot ho.n khong co gı.Khong co gı thı tot ho.n co mo.t dıa cu.ng.∴ Co dıa mem thı tot ho.n co mo.t dıa cu.ng.

2.5 Quy na.p toan ho.c

D- i.nh nghıa 2.5.1. Gia’ su.’ vo.i moi so nguyen du.o.ng n ta co mo.t phat bie’u S(n) sao cho

Bu.o.c co. ba’n: S(1) dung;

Bu.o.c quy na. p: neu S(i) dung vo.i mo.i i = 1, 2, . . . , n, thı S(n + 1) dung.

Khi do S(n) dung vo.i mo.i n nguyen du.o.ng.

Vı du. 2.5.1. Ta con! ≥ 2n−1 vo.i n = 1, 2, . . . .

Tha. t va.y, kha’ng di.nh dung vo.i n = 1 vı

1! = 1 ≥ 1 = 21−1.

Bay gio. gia’ su.’ rang n ≥ 2 va

i! ≥ 2i−1 vo.i i = 1, 2, . . . , n.

Khi do

(n + 1)! = (n + 1)(n!)

≥ (n + 1)2n−1

≥ 2 · 2n−1 = 2(n+1)−1.

Theo nguyen ly quy na.p, n! ≥ 2n−1 vo.i mo.i n nguyen du.o.ng.

Cho X la mo.t ta.p ho.. p. Ky hie.u P(X) (hoa. c 2X) la ho. cac ta.p ho.. p con (thu.. c su.. hoa. ckhong) cu’a X. Ta co

D- i.nh ly 2.5.2. Neu ta. p ho.. p hu.u ha. n X gom n phan tu.’ thı

#P(X) = 2n.

Chu.ng minh. Su.’ du.ng quy na.p toan ho.c. 2

31

Bai ta.p

1. Dung quy na.p toan ho.c, chu.ng minh cac da’ng thu.c sau vo.i mo.i n nguyen du.o.ng:

(a) 1 + 3 + 5 + · · · + (2n − 1) = n2.

(b) 1 · 2 + 2 · 3 + 3 · 4 + · · · + n(n + 1) = n(n+1)(n+2)3

.

(c) 1(1!) + 2(2!) + · · · + n(n!) = (n + 1)! − 1.

(d) 12 + 22 + 32 + · · · + n2 = n(n+1)(2n+1)6

.

(e) 12 − 22 + 32 − · · · + (−1)n+1n2 = (−1)n+1n(n+1)2

.

(f) 13 + 23 + 33 + · · · + n3 =[

n(n+1)2

]2.

(g) 11·3 + 1

3·5 + 15·7 + · · · + 1

(2n−1)(2n+1)= n

2n+1.

(h) 122−1

+ 132−1

+ · · · + 1(n+1)2−1

= 34− 1

2(n+1)− 1

2(n+2).

(i) cos x + cos 2x + · · · + cos nx = cos[(x/2)(n+1)]sin(nx/2)sin(x/2)

neu sin(x/2) 6= 0.

(j) 1 sin x + 2 sin 2x + · · · + n sin nx = sin[(n+1)x]

4 sin2(x/2)− (n+1) cos( 2n+1

2x)

2 sin(x/2)neu sin(x/2) 6= 0.

2. Dung quy na.p toan ho.c, chu.ng minh cac bat da’ng thu.c sau

(a) 12n

≤ 1·3·5···(2n−1)2·4·6···(2n)

, vo.i n = 1, 2, · · · .

(b) 1√n+1

≥ 1·3·5···(2n−1)2·4·6···(2n)

, vo.i n = 1, 2, · · · .

(c) 2n + 1 ≤ 2n vo.i n = 3, 4, . . . .

(d) 2n ≥ n2 vo.i n = 4, 5, . . . .

(e) (a1a2 . . . a2n)1/2n ≤ a1+a2+···+a2n

2n vo.i n = 1, 2, . . . , va cac so khong am ai.

(f) (1 + x)n ≥ 1 + nx vo.i x ≥ −1 va n = 1, 2, . . . .

3. Dung quy na.p toan ho.c, chu.ng minh cac kha’ng di.nh sau:

(a) 7n − 1 chia het cho 6 vo.i mo.i n = 1, 2, . . . .

(b) 11n − 6 chia het cho 5 vo.i mo.i n = 1, 2, . . . .

(c) 6 · 7n − 2 · 3n chia het cho 4 vo.i mo.i n = 1, 2, . . . .

(d) 3n + 7n − 2 chia het cho 8 vo.i mo.i n = 1, 2, . . . .

4. Dung quy na.p toan ho.c, chu.ng minh rang n du.o.ng tha’ng trong ma.t pha’ng chia ma.tpha’ng thanh (n2 + n − 2)/2 vung. Gia’ su.’ hai du.o.ng tha’ng bat ky khong song songva khong co ba du.o.ng tha’ng cat nhau ta. i mo.t die’m.

32

Chu.o.ng 3

THUA. T TOAN

3.1 Mo.’ dau

Co the’ di.nh nghıa thua.t toan theo nhieu cach khac nhau. Chung ta se khong trınh bay cha.tche ve thua.t toan nhu. trong cac giao trınh logic, ma se hie’u khai nie.m thua.t toan theo cachthong thu.o.ng nhat.

Co the’ xem thua. t toan la mo. t quy tac de’, vo.i nhu.ng du. lie. u ban dau da cho, tım du.o.. clo.i gia’i cu’a bai toan dang xet sau mo. t khoa’ng tho.i gian hu.u ha. n.

D- e’ minh ho.a cach ghi mo.t thua.t toan, cung nhu. tım hie’u nhu.ng yeu cau de ra cho thua.ttoan, ta xet tren cac vı du. cu. the’ sau day.

3.1.1 Tım so lo.n nhat trong ba so

Thua.t toan nay tım so lo.n nhat trong ba so thu.. c a, b va c.

Vao: a, b va c.Ra: x la so lo.n nhat trong ba so a, b, c.

Bu.o.c 1. Neu a > b thı da. t x := a; ngu.o.. c la. i, da. t x := b.

Bu.o.c 2. Neu c > x thı da. t x := c.

3.1.2 Tım so lo.n nhat trong day hu.u ha.n cac so thu.. c

Thua.t toan nay tım so lo.n nhat trong day hu.u ha.n cac so thu.. c s1, s2, . . . , sn.

33

Vao: day hu.u ha.n cac so thu.. c s1, s2, . . . , sn.Ra: x = max{si | i = 1, 2, . . . , n}.

Bu.o.c 1. D- a.t x := s1.

Bu.o.c 2. Vo.i i := 2 den n thu.. c hie.n Bu.o.c 3.

Bu.o.c 3. Neu si > x thı gan x := si.

Tren day ta da ghi mo.t thua. t toan bang ngon ngu. thong thu.o.ng. Trong tru.o.ng ho.. p thua.ttoan du.o.. c viet bang ngon ngu. cu’a may tınh, ta co mo.t chu.o.ng trınh.

D- e’ ket thuc, chung ta hay tha’o lua.n them mo.t vai tınh chat cu’a cac thua.t toan.

Mo.t thua. t toan la mo.t ta.p ho.. p cac chı’ thi. co nhu.ng da. c tru.ng sau:

• Tınh chınh xac. Cac bu.o.c du.o.. c phat bie’u mo.t cach chınh xac.

• Tınh duy nhat. Cac ket qua’ trung gian trong moi bu.o.c thu.. c hie.n du.o.. c xac di.nh mo.tcach duy nhat va chı’ phu. thuo.c vao du. lie.u du.a vao va cac ket qua’ cu’a bu.o.c tru.o.c.

• Tınh hu.u ha.n. Thua.t toan du.ng sau hu.u ha.n bu.o.c.

• D- au vao. Thua.t toan co du. lie.u vao.

• D- au ra. Thua.t toan co du. lie.u ra.

• Tınh to’ng quat. Thua. t toan thu.. c hie.n tren mo.t ta.p cac du. lie.u vao.

Ngoai nhu.ng yeu to ke’ tren, ta con pha’i xet den tınh hie.u qua’ cu’a thua. t toan. Co ratnhieu thua. t toan, ve ma.t ly thuyet la ket thuc sau hu.u ha.n bu.o.c, tuy nhien tho.i gian “hu.uha.n” do vu.o.. t qua kha’ nang lam vie.c cu’a chung ta. Nhu.ng thua. t toan do se khong du.o.. c xeto.’ day, vı chung ta chı’ quan tam nhu.ng thua. t toan co the’ su.’ du.ng thu.. c su.. tren may tınh.

Cung do mu.c tieu noi tren, ta con pha’i chu y den do. phu.c ta.p cu’a cac thua.t toan. D- o.phu.c ta.p cu’a mo.t thua.t toan co the’ do bang khong gian, tu.c la dung lu.o.. ng bo. nho. cu’a maytınh can thiet de’ thu.. c hie.n thua. t toan, va bang tho.i gian, tu.c la tho.i gian may tınh lamvie. c.

Bai ta.p

1. Viet thua.t toan tım gia tri. nho’ nhat cu’a day

s1, s2, . . . , sn.

34

2. Viet thua.t toan tım vi. trı dau tien cu’a phan tu.’ lo.n nhat trong day

s1, s2, . . . , sn.

Cha’ng ha.n, vi. trı dau tien cu’a phan tu.’ lo.n nhat trong day

6.2, 8.9, 4.2, 8.9

la 2.

3. Viet thua.t toan tım vi. trı sau cung cu’a phan tu.’ lo.n nhat trong day

s1, s2, . . . , sn.

Cha’ng ha.n, vi. trı sau cung cu’a phan tu.’ lo.n nhat trong day

6.2, 8.9, 4.2, 8.9

la 4.

4. Viet thua.t toan da’o ngu.o.. c vi. trı cu’a day

s1, s2, . . . , sn.

5. Viet thua.t toan co.ng hai so nguyen du.o.ng.

6. Viet thua.t toan nhan hai so nguyen du.o.ng.

7. Viet thua.t toan kie’m tra tınh doi xu.ng cu’a ma tra.n vuong.

8. Viet thua.t toan kie’m tra tınh pha’n doi xu.ng cu’a ma tra.n vuong.

3.2 Thua. t toan Euclid

Phan nay trınh bay thua.t toan Euclid tım u.o.c so chung lo.n nhat cu’a hai so nguyen. U.o.c

so chung lo.n nhat cu’a hai so nguyen n va m (khong dong tho.i bang khong) la so nguyendu.o.ng lo.n nhat va la u.o.c so cu’a m va n. Cha’ng ha.n, u.o.c so chung lo.n nhat cu’a 4 va 6 la2 va u.o.c so chung lo.n nhat cu’a 3 va 8 la 1.

Neu a, b, q ∈ Z, b 6= 0, sao cho a = bq, ta noi a chia het cho b va ky hie.u b | a; trong tru.o.ngho.. p nay, ta noi q la thu.o.ng va b la u.o.c so cu’a a. Neu a khong chia het cho b, ta viet b - a.

Vı du. 3.2.1. Vı 21 = 3 · 7 nen 3 | 21. Thu.o.ng la 7.

D- i.nh nghıa 3.2.1. Gia’ su.’ n va m la hai so nguyen khong dong tho.i bang khong. So nguyenx go. i la u.o.c so chung cu’a m va n neu x la u.o.c so cu’a m va n. So nguyen

USCLN(m,n) := max{x | x la u.o.c so chung cu’a m va n}

go. i la u.o.c so chung lo.n nhat.

35

Vı du. 3.2.2. Cac u.o.c so nguyen du.o.ng cu’a so 30 la

1, 2, 3, 5, 6, 10, 15, 30

va cac u.o.c so nguyen du.o.ng cu’a so 105 la

1, 3, 5, 7, 15, 21, 35, 105;

do va.y cac u.o.c so chung du.o.ng cu’a 30 va 105 la

1, 3, 5, 15.

Suy ra u.o.c so chung lo.n nhat cu’a 30 va 105 la USCLN(30, 105) = 15.

D- i.nh ly 3.2.2. Gia’ su.’ m,n va c la cac so nguyen. Khi do

(a) Neu c la u.o.c so chung cu’a m va n thı

c | (m + n).

(b) Neu c la u.o.c so chung cu’a m va n thı

c | (m − n).

(c) Neu c | m thı c | mn.


Tınh chat 3.2.3. Gia’ su.’ a, b ∈ N, b > 0. Khi do ton ta. i cac so nguyen q va r sao cho

a = bq + r, 0 ≤ r < b, q ≥ 0.



22 = 7 × 3 + 1,

24 = 8 × 3 + 0,

103 = 21 × 4 + 19,

0 = 47 × 0 + 0.

D- i.nh ly 3.2.4. Cho a, b ∈ N, b > 0. Gia’ su.’ q va r la cac so nguyen sao cho

a = bq + r, 0 ≤ r < b, q ≥ 0.

Khi doUSCLN(a, b) = USCLN(b, r).

36



105 = 30 × 3 + 15.

Suy raUSCLN(105, 30) = USCLN(30, 15).

La. i co

30 = 15 × 2 + 0.

NenUSCLN(105, 30) = USCLN(30, 15) = USCLN(15, 0) = 15.

3.2.1 Thua. t toan Euclid

Thua.t toan nay tım u.o.c so chung lo.n nhat cu’a hai so tu.. nhien a va b, trong do a, b khongdong tho.i bang 0.

Vao: a, b la so tu.. nhien khong dong tho.i bang 0.Ra: USCLN la u.o.c so chung lo.n nhat cu’a a va b.

Bu.o.c 1. Neu a < b thı hoan do’i a va b.

Bu.o.c 2. Neu b = 0 thı thu.. c hie.n USCLN := a va du.ng.

Bu.o.c 3. Chia a cho b va nha.n du.o.. c a = bq + r vo.i 0 ≤ r < b.

Bu.o.c 4. Thu.. c hie.n a := b, b := r va chuye’n den Bu.o.c 2.

Vı du. 3.2.5. Ta se ap du.ng thua.t toan Euclid de’ tınh USCLN(504, 396).

D- a.t a := 504, b := 396. Vı a > b nen ta chuye’n den Bu.o.c 2. Vı b 6= 0 nen chuye’n denBu.o.c 3. Thu.. c hie.n Bu.o.c 3 ta co

504 = 396 × 1 + 108.

Ke tiep ta thu.. c hie.n Bu.o.c 4: da. t a := 396, b := 108 va chuye’n den Bu.o.c 2.

Vı b 6= 0 nen thu.. c hie.n Bu.o.c 3:

396 = 108 × 3 + 72.

Thu.. c hie.n Bu.o.c 4: da. t a := 108, b := 72 va chuye’n den Bu.o.c 2.

37


108 = 72 × 1 + 36.



72 = 36 × 2 + 0.


Vı b = 0 nen ap du.ng Bu.o.c 2 co USCLN := a = 36 va thua.t toan du.ng. Va.yUSCLN(504, 396) = 36.

Bai ta.p

1. Tım cac so nguyen q va r sao cho a = bq + r vo.i 0 ≤ r < b va

(a) a := 45, b := 6.

(b) a := 106, b := 12.

(c) a := 66, b := 11.

(d) a := 221, b := 17.

(e) a := 0, b := 31.

2. Dung thua.t toan Euclid de’ tım u.o.c so chung lo.n nhat cu’a ca.p cac so nguyen:

(60, 90), (110, 273), (220, 1400), (315, 825), (20, 40).

3. Gia’ su.’ a, b, c la cac so nguyen du.o.ng. Chu.ng minh neu a | b va b | c thı a | c.

4. Gia’ su.’ a, b la cac so nguyen du.o.ng. Chu.ng minh USCLN(a, b) = USCLN(a, a + b).

5. Gia’ su.’ a, b la cac so nguyen du.o.ng va p la so nguyen to. Chu.ng minh neu p | ab thıhoa. c p | a hoa.c p | b.

6. Tım cac so nguyen du.o.ng a, b, c sao cho a | bc, a - b va a - c.

7. Gia’ su.’ a > b ≥ 0. Chu.ng minh rang

USCLN(a, b) = USCLN(a − b, b).

8. Hay viet mo.t thua.t toan tım u.o.c so chung lo.n nhat cu’a hai so nguyen khong dongtho.i bang khong su.’ du.ng phep toan tru. thay cho phep toan chia.

38

3.3 Thua. t toan de. quy

Thua. t toan de. quy la mo.t thua.t toan go. i la. i chınh no. D- e. quy la mo.t cong cu. hu.u du.ng vatu.. nhien de’ gia’i quyet mo.t lo.p lo.n cac bai toan. D- e’ gia’i nhu.ng bai toan trong lo.p nay taco the’ su.’ du.ng ky thua.t chia de’ tri.: Bai toan can gia’i quyet du.o.. c chia thanh nhu.ng baitoan con co da.ng nhu. bai toan ban dau. Moi bai toan con la. i du.o.. c phan ra them. Quatrınh phan ra cho den khi nha.n du.o.. c nhu.ng bai toan con vo.i lo.i gia’i de dang. Cuoi cung,to’ ho.. p cac lo.i gia’i cu’a cac bai toan con ta du.o.. c lo.i gia’i cu’a bai toan ban dau.

Vı du. 3.3.1. n giai thu.a cu’a so tu.. nhien n la so nguyen du.o.ng xac di.nh bo.’ i

n! :=

{1 neu n = 0,

n(n − 1)(n − 2) · · · 2 · 1 neu n ≥ 1.

Tu.c la neu n ≥ 1 thı n! bang tıch cu’a tat ca’ cac so tu.. nhien tu. 1 den n. Cha’ng ha.n,

3! = 3 · 2 · 1 = 6,

6! = 6 · 5 · 4 · 3 · 2 · 1 = 720.

Tu. di.nh nghıa suy ra

n! = n · (n − 1)!

vo.i mo.i n ≥ 1. Vı va.y, bai toan ban dau (tınh n!) du.o.. c phan ra thanh cac bai toan con(tınh (n − 1)!, tınh (n − 2)!, · · · ) cho den bai toan con du.o.. c gia’i de dang la tınh 0!. Cuoicung, lo.i gia’i cu’a cac bai toan con du.o.. c to’ ho.. p la. i bang phep nhan de’ nha.n du.o.. c lo.i gia’ibai toan ban dau.

Thua.t toan de. quy du.o.i day tınh cac giai thu.a.

3.3.1 Tınh n giai thu.a

Thua.t toan nay tınh n!.

Vao: n la so tu.. nhien.Ra: n!.

Bu.o.c 1. Neu n = 0 thı xuat 1 va du.ng.

Bu.o.c 2. Su.’ du.ng thua.t toan nay de’ tınh (n − 1)!. Xuat (n − 1)!n.

D- i.nh ly 3.3.1. Thua. t toan 3.3.1 tınh gia tri. cu’a n! vo.i mo. i n ∈ N.

39

Chu.ng minh. Chu.ng minh su.’ du.ng quy na.p toan ho.c. 2

Ke tiep ta trınh bay thua.t toan de. quy tım u.o.c so chung lo.n nhat cu’a hai so tu.. nhienkhong dong tho.i bang khong.

Ta biet rang neu a la so nguyen khong am, b la so nguyen du.o.ng va

a = bq + r, 0 ≤ r < b,

thı

USCLN(a, b) = USCLN(b, r).

D- ieu nay de dang suy ra thua.t toan de. quy tım u.o.c so chung lo.n nhat cu’a hai so tu.. nhienkhong dong tho.i bang khong.

3.3.2 Tım u.o.c so chung lo.n nhat

Thua.t toan nay tım u.o.c so chung lo.n nhat cu’a hai so tu.. nhien a va b, trong do a, b khongdong tho.i bang 0.

Vao: a, b la so tu.. nhien khong dong tho.i bang 0.Ra: x la u.o.c so chung lo.n nhat cu’a a va b.


Bu.o.c 2. Neu b = 0 thı thu.. c hie.n x := a va du.ng.


Bu.o.c 4. Go.i thua.t toan nay de’ tınh u.o.c so chung lo.n nhat cu’a b va r. Lu.u tru. gia tri. naytrong x.

Vı du. cuoi cung la thua.t toan de. quy xac di.nh bu.o.c di cu’a ngu.o.i may.

Vı du. 3.3.2. Mo.t ngu.o.i may co the’ bu.o.c 1 hoa. c 2 meter. Hay tınh so cach de’ ngu.o.i mayco the’ bu.o.c n meter. Cha’ng ha.n:

Khoa’ng cach Day cac bu.o.c So cach de’ bu.o.c1 1 12 1, 1, hoa. c 2 23 1, 1, 1, hoa. c 1, 2 hoa. c 2, 1 34 1, 1, 1, 1, hoa. c 1, 1, 2 hoa. c 5

1, 2, 1, hoa. c 2, 1, 1 hoa. c 2, 2

40

Go.i fn la so cach de’ ngu.o.i may co the’ bu.o.c n meter. Ta co

f1 = 1, f2 = 2.

Ho.n nu.a, co the’ chu.ng minh cong thu.c truy hoi sau:

fn = fn−1 + fn−2, n ≥ 3.

3.3.3 Thua. t toan xac di.nh day Fibonacci

Thua.t toan nay tınh ham xac di.nh bo.’ i

fn :=

1, neu n = 1

2, neu n = 2

fn−1 + fn−2, neu n > 2.

Vao: n la so tu.. nhien.Ra: fn.

Bu.o.c 1. Neu n = 1 hoa. c n = 2 thı xuat n va du.ng.

Bu.o.c 2. Tınh fn−1 va fn−2 va xuat fn−1 + fn−2.

Dayf1, f2, f3, . . . ,

co cac gia tri. dau tien1, 2, 3, 5, 8, 13, . . . ,

go. i la day Fibonacci.

Bai ta.p

1. (a) Su.’ du.ng cong thu.c

s1 = 1,

sn = sn−1 + n, n ≥ 2,

hay viet thua.t toan da.ng de. quy tınh to’ng

sn := 1 + 2 + 3 + · · · + n.

(b) Su.’ du.ng quy na.p toan ho.c, chu.ng minh tınh dung cu’a thua.t toan trong cau (a).

41

2. (a) Su.’ du.ng cong thu.c

s1 = 2,

sn = sn−1 + 2n, n ≥ 2,

hay viet thua.t toan da.ng de. quy tınh to’ng

sn := 2 + 4 + 6 + · · · + 2n.


3. (a) Mo. t ngu.o.i may co the’ bu.o.c 1 meter, 2 meter hoa. c 3 meter. Hay viet thua.t toanda.ng de. quy tınh so cach ngu.o.i may co the’ bu.o.c n meter.


4. Hay viet mo.t thua. t toan da.ng de. quy tım u.o.c so chung lo.n nhat cu’a hai so nguyenkhong dong tho.i bang khong su.’ du. ng phep toan tru. thay cho phep toan chia.

5. Viet thua.t toan khong de. quy tınh n giai thu.a.

6. Mo.t ngu.o.i may co the’ bu.o.c 1 hoa. c 2 meter. Hay viet thua.t toan lie.t ke tat ca’ caccach ngu.o.i may co the’ bu.o.c n meter.

7. Mo.t ngu.o.i may co the’ bu.o.c 1, 2 hoa. c 3 meter. Hay viet thua.t toan lie.t ke tat ca’ caccach ngu.o.i may co the’ bu.o.c n meter.

8. Ky hie.u fn la day Fibonacci. Su.’ du. ng quy na.p toan ho.c, chu.ng minh cac quan he. sau:

(a)∑n

k=1 fk = fn+2 − 2, n ≥ 1.

(b) f2n = fn−1fn+1 + (−1)n, n ≥ 2.

(c) f2n+2 − f2

n+1 = fnfn+3, n ≥ 1.

(d)∑n

k=1 f2k = fnfn+1 − 1, n ≥ 1.

(e) fn chan neu va chı’ neu n + 1 chia het cho 3.

(f) vo.i mo.i n ≥ 5 co

fn >

(3

2

)n

.

(g) vo.i mo.i n ≥ 1 cofn < 2n.

(h) vo.i mo.i n ≥ 1 co

n∑

k=1

f2k−1 = f2n − 1,

n∑

k=1

f2k = f2n+1 − 1.

42

(i) Mo.i so nguyen du.o.ng co the’ viet da.ng to’ng cu’a cac so Fibonacci phan bie.t vakhong co hai so nao la lien tiep. Chu.ng minh cach viet nay la duy nhat.

9. Gia’ su.’ co cong thu.c da.o ham cu’a tıch

d(fg)

dx= f

dg

dx+ g

df

dx.

Dung quy na.p toan ho.c, chu.ng minh cong thu.c

dxn

dx= nxn−1, n ≥ 1.

3.4 D- o. phu.c ta.p cu’a thua. t toan

Mo.t chu.o.ng trınh may tınh, tha.m chı du.. a vao mo.t thua.t toan dung, co the’ khong hu.u du.ngdoi vo.i mo.t lo.p cac du. lie.u vao do tho.i gian can thiet cha.y chu.o.ng trınh hoa. c khong gianlu.u tru. du. lie.u, cac bien... qua lo.n. Phan tıch thua. t toan de ca.p den qua trınh u.o.c lu.o.. ngtho.i gian va khong gian can thiet de’ thu.. c hie.n thua.t toan. D- o. phu.c ta. p cu’a thua. t toan amchı’ den so lu.o.. ng tho.i gian va khong gian doi ho’i de’ thu.. c hie.n thua.t toan.

Tho.i gian can thiet de’ thu.. c hie.n mo.t thua.t toan la mo.t ham phu. thuo.c du. lie.u dau vao.Thu.o.ng thı kho co the’ xac di.nh chınh xac ham nay. Vı va.y, chung ta se su.’ du.ng cac thamso da.c tru.ng kıch thu.o.c cu’a du. lie.u du.a vao. Co ba khai nie.m:

(a) Tho.i gian tru.o.ng ho.. p tot nhat la tho.i gian ıt nhat de’ thu.. c hie.n thua.t toan.

(b) Tho.i gian tru.o.ng ho.. p xau nhat la tho.i gian nhieu nhat de’ thu.. c hie.n thua.t toan.

(c) Tho.i gian tru.o.ng ho.. p trung bınh la tho.i gian trung bınh de’ thu.. c hie.n thua.t toan.

D- i.nh nghıa 3.4.1. Gia’ su.’ f, g : N→ R la hai ham so. Ta viet

f(n) = O(g(n))

va noi f(n) co ba. c nhieu nhat g(n) neu ton ta. i hang so du.o.ng C sao cho ngoai mo.t ta.p hu.uha.n cac so tu.. nhien ta luon co

|f(n)| ≤ C|g(n)|.

Khi do ta noi f(n) la O-lo.n cu’a g(n).

Vı du. 3.4.1. Ta co cac quan he. sau

70n2 + 5n + 1 = O(n2),

2n + 3 ln n = O(n),

1 + 2 + · · · + n = O(n2).

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D- i.nh ly 3.4.2. Gia’ su.’

f(n) := aknk + ak−1n

k−1 + · · · + a1n + a0

la da thu.c ba. c k theo bien n. Khi do

f(n) = O(nk).


Vı du. 3.4.2. Vı 3n4 − 7n2 + 4n la da thu.c ba.c 4 theo bien n nen

3n4 − 7n2 + 4n = O(n4).

D- i.nh nghıa 3.4.3. Neu mo.t thua.t toan doi ho’i t(n) do.n vi. tho.i gian trong tru.o.ng ho.. p totnhat (tu.o.ng u.ng, xau nhat hoa. c trung bınh) vo.i du. lie.u vao co kıch thu.o.c n va

t(n) = O(g(n))

thı ta noi tho.i gian tru.o.ng ho.. p tot nhat (tu.o.ng u.ng, xau nhat hoa. c trung bınh) thu.. c hie. nthua. t toan la O(g(n)).

Gia’ su.’ t(n) = O(g(n)). Ta noi thua. t toan co do. phu.c ta. p da thu.c hoa. c co tho.i gian dathu.c neu g(n) la da thu.c theo bien n.

Vı du. 3.4.3. Ky hie.u t(n) la so lan cau le.nh x := x + 1 du.o.. c thu.. c hie.n trong thua.t toansau:


Bu.o.c 2. Vo.i j := 1 den i thu.. c hie.n Bu.o.c 3.

Bu.o.c 3. x := x + 1.

De thay

t(n) = 1 + 2 + · · · + n =n(n + 1)

2.

Suy rat(n) = O(n2).

Vı du. 3.4.4. Ky hie.u t(n) la so lan cau le.nh x := x + 1 du.o.. c thu.. c hie.n trong thua.t toansau:

Bu.o.c 1. D- a.t j := n.

Bu.o.c 2. Neu j < 1 thı du.ng thua.t toan.

44

Bu.o.c 3. Vo.i i := 1 den j thu.. c hie.n Bu.o.c 4.

Bu.o.c 4. D- a.t x := x + 1.

Bu.o.c 5. D- a.t j := bj/2c.

Bu.o.c 6. Chuye’n den Bu.o.c 2.

Co the’ chı’ rat(n) = O(n).

Vı du. 3.4.5. Xet thua. t toan tım kiem trong mo.t day khong du.o.. c sap thu. tu.. sau:

Vao: s, s1, s2, . . . , sn.Ra: j = 0 neu s 6= si vo.i mo.i i; ngu.o.. c la. i, j la chı’ so nho’ nhat sao cho s = sj.


Bu.o.c 2. Neu s = si thı da. t j := i va du.ng thua.t toan.

Bu.o.c 3. D- a.t j := 0.

Co the’ chu.ng minh tho.i gian tru.o.ng ho.. p tot nhat bang O(1), tho.i gian tru.o.ng ho.. p xaunhat bang tho.i gian tru.o.ng ho.. p trung bınh va bang O(n).

Bai ta.p

1. Xac di.nh ky hie.u O lo.n doi vo.i f(n) + g(n) neu

(a) f(n) := O(1), g(n) := O(n2).

(b) f(n) := 6n3 − 2n2 + 4, g(n) := O(n ln n).

(c) f(n) := O(n3/2), g(n) := O(n5/2).

2. Xac di.nh do. phu.c ta.p tınh toan vo.i cac thua. t toan sau (xet so lan thu.. c hie.n cau le.nhx := x + 1):

Bu.o.c 1. Vo.i i := 1 den 2n thu.. c hie.n x := x + 1.


Bu.o.c 1. i := 1.

Bu.o.c 2. Neu i > 2n thı du.ng.

Bu.o.c 3. x := x + 1.

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Bu.o.c 4. i := i + 2.



Bu.o.c 1. Vo.i i := 1 den n va vo.i j := 1 den n thu.. c hie.n x := x + 1.


Bu.o.c 1. Vo.i i := 1 den 2n va vo.i j := 1 den n thu.. c hie.n x := x + 1.


Bu.o.c 1. Vo.i i := 1 den 2n thu.. c hie.n Bu.o.c 2.

Bu.o.c 2. Vo.i j := 1 den bi/2c thu.. c hie.n x := x + 1.

7. Xac di.nh so phep toan so sanh va ky hie.u O lo.n trong thua.t toan sau

Vao: s1, s2, . . . , sn.Ra: M := maxi si va m := mini si.

Bu.o.c 1. t := 2bn/2c.Bu.o.c 2. i := 1.

Bu.o.c 3. Neu i > t − 1 thı chuye’n den Bu.o.c 7.

Bu.o.c 4. Neu si > si+1 thı hoan do’i si va si+1.

Bu.o.c 5. i := i + 2.


Bu.o.c 7. Neu n ≤ t thı chuye’n den Bu.o.c 10.

Bu.o.c 8. Neu sm−1 > sn thı hoan do’i sm−1 va sn.

Bu.o.c 9. Neu sn > sm thı hoan do’i sm va sn.

Bu.o.c 10. m := s1.

Bu.o.c 11. M := s2.

Bu.o.c 12. i := 3.

Bu.o.c 13. Neu i > t − 1 thı du.ng.

Bu.o.c 14. Neu si < m thı gan m := si.

Bu.o.c 15. Neu si+1 > M thı gan M := si+1.

Bu.o.c 16. i := i + 1.


46

8. Gia’ su.’ a > 1 va f(n) := O(loga n). Chu.ng minh rang f(n) = O(lnn).

9. Gia’ su.’ g(n) > 0 vo.i mo.i n ∈ N. Chu.ng minh f(n) = O(g(n)) neu va chı’ neu ton ta. ihang so du.o.ng c sao cho

|f(n)| ≤ cg(n)

vo.i mo.i n ∈ N.

10. Chu.ng minh rang neu

f(n) = O(h(n)) va g(n) = O(h(n))

thı

f(n) + g(n) = O(h(n)) va cf(n) = O(h(n))

vo.i mo.i c ∈ R.

11. Chu.ng minh n! = O(nn).

12. Chu.ng minh 2n = O(n!).

13. Chu.ng minh n lnn = O(ln(n!)).

14. Chu.ng minh ln(n!) = O(n ln n).

15. Tım cac ham f va g sao cho

f(n) 6= O(g(n)) va g(n) 6= O(f(n)).

16. Tım cac ham f, g, h va k sao cho

f(n) = O(g(n)), h(n) = O(k(n)), f(n) − h(n) 6= O(g(n) − k(n)).

17. Ta viet f(n) = Θ(g(n)) neu ton ta. i cac hang so du.o.ng C1, C2 sao cho

C1|g(n)| ≤ |f(n)| ≤ C2|g(n)|

ngoai mo.t ta.p ho.. p con hu.u ha.n cu’a ta.p cac so tu.. nhien N. Chu.ng minh rang

(a) 2n − 1 = Θ(n).

(b) 3n2 − 1 = Θ(n2).

(c) (4n − 1)2 = Θ(n2).

(d) (2n − 1)(7n + 1)/(n − 1) = Θ(n).

(e) Quan he. f(n) = Θ(g(n)) la quan he. tu.o.ng du.o.ng?

18. Ta viet f ∼ g neu f(n) = O(g(n)). Quan he. ∼ la quan he. tu.o.ng du.o.ng tren ta.p cacso tu.. nhien N?

47

19. Su.’ du.ng tıch phan xac di.nh, chu.ng minh bat da’ng thu.c sau

1

2+

1

3+ · · · + 1

n< lnn.

Tu. do suy ra

1 +1

2+

1

3+ · · · + 1

n< O(ln n).

20. Su.’ du.ng tıch phan xac di.nh, chu.ng minh bat da’ng thu.c sau

1m + 2m + · · · + nm <(n + 1)m+1

m + 1,

trong do m la so nguyen du.o.ng.

21. Chu.ng minh hoa. c cho pha’n vı du. cac kha’ng di.nh sau:

(a) Neu ton ta. i gio.i ha.n hu.u ha.n

limn→∞

f(n)

g(n)

thı f(n) = O(g(n)).

(b) Neu f(n) = O(g(n)) thı ton ta. i gio.i ha.n hu.u ha.n

limn→∞

f(n)

g(n).

(c) Neu ton ta. i gio.i ha.n hu.u ha.n

limn→∞

f(n)

g(n)

thı f(n) = Θ(g(n)).

(d) Neu

limn→∞

f(n)

g(n)= 1

thı f(n) = Θ(g(n)).

(e) Neu f(n) = Θ(g(n)) thı ton ta. i gio.i ha.n hu.u ha.n

limn→∞

f(n)

g(n).

3.5 Phan tıch thua. t toan Euclid

Phan nay phan tıch tru.o.ng ho.. p xau nhat cu’a thua. t toan Euclid tım u.o.c so chung lo.n nhatcu’a hai so tu.. nhien a va b, trong do a, b khong dong tho.i bang 0. Tru.o.c het ta nhac la.ithua.t toan nay:

Vao: a, b la so tu.. nhien khong dong tho.i bang 0.Ra: USCLN la u.o.c so chung lo.n nhat cu’a a va b.

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Bu.o.c 2. Neu b = 0 thı thu.. c hie.n USCLN := a va du.ng.


Bu.o.c 4. Thu.. c hie.n a := b, b := r va chuye’n den Bu.o.c 2.

Ta di.nh nghıa tho.i gian de’ thu.. c hie.n thua.t toan Euclid la so phep toan chia trong Bu.o.c 3.Tru.o.ng ho.. p xau nhat doi vo.i thua.t toan Euclid xa’y ra khi so phep chia nhieu nhat. Nhacla. i rang, day Fibonacci {fn} xac di.nh bo.’ i

f1 := 1, f2 := 2; fn := fn−1 + fn−2, n ≥ 3.

Ta co

D- i.nh ly 3.5.1. Gia’ su.’ thua. t toan Euclid doi vo.i ca. p cac so tu.. nhien a, b vo.i a > b can nphep toan chia. Khi do a ≥ fn+1 va b > fn trong do {fn} la day Fibonacci.

Chu.ng minh. Su.’ du.ng quy na.p toan ho.c. 2

D- i.nh ly nay de dang suy ra

D- i.nh ly 3.5.2. Gia’ su.’ thua. t toan Euclid doi vo.i ca. p cac so tu.. nhien khong dong tho.i bangkhong va thuo. c khoa’ng [0,m],m ≥ 8. Khi do so phep chia can thiet trong thua. t toan Euclidkhong vu.o.. t qua

log3/2

2m

3.


Vı ham logarithm co cap tang cha.m, ket qua’ tren chu.ng to’ thua.t toan Euclid rat hie.u qua’

tha.m chı doi vo.i cac gia tri. dau vao rat lo.n.

Bai ta.p

1. Co nhieu nhat bao nhieu phep toan chia trong thua.t toan Euclid doi vo.i ca.p cac sothay do’i trong khoa’ng tu. 0 den 1000000?

2. Chu.ng minh co chınh xac n phep toan chia trong thua.t toan Euclid doi vo.i ca.p so(fn, fn+1), n ≥ 1.

3. Chu.ng minh vo.i mo.i so nguyen k > 1 ta co so phep toan chia trong thua.t toan Eucliddoi vo.i hai ca.p so (a, b) va (ka, kb) la bang nhau.

4. Chu.ng minh rang USCLN(fn, fn+1) = 1, n ≥ 1.

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Chu.o.ng 4

PHEP D- EM

Toan to’ ho.. p nghien cu.u chu’ yeu ve cach sap xep cac doi tu.o..ng. D- ay la mo.t bo. pha.n quantro.ng cu’a toan ho.c ro.i ra.c. Nhu.ng van de cu’a to’ ho.. p du.o.. c nghien cu.u tu. The ky’ 17, lienquan tru.o.c tien den cac tro cho.i may ru’i. Ngay nay toan to’ ho.. p du.o.. c dung ro.ng rai trongtin ho.c.

Mu.c dıch cu’a chu.o.ng nay la thiet la.p mo.t so phu.o.ng phap dem cac ta.p ho.. p gom hu.uha.n cac phan tu.’ ma khong pha’i lie.t ke cac phan tu.’ cu’a chung.

4.1 Cac nguyen ly co. ba’n cu’a phep dem

Nhac la. i: #S la so phan tu.’ cu’a ta.p ho.. p S. Do do #S = #T neu hai ta.p S va T co cung socac phan tu.’ . Chu y rang

#∅ = 0, #{1, 2, . . . , n} = n vo.i n ∈ N.

Chung ta bat dau vo.i mo.t so nguyen ly dem.

4.1.1 Nguyen ly to’ng

Gia’ su.’ A1, A2, . . . , Am la cac su.. kie.n doi mo. t loa. i tru. nhau. Gia’ su.’ cac su.. kie.n A1, A2, . . . , Am

co tu.o.ng u.ng n1, n2, . . . , nm cach xa’y ra. Khi do su.. kie.n (hoa. c A1, hoa. c A2, . . . , hoa. c Am)co n1 + n2 + · · · + nm cach xa’y ra.

Vı du. 4.1.1. Lo.p tru.o.’ ng hoa. c la mo.t nu. sinh, hoa. c la mo.t nam sinh. Co bao nhieu cachcho.n lo.p tru.o.’ ng khac nhau neu so ho.c sinh nu. la 36 va so ho.c sinh nam la 20?

51

Go.i A1 (tu.o.ng u.ng, A2) la su.. kie.n lo.p tru.o.’ ng la nu. sinh (tu.o.ng u.ng, nam sinh). Ta co36 cach cho.n lo.p tru.o.’ ng la nu. sinh va 20 cach cho.n lo.p tru.o.’ ng la nam sinh. Theo nguyenly to’ng, su.. kie.n (A1 hoa. c A2) co (36 + 20) = 56 cach cho.n.

Vı du. 4.1.2. Mo.t sinh vien co the’ cho.n dung mo.t chuyen de tu.. cho.n thuo.c mo.t trong badanh sach. Ba danh sach nay gom 3, 5 va 9 chuyen de. Ho’i sinh vien do co bao nhieu cachlu.. a cho.n?

Theo nguyen ly to’ng, co 3 + 5 + 9 = 17 cach.

Nha.n xet 3. Co the’ phat bie’u nguyen ly to’ng theo thua.t ngu. cu’a ly thuyet ta.p ho.. p nhu.

sau: Neu cac ta.p T1, T2, . . . , Tm doi mo.t ro.i nhau thı so cac phan tu.’ cu’a ta.p T1∪T2∪· · ·∪Tm

bang to’ng so cac phan tu.’ cu’a cac ta.p nay; tu.c la

#(T1 ∪ T2 ∪ . . . ∪ Tm) =m∑

i=1

#Ti.

4.1.2 Nguyen ly tıch

Gia’ su.’ A1, A2, . . . , Am la cac su.. kie.n doi mo. t loa. i tru. nhau. Gia’ su.’ cac su.. kie.n A1, A2, . . . , Am

co tu.o.ng u.ng n1, n2, . . . , nm cach xa’y ra. Khi do su.. kie.n (A1 va A2 va . . . va Am) con1 × n2 × · · · × nm cach xa’y ra.

Vı du. 4.1.3. Gia’ su.’ co hai ma.t na. , ba mu. Ho’i co may cach hoa trang?

Dung nguyen ly tıch, co 3× 2 = 6 cach hoa trang khac nhau. Cung co the’ dung ly thuyetta.p ho.. p nhu. sau: Moi cach hoa trang la mo.t cach cho.n x ∈ X va mo.t cach cho.n y ∈ Y. Dodo so cach hoa trang la so cac ca.p (x, y) thuo. c X ×Y va do do bang #X ×#Y = 2×3 = 6.

Nha.n xet 4. Nguyen ly nay cung thu.o.ng du.o.. c phat bie’u du.o.i da.ng ta.p ho.. p nhu. sau: Gia’

su.’ cac ta.p T1, T2, . . . , Tm co hu.u ha.n phan tu.’ va doi mo.t ro.i nhau. Khi do so phan tu.’ cu’ata.p tıch Descartes T1 × T2 × · · · × Tm bang

#T1 ×#T2 × · · · × #Tm.

Vı du. 4.1.4. Co bao nhieu chuoi bit khac nhau co do. dai 8? Moi bit co hai cach cho.n, hoa. c0 hoa. c 1. Do do theo nguyen ly tıch, co 28 = 256 chuoi bit co do. dai 8.

Vı du. 4.1.5. Co bao nhieu ba’ng so xe khac nhau, neu moi ba’ng gom ba chu. cai va theosau la ba con so (gia’ thiet ba’ng chu. cai gom 26 ky tu.. )?

Moi chu. cai co 26 cach cho.n; moi so co 10 cach cho.n. Do do theo nguyen ly tıch, so cacba’ng so xe khac nhau la:

26 × 26 × 26 × 10 × 10 × 10 = 17.576.000.

52

Vı du. 4.1.6. Co bao nhieu anh xa. khac nhau tu. ta.p X co m phan tu.’ vao ta.p Y co n phantu.’ ?

Moi anh xa. la mo.t bo. m cach cho.n mo.t trong n phan tu.’ cu’a Y cho moi mo.t trong m phantu.’ cu’a X. Theo nguyen ly tıch, so anh xa. nay bang

n × n × · · · × n︸︷︷︸m lan

= nm.

Vı du. 4.1.7. Co bao nhieu anh xa. mo.t-mo.t (do.n anh) khac nhau tu. ta.p X co m phan tu.’

vao ta.p Y co n phan tu.’ ?

Neu m > n : khong co anh xa. mo.t-mo. t tu. X vao Y.

Gia’ su.’ m ≤ n va X := {a1, a2, . . . , am}.

+ Vo.i phan tu.’ a1 co n cach cho.n phan tu.’ tu.o.ng u.ng trong Y.

+ Vı anh xa. la mo. t-mo.t, nen doi vo.i a2 chı’ con (n − 1) cach cho.n.

...

+ Tu.o.ng tu.. , am chı’ con (n − m + 1) cach cho.n.

Theo nguyen ly tıch, so anh xa. mo.t-mo.t khac nhau bang

n(n − 1)(n − 2) · · · (n − m + 1).

Vı du. 4.1.8. D- em so ta.p con cu’a mo.t ta.p hu.u ha.n S.

Gia’ su.’ S := {a1, a2, . . . , an}. De dang thiet la.p mo.t tu.o.ng u.ng mo.t-mo.t giu.a ta.p con Pcu’a S vo.i cac chuoi bit do. dai n : bit thu. i bang 1 neu va chı’ neu ai ∈ P. Ma.t khac, so cacchuoi bit do. dai n la 2n nen so cac ta.p con cu’a S la 2n.

Vı du. 4.1.9. Cho hai doa.n chu.o.ng trınh sau:

Chu.o.ng trınh 1: Chu.o.ng trınh 2:k := 0; k := 0;for i1 := 1 to n1 do k := k + 1; for i1 := 1 to n1 dofor i2 := 1 to n2 do k := k + 1; for i2 := 1 to n2 do... ...for im := 1 to nm do k := k + 1; for im := 1 to nm do k := k + 1;

Ho’i k se lay gia tri. bao nhieu sau khi moi doa.n chu.o.ng trınh tren du.o.. c thu.. c hie.n?

+ Chu.o.ng trınh 1: Cu. moi vong la.p di.a phu.o.ng, k tang len mo.t do.n vi..

Go.i Ai la so lan la.p cu’a vong la.p thu. i. Ai co ni kha’ nang. Ho.n nu.a Ai va Aj, i 6= j, loa. itru. nhau. Do do theo nguyen ly to’ng, so vong la.p la n1 + n2 + · · · + nm.

53

+ Chu.o.ng trınh 2: Cu. moi vong la.p toan cu.c, k tang len mo.t do.n vi.. Moi vong la.p toancu.c do m vong la.p di.a phu.o.ng ghep la. i. Theo nguyen ly tıch so vong la.p toan cu.c bangn1 × n2 × · · · × nm.

Trong nhieu tru.o.ng ho.. p ta can pha’i phoi ho.. p ca’ hai nguyen ly to’ng va tıch; cha’ng ha.n,xet vı du. sau:

Vı du. 4.1.10. Gia’ su.’ moi ngu.o.i su.’ du.ng may tınh co mo.t ma.t ma, gom tu. 6 den 8 ky tu.. ;moi ky tu.. la mo.t chu. cai hoa hoa. c la mo.t con so. Moi ma.t ma nhat thiet pha’i chu.a ıt nhatmo.t con so. Ho’i co bao nhieu ma.t ma co the’ co?

Go.i P la to’ng so cac ma.t ma co the’ co va P6, P7, P8 la so cac ma.t ma co the’ co vo.i do. daitu.o.ng u.ng bang 6, 7, 8.

Theo nguyen ly to’ng: P = P6 + P7 + P8.

Vie.c tınh tru.. c tiep P6 la kho. Ta tınh gian tiep nhu. sau:

+ So cac chuoi co do. dai 6, gom chu. va so, bao gom ca’ tru.o.ng ho.. p khong co con so naotheo nguyen ly tıch la (26 + 10)6 = 366.

+ So cac chuoi do. dai 6, khong chu.a con so nao la 266.

+ Do do P6 = 366 − 266 = 1.867.866.560.

Tu.o.ng tu.. cho P7 va P8 :

P7 = 367 − 267 = 70.332.353.920,P8 = 368 − 268 = 2.612.282.842.880.

Cuoi cung

P = P6 + P7 + P8 = 2.684.483.063.360.

Nha.n xet 5. Khi cac su.. kie.n A1 va A2 co the’ xa’y ra dong tho.i ta khong the’ dung nguyenly to’ng. Tru.o.ng ho.. p nay can su.’ a do’i nhu. sau: Neu van co.ng (n1 + n2) ta da dem thu.a,vı co tru.o.ng ho.. p da dem hai lan cung mo.t su.. kie.n (mo.t lan trong A1, mo.t lan trong A2).Tru.o.ng ho.. p nay chı’ xa’y ra khi no dong tho.i co the’ xa’y ra A1 va A2. Vı va.y can tru. di sotru.o.ng ho.. p doi thu.a nay.

4.1.3 Nguyen ly bao ham-loa.i tru.

Gia’ su.’ A1 va A2 la hai su.. kie.n bat ky. Neu su.. kie.n A1 co the’ xa’y ra n1 cach, su.. kie.n A2 cothe’ xa’y ra n2 cach, thı su.. kie.n (A1 hoa. c A2) co the’ xa’y ra [(n1 + n2)− so cach (A1 va A2)]cach.

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Bang thua.t ngu. ta.p ho.. p, nguyen ly bao ham-loa. i tru. tro.’ thanh:

#(A1 ∪ A2) = #A1 + #A2 −#(A1 ∩ A2).

Vı du. 4.1.11. Co bao nhieu chuoi bit do. dai 8 hoa. c bat dau bang 1, hoa. c ket thuc bang00? (Co the’ co chuoi vu.a bat dau bang 1, vu.a ket thuc bang 00).

Go.i P1 la so cac chuoi bit do. dai 8 bat dau bang 1. Nhu. va.y, phan tu.’ thu. nhat da du.o.. ccho.n, chı’ con la. i 7 bit. Theo nguyen ly tıch,

P1 = 27 = 128.

Go. i P2 la so cac chuoi bit do. dai 8 ket thuc bang 00. Theo nguyen ly tıch

P2 = 26 = 64.

Go. i P3 la so cac chuoi bit do. dai 8 bat dau bang 1 va ket thuc bang 00. Theo nguyen lytıch

P3 = 25 = 32.

Ap du.ng nguyen ly bao ham-loa. i tru. ta co

P = P1 + P2 − P3 = 160.

Nguyen ly bao ham-loa. i tru. co the’ mo.’ ro.ng cho tru.o.ng ho.. p m su.. kie.n, nhu.ng phu.c ta.pho.n, ta se de ca.p o.’ phan sau.

Su.. cong nha.n ba nguyen ly du.o.. c de ca.p tren day nhu. la xuat phat die’m cu’a ly thuyetto’ ho.. p:

+ Tınh dung dan cu’a ba nguyen ly tren la “dung hie’n nhien”. Quan die’m cu’a chung tala cong nha.n 3 nguyen ly tren, coi nhu. xuat phat die’m cu’a ly thuyet to’ ho.. p. Cac ket qua’

khac se lan lu.o.. t du.o.. c suy ra tru.. c tiep hoa. c gian tiep tu. ba nguyen ly nay.

+ Neu khong thoa’ man, cung co the’ tım cach chu.ng minh ba nguyen ly nay, nhu. va.y tala. i pha’i can den cac cong cu. khac, thu.. c chat ta la. i cong nha.n mo.t dieu gı khac la “dunghie’n nhien” de’ roi suy lua.n ra ba nguyen ly tren.

Bai ta.p

1. Co bao nhieu chuoi 8 bit bat dau bang 1100?

2. Co bao nhieu chuoi 8 bit bat dau va ket thuc bang 1?

3. Co bao nhieu chuoi 8 bit co dung mo.t bit bang 1? D- ung hai bit bang 1? Co ıt nhatmo.t bit bang 1?

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4. Co bao nhieu chuoi 8 bit do.c xuoi va do.c ngu.o.. c deu giong nhu. nhau?

5. Cac ky tu.. ABCDE du.o.. c su.’ du.ng de’ ta.o thanh cac chuoi do. dai 3.

(a) Co bao nhieu chuoi du.o.. c ta.o ra neu cho phep la.p?

(b) Co bao nhieu chuoi du.o.. c ta.o ra neu khong cho phep la.p?

(c) Co bao nhieu chuoi bat dau bang A du.o.. c ta.o ra neu cho phep la.p?

(d) Co bao nhieu chuoi bat dau bang A du.o.. c ta.o ra neu khong cho phep la.p?

(e) Co bao nhieu chuoi khong chu.a ky tu.. A du.o.. c ta.o ra neu cho phep la.p?

(f) Co bao nhieu chuoi khong chu.a ky tu.. A du.o.. c ta.o ra neu khong cho phep la.p?

6. Tren ta.p X := {5, 6, . . . , 200} :

(a) Co bao nhieu so chan, (tu.o.ng u.ng, le’)?

(b) Co bao nhieu so chia het cho 5?

(c) Co bao nhieu so gom nhu.ng chu. so phan bie.t?

(d) Co bao nhieu so khong chu.a chu. so 0?

(e) Co bao nhieu so lo.n ho.n 101 va khong chu.a chu. so 6?

(f) Co bao nhieu so co cac chu. so du.o.. c sap theo thu. tu.. tang thu.. c su.. ?

(g) Co bao nhieu so co da.ng xyz vo.i 0 6= x < y va y > z?

7. Gia’ su.’ co 5 sach tin ho.c, 3 sach may tınh, 2 sach va. t ly.

(a) Co bao nhieu cach sap xep chung len gia sach?

(b) Co bao nhieu cach sap xep sao cho 5 sach tin ho.c o.’ phıa trai, con 2 sach va.t ly o.’

ben pha’i?

(c) Co bao nhieu cach sap chung len gia sao cho tat ca’ cac sach theo cung nhom du.o.. csap ke nhau?

(d) Co bao nhieu cach sap chung len gia sao cho hai sach va. t ly khong ke nhau?

8. Co 10 ba’n ba’n sao (copy) cu’a mo.t cuon sach va co mo.t ba’n sao cu’a 10 cuon sach khac.Co bao nhieu cach co the’ cho.n 10 cuon sach?

9. Co bao nhieu ta.p con co nhieu nhat n phan tu.’ cu’a ta.p gom (2n + 1) phan tu.’ ?

10. Ap du.ng nguyen ly bao ham-loa. i tru. de’ gia’i:

(a) Co bao nhieu chuoi 8 bit hoa. c bat dau bang 100 hoa. c co bit thu. tu. bang 1?

(b) Co bao nhieu chuoi 8 bit hoa. c bat dau bang 1 hoa. c ket thuc bang 1?

(c) Co bao nhieu chuoi 8 bit trong do hoa. c bit thu. hai, hoa.c bit thu. tu. bang 1?

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4.2 Hoan vi. va to’ ho.. p

D- i.nh nghıa 4.2.1. Hoan vi. cu’a n phan tu.’ x1, x2, . . . , xn la mo.t sap xep co thu. tu.. n phantu.’ nay.

Vı du. 4.2.1. Co sau hoan vi. cu’a ba phan tu.’ . Neu cac phan tu.’ du.o.. c ky hie.u la A,B,C thısau hoan vi. la

ABC,ACB,BAC,BCA,CAB,CBA.

D- i.nh ly 4.2.2. Co n! hoan vi. cu’a n phan tu.’ .

Chu.ng minh. Ta chu.ng minh theo quy na.p. Mo.t hoan vi. cu’a n phan tu.’ co the’ du.o.. c xaydu.. ng theo n bu.o.c lien tiep: Cho.n phan tu.’ dau tien, cho.n phan tu.’ thu. hai, ..., cho.n phantu.’ cuoi cung. Phan tu.’ dau tien co the’ cho.n n cach. Ngay khi phan tu.’ dau tien du.o.. c cho.n,phan tu.’ thu. hai co the’ du.o.. c cho.n n− 1 cach. Khi phan tu.’ thu. hai da du.o.. c cho.n, phan tu.’

thu. ba co the’ du.o.. c cho.n n−2 cach, va van van. Theo nguyen ly quy na.p va sau do nguyenly tıch, ton ta. i

n(n − 1)(n − 2) · · · 2 · 1 = n!

hoan vi. cu’a n phan tu.’ . 2

Vı du. 4.2.2. Co

10! = 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 3.628.800

hoan vi. cu’a 10 phan tu.’ .

Vı du. 4.2.3. Co bao nhieu hoan vi. cu’a cac ky tu.. ABCDEF chu.a chuoi con DEF ?

Co the’ xem chuoi con DEF nhu. mo.t ky tu.. . Theo D- i.nh ly 4.2.2 co 4! = 24 hoan vi. cu’acac ky tu.. ABCDEF chu.a chuoi con DEF.

Vı du. 4.2.4. Co bao nhieu hoan vi. cu’a cac ky tu.. ABCDEF chu.a cac ky tu.. DEF theothu. tu.. bat ky?

Ta co the’ gia’i bai toan qua hai bu.o.c: Cho.n mo. t thu. tu.. cu’a cac ky tu.. DEF ; va xay du.. ngmo.t hoan vi. cu’a ABC chu.a thu. tu.. da cho cu’a cac ky tu.. DEF. Theo D- i.nh ly 4.2.2, bu.o.cdau tien co 3! = 6 cach; theo Vı du. 4.2.3 bu.o.c thu. hai co 4! = 24 cach. Theo nguyen lytıch, so cac hoan vi. cu’a ABCDEF chu.a cac ky tu.. DEF theo thu. tu.. bat ky la 6 · 24 = 144.

Trong mo.t so tru.o.ng ho.. p ta muon kha’o sat mo.t thu. tu.. cu’a r phan tu.’ du.o.. c cho.n tu. nphan tu.’ . Mo.t thu. tu.. nhu. the go. i la “r-hoan vi.”.

D- i.nh nghıa 4.2.3. r-hoan vi. cu’a n phan tu.’ (phan bie.t) x1, x2, . . . , xn la mo.t sap xep r-phantu.’ co thu. tu.. tu. n phan tu.’ nay. Ky hie.u P (n, r) la so cac r-hoan vi. cu’a ta.p n phan tu.’ phanbie.t.

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Vı du. 4.2.5. Ta co mo.t so 2-hoan vi. cu’a a, b, c la

ab, bc, ac.

Neu r = n trong D- i.nh nghıa 4.2.3, chung ta nha.n du.o.. c mo.t thu. tu.. cu’a tat ca’ n phan tu.’ .Theo D- i.nh ly 4.2.2 thı P (n, n) = n!. To’ng quat ta co

D- i.nh ly 4.2.4. So cac r-hoan vi. cu’a ta. p n phan tu.’ phan bie. t la

P (n, r) = n(n − 1)(n − 2) · · · (n − r + 1), r ≤ n.

Chu.ng minh. Chung ta dem so cac cach co thu. tu.. cu’a r phan tu.’ du.o.. c cho.n tu. ta.p gomn phan tu.’ . Co n cach cho.n phan tu.’ dau tien. Ke tiep, co n − 1 cach cho.n phan tu.’ thu.

hai, n − 2 cach cho.n phan tu.’ thu. ba, ..., co n − r + 1 cach cho.n phan tu.’ thu. r. Do do theonguyen ly tıch, so cac r-hoan vi. cu’a ta.p n phan tu.’ phan bie.t la

n(n − 1)(n − 2) · · · (n − r + 1).

2

Vı du. 4.2.6. Theo D- i.nh ly 4.2.4, so cac 2-hoan vi. cu’a X = {a, b, c} la

P (3, 2) = 3 · 2 = 6.

Sau hoan vi. nay la

ab, ac, ba, bc, ca, cb.

Vı du. 4.2.7. Co bao nhieu cach cho.n mo.t chu’ ti.ch, mo.t pho chu’ ti.ch, mo.t thu. ky va mo.tthu’ quy tu. mo.t nhom gom 10 ngu.o.i?

Chung ta can dem so cac cach co thu. tu.. cu’a 4 ngu.o.i du.o.. c cho.n tu. mo.t nhom gom 10ngu.o.i. Theo D- i.nh ly 4.2.4 so cac cach cho.n la

P (10, 4) = 10 · 9 · 8 · 7 = 5040.

Chu y rang cung co the’ suy ra ket qua’ tru.. c tiep tu. nguyen ly tıch (ta. i sao?).

Vı du. 4.2.8. Mo.t ngu.o.i ban hang rong can di qua 7 di.a die’m khac nhau. Ong ta co the’

di theo thu. tu.. bat ky. Co bao nhieu hanh trınh khac nhau?

So cac hanh trınh co the’ co la so cac hoan vi. tu. ta.p gom 7 phan tu.’ :

P (7, 7) = 7! = 5040.

Neu cha’ng ha.n ong ta muon tım hanh trınh co do. dai ngan nhat, ong ta can tınh toan vaso sanh 5040 hanh trınh ca’ tha’y!(?).

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Ta co the’ viet

P (n, r) = n(n − 1)(n − 2) · · · (n − r + 1)

=n(n − 1)(n − 2) · · · (n − r + 1)(n − r) · · · 2 · 1

(n − r) · · · 2 · 1=

n!

(n − r)!.

D- i.nh nghıa 4.2.5. Xet ta.p X chu.a n phan tu.’ phan bie.t. Mo.t r-to’ ho.. p cu’a ta.p X la mo.tbo. r phan tu.’ , khong phan bie.t thu. tu.. , lay tu. ta.p nay. So cac r-to’ ho.. p cu’a ta.p gom n phan

tu.’ phan bie.t ky hie.u la C(n, r) hay

(nr

)va go. i C(n, r) la to’ ho.. p cha.p r cu’a n phan tu.’ .

Chung ta se xac di.nh cong thu.c cho C(n, r) bang cach dem so cac r-hoan vi. cu’a ta.p gomn phan tu.’ theo hai cach. Thu. nhat, su.’ du.ng cong thu.c P (n, r). Cach thu. hai la dem so cacr-hoan vi. cu’a ta.p gom n phan tu.’ co lien quan vo.i C(n, r). Tu. do se suy ra ket qua’.

Ta co the’ xay du.. ng r-hoan vi. cu’a ta.p n phan tu.’ phan bie.t qua hai bu.o.c lien tiep: D- autien, cho.n mo.t r-to’ ho.. p cu’a X (ta.p con r phan tu.’ khong phan bie.t thu. tu.. ) va sau do sapthu. tu.. no. Cha’ng ha.n, de’ xay du.. ng mo.t 2-hoan vi. cu’a {a, b, c, d} ta co the’ cho.n 2-to’ ho.. pva sau do sap thu. tu.. no. Theo nguyen ly tıch, so cac r-hoan vi. bang tıch cu’a so cac r-to’

ho.. p va so cac cach sap thu. tu.. cu’a r phan tu.’ . Tu.c la

P (n, r) = C(n, r)r!.

Va.y

C(n, r) =P (n, r)

r!.

Do do theo D- i.nh ly 4.2.4 ta co

D- i.nh ly 4.2.6. So cac r-hoan vi. cu’a ta. p n phan tu.’ phan bie. t la

C(n, r) =n!

(n − r)!r!, r ≤ n.

Vı du. 4.2.9. Co bao nhieu cach cho.n 5 ngu.o.i tu. 10 ngu.o.i de’ la.p thanh mo.t do. i bong(khong phan bie.t thu. tu.. )?

Cau tra’ lo.i la bang so to’ ho.. p cha.p 5 cu’a 10 phan tu.’

C(10, 5) =10!

5!5!= 252.

Vı du. 4.2.10. Co bao nhieu cach cho.n mo.t ho. i dong gom hai ngu.o.i nu. va ba ngu.o.i namtu. mo.t nhom nam ngu.o.i nu. va sau ngu.o.i nam?

So cach cho.n hai ngu.o.i nu. va ba ngu.o.i nam tu.o.ng u.ng la C(5, 2) = 10 va C(6, 3) = 20.

Ho.i dong du.o.. c xay du.. ng qua hai bu.o.c lien tiep: Cho.n ngu.o.i nu.; cho.n ngu.o.i nam. Theonguyen ly tıch, to’ng so cac ho. i dong la 10 · 20 = 200.

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Hınh 4.1:

Vı du. 4.2.11. Co bao nhieu chuoi tam bit chu.a chınh xac bon bit 1?

Mo.t chuoi tam bit chu.a bon bit 1 du.o.. c xac di.nh duy nhat ngay khi chung ta biet cac bitnao bang 1. Nhu.ng dieu nay co the’ thu.. c hie.n bo.’ i C(8, 4) cach.

Vı du. 4.2.12. Co bao nhieu hanh trınh tu. goc du.o.i ben trai cu’a mo.t ban co. vuong kıchthu.o.c n×n den goc tren ben pha’i neu chung ta chı’ di theo cach sang pha’i va len tren? Mo.thanh trınh nhu. va.y tren ban co. 4 × 4 du.o.. c cho trong Hınh 4.1.

Moi hanh trınh co the’ du.o.. c mo ta’ bo.’ i mo.t chuoi do. dai 2n cu’a n ky tu.. R va n ky tu.. U.Cha’ng ha.n, hanh trınh trong Hınh 4.1 tu.o.ng u.ng chuoi RUURRURU. Mo.t chuoi nhu. va.yco the’ nha.n du.o.. c bang cach cho.n n vi. trı doi vo.i R (khong phan bie.t thu. tu.. ) trong so 2nvi. trı cho phep cu’a chuoi va sau do chen n ky tu.. U vao nhu.ng vi. trı con la. i. Do do so hanhtrınh la C(2n, n).

Bai ta.p

1. Co bao nhieu hoan vi. cu’a a, b, c, d? Lie.t ke cac hoan vi. nay.

2. Co bao nhieu 3-hoan vi. cu’a a, b, c, d? Lie.t ke cac hoan vi. nay.

3. Co bao nhieu hoan vi., 5-hoan vi. cu’a 11 doi tu.o.. ng khac nhau?

4. Co bao nhieu cach cho.n mo.t chu’ ti.ch, mo. t pho chu’ ti.ch va mo.t thu. ky tu. mo.t nhom11 ngu.o.i?

5. Co bao nhieu cach cho.n mo.t chu’ ti.ch, mo.t pho chu’ ti.ch, mo.t ke toan va mo.t thu. kytu. mo.t nhom 12 ngu.o.i?

6. Co bao nhieu chuoi do. dai 5 co phan bie.t thu. tu.. du.o.. c ta.o ra tu. cac ky tu.. A,B,C,D,Eneu:

(a) Chu.a chuoi con ACE.

(b) Chu.a cac ky tu.. ACE theo thu. tu.. tuy y.

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(c) Chu.a cac chuoi con DB va AE.

(d) Chu.a hoa. c chuoi con AE hoa. c EA.

(e) Ky tu.. A xuat hie.n tru.o.c ky tu.. D. Cha’ng ha.n BCAED,BCADE.

(f) Khong chu.a cac chuoi con AB,CD.

(g) Ky tu.. A xuat hie.n tru.o.c ky tu.. C va C xuat hie.n tru.o.c E.

7. D- a.t X := {a, b, c, d}.(a) Tım so cac 3-to’ ho.. p cu’a X. Lie.t ke cac to’ ho.. p nay.

(b) Tım moi quan he. giu.a cac 3-to’ ho.. p va 3-hoan vi. cu’a X.

8. Co bao nhieu cach cho.n mo.t ho. i dong gom ba ngu.o.i tu. nhom 11 ngu.o.i?

9. Co bao nhieu cach cho.n mo.t ho. i dong gom bon ngu.o.i tu. nhom 12 ngu.o.i?

10. Mo.t cau la.c bo. gom sau ngu.o.i nam va ba’y ngu.o.i nu..

(a) Co bao nhieu cach cho.n mo.t ho. i dong gom nam ngu.o.i?

(b) Co bao nhieu cach cho.n mo.t ho. i dong gom ba nam va bon nu.?

(c) Co bao nhieu cach cho.n mo. t ho. i dong gom bon ngu.o.i va ıt nhat mo.t nu.?

(d) Co bao nhieu cach cho.n mo.t ho. i dong gom bon ngu.o.i vo.i nhieu nhat mo.t nam?

(e) Co bao nhieu cach cho.n mo. t ho. i dong gom bon ngu.o.i co ca’ nam va nu.?

11. (a) Co bao nhieu chuoi 8 bit chu.a chınh xac ba bit 0?

(b) Co bao nhieu chuoi 8 bit chu.a ba bit 0 va 5 bit 1?

(c) Co bao nhieu chuoi 8 bit chu.a ıt nhat hai bit 0?

12. Mo.t cu.’ a hang co 50 may tınh trong do co bon bi. ho’ng.

(a) Co bao nhieu cach cho.n bon may tınh?

(b) Co bao nhieu cach cho.n bon may tınh khong ho’ng?

(c) Co bao nhieu cach cho.n bon may tınh trong do co hai chiec bi. ho’ng?

(d) Co bao nhieu cach cho.n bon may tınh trong do co ıt nhat mo.t chiec bi. ho’ng?

13. Xet mo.t hanh trınh tren ban co. kıch thu.o.c m × n tu. goc trai ben du.o.i den goc trenben pha’i va theo hu.o.ng hoa. c sang pha’i hoa. c len tren.

(a) So hanh trınh co the’ la bao nhieu?

(b) Ap du.ng de’ chu.ng minh da’ng thu.c

n∑

k=0

C(k + m− 1, k) = C(m + n,m).

14. Chu.ng minh rang so cac chuoi bit do. dai n ≥ 4 chu.a chınh xac hai lan xuat hie.n cu’achuoi bit 10 la C(n + 1, 5).

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15. Chu.ng minh rang so cac chuoi bit do. dai n chu.a chınh xac k bit 0 sao cho hai bit 0khong xuat hie.n lien tiep la C(n − k + 1, k).

16. Chu.ng minh rang tıch cu’a k so nguyen lien tiep chia het cho k!.

17. Chu.ng minh rang co (2n − 1)(2n − 3) · . . . · 3 · 1 cach cho.n n ca.p tu. 2n phan tu.’ phanbie.t.

18. Gia’ su.’ co n doi tu.o.. ng trong do co r doi tu.o..ng phan bie.t va n−r la dong nhat. Chu.ngminh cong thu.c

P (n, r) = r!C(n, r)

bang cach dem so co phan bie.t thu. tu.. cu’a n doi tu.o.. ng theo hai cach:

+ D- au tien dem so co phan bie.t thu. tu.. cac vi. trı cu’a r doi tu.o.. ng phan bie.t.

+ D- au tien dem so co phan bie.t thu. tu.. cac vi. trı cu’a n − r doi tu.o.. ng dong nhat.

4.3 Cac thua. t toan sinh ra hoan vi. va to’ ho.. p

Nhom nha.c rock cu’a tru.o.ng D- a. i ho.c D- a La. t co n bai hat can ghi len mo.t dıa CD. Cac baihat chiem tho.i gian (tınh bang giay) tu.o.ng u.ng la

t1, t2, . . . , tn.

D- ıa CD co the’ lu.u tru. nhieu nhat la C giay. Vı day la dıa CD dau tien cu’a nhom, nen ho.muon ghi cac bai hat vo.i tho.i lu.o.. ng cang nhieu cang tot. Do do bai toan la cho.n mo.t ta.pcon {i1, i2, . . . , ik} cu’a {1, 2, . . . , n} sao cho to’ng

k∑

j=1

tij (4.1)

khong vu.o.. t qua C va lo.n nhat co the’. Cach tiep ca.n la kie’m tra tat ca’ cac ta.p con cu’a{1, 2, . . . , n} va cho.n mo.t ta.p con sao cho to’ng (4.1) lo.n nhat co the’. D- e’ thu.. c hie.n chungta can mo.t thua.t toan ta.o ra tat ca’ cac to’ ho.. p cu’a ta.p gom n phan tu.’ . Phan nay trınh baycac thua.t toan sinh ra cac hoan vi. va to’ ho.. p.

Do co 2n ta.p con cu’a ta.p gom n phan tu.’ nen tho.i gian thu.. c hie.n cu’a thua.t toan kie’m tratat ca’ cac ta.p con ıt nhat la O(2n). Nhu.ng thua.t toan nhu. va.y la khong ho.. p ly ngoa. i tru.

vo.i nhu.ng gia tri. n nho’. Tuy nhien co nhu.ng bai toan ma de’ gia’i no khong co cach nao totho.n la “lie.t ke” tat ca’ cac tru.o.ng ho.. p.

Phu.o.ng phap lie.t ke tat ca’ cac to’ ho.. p va cac hoan vi. theo “thu. tu.. tu. die’n”: Vo.i hai tu.

da cho, de’ xac di.nh tu. nao du.ng tru.o.c trong tu. die’n, chung ta so sanh cac ky tu.. trong tu..Co hai kha’ nang:

(a) Moi ky tu.. trong tu. ngan ho.n trung vo.i ky tu.. tu.o.ng u.ng trong tu. dai ho.n.

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(b) Ta.i mo.t vi. trı nao do, cac ky tu.. trong hai tu. khac nhau.

Neu (a) dung, tu. ngan ho.n se du.ng tru.o.c. Cha’ng ha.n, “dog” du.ng tru.o.c “doghouse”trong tu. die’n. Neu (b) dung chung ta xac di.nh vi. trı ben trai nhat p ma ta. i do cac ky tu..khac nhau. Thu. tu.. cu’a cac tu. du.o.. c xac di.nh bo.’ i thu. tu.. cu’a cac ky tu.. ta. i vi. trı p. Cha’ngha.n, “nha” du.ng tru.o.c “nhanh” trong tu. die’n.

D- e’ do.n gia’n ta se di.nh nghıa thu. tu.. tu. die’n tren ta.p cac ky hie.u la cac so tu.. nhien.

D- i.nh nghıa 4.3.1. Gia’ su.’ α = s1s2 . . . sp va β = t1t2 . . . tq la cac chuoi tren ta.p {1, 2, . . . , n}.Ta noi α co thu. tu.. tu. die’n nho’ ho.n β, ky hie.u α < β, neu hoa. c

(a) p < q va si = ti vo.i i = 1, 2, . . . , p; hoa. c

(b) Ton ta. i i sao cho si 6= ti, va vo.i chı’ so i nho’ nhat nhu. va.y, ta co si < ti.

Vı du. 4.3.1. Tren ta.p {1, 2, 3, 4} ta co α = 132 < β = 1324. Tren ta.p {1, 2, 3, 4, 5, 6} ta coα = 13246 < β = 1342.

D- au tien ta xet bai toan lie.t ke tat ca’ cac r-to’ ho.. p cu’a ta.p {1, 2, . . . , n}. Trong thua.ttoan, chung ta se lie.t ke r-to’ ho.. p {x1, x2, . . . , xr} tu.o.ng u.ng chuoi s1s2 . . . sr trong dos1 < s2 < · · · < sr va {x1, x2, . . . , xr} = {s1, s2, . . . , sr}. Cha’ng ha.n, 3-to’ ho.. p {6, 2, 4} setu.o.ng u.ng chuoi 246.

Ta se lie.t ke cac r-to’ ho.. p cu’a ta.p {1, 2, . . . , n} theo thu. tu.. tu. die’n. Do do, cac chuoi du.o.. clie.t ke dau tien va cuoi cung tu.o.ng u.ng la 12 . . . r va (n − r + 1) . . . n.

Vı du. 4.3.2. Lie.t ke tat ca’ 5-to’ ho.. p cu’a {1, 2, 3, 4, 5, 6, 7}.

Chuoi dau tien la 12345, theo sau la 12346 va 12347. Chuoi ke tiep la 12356 va sau do12357. Chuoi cuoi cung la 34567.

Vı du. 4.3.3. Tım chuoi tiep theo 13467 khi chung ta lie.t ke 5-to’ ho.. p cu’a ta.p ho.. p X :={1, 2, 3, 4, 5, 6, 7}.

Khong co chuoi nao bat dau vo.i 134 va cac bie’u dien cu’a mo.t to’ ho.. p 5 phan tu.’ cu’a Xpha’i lo.n ho.n 13467. Do do chuoi tiep theo 13467 pha’i bat dau la 135. Vı 13567 la chuoinho’ nhat bat dau bang 135 va la mo.t to’ ho.. p cu’a 5 phan tu.’ cu’a X nen 13567 la to’ ho.. p pha’itım.

Vı du. 4.3.4. Tım chuoi tiep theo 2367 khi chung ta lie.t ke 4-to’ ho.. p cu’a ta.p ho.. p X :={1, 2, 3, 4, 5, 6, 7}.

Khong co chuoi nao bat dau vo.i 23 va cac bie’u dien cu’a mo.t to’ ho.. p 4 phan tu.’ cu’a Xpha’i lo.n ho.n 2367. Do do chuoi tiep theo 2367 pha’i bat dau la 24. Vı 2456 la chuoi nho’

nhat bat dau bang 24 va la mo.t to’ ho.. p cu’a 5 phan tu.’ cu’a X nen 2456 la to’ ho.. p pha’i tım.

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Xet chuoi α = s1s2 . . . sr bie’u dien to’ ho.. p {x1, x2, . . . , xr}. D- e’ tım chuoi ke tiep β =t1t2 . . . tr ta tım phan tu.’ ben pha’i nhat sm ma khong pha’i la gia tri. cu.. c da. i cu’a no ta. i do.(sr co the’ lay gia tri. cu.. c da. i n, sr−1 co the’ lay gia tri. cu.. c da. i n − 1, . . . ). Khi do

ti = si, vo.i i = 1, 2, . . . ,m − 1.

Phan tu.’ tm bang sm + 1. Nhu.ng phan tu.’ con la. i cu’a chuoi β xac di.nh bo.’ i

tm+1 = sm + 2, tm+2 = sm + 3, . . . .

Thua.t toan sinh cac to’ ho.. p

Bu.o.c 1. [Kho.’ i ta.o chuoi] D- a.t si = i, i = 1, 2 . . . , r.

Bu.o.c 2. [Xuat to’ ho.. p dau tien] Xuat chuoi s = s1s2 . . . sr.

Bu.o.c 3. [La.p] Vo.i moi i = 2, 3, . . . , C(n, r) thu.. c hie.n cac bu.o.c sau:

3.1. Tım phan tu.’ ben pha’i nhat khong pha’i la gia tri. cu.. c da. i cu’a no.

3.2. (Gia tri. cu.. c da. i cu’a sk du.o.. c di.nh nghıa la n − r + k).

3.3. D- a.t sm = sm + 1.

3.4. Vo.i moi j = m + 1, . . . , r, da. t sj = sj−1 + 1.

3.5. Xuat s.

Vı du. 4.3.5. Xet ta.p {1, 2, 3, 4, 5, 6, 7}. Gia’ su.’

s1 = 2, s2 = 3, s3 = 4, s4 = 6, s5 = 7.

Ta co s3 la phan tu.’ ben pha’i nhat khong pha’i la gia tri. cu.. c da. i cu’a no ta. i do. Ap du.ngthua.t toan tren, ta co chuoi tiep theo 23467 la 23567.

Vı du. 4.3.6. Thua.t toan ta.o 4-to’ ho.. p cu’a {1, 2, 3, 4, 5, 6} cho ta

1234, 1235, 1236, 1246, 1256, 1345, 1346,

1356, 1456, 2345, 2346, 2356, 3456.

Tu.o.ng tu.. thua.t toan sinh cac to’ ho.. p, thua. t toan sinh cac hoan vi. se lie.t ke theo thu. tu..tu. die’n.

Vı du. 4.3.7. D- e’ xay du.. ng hoan vi. cu’a ta.p {1, 2, 3, 4, 5, 6} sau hoan vi. 163542, chung ta canco di.nh cac chu. so ben trai nhieu nhat co the’.

Ton ta. i hoan vi. tiep theo hoan vi. 1635 ? Vı hoan vi. co da.ng 1635 khac hoan vi. da chola 163524 va 163524 nho’ ho.n 163542 nen hoan vi. sau 163542 khong the’ co da.ng 1635 .

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Ton ta. i hoan vi. tiep theo hoan vi. 163 ? Ba chu. so cuoi cung pha’i la mo.t hoan vi. cu’a{2, 4, 5}. Vı 542 la hoan vi. lo.n nhat cu’a {2, 4, 5} nen hoan vi. bat ky vo.i ba chu. so bat dau163 nho’ ho.n hoan vi. 63542. Va.y hoan vi. sau hoan vi. da cho khong the’ co da.ng 163 .

Hoan vi. tiep theo cu’a 163542 khong the’ bat dau la 1635 hay 163 do hoa. c cac chu. so conla. i trong hoan vi. da cho (42 va 542, tu.o.ng u.ng) la gia’m. Do do, bat dau tu. ben pha’i, chungta can tım chu. so dau tien d ma lan ca.n ben pha’i cu’a no la r thoa’ man d < r. Trong tru.o.ngho.. p tren, chu. so thu. ba: 3 co tınh chat nay. Va.y hoan vi. tiep theo hoan vi. da cho se batdau la 16. Chu. so tiep theo khong the’ nho’ ho.n 3. Vı ta muon hoan vi. tiep theo nho’ nhat,nen chu. so ke tiep la 4. Do do hoan vi. tiep theo bat dau vo.i 164. Cac chu. so con la. i: 235can tang vo.i gia tri. nho’ nhat. Va.y hoan vi. tiep theo hoan vi. da cho la 164235.

Nha.n xet rang de’ ta.o tat ca’ cac hoan vi. cu’a ta.p {1, 2, . . . , n} chung ta co the’ bat dau vo.ihoan vi. 12 . . . n va la.p la. i phu.o.ng phap cu’a Vı du. 4.3.7 de’ ta.o hoan vi. ke tiep. Thua.t toanket thuc khi ta.o ra hoan vi. n(n − 1) . . . 21.

Vı du. 4.3.8. Ap du.ng phu.o.ng phap cu’a Vı du. 4.3.7, ta co the’ lie.t ke tat ca’ cac hoan vi.cu’a {1, 2, 3, 4} theo thu. tu.. tu. die’n nhu. sau:

1234, 1243, 1324, 1342, 1423, 1432, 2134, 2143,2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241,3412, 3421, 4123, 4132, 4213, 4231, 4312, 4321.

Thua.t toan sinh cac hoan vi.

Bu.o.c 1. [Kho.’ i ta.o chuoi] D- a.t si = i, i = 1, 2 . . . , n.

Bu.o.c 2. [Xuat hoan vi. dau tien] Xuat chuoi s = s1s2 . . . sn.

Bu.o.c 3. [La.p] Vo.i moi i = 2, 3, . . . , n! thu.. c hie.n cac bu.o.c sau:

3.1. Tım chı’ so lo.n nhat m thoa’ man sm < sm+1.

3.2. Tım chı’ so lo.n nhat k thoa’ man sk > sm.

3.3. Hoan vi. hai phan tu.’ sm va sk.

3.4. D- a’o ngu.o.. c thu. tu.. cu’a cac phan tu.’ sm+1, . . . , sn.

3.5. Xuat s.

Vı du. 4.3.9. Ap du.ng thua.t toan tren tım hoan vi. tiep theo 163542: Gia’ su.’

s1 = 1, s2 = 6, s3 = 3, s4 = 5, s5 = 4, s6 = 2.

Chı’ so m lo.n nhat thoa’ sm < sm+1 la 3. Chı’ so k lo.n nhat thoa’ sk > sm la 5. Hoan vi. sm

va sk ta co s3 = 4, s5 = 3. D- a’o ngu.o.. c thu. tu.. cac phan tu.’ s4, s5, s6 ta nha.n du.o.. c hoan vi.tiep theo la 164235.

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Bai ta.p

1. Tım r-to’ ho.. p sinh ra bo.’ i thua.t toan sinh to’ ho.. p vo.i n = 7 sau khi r-to’ ho.. p du.o.. c cho:1356, 12367, 14567.

2. Tım hoan vi. sinh ra bo.’ i thua.t toan sinh hoan vi. sau hoan vi. du.o.. c cho: 12354, 625431,12876543.

3. Tım tat ca’ r-to’ ho.. p tu. ta.p n phan tu.’ neu

(a) n = 6, r = 3.

(b) n = 6, r = 2.

(c) n = 7, r = 5.

4. Tım cac hoan vi. cu’a ta.p hai, ba phan tu.’ .

5. Viet thua.t toan de. quy sinh ra tat ca’ cac r-to’ ho.. p cu’a ta.p {s1, s2, . . . , sn}. Chia baitoan thanh hai bai toan con:

+ Lie.t ke cac r-to’ ho.. p chu.a s1.

+ Lie.t ke cac r-to’ ho.. p khong chu.a s1.

6. Viet thua.t toan de. quy sinh ra tat ca’ cac hoan vi. cu’a ta.p {s1, s2, . . . , sn}. Chia baitoan thanh n bai toan con:

+ Lie.t ke cac hoan vi. bat dau vo.i s1.

+ Lie.t ke cac hoan vi. bat dau vo.i s2.

...

+ Lie.t ke cac hoan vi. bat dau vo.i sn.

4.4 Hoan vi. va to’ ho.. p suy ro.ng

Trong cac mu.c tru.o.c, chung ta da nghien cu.u cac hoan vi. va to’ ho.. p khong cho phep la.p la.icac phan tu.’ . Phan nay tım hie’u cac hoan vi. cu’a cac day chu.a nhu.ng phan tu.’ la.p la. i va cacphep cho.n khong phan bie.t thu. tu.. co la.p la. i. Tru.o.c het ta xet vı du. sau.

Vı du. 4.4.1. Trong nhieu van de dem, cac phan tu.’ co the’ la.p la. i; cha’ng ha.n co bao nhieuxau khac nhau co do. dai n tu. ba’ng 26 chu. cai?

Hie’n nhien o.’ day, co the’ coi cac chu. cai du.o.. c rut ra co hoan la. i. Mo.t xau do. dai n gomn chu. cai. Moi chu. cai co 26 cach cho.n lu.. a. Theo nguyen ly tıch, so xau co the’ la

26 × 26 × · · · × 26︸︷︷︸n lan

= 26n.

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D- i.nh ly 4.4.1. So cac r-hoan vi. co la. p la. i cu’a ta. p n phan tu.’ bang nr.

Chu.ng minh. Co n cach cho.n cho moi vi. trı trong r-hoan vi. (vı co la.p la. i). Ap du.ngnguyen ly tıch, so cac r-hoan vi. co la.p la. i bang nr. 2

Vı du. 4.4.2. Xet chuoi SUCCESS. Co bao nhieu chuoi khac nhau co the’ co khi sap xepla. i cac ky tu.. cu’a chuoi nay?

Tru.o.c het chu y rang trong chuoi SUCCESS do. dai 7 co ba ky tu.. S, hai ky tu.. C, mo.tky tu.. U va mo.t ky tu.. E. Ba ky tu.. S (tu.o.ng u.ng, hai ky tu.. C) la khong phan bie.t, nenhoan vi. chung khong ta.o ra chuoi mo.i.

Co tat ca’ 7! chuoi la hoan vi. cu’a chuoi SUCCESS. Ba ky tu.. S hoan vi. ta.o ra 3! chuoi;hai ky tu.. C hoan vi. ta.o ra 2! chuoi; mo.t ky tu.. U hoan vi. ta.o ra 1! chuoi; va mo.t ky tu.. Ehoan vi. ta.o ra 1! chuoi. Va.y so chuoi tha.t su.. khac nhau la

7!

3!2!1!1!.

Vı du. 4.4.3. Xet chuoi MISSISSIPPI. Co bao nhieu chuoi khac nhau co the’ co khi sapxep la. i cac ky tu.. cu’a chuoi nay?

Xet bai toan dien vao 11 cho trong

−−−−−−−−−−−,

vo.i cac ky tu.. da cho. Co C(11, 2) cach cho.n cac vi. trı doi vo.i P. Khi da cho.n xong P, ta coC(9, 4) cach cho.n cac vi. trı doi vo.i S. Khi da cho.n S, co C(5, 4) cach cho.n cac vi. trı doi vo.iI. Cuoi cung chı’ con mo.t cach cho.n M. Theo nguyen ly tıch, so cac cach de’ dien cac ky tu..la

C(11, 2)C(9, 4)C(5, 4) =11!

2!9!

9!

4!5!

5!

4!1!

=11!

2!4!4!1!= 34.650.

To’ng quat ta co

D- i.nh ly 4.4.2. Gia’ su.’ day n phan tu.’ S co n1 doi tu.o.. ng loa. i 1, n2 doi tu.o.. ng loa. i 2, ...,va nt doi tu.o.. ng loa. i t. Khi do so cac cach cho. n day S la

n!

n1!n2! . . . nt!.

Chu.ng minh. Ta gan cac vi. trı doi vo.i moi day do. dai n cac doi tu.o.. ng de’ ta.o ra mo.t thu.

tu.. trong S. Co C(n, n1) cach cho.n cac vi. trı doi vo.i cac doi tu.o.. ng loa. i 1. Khi da cho.n xong

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cac doi tu.o.. ng nay, ta co C(n − n1, n2) cach cho.n cac vi. trı doi vo.i cac doi tu.o.. ng loa. i 2, vavan van. Theo nguyen ly tıch, so cac cach de’ thu.. c hie.n la

C(n, n1)C(n − n1, n2) · · ·C(n− n1 − n2 − · · · − nt−1, nt)

va do do co dieu can chu.ng minh. 2

Ke tiep chung ta kha’o sat bai toan dem cac phep cho.n khong phan bie.t thu. tu.. co la.p la. i.

Vı du. 4.4.4. Xet ba loa. i sach: sach may tınh, sach va.t ly va sach li.ch su.’ . Gia’ su.’ thu. vie.nco ıt nhat sau cuon sach moi loa. i. Co bao nhieu cach co the’ cho.n sau cuon sach?

Bai toan la lay sau phan tu.’ khong phan bie.t thu. tu.. tu. ta.p {may tınh, va.t ly, li.ch su.’} chophep la.p la. i. Mo.t phep cho.n du.o.. c xac di.nh duy nhat bo.’ i so moi kie’u sach du.o.. c cho.n. Kyhie.u

May tınh Va.t ly Li.ch su.’

× × × | × × | ×

co nghıa la phep cho.n ba cuon sach may tınh, hai sach va.t ly va mo.t sach li.ch su.’ . Nha.n xetrang moi thu. tu.. cu’a sau ky hie.u × va hai ky hie.u | tu.o.ng u.ng mo.t phep cho.n. Do do baitoan la dem so cac thu. tu.. . Va.y co the’ thu.. c hie.n bang C(8, 2) = 28 cach.

D- i.nh ly 4.4.3. Neu X la ta. p gom t phan tu.’ thı so phep cho. n k phan tu.’ khong phan bie. tthu. tu.. tu. X cho phep la. p la

C(k + t− 1, t − 1) = C(k + t− 1, k).

Chu.ng minh. D- a.t X := {a1, a2, . . . , at}. Xet k + t − 1 khoa’ng trang

. . .

gom k ky hie.u × va t − 1 ky hie.u |. Moi vi. trı cu’a ky hie.u nay tren cac khoa’ng trang xacdi.nh mo. t phep cho.n. n1 ky hie.u × den ky hie.u | dau tien tu.o.ng u.ng phep cho.n n1 phan tu.’

a1; n2 ky hie.u × den ky hie.u | thu. hai tu.o.ng u.ng phep cho.n n2 phan tu.’ a2; va van van. Taco C(k + t− 1, t− 1) cach cho.n cac vi. trı cho | nen co C(k + t− 1, t− 1) cach cho.n. Gia tri.nay bang C(k + t− 1, k), so cach cho.n cac vi. trı cu’a ×; do do co

C(k + t− 1, t − 1) = C(k + t − 1, k)

cach cho.n k phan tu.’ khong phan bie.t thu. tu.. tu. ta.p X cho phep la.p la. i. 2

Vı du. 4.4.5. Co cac ho.p chu.a cac qua’ bong mau do’, xanh va vang. Moi ho.p chu.a ıt nhattam qua’ bong. Co bao nhieu cach cho.n tam qua’ bong? Co bao nhieu cach cho.n tam qua’

bong, moi mau ıt nhat mo.t qua’ bong?

(a) Theo D- i.nh ly 4.4.3, so cach cho.n tam qua’ bong la

C(8 + 3 − 1, 3 − 1) = C(10, 2) = 45.

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(b) D- au tien cho.n mo. t qua’ bong moi mau; sau do cho.n them nam qua’ bong. Theo D- i.nh ly4.4.3 ta co

C(5 + 3 − 1, 3 − 1) = C(7, 2) = 21

cach.

Vı du. 4.4.6. (a) Co bao nhieu nghie.m nguyen khong am cu’a phu.o.ng trınh

x1 + x2 + x3 + x4 = 29? (4.2)

Moi nghie.m cu’a phu.o.ng trınh (4.2) tu.o.ng du.o.ng vo.i phep cho.n 29 phan tu.’ xi co kie’u ivo.i i := 1, 2, 3, 4. Theo D- i.nh ly 4.4.3, so phep cho.n la

C(29 + 4 − 1, 4 − 1) = C(32, 3) = 4960.

(b) Co bao nhieu nghie.m nguyen cu’a phu.o.ng trınh (4.2) thoa’ man

x1 > 0, x2 > 1, x3 > 2, x4 ≥ 0?

Moi nghie.m cu’a (4.2) thoa’ dieu kie.n da cho tu.o.ng du.o.ng vo.i phep cho.n 29 phan tu.’ xi

co kie’u i, i = 1, 2, 3, 4, sao cho can ıt nhat mo.t phan tu.’ co kie’u 1, ıt nhat hai phan tu.’ cokie’u 2, ıt nhat ba phan tu.’ co kie’u 3. D- au tien cho.n mo.t phan tu.’ co kie’u 1, hai phan tu.’ cokie’u 2 va ba phan tu.’ co kie’u 3. Sau do cho.n them 23 phan tu.’ con la.i. Theo D- i.nh ly 4.4.3so phep cho.n la

C(23 + 4 − 1, 4 − 1) = C(26, 3) = 2600.

Chung ta ket thuc phan nay vo.i vie.c mo.’ ro.ng nguyen ly bao ham-loa. i tru..

Xet tru.o.ng ho.. p co ba su.. kie.n A,B,C. Ta can tınh #(A ∪ B ∪ C). Nha.n xet la

(a) Neu lay #A + #B + #C : co phan du.o.. c tınh mo.t lan, hai lan va ba lan (Hınh 4.2(a));

(b) Neu lay #A + #B + #C − #(A ∩ B) − #(A ∩ C)− #(B ∩ C) : co phan khong du.o.. ctınh lan nao (Hınh 4.2(b));

(c) Neu lay #A + #B + #C − #(A ∩ B) − #(A ∩ C) − #(B ∩ C) + #(A ∩ B ∩ C) : moiphan du.o.. c tınh dung mo.t lan (Hınh 4.2(c)).

Va.y

#(A ∪ B ∪ C) = #A + #B + #C −#(A ∩ B) −#(A ∩ C)− #(B ∩ C) + #(A ∩ B ∩ C).

To’ng quat ta co

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Hınh 4.2:

D- i.nh ly 4.4.4. Gia’ su.’ co m su.. kie. n A1, A2, . . . Am. Khi do

#(A1 ∪ A2 ∪ · · · ∪ Am) =m∑

i=1

#Ai −∑

1≤i<j≤m

#(Ai ∩ Aj) +∑

1≤i<j<k≤m

#(Ai ∩ Aj ∩ Ak)

+ · · · + (−1)m+1#(A1 ∩ A2 ∩ · · · ∩ Am).

Chu.ng minh. Ta se chu.ng minh rang lay mo.t phan tu.’ a bat ky thuo.c ta.p A1∪A2∪· · ·∪Am

thı a cung du.o.. c ke’ den dung mo.t lan o.’ ve pha’i.

Gia’ su.’ a thuo.c dung r ta.p, cha’ng ha.n trong A1 ∩ A2 ∩ · · · ∩ Ar, r ≤ m. Phan tu.’ nay dadu.o.. c tınh

+ C(r, 1) lan trong∑m

i=1 #Ai;

+ C(r, 2) lan trong∑m

i=1 #(Ai ∩ Aj);

...

+ C(r,m) lan trong∑m

i=1 #(Ai1 ∩ Ai2 ∩ · · ·Aim).

Va.y no da du.o.. c tınh to’ng co.ng so lan la

C(r, 1) − C(r, 2) + C(r, 3)− · · · + (−1)m+1C(r, r).

Nhu.ngC(r, 0)− C(r, 1) + C(r, 2) − · · · + (−1)rC(r, r) = 0

va C(r, 0) = 1. Va.y phan tu.’ a da du.o.. c tınh

C(r, 1) − C(r, 2) + C(r, 3)− · · · + (−1)r+1C(r, r) = 1

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lan. 2

Vı du. 4.4.7. Co bao nhieu nghie.m nguyen khong am cu’a phu.o.ng trınh

x1 + x2 + x3 = 11 (4.3)

vo.i dieu kie.n x1 ≤ 3, x2 ≤ 4 va x3 ≤ 6?

Tu.o.ng tu.. nhu. Vı du. 4.4.6, ta co

+ To’ng so nghie.m nguyen khong am cu’a phu.o.ng trınh (4.3) la

C(11 + 3 − 1, 11) = C(13, 11) = 78.

+ So nghie.m vo.i dieu kie.n x1 ≥ 4 la

C(7 + 3 − 1, 7) = C(9, 7) = 36.


C(6 + 3 − 1, 6) = C(8, 6) = 28.


C(4 + 3 − 1, 4) = C(6, 4) = 15.

+ So nghie.m vo.i dieu kie.n x1 ≥ 4, x2 ≥ 5 la

C(2 + 3 − 1, 2) = C(4, 2) = 6.

+ So nghie.m vo.i dieu kie.n x1 ≥ 4, x3 ≥ 7 la

C(0 + 3 − 1, 0) = C(2, 0) = 1.

+ So nghie.m vo.i dieu kie.n x2 ≥ 5, x3 ≥ 7 bang 0.

+ So nghie.m vo.i dieu kie.n x1 ≥ 4, x2 ≥ 4, x3 ≥ 7 bang 0.

Theo D- i.nh ly 4.4.4, so nghie.m doi ho’i la

78 − 36 − 28 − 15 + 6 + 1 + 0 − 0 = 6.

D- i.nh ly 4.4.5. Gia’ su.’ m,n la cac so nguyen du.o.ng khac nhau, m ≤ n. Khi do co

nm − C(n, 1)(n − 1)m + C(n, 2)(n − 2)m − · · · + (−1)n−1C(n, n− 1)1m

anh xa. len khac nhau tu. ta. p m phan tu.’ den ta. p co n phan tu.’ .


Vı du. 4.4.8. Gia’ su.’ co nam cong vie.c va bon ngu.o.i xin vie.c. Co bao nhieu cach phan congvie.c khac nhau neu moi ngu.o.i pha’i du.o.. c phan cong ıt nhat mo.t cong vie.c?

Moi phu.o.ng phap phan cong tu.o.ng u.ng mo.t anh xa. len tu. ta.p cac cong vie.c den ta.pngu.o.i. Theo gia’ thiet, moi ngu.o.i deu du.o.. c phan cong ıt nhat mo.t cong vie.c, cac anh xa. lalen. Ap du.ng D- i.nh ly 4.4.5 vo.i m = 5, n = 4 ta co so cach phan cong cong vie.c bang so cacanh xa. len khac nhau va bang

45 − C(4, 1)35 + C(4, 2)25 − C(4, 3)15 = 1024 − 972 + 192 − 4 = 240.

71

Bai ta.p

1. Co bao nhieu chuoi khac nhau co the’ co khi sap xep la. i cac ky tu.. cu’a cac chuoi sau:

(a) GUIDE.

(b) SCHOOL.

(c) SALEPERSONS.

2. Co bao nhieu cach chia 10 cuon sach cho ba sinh vien sao cho sinh vien thu. nhat conam cuon, sinh vien thu. hai co ba cuon va sinh vien thu. ba co hai cuon?

3. Gia’ su.’ co cac ho.p chu.a cac qua’ bong mau xanh, do’ va vang. Moi ho.p chu.a ıt nhat10 qua’.

(a) Co bao nhieu cach cho.n 10 qua’ bong?

(b) Co bao nhieu cach cho.n 10 qua’ bong vo.i ıt nhat mo.t qua’ mau do’?

(c) Co bao nhieu cach cho.n 10 qua’ bong vo.i ıt nhat mo.t qua’ mau do’, ıt nhat hai qua’

mau xanh va ıt nhat ba qua’ mau vang?

(d) Co bao nhieu cach cho.n 10 qua’ bong vo.i dung mo.t qua’ mau do’?

(e) Co bao nhieu cach cho.n 10 qua’ bong vo.i dung mo.t qua’ mau do’ va ıt nhat mo.tqua’ mau xanh?

4. Tım so nghie.m nguyen cu’a phu.o.ng trınh

x1 + x2 + x3 = 15

neu

(a) x1 ≥ 0, x2 ≥ 0, x3 ≥ 0.

(b) x1 = 1, x2 ≥ 0, x3 ≥ 0.

(c) 6 ≥ x1 ≥ 0, x2 ≥ 0, x3 ≥ 0.

(d) x1 ≥ 1, x2 ≥ 1, x3 ≥ 1.

(e) x1 ≥ 0, x2 > 0, x3 = 1.

(f) 6 > x1 ≥ 0, 9 > x2 ≥ 1, x3 ≥ 0.

5. Tım so nghie.m nguyen cu’a phu.o.ng trınh

x1 + x2 + x3 + x4 = 15

neu 0 ≤ x1 ≤ 4, 0 ≤ x2 ≤ 5, 0 ≤ x3 ≤ 9.

6. Co bao nhieu so nguyen trong ta.p {1, 2, . . . , 1000000} co to’ng cac chu. so bang 15?

7. Co bao nhieu so nguyen trong ta.p {1, 2, . . . , 1000000} co to’ng cac chu. so bang 20?

72

8. Co bao nhieu cach cho.n ba do. i: mo.t do. i bon ngu.o.i, hai do. i hai ngu.o.i tu. mo.t nhomtam ngu.o.i?

9. Mo.t tui sach chu.a 20 qua’ bong: sau do’, sau xanh va tam tım.

(a) Co bao nhieu cach cho.n nam qua’ bong neu cac qua’ bong du.o.. c xem la phan bie.t?

(b) Co bao nhieu cach cho.n nam qua’ bong neu cac qua’ bong cung mau du.o.. c xem ladong nhat?

10. Chu.ng minh rang (n!)k chia het (kn)!.

11. Chu.ng minh rangn+k−2∑

i=k−1

C(i, k − 1) = C(n + k − 1, k − 1).

12. Viet thua.t toan tım tat ca’ cac nghie.m nguyen khong am cu’a phu.o.ng trınh

x1 + x2 + x3 = n (n ∈ N).

4.5 He. so cu’a nhi. thu.c va cac dong nhat thu.c

D- i.nh ly 4.5.1. (D- i.nh ly nhi. thu.c) Neu a va b la cac so thu.. c va n la so tu.. nhien thı

(a + b)n =

n∑

k=0

C(n, k)an−kbk.

Chu.ng minh. Khi khai trie’n (a+b)n cac tu. co da.ng an−kbk, k = 0, 1, . . . , n. D- e’ co mo.t thanhphan an−kbk can co dung n − k chu. a trong to’ng so n vi. trı (va keo theo co dung k chu. b).D- ieu nay co the’ thu.. c hie.n bang C(n, k) cach. Do do an−kbk xuat hie.n C(n, k) lan. Suy ra

(a + b)n = C(n, 0)anb0 + C(n, 1)an−1b1 + · · · + C(n, n)a0bn.

2

Chınh vı ly do tren ma C(n, r) du.o.. c go. i la he. so nhi. thu.c.

Vı du. 4.5.1. Tım he. so cu’a a5b4 trong khai trie’n cu’a (a + b)9.

Theo D- i.nh ly nhi. thu.c, he. so cu’a a5b4 trong khai trie’n (a + b)9 la

C(9, 4) =9!

4!5!= 126.

73

Vı du. 4.5.2. Chu.ng minh rang

n∑

k=0

(−1)kC(n, k) = 0.

Ta co

0 = [1 + (−1)]n =n∑

k=0

C(n, k)1n−k(−1)k =n∑

k=0

(−1)kC(n, k).

Vı du. 4.5.3. Su.’ du.ng D- i.nh ly nhi. thu.c ta co

2n = (1 + 1)n =n∑

k=0

C(n, k).

D- i.nh ly 4.5.2. (D- a’ng thu.c Pascal)

C(n + 1, k) = C(n, k − 1) + C(n, k)

vo.i 1 ≤ k ≤ n.

Chu.ng minh. Gia’ su.’ X la ta.p gom n phan tu.’ . Cho.n a /∈ X. Ta co C(n + 1, k) la so cacta.p con k phan tu.’ cu’a ta.p Y := X ∪ {a}. Moi ta.p con k phan tu.’ cu’a Y co the’ chia thanhhai lo.p:

+ Cac ta.p con cu’a Y khong chu.a a.

+ Cac ta.p con cu’a Y chu.a a.

Cac ta.p con thuo.c nhom thu. nhat la cac ta.p con cu’a X gom k phan tu.’ va do do co C(n, k)ta.p con nhu. va.y.

Cac ta.p con thuo.c nhom thu. hai la cac ta.p la ho.. p cu’a ta.p con (k − 1) phan tu.’ cu’a X vo.ita.p gom mo.t phan tu.’ a va do do co C(n, k − 1) ta.p con nhu. va.y. Suy ra

C(n + 1, k) = C(n, k − 1) + C(n, k).

2

Vı du. 4.5.4. Chu.ng minh da’ng thu.c

n∑

i=k

C(i, k) = C(n + 1, k + 1).

Theo D- i.nh ly 4.5.2

C(i, k) = C(i + 1, k + 1) − C(i, k + 1), i ≥ k.

74

Va.y

n∑

i=k

C(i, k) =n∑

i=k

C(i + 1, k + 1) −n∑

i=k

C(i, k + 1)

= C(n + 1, k + 1).

Vı du. 4.5.5. Tu. da’ng thu.c (4.5.4) ta co

1 + 2 + · · · + n = C(1, 1) + C(2, 1) + · · · + C(n, 1)

= C(n + 1, 2)

=(n + 1)n

2.

D- i.nh ly 4.5.3. (D- a’ng thu.c Vandermonde)

C(m + n, r) =r∑

k=0

C(m,k)C(n, r − k)

vo.i r ≤ min(m,n).

Chu.ng minh. Gia’ su.’ cac ta.p T1, T2 tu.o.ng u.ng gom m,n phan tu.’ phan bie.t. Lay ta.p S gomr phan tu.’ tu. hai ta.p nay. So cac ta.p S nhu. va.y bang C(m + n, r).

Ma.t khac, ta.p S co the’ gom

+ k phan tu.’ thuo.c ta.p T1. So cac ta.p con nhu. va.y bang C(m,k);

+ (r − k) phan tu.’ thuo.c ta.p T2. So cac ta.p con nhu. va.y bang C(n, r − k);

vo.i 0 ≤ k ≤ r.

Theo nguyen ly tıch, sau do nguyen ly to’ng ta co dieu can chu.ng minh. 2

Bai ta.p

1. Su.’ du.ng D- i.nh ly nhi. thu.c khai trie’n cac bie’u thu.c

(a) (x + y)4.

(b) (2c − 3d)5.

2. Tım he. so cu’a so ha.ng khi bie’u thu.c du.o.. c khai trie’n:

(a) x4y7; (x + y)11.

(b) x2y3z5; (x + y + z)10.

(c) a2x3; (a + x + c)2(a + x + d)3.

75

(d) a3x4; (a +√

ax + x)2(a + x)5.

(e) a2x3; (a + ax + x)(a + x)4.

3. Tım so cac so ha.ng khi khai trie’n bie’u thu.c

(a) (x + y + z)10.

(b) (w + x + y + z)12.

(c) (x + y + z)10(w + x + y + z)2.

4. (a) Chu.ng minh rang C(n, k) < C(n, k + 1) neu va chı’ neu k < (n − 1)/2.

(b) Suy ra max{C(n, k) | k = 0, 1, . . . , n} = C(n, [n/2]).

5. Chu.ng minh D- i.nh ly nhi. thu.c bang quy na.p toan ho.c.

6. Su.’ du.ng ly lua.n to’ ho.. p chu.ng minh rang

C(n, k) = C(n, n − k).

7. Tınh to’ngn−1∑

k=1

k(k + 1).

8. Tınh to’ngn∑

k=1

k2.

9. Dung D- i.nh ly nhi. thu.c chu.ng minh

n∑

k=0

2kC(n, k) = 3n.

10. Gia’ su.’ n chan. Chu.ng minh rang

n/2∑

k=0

C(n, 2k) = 2n−1 =

n/2∑

k=1

C(n, 2k − 1).

11. Chu.ng minh rang

(a + b + c)n =∑

0≤i+j≤n

n!

i!j!(n− i− j)!aibjcn−i−j .


3n =∑

0≤i+j≤n

n!

i!j!(n− i − j)!.

76

13. Dung ly lua.n to’ ho.. p chu.ng minh rang

n∑

k=0

C(n, k)2 = C(2n, n).

14. (a) Chu.ng minh rang

n(1 + x)n−1 =

n∑

k=1

C(n, k)kxk−1.

(b) Tu. do suy ra

n2n−1 =n∑

k=1

kC(n, k).

4.6 Nguyen ly chuong chim bo cau

Nguyen ly chuong chim bo cau (con go. i la nguyen ly Dirichlet) thu.o.ng dung nham tra’ lo.icau ho’i: Co ton ta. i mo.t phan tu.’ thoa’ tınh chat cho tru.o.c? Khi ap du.ng thanh cong, nguyenly nay chı’ ra rang doi tu.o.. ng ton ta. i; tuy nhien khong chı’ ra cach tım no nhu. the nao va cobao nhieu phan tu.’ ton ta. i.

Da.ng dau tien cu’a nguyen ly chuong chim bo cau kha’ng di.nh rang neu co n va.t can xepvao k ho.p va n > k thı co ıt nhat co mo.t ho.p chu.a hai hoa. c nhieu ho.n hai va.t. Ly dokha’ng di.nh nay dung co the’ chu.ng minh bang pha’n chu.ng: Neu ket lua.n la sai, moi ho.pchu.a nhieu nhat mo.t va.t va do do trong tru.o.ng ho.. p nay co nhieu nhat k va.t. Nhu.ng co nva.t nen n ≤ k vo ly.

4.6.1 Nguyen ly chuong chim bo cau (da.ng thu. nhat)

Neu co n va. t can xep vao k ho.p va n > k thı ton ta. i ıt nhat mo. t ho.p co chu.a hai hoa. c nhieuho.n hai va. t.

Chu y rang, nguyen ly chuong chim bo cau khong chı’ ra ho.p nao chu.a ho.n hai va.t. Nochı’ kha’ng di.nh su.. ton ta. i cu’a mo.t ho.p vo.i ıt nhat hai va.t trong do.

Vı du. 4.6.1. So cac ho.c vien cu’a mo.t lo.p ho.c ıt nhat la bao nhieu de’ co ıt nhat hai ho.cvien co so die’m nhu. nhau trong ky thi mon Toan ho.c ro.i ra. c, neu du.. di.nh thang die’m la0-10?

Co 11 thang die’m. Theo nguyen ly chuong chim bo cau, can co ıt nhat 11 + 1 = 12 ho.cvien.

77

Vı du. 4.6.2. Chu.ng minh rang vo.i n + 1 so nguyen du.o.ng khac nhau khong vu.o.. t qua 2nthı pha’i co hai so chia het cho nhau.

Gia’ su.’ n + 1 so nguyen du.o.ng la a1, a2, . . . , an+1, vo.i 0 < ai ≤ 2n. Ta co the’ viet

ai = 2kiqi, i = 1, 2, . . . , n + 1,

trong do ki la so nguyen khong am va qi la so nguyen le’ khong am va khong vu.o.. t qua 2n.Vı du. 1 = 20, 14 = 21 × 7, 40 = 23 × 5, . . . .

Vı chı’ co n so le’ khong vu.o.. t qua 2n nen trong n + 1 so le’ q1, q2, . . . , qn+1 pha’i co ıt nhathai so bang nhau, cha’ng ha.n qi = qj = q vo.i i 6= j.

Khi doai = 2kiqi = 2kiq, aj = 2kjqj = 2kjq,

vo.i ki 6= kj . Suy ra ai | aj neu ki > kj va aj | ai neu kj > ki.

Ket qua’ tren la tot nhat theo nghıa neu ta gia’m nhe. gia’ thiet di bang cach thay n chon + 1 thı ket qua’ khong con dung nu.a. Tha.t va.y chı’ can lay ta.p cac so

{n + 1, n + 2, . . . , 2n}.

Vı du. 4.6.3. Chu.ng minh rang trong mo.i day gom n2 + 1 so thu.. c phan bie.t deu chu.a mo.tday con do. dai n + 1 hoa. c tang thu.. c su.. , hoa. c gia’m thu.. c su.. .

Gia’ su.’ n2 + 1 so thu.. c phan bie.t la a1, a2, . . . , an2+1. Vo.i moi so ai ta gan cho no ca.p so(ki, di) nhu. sau:

+ ki la do. dai cu’a day con tang dai nhat xuat phat tu. ai.

+ di la do. dai cu’a day con gia’m dai nhat xuat phat tu. ai.

Bang pha’n chu.ng gia’ su.’ khong co day con nao co do. dai n+1 la. i tang thu.. c su.. hoa.c gia’mthu.. c su.. . Khi do ki, di ≤ n, i = 1, 2, . . . , n2 + 1.

Nha.n xet rang co n2 ca.p (ki, di) khac nhau vo.i ki, di ≤ n. Nen ton ta. i cac chı’ so s, t saocho (ks, ds) = (kt, dt).

Nhu.ng cac so lay la phan bie.t, nen as 6= at. Khong mat tınh to’ng quat gia’ su.’ as < at.Bay gio. them as vao day con xuat phat tu. at de’ du.o.. c mo.t day con mo.i tang co do. dai1 + kt = 1 + ks trai vo.i gia’ thiet ks la do. dai cu’a day con tang dai nhat.

4.6.2 Nguyen ly chuong chim bo cau (da.ng thu. hai)

Neu f la anh xa. tu. ta.p hu.u ha.n X den ta.p hu.u ha.n Y va #X > #Y thı ton ta. i x1, x2 ∈X,x1 6= x2, sao cho f(x1) = f(x2).

78

Tha.t va.y, da. t X la ta.p cac va.t va Y la ta.p cac ho.p. Gan moi va.t x vo.i mo.t ho.p f(x).Theo nguyen ly chuong chim bo cau da.ng thu. nhat, co ıt nhat hai va.t khac nhau x1, x2 ∈ Xdu.o.. c gan cung mo.t ho.p; tu.c la f(x1) = f(x2).

Vı du. 4.6.4. Neu 20 bo. vi xu.’ ly du.o.. c noi vo.i nhau thı co ıt nhat hai bo. vi xu.’ ly du.o.. c noitru.. c tiep to.i cung so cac bo. vi xu.’ ly.

Ky hie.u cac bo. vi xu.’ ly la 1, 2, . . . , 20. D- a.t ai la so cac bo. vi xu.’ ly du.o.. c noi tru.. c tiep vo.ibo. vi xu.’ ly i. Chung ta can chu.ng minh rang ai = aj vo.i i 6= j nao do. Mien xac di.nh vamien gia tri. cu’a bai toan tu.o.ng u.ng la X := {1, 2, . . . , 20} va {0, 1, . . . , 19}. Tuy nhien sophan tu.’ cu’a hai ta.p ho.. p nay bang nhau, nen khong the’ ap du.ng tru.. c tiep nguyen ly chuongchim bo cau da.ng hai.

Chu y rang ta khong the’ co ai = 0 va aj = 19 vo.i i, j nao do, vı neu ngu.o.. c la. i ta co mo.tbo. vi xu.’ ly (thu. i) khong du.o.. c noi vo.i bat cu. bo. vi xu.’ ly nao trong khi la. i co mo. t bo. vixu.’ ly (thu. j) du.o.. c noi vo.i tat ca’ cac bo. vi xu.’ ly khac (ke’ cac bo. vi xu.’ ly thu. i). Do do Yla ta.p con cu’a ta.p {0, 1, . . . , 18} hoa. c {1, 2, . . . , 19}. Va.y #Y < 20 = #X. Theo nguyen lychuong chim bo cau da.ng hai ta co ai = aj vo.i i 6= j nao do.

Vı du. 4.6.5. Chu.ng minh rang neu cho.n 151 giao trınh may tınh phan bie.t du.o.. c danh sothu. tu.. tu. 1 den 300 thı co ıt nhat hai giao trınh co so thu. tu.. lien tiep.

Gia’ su.’ cac giao trınh du.o.. c danh so la

c1, c2, . . . , c151. (4.4)

Cac so nay cung vo.ic1 + 1, c2 + 1, . . . , c151 + 1 (4.5)

ta.o thanh 302 so thay do’i tu. 1 den 301. Theo nguyen ly chuong chim bo cau da.ng thu. haico ıt nhat hai gia tri. bang nhau. Cac so trong (4.4) la phan bie.t va do do cac so trong (4.5)cung khac nhau. Vı va.y pha’i co mo.t so trong day (4.4) bang mo.t so trong day (4.5). Do do

ci = cj + 1

(hie’n nhien i 6= j) va ta co hai giao trınh ci va cj du.o.. c danh so lien tiep.

Vı du. 4.6.6. Ba’n ke tai khoa’n gom 80 khoa’n mu.c, moi khoa’n mu.c du.o.. c danh dau “ho.. ple.” hoa. c “khong ho.. p le.”. Co 45 khoa’n mu.c ho.. p le.. Chu.ng minh rang co ıt nhat hai khoa’nmu.c ho.. p le. trong danh sach cach nhau chınh xac chın khoa’n mu.c. (Cha’ng ha.n cac khoa’nmu.c ta. i cac vi. trı 13 va 22 hoa.c ta. i vi. trı 69 va 78 cach nhau dung 9 khoa’n mu.c).

Ky hie.u ai la vi. trı cu’a khoa’n mu.c ho.. p le. thu. i. Ta can chı’ ra ai − aj = 9 vo.i i, j nao do.Xet cac so

a1, a2, . . . , a45 (4.6)

vaa1 + 9, a2 + 9, . . . , a45 + 9. (4.7)

79

90 so trong (4.6) va (4.7) lay cac gia tri. tu. 1 den 89. Do do theo nguyen ly chuong chim bocau da.ng thu. hai, co ıt nhat hai so trung nhau. Hie’n nhien khong the’ co hai so trong day(4.6) hoa. c (4.7) bang nhau; nen ton ta. i mo.t so trong day (4.6) bang mo.t so trong day (4.7).Va.y ai − aj = 9 vo.i i, j nao do.

4.6.3 Nguyen ly chuong chim bo cau (da.ng thu. ba)

Cho f la anh xa. tu. ta. p hu.u ha. n X den ta.p hu.u ha.n Y. Gia’ su.’ n := #X,m := #Y, k :=dn/me. Khi do ton ta. i ıt nhat k gia tri. a1, a2, . . . , ak sao cho

f(a1) = f(a2) = · · · = f(ak).

Chu.ng minh. D- a.t Y := {y1, y2, . . . , ym}. Gia’ su.’ kha’ng di.nh la sai. Khi do ton ta. i nhieu nhatk − 1 gia tri. x ∈ X vo.i f(x) = y1; ton ta. i nhieu nhat k − 1 gia tri. x ∈ X vo.i f(x) = y2; . . . ;ton ta. i nhieu nhat k − 1 gia tri. x ∈ X vo.i f(x) = ym. Do do ton ta. i nhieu nhat m(k − 1)phan tu.’ trong mien xac di.nh cu’a f. Nhu.ng

m(k − 1) < mn

m= n,

vo ly. Do do ton ta. i ıt nhat k gia tri. a1, a2, . . . , ak ∈ X sao cho

f(a1) = f(a2) = · · · = f(ak).

2

Vı du. 4.6.7. Mo.t da. c tru.ng hu.u ıch cu’a cac a’nh den trang la do. sang trung bınh cu’a a’nh.Ta noi rang hai a’nh la tu.o.ng tu.. neu do. sang trung bınh cu’a chung khac nhau khong vu.o.. tqua mo.t ngu.o.ng nao do. Chu.ng minh rang trong so sau a’nh, hoa.c co ba a’nh dong tho.itu.o.ng tu.. , hoa. c co ba a’nh dong tho.i khong tu.o.ng tu.. .

Ky hie.u cac a’nh la P1, P2, . . . , P6. Moi ca.p (P1, Pi), i = 2, 3, . . . , 6, co gia tri. “tu.o.ng tu.. ”hoa. c “khong tu.o.ng tu.. ”. Theo nguyen ly chuong chim bo cau da.ng thu. ba, ton ta. i ıt nhatd5/2e = 3 ca.p vo.i cung gia tri.; tu.c la ton ta. i cac ca.p

(P1, Pi), (P1, Pj), (P1, Pk)

hoa. c tu.o.ng tu.. , hoa. c khong tu.o.ng tu.. . Gia’ su.’ moi ca.p la tu.o.ng tu.. (trong tru.o.ng ho.. p ngu.o.. cla. i, xem Bai ta.p 5). Neu mo.t trong cac ca.p

(Pi, Pj), (Pi, Pk), (Pj , Pk) (4.8)

la tu.o.ng tu.. , thı hai hınh a’nh nay cung vo.i P1 doi mo.t tu.o.ng tu.. va do do ta co ba hınhtu.o.ng tu.. . Ngu.o.. c la. i, neu cac ca.p trong (4.8) khong tu.o.ng tu.. thı ta co ba a’nh tu.o.ng u.ngkhong tu.o.ng tu.. .

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Vı du. 4.6.8. So ho.c vien toi thie’u la bao nhieu de’ da’m ba’o ıt nhat co 6 ngu.o.i cung thangdie’m, neu giao vien cho die’m theo thang die’m A,B,C,D,F ?

Ta co N la so nho’ nhat thoa’ dN/5e = 6. Suy ra N = 5 × 5 + 1 = 26 ho.c vien.

Vı du. 4.6.9. Gia’ su.’ nhom co sau ngu.o.i; cu. lay mo.t ca.p bat ky, thı hai ngu.o.i nay hoa.c laba.n, hoa. c la thu. Chu.ng minh rang se co cac bo. ba hoa. c deu la ba.n cu’a nhau, hoa. c deu lathu cu’a nhau.

Lay x la ngu.o.i bat ky trong nhom; nam ngu.o.i con la.i la.p thanh nhom rieng. Ta ta.o haiho.p B va T. Nam ngu.o.i nay se du.o.. c phan loa. i (theo quan he. vo.i x) :

(a) hoa. c la ba.n cu’a x : tu.o.ng u.ng ngu.o.i trong ho.p B;

(b) hoa. c la thu cu’a x : tu.o.ng u.ng ngu.o.i trong ho.p T.

Theo nguyen ly chuong chim bo cau da.ng thu. ba, se co mo.t ho.p co ıt nhat d5/2e = 3ngu.o.i. Gia’ su.’ do la ho.p B vo.i ba ngu.o.i y, z, u.

Neu ton ta. i ca.p trong nhom ba ngu.o.i nay la ba.n cu’a nhau, cha’ng ha.n y va z, khi do{x, y, z} la bo. ba can tım. Ngu.o.. c la. i, tu.c la y, z, u moi ca.p doi mo.t la thu cu’a nhau, khi do{y, z, u} la bo. ba can tım.

Cac tru.o.ng ho.. p con la. i chu.ng minh tu.o.ng tu.. .

Bai ta.p

1. Co the’ noi nam may tınh vo.i nhau sao cho co chınh xac hai may tınh du.o.. c noi tru.. ctiep den cung mo.t so may? Gia’i thıch.

2. Ba’n ke tai khoa’n gom 115 khoa’n mu.c, moi khoa’n mu.c du.o.. c danh dau “ho.. p le.” hoa. c“khong ho.. p le.”. Co 60 khoa’n mu.c ho.. p le.. Chu.ng minh rang co ıt nhat hai khoa’n mu.cho.. p le. trong danh sach cach nhau chınh xac bon khoa’n mu.c.

3. Ba’n ke tai khoa’n gom 100 khoa’n mu.c, moi khoa’n mu.c du.o.. c danh dau “ho.. p le.” hoa. c“khong ho.. p le.”. Co 55 khoa’n mu.c ho.. p le.. Chu.ng minh rang co ıt nhat hai khoa’n mu.cho.. p le. trong danh sach cach nhau chınh xac chın khoa’n mu.c.

4. Ba’n ke tai khoa’n gom 80 khoa’n mu.c, moi khoa’n mu.c du.o.. c danh dau “ho.. p le.” hoa. c“khong ho.. p le.”. Co 50 khoa’n mu.c ho.. p le.. Chu.ng minh rang co ıt nhat hai khoa’n mu.ctrong danh sach cach nhau chınh xac hoa. c ba hoa. c sau khoa’n mu.c.

5. Hoan chı’nh Vı du. 4.6.7 bang cach chı’ ra rang neu cac ca.p (P1, Pi), (P1, Pj), (P1, Pk) lakhong tu.o.ng tu.. thı ton ta. i ba a’nh doi mo.t tu.o.ng tu.. hoa. c doi mo.t khong tu.o.ng tu.. .

6. Ket lua.n cu’a Vı du. 4.6.7 nhu. the nao neu:

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(a) Co ıt ho.n sau a’nh?

(b) Co ho.n sau a’nh?

7. Gia’ su.’ X gom (n + 2) phan tu.’ la ta.p con cu’a {1, 2, . . . , 2n + 1} va m := maxX. Vo.imoi k ∈ X \ {m} da. t

ak :=

{k neu k ≤ m

2,

m− k neu k > m2.

(a) Chu.ng minh mien gia tri. cu’a a chu.a trong {1, 2, . . . , n}.(b) Suy ra ton ta. i i 6= j sao cho ai = aj.

(c) Chu.ng minh ton ta. i hai phan tu.’ phan bie.t i, j ∈ X sao cho m = i + j.

(d) Cho vı du. ta.p X gom (n + 1) phan tu.’ la ta.p con cu’a {1, 2, . . . , 2n + 1} co tınhchat: Khong ton ta. i i, j ∈ X sao cho i + j ∈ X.

8. Xet mo.t nhom 10 ngu.o.i vo.i cac tuo’i (du.o.. c tınh la so nguyen) la a1, a2, . . . , a10. D- a.tri := ai mod 16 va

si :=

{ri neu ri ≤ 8,

16 − ri neu ri > 8.

(a) Chu.ng minh rang 0 ≤ si ≤ 8 vo.i mo.i i := 1, 2, . . . , 10.

(b) Chu.ng minh ton ta. i j 6= k sao cho sj 6= sk.

(c) Chu.ng minh rang neu (sj = rj va sk = rk) hoa.c (sj = 16 − rj va sk = 16 − rk)thı 16 chia het aj − ak.

(d) Chu.ng minh neu cac dieu kie.n trong (c) sai thı 16 chia het aj + ak.

9. Chu.ng minh rang trong khai trie’n tha.p phan cu’a thu.o.ng cu’a hai so nguyen, khoi cacchu. so cuoi cung la la.p la. i. Vı du.

1/6 = 0.1666 . . . , 217/660 = 0.32878787 . . . .

10. Mu.o.i sau cau thu’ bong ro’ ma.c ao mang cac so tu. 1 den 12 du.ng thanh vong tron trensan dau theo thu. tu.. tuy y. Chu.ng minh rang ton ta. i ba cau thu’ lien tiep co to’ng cacso ıt nhat 26.

11. Gia’ su.’ f la anh xa. mo.t-mo.t len tu. X := {1, 2, . . . , n} len X. Ky hie.u fk la anh xa. ho.. pk lan cu’a f :

fk := f ◦ f ◦ · · · ◦ f︸︷︷︸k lan

.

Chu.ng minh rang ton ta. i cac so nguyen phan bie.t i 6= j sao cho f i(x) 6= f j(x) vo.i mo.ix ∈ X. Chu.ng minh rang ton ta. i so nguyen k sao cho fk(x) = x vo.i mo.i x ∈ X.

12. Mo.t hınh chu. nha.t kıch thu.o.c 3 × 7 du.o.. c chia thanh 21 hınh vuong; moi hınh vuongdu.o.. c to mau den hoa. c trang. Chu.ng minh rang ban co. chu.a mo.t hınh chu. nha.t khongtam thu.o.ng (khong co kıch thu.o.c 1× k hoa. c k× 1) sao cho bon hınh vuong o.’ moi gochoa. c tat ca’ to mau den hoa.c tat ca’ to mau trang.

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13. Chu.ng minh rang neu p bit 1 va q bit 0 du.o.. c da. t xung quanh mo.t vong tron theo thu.

tu.. tuy y, trong do p, q, k la cac so nguyen thoa’ p ≥ kq thı ton ta. i k bit 1 du.ng lientiep.

14. Viet thua. t toan tım do. dai cu’a day con do.n die.u tang dai nhat cu’a mo.t day so chotru.o.c.

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Chu.o.ng 5

QUAN HE.

Nhu. da biet, tat ca’ cac doi tu.o.. ng trong the gio.i xung quanh ta deu co nhu.ng moi quan he.nhat di.nh vo.i nhau. Ro rang khong co mo.t doi tu.o.. ng nao co the’ ton ta. i tach ro.i (khonglien quan) vo.i the gio.i ben ngoai. Ma. t khac, moi doi tu.o.. ng la. i chu.a du.. ng rat nhieu moiquan he. no. i ta. i cu’a ba’n than no. Xet mo.t nhom sinh vien trong cung mo.t lo.p, ta co the’ noirang hai sinh vien co quan he. vo.i nhau neu ho. co cung que. Xet mo.t ta.p ho.. p cac so nguyen{1, 2, . . . , 15}, ta co the’ noi rang ba phan tu.’ nao do cu’a ta.p ho.. p nay co quan he. vo.i nhauneu to’ng cu’a chung chia het cho 4. Noi mo.t cach khac, cac phan tu.’ hay cac doi tu.o.. ng coquan he. cha.t che vo.i nhau, nhu.ng moi quan he. du.o.. c hie’u nhu. the nao la phu. thuo.c vaodi.nh nghıa cu’a chung ta. Mo hınh co. so.’ du. lie.u quan he. , du.o.. c du.a ra bo.’ i E. F. Codd vaonam 1970, du.. a tren khai nie.m cu’a quan he. n ngoi la mo. t trong nhu.ng u.ng du.ng cu’a quanhe. trong Tin ho.c.

Trong chu.o.ng nay, chung ta se nghien cu.u cac moi quan he. tren co. so.’ ly thuyet ta.p ho.. p.Tru.o.c het ta nghien cu.u cac quan he. hai ngoi tren hai ta.p ho.. p va tren cung mo.t ta.p ho.. p,cung vo.i cac tınh chat cu’a cac quan he. do. Tiep theo, chung ta se xet den quan he. thu. tu.. ,quan he. tu.o.ng du.o.ng va cac moi lien quan.

5.1 Quan he. hai ngoi

D- i.nh nghıa 5.1.1. Quan he. hai ngoi R tu. ta.p S len ta.p T la mo.t ta.p ho.. p con cu’a S × T.Ta.p S du.o.. c go. i la mien xac di.nh con T la doi mien xac di.nh. Neu S ≡ T ta noi R la quanhe. hai ngoi tren S.

Vı du. 5.1.1. Gia’ su.’ S la danh sach cac sinh vien cu’a tru.o.ng da. i ho.c. T la danh sach cacchu.ng chı’ ho.c. Ta.p R ⊂ S × T gom cac ca.p (a, b), trong do a la sinh vien con b la chu.ngchı’ ma sinh vien ghi danh ho.c. Vo.i moi a ∈ S, ta.p {b ∈ T | (a, b) ∈ R} la danh sach cacchu.ng chı’ ma sinh vien a theo ho.c. Ta.p {a | (a, b) ∈ R} la danh sach cac sinh vien theo ho.c

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a2

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b3

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Hınh 5.1:

chu.ng chı’ b.

Vı du. 5.1.2. Gia’ su.’ P la ta.p cac chu.o.ng trınh du.o.. c thu.. c hie.n tren may tınh va mo.t do.nvi. C cac chu.o.ng trınh co san cho phep de’ su.’ du.ng. Ta da. t mo.t quan he. R tu. C len P nhu.

sau: (c, p) ∈ R neu chu.o.ng trınh p su.’ du.ng thu’ tu.c c.

Vı du. 5.1.3. ChoS := {a1, a2, a3, a4}

la ta.p cac sinh vien tot nghie.p con

T := {b1, b2, b3, b4, b5, b6}

la ta.p cac co. quan can nha.n sinh vien tot nghie.p. Quan he.

R := {(a1, b2), (a2, b1), (a3, b6), (a4, b4)}

tu. S len T mo ta’ cac ca.p sap xep no.i cong tac cho moi sinh vien.

Trong chu.o.ng nay, ngoa. i tru. nhu.ng tru.o.ng ho.. p ngoa. i le. ma se noi ro, ta se luon luon gia’

thiet rang cac quan he. du.o.. c xet tren cac ta. p hu.u ha. n. Khi do co the’ mo ta’ quan he. R tu.

S len T bang phu.o.ng phap do thi. nhu. sau: cac dı’nh cu’a do thi. bie’u thi. cac phan tu.’ cu’a Sva T, con cac cung la cac du.o.ng co hu.o.ng noi cac ca.p (a, b) ∈ R (co khi ngu.o.i ta viet tatdu.o.i da.ng aRb); cha’ng ha.n quan he. trong Vı du. 5.1.3 co do thi. trong Hınh 5.1.

Ngoai ra, ngu.o.i ta cung thu.o.ng dung ma tra.n cap m × n de’ bie’u thi. moi quan he. R tu.

S = {a1, a2, . . . , am} len T = {b1, b2, . . . , bn}, trong do m := #S, n := #T. Phan tu.’ mij cu’ama tra.n du.o.. c xac di.nh nhu. sau

mij :=

{1 neu (ai, bj) ∈ R,

0 neu ngu.o.. c la. i.

86

Vı du. 5.1.4. Ma tra.n bie’u dien quan he. R trong Vı du. 5.1.3 la

b1 b2 b3 b4 b5 b6

a1 0 1 0 0 0 0a2 1 0 0 0 0 0a3 0 0 0 0 0 1a4 0 0 0 1 0 0

.

Ba’n than cac quan he. la. i lien quan vo.i nhau ta.o nen cac quan he. mo.i. Cha’ng ha.n, ho.. pgiu.a cac quan he. la mo. t hınh thu.c ta.o nen cac quan he. mo.i.

D- i.nh nghıa 5.1.2. Gia’ su.’ R1 la quan he. tu. S1 len S2;R2 la quan he. tu. S2 len S3. Ho.. p cu’ahai quan he. R1 va R2 la mo.t quan he. tu. S1 len S3 xac di.nh bo.’ i

R2 ◦ R1 := {(x, z) ∈ S1 × S3 | ton ta. i y ∈ S2 de’ (x, y) ∈ R1, (y, z) ∈ R2}.

Vı du. 5.1.5. Gia’ su.’ T la ta.p cac chu.ng chı’, U la ta.p cac khoa. Xet quan he. R nhu. trongVı du. 5.1.1. Quan he. R′ ⊂ S × U gom cac ca.p (b, c) sao cho chu.ng chı’ b ∈ T la bat buo.cghi danh ho.c khoa U. The thı R′ ◦R la ta.p cac ca.p (a, c) sao cho ton ta. i chu.ng chı’ bat buo.cma sinh vien a pha’i ho.c khi ghi danh vao khoa c. Chu y rang trong tru.o.ng ho.. p nay R ◦ R′

la khong co nghıa!

Tınh chat 5.1.3. Ho.. p cac quan he. co cac tınh chat sau:

(a) Tınh ket ho.. p

R3 ◦ (R2 ◦ R1) = (R3 ◦ R2) ◦ R1.

(b) Tınh phan bo

R3 ◦ (R1 ∪ R2) = (R3 ◦ R1) ∪ (R3 ◦R2),

(R2 ∪ R3) ◦ R1 = (R2 ◦ R1) ∪ (R3 ◦R1).

(c) Neu R1 ⊂ R2 va R3 ⊂ R4 thı

R1 ∪ R3 ⊂ R2 ∪ R4,

R1 ∩ R3 ⊂ R2 ∩ R4.

(d) Neu R1 ⊂ R2 va R3 ⊂ R4 thı

R3 ◦ R1 ⊂ R4 ◦ R2.


87

Vı du. 5.1.6. D- a.t S1 := {1, 2, 3, 4, 5}, S2 := {a, b, c} va S3 := [e, f, g, h}. Xet cac quan he. tu.

S1 len S2 va tu. S2 len S3 xac di.nh tu.o.ng u.ng bo.’ i

R1 := {(1, a), (2, a), (2, c), (3, a), (3, b), (4, a), (4, b), (4, c), (5, b)},R2 := {(a, e), (a, g), (b, f), (b, g), (b, h), (c, e), (c, g), (c, h)}.

Khi doR2 ◦ R1 = {(1, e), (1, g), (2, e), (2, g), (2, h), (3, e), (3, f), (3, g), (3, h),

(4, e), (4, f), (4, g), (4, h), (5, f), (5, g), (5, h)},va cac ma tra.n A1, A2 va A tu.o.ng u.ng cac quan he. R1, R2 va R2 ◦ R1 la

A1 =

1 0 01 0 11 1 01 1 10 1 0

, A2 =

1 0 1 00 1 1 11 0 1 1

, A =

1 0 1 01 0 1 11 1 1 11 1 1 10 1 1 1

.

So sanh A va ma tra.n tıch cu’a A1 va A2

A1A2 =

1 0 1 02 0 2 11 1 2 12 1 3 20 1 1 1

,

ta thay so 1 trong ma tra.n A tu.o.ng u.ng vo.i phan tu.’ khac 0 trong ma tra.n A1A2! D- ieu nayse du.o.. c gia’i thıch trong phan sau.

D- i.nh nghıa 5.1.4. Gia’ su.’ R la quan he. tu. S len T. Quan he. ngu.o.. c cu’a R, ky hie.u R−1,la mo.t quan he. tu. T len S xac di.nh bo.’ i

R−1 := {(x, y) ∈ T × S | (y, x) ∈ R}.

Tınh chat 5.1.5. Gia’ su.’ R la quan he. tren S. Khi do

(a) R = R−1 neu va chı’ neu R doi xu.ng, tu.c la

R = R−1 ⇔ xRy suy ra yRx.

(b) R ∩ R−1 ⊂ E := {(x, x) | x ∈ S} neu va chı’ neu R pha’n doi xu.ng, tu.c la

R ∩ R−1 ⊂ E ⇔ xRy va yRx thı x = y.

Chu.ng minh. (a) Hie’n nhien theo di.nh nghıa.

(b) Gia’ su.’ R pha’n doi xu.ng, va (x, y) ∈ R ∩ R−1. Khi do xRy va yRx. Suy ra xRx. Hay(x, x) ∈ E.

Ngu.o.. c la. i gia’ su.’ R ∩R−1 ⊂ E, xRy va yRx. Thı (x, y) ∈ R ∩R−1 ⊂ E. Do do (x, y) ∈ E.2

88

Bai ta.p

1. D- a.t S := {0, 1, 2}. Moi phat bie’u sau xac di.nh mo.t quan he. R tren S bo.’ i mRn neukha’ng di.nh la dung doi vo.i m,n ∈ S. Viet moi quan he. nhu. mo.t ta.p cac ca.p co thu.

tu.. .(a) m ≤ n.(b) m < n.(c) m = n.

(d) mn = 0.(e) mn = m.(f) m + n ∈ S.

(g) m2 + n2 = 2.(h) m2 + n2 = 3.(i) m = max{n, 1}.

Cac quan he. nao la doi xu.ng? pha’n doi xu.ng? Viet ma tra.n va ve cac do thi. tu.o.ngu.ng.

2. Cac quan he. hai ngoi sau xac di.nh tren N.

(a) Viet quan he. hai ngoi R1 xac di.nh bo.’ i m + n = 5 da.ng cac ca.p thu. tu.. .

(b) Nhu. tren vo.i R2 xac di.nh bo.’ i max{m,n} = 2.

(c) Quan he. hai ngoi R3 xac di.nh bo.’ i min{m,n} = 2 gom vo ha.n cac ca.p thu. tu.. . Hayviet nam ca.p trong do.

3. Neu A la ma tra.n cu’a quan he. R tu. S len T (gia’ thiet S va T la cac ta.p hu.u ha.n).Tım ma tra.n cu’a quan he. ngu.o.. c R−1.

4. Gia’ su.’ R la quan he. hai ngoi tren ta.p S. Chu.ng minh rang R la doi xu.ng neu va chı’

neu R = R−1.

5. Gia’ su.’ R1, R2 la cac quan he. tu. S len T.

(a) Chu.ng minh rang (R1 ∪ R2)−1 = R−1

1 ∪ R−12 .

(b) Chu.ng minh rang (R1 ∩ R2)−1 = R−1

1 ∩ R−12 .

(c) Chu.ng minh rang neu R1 ⊆ R2 thı R−11 ⊆ R−1

2 .

6. Gia’ su.’ G la do thi. cu’a quan he. R tren ta.p hu.u ha.n S. Mo ta’ do thi. cu’a quan he. R−1.

7. Tren ta.p S := {1, 2, 3, 4} xet cac quan he. hai ngoi sau:

R1 := {(1, 1), (1, 2), (3, 4), (4, 2)},R2 := {(1, 1), (2, 1), (3, 1), (4, 4), (2, 2)}.

Lie.t ke cac phan tu.’ cu’a R1 ◦ R2 va R2 ◦ R1.

8. Kha’o sat cac quan he. R1 va R2 tu. S len T va cac quan he. R3 va R4 tu. T len U.

(a) Chu.ng minh rang (R3 ∪ R4) ◦ R1 = R3 ◦ R1 ∪ R4 ◦ R1.

(b) Chu.ng minh rang R3 ◦ (R1 ∩ R2) ⊆ R3 ◦ R1 ∩ R3 ◦ R2 va da’ng thu.c khong nhatthiet dung.

(c) Cac quan he. (R3 ∩ R4) ◦ R1 va R3 ◦ R1 ∩ R4 ◦ R1 co lien he. nhu. the nao?

89

5.2 Quan he. va ma tra.n

Nhu. Vı du. 5.1.6 chı’ ra, ma tra.n cu’a quan he. R2 ◦ R1 khong pha’i la tıch A1A2 cu’a cac matra.n R1 va R2. Tuy nhien chung co moi lien he.: phan tu.’ bang 1 trong A tu.o.ng u.ng mo.t-mo.tvo.i phan tu.’ khac khong trong A1A2.

Xet B := {0, 1} va hai phep toan Boole ∧,∨ di.nh nghıa nhu. sau:

∨ 0 10 0 11 1 1

∧ 0 10 0 01 0 1

Ta co

Tınh chat 5.2.1. Vo.i mo. i x, y ∈ B ta co

x ∨ y = max{x, y}, x ∧ y = min{x, y}.


D- i.nh nghıa 5.2.2. (a) A du.o.. c go. i la ma tra.n Boole neu cac phan tu.’ cu’a no thuo.c B.

(b) Tıch hai ma tra. n Boole A1 va A2 cap m× n va n× p tu.o.ng u.ng la ma tra.n Boole capm × p, kı hie.u A1 ∗ A2, xac di.nh bo.’ i

(A1 ∗ A2)[i, j] := ∨nk=1(A1[i, k] ∧ A2[k, j]), i = 1, 2, . . . ,m, j = 1, 2, . . . , p.

(c) Ho. i hai ma tra.n Boole A1 va A2 cap m×n la ma tra.n Boole cap m×n, kı hie.u A1∧A2,co cac phan tu.’ la

(A1 ∧ A2)[i, j] := A1[i, j] ∧ A2[i, j], i = 1, 2, . . . ,m, j = 1, 2, . . . , n.

(d) Tuye’n hai ma tra.n Boole A1 va A2 cap m × n la ma tra.n Boole cap m × n, kı hie.uA1 ∨ A2, co cac phan tu.’ la

(A1 ∨ A2)[i, j] := A1[i, j] ∨ A2[i, j], i = 1, 2, . . . ,m, j = 1, 2, . . . , n.

Vı du. 5.2.1. Trong Vı du. 5.1.6 thı A = A1 ∗ A2.

D- i.nh ly 5.2.3. Neu A1 va A2 la cac ma tra. n tu.o.ng u.ng quan he. R1 tu. A len B va R2 tu.

B len C thı A1 ∗ A2 la ma tra. n cu’a quan he. ho.. p R2 ◦ R1.

Chu.ng minh. Ta co

(A1 ∗ A2)[i, j] = 0 ⇔ A1[i, k] ∧ A2[k, j] = 0,∀k = 1, 2, . . . , n,

⇔ A1[i, k] = A2[k, j] = 0,∀k = 1, 2, . . . , n.

2

90

Vı du. 5.2.2. Gia’ su.’ R la quan he. tren {1, 2, 3} vo.i ma tra.n

A =

1 0 01 0 11 1 0

.

Quan he. R2 := R ◦ R co ma tra.n

A ∗ A =

1 0 01 1 01 0 1

.

Va quan he. R3 := R2 ◦ R co ma tra.n

A ∗ A ∗ A =

1 0 01 0 11 1 0

= A.

Suy ra R = R3. Ho.n nu.a vo.i mo.i n ≥ 1 ta co

Rn+2 = R(n−1)+3 = Rn−1 ◦ R3 = Rn−1 ◦ R = Rn!

Vı du. 5.2.3. Gia’ su.’ R1 va R2 la cac quan he. tren {1, 2} co cac ma tra.n Boole tu.o.ng u.ng

A1 =

(1 11 0

), A2 =

(1 10 1

).

Do

A1 ∗ A2 =

(1 11 1

)6=(

1 11 0

)= A2 ∗ A1,

nen (2, 2) ∈ R2 ◦ R1 nhu.ng (2, 2) /∈ R1 ◦ R2. Suy ra R1 ◦ R2 6= R2 ◦ R1.

D- i.nh ly 5.2.4. Ta. p P(S × S), tat ca’ cac quan he. tren S, vo.i phep toan ho.. p la nu.’a nhom,tu.c la co cac tınh chat sau: phep toan ho.. p co tınh ket ho.. p va P(S × S) chu.a phan tu.’ do.nvi..

Chu.ng minh. Tha.t va.y, phep ho.. p co tınh chat ket ho.. p do Tınh chat 5.1.3(a); va do.n vi. laquan he. “dong nhat”:

E := {(x, x) ∈ S | x ∈ S}.2

D- i.nh ly sau chı’ ra vie.c nghien cu.u quan he. chuye’n ve nghien cu.u cac ma tra.n cu’a chung.

D- i.nh ly 5.2.5. Gia’ su.’ S la ta. p n phan tu.’ . Khi do ton ta. i anh xa. mo. t-mo. t len giu.a ta. pP(S ×S) cac quan he. tren S va ta. p cac ma tra. n Boole cap n× n. Anh xa. nay ba’o toan cacphep toan nu.’a nhom: neu R1, R2 va R la cac quan he. vo.i cac ma tra. n Boole A1, A2 va Atu.o.ng u.ng, thı

R2 ◦ R1 = R ⇔ A1 ∗ A2 = A.

91

Chu.ng minh. Hie’n nhien theo cac ket qua’ tren. 2

D- i.nh nghıa 5.2.6. Quan he. hai ngoi R tren S du.o.. c go. i la

(a) Pha’n xa. neu xRx vo.i mo.i x ∈ S;

(b) Bac cau neu xRy va yRz thı xRz.

Vı du. 5.2.4. Xet cac quan he. R1, R2, R3 va E tren S := {1, 2, 3, 4} tu.o.ng u.ng vo.i cac matra.n

A1 :=

0 1 1 10 0 1 10 0 0 10 0 0 0

, A2 :=

1 1 0 01 1 1 00 1 1 10 0 1 1

,

A3 :=

1 0 0 10 1 0 00 0 1 01 0 0 1

, I4 :=

1 0 0 00 1 0 00 0 1 00 0 0 1

.

(a) R1 la quan he. du.o.. c xac di.nh bo.’ i mR1n neu m ≤ n. Quan he. R1 la pha’n xa. va baccau.

(b) R2 la quan he. du.o.. c xac di.nh bo.’ i mR2n neu |m− n| ≤ 1. Quan he. R2 la pha’n xa. , doixu.ng nhu.ng khong bac cau.

(c) R3 la quan he. du.o.. c xac di.nh bo.’ i mR3n neu va chı’ neu m = n (mod 3). Ta co R3 laquan he. pha’n xa. , doi xu.ng va bac cau.

(d) Quan he. E := {(m,n) ∈ S × S | m = n} tren S la pha’n xa. , doi xu.ng va bac cau.

Vı du. 5.2.5. (a) Quan he. R tren Z di.nh nghıa bo.’ i

mRn neu va chı’ neu m + n = 0 (mod 3)

la doi xu.ng, khong pha’n xa. do (1, 1) /∈ R va khong bac cau do (4, 2), (2, 1) ∈ R nhu.ng(4, 1) /∈ R.

(b) Vo.i m,n ∈ Z di.nh nghıa mRn neu m− n le’. Quan he. la doi xu.ng nhu.ng khong pha’nxa. va khong bac cau.

Tınh chat 5.2.7. Gia’ su.’ R la quan he. tren ta. p A. Khi do

(a) R pha’n xa. neu va chı’ neu E ⊂ R.

(b) R bac cau neu va chı’ neu R2 ⊂ R.

92

Chu.ng minh. (a) Hie’n nhien.

(b) Gia’ su.’ R la bac cau va (x, z) ∈ R2. Khi do ton ta. i y ∈ A sao cho (x, y), (y, z) ∈ R. VıR bac cau nen (x, z) ∈ R. Ngu.o.. c la. i, gia’ su.’ R2 ⊂ R. Xet (x, y), (y, z) ∈ R. Thı (x, z) ∈ R2.Va.y (x, z) ∈ R. 2

Gia’ su.’ A1, A2 la hai ma tra.n Boole cung cap m × n. Ky hie.u A1 ≤ A2 nghıa la

A1[i, j] ≤ A2[i, j]

vo.i mo.i i = 1, 2, . . . ,m, j = 1, 2, . . . , n.

Tınh chat 5.2.8. Gia’ su.’ R1, R2 la hai quan he. tu. S len T tu.o.ng u.ng cac ma tra. n A1, A2.Ta co

(a) R1 ⊆ R2 neu va chı’ neu A1 ≤ A2.

(b) R1 ∪ R2 co ma tra. n Boole A1 ∨ A2.

(c) R1 ∩ R2 co ma tra. n Boole A1 ∧ A2.


He. qua’ 5.2.9. Gia’ su.’ R la quan he. tren ta. p S tu.o.ng u.ng ma tra. n Boole A := (aij)n×n, n =#S. Khi do

(a) R2 ⊆ R neu va chı’ neu A ∗ A ≤ A.

(b) R pha’n xa. neu va chı’ neu aii = 1, i = 1, 2, . . . , n.

(c) R doi xu.ng neu va chı’ neu A = At.

(d) R pha’n doi xu.ng neu va chı’ neu A ∧ At ≤ In.

(e) R bac cau neu va chı’ neu A ∗ A ≤ A.


Bai ta.p

1. Vo.i moi ma tra.n Boole sau, xet quan he. tu.o.ng u.ng R tren {1, 2, 3}. Tım ma tra.nBoole cu’a R2 va xac di.nh quan he. nao la bac cau.

1 1 00 1 11 0 1

,

1 0 10 1 01 0 1

,

0 0 10 1 01 0 0

.

Ve do thi. cu’a cac quan he. tren.

93

2. Gia’ su.’ S := {1, 2, 3}, T := {a, b, c, d} va R1, R2 la cac quan he. tu. S len T vo.i cac matra.n Boole

A1 :=

1 0 1 00 1 0 01 0 0 1

, A2 :=

0 1 0 01 0 0 10 1 1 0

.

(a) Tım cac ma tra.n Boole cu’a R−11 , R−1

2 .

(b) Tım cac ma tra.n Boole cu’a R−11 ◦ (R1 ∩ R2), (R

−11 ◦ R1) ∩ (R−1

1 ◦ R2).

(c) Tım cac ma tra.n Boole cu’a (R−11 ∪ R−1

2 ) ◦ R2, (R−11 ◦ R2) ∪ (R−1

2 ◦ R2).

(d) So sanh cac cau tra’ lo.i trong phan (b) va (c) vo.i cac kha’ng di.nh trong Bai ta.p10.

3. Gia’ su.’ S := {1, 2, 3} va R := {(1, 1), (1, 2), (1, 3), (3, 2)}.

(a) Tım cac ma tra.n cu’a R,R ◦ R−1 va R−1 ◦ R.

(b) Ve cac do thi. cu’a cac quan he. trong phan (a).

(c) Chu.ng minh rang R la bac cau (tu.c la R2 ⊆ R), nhu.ng R2 6= R.

(d) R ∪ R−1 la quan he. bac cau? Gia’i thıch.

(e) Tım Rn vo.i n = 2, 3, . . . .

4. Gia’ su.’ S := {1, 2, 3} va R := {(2, 1), (2, 3), (3, 2)}.

(a) Tım cac ma tra.n cu’a R,R−1 va R ◦ R2.

(b) Ve cac do thi. cu’a cac quan he. trong phan (a).

(c) R la bac cau?

(d) R2 la bac cau?

(e) R ∪ R2 la bac cau?

5. Gia’ su.’ R la quan he. tren S := {1, 2, 3} vo.i ma tra.n Boole

A :=

0 1 01 1 10 1 0

.

(a) Tım ma tra.n Boole cu’a Rn, n ∈ N.

(b) R la pha’n xa.? D- oi xu.ng? Bac cau?

6. La.p la. i Bai ta.p 5 vo.i

A :=

1 0 00 1 11 0 1

.

7. Gia’ su.’ P la ta.p tat ca’ cac ngu.o.i va kha’o sat quan he. R, trong do pRq neu p “thıch”q.

94

(a) Mo ta’ cac quan he. R ∩ R−1, R ∪ R−1, va R2.

(b) R la pha’n xa.? D- oi xu.ng? Bac cau?

8. Cho vı du. quan he. ma

(a) Pha’n doi xu.ng, bac cau nhu.ng khong pha’n xa. .

(b) D- oi xu.ng nhu.ng khong pha’n xa. hay bac cau.

9. Vo.i anh xa. f : S → T ta di.nh nghıa quan he.

Rf := {(x, y) ∈ S × T |y = f(x)}.

Xet cac anh xa. f, g : {1, 2, 3, 4} → {1, 2, 3, 4} xac di.nh bo.’ i f(m) := max{2, 4 − m} vag(m) := 5 − m.

(a) Tım cac ma tra.n Boole Af , Ag cu’a cac quan he. Rf va Rg tu.o.ng u.ng vo.i cac anhxa. f, g.

(b) Tım cac ma tra.n Boole cu’a Rf , Rg, va Rf◦g va so sanh.

(c) Tım cac ma tra.n Boole cu’a R−1f , R−1

g . Cac quan he. nay tu.o.ng u.ng vo.i cac anh xa.nao?

10. Kha’o sat cac quan he. R1 va R2 tren ta.p S. Chu.ng minh hoa. c cho pha’n vı du. :

(a) Neu R1 va R2 pha’n xa. thı R2 ◦ R1 pha’n xa. .

(b) Neu R1 va R2 doi xu.ng thı R2 ◦ R1 doi xu.ng.

(c) Neu R1 va R2 bac cau thı R2 ◦ R1 bac cau.

11. Gia’ su.’ R1, R2 la cac quan he. hai ngoi tren ta.p S.

(a) Chu.ng minh rang R1 ∩ R2 la pha’n xa. neu R1 va R2 la pha’n xa. .

(b) Chu.ng minh rang R1 ∩ R2 la doi xu.ng neu R1 va R2 la doi xu.ng.

(c) Chu.ng minh rang R1 ∩ R2 la bac cau neu R1 va R2 la bac cau.

12. Gia’ su.’ R1, R2 la cac quan he. hai ngoi tren ta.p S.

(a) R1 ∪ R2 la pha’n xa. neu R1 va R2 la pha’n xa.?

(b) R1 ∪ R2 la doi xu.ng neu R1 va R2 la doi xu.ng?

(c) R1 ∪ R2 la bac cau neu R1 va R2 la bac cau?

13. Gia’ su.’ R la quan he. tu. S := {1, 2, 3, 4} len T := {a, b, c} vo.i ma tra.n Boole

A :=

a b c

1 1 0 12 0 0 13 1 0 04 0 1 0

.

95

(a) Chu.ng minh rang R−1 ◦ R la quan he. doi xu.ng tren S.

(b) Chu.ng minh rang R ◦ R−1 la quan he. doi xu.ng tren S.

(c) Cac quan he. R ◦ R−1, R−1 ◦ R la pha’n xa.? Bac cau?

14. Gia’ su.’ R la quan he. tu. S len T.

(a) Chu.ng minh rang R−1 ◦R la quan he. doi xu.ng tren S. (Khong su.’ du.ng ma tra.nBoole do S hoa. c T co the’ khong hu.u ha.n).

(b) Suy ra R ◦ R−1 la doi xu.ng tren T.

(c) Khi nao thı R−1 ◦ R la pha’n xa.?

15. Gia’ su.’ R1 la quan he. tu. S len T, R2 la quan he. tu. T len U, trong do S, T, U la cacta.p hu.u ha.n. Du.. a vao ma tra.n Boole bie’u dien quan he., chu.ng minh rang

(R2 ◦ R1)−1 = R−1

1 ◦ R−12 .

16. Gia’ su.’ R1, R2 la cac quan he. tu. S := {1, 2, . . . ,m} len T := {1, 2, . . . , n}, tu.o.ng u.ngvo.i cac ma tra.n A1, A2. Chu.ng minh rang R1 ⊆ R2 neu va chı’ neu A1 ≤ A2.

17. Su.’ du. ng tınh ket ho.. p cu’a cac quan he., chu.ng minh rang tıch Boole la mo.t phep toanco tınh ket ho.. p.

18. Gia’ su.’ S la ta.p khac trong. P(S × S) la mo.t nhom vo.i phan tu.’ ngu.o.. c R−1? Gia’ithıch.

19. Gia’ su.’ R la quan he. tren S va R∗ := ∪n≥0Rn la bao dong truyen u.ng cu’a R. Chu.ng

minh rang R∗ la pha’n xa. va bac cau. Ho.n nu.a, neu R ⊂ R′, trong do R′ la bac cauva doi xu.ng, thı R∗ ⊂ R′.

5.3 Quan he. thu. tu..

D- i.nh nghıa 5.3.1. Quan he. hai ngoi R tren ta.p S du.o.. c go. i la quan he. thu. tu.. (hay ro ho.n,quan he. thu. tu.. bo. pha.n) neu no co cac tınh chat: pha’n xa. , pha’n doi xu.ng va bac cau. Khido thay cho cach viet aRb, ngu.o.i ta thu.o.ng viet a ≤ b hoa. c b ≥ a va noi rang a di tru.o.c b,hoa. c b di sau a. Nhu. va.y

(a) a ≤ a vo.i mo.i a ∈ S.

(b) Neu a ≤ b va b ≤ a thı a = b.

(c) Neu a ≤ b va b ≤ c thı a ≤ c.

96

Neu a ≤ b va a 6= b ta ky hie.u a < b hoa. c b > a va noi rang a thu.. c su.. di tru.o.c b hoa.c bthu.. c su.. di sau a.

Kı hie.u (S,≤) co nghıa ≤ la quan he. thu. tu.. tren ta.p S; va (S,≤) du.o.. c go. i la ta.p co thu.

tu.. bo. pha. n.

Nha.n xet rang vo.i hai phan tu.’ a, b ∈ S thı khong nhat thiet pha’i co a ≤ b hoa. c b ≤ a.Neu hoa. c a ≤ b hoa. c b ≤ a thı cac phan tu.’ a va b go. i la so sanh du.o.. c vo.i nhau. Neu A ⊂ Sva hai phan tu.’ bat ky cu’a A la so sanh du.o.. c vo.i nhau thı A go. i la ta.p con sap tha’ng cu’a S.

Vı du. 5.3.1. Xet tru.o.ng so phu.c C va quan he. x ≤ y, trong do x = a+ ib va y = c+ id, i =√−1, neu a ≤ c va b ≤ d. Hie’n nhien ≤ la quan he. thu. tu.. . D- a.t

A := {x ∈ C | x = a + i0, a ∈ R}.

Vo.i quan he. ≤ ta.p A la ta.p con sap tha’ng cu’a C. Ta co 2+ i3 < 2+ i5. Nhu.ng 2+ i3 khongso sanh du.o.. c vo.i 1 + i5.

D- i.nh nghıa 5.3.2. (a) Gia’ su.’ ≤ la quan he. thu. tu.. tren ta.p S. Ta noi rang t phu’ s neus < t va khong ton ta. i u ∈ S sao cho s < u < t.

(b) Lu.o.. c do Hasse cu’a (S,≤) la mo.t do thi. co hu.o.ng gom cac dı’nh la cac phan tu.’ cu’a Sva neu t phu’ s thı co mo.t cung noi tu. s den t.

Vı du. 5.3.2. (a) D- a.t S := {1, 2, 3, 4, 5, 6}. Ta viet m|n neu n la bo. i nguyen cu’a m. Khi do(S, |) la ta.p du.o.. c sap thu. tu.. bo. pha.n. Ta co lu.o.. c do Hasse trong Hınh 5.2.

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•1

2

4

3

5

6

Hınh 5.2:

(b) Tren S := P({a, b, c}) xet quan he. bao ham ⊂ . Khi do (S,⊂) la ta.p du.o.. c sap thu. tu..bo. pha.n va co lu.o.. c do Hasse trong Hınh 5.3.

(c) Lu.o.. c do trong Hınh 5.4 khong pha’i la lu.o.. c do Hasse (ta. i sao?):

(d) Cac lu.o.. c do trong Hınh 5.5 la lu.o.. c do Hasse cu’a cac ta.p du.o.. c sap thu. tu.. bo. pha.n(du.o.. c suy tru.. c tiep tu. do thi.).

Noi chung vo.i lu.o.. c do Hasse cu’a ta.p du.o.. c sap thu. tu.. bo. pha.n, ta co s ≤ t neu va chı’ neuhoa. c s = t hoa. c co mo.t du.o.ng di di.nh hu.o.ng tu. s den t.

Gia’ su.’ (S,≤) la ta.p du.o.. c sap thu. tu.. bo. pha.n va A ⊂ S,A 6= ∅.

97

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{∅}

{a}{b}

{c}

{a, b} {b, c}

{a, b, c}

{a, c}

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Hınh 5.3:

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a b

cde

Hınh 5.4:

D- i.nh nghıa 5.3.3. (a) Phan tu.’ x ∈ S du.o.. c go. i la ca.n tren cu’a A neu a ≤ x vo.i mo.ia ∈ A; khi do A du.o.. c go. i la bi. cha.n tren. Neu x la ca.n tren cu’a A va x ∈ A thı xdu.o.. c go. i la phan tu.’ lo.n nhat cu’a A, ky hie.u

maxA := max{a | a ∈ A}.

(b) Phan tu.’ y ∈ S du.o.. c go. i la ca.n du.o.i cu’a A neu y ≤ a vo.i mo.i a ∈ A; khi do A du.o.. cgo. i la bi. cha.n du.o.i. Neu y la ca.n du.o.i cu’a A va y ∈ A thı y du.o.. c go. i la phan tu.’ nho’

nhat cu’a A, ky hie.uminA := min{a | a ∈ A}.

(c) Ky hie.u As la ta.p ho.. p tat ca’ cac ca.n tren cu’a A. Neu As 6= ∅ (tu.c la neu A bi. cha.ntren) va neu As co phan tu.’ nho’ nhat x∗ thı x∗ du.o.. c go. i la ca.n tren nho’ nhat hoa. c ca.ntren dung cu’a A, ky hie.u

sup A := sup{a | a ∈ A}.

(d) Ky hie.u Ai la ta.p ho.. p tat ca’ cac ca.n du.o.i cu’a A. Neu Ai 6= ∅ (tu.c la neu A bi. cha.ndu.o.i) va neu Ai co phan tu.’ lo.n nhat y∗ thı y∗ du.o.. c go. i la ca.n du.o.i lo.n nhat hoa. c ca.ndu.o.i dung cu’a A, ky hie.u

inf A := inf{a | a ∈ A}.

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Hınh 5.5:

Vı du. 5.3.3. Xet ta.p du.o.. c sap thu. tu.. bo. pha.n trong Vı du. 5.3.2(a). Ta.p S khong co phantu.’ lo.n nhat; 1 la phan tu.’ nho’ nhat.

Nha.n xet 6. (a) Phan tu.’ lo.n nhat (nho’ nhat) cu’a A, neu ton ta. i, la duy nhat.

(b) Neu ton ta. i x = maxA (tu.o.ng u.ng y = minA) thı x = sup A (tu.o.ng u.ng y = inf A).D- ieu ngu.o.. c la. i khong dung (cho vı du. ).

Vı du. 5.3.4. (a) Trong ta.p du.o.. c sap thu. tu.. bo. pha.n ({1, 2, 3, 4, 5, 6}, |) ta.p con {2, 3} codung mo.t ca.n tren la 6, va do do sup{2, 3} = 6. Tu.o.ng tu.. inf{2, 3} = 1. Ta.p con {4, 6}khong co ca.n tren; inf{4, 6} = 2. Ta.p con {3, 6} co ca.n tren 6 va hai ca.n du.o.i la 1 va 3;do do sup{3, 6} = 6 va inf{3, 6} = 3. Va.y cac ca.n tren dung va ca.n du.o.i dung cu’a A chu.achac ton ta. i, va neu chung ton ta. i chu.a chac chung thuo.c ta.p con A.

(b) Xet ta.p con du.o.. c sap thu. tu.. bo. pha.n co lu.o.. c do Hasse trong Hınh 5.6. Ta cosup{d, f} = g va inf{b, d, e, f} = a. Nhu.ng sup{b, c} va inf{d, e, f} khong ton ta. i.

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Hınh 5.6:

D- i.nh nghıa 5.3.4. Ta.p du.o.. c sap thu. tu.. bo. pha.n (S,≤) du.o.. c go. i la lattice (dan) neu tonta. i sup{x, y} va inf{x, y} vo.i mo.i x, y ∈ S. Khi do ta di.nh nghıa hai phep toan

x ∨ y := sup{x, y}, x ∧ y := inf{x, y}.

Hie’n nhien ∨ va ∧ la cac phep toan hai ngoi tren S. Ho.n nu.a

x ∧ y = x ⇔ x ≤ y ⇔ x ∨ y = y.

Bang quy na.p, chung ta co the’ chu.ng minh mo.i ta.p con A hu.u ha.n phan tu.’ cu’a lattice Lluon ton ta. i sup A, inf A.

99

Vı du. 5.3.5. (a) Ta.p du.o.. c sap thu. tu.. trong Vı du. 5.3.2(b) la lattice.

(b) Ta.p du.o.. c sap thu. tu.. trong Vı du. 5.3.2(a) khong la lattice do ta.p {3, 4} khong co ca.ntren trong S.

Tu. di.nh nghıa cu’a cac phep toan ∧ va ∨ ta co cac da’ng thu.c sau:

x ∧ x = x,x ∧ y = y ∧ x,(x ∧ y) ∧ z = x ∧ (y ∧ z),

x ∨ x = x,x ∨ y = y ∨ x,(x ∨ y) ∨ z = x ∨ (y ∨ z).

Bai ta.p

1. Ve lu.o.. c do Hasse cu’a cac ta.p du.o.. c sap thu. tu.. bo. pha.n sau:

(a) ({1, 2, 3, 4, 6, 8, 12, 24}, |), trong do m|n nghıa la n chia het cho m.

(b) Ta.p cac ta.p con cu’a {3, 7} vo.i quan he. ⊆ .

2. Tım cac ta.p con thu.. c su.. cu.. c da. i cu’a ta.p {a, b, c}. Tu.c la tım cac phan tu.’ cu.. c da. icu’a ta.p con du.o.. c sap thu. tu.. bo. pha.n cu’a P({a, b, c}) la nhu.ng ta.p con thu.. c su.. cu’a{a, b, c}.

3. Tren R× R xet cac quan he. <,≺,� xac di.nh bo.’ i

(x, y) < (z,w) neu x2 + y2 < z2 + w2,

(x, y) ≺ (z,w) neu (x, y) < (z,w) hoa. c (x, y) = (z,w),

(x, y) � (z,w) neu x2 + y2 ≤ z2 + w2.

(a) Quan he. nao la quan he. thu. tu.. bo. pha.n?

(b) Ve mo.t phan cu’a {(x, y) | (x, y) ≺ (3, 4)} trong R2.

(c) Ve mo.t phan cu’a {(x, y) | (x, y) � (3, 4)} trong R2.

4. Gia’ su.’ E(N) la ta.p tat ca’ cac ta.p con hu.u ha.n cu’a N ma co mo.t so chan phan tu.’ , vo.iquan he. thu. tu.. bo. pha.n ⊆ .

(a) D- a.t A := {1, 2} va B := {1, 3}. Tım bon ca.n tren cu’a {A,B}.

(b) {A,B} co ca.n tren nho’ nhat trong E(N)? Gia’i thıch.

(c) E(N) la lattice?

5. Mo.i ta.p con du.o.. c sap thu. tu.. bo. pha.n cu’a mo.t lattice la mo.t lattice? Gia’i thıch.

100

6. Ba’ng trong hınh sau cho quan he. thu. tu.. bo. pha.n. No cho gia tri. x ∨ y doi vo.i lattice(L,≤). Cha’ng ha.n b ∨ c = d.

∨ a b c d e fa e a e e ab d d e bc d e cd e de ef

(a) Viet cac cho trong con la. i cu’a ba’ng.

(b) Tım cac phan tu.’ lo.n nhat va nho’ nhat cu’a L.

(c) Chu.ng minh rang f ≤ c ≤ d ≤ e.

(d) Ve lu.o.. c do Hasse cu’a L.

7. Xet R vo.i thu. tu.. ≤ thong thu.o.ng.

(a) R la lattice? Neu dung thı y nghıa cu’a a ∨ b, a ∧ b trong R.

(b) Tım vı du. cu’a ta.p con khac trong cu’a R ma khong co ca.n tren nho’ nhat.

(c) Tım sup{x ∈ R | x < 73}, sup{x ∈ R | x ≤ 73}, sup{x ∈ R | x2 ≤ 73},inf{x ∈ R | x2 < 73}.

8. (a) Dung quy na.p, chu.ng minh rang mo.i ta.p du.o.. c sap thu. tu.. bo. pha.n hu.u ha.n cophan tu.’ nho’ nhat.

(b) Cho vı du. ta.p du.o.. c sap thu. tu.. bo. pha.n co phan tu.’ lo.n nhat nhu.ng khong cophan tu.’ nho’ nhat.

9. Kha’o sat ta.p du.o.. c sap thu. tu.. bo. pha.n C co lu.o.. c do Hasse sau:

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101

Chu.ng minh cac bat da’ng thu.c:

w ∨ (x ∧ y) 6= (w ∨ x) ∧ (w ∨ y),

w ∧ (x ∨ y) 6= (w ∧ x) ∨ (w ∧ y).

Tu. do suy ra lattice C khong thoa’ man lua. t phan phoi.

10. (a) Chu.ng minh rang neu � la mo.t thu. tu.. bo. pha.n tren S thı quan he. ngu.o.. c � cungla thu. tu.. bo. pha.n tren S.

(b) Chu.ng minh rang neu ≺ la quan he. tren S thoa’ tınh chat (T) va s ≺ s sai vo.imo.i s ∈ S thı quan he. � xac di.nh bo.’ i

x � y neu va chı’ neu x ≺ y hoa. c x = y,

la quan he. thu. tu.. bo. pha.n.

11. Gia’ su.’ R la quan he. pha’n doi xu.ng va bac cau tren ta.p S.

(a) Chu.ng minh rang R ∪ E la thu. tu.. bo. pha.n tren S.

(b) R \ E la thu. tu.. bo. pha.n tren S?

12. Gia’ su.’ Σ la mo.t ba’ng cac ky tu.. va Σ∗ la ta.p cac chuoi ky tu.. . Xet quan he. � tren Σ∗

nhu. sau. Vo.i moi x, y ∈ Σ∗ ky hie.u x � y neu x la mo.t doa.n kho.’ i dau cu’a y, tu.c laton ta. i z ∈ Σ∗ sao cho xz = y. Ky hie.u length(w), w ∈ Σ∗, la do. dai cu’a chuoi w.

(a) Chu.ng minh rang � co tınh pha’n xa. , pha’n doi xu.ng va bac cau.

(b) Chu.ng minh rang neu x phu’ y thı length(x) = 1 + length(y).

13. Gia’ su.’ Σ la mo.t ba’ng cac ky tu.. . Ky hie.u w1 � w2, w1, w2 ∈ Σ∗, nghıa la length(w1) ≤length(w2). Quan he. � la quan he. thu. tu.. bo. pha.n tren Σ∗? Ta. i sao?

14. Gia’ su.’ Σ la mo.t ba’ng cac ky tu.. .

(a) Vo.i moi x, y ∈ Σ∗, di.nh nghıa x � y neu ton ta. i v, v′ ∈ Σ∗ sao cho y = vxv′.Quan he. � la thu. tu.. bo. pha.n tren Σ∗? Gia’i thıch.

(b) Tra’ lo.i cau ho’i tren neu ha.n che x, y ∈ Σ.

15. Ky hie.u Σ∗ la ta.p tat ca’ cac chuoi ky tu.. tren ba’ng chu. cai Σ := {a, b}. Xet quan he.hai ngoi � tren P(Σ∗) bo.’ i A � B neu va chı’ neu A∗ ⊆ B∗. Ky hie.u (R), (S), (AS) va(T) la cac tınh chat pha’n xa. , doi xu.ng, pha’n doi xu.ng va bac cau.

(a) Cac tınh chat nao trong so (R), (S), (AS), (T) ma quan he. � thoa’?

(b) � la thu. tu.. bo. pha.n?

16. Gia’ su.’ x, y, z la cac chuoi tren ba’ng ky tu.. khac trong Σ nao do. Quan he. hai ngoiP (x, y) sau co tınh chat gı:

P (x, y) ⇔ (∃z)(concat(x, z) = y),

trong do concat(x, z) la chuoi nha.n du.o.. c bang cach noi chuoi z sau chuoi x. Cha’ngha.n, neu x = “ANH”, z = “ EM.”, thı concat(x, z) = “ANH EM.”.

102

17. Ky hie.u J (N) la ho. tat ca’ cac ta.p con hu.u ha.n cu’a N. Khi do (J (N),⊆) la mo.t ta.pdu.o.. c sap thu. tu.. bo. pha.n.

(a) J (N) co phan tu.’ lo.n nhat? Neu co, hay tım. Neu khong, gia’i thıch.

(b) J (N) co phan tu.’ nho’ nhat? Neu co, hay tım. Neu khong, gia’i thıch.

(c) Gia’ su.’ A,B ∈ J (N), {A,B} co ca.n tren nho’ nhat trong J (N)? Neu co, hay tım.Neu khong, cho vı du. .

(d) Gia’ su.’ A,B ∈ J (N), {A,B} co ca.n du.o.i lo.n nhat trong J (N)? Neu co, hay tım.Neu khong, cho vı du. .

(e) J (N) la lattice? Gia’i thıch.

18. La.p la. i bai ta.p tren, neu thay J (N) la ho. cac ta.p con vo ha.n cu’a N.

19. Gia’ su.’ x, y, z la cac chuoi. Y nghıa cu’a bie’u thu.c sau

Q(x, y) ⇔ (∃z)(concat(z, x) = y)?

20. Gia’ su.’ (A,≤) la ta.p du.o.. c sap thu. tu.. bo. pha.n. Xet quan he. ≥ tren A : a ≥ b neu vachı’ neu b ≤ a. Chu.ng minh rang (A,≥) cung la mo.t ta.p du.o.. c sap thu. tu.. .

21. Gia’ su.’ Σ∗ la ta.p cac day hu.u ha.n phan tu.’ cu’a ta.p Σ bao gom ca’ ta.p trong. Chı’ ra(Σ∗,≤) co pha’i la ta.p du.o.. c sap thu. tu.. bo. pha.n neu

(a) w ≤ w′ neu va chı’ neu l(w) ≤ l(w′), trong do l(w) la do. dai cu’a chuoi w.

(b) w ≤ w′ neu va chı’ neu ton ta. i cac chuoi w1, w2 ∈ Σ∗ sao cho w1w′w2 = w.

22. (a) Kha’o sat cac phan tu.’ x, y, z trong mo.t ta.p du.o.. c sap thu. tu.. bo. pha.n. Chu.ngminh rang neu sup{x, y} = a va inf{a, z} = b, thı sup{x, y, z} = b.

(b) Chu.ng minh rang mo.i ta.p con hu.u ha.n cu’a mo.t lattice co ca.n tren nho’ nhat.

(c) Chu.ng minh rang neu x, y, z la cac phan tu.’ cu’a mo.t lattice, thı (x ∨ y) ∨ z =x ∨ (y ∨ z).

23. Gia’ su.’ Left la quan he. hai ngoi tren ta.p cac node cu’a cay nhi. phan T xac di.nh nhu.

sau: Left(x,y) neu va chı’ neu x va y co chung mo.t to’ tien z sao cho x la node tren caycon ben trai tu. z; va y la node tren cay con ben pha’i tu. z.

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x

z

y•

•

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Chu.ng minh rang Left(x, y) va Left(y,w) suy ra Left(x,w).

103

24. Gia’ su.’ (a1, . . . , an) la day gom n phan tu.’ cu’a ta.p hu.u ha.n A; R la mo.t thu. tu.. bo.pha.n tren A. Ta noi rang (a1, . . . , an) la sap xep to po cu’a A doi vo.i R neu vo.i mo.iai, aj ∈ A, (ai, aj) ∈ R suy ra i < j.

(a) Chu.ng minh rang neu R la thu. tu.. bo. pha.n tren A, thı ton ta. i cac phan tu.’ x, y ∈ Asao cho khong co z ∈ A thoa’ (z, x) ∈ R va (y, z) ∈ R. (x, y go. i la cac phan tu.’

nho’ nhat va lo.n nhat tu.o.ng u.ng).

(b) Chu.ng minh rang neu R la thu. tu.. bo. pha.n tren A, thı A co the’ sap xep to podoi vo.i R.

25. Gia’ su.’ (A,≥) nha.n du.o.. c tu. (A,≤) nhu. trong Bai ta.p 19. Chu.ng minh rang b laca.n du.o.i lo.n nhat doi vo.i (ai|i ∈ I) trong (A,≤) neu b la ca.n tren nho’ nhat doi vo.i(ai|i ∈ I) trong (A,≥).

5.4 Quan he. tu.o.ng du.o.ng

Trong mu.c nay chung ta se nghien cu.u cac quan he. tu.o.ng du.o.ng: la quan he. ma nhom cacphan tu.’ co cung mo.t da. c tru.ng hay tınh chat.

D- i.nh nghıa 5.4.1. Quan he. R tren S du.o.. c go. i la quan he. tu.o.ng du.o.ng neu no co cac tınhchat: pha’n xa. , doi xu.ng va bac cau. Khi do thay cho cach viet aRb, ta thu.o.ng viet a ∼ bhoa. c a ≡ b.

Vı du. 5.4.1. (a) Gia’ su.’ p la so tu.. nhien lo.n ho.n 2. Tren ta.p cac so tu.. nhien N, quan he.sau la quan he. tu.o.ng du.o.ng:

mRn ⇔ m − n... p ⇔ m = n (mod p), ∀ m,n ∈ N.

(b) Gia’ su.’ S la ta.p cac tam giac trong ma.t pha’ng. Xet quan he. R tren S: T1RT2 neu vachı’ neu ton ta. i anh xa. mo.t-mo.t tu. tam giac T1 len tam giac T2 sao cho cac goc tu.o.ngu.ng bang nhau. Thı R la quan he. tu.o.ng du.o.ng.

(c) Hai ma tra.n vuong cap n : A va B du.o.. c go. i la tu.o.ng du.o.ng, kı hie.u A ∼ B, neu tonta. i cac ma tra.n vuong cap n kha’ nghi.ch P,Q sao cho B = PAQ. Khi do “∼” la quanhe. tu.o.ng du.o.ng.

(d) Xet P(S) cac ta.p con cu’a ta.p S. Vo.i A,B ∈ P(S), ta di.nh nghıa A ∼ B neu hie.u doixu.ng cu’a chung A⊕B := (A \B) ∪ (B \A) la mo.t ta.p hu.u ha.n. Thı “∼” la quan he.tu.o.ng du.o.ng.

Gia’ su.’ ∼ la quan he. tu.o.ng du.o.ng tren ta.p S. Ta.p ho.. p

[s] := {t ∈ S | s ∼ t}

104

du.o.. c go. i la lo.p tu.o.ng du.o.ng cu’a s, va

[S] := {[s] | s ∈ S}

la ta.p cac lo.p tu.o.ng du.o.ng.

Vı du. 5.4.2. Gia’ su.’ ∼ la quan he. tu.o.ng du.o.ng trong Vı du. 5.4.1(a) thı

[m] = {n ∈ Z | m = n (mod p)}.

Do do vo.i p = 3 ta co ba lo.p tu.o.ng du.o.ng: [0], [1] va [2].

Bo’ de 5.4.2. Gia’ su.’ ∼ la quan he. tu.o.ng du.o.ng tren S; va s, t ∈ S. Cac kha’ng di.nh saula tu.o.ng du.o.ng

(a) s ∼ t;

(b) [s] = [t];

(c) [s] ∩ [t] 6= ∅.

Chu.ng minh. (a) ⇒ (b) Gia’ su.’ s ∼ t va xet s′ ∈ [s]. Thı s ∼ s′. Ta co t ∼ s (doi xu.ng).Suy ra t ∼ s′ (bac cau). Do do s′ ∈ [t]. Va.y [s] ⊆ [t]. Tu.o.ng tu.. cung co [t] ⊆ [s].

(b) ⇒ (c) Hie’n nhien.

(c) ⇒ (a) Lay u ∈ [s] ∩ [t]. Thı s ∼ u va u ∼ t. Va.y s ∼ t. 2

Nhac la. i rang phan hoa. ch cu’a ta.p A la mo.t ho. cac ta.p con A1, A2, . . . , Ak cu’a A sao cho

(a)⋃k

i=1 Ai = A; va

(b) Ai ∩ Aj = ∅ vo.i mo.i i, j = 1, 2, . . . , k, i 6= j.

Vı du. 5.4.3. Ho.Σ := {{1, 2, 4}, {3, 5, 7}, {6, 8}}

la mo.t phan hoa.ch cu’a ta.p A := {1, 2, . . . , 8}.

D- i.nh ly sau cho chung ta moi quan he. giu.a phan hoa.ch va quan he. tu.o.ng du.o.ng.

D- i.nh ly 5.4.3. (a) Neu ∼ la quan he. tu.o.ng du.o.ng tren ta. p khac trong S thı [S] la mo. tphan hoa. ch cu’a ta. p S.

(b) Ngu.o.. c la. i, neu {Ai | i ∈ I} la mo. t phan hoa. ch cu’a ta. p S thı cac ta. p Ai la cac lo.ptu.o.ng du.o.ng u.ng vo.i quan he. tu.o.ng du.o.ng nao do tren S.

105

Chu.ng minh. (a) Ta can chu.ng to’

(i)⋃

s∈S[s] = S.

(ii) Vo.i mo.i s, t ∈ S ta co hoa. c [s] = [t] hoa. c [s] ∩ [t] = ∅.

Tha.t va.y, hie’n nhien rang⋃

s∈S[s] ⊂ S. Lay s0 ∈ S ta co s0 ∈ [s0]. Do do S ⊂⋃

s∈S[s].Va.y (i) dung.

Kha’ng di.nh (ii) suy tu. Bo’ de 5.4.2.

(b) Gia’ su.’ {Ai | i ∈ I} la mo.t phan hoa.ch cu’a S. Tren S xet quan he. ∼:

s ∼ t ⇔ ton ta. i i ∈ I sao cho s, t ∈ Ai.

De dang kie’m tra “ ∼′′ la quan he. tu.o.ng du.o.ng. 2

Vı du. 5.4.4. (a) Gia’ su.’ J la ho. cac ta.p nao do va vo.i moi S, T ∈ J ta di.nh nghıa S ∼ Tneu ton ta. i anh xa. mo. t-mo.t tu. S len T. Thı ‘ ∼′ la quan he. tu.o.ng du.o.ng tren J . Hie’nnhien rang neu S la ta.p hu.u ha.n, thı [S] gom tat ca’ cac ta.p con cu’a J co cung sophan tu.’ vo.i ta.p S. Neu S la ta.p dem du.o.. c thı [S] gom tat ca’ cac ta.p con dem du.o.. c(cu’a J ).

(b) Tren N× N ta di.nh nghıa

(m,n) ∼ (j, k) neu m2 + n2 = j2 + k2.

Hie’n nhien ∼ la quan he. tu.o.ng du.o.ng. Bang cach xet anh xa.

f : N× N −→ N, (m,n) 7→ m2 + n2,

thı cac lo.p tu.o.ng du.o.ng chınh la cac ta.p con khac trong f−1(u), u ∈ N.

D- i.nh ly 5.4.4. (a) Gia’ su.’ S 6= ∅ va anh xa. f : S −→ T. Ta di.nh nghıa s ∼ t (s, t ∈ S)neu f(s) = f(t). Thı ∼ la quan he. tu.o.ng du.o.ng tren S va cac lo.p tu.o.ng du.o.ng lacac ta. p khac trong f−1(u), trong do u ∈ T.

(b) Moi quan he. tu.o.ng du.o.ng tren S du.o.. c xac di.nh bo.’ i mo. t anh xa. f thıch ho.. p nhu. trongphan (a).

Chu.ng minh. (a) Chung ta kie’m tra ∼ la quan he. tu.o.ng du.o.ng:

[Pha’n xa. ]. Ta co f(s) = f(s). Va.y s ∼ s vo.i mo.i s ∈ S.

[D- oi xu.ng]. f(s1) = f(s2) ⇔ f(s2) = f(s1). Va.y quan he. s1 ∼ s2 suy ra s2 ∼ s1.

[Bac cau]. Neu f(s1) = f(s2) va f(s2) = f(s3) thı f(s1) = f(s3).

(b) Xet anh xa. tu.. nhienf : S −→ [S], s 7→ [s].

De dang kie’m tra f tho’a cac dieu kie.n doi ho’i. 2

106

Bai ta.p

1. Ky hie.u (R), (S), (AS) va (T) la cac tınh chat pha’n xa. , doi xu.ng, pha’n doi xu.ng vabac cau. Tım cac ma tra.n cu’a cac quan he. tren S := {0, 1, 2, 3} va kie’m tra cac tınhchat (R), (S), (AS) va (T), neu

(a) mR1n neu m + n = 3.

(b) mR2n neu m = n (mod 2).

(c) mR3n neu m ≤ n.

(d) mR4n neu m + n ≤ 4.

(e) mR5n neu max{m,n} = 3.

Cac quan he. nao la thu. tu.. bo. pha.n, quan he. tu.o.ng du.o.ng?

2. Cac quan he. sau tren Z, quan he. nao la tu.o.ng du.o.ng, khi do lie.t ke cac lo.p tu.o.ngdu.o.ng.

(a) n ≡ m (mod 4).(b) mn = 0.

(c) mn > 0.(d) n ≤ m.

3. Xet quan he. R tren Z xac di.nh bo.’ i mRn neu va chı’ neu m3 − n3 ≡ 0 (mod 5).

(a) R thoa’ cac tınh chat nao trong so (R), (S), (AS), (T).

(b) R la quan he. tu.o.ng du.o.ng? thu. tu.. bo. pha.n?

4. D- a.t Σ := {a, b, c, d, e, f, g}. Viet ma tra.n tu.o.ng u.ng quan he. tren Σ xac di.nh bo.’ i phanhoa.ch {{a, d}, {c, e, f}, {b, g}}.

5. (a) Cho ta.p khac trong S. Kha’o sat quan he. trong ∅ ⊂ S × S tren S. Cac tınh chatnao trong so (R), (S), (AS), (T) ma quan he. ∅ thoa’?

(b) Nhu. tren doi vo.i quan he. U := S × S tren S.

6. (a) Chu.ng minh rang giao cu’a hai quan he. tu.o.ng du.o.ng la quan he. tu.o.ng du.o.ng.

(b) Ho.. p hai quan he. tu.o.ng du.o.ng la quan he. tu.o.ng du.o.ng?

7. Gia’ su.’ S la ta.p cac ta.p con vo ha.n cu’a N. Vo.i A,B trong S, xet quan he.

A ∼ B ⇔ A ∩ B la ta.p hu.u ha.n.

D- ay la quan he. tu.o.ng du.o.ng?

8. Chu.ng minh rang quan he. R tren ta.p S la quan he. tu.o.ng du.o.ng, neu va chı’ neu thoa’

man ba dieu kie.n

(a) E := {(x, x) ∈ S × S} ⊂ R,

(b) R = R−1,

(c) R ◦ R ⊂ R.

107

9. Ta noi mo.t ho. cac ta.p con khac trong ro.i nhau cu’a ta.p S la mo.t phan hoa.ch cu’a ta.pS neu ho.. p cu’a cac ta.p nay bang S. Chu.ng minh rang cac lo.p tu.o.ng du.o.ng cu’a mo.tquan he. tren S la.p thanh phan hoa.ch cu’a ta.p S. Ngu.o.. c la. i, gia’ su.’ (Ai|i ∈ I) cac ta.pcon cu’a S sao cho Ai ∩ Aj = ∅, i 6= j, va ∪(Ai|i ∈ I) = S. Xet quan he. ∼S tren S :a ∼S b neu va chı’ neu ton ta. i chı’ so i ∈ I sao cho a, b ∈ Ai.

(a) Chu.ng minh rang ∼S la quan he. tu.o.ng du.o.ng.

(b) Chu.ng minh rang cac lo.p tu.o.ng du.o.ng cu’a ∼S la cac khoi Ai cu’a phan hoa.ch(Ai|i ∈ I).

10. Gia’ su.’ F la ta.p tat ca’ cac ham tu. N len N. Vo.i f, g ∈ F, xet quan he. f ∼ g neuf(n) = g(n) ngoai mo. t ta.p con hu.u ha.n cu’a ta.p cac so tu.. nhien N.

(a) Chu.ng minh rang ∼ la quan he. tu.o.ng du.o.ng.

(b) Ky hie.u lo.p tu.o.ng du.o.ng cu’a f bo.’ i [f ]. Ta noi, [f ] ≤ [g] neu f(n) ≤ g(n) ngoaimo.t ta.p con hu.u ha.n cu’a ta.p cac so tu.. nhien N.

(c) Ky hie.u [k] la lo.p tu.o.ng du.o.ng cu’a f(n) = k vo.i mo.i n ∈ N. Chu.ng minh rangton ta. i vo ha.n cac phan tu.’ [f1], [f2], . . . , [fm], . . . , sao cho

[k] ≤ [f1] ≤ [f2] ≤ · · · ≤ [fm] ≤ · · · ≤ [k + 1].

11. Cac quan he. sau, quan he. nao la quan he. tu.o.ng du.o.ng?

(a) L1||L2, doi vo.i cac du.o.ng tha’ng trong ma. t pha’ng, neu L1 va L2 la trung nhauhoa. c song song.

(b) L1 ⊥ L2, doi vo.i cac du.o.ng tha’ng trong ma.t pha’ng, neu L1 va L2 vuong goc.

(c) p1 ∼ p2, doi vo.i ngu.o.i Vie.t Nam, neu p1 va p2 song trong cung mo.t thanh pho.

(d) p1 ∼ p2, doi vo.i ngu.o.i, neu p1 va p2 co chung cha me..

(e) p1 ∼ p2, doi vo.i ngu.o.i, neu p1 va p2 co chung me..

12. (a) Lie.t ke tat ca’ cac lo.p tu.o.ng du.o.ng cu’a Z doi vo.i quan he. tu.o.ng du.o.ng dong du.

modulo cho 4.

(b) Co bao nhieu lo.p tu.o.ng du.o.ng khac nhau cu’a Z tu.o.ng u.ng vo.i quan he. tu.o.ngdu.o.ng dong du. modulo cho 73.

13. Gia’ su.’ S la ta.p ho.. p. Quan he. = la quan he. tu.o.ng du.o.ng?

14. Cac ma tra.n A va B trong Mat(n, n) la dong da. ng (similar) neu ton ta. i ma tra.n kha’

nghi.ch P sao cho B = PAP−1; khi do ta ky hie.u A ≈ B. Chu.ng minh rang ≈ la quanhe. tu.o.ng du.o.ng tren Mat(n, n).

15. Gia’ su.’ S la ta.p tat ca’ cac day (sn) ⊂ R, va di.nh nghıa (sn) ≈ (tn) neu {n ∈ N | sn 6= tn}la ta.p hu.u ha.n. Chu.ng minh rang ≈ la quan he. tu.o.ng du.o.ng tren S.

16. Quan he. “quen biet” la quan he. tu.o.ng du.o.ng?

108

17. Gia’ su.’ S la ta.p ho.. p va G la nhom cac ham mo.t-mo.t len f : S → S, tu.c la,

(a) ham dong nhat 1S thuo.c G;

(b) neu f, g ∈ G thı f ◦ g ∈ G;

(c) neu f ∈ G thı f−1 ∈ G.

Vo.i x, y ∈ S, di.nh nghıa x ∼ y neu ton ta. i f ∈ G sao cho f(x) = y. Chu.ng minh rang∼ la quan he. tu.o.ng du.o.ng tren S.

18. Tren Z xet quan he. ≈ di.nh nghıa bo.’ i m ≈ n neu m2 = n2.

(a) Chu.ng minh rang ≈ la quan he. tu.o.ng du.o.ng tren Z.

(b) Mo ta’ cac lo.p tu.o.ng du.o.ng doi vo.i ≈ . Co bao nhieu lo.p tu.o.ng du.o.ng?

19. Tren N xet quan he. ∼ di.nh nghıa bo.’ i m ∼ n neu m2 − n2 la bo. i cu’a 3.

(a) Chu.ng minh rang ∼ la quan he. tu.o.ng du.o.ng tren N.

(b) Lie.t ke bon phan tu.’ trong lo.p tu.o.ng du.o.ng [0].

(c) Lie.t ke bon phan tu.’ trong lo.p tu.o.ng du.o.ng [1].

(d) Co lo.p tu.o.ng du.o.ng nao khac?

20. Kha’o sat ta.p P(S) cac ta.p con cu’a ta.p S. Vo.i A,B ∈ P(S), ky hie.u A ∼ B neu hie.udoi xu.ng A⊕B := (A \B)∪ (B \A) la ta.p hu.u ha.n. Chu.ng minh rang ∼ la quan he.tu.o.ng du.o.ng tren P(S).

(a) Mo ta’ cac ta.p trong lo.p tu.o.ng du.o.ng chu.a ta.p trong ∅.

(b) Mo ta’ cac ta.p trong lo.p tu.o.ng du.o.ng chu.a S.

21. Tren N× N di.nh nghıa (m,n) ∼ (k, l) neu m + l = n + k.

(a) Chu.ng minh rang ∼ la quan he. tu.o.ng du.o.ng tren N× N.

(b) Ve mo.t phan cu’a N ×N de’ chı’ ra cac lo.p tu.o.ng du.o.ng.

22. D- i.nh nghıa m ≡ n (mod p) van co nghıa khi p = 1 hoa. c p = 0.

(a) Mo ta’ quan he. tu.o.ng du.o.ng nay doi vo.i p = 1 va cac lo.p tu.o.ng du.o.ng tu.o.ng u.ngtrong Z.

(b) Nhu. tren vo.i p = 0.

23. Gia’ su.’ P la ta.p tat ca’ cac chu.o.ng trınh may tınh va hai chu.o.ng trınh p1 va p2 latu.o.ng du.o.ng neu chung cho cung mo.t ket qua’ vo.i cung du. lie.u ban dau. D- ay la quanhe. tu.o.ng du.o.ng? Ta.i sao?

24. Gia’ su.’ Σ la ba’ng chu. cai, va vo.i x, y ∈ Σ∗, ky hie.u x ∼ y neu length(x) = length(y).Chu.ng minh rang ∼ la quan he. tu.o.ng du.o.ng va mo ta’ cac lo.p tu.o.ng du.o.ng.

109

25. Kha’o sat P× P va di.nh nghıa (m,n) ∼ (p, q) neu mq = np.

(a) Chu.ng minh rang ∼ la quan he. tu.o.ng du.o.ng tren P× P.(b) Chu.ng minh rang ∼ la quan he. tu.o.ng du.o.ng tu.o.ng u.ng vo.i ham P× P→ Q chobo.’ i f((m,n)) = m/n.

26. Co bao nhieu quan he. tu.o.ng du.o.ng tren ta.p {0, 1, 2, 3}?

27. Tren Z xet quan he. ≈ di.nh nghıa bo.’ i m ≈ n neu m2 = n2.

(a) Sai (vı sao) neu di.nh nghıa ≤ tren [Z] bo.’ i [m] ≤ [n] neu va chı’ neu m ≤ n?

(b) Sai (vı sao) neu di.nh nghıa ham f : [Z] → Z, f([m]) = m2 + m + 1?

(c) Nhu. tren, vo.i g([m]) = m4 + m2 + 1.

(d) Sai (vı sao) neu di.nh nghıa tren Z, [m]⊕ [n] = [m + n]?

5.5 Bao dong cu’a quan he.

D- oi khi chung ta muon xay du.. ng mo.t quan he. mo.i tu. mo.t quan he. da co. Cha’ng ha.n, ta cohai quan he. tu.o.ng du.o.ng R1 va R2 tren S va chung ta muon tım quan he. nho’ nhat chu.aca’ R1 va R2. Quan he. nay co the’ khong pha’i la quan he. tu.o.ng du.o.ng. D- ieu nay xa’y ra doR1 ∪ R2 co the’ khong pha’i la bac cau. Va.y quan he. bac cau nho’ nhat chu.a R1 ∪ R2 la gı?Trong phan nay chung ta se tra’ lo.i cau ho’i nay.

Gia’ su.’ R la quan he. tren S. Kı hie.u r(R), s(R) va t(R) la cac quan he. pha’n xa. , doi xu.ngva bac cau nho’ nhat chu.a R. Cac quan he. nay du.o.. c go. i tu.o.ng u.ng la cac bao dong pha’nxa. , doi xu.ng va bac cau cu’a quan he. R.

Me.nh de 5.5.1. (a) R = r(R) neu va chı’ neu R pha’n xa. .

(b) R = s(R) neu va chı’ neu R doi xu.ng.

(c) R = t(R) neu va chı’ neu R bac cau.

Ho.n nu.ar(r(R)) = r(R), s(s(R)) = s(R), t(t(R)) = t(R).

Chu.ng minh. Suy tu. di.nh nghıa. 2

Vı du. 5.5.1. Xet quan he. R tren {1, 2, 3, 4} tu.o.ng u.ng ma tra.n Boole:

A =

0 0 1 10 1 0 00 0 1 01 0 0 0

.

110

(a) Quan he. R khong pha’n xa. . Ma tra.n Boole r(A) cu’a quan he. r(R) nha.n du.o.. c tu. matra.n A vo.i tat ca’ cac phan tu.’ tren du.o.ng cheo bang mo.t va do thi. cu’a r(R) suy tu.

do thi. cu’a R bang cach them cac khuyen ta. i cac dı’nh:

r(A) =

1 0 1 10 1 0 00 0 1 01 0 0 1

.

(b) Quan he. R khong doi xu.ng. D- o thi. cu’a quan he. s(R) suy tu. do thi. cu’a quan he. Rbang cach them (neu chu.a co) cac cung (j, i) neu ton ta. i cung (i, j). Ma tra.n Booles(A) co da.ng

s(A) =

0 0 1 10 1 0 01 0 1 01 0 0 0

.

(c) Quan he. R khong bat cau. Ma tra.n Boole t(A) co da.ng

t(A) =

1 0 1 10 1 0 00 0 1 01 0 1 1

.

D- i.nh ly 5.5.2. Neu R va E := {(x, x) | x ∈ S} la cac quan he. tren S thı

(a) r(R) = R ∪ E.

(b) s(R) = R ∪ R−1.

(c) t(R) =⋃∞

i=1 Ri.

Chu.ng minh. (a) Ta biet rang quan he. la pha’n xa. neu va chı’ neu no chu.a E. Do do R ∪Ela pha’n xa. va do mo.i quan he. pha’n xa. chu.a R pha’i chu.a R ∪ E nen r(R) = R ∪ E.

(b) Nhac la.i rang R1 la doi xu.ng neu va chı’ neu R−11 = R1. Neu (x, y) ∈ R ∪ R−1 thı

(y, x) ∈ R−1 ∪ R = R ∪ R−1. Do do R ∪ R−1 la quan he. doi xu.ng.

Xet R1 la quan he. doi xu.ng chu.a R. Neu (x, y) ∈ R−1 thı (y, x) ∈ R ⊂ R1, va do R1 doixu.ng nen (x, y) ∈ R1. Suy ra R−1 ⊂ R1. Va.y R ∪ R−1 ⊂ R1.

(c) D- au tien ta chu.ng minh ho.. p U :=⋃∞

i=1 Ri la bac cau. Tha.t va.y lay x, y, z ∈ S saocho (x, y) ∈ U, (y, z) ∈ U. Khi do ton ta. i i, j ∈ N sao cho (x, y) ∈ Ri va (y, z) ∈ Rj . Do do

(x, z) ∈ Ri+j ⊂ U.

111

Va.y U la quan he. bac cau. Bay gio. lay R1 la quan he. bac cau chu.a R. Ta chu.ng minhquy na.p theo k : Rk ⊂ R1.

Vo.i k = 1 la hie’n nhien. Gia’ su.’ dung den k. Ta co

Rk+1 = Rk ◦ R ⊂ R1 ◦ R1 ⊂ R1

(bao ham thu.c cuoi co du.o.. c do R1 bac cau).

Va.y vo.i mo.i k > 1 ta coRk ⊂ R1.

Hay U ⊂ R1. 2

Vı du. 5.5.2. (a) Gia’ su.’ R la quan he. tren ta.p S co n phan tu.’ va A la ma tra.n Boole cu’aR. Nhu. trong D- i.nh ly 5.5.3 du.o.i day chı’ ra:

t(R) =n⋃

i=1

Ri.

Du.. a vao cac ket qua’ tru.o.c ta co

t(A) = A ∨ A2 ∨ . . . ∨ An, s(A) = A ∨ At, r(A) = A ∨ In.

Trong do In la ma tra.n do.n vi. cap n. Tu. cac ma tra.n nay chung ta de dang xac di.nhcac quan he. t(R), s(R) va r(R).

(b)Xet quan he. R trong Vı du. 5.5.1, de dang kie’m tra la. i cac da’ng thu.c tren.

D- i.nh ly 5.5.3. Neu R la quan he. tren S co n phan tu.’ , thı

t(R) =n⋃

i=1

Ri.

Chu.ng minh. Chı’ can chu.ng minh t(R) ⊂⋃n

i=1 Ri. Lay (x, y) ∈ t(R). Go. i m la so nguyendu.o.ng nho’ nhat sao cho (x, y) ∈ Rm. Chı’ can xet tru.o.ng ho.. p m > 2. Khi do ton ta. i daycac phan tu.’ trong S

x1, x2, . . . , xm−1, xm = y,

sao choxRx1, x1Rx2, . . . , xm−1Rxm.

Neu m > n thı ton ta. i i, j (i < j) sao cho xi = xj thı ta co the’ bo’ qua xi, xi+1, . . . , xj−1 vadu.o.. c mo.t xıch ngan ho.n:

xRx1, . . . , xi−1Rxj, . . . , xm−1Ry.

Mau thuan vı m nho’ nhat. 2

112

Chung ta da xet cac bao dong tren quan he. R, bay gio. ta se lay mo.t quan he. mo.i co da.ngla to’ ho.. p cu’a bao dong.

Vı du. 5.5.3. Xet R trong Vı du. 5.5.1. Ma tra.n Boole cu’a s(r(R)) la

1 0 1 10 1 0 01 0 1 01 0 0 1

.

D- o cung la ma tra.n Boole r(s(A)) cu’a quan he. r(s(R)) va vı va.y s(r(R)) = r(s(R)). Ho.nnu.a quan he. trs(R) = tsr(R) la quan he. tu.o.ng du.o.ng vo.i ma tra.n Boole

tsr(A) = trs(A) =

1 0 1 10 1 0 01 0 1 11 0 1 1

.

Vı du. 5.5.4. Xet quan he. R tren {1, 2, 3} co ma tra.n Boole

A =

1 1 10 0 00 0 1

.

Vı A ∗ A = A nen R la bac cau. Ma tra.n tu.o.ng u.ng quan he. s(R) co da.ng

s(A) =

1 1 11 0 01 0 1

.

De thay rang quan he. s(R) khong bac cau!

Vı du. nay chu.ng to’ bao dong doi xu.ng cu’a bao dong bac cau chu.a chac la quan he. baccau. Noi mo.t cach khac cac phep lay bao dong co the’ pha hu’y tınh pha’n xa. , doi xu.ng haybac cau.

Bo’ de 5.5.4. (a) Neu R pha’n xa. thı s(R) va t(R) cung pha’n xa. .

(b) Neu R doi xu.ng thı r(R) va t(R) cung doi xu.ng.

(c) Neu R bac cau thı r(R) cung bac cau.

Chu.ng minh. (a) Hie’n nhien vı neu E ⊆ R thı E ⊆ s(R) va E ⊆ t(R).

(b) Bai ta.p.

(c) Gia’ su.’ R bac cau va (x, y), (y, z) ∈ r(R) = R ∪ E.

113

+ Neu (x, y) ∈ E thı x = y va do do (x, z) = (y, z) ∈ R ∪ E.

+ Neu (y, z) ∈ E thı y = z va do do (x, z) = (x, y) ∈ R ∪ E.

+ Neu (x, y) /∈ E va (y, z) /∈ E thı (x, y), (y, z) ∈ R, do do (x, z) ∈ R ⊆ R ∪ E.

Va.y ta luon luon co (x, z) ∈ R ∪ E. 2

D- i.nh ly 5.5.5. Gia’ su.’ R la quan he. tren S thı tsr(R) la quan he. tu.o.ng du.o.ng nho’ nhatchu.a R.

Chu.ng minh. + Tu. r(R) pha’n xa. va Bo’ de 5.5.4(a), suy ra tsr(R) pha’n xa. .

+ Tu. sr(R) doi xu.ng va Bo’ de 5.5.4(b), suy ra tsr(R) doi xu.ng.

+ Hie’n nhien tsr(R) bac cau.

Va.y tsr(R) la quan he. tu.o.ng du.o.ng. Lay R1 la quan he. tu.o.ng du.o.ng chu.a R. Thı

r(R) ⊆ r(R1) = R1.

Do do

sr(R) ⊆ s(R1) = R1.

Suy ra

tsr(R) ⊆ t(R1) = R1.

Noi cach khac tsr(R) la quan he. tu.o.ng du.o.ng nho’ nhat chu.a R. 2

Vı du. 5.5.5. Gia’ su.’ R la quan he. tren {1, 2, 3} trong Vı du. 5.5.4. Ta co

r(A) =

1 1 10 1 00 0 1

, s(r(A)) =

1 1 11 1 01 0 1

, t(s(r(A))) =

1 1 11 1 11 1 1

.

Va.y trs(R) = {1, 2, 3} × {1, 2, 3}.

Bai ta.p

1. Xet quan he. R tren ta.p S := {1, 2, 3} tu.o.ng u.ng ma tra.n Boole

A :=

0 1 00 0 00 0 1

.

Tım cac ma tra.n Boole cu’a r(R), s(R), rs(R), sr(R) va tsr(R).

114

2. La.p la. i Bai ta.p 1 vo.i

A :=

0 1 10 0 10 0 0

.

3. Vo.i Bai ta.p 1, lie.t ke cac lo.p tu.o.ng du.o.ng cu’a tsr(R).


5. La.p la. i Bai ta.p 1 vo.i quan he. R tren {1, 2, 3, 4} tu.o.ng u.ng vo.i ma tra.n Boole

A :=

0 1 0 11 0 1 00 1 1 01 0 1 0

.


7. Gia’ su.’ R la quan he. “tu.. a thu. tu.. ” tren ta.p cac so nguyen du.o.ng P : mRn neu m < n.Tım hoa.c mo ta’ r(R), sr(R), rs(R), tsr(R), t(R) va st(R).

8. La.p la. i Bai ta.p 7 vo.i mRn neu m la u.o.c cu’a n.

9. (a) Chu.ng minh rang neu (Rk) la mo.t day cac quan he. doi xu.ng tren S thı ho.. p ∪∞k=1Rk

la doi xu.ng.

(b) Gia’ su.’ R la quan he. doi xu.ng tren S. Chu.ng minh rang Rn, n ∈ P, la quan he. doixu.ng.

(c) Chu.ng minh rang neu R la quan he. doi xu.ng tren S thı r(R), t(R) la cac quan he.doi xu.ng tren S.

10. Xet quan he. R tren ta.p S.

(a) Chu.ng minh rang sr(R) = rs(R).

(b) Chu.ng minh rang tr(R) = rt(R).

11. Bang pha’n vı du. , chu.ng minh rang st(R) 6= ts(R).

12. Gia’ su.’ R la quan he. tren S := {1, 2} tu.o.ng u.ng ma tra.n Boole

(1 01 0

). Chu.ng minh

rang khong ton ta. i quan he. R′ nho’ nhat chu.a R sao cho sR′s sai vo.i mo.i s ∈ S.

13. Ta noi quan he. R tren S la quan he. len neu vo.i mo.i y ∈ S ton ta. i x ∈ S sao cho(x, y) ∈ R. Chu.ng minh rang khong ton ta. i mo.t quan he. len nho’ nhat chu.a quan he.R tren {1, 2} du.o.. c xac di.nh trong Bai ta.p 12.

115

14. Gia’ su.’ tınh chat p cu’a quan he. tren ta.p khac trong S thoa’ man

(i) Quan he. pho’ du. ng U := S × S co tınh chat p;

(ii) p dong doi vo.i phep giao, tu.c la, neu {Ri | i ∈ I} la mo.t ho. cac quan he. tren S cotınh chat p thı giao ∩i∈IRi cung co tınh chat p.

(a) Chu.ng minh rang vo.i mo.i quan he. R tren S ton ta. i mo.t quan he. nho’ nhat chu.aR va co tınh chat p.

(b) Nha.n xet rang cac tınh chat pha’n xa. , doi xu.ng va bac cau thoa’ man ca’ hai tınhchat (i) va (ii).

(c) Tınh chat quan he. len trong Bai ta.p 13 khong thoa’ tınh chat nao trong so (i), (ii)?

5.6 Lattice cu’a cac phan hoa.ch

Nha.n xet rang ho. C = P(S × S) tat ca’ cac quan he. tu.o.ng du.o.ng tren S la ta.p du.o.. c sapthu. tu.. vo.i quan he. bao ham. Phan tu.’ nho’ nhat la quan he. dong nhat E va phan tu.’ lo.nnhat la quan he. pho’ du.ng U vı E ⊂ R ⊂ U vo.i mo.i quan he. R tren S. Ho. C la lattice vo.ihai phep toan

R1 ∨ R2 := tsr(R1 ∪ R2), R1 ∧ R2 := R1 ∩ R2.

Chu y rang tsr(R1 ∪R2) = t(R1 ∪R2) vı R1 ∪R2 la quan he. co tınh pha’n xa. va doi xu.ng.

Vı du. 5.6.1. Xet ho.p S chu.a cac vien bi va hai quan he. tu.o.ng du.o.ng:

• (s, t) ∈ R1 neu s va t co cung mau;

• (s, t) ∈ R2 neu s va t co cung kıch thu.o.c.

Khi do (s, t) ∈ R1 ∧ R2 neu va chı’ neu s va t co cung mau va cung kıch thu.o.c. Ca.p (s, t)thuo.c R1 ∨ R2 neu ton ta. i day cac vien bi t1, t2, . . . , tm−1 ∈ S sao cho

(s, t1), (t1, t2), . . . , (tm−1, t) ∈ R1 ∪ R2.

Vı du. 5.6.2. Tren ta.p cac so nguyen du.o.ng P xet cac quan he. tu.o.ng du.o.ng R6 va R8 trongdo (m,n) ∈ R6 neu m = n (mod 6) va (m,n) ∈ R8 neu m = n (mod 8).

(a) Neu (m,n) ∈ R6 ∧R8 thı m− n la bo. i so cu’a 6 va 8, tu.c la m− n la bo. i so cu’a 24. Dodo (m,n) ∈ R6 ∧ R8 neu va chı’ neu m = n (mod 24).

(b) Ta se chu.ng minh R6 ∨ R8 = R2 trong do (m,n) ∈ R2 neu va chı’ neu m = n (mod 2).Chu y rang 2 la u.o.c so chung lo.n nhat cu’a 6 va 8. De dang thay rang R6 ∪ R8 ⊂ R2

va do R2 la quan he. tu.o.ng du.o.ng nen R6 ∨ R8 ⊂ R2. Ta can chı’ ra

R2 ⊂ R6 ∨ R8 = t(R6 ∪ R8) ⊂ R2.

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Nha.n xet la(k, k + 2) ∈ R6 ∨ R8 (5.1)

vo.i mo.i k ∈ P vı ca’ hai (k, k + 8) va (k + 8, k + 2) thuo.c R6 ∪ R8. Gia’ su.’ (m,n) ∈R2,m < n. Ta co the’ viet n = m+2r vo.i r ∈ P nao do. Tu. (5.1) suy ra tat ca’ cac ca.p

(m,m + 2), (m + 2,m + 4), . . . , (m + 2r − 2,m + 2r)

thuo.c R6 ∨ R8. Do do theo tınh bac cau, (m,m + 2r) = (m,n) cung thuo.c R6 ∨ R8,suy ra dieu can chu.ng minh.

Ta biet rang, co mo.t tu.o.ng u.ng giu.a cac quan he. tu.o.ng du.o.ng tren S va ta.p Π(S) tatca’ cac phan hoa.ch cu’a S. Ma.t khac ta.p cac quan he. tu.o.ng du.o.ng ta.o thanh lattice. Vı va.yton ta. i cau truc lattice tren Π(S).

Tha.t va.y, xet cac quan he. tu.o.ng du.o.ng R1 va R2 tu.o.ng u.ng cac phan hoa.ch π1 va π2.Khi do R1 ⊂ R2 neu va chı’ neu (s, t) ∈ R1 thı (s, t) ∈ R2. Noi cach khac, R1 ⊂ R2 neu vachı’ neu moi ta.p trong π1 la ta.p con nao do trong π2; trong tru.o.ng ho.. p nay ta noi π1 mi.nho.n π2 va ky hie.u π1 ≤ π2. Ta co ≤ la quan he. thu. tu.. tren Π(S) va Π(S) la lattice vo.i cacphep toan π1∧π2 va π1∨π2 tu.o.ng u.ng cac quan he. R1 ∩R2 va R1 ∨R2. Phan hoa.ch π1∧π2

de dang tım: gom tat ca’ cac ta.p con khac trong nha.n du.o.. c bang cach giao mo.t ta.p trongπ1 vo.i mo.t ta.p trong π2. Vie.c xac di.nh π1 ∨ π2 kho ho.n.

Vı du. 5.6.3. (a) Xet ho.p du.. ng cac vien bi trong Vı du. 5.6.1, moi ta.p trong phan hoa.chπ1 ∧ π2 gom tat ca’ cac vien bi co cung mau va cung kıch thu.o.c. Phan hoa.ch π1 ∨ π2

phu. thuo.c vao cac vien bi trong S va moi quan he. giu.a chung (xem cac Bai ta.p tu. 1den 4).

(b) Phan hoa.ch π6 ∧ π8 cu’a P tu.o.ng u.ng R6 ∧ R8 trong Vı du. 5.6.2 gom cac lo.p tu.o.ngdu.o.ng du.o.. c xac di.nh theo quan he. m = n (mod 24) (co 24 lo.p).

Trong tru.o.ng ho.. p nay, phan hoa.ch π1∨π2 tu.o.ng u.ng quan he. tu.o.ng du.o.ng R6∨R8 =R2 va do do co hai lo.p tu.o.ng du.o.ng la [0] va [1].

Phan cuoi trınh bay thua.t toan xac di.nh cac phan hoa.ch π1 ∨ π2 va π1 ∧ π2 khi S hu.uha.n. Gia’ su.’ S := {1, 2, . . . , n} va π la phan hoa.ch cu’a S. Vo.i moi A ∈ π, cho.n mo.t phan tu.’

mA ∈ A va di.nh nghıa α(k) := mA, vo.i mo.i k ∈ A (cha’ng ha.n mA la so nho’ nhat cu’a A).Moi phan tu.’ cu’a S thuo.c mo.t ta.p A nao do, bo.’ i va.y ta co ham α : S → S thoa’ man

(i) α(j) = α(k) neu va chı’ neu j, k thuo.c cung mo.t ta.p cu’a phan hoa.ch π;

(ii) α(α(k)) = α(k), vo.i mo.i k.

Vo.i moi k ∈ S, ta.p A ∈ π sao cho k ∈ A thı α(k) ∈ A. Va.y A = α−1(α(k)). Do do π du.o.. cxac di.nh bo.’ i α. Nhu. va.y, de’ xac di.nh π1 ∨ π2 va π1 ∧ π2, ta can tım cac ham tu.o.ng u.ngthoa’ (i) va (ii). Neu R la quan he. tu.o.ng du.o.ng tu.o.ng u.ng vo.i phan hoa.ch π thı tınh chat(i) suy ra

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(i’) α(j) = α(k) neu va chı’ neu jRk.

Vı du. 5.6.4. Gia’ su.’ R la quan he. tu.o.ng du.o.ng tren S := {1, 2, . . . , 10} ma phan hoa.ch πcu’a no la

{{1, 4, 6}, {2}, {3, 7, 10}, {5, 9}, {8}}.Ham α cho.n so nho’ nhat trong moi lo.p la

k 1 2 3 4 5 6 7 8 9 10α(k) 1 2 3 1 5 1 3 8 5 3

Chu y rang α thoa’ (i) va (ii). Cung co the’ cho.n ham

k 1 2 3 4 5 6 7 8 9 10α(k) 4 2 3 4 9 4 3 8 9 3

Vo.i hai phan hoa.ch π1, π2 cu’a ta.p S := {1, 2, . . . , n}, gia’ su.’ α, β la hai ham thoa’ man (i)va (ii). Chu y rang, π1 mi.n ho.n π2 neu

α(i) = α(j) suy ra β(i) = β(j), vo.i mo.i i, j ∈ S.

Bay gio. ta trınh bay thua.t toan tım ham γ tu.o.ng u.ng vo.i π1 ∧ π2. Thua.t toan duye.t moiphan tu.’ cu’a S mo.t lan. Khi ga.p phan tu.’ s trong mo.t khoi mo.i cu’a phan hoa.ch π1 ∧ π2 tagan nhan γ cu’a khoi nay la s.

5.6.1 Thua.t toan xac di.nh ho. i cu’a hai phan hoa.ch

Bu.o.c 1. D- a.t γ(k) = 0, k = 1, 2, . . . , n.

Bu.o.c 2. Cho.n k = 1.

Bu.o.c 3. Neu γ(k) 6= 0 thı chuye’n sang Bu.o.c 4; ngu.o.. c la. i, vo.i moi j = k, k + 1, . . . , n thoa’

α(j) = α(k) va β(j) = β(k), da. t γ(j) = k.

Bu.o.c 4. Neu k = n, du.ng; ngu.o.. c la. i, k := k + 1 va chuye’n sang Bu.o.c 3.

Vı du. 5.6.5. Gia’ su.’ π1 va π2 la cac phan hoa.ch cu’a S := {1, 2, . . . , 8} tu.o.ng u.ng cac hamα va β :

k 1 2 3 4 5 6 7 8α(k) 3 2 3 2 3 7 7 2β(k) 5 4 4 4 5 5 5 4

Ta co π1 = {{1, 3, 5}, {2, 4, 8}, {6, 7}}; π2 gom hai ta.p. Ba’ng 5.1 minh ho.a thua.t toan cha.ytu.ng bu.o.c. Phan hoa.ch π1 ∧ π2 tu.o.ng u.ng hang cuoi trong ba’ng va do do co bon ta.p.

Thua.t toan ke tiep tım ham γ tu.o.ng u.ng vo.i π1 ∨ π2.

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k γ(1) γ(2) γ(3) γ(4) γ(5) γ(6) γ(7) γ(8)0 0 0 0 0 0 0 0 01 1 0 0 0 1 0 0 02 1 2 0 2 1 0 0 23 1 2 3 2 1 0 0 24 1 2 3 2 1 0 0 25 1 2 3 2 1 0 0 26 1 2 3 2 1 6 6 27 1 2 3 2 1 6 6 28 1 2 3 2 1 6 6 2

Ba’ng 5.1:

k γ(1) γ(2) γ(3) γ(4) γ(5) γ(6) γ(7) γ(8)0 1 2 5 4 5 1 7 4 [ham α]1 5 2 5 4 5 5 7 4

2, 3, 4, 5, 6 5 4 5 4 5 5 7 47, 8 5 4 5 4 5 5 5 4

Ba’ng 5.2:

5.6.2 Thua. t toan xac di.nh tuye’n cu’a hai phan hoa.ch

Bu.o.c 1. D- a.t γ(k) = α(k), k = 1, 2, . . . , n.

Bu.o.c 2. Vo.i k = 1, . . . , n neu γ(k) 6= γ(β(k)) thı tım tat ca’ j vo.i γ(j) = γ(k) thay γ(j)bang γ(β(k)) vo.i mo.i j nhu. the.

Vı du. 5.6.6. Gia’ su.’ π1 va π2 la cac phan hoa.ch cu’a S := {1, 2, . . . , 8} tu.o.ng u.ng cac hamα va β :

k 1 2 3 4 5 6 7 8α(k) 1 2 5 4 5 1 7 4β(k) 3 4 3 4 5 6 6 8

Moi phan hoa.ch π1 va π2 co nam ta.p. Ba’ng 5.2 minh ho.a thua.t toan cha.y tu.ng bu.o.c. Phanhoa.ch π1 ∨ π2 tu.o.ng u.ng hang cuoi trong ba’ng va do do co hai ta.p.

Bai ta.p

1. Gia’ su.’ mo. t ho.p du.. ng 10 vien bi, trong do 6 vien nho’ mau xanh, 3 vien lo.n mau do’

va 1 vien lo.n mau xanh. Mo ta’ π1 ∨ π2 va π1 ∧ π2. Co bao nhieu ta.p trong moi phanhoa.ch nay?

2. Cau tra’ lo.i cu’a ba.n nhu. the nao doi vo.i Bai ta.p 1, neu vien bi lo.n mau xanh bienmat?

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3. La.p la. i Bai ta.p 1, neu ho.p bi co 10 vien, trong do 4 vien nho’ mau vang, 3 vien vu.amau xanh, 2 vien vu.a mau trang va 1 vien lo.n mau vang.

4. Cau tra’ lo.i cu’a ba.n nhu. the nao doi vo.i Bai ta.p 3, neu mo.t vien bi lo.n mau xanh ro.ivao ho.p?

5. Kha’o sat cac quan he. tu.o.ng du.o.ng R3, R5 tren P, trong do (m,n) ∈ R3 neu m ≡n (mod 3) va (m,n) ∈ R5 neu m ≡ n (mod 5) tu.o.ng u.ng vo.i cac phan hoa.ch π3, π5.

(a) Mo ta’ quan he. tu.o.ng du.o.ng R3 ∧ R5.

(b) Mo ta’ phan hoa.ch π3 ∧ π5.

(c) Suy ra rang, R3 ∨ R5 la quan he. pho’ du.ng tren P. Kie’m tra la. i

(1, 2), (1, 30), (1, 73), (47, 73), (72, 73) ∈ R3 ∨ R5.

(d) Mo ta’ quan he. phan hoa.ch π3 ∨ π5.

6. Vo.i moi phan hoa.ch du.o.i day cu’a S := {1, 2, . . . , 6} tım ham α thoa’ man cac tınhchat (i) va (ii):

(a) π1 = {{1, 3, 5}, {2, 6}, {4}}.(b) π2 = {{1, 2, 4}, {3, 6}, {5}}.(c) π3 = {{1}, {2}, {3}, {4}, {5}, {6}}.(d) π4 = {{1, 2, 3, 4, 5, 6}}.(e) Quan he. tu.o.ng du.o.ng nao tu.o.ng u.ng vo.i π3, π4?

7. Tım cac phan hoa.ch π1, π2, π3, π4 cu’a {1, 2, . . . , 8} du.o.. c xac di.nh bo.’ i cac ham α1, α2, α3, α4

sauk 1 2 3 4 5 6 7 8α1(k) 1 1 3 1 5 6 3 5α2(k) 2 2 6 8 5 6 7 8α3(k) 4 4 3 4 5 3 3 4α4(k) 3 2 3 8 2 3 7 8

8. Tım cac ham tu.o.ng u.ng vo.i phan hoa.ch π1 ∨ π2 va π1 ∧ π2 trong Bai ta.p 7.

9. Nhu. Bai ta.p 8 cho π3 va π4.

10. Nhu. Bai ta.p 8 cho π2 va π3.

11. Thua.t toan tro.n cac phan hoa.ch van cha.y dung neu hoan do’i vai tro cu’a α va β?

12. (a) Chu.ng minh quan he. ≤ xac di.nh tren Π(S) bo.’ i “π1 ≤ π2 neu va chı’ neu π1 mi.nho.n π2” la thu. tu.. bo. pha.n tren Π(S).

(b) Chu.ng minh rang neu π1, π2, π3 ∈ Π(S) va neu π3 ≤ π1, π3 ≤ π2 thı π3 ≤ π1 ∧ π2.

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13. Phan tıch thua.t toan tım giao va tro.n cac phan hoa.ch trong tru.o.ng ho.. p π1 mi.n ho.nπ2 qua vı du. S := {1, 2, 3, 4, 5, 6, 7} va

k 1 2 3 4 5 6 7α(k) 1 4 3 4 1 6 7β(k) 5 4 5 4 5 4 7

14. Kie’m tra tınh dung dan cu’a thua.t toan giao cac phan hoa.ch bang cach chı’ ra rang

(a) Gia tri. γ(j) thay do’i ıt nhat mo.t lan vo.i moi j trong suot qua trınh thu.. c hie.nthua.t toan;

(b) Neu gia tri. γ(j) thay do’i khi k ≤ k0 va neu α(k′) = α(j) va β(k′) = β(j) thı γ(k′)thay bang k0;

(c) Gia tri. γ(j) thay do’i dung mo.t lan vo.i moi j trong suot qua trınh thu.. c hie.n thua.ttoan;

(d) Neu 0 6= γ(a) = γ(j) thı α(a) = α(j) va β(a) = β(j);

(e) Neu α(a) = α(j) va β(a) = β(j) thı γ(a) = γ(j) vao luc ket thuc thua. t toan.

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Chu.o.ng 6

D- A. I SO BOOLE

D- e’ tu.o.’ ng nho. nha toan ho.c G. Boole, mo.t vai khai nie.m du.o.. c mang ten ong: da. i so Boole,ham Boole, bie’u thu.c Boole va vanh Boole. G. Boole la mo.t trong nhu.ng nha toan ho.cquan tam den vie.c hınh thu.c hoa va co. che hoa tu. duy logic (xem The law of thought cu’aong xuat ba’n nam 1854). G. Boole co nhieu dong gop trong vie.c phat trie’n ly thuyet logicsu.’ du. ng cac ky hie.u thay cho cac tu..

Mo.t the ky’ sau, nhieu nha toan ho.c (da.c bie.t C. E. Shannon) da nha.n ra rang da. i soBoole co the’ su.’ du.ng de’ phan tıch cac ma.nh die.n tu.’ . Do do da. i so Boole tro.’ thanh mo.tcong cu. khong the’ thieu du.o.. c trong vie.c phan tıch va thiet ke cac may tınh die.n tu.’ , cha’ngha.n trong vie.c thiet ke cac ma.ch die.n tu.’ vo.i so linh kie.n ıt nhat.

6.1 Lattice

D- i.nh nghıa 6.1.1. Gia’ su.’ L la ta.p khac rong va ∨,∧ la cac phep toan hai ngoi tren L. Bo.(L,∨,∧) du.o.. c go. i la lattice da. i so neu vo.i mo.i x, y, z ∈ L cac tien de sau tho’a man

1L. Tınh giao hoan.

(a) x ∨ y = y ∨ x;

(b) x ∧ y = y ∧ x.

2L. Tınh ket ho.. p

(a) (x ∨ y) ∨ z = x ∨ (y ∨ z);

(b) (x ∧ y) ∧ z = x ∧ (y ∧ z).

3L. Tınh hap thu. cu’a cac phep toan:

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(a) x ∨ (x ∧ y) = x;

(b) x ∧ (x ∨ y) = x.

• x ∨ y do.c la x tuye’n y hoa.c to’ng cu’a x va y.

• x ∧ y do.c la x ho. i y hoa. c tıch cu’a x va y.

Nha.n xet 7. (a) (1La)-(3La) doi ngau vo.i (1Lb)-(3Lb) theo nghıa neu ta hoan vi. vai trocu’a hai phep toan ∨,∧ trong (1La)-(3La) thı ta se du.o.. c (1Lb)-(3Lb) va ngu.o.. c la. i.

(b) Do tınh ket ho.. p cu’a cac phep toan ∨,∧ ta co the’ viet

x ∨ y ∨ z va x ∧ y ∧ z.

To’ng quat, co the’ viet cho tru.o.ng ho.. p n phan tu.’

x1 ∨ x2 ∨ . . . ∨ xn va x1 ∧ x2 ∧ . . . ∧ xn.

Tınh chat 6.1.2. Neu (L,∨,∧) la lattice da. i so thı

(a) x ∨ x = x.

(b) x ∧ x = x.

(c) x ∨ y = y neu va chı’ neu x ∧ y = x.

Chu.ng minh. (a) D- a.t y := x ∨ x. Khi do

x = x ∨ (x ∧ y) (theo 3La)= x ∨ [x ∧ (x ∨ x)]= x ∨ x (theo 3Lb).

(b) Su.’ du.ng tınh chat doi ngau.

(c) Gia’ su.’ rang x ∨ y = y. Ta co

x = x ∧ (x ∨ y) (theo 3Lb)

= x ∧ y (theo gia’ thiet).

Ngu.o.. c la. i: bai ta.p. 2

D- i.nh ly 6.1.3. Gia’ su.’ (L,≤) la lattice. D- a. t

x ∨ y := sup(x, y),

x ∧ y := inf(x, y).

Khi do (L,∨,∧) la lattice da. i so.

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Chu.ng minh. Ta kie’m tra (L,∨,∧) tho’a man cac tien de cu’a lattice da. i so.

+ (1L). Hie’n nhien.

+ (2La). Lay x, y, z ∈ L. D- a.t {u := (x ∨ y) ∨ z,

v := x ∨ (y ∨ z).

Vıy ≤ x ∨ y ≤ u, z ≤ u.

Suy ra u la mo.t ca.n tren cu’a y, z. Nhu.ng y ∨ z la ca.n tren nho’ nhat cu’a y va z nen

y ∨ z ≤ u.

Ma.t khacx ≤ x ∨ y ≤ u.

Nen u la ca.n tren cu’a x va y ∨ z. Va.y

x ∨ (y ∨ z) ≤ u.

Tu.c la v ≤ u.

Chu.ng minh tu.o.ng tu.. ta cung co u ≤ v. Va.y u = v.

+ (2Lb). Tu.o.ng tu.. nhu. chu.ng minh (2La).

+ (3La). Lay x, y ∈ L. Vı x ≤ x ∨ w vo.i w tuy y. D- a.c bie.t, vo.i w := x ∧ y, ta co

x ≤ x ∨ (x ∧ y).

Vı x ≤ x va x ∧ y ≤ x nen x la ca.n tren cu’a x va x ∧ y. Do do

x ∨ (x ∧ y) ≤ x.

Va.yx ∨ (x ∧ y) = x.

+ (3Lb). Tu.o.ng tu.. nhu. chu.ng minh (3La). 2

D- i.nh ly 6.1.3 chı’ ra rang lattice da. i so (L,∨,∧) ca’m sinh tu. lattice (L,≤). D- i.nh ly saucho chung ta kha’ng di.nh ngu.o.. c la. i.

D- i.nh ly 6.1.4. Gia’ su.’ (L,∨,∧) la mo. t lattice da. i so. Ky hie.u

x ≤ y ⇔ x ∨ y = y,

vo.i mo. i x, y ∈ L. Khi do ≤ la quan he. thu. tu.. tren L va (L,≤) la lattice tho’a

x ∨ y = sup(x, y), x ∧ y = inf(x, y).

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Chu.ng minh. • Ta chu.ng minh ≤ la quan he. thu. tu.. tren L.

+ Tınh pha’n xa. : vı x ∨ x = x nen x ≤ x.

+ Tınh pha’n doi xu.ng: gia’ su.’ x ≤ y va y ≤ x, tu.c la

x ∨ y = y va y ∨ x = x.

Do phep toan ∨ giao hoan, nen x = y.

+ Tınh bac cau: gia’ su.’ x ≤ y va y ≤ z, tu.c la

x ∨ y = y va y ∨ z = z.

Suy rax ∨ z = x ∨ (y ∨ z)

= (x ∨ y) ∨ z

= y ∨ z

= z.

• Chu.ng minh x ∨ y = sup(x, y). Vı

x ∨ (x ∨ y) = (x ∨ x) ∨ y

= x ∨ y.

Nenx ≤ x ∨ y.

Tu.o.ng tu.. , ta coy ≤ x ∨ y.

Va.y x ∨ y la mo.t ca.n tren cu’a x va y.

Gia’ su.’ u la mo.t ca.n tren cu’a x va y. Khi do

x ≤ u va y ≤ u.

Hayx ∨ u = u va y ∨ u = u.

Vı va.y(x ∨ y) ∨ u = x ∨ (y ∨ u)

= x ∨ u

= u.

Suy rax ∨ y ≤ u.

Do dox ∨ y = sup(x, y).

• Chu.ng minh tu.o.ng tu.. cho x ∧ y = inf(x, y). 2

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Nha.n xet 8. (a) Tu. Tınh chat 6.1.2, chung ta co the’ di.nh nghıa

x ≤ y ⇔ x ∧ y = x, vo.i mo.i x, y ∈ L.

(b) Hai D- i.nh ly 6.1.3 va 6.1.4 cho ta moi quan he. giu.a lattice da. i so va lattice. Ho.n nu.a,neu cho lattice (L,≤) thı quan he. thu. tu.. bo. pha.n ca’m sinh bo.’ i lattice da. i so (L,∨,∧) trungvo.i quan he. thu. tu.. ≤ ban dau; ngu.o.. c la. i neu cho lattice da. i so (L,∨,∧) thı cac phep toanhai ngoi ca’m sinh bo.’ i lattice (L,≤) trung vo.i cac phep toan ∨,∧ ban dau.

Vı du. 6.1.1. Cho S la ta.p bat ky. Vo.i cac phep toan ho.. p va giao, (P(S),∪,∩) la mo.tlattice da. i so. Theo D- i.nh ly 6.1.4, vo.i moi ta.p A,B trong P(S), ta di.nh nghıa A ≤ B neuva chı’ neu A ∪ B = B. Thı (P(S),≤) la ta.p du.o.. c sap thu. tu.. bo. pha.n va

{sup(A,B) = A ∪ B,

inf(A,B) = A ∩ B.

Vı du. 6.1.2. Gia’ su.’ L la ta.p ho.. p cac me.nh de. Tren L ta xet quan he. “ ≡ ” du.o.. c di.nhnghıa nhu. sau: p ≡ q neu va chı’ neu “p ⇔ q” la mo.t me.nh de logic. Khi do “ ≡ ” la quanhe. tu.o.ng du.o.ng tren L. Go. i Σ la ta.p ho.. p cac lo.p tu.o.ng du.o.ng tren L xac di.nh bo.’ i “ ≡ ”.Tu.c la

Σ := {[p] | p ∈ L}.D- a.t

[p] ∨ [q] := [p or q], [p] ∧ [q] := [p and q].

Khi do (Σ,∨,∧) la mo.t lattice da. i so. Ky hie.u [p] ≤ [q] neu va chı’ neu [p] ∨ [q] = [q]. Ta co(Σ,≤) la ta.p du.o.. c sap thu. tu.. va

sup([p], [q]) = [p] ∨ [q], inf([p], [q]) = [p] ∧ [q].

Vı du. 6.1.3. Ky hie.u Fun(R,R) la ta.p tat ca’ cac ham so thu.. c xac di.nh tren R. TrenFun(R,R) ta xet quan he. “≤” du.o.. c di.nh nghıa nhu. sau: f ≤ g, f, g ∈ Fun(R,R), neu vachı’ neu f(x) ≤ g(x) vo.i mo.i x ∈ R.

Khi do “≤” la quan he. thu. tu.. tren Fun(R,R). De thay rang sup(f, g) va inf(f, g) ton ta. ivo.i mo.i f, g ∈ Fun(R,R). Vo.i mo. i x ∈ R, da. t

(f ∨ g)(x) := max(f(x), g(x)),

(f ∧ g)(x) := min(f(x), g(x)).

Khi do (Fun(R,R),∨,∧) la lattice da. i so va

f ∨ g = sup(f, g), f ∧ g = inf(f, g).

D- i.nh nghıa 6.1.5. Cho lattice da. i so (L,∨,∧) tu.o.ng u.ng vo.i ta.p du.o.. c sap thu. tu.. (L,≤).

(a) Phan tu.’ trong L, kı hie.u 1, tho’a man

x ≤ 1 vo.i mo.i x ∈ L, (6.1)

127

go. i la phan tu.’ lo.n nhat.

(b) Phan tu.’ trong L, ky hie.u 0, tho’a man

0 ≤ x vo.i mo.i x ∈ L, (6.2)

go. i la phan tu.’ nho’ nhat.

(c) Neu ton ta. i phan tu.’ nho’ nhat, thı cac phan tu.’ phu’ 0 go. i la cac nguyen tu.’ (atom).

(d) Phan tu.’ x trong lattice da. i so du.o.. c go. i la bat kha’ quy (hay toi gia’n) doi vo.i pheptuye’n neu x = y ∨ z thı hoa.c x = y hoa. c x = z.

Hie’n nhien (6.1) tu.o.ng du.o.ng vo.i

x ∨ 1 = 1 va x ∧ 1 = x.

Va (6.2) tu.o.ng du.o.ng vo.i0 ∨ x = x va 0 ∧ x = 0.

Nha.n xet 9. (a) Trong lattice, cac phan tu.’ nho’ nhat va lo.n nhat co the’ ton ta. i hoa. c khongton ta. i. Trong tru.o.ng ho.. p ton ta. i thı chung duy nhat.

(b) Cac nguyen tu.’ la bat kha’ quy. Ho.n nu.a ta co

D- i.nh ly 6.1.6. Trong lattice da. i so hu.u ha. n phan tu.’ , mo. i phan tu.’ co the’ bie’u dien o.’ da. ngtuye’n cac phan tu.’ bat kha’ quy.

Chu.ng minh. Bai ta.p! 2

Vı du. 6.1.4. (a) Lattice da. i so cho trong Vı du. 6.1.1 co

+ ta.p trong la phan tu.’ nho’ nhat;

+ S la phan tu.’ lo.n nhat;

+ {x}, x ∈ S, la cac nguyen tu.’ .

(b) Gia’ su.’ S la ta.p bat ky khac trong. Xet lattice Fun(S,B). Theo Vı du. 6.1.3.

(f ∨ g)(x) = max(f(x), g(x)),

(f ∧ g)(x) = min(f(x), g(x)),

vo.i mo.i x ∈ S. Nen

(f ∨ g)(x) =

{1 neu f(x) = 1 hoa. c g(x) = 1,

0 neu ngu.o.. c la. i,

(f ∧ g)(x) =

{1 neu f(x) = g(x) = 1,

0 neu ngu.o.. c la. i.

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Cac nguyen tu.’ trong Fun(S,B) la cac ham da.c tru.ng 1{x}, x ∈ S.

Vı du. 6.1.5. Cac lu.o.. c do Hasse trong Hınh 6.1 tu.o.ng u.ng cac lattice da. i so.

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Hınh 6.1:

Lu.o.. c do Hasse trong Hınh 6.2 khong tu.o.. ng tru.ng cho mo.t lattice nao vı hai phan tu.’ a, bkhong co ca.n tren nho’ nhat, ma.c du chung co ca.n du.o.i lo.n nhat la e :

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Hınh 6.2:

D- i.nh nghıa 6.1.7. Gia’ su.’ (L,∨,∧) la lattice da. i so. Ta.p con M khac rong cu’a L du.o.. c go. ila lattice con (sublattice) cu’a L neu vo.i mo.i x, y ∈ M, ta co

x ∨ y ∈ M, x ∧ y ∈ M.

Tu.c la M dong doi vo.i cac phep toan tuye’n va ho. i.

Vı du. 6.1.6. Xet lattice da. i so (L,∨,∧) co lu.o.. c do Hasse trong Hınh 6.3(a). Ta co M1

trong Hınh 6.3(b) la lattice con cu’a L; con M2 trong Hınh 6.3(c) khong pha’i la lattice concu’a L.

Bai ta.p

1. Viet du.o.i da.ng doi ngau cac phu.o.ng trınh sau (ma khong pha’i luc nao cung dung):

(a) x ∨ (y ∧ z) = (x ∨ y) ∧ z.

(b) x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z).

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Hınh 6.3:

2. Gia’ su.’ L la lattice da. i so vo.i phan tu.’ lo.n nhat 1 va phan tu.’ nho’ nhat 0.

(a) 1 la bat kha’ quy? Gia’i thıch.

(b) 0 la bat kha’ quy? Gia’i thıch.

3. Xet lattice hu.u ha.n (P,≤) va lu.o.. c do Hasse cu’a no. Gia’i thıch ta. i sao mo. t phan tu.’

la bat kha’ quy neu va chı’ neu no phu’ nhieu nhat mo.t phan tu.’ .

4. Xet lattice trong cac hınh du.o.i:

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(a) Lie.t ke cac nguyen tu.’ cu’a lattice.

(b) Lie.t ke cac phan tu.’ bat kha’ quy.

(c) Viet cac phan tu.’ cu’a lattice du.o.i da.ng tuye’n cu’a cac phan tu.’ bat kha’ quy.

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5. La.p la. i Bai ta.p 4 cho hınh du.o.i:

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6. Ky hie.u P la ta.p cac so nguyen du.o.ng. Xet lattice (P, |), trong do m|n neu m la u.o.cso cu’a n.

(i) Ca.n du.o.i dung cu’a P bang may?

(b) Ton ta. i ca.n tren dung cu’a P?(c) Mo ta’ cac nguyen tu.’ cu’a P.(d) Mo ta’ cac phan tu.’ bat kha’ qui theo phep tuye’n cu’a P.

7. Gia’ su.’ D90 la ta.p tat ca’ cac u.o.c so cu’a 90 bao gom 1 va 90. Chu.ng minh rang D90 lalattice vo.i thu. tu.. |.(a) Ve lu.o.. c do Hasse cu’a lattice nay.

(b) Tınh 6 ∨ 10, 6 ∧ 10, 9 ∨ 30, 9 ∧ 30.

(c) Lie.t ke cac nguyen tu.’ cu’a D90.

(d) Lie.t ke cac phan tu.’ bat kha’ qui cu’a D90.

(e) Viet 90, 18, 5 da.ng tuye’n cu’a cac phan tu.’ bat kha’ qui.

8. Tım tat ca’ cac lattice con cu’a D90 ma co bon phan tu.’ bao gom 1 va 90.

9. Vo.i moi x, y ∈ R, di.nh nghıa x ∨ y := max{x, y} va x ∧ y := min{x, y}.(a) Chu.ng minh rang (R,∨,∧) la lattice da. i so.

(b) Thu. tu.. ca’m sinh bo.’ i lattice nay la gı?

(c) Ta.i sao cac phan tu.’ cu’a R la bat kha’ qui theo phep tuye’n?

10. Hai lattice (L1,∨,∧) va (L2,∪,∩) du.o.. c go. i la da’ng cau neu ton ta. i mo.t tu.o.ng u.ngmo.t-mo.t len ϕ : L1 → L2 sao cho

ϕ(x ∨ y) = ϕ(x) ∪ ϕ(y), ϕ(x ∧ y) = ϕ(x) ∩ ϕ(y)

vo.i mo.i x, y ∈ L1.

(a) Chu.ng minh rang trong tru.o.ng ho.. p nay ϕ(x) ≤ ϕ(y) neu va chı’ neu x ≤ y.

131

(b) Chu.ng minh rang neu (L1,∨,∧) va (L2,∪,∩) da’ng cau thı x la nguyen tu.’ cu’a L1

neu va chı’ neu ϕ(x) la nguyen tu.’ cu’a L2.

(c) Chu.ng minh rang hai lattice trong hınh sau khong da’ng cau:

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(d) Chu.ng minh rang lattice D30 gom cac u.o.c so cu’a 30 (ke’ ca’ 1 va 30) da’ng cau vo.ilattice P(S), trong do |S| = 3. (HD. Su.’ du.ng S = {2, 3, 5}).

11. Gia’ su.’ (L,≤) la lattice. Chu.ng minh rang neu x ≤ y, thı x∨ (z ∧ y) ≤ (x∨ z) ∧ y vo.imo.i z ∈ L.

6.2 Lattice phan bo

D- i.nh nghıa 6.2.1. Lattice da. i so (L,∨,∧) du.o.. c go. i la lattice phan bo (distributive) neucac phep toan ∨,∧ phan phoi doi vo.i nhau, tu.c la vo.i mo.i x, y, z ∈ L ta co

(a) x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z);

(b) x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z).

Vı du. 6.2.1. (a) Lattice da. i so trong Vı du. 6.1.1 la phan bo.

(b) Lattice da. i so trong Vı du. 6.1.2 la lattice phan bo.

(c) R hoa. c N la ta.p du.o.. c sap thu. tu.. doi vo.i quan he. thu. tu.. tuyen tınh thong thu.o.ng.D- o la cac lattice phan bo, vo.i

x ∨ y = max(x, y) va x ∧ y = min(x, y).

Vı du. 6.2.2. Cac lattice da. i so trong Hınh 6.4 khong phan bo.

Cha’ng ha.n, chu.ng minh (a). Ta co

B ∨ (C ∧ D) = B ∨ A = B.

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Hınh 6.4: Cac lattice khong phan bo.

Ma.t khac(B ∨ C) ∧ (B ∨ D) = E ∧ E = E.

Va.yB ∨ (C ∧ D) = B 6= E = (B ∨ C) ∧ (B ∨ D).

Mo.t dieu thu vi. la co the’ chu.ng minh du.o.. c mo. t lattice la phan bo neu va chı’ neu no khongchu.a lattice con giong nhu. cac lattice trong Vı du. 6.2.2. D- i.nh ly sau chı’ ra rang chı’ cankie’m tra mo.t tieu chua’n cu’a lua.t phan bo.

D- i.nh ly 6.2.2. Gia’ su.’ L la lattice da. i so. Hai kha’ng di.nh sau la tu.o.ng du.o.ng

(a) x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z), vo.i mo. i x, y ∈ L.

(b) x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z), vo.i mo. i x, y ∈ L.

Chu.ng minh. (a) ⇒ (b).

(x ∧ y) ∨ (x ∧ z) = [(x ∧ y) ∨ x] ∧ [(x ∧ y) ∨ z] (do (a))= [x ∨ (x ∧ y)] ∧ [z ∨ (x ∧ y)] (tınh giao hoan)= x ∧ [z ∨ (x ∧ y)] (tınh hap thu. )= x ∧ [(z ∨ x) ∧ (z ∨ y)] (do (a))= [x ∧ (z ∨ x)] ∧ (z ∨ y) (tınh ket ho.. p)= [x ∧ (x ∨ z)] ∧ (y ∨ z) (tınh giao hoan)= x ∧ (y ∨ z) (tınh hap thu. )

(b) ⇐ (a). Do nguyen ly doi ngau. 2

D- i.nh ly 6.2.3. Gia’ su.’ (L,∨,∧) la lattice phan bo va x, y, a ∈ L sao cho

x ∨ a = y ∨ a va x ∧ a = y ∧ a.

Thı x = y.

133

Chu.ng minh. Ta cox = x ∨ (x ∧ a)

= x ∨ (y ∧ a)= (x ∨ y) ∧ (x ∨ a)= (y ∨ x) ∧ (y ∨ a)= y ∨ (x ∧ a)= y ∨ (y ∧ a)= y.

2

Tu. day ve sau chung ta luon gia’ su.’ (L,∨,∧) la lattice co phan tu.’ lo.n nhat la 1 va phantu.’ nho’ nhat la 0 (1 khac 0).

D- i.nh nghıa 6.2.4. Hai phan tu.’ x, y ∈ L du.o.. c go. i la bu nhau (complement) neu

x ∨ y = 1 va x ∧ y = 0.

Lattice L du.o.. c go. i la kha’ bu (complemented) neu mo.i phan tu.’ cu’a L deu ton ta. i phan tu.’

bu.

Nha.n xet 10. (a) Mo.t phan tu.’ cu’a lattice L co the’ co hoa. c khong co phan tu.’ bu. Trongtru.o.ng ho.. p ton ta. i, co the’ duy nhat hoa. c khong duy nhat.

(b) Cac phan tu.’ 1 va 0 la bu nhau va la phan tu.’ bu duy nhat cu’a nhau.

Vı du. 6.2.3. (a) Lattice trong Vı du. 6.2.2(a) la kha’ bu. Cac phan tu.’ bu khong nhat thietduy nhat: B la phan tu.’ bu cu’a ca’ C va D.

(b) Lattice L trong Vı du. 6.2.2(b) cung kha’ bu. Cac phan tu.’ bu cung khong duy nhat.Cha’ng ha.n, ca’ A va C deu la bu cu’a B.

Vı du. 6.2.4. Lattice trong Hınh 6.5 la phan bo nhu.ng khong kha’ bu. Vı neu co

x ∨ y = 1 va x ∧ y = 0.

Thı da’ng thu.c dau cho y = 1 con da’ng thu.c sau cho y = 0. Vo ly.

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Hınh 6.5:

D- i.nh ly 6.2.5. Trong lattice phan bo L co phan tu.’ 1 va 0, bu cu’a phan tu.’ x (neu co) laduy nhat.

134

Chu.ng minh. Gia’ su.’ rang

x ∨ y = 1, x ∧ y = 0, x ∨ z = 1, x ∧ z = 0.

Ta coy = y ∨ 0 (vı 0 ≤ y)

= y ∨ (x ∧ z) (vı x ∧ z = 0)= (y ∨ x) ∧ (y ∨ z) (vı tınh phan bo)= 1 ∧ (y ∨ z) (vı y ∨ x = x ∨ y = 1)= (x ∨ z) ∧ (y ∨ z) (vı x ∨ z = 1)= (x ∧ y) ∨ z (vı tınh phan bo)= 0 ∨ z (vı x ∧ y = 0)= z (vı 0 ≤ z).

2

Nha.n xet 11. Neu phan tu.’ trong lattice ton ta. i duy nhat phan tu.’ bu, thı phan tu.’ bu cu’ax du.o.. c ky hie.u la x′.

Bai ta.p

1. Xet lattice da. i so L1 vo.i lu.o.. c do Hasse:

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(a) Lie.t ke cac nguyen tu.’ cu’a L1.

(b) Lie.t ke cac phan tu.’ bat kha’ qui cu’a L1.

(c) Viet 1 du.o.i da.ng tuye’n cu’a cac phan tu.’ bat kha’ qui.

(d) Tım cac phan tu.’ bu, neu ton ta. i, cu’a a, b, d, 0.

135

(e) L1 la lattice kha’ bu? Gia’i thıch.

(f) L1 la lattice phan bo?

2. Xet lattice L2 vo.i lu.o.. c do Hasse trong hınh:

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(a) Tım cac phan tu.’ lo.n nhat va phan tu.’ nho’ nhat cu’a L2.

(b) Tım v ∨ x, s ∨ v va u ∧ v.

(c) L2 la lattice kha’ bu? Gia’i thıch.

(d) Tım phan tu.’ co hai phan tu.’ bu.

(e) L2 la lattice phan bo?

3. (a) Chu.ng minh rang cac phan tu.’ 2 va 6 trong lattice D12 khong co phan tu.’ bu.

(b) Chu.ng minh rang Dm,m ≥ 2, la kha’ bu neu va chı’ neu m la tıch cu’a cac so nguyento phan bie.t, tu.c la neu phan tıch thanh cac thu.a so nguyen to m = pα1

1 .pα22 · · · pαk

k ,thı α1 = α2 = · · · = αk = 1.

4. (a) Ve lu.o.. c do Hasse cu’a lattice (D24, |).(b) Tım cac phan tu.’ bu, neu ton ta. i, cu’a 2, 3, 4, 6.

(c) D24 la lattice kha’ bu? Gia’i thıch.

(d) D24 la lattice phan bo? Gia’i thıch.

5. (a) Ve lu.o.. c do Hasse cu’a lattice (D36, |).(b) D36 la lattice kha’ bu? Gia’i thıch.

(d) D36 la lattice phan bo? Gia’i thıch.

6. Chu.ng minh cac do thi. trong hınh sau la lu.o.. c do Hasse cu’a lattice phan bo. No lakha’ bu?

7. Ky hie.u D70 la ta.p tat ca’ cac u.o.c so cu’a 70 bao gom 1 va 70.

(a) Ve lu.o.. c do Hasse cu’a lattice (D70, |).(b) Tınh 10 ∨ 14, 10 ∧ 14.

136

(c) Lie.t ke cac nguyen tu.’ cu’a D70.

(d) Lie.t ke cac phan tu.’ bat kha’ qui cu’a D70.

(e) Viet 70, 10, 5 da.ng tuye’n cu’a cac phan tu.’ bat kha’ qui.

(f) D70 la lattice kha’ bu? Tım cac phan tu.’ bu cu’a 2, 5.

8. La.p la. i Bai ta.p tren doi vo.i (D36, |).

9. (a) Cac xıch1 nao co ca.n tren va du.o.i?

(b) Cac xıch nao la phan bo?

(c) Cac xıch nao la kha’ bu?

10. Su.’ du.ng D- i.nh ly 6.2.5 chu.ng minh cac lattice trong hınh sau khong phan bo:

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11. Gia’ su.’ L la lattice vo.i phan tu.’ lo.n nhat 1, phan tu.’ nho’ nhat 0. Chu.ng minh rang 0la bu duy nhat cu’a 1 va ngu.o.. c la. i.

12. Chu.ng minh hoa. c cho pha’n vı du. :

(a) Mo.i lattice hu.u ha.n la phan bo.

(b) Mo.i lattice hu.u ha.n co ca.n tren.

13. Gia’ su.’ (L,∧,∨) la lattice phan bo kha’ bu.

(a) Chu.ng minh rang neu x � y thı y′ � x′.

(b) Chu.ng minh rang neu y ∧ z = 0 thı y � z′.

(c) Chu.ng minh rang neu x � y va y ∧ z = 0 thı z � x′.

6.3 D- a. i so Boole

D- i.nh nghıa 6.3.1. D- a. i so Boole (con go. i la lattice Boole) la mo.t lattice phan bo kha’ bu.

1Xıch la ta.p du.o.. c sap thu. tu.. ma hai phan tu.’ bat ky co the’ so sanh du.o.. c vo.i nhau.

137

Nha.n xet 12. (a) D- a. i so Boole la mo.t lattice phan bo co phan tu.’ lo.n nhat 1, phan tu.’ nho’

nhat 0 (1 6= 0), va mo.i phan tu.’ cu’a no luon ton ta. i duy nhat phan tu.’ bu. Cac phep toanhai ngoi

(x, y) 7→ x ∨ y, (x, y) 7→ x ∧ y

va phep toan mo.t ngoix 7→ x′

du.o.. c go. i la cac phep toan Boole.

(b) Ta thu.o.ng ky hie.u (x′)′ = x′′.

(c) Trong da. i so Boole : (x′)′ = x.

Vı du. 6.3.1. Lattice P(S) trong Vı du. 6.1.1 la lattice phan bo, trong do 1 = S, 0 = ∅ vavo.i mo.i A ⊂ S ta co

A ∪ Ac = S, A ∩ Ac = ∅.

Nen P(S) la da. i so Boole.

Vı du. 6.3.2. Lattice Σ trong Vı du. 6.1.2 la lattice phan bo trong do

+ phan tu.’ lo.n nhat la 1 = [True].

+ phan tu.’ phan tu.’ nho’ nhat la 0 = [False].

+ vo.i mo.i me.nh de p,

[p] or [not p] = [True]; [p] and [not p] = [False].

Tu.c lap′ = not p.

Nen la da. i so Boole.

Vı du. 6.3.3. (a) Xet lattice Fun(S,B) trong Vı du. 6.1.4 (b).

+ Fun(S,B) la lattice phan bo vı max,min phan bo vo.i nhau.

+ Co phan tu.’ lo.n nhat la 1 di.nh nghıa bo.’ i 1(x) = 1 vo.i mo.i x ∈ S.

+ Co phan tu.’ nho’ nhat la 0 di.nh nghıa bo.’ i 0(x) = 0 vo.i mo.i x ∈ S.

+ Vo.i mo.i f ∈ Fun(S,B) ta co [f(x)]′ = 1 neu va chı’ neu f(x) = 0 vo.i mo.i x ∈ S. Vı

(f ′ ∨ f)(x) = max([f(x)]′, f(x)) = 1,

(f ′ ∧ f)(x) = min([f(x)]′, f(x)) = 0.

138

Nen Fun(S,B) la da. i so Boole.

(b) Tren Bn := {(x1, x2, . . . , xn) | xi ∈ B, i = 1, 2, . . . , n} xet cac phep toan

x ∨ y := (max(x1, y1),max(x2, y2), . . . ,max(xn, yn)),x ∧ y := (min(x1, y1),min(x2, y2), . . . ,min(xn, yn)).

Khi do Bn la da. i so Boole vo.i phan tu.’ lo.n nhat la 1 = (1, 1, . . . , 1) va phan tu.’ nho’ nhat la0 = (0, 0, . . . , 0).

D- i.nh ly 6.3.2. (Lua.t de Morgan) Neu A la da. i so Boole thı vo.i mo. i x, y ∈ A ta co

(a) (x ∨ y)′ = x′ ∧ y′;

(b) (x ∧ y)′ = x′ ∨ y′.

Chu.ng minh. (a) Ta co

(x ∨ y) ∨ (x′ ∧ y′) = [(x ∨ y) ∨ x′] ∧ [(x ∨ y) ∨ y′] (do tınh phan bo)= [y ∨ (x ∨ x′)] ∧ [x ∨ (y ∨ y′)] (do tınh ket ho.. p va giao hoan)= [y ∨ 1] ∧ [x ∨ 1]= 1 ∧ 1= 1.

Tu.o.ng tu..(x ∨ y) ∧ (x′ ∧ y′) = 0.

Tu. do co (a). (b) Vı

x ∧ y = (x′)′ ∧ (y′)′

= (x′ ∨ y′)′.

Nen(x ∧ y)′ = (x′ ∨ y′)′′

= x′ ∨ y′.

2

D- i.nh ly 6.3.3. Gia’ su.’ A la da. i so Boole hu.u ha. n vo.i ta. p cac nguyen tu.’ S := {a1, a2, . . . , an}.Vo.i moi x ∈ A,x 6= 0, ta co the’ viet du.o.i da. ng tuye’n cac nguyen tu.’ khac nhau nhu. sau

x = ai1 ∨ ai2 ∨ . . . ∨ aik . (6.3)

Ho.n nu.a bie’u thu.c tren la duy nhat khong ke’ thu. tu.. cu’a cac nguyen tu.’ trong bie’u thu.c,va ai1, ai2, . . . , aik la cac nguyen tu.’ ≤ x.

Chu.ng minh. + D- au tien ta chu.ng minh tınh ton ta. i.

139

Neu x = 0 hoa.c x la nguyen tu.’ thı hie’n nhien. Ngu.o.. c la. i, ton ta. i y ∈ A sao cho 0 < y < x.Ta co

x = x ∨ y

= (x ∨ y) ∧ 1

= (x ∨ y) ∧ (y′ ∨ y)

= (x ∧ y′) ∨ y.

Ma.t khac x ∧ y′ < x. Vı neu ngu.o.. c la. i, thı x ∧ y′ = x. Do do

y < x = x ∧ y′ ≤ y′.

Va.y0 < y = y ∧ y′.

Ma khong the’.

Va.y ta phan tıch x da.ng tuye’n cac phan tu.’ nho’ ho.n la x ∧ y′ va y. (Ly lua.n nay chu.ngto’ chı’ co cac nguyen tu.’ va phan tu.’ nho’ nhat 0 la bat kha’ quy). Neu ca’ hai y va x ∧ y′

la nguyen tu.’ , chu.ng minh xong. Ngu.o.. c la. i, bang phu.o.ng phap tren ta phan tıch chung o.’

da.ng tuye’n cac phan tu.’ nho’ ho.n.

Vı A hu.u ha.n, nen cuoi cung qua trınh tren pha’i du.ng va phan tıch x da.ng tuye’n cacnguyen tu.’ .

Chu y rang phu.o.ng phap tren cho chung ta thua. t toan de. quy tım bie’u dien cu’a mo.t phantu.’ qua cac nguyen tu.’ .

+ Ta chu.ng minh rang vo.i mo.i x ∈ A deu tho’a

x = ∨{a ∈ S | a ≤ x}. (6.4)

Ky hie.u ben pha’i chı’ phan tu.’ la tuye’n cac phan tu.’ trong ta.p {a ∈ S | a ≤ x}. Vı co the’

xem phan tu.’ 0 la tuye’n cu’a ta.p trong cu’a cac nguyen tu.’ , nen co the’ gia’ su.’ x 6= 0. Tu. (6.3)ta de dang suy ra

1 = ∨{a ∈ S | a ≤ 1} = a1 ∨ a2 ∨ . . . ∨ an.

Nenx = x ∧ 1

= x ∧ (a1 ∨ a2 ∨ . . . ∨ an)

= (x ∧ a1) ∨ (x ∧ a2) ∨ . . . ∨ (x ∧ an).

Ma.t khac

x ∧ ai =

{ai neu ai ≤ x,


do ai la nguyen tu.’ . Va.y x co bie’u dien da.ng (6.4).

+ Tınh duy nhat. Gia’ su.’ rang

140

x = b1 ∧ b2 ∧ . . . ∧ bm,

trong do bi la cac nguyen tu.’ . Khi do bi ≤ x, i = 1, 2, . . . ,m.

Va.ybi ∈ {a ∈ S | a ≤ x}, i = 1, 2, . . . ,m.

Ma.t khac, neu a ∈ S, a ≤ x, thı

0 6= a = a ∧ x

= a ∧ (b1 ∨ b2 ∨ . . . ∨ bm)

= (a ∧ b1) ∨ (a ∧ b2) ∨ . . . ∨ (a ∧ bm).

Va.y ton ta. i chı’ so i sao choa ∧ bi 6= 0.

Do a va bi la cac nguyen tu.’ , nena ∧ bi = a = bi.

Noi cach khac, a la phan tu.’ bi nao do. D- ieu pha’i chu.ng minh. 2

Ket qua’ sau day se chu.ng to’ da. i so Boole du.o.. c hoan toan xac di.nh bo.’ i so cac nguyen tu.’

cu’a no.

D- i.nh ly 6.3.4. Cho A,B la cac da. i so Boole hu.u ha. n vo.i ta. p cac nguyen tu.’ S :={a1, a2, . . . , an} va T := {b1, b2, . . . , bn} tu.o.ng u.ng. Khi do ton ta. i mo. t da’ng cau da. i soBoole tu. A len B; tu.c la ton ta. i anh xa. mo. t-mo. t len f : A → B sao cho

(a) f(x ∨ y) = f(x) ∨ f(y);

(b) f(x ∧ y) = f(x) ∧ f(y);

(c) f(x′) = [f(x)]′.

Ngoai raf(ai) = bi, i = 1, 2, . . . , n.

Chu.ng minh. Theo D- i.nh ly 6.3.3, mo.i x ∈ A co the’ bie’u dien duy nhat du.o.i da.ng

x = ai1 ∨ ai2 ∨ . . . ∨ aik .

Ta di.nh nghıaf(x) = bi1 ∨ bi2 ∨ . . . ∨ bik .

D- a.c bie.tf(ai) = bi, i = 1, 2, . . . , n.

141

Theo di.nh nghıa cu’a f va do D- i.nh ly 6.3.3, ta co

f(x) = ∨{f(a) | a ∈ S, a ≤ x}

vaf(x) = ∨{b ∈ T | b ≤ f(x)}.

Vı bie’u dien cu’a f(x) la duy nhat, nen vo.i mo.i a ∈ S ta co

a ≤ x neu va chı’ neu f(a) ≤ f(x).

D- e’ chu.ng minh (a), lay x, y ∈ A va chu y rang a ∈ S, ta co

f(a) ≤ f(x ∨ y) ⇔ a ≤ (x ∨ y)

⇔ a ≤ x hoa. c a ≤ y

⇔ f(a) ≤ f(x) hoa. c f(a) ≤ f(y).

Tu.c la, vo.i moi b ∈ T ta co

b ≤ f(x ∨ y) ⇔ b ≤ f(x) hoa. c b ≤ f(y)

⇔ b ≤ f(x) ∨ f(y).

Ap du.ng D- i.nh ly 6.3.3, suy ra

f(x ∨ y) = f(x) ∨ f(y).

Va.y kha’ng di.nh (a) du.o.. c chu.ng minh. Tu.o.ng tu.. ta cung co (b).

Chu.ng minh (c). Ta co

f(x) ∨ f(x′) = f(x ∨ x′) = f(1) = 1,

f(x) ∧ f(x′) = f(x ∧ x′) = f(0) = 0.

Va.y [f(x)]′ = f(x′). 2

Neu S la ta.p co n phan tu.’ thı P(S) la mo.t da. i so Boole (tu.o.ng u.ng vo.i cac phep toanho.. p, giao va lay phan bu) co n nguyen tu.’ , cu. the’ {x}, x ∈ S. Va.y

He. qua’ 6.3.5. Mo. t da. i so Boole hu.u ha. n co n nguyen tu.’ thı da’ng cau da. i so Boole vo.iP(S),#S = n, va vı va. y co dung 2n phan tu.’ .

Bai ta.p

1. (a) Kie’m tra B := {1, 0} vo.i hai phep toan ∨,∧ thong thu.o.ng va 0′ = 1, 1′ = 0, la da. iso Boole.

142

(b) Kie’m tra ta.p Fun(S,B) cac ham tu. S len B vo.i hai phep toan

(f ∨ g)(x) := f(x) ∨ g(x),

(f ∧ g)(x) := f(x) ∧ g(x),

(f ′)(x) := [f(x)]′,

la da. i so Boole.

2. (a) D- a.t S := {a, b, c, d, e}. Viet {a, c, d} nhu. tuye’n cu’a cac nguyen tu.’ trong P(S).

(b) Bie’u dien phan tu.’ (1, 0, 1, 1, 0) da.ng tuye’n cac nguyen tu.’ trong B5.

(c) Gia’ su.’ f ∈ Fun(S,B) sao cho f(a) = f(c) = f(d) = 1, f(b) = f(e) = 0. Bie’u dienf da.ng tuye’n cac nguyen tu.’ trong Fun(S,B).

3. Tren ta.p D6 := {1, 2, 3, 6} xet cac phep toan:

x + y := BSCNN(x, y), x.y := USCLN(x, y), x′ :=6

x.

Chu.ng minh rang (D6,+, ·,′ ) la da. i so Boole. Tım cac phan tu.’ nho’ nhat va phan tu.’

lo.n nhat.

4. Tren ta.p D8 := {1, 2, 4, 8} xet cac phep toan + va · nhu. trong Bai ta.p 3 va x′ = 8/x.Chu.ng minh (D8,+, ·,′ ) khong pha’i da. i so Boole.

5. Lattice (D30, |) la da. i so Boole.

(a) Ve lu.o.. c do Hasse cu’a lattice nay.

(b) Lie.t ke cac nguyen tu.’ cu’a D30.

(c) Tım tat ca’ cac da. i so Boole con cu’a D30. Chu y rang, cac da. i so con can chu.a 1va 30.

(d) Tım lattice con co bon phan tu.’ nhu.ng khong pha’i la da. i so Boole con.

6. Lattice (D210, |) la da. i so Boole. Tım ta.p S sao cho P(S) va D210 la da’ng cau da. i soBoole va tım da’ng cau nay.

7. Vo.i nhu.ng gia tri. m nao thı lattice (Dm, |) la da. i so Boole?

8. Tren ta.p Sn := {1, 2, . . . , n} xet cac phep toan:

x + y := max(x, y), x.y := min(x, y).

(a) Chu.ng minh tren Sn cac phep toan nay tho’a man cac tınh chat giao hoan, ket ho.. pva hap thu. .

(b) Chu.ng minh co the’ di.nh nghıa phan tu.’ nho’ nhat 0, phan tu.’ lo.n nhat 1 va pheptoan phu’ di.nh ′ sao cho Sn vo.i cac phep toan nay la da. i so Boole neu va chı’ neu n = 2.

9. Gia’ su.’ (A,∨,∧) la da. i so Boole va S la ta.p con cu’a A. Chu.ng minh S vo.i cac pheptoan ∨,∧ ca’m sinh la da. i so Boole neu va chı’ neu 1 ∈ S va x∧y′ ∈ S vo.i mo.i x, y ∈ S.

143

10. (a) Chu.ng minh trong da. i so Boole, [x(x′ + y))]′ = x′ + y′ vo.i mo.i x, y.

(b) Viet doi ngau va chu.ng minh bie’u thu.c tren.

11. Gia’ su.’ P la ta.p cac so nguyen du.o.ng va S la ho. cac ta.p con hu.u ha.n cu’a P. Gia’i thıchta. i sao S vo.i cac phep ho.. p, giao va lay phan bu khong la da. i so Boole.

12. Tım ta.p S sao cho P(S) va B5 la da’ng cau da. i so Boole va tım da’ng cau nay.

13. Mo ta’ cac nguyen tu.’ cu’a Fun(S,B), S := N. D- ieu nay con dung neu S := R?

14. (a) Ton ta. i da. i so Boole vo.i 6 phan tu.’ ? Gia’i thıch.

(b) Mo.i da. i so Boole hu.u ha.n phan tu.’ da’ng cau vo.i da. i so Boole Jn cu’a cac hamBoole? Gia’i thıch.

15. (a) Mo ta’ cac nguyen tu.’ cu’a lattice P(N).

(b) Moi phan tu.’ cu’a lattice la tuye’n cu’a cac nguyen tu.’ ? Tha’o lua.n.

16. Gia’ su.’ x, y la cac phan tu.’ cu’a da. i so Boole, va a la mo.t nguyen tu.’ .

(a) Chu.ng minh rang a ≤ x ∨ y neu va chı’ neu a ≤ x hoa. c a ≤ y.

(b) Chu.ng minh rang a ≤ x ∧ y neu va chı’ neu a ≤ x va a ≤ y.

(c) Chu.ng minh rang hoa. c a ≤ x hoa. c a ≤ x′ va khong dong tho.i ca’ hai.

17. Gia’ su.’ x, y la cac phan tu.’ cu’a da. i so Boole hu.u ha.n ma du.o.. c viet du.o.i da.ng tuye’ncac nguyen tu.’

x = a1 ∨ a2 ∨ · · · ∨ an, va y = b1 ∨ b2 ∨ · · · ∨ bm.

18. (a) Gia’i thıch cach viet x∨ y va x∧ y da.ng tuye’n cac nguyen tu.’ phan bie.t. Minh ho.abang vı du. .

(b) Viet s′ da.ng tuye’n cac nguyen tu.’ phan bie.t.

19. Chu.ng minh rang neu Φ la da’ng cau da. i so Boole giu.a cac da. i so Boole A va B thıx ≤ y neu va chı’ neu Φ(x) ≤ Φ(y).

20. Gia’ su.’ S := [0, 1] va A gom ta.p trong va tat ca’ cac ta.p con cu’a S sao cho co the’ vieto.’ da.ng ho.. p hu.u ha.n cac khoa’ng co da.ng [a, b).

(a) Chu.ng minh rang moi phan tu.’ cu’a A co the’ viet nhu. ho.. p hu.u ha.n cu’a cac khoa’ngro.i nhau da.ng [a, b).

(b) Chu.ng minh A la da. i so Boole tu.o.ng u.ng vo.i cac phep toan giao (∩), ho.. p (∪) valay phan bu.

(c) Chu.ng minh A khong co nguyen tu.’ .

21. Gia’ su.’ a → a, a → a la hai phep toan lay phan bu tu.o.ng u.ng vo.i da. i so Boole(A,∨,∧). Chu.ng minh rang a = a, vo.i mo.i a ∈ A.

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6.4 Ham Boole

Phan nay chung ta se di.nh nghıa mo.t cach to’ng quat ve “ham Boole”, dong tho.i mo ta’ cacda.ng “chınh quy” cu’a chung. Nghien cu.u ham Boole tu.c la nghien cu.u cac anh xa. Boole tu.

mo.t da. i so Boole vao chınh ba’n than no. Moi phan tu.’ cu’a da. i so Boole go. i la “hang so”.Moi mo.t ky hie.u bie’u dien mo.t trong cac phan tu.’ cu’a da. i so Boole go. i la “bien Boole”.

D- i.nh nghıa 6.4.1. Anh xa.

f : Bn −→ B, (x1, x2, . . . , xn) 7→ f(x1, x2, . . . , xn),

du.o.. c go. i la ham Boole n bien neu no du.o.. c cau ta.o theo nguyen tac sau day

(a) Ham hang f(x) = a, a ∈ B, va phep chieu len thanh phan thu. i : f(x) = xi la hamBoole.

(b) Neu f la ham Boole thı ham phu’ di.nh f ′ cung la ham Boole.

(c) Neu f va g la cac ham Boole thı f ∨ g va f ∧ g cung la ham Boole.

(d) Mo.i ham so du.o.. c cau ta.o bang cach ap du.ng mo.t so hu.u ha.n lan cac quy lua.t ke’ trendeu la ham Boole.

Nha.n xet 13. Theo di.nh nghıa tren thı ham Boole la mo.t ham so du.o.. c cau ta.o tu. cachang so va cac phep chieu bang cach u.ng du.ng mo.t so hu.u ha.n lan cac phep toan ho. i, tuye’nva phu’ di.nh.

Vı du. 6.4.1. (a) Cac ham du.o.i day la cac ham Boole theo ba bien x, y, z :

(x ∨ y) ∧ (x′ ∨ z) ∧ y, y′ ∨ (x ∨ z′), x ∨ y, z.

(b) Ham Boole n bien

(x1 ∧ x2 ∧ . . . ∧ xn) ∨ (x′1 ∧ x2 ∧ . . . ∧ xn) ∨ (x1 ∧ x′

2 ∧ . . . ∧ xn).

D- e’ gia’n tie.n, ta su.’ du. ng cac ky hie.u + (co.ng) va . (nhan) thay cho ∨ va ∧.

Mo.t trong nhu.ng cach thua.n tie.n nhat de’ mo ta’ ham Boole la cho tu.o.ng u.ng mo.t-mo.tvo.i ba’ng chan tri. (hay ba’ng gia tri. tha. t), tu.c la ba’ng gia tri. cu’a ham so u.ng vo.i nhu.ng to’

ho.. p gia tri. khac nhau cu’a cac bien.

Vı du. 6.4.2. Ba’ng chan tri. cu’a ham

f(x, y, z) = y′ ∧ (x ∨ z)

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lax y z y′ x ∨ z f0 0 0 1 0 00 0 1 1 1 10 1 0 0 0 00 1 1 0 1 01 0 0 1 1 11 0 1 1 1 11 1 0 0 1 01 1 1 0 1 0

Nha.n xet 14. Moi ham Boole co duy nhat mo.t ba’ng chan tri.. Ngu.o.. c la. i, ta luon luon cothe’ xay du.. ng du.o.. c vo so ham Boole n bien co ba’ng chan tri. gom 2n hang cho truo.c.

Vı du. 6.4.3. Xet ba’ng chan tri.

x y z f0 0 0 1 x0 0 1 00 1 0 00 1 1 1 x1 0 0 01 0 1 1 x1 1 0 01 1 1 1 x

D- e’ tım ham Boole f(x, y, z) co ba’ng chan tri. tren, chung ta tien hanh theo cac bu.o.c sau

+ D- au tien, danh dau moi hang ma co co.t cuoi bang 1.

+ Vo.i moi hang du.o.. c danh dau, ta da. t tu.o.ng u.ng mo.t so ha.ng da.ng:

e1 ∧ e2 ∧ e3,

trong do e1 = x neu phan tu.’ trong co.t dau cu’a hang nay bang mo.t va e1 = x′ neu ngu.o.. cla. i. Tu.o.ng tu.. e2 = y neu phan tu.’ trong co.t thu. hai cu’a hang nay bang 1 va e2 = y′ neungu.o.. c la. i. Cuoi cung e3 = z neu phan tu.’ trong co.t thu. ba cu’a hang nay bang 1 va e3 = z′

neu ngu.o.. c la. i.

Do do cac phan tu.’ tu.o.ng u.ng vo.i bon hang du.o.. c danh dau la

x ∧ y ∧ z, x ∧ y′ ∧ z, x′ ∧ y ∧ z, x′ ∧ y′ ∧ z′.

+ Cuoi cung, ta tuye’n cac bie’u thu.c nay de’ co ham

f(x, y, z) = (x ∧ y ∧ z) ∨ (x ∧ y′ ∧ z) ∨ (x′ ∧ y ∧ z) ∨ (x′ ∧ y′ ∧ z′).

Neu co.t cuoi cu’a ba’ng chan tri. gom toan so 0, thı phu.o.ng phap tren khong lam vie.c; tuynhien, ham Boole f ≡ 0 la ham co ba’ng chan tri. nhu. va.y.

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D- i.nh nghıa 6.4.2. Hai ham Boole du.o.. c go. i la tu.o.ng du.o.ng vo.i nhau neu chung co cungmo.t ba’ng chan tri..

Vı du. 6.4.4. Cac bie’u thu.c x(y ∨ z) va xy ∨ xz la tu.o.ng du.o.ng.

D- i.nh ly sau cho chung ta so cac phan tu.’ cu’a ta.p tat ca’ cac ham Boole n bien: Fun(Bn,B) :={f : Bn → B}.

D- i.nh ly 6.4.3. Co 22nanh xa. tu. Bn vao B.

Chu.ng minh. Ro rang #Bn = 2n. Moi ham tu. Bn vao B co the’ lay mo.t trong hai gia tri.do.c la.p la 0 va 1. Do va.y ta co 22n

to’ ho.. p kha’ nang khac nhau; nghıa la co 22nanh xa. khac

nhau. 2

Vı du. 6.4.5. (a) Tru.o.ng ho.. p n = 1 ta co bon ham Boole:

f1 = 0, f2 = x, f3 = x′, f4 = 1.

(b) Tru.o.ng ho.. p n = 2 ta co 16 ham so Boole du.o.. c lie.t ke trong ba’ng sau

STT f Ten go. i1 0 Ham hang 02 x1x2 Ham AND3 x1x

′2 Ham keo theo khong dieu kie.n

4 x1 Phep chieu len bien thu. nhat5 x′

1x2 Ham keo theo khong da’o6 x2 Phep chieu len bien thu. hai7 x1x

′2 + x′

1x2 Ham co.ng modulo 28 x1 + x2 Ham OR9 x′

1x′2 Ham NOR

10 x1x2 + x′1x

′2 Ham tu.o.ng du.o.ng

11 x′2 Ham phu’ di.nh x2

12 x1 + x′2 Ham keo theo da’o

13 x′1 Ham phu’ di.nh x1

14 x′1 + x2 Ham keo theo co dieu kie.n

15 x′1 + x′

2 Ham NAND (Sheffer)16 1 Ham hang 1

He. qua’ 6.4.4. Fun(Bn,B) vo.i cac phep toan +, .,− la mo. t da. i so Boole da’ng cau vo.i B2n.

Bai ta.p

1. Chu.ng minh cac bie’u thu.c du.o.i day la cac ham Boole va tım gia tri. cu’a cac ham naykhi x = 1, y = 1, z = 0 :

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(a) (x ∧ y) ∨ (y′ ∧ z).

(b) (x ∧ y)′.

(c) x ∨ (y′ ∧ z).

(d) (x ∧ y′) ∨ (y ∧ z′).

(e) (x ∧ (y ∨ (x ∧ y′))) ∨ ((x ∧ y′) ∨ (x ∧ z′)′).

2. Cac bie’u thu.c nao la ham Boole:

(a) x ∧ (y ∧ z).

(b) x ∧ (y′ ∧ z).

(c) (x).

(d) (x ∧ y) ∨ z′.

(e) ((x)).

3. Tım ham Boole f : B3 → B neu f(0, 0, 0) = f(0, 0, 1) = f(1, 1, 0) = 1 va f(a, b, c) = 0vo.i tat ca’ (a, b, c) ∈ B3 khac.

4. Kie’m tra cac da’ng thu.c sau:

(a) x ∨ x = x.

(b) x ∨ (x ∧ y) = x.

(c) x ∧ y′ = (x′ ∨ y)′.

(d) x ∧ (y ∧ z)′ = (x ∧ y′) ∨ (x ∧ z′).

(e) x′ ∧ ((y ∧ z) ∨ (x ∧ y ∧ z)) = x ∧ z.

5. D- ung hay sai:

(a) (x ∧ y) ∨ (x′ ∧ z) ∨ (x′ ∧ y ∧ z′) = y ∨ (x′ ∧ z).

(b) (x ∧ y ∧ z) ∨ (x ∧ z)′ = (x ∧ z) ∨ (x′ ∧ z′).

6. Chu.ng minh neu f1 va f2 la cac ham Boole theo cac bien x1, x2, . . . , xn thı f1 ∨ f2

tu.o.ng du.o.ng vo.i f2 ∨ f1.

7. Cac ham Boole nhu. x hay y′ gom mo.t bien do.n hoa.c phan bu cu’a no du.o.. c go. i laliteral.

(a) Chu.ng minh x′z ∨ y′z khong tu.o.ng du.o.ng vo.i tıch cac literal.

(b) Chu.ng minh x′z ∨ y′z khong tu.o.ng du.o.ng vo.i tuye’n cu’a cac tıch cu’a cac literalma trong do mo.t tıch la mo.t literal do.n. (Phan (a) va (b) chı’ ra rang x′z ∨ y′z la toiu.u).

(c) Nhom ba so ha.ng xyz ∨ xyz′ ∨ xy′z da.ng cac ca.p de’ nha.n du.o.. c mo.t bie’u thu.ctu.o.ng du.o.ng da.ng tuye’n cu’a hai tıch ma moi tıch gom hai literal.

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6.5 Bie’u dien cac ham Boole qua he. tuye’n, ho. i va phu’

di.nh

Nhu. chung ta da biet, mo.t trong nhu.ng cach cho ham Boole la dung ba’ng chan tri.. Moiba’ng chan tri. co the’ bie’u dien nhieu ham so khac nhau, nhu.ng cac ham so nay pha’i tu.o.ngdu.o.ng vo.i nhau. Noi mo.t cach khac co the’ dung ba’ng chan tri. de’ kie’m tra cac ham Booleco tu.o.ng du.o.ng vo.i nhau hay khong?

Ngoai ra, de’ so sanh cac ham Boole vo.i nhau ngu.o.i ta du.a ra da.ng chınh quy (hay da.ngchua’n). Hai cach bie’u dien khac nhau cu’a ham Boole co cung mo.t da.ng chınh quy neu vachı’ neu chung tu.o.ng du.o.ng vo.i nhau. Noi cach khac, da.ng chınh quy cu’a mo.t cach bie’udien ham Boole la duy nhat. Co hai da.ng chınh quy thu.o.ng dung, do la da.ng tuye’n chınhquy (hay da.ng to’ng cu’a cac tıch) va da.ng ho. i chınh quy (hay da.ng tıch cu’a cac to’ng).

D- e’ tie.n trınh bay, ta du.a vao quy u.o.c sau. Gia’ su.’ x la mo.t bien va e ∈ B. Ky hie.u

xe :=

{x neu e = 1,

x′ neu ngu.o.. c la. i.

Tu. di.nh nghıa ta co

xe = 1 neu va chı’ neu x = e.

D- i.nh nghıa 6.5.1. Gia’ su.’ f la ham Boole n bien. Ta.p

Tf := {x = (x1, x2, . . . , xn) ∈ Bn | f(x) = 1}

du.o.. c go. i la ta. p da. c tru.ng cu’a f.

Tınh chat 6.5.2. (a) Tf ′ = [Tf ]′ = {x = (x1, x2, . . . , xn) ∈ Bn | f(x′) = 1}.

(b) Tf+g = Tf ∪ Tg.

(c) Tfg = Tf ∩ Tg.

Chu.ng minh. Hie’n nhien theo di.nh nghıa. 2

Ho.n nu.a co mo.t tu.o.ng u.ng mo.t-mo.t giu.a cac ham Boole va ta.p da.c tru.ng cu’a no. Cactınh chat nay cho phep chuye’n chu.ng minh tren da. i so logic sang cac chu.ng minh tu.o.ngu.ng tren da. i so ta.p ho.. p.

D- i.nh ly 6.5.3. Co di.nh i ∈ {1, 2, . . . , n}. Khi do mo. i ham Boole n bien f deu co the’ bie’udien du.o.i da. ng tuye’n chınh quy

f(x) =∑

f(e1, e2, . . . , ei, xi+1, xi+2, . . . , xn)xe11 ∧ xe2

2 ∧ . . . ∧ xeii , (6.5)

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hoa. c du.o.i da. ng ho. i chınh quy

f(x) =∏

f(e1, e2, . . . , ei, xi+1, xi+2, . . . , xn)xe11 ∨ xe2

2 ∨ . . . ∨ xeii , (6.6)

trong do tuye’n, ho. i lay tren ta. p (e1, e2, . . . , ei) ∈ Bi.

Chu.ng minh. Bang lua.t doi ngau, ta chı’ can chu.ng minh bie’u dien da.ng (6.5). Gia’ su.’

(x1, x2, . . . , xn) ∈ Tf . Khi do so ha.ng u.ng vo.i bo. gia tri. e1 = x1, e2 = x2, . . . , ei = xi trongtuye’n ve pha’i cu’a (6.5)

xe11 xe2

2 . . . xeii f(e1, e2, . . . , ei, xi+1, xi+2, . . . , xn)

se bang 1. D- ieu nay keo theo toan bo. ve pha’i bang 1.

Ngu.o.. c la. i, neu ve pha’i bang 1 thı pha’i xa’y ra ta. i so ha.ng nao do, cha’ng ha.n ta. i so ha.ngtu.o.ng u.ng vo.i bo. gia tri. (e1, e2, . . . , ei) va do do (x1, x2, . . . , xn) ∈ Tf . 2

Cho i = 1 trong di.nh ly va nha.n xet rang vai tro cu’a cac bien xi la nhu. nhau, ta du.o.. c

He. qua’ 6.5.4. Ham Boole f co the’ du.o.. c khai trie’n theo mo. t doi so xi

f(x) = x′if(x1, . . . , xi−1, 0, xi+1, . . . , xn) ∨ xif(x1, . . . , xi−1, 1, xi+1, . . . , xn), (6.7)

hoa. c

f(x) = x′if(x1, . . . , xi−1, 0, xi+1, . . . , xn) ∧ xif(x1, . . . , xi−1, 1, xi+1, . . . , xn). (6.8)

Cho i = n trong di.nh ly va bo’ di cac phan tu.’ bang 1 trong mo.t tıch, ta du.o.. c

He. qua’ 6.5.5. Mo. i ham Boole co the’ du.o.. c khai trie’n du.o.i da. ng tuye’n chınh quy

f(x) =∑

e∈Tf

xe11 xe2

2 . . . xenn (6.9)

hoa. c du.o.i da. ng ho. i chınh quy

f(x) =∏

e∈Tf

xe11 ∨ xe2

2 ∨ . . . ∨ xenn (6.10)

Cong thu.c khai trie’n (6.9) con du.o.. c go. i la da.ng tuye’n chua’n tac hoan toan cu’a f va moiso ha.ng cu’a no du.o.. c go. i la mo.t cau ta.o do.n vi. (hay phan tu.’ toi thie’u) cu’a f.

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Vı du. 6.5.1. Da.ng tuye’n chınh quy va da.ng ho. i chınh quy cu’a ham Boole co ba’ng chan tri.

x1 x2 x3 f(x1, x2, x3)0 0 0 10 0 1 00 1 0 10 1 1 11 0 0 01 0 1 11 1 0 01 1 1 1

tu.o.ng u.ng la

fΣ = x′1x

′2x

′3 + x′

1x2x′3 + x′

1x2x3 + x1x′2x3 + x1x2x3,

fΠ = (x1 + x2 + x′3)(x

′1 + x2 + x3)(x

′1 + x′

2 + x3).

Nhu. va.y da.ng chınh quy khong nhu.ng giup chung ta so sanh cac ham so ma con giupchung ta trong vie.c bie’u dien ham Boole du.o.i da.ng bie’u thu.c da. i so tu. ba’ng chan tri. vatrong vie.c do.n gia’n hoa toi thie’u cac ham Boole. Tu. He. qua’ 6.5.5, ta nha.n du.o.. c

He. qua’ 6.5.6. Mo. i ham Boole deu co the’ xay du.. ng tu. cac bien nho. cac ham OR, AND,va NOT.

Ngoai he. tuye’n, ho. i va phu’ di.nh, ton ta. i nhieu he. khac cung co tınh chat mo. i ham Booledeu bie’u dien qua cac thanh vien cu’a he. . Mo.t he. ham nhu. va.y du.o.. c go. i la he. day du’.

He. qua’ 6.5.7. Cac he.

(a) {AND, NOT}; va

(b) {OR, NOT} la nhu.ng he. ham day du’ hai bien.

Chu.ng minh. (a) Tha. t va.y, do

x ∨ y = (x′)′ ∨ (y′)′

= (x′y′)′

nen ham OR du.o.. c thay bang hai ham AND va NOT. Ket lua.n du.o.. c suy tu. He. qua’ 6.5.6.

(b) Bai ta.p. 2

Vie.c nghien cu.u tınh day du’ cu’a mo.t he. ham co y nghıa thu.. c tien quan tro.ng, no tra’ lo.icau ho’i co the’ xay du.. ng mo.t ham Boole tu. mo.t so ham do.n gia’n cho.n tru.o.c hay khong?

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Bai ta.p

1. Chu.ng minh cac khai trie’n trong He. qua’ 3.5.5 la duy nhat.

2. Tım da.ng tuye’n chınh quy cu’a ham Boole ba bien:

(a) xy.(b) z′.(c) xz ∨ (y′ ∨ y′z) ∨ xy′z′.(d) x ∨ yz.

(e) [(xy ∨ xyz) ∨ xz] ∨ z.(f) xy ∨ z′.(g) [(x ∨ y)′ ∨ z]′.(h) (x ∨ y)′ ∨ z ∨ x(yz ∨ y′z′).

3. Trınh bay phu.o.ng phap tım da.ng ho. i chınh quy. Cho vı du. minh ho.a.

4. Su.’ du.ng cac phu.o.ng phap da. i so, tım da.ng tuye’n chınh quy cu’a cac ham Boole sau:

(a) x ∨ xy.

(b) (x ∨ y)(x′ ∨ y′).

(c) (yz ∨ xz′)(xy′ ∨ z)′.

(d) (x′y ∨ x′z′)(x ∨ yz)′.

(e) x ∨ (y′ ∨ (xy′ ∨ xz′)).

5. Chu.ng minh neu m1∨m2∨· · ·∨mk la da.ng tuye’n chınh quy cu’a f thı m′1∧m′

2∧· · ·∧m′k

la da.ng ho. i chınh quy cu’a f ′. Cho vı du. minh ho.a.

6. Chu.ng minh cac he. ham sau la day du’: {OR, NOT}, {NOR}, va {NAND}. (HamNAND va NOR con ky hie.u tu.o.ng u.ng la ↑ va ↓).

7. Chu.ng minh cac he. ham sau khong day du’: {AND}, {OR }, {NOT}, va {AND, OR}.

8. Chu.ng minh hoa. c tım pha’n vı du. : x ↑ (y ↑ z) = (x ↑ y) ↑ z vo.i mo.i x, y, z ∈ B.

9. Bie’u dien ham XOR qua he. ham NAND.

6.6 Bie’u dien toi thie’u cu’a ham Boole

6.6.1 Khai nie.m

Bie’u dien ham Boole qua mo.t he. ham day du’ H la khong duy nhat. Vı du. ham Shefferdi.nh nghıa bo.’ i

x ↑ y :=

{0 neu x = y = 1,


khi bie’u dien qua he. tuye’n, ho. i va phu’ di.nh, co the’ co cac cach

x ↑ y = x′y′ ∨ x′y ∨ xy′ = x′ ∨ y′.

152

Moi mo.t bie’u dien f tu.o.ng u.ng vo.i mo.t cach “ghep” cac thanh vien cu’a H (ma ta go. i lacac yeu to co. ba’n) de’ thu du.o.. c f. Hie’n nhien, mo.t van de co y nghıa thu.. c tien quan tro.ngla tım mo.t bie’u dien sao cho vie.c ghep nhu. the ton ıt yeu to co. ba’n nhat. Theo mo.t nghıanao do, dieu nay dan ve vie.c tım mo.t cong thu.c tren he. H bie’u dien ham f vo.i so ky hie.ucac yeu to nay la ıt nhat. Mo.t cong thu.c nhu. va.y, du.o.. c go. i la mo.t bie’u dien toi thie’u cu’aham f trong he. H.

Ve nguyen tac, so cong thu.c bie’u dien f la hu.u ha.n, nen bang cach duye.t tat ca’ cac kha’

nang, ta luon tım du.o.. c bie’u dien toi thie’u cu’a f. Tuy nhien, so kha’ nang nay la rat lo.n vavie.c duye.t no doi ho’i mo.t khoi lu.o.. ng tınh toan kho’ng lo, do do tren thu.. c te kho ma thu.. chie.n du.o.. c du rang ngay ca’ vo.i nhu.ng sieu may tınh. Vie.c xay du.. ng nhu.ng thua.t toan hu.uhie.u tım bie’u dien toi thie’u cu’a cac ham Boole, vı the cang tro.’ nen cap bach. Nhu.ng dongtho.i no cung la bai toan rat kho. Cho den nay van chu.a du.o.. c gia’i quyet tho’a dang ngayca’ trong mo.t so tru.o.ng ho.. p do.n gia’n va con dang du.o.. c tiep tu. c nghien cu.u.

Mo.t he. day du’ du.o.. c nghien cu.u nhieu nhat la he. tuye’n, ho. i va phu’ di.nh. Bai toan tımbie’u dien toi thie’u cu’a cac ham Boole trong he. nay da du.o.. c nghien cu.u trong vai chu.c namgan day. Nhu. da biet, mo.t ham Boole noi chung co the’ bie’u dien theo nhieu bie’u thu.cBoole khac nhau, vo.i do. phu.c ta.p nhieu ıt cung khac nhau. Thu.. c chat cu’a van de toi thie’uhoa la tım da.ng bie’u dien do.n gia’n nhat cho mo.t bie’u thu.c Boole. Nhu. va.y bai toan toithie’u cac bie’u thu.c Boole tro.’ thanh bai toan so sanh mu.c do. phu.c ta.p cu’a cac bie’u thu.ctu.o.ng du.o.ng.

Noi chung, co hai nhom phu.o.ng phap de’ toi thie’u hoa cac bie’u thu.c Boole. Nhom thu.

nhat bao gom cac phu.o.ng phap bien do’i da. i so cac bie’u thu.c Boole du.. a tren co. so.’ cac da’ngthu.c da gio.i thie.u trong phan cac tınh chat cu’a da. i so Boole. Cac phu.o.ng phap nay khongtie.n lo.. i, doi ho’i nhieu tho.i gian, da.c bie.t trong tru.o.ng ho.. p co nhieu bien. Nhom thu. haibao gom cac phu.o.ng phap thua.t toan, cac phu.o.ng phap nay cho phep de dang tu.. do.ng hoabie’u thu.c Boole.

6.6.2 Phu.o.ng phap ba’n do Karnaugh

Nhu. da biet, thong qua Vı du. 6.4.3, chung ta co the’ xay du.. ng du.o.. c mo.t ham Boole da.ngtuye’n chınh quy tu.o.ng u.ng ba’ng chan tri. do. Kho khan chınh la tım mo.t ham Boole dacho co da.ng toi thie’u. Du.o.i day chung ta se du.a ra phu.o.ng phap ba’n do Karnaugh de’ gia’iquyet kho khan nay. Phu.o.ng phap nay chı’ hu.u ıch vo.i so bien ıt, va chung ta se ha.n checho cac tru.o.ng ho.. p hai va ba bien.

Ba’n do Karnaugh hai bien

Ba’n do Karnaugh hai bien la mo.t hınh vuong du.o.. c chia thanh bon hınh vuong nho’ ho.nnhu. trong Hınh 6.6.

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x′ ∧ y′ x′ ∧ y

x ∧ y′ x ∧ y

x′

x

y′ y

Hınh 6.6: Ba’n do Karnaugh hai bien

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Hınh 6.7: Ket ho.. p cac hınh vuong ke nhau

Nha.n xet la moi hınh vuong con tu.o.ng u.ng mo.t-mo.t vo.i mo.t phan tu.’ toi thie’u va codung bon phan tu.’ toi thie’u trong tru.o.ng ho.. p hai bien.

Ta noi rang hai hınh vuong con la ke nhau neu chung co chung mo.t ca.nh. Vı mo.t hınhvuong con tu.o.ng u.ng mo.t phan tu.’ toi thie’u (la bie’u thu.c Boole hai bien) nen cac hınhvuong con ke nhau la bie’u thu.c Boole mo.t bien nhu. Hınh 6.7.

Ta minh ho.a phu.o.ng phap qua vı du. sau.

Vı du. 6.6.1. Xet ham Boole

f(x, y) = (x′ ∧ y) ∨ (x ∧ y) ∨ (x ∧ y′).

Ta chia lam ba bu.o.c.

Bu.o.c 1. Ve mo.t ba’n do Karnaugh va da. t 1 vao moi hınh vuong con tu.o.ng u.ng vo.i mo.tphan tu.’ toi thie’u cu’a f. Ta co Hınh 6.8.

Bu.o.c 2. Bay gio. ve cac ellipse chu.a cac so 1 ke nhau sao cho cac ellipse nay chu.a tat ca’

cac so 1. Chu y la khong ve nhieu ho.n can thiet. Ta co Hınh 6.9.

Bu.o.c 3. Vo.i moi ellipse co du.o.. c trong bu.o.c tru.o.c, chung ta to’ ho.. p la. i thanh mo.t bie’uthu.c Boole mo.t bien, va roi tuye’n cac bien nay la. i de’ co da.ng do.n gia’n g(x, y). Trong vı du.

154

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1 1

1

Hınh 6.8:

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1 1

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Hınh 6.9:

nay ta co

g(x, y) = x ∨ y.

Ba’n do Karnaugh ba bien

Ba’n do Karnaugh ba bien la mo. t hınh chu. nha. t du.o.. c chia thanh tam hınh vuong con nhu.

Hınh 6.10. Nhu. tru.o.ng ho.. p hai bien, moi hınh vuong con du.o.. c gan vo.i mo.t trong tam kha’

nang cu’a cac phan tu.’ toi thie’u ba bien. Mo.t trong nhu.ng ly do de’ thua.t toan Karnaughthu.. c hie.n la hai hınh vuong con ke nhau tu.o.ng u.ng hai phan tu.’ toi thie’u chı’ khac nhaumo.t bien. Tuy nhien can chu y rang, cac hınh vuong con o.’ co.t dau va co.t cuoi (trong cungmo.t hang) la ke nhau. Trong tru.o.ng ho.. p ba bien, moi hınh vuong con tu.o.ng u.ng mo.t phantu.’ toi thie’u ma la bie’u thu.c Boole ba bien. Do do hai hınh vuong con ke nhau tu.o.ng u.ngmo.t bie’u thu.c Boole hai bien, cha’ng ha.n nhu. Hınh 6.11. Ho.n nu.a bon hınh vuong ke nhau(go. i la quadruple) tu.o.ng u.ng bie’u thu.c mo.t bien nhu. Hınh 6.12.

Ta minh ho.a phu.o.ng phap qua cac vı du. sau.

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x′ ∧ y′ ∧ z′

x ∧ y′ ∧ z′

x′ ∧ y′ ∧ z

x ∧ y′ ∧ z

x′ ∧ y ∧ z

x ∧ y ∧ z

x′ ∧ y ∧ z′

x ∧ y ∧ z′

x′

x

y′z′ y′z yz yz′

Hınh 6.10:

155

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1 1

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x′

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xyz ∨ x′y′

(a)

1 1 1 1

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x

(b)

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(c)

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x′y ∨ z′

(d)

Hınh 6.11:

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z′

Hınh 6.12:


f(x, y, z) = (x′ ∧ y′ ∧ z) ∨ (x ∧ y′ ∧ z′) ∨ (x ∧ y ∧ z) ∨ (x′ ∧ y ∧ z).

Bu.o.c 1. D- au tien ve ba’n do Karnaugh va da. t trong moi hınh vuong mo.t so 1 tu.o.ng u.ngphan tu.’ toi thie’u trong f. Ta du.o.. c Hınh 6.13

Bu.o.c 2. Ve cac ellipse hay quadruple chu.a cac so 1 ke nhau sao cho phu’ tat ca’ cac so 1va khong su.’ du.ng cac ellipse hay quadruple ho.n so can thiet. Ta co Hınh 6.14.

(Chu y rang, neu co the’, hay su.’ du. ng cac quadruple nhu. Vı du. 6.6.3 du.o.i day).

Bu.o.c 3. Bay gio. vo.i moi ellipse (hoa.c quadruple) ta co tu.o.ng u.ng mo.t bie’u thu.c mo.thoa. c hai bien. Tuye’n cac bie’u thu.c nay ta du.o.. c ham toi thie’u

g(x, y, z) = (x ∧ y′ ∧ z′) ∨ (x′ ∧ z) ∨ (y ∧ z).

156

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Hınh 6.13:

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Hınh 6.14:


f(x, y, z) = (x′ ∧ y′ ∧ z′) ∨ (x∧ y ∧ z)∨ (x∧ y′ ∧ z)∨ (x′ ∧ y′ ∧ z)∨ (x′ ∧ y ∧ z)∨ (x′ ∧ y ∧ z′).

Bu.o.c 1. Ta co ba’n do Karnaugh va da. t so 1 vao cac hınh vuong tu.o.ng u.ng cac phan tu.’

toi thie’u (Hınh 6.15).

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1 1

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Hınh 6.15:

Bu.o.c 2. Ve cac ellipse hay cac quadruple cu’a cac so 1 ke nhau sao cho phu’ tat ca’ cac so1 va khong ve thu.a. Co the’ lam ba cach nhu. sau

Su.’ du.ng Hınh 6.16(a) ta co

g1(x, y, z) = x′ ∨ (x ∧ z).

Su.’ du.ng Hınh 6.16(b) ta cog2(x, y, z) = z ∨ (x′ ∧ z′).

Su.’ du.ng Hınh 6.16(c) ta cog3(x, y, z) = x′ ∨ z.

Hie’n nhien ham g3 la ham do.n gia’n nhat!

157

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(c)

Hınh 6.16:

Bai ta.p

1. Ve cac ba’n do Karnaugh va tım da.ng tuye’n chınh tac toi thie’u cu’a cac ham Boole haibien:

(a) xy + xy′. (b) xy + x′y + x′y′. (c) xy + x′y′.

2. Ve cac ba’n do Karnaugh va tım da.ng tuye’n chınh tac toi thie’u cu’a cac ham Boole babien:

(a) x ∨ x′yz. (b) (x ∨ yz)′. (c) y′z ∨ xyz.(d) (y ∨ z). (e) xz ∨ yz. (f) xy ∨ xz ∨ yz.(g) xyz ∨ xy′z′ ∨ x′yz′ ∨ x′y′z. (h) xy ∨ yz ∨ zx. (h) xy ∧ yz ∧ zx.

158

Chu.o.ng 7

MA TUYEN TINH

Ly thuyet ma bat dau hınh thanh va phat trie’n tu. nam 1940 vo.i nhu.ng ket qua’ rat co. ba’ncu’a M. J. E. Golay, R. W. Hamming va C. E. Shannon. Ma.c du ban dau la bai toan cu’aky su., nhu.ng van de da du.o.. c phat trie’n su.’ du. ng rat nhieu cong cu. toan ho.c. Chu.o.ng naytrınh bay ly thuyet cac ma phat hie.n va su.’ a sai o.’ mu.c do. do.n gia’n nhat. Qua do ngu.o.ido.c co the’ thay ro moi lien he. ma.t thiet vo.i nhu.ng bai toan da.t ra do su.. phat trie’n congnghe. vien thong.

7.1 Mo.’ dau

7.1.1 Khai nie.m

Cach thong thu.o.ng de’ bie’u dien, lu.u tru. va truyen thong tin la su.’ du.ng chuoi cac bit, tu.cla day cac so 0 va 1. Tha. t la kho khan va thu.o.ng khong the’ ngan ngu.a cac loi xa’y ra khidu. lie.u du.o.. c lu.u tru., phu.c hoi, xu.’ ly hay du.o.. c truyen tu. no.i nay sang no.i nay khac. Cacloi co the’ xuat hie.n do tieng on cu’a kenh thong tin, do nhieu, do con ngu.o.i hay do thiet bi..Cac loi cung co the’ xa’y ra khi du. lie.u du.o.. c lu.u tru. trong tho.i gian dai tren cac bang tu.

D- o. tin ca.y cu’a du. lie.u nha.n du.o.. c tu. cac ta.p tin lo.n hay khi du. lie.u du.o.. c gu.’ i tu. mo.t no.irat xa la quan tro.ng. Tu.o.ng tu.. , vie.c phu. c hoi du. lie.u du.o.. c lu.u tru. khap no.i tren bang tu.

cung la van de dang quan tam.

Ly thuyet ma na’y sinh tu. bai toan da’m ba’o do. tin ca.y hay phu.c hoi du. lie.u. Cac ba’n tino.’ da.ng chuoi bit du.o.. c ma hoa thanh chuoi bit dai ho.n go. i la tu. ma. Bo. ma la ta.p ho.. p cactu. ma.

Chung ta co the’ phat hie.n cac loi khi su.’ du.ng cac bo. ma nao do. Tu.c la, neu khong coqua nhieu loi, chung ta co the’ xac di.nh du.o.. c cac loi xa’y ra khi truyen du. lie.u. Ho.n nu.a,

159

vo.i mo.t vai bo. ma, chung ta co the’ su.’ a du.o.. c cac loi do. Noi cach khac, neu khong co quanhieu loi xa’y ra trong du.o.ng truyen, chung ta co the’ phu.c hoi tu. ma tu. chuoi bit nha.ndu.o.. c.

Ly thuyet ma ra do.i tu. nam 1940 nham nghien cu.u cac bo. ma, bao gom phat hie.n va su.’ asai cac loi. Su.. phat trie’n cong nghe. mo.i nham truyen va lu.u du. lie.u khien cho vie.c nghiencu.u ly thuyet ma cang tro.’ nen quan tro.ng. Chu.o.ng nay gio.i thie.u so. lu.o.. c ve vie.c phat hie.nloi va su.’ a sai loi vo.i hai gia’ thiet:

1. Xac suat truyen bit 1 va nha.n du.o.. c bit 0 bang xac suat truyen bit 0 nha.n bit 1 vabang p vo.i 0 ≤ p < 1

2(go. i la kenh doi xu.ng nhi. phan).

2. Cac bit du.o.. c truyen mo.t cach do.c la.p.

7.1.2 Ma phat hie.n loi

Cach do.n gia’n de’ phat hie.n cac loi khi mo.t chuoi bit du.o.. c truyen la them mo.t bit kie’mtra chan le’ vao cuoi chuoi: chung ta ma hoa ba’n tin x1x2 . . . xn thanh tu. ma x1x2 . . . xn+1,trong do

xn+1 = (x1 + x2 + · · · + xn) mod 2.

Vie.c them bit chan le’ ba’o da’m rang so cac so 1 trong tu. ma pha’i la so chan. De dang thayrang trong bo. ma nay, cac tu. ma la cac chuoi bit vo.i mo.t so chan cac so 1.

Nha.n xet 15. Neu mo.t loi xuat hie.n, so cac so 1 trong chuoi nha.n du.o.. c la mo.t so le’, dodo loi nay du.o.. c phat hie.n. Neu hai loi xuat hie.n, so cac so 1 trong chuoi nha.n du.o.. c la mo.tso chan, do do cac loi nay khong du.o.. c phat hie.n. To’ng quat mo.t so le’ cac loi co the’ du.o.. cphat hie.n, trong khi mo.t so chan cac loi thı khong.

Vı du. 7.1.1. Neu nha.n du.o.. c chuoi bit 1110011 thı day la tu. ma khong ho.. p le..

Vı du. 7.1.2. Neu nha.n du.o.. c chuoi bit y = 10111101 thı hoa. c y la tu. ma ho.. p le., hoa. c como.t so chan loi xa’y ra.

Mo.t cach do.n gia’n khac de’ phat hie.n loi la la.p moi bit trong mo.t thong bao hai lan nhu.

vı du. sau.

Vı du. 7.1.3. Chuoi 011001 du.o.. c ma hoa thanh tu. ma 001111000011.

Nha.n xet 16. Chung ta co the’ phat hie.n cac loi trong bit thu. 2, 3 va thu. 8 cu’a cac tu.

ma co 8 bit (nhu. khi tu. ma 00001111 go.’ i va nha.n du.o.. c 01101110 la co loi). Ma.t khac,khong the’ phat hie.n ra loi neu bit thu. 3, 4 bi. thay do’i (nhu. khi 00111111 nha.n du.o.. c tu. ma00001111 la co loi).

Chung ta da tha’o lua.n hai bo. ma co the’ dung de’ phat hie.n loi. Khi cac loi du.o.. c phathie.n, chung ta co the’ yeu cau truyen la. i va hy vo.ng rang khong co loi nao xuat hie.n. Tuynhien, co cac bo. ma khong chı’ phat hie.n sai ma con su.’ a chu.a cac loi sai (neu co).

160

7.1.3 Ma su.’ a sai

D- e’ phat hie.n loi, trong cac vı du. tru.o.c, chung ta xay du.ng tu. ma bang cach them cac bitthıch ho.. p vao ba’n tin. Chung ta khong chı’ phat hie.n cac loi ma con su.’ a chung neu themnhieu bit ho.n vao ba’n tin. Chınh xac ho.n, neu cac loi la du’ ıt, chung ta co the’ xac di.nh tu.

ma nao du.o.. c truyen.

Vı du. 7.1.4. Ma hoa mo.t ba’n tin, chung ta co the’ dung ma la.p ba lan. Cha’ng ha.n,neu thong bao la x1x2x3, chung ta ma hoa no thanh tu. ma x1x2x3x4x5x6x7x8x9, trong dox1 = x4 = x7, x2 = x6 = x8, x3 = x5 = x9.

Cac tu. ma ho.. p le. la

000000000, 001001001, 010010010, 011011011,100100100, 101101101, 111111111.

Chung ta phat hie.n mo.t chuoi bit nha.n du.o.. c co loi bang cach su.’ du. ng “lua.t so lo.n”.Cha’ng ha.n de’ xac di.nh x1, xet cac bit x1, x4, x7. Neu hai trong ba bit bang 1, ta ket lua.nx1 = 1, ngu.o.. c la. i ket lua.n x1 = 0.

Bai ta.p

1. Cac chuoi bit nha.n du.o.. c sau co the’ la dung (su.’ du.ng bit kie’m tra chan le’):

(a) 1000011.

(b) 111111000.

(c) 10101010101.

(d) 110111011100.

2. Cac chuoi bit nha.n du.o.. c sau co the’ la dung (la.p moi bit trong thong bao hai lan):

(a) 110011.

(b) 1100000011.

(c) 101111.

3. Cac ba’n tin du.o.. c la.p ba lan. Su.’ a sai cac chuoi bit nha.n du.o.. c sau (neu sai):

(a) 111000101.

(b) 110000001.

(c) 111011111000.

161

7.2 Cac khai nie.m

Trong chu.o.ng nay, gia’ thiet moi ba’n tin u ∈ Bk du.o.. c ma hoa thanh cac “tu. ma” x ∈ Bn, n >k. D- e’ do.n gia’n, ta se dong nhat vector co.t x = (x1, x2, . . . , xn)t vo.i chuoi bit x1x2 . . . xn.

D- i.nh nghıa 7.2.1. Khong gian vector con k chieu C cu’a khong gian vector Bn tren tru.o.ngB go. i la [n, k]-ma tuyen tınh. n du.o.. c go. i la do. dai cu’a bo. ma va dimC := k la chieu. He.so cu’a bo. ma la tı’ so k/n. Cac phan tu.’ cu’a C go. i la cac tu. ma.

Noi cach khac, ta.p con C cu’a Bn la mo.t ma tuyen tınh neu

(a) x + y ∈ C vo.i mo.i x, y ∈ C; va

(b) αx ∈ C vo.i mo.i x ∈ C,α ∈ B.

Tu. di.nh nghıa ta thay rang, [n, k] ma tuyen tınh C hoan toan du.o.. c xac di.nh bo.’ i ta.p batky cac tu. ma do. c la.p tuyen tınh x1, x2, . . . , xk vı moi tu. ma x ∈ C deu co the’ bie’u dienda.ng

x =k∑

i=1

αixi (mod 2),

trong do αi ∈ B. Neu chung ta sap xep cac tu. ma nay thanh mo.t ma tra.n Boole G cap k×nta se du.o.. c mo.t ma tra.n sinh cu’a ma C. Chınh xac ho.n:

D- i.nh nghıa 7.2.2. Gia’ su.’ C la [n, k]-ma tuyen tınh. Ma tra.n Boole G cap k × n ma cachang cu’a no sinh ra khong gian vector C go. i la ma tra. n sinh cu’a C. Ngu.o.. c la. i, neu G la matra.n Boole cap k × n thı khong gian vector sinh bo.’ i cac hang cu’a no go. i la ma sinh bo.’ i G.

Nha.n xet 17. Mo.t ma co the’ co nhieu ma tra.n sinh khac nhau. Cha’ng ha.n cac ma tra.n

(1 1 1 00 1 0 1

),

(1 0 1 10 1 0 1

)

cung la cac ma tra.n sinh cu’a ma vo.i cac phan tu.’ :

c1 = 0 0 0 0c2 = 0 1 0 1c3 = 1 1 1 0c4 = 1 0 1 1

Vı du. 7.2.1. [5, 1]-ma tuyen tınh C1 vo.i ma tra.n sinh

G1 :=(1 1 1 1 1

),

chu.a hai tu. ma la 00000 va 11111.

162


G2 :=

1 1 1 0 00 0 1 1 01 1 1 1 1

.


G3 :=

1 0 0 0 0 1 10 1 0 0 1 0 10 0 1 0 1 1 00 0 0 1 1 1 1

.

Do [n, k]-ma tuyen tınh C co 2k tu. ma nen ta co the’ truyen di toi da 2k ba’n tin khacnhau; neu gia’ thiet cac hang cu’a ma tra.n G do.c la.p tuyen tınh thı ba’n tin u ∈ Bk se du.o.. cma hoa thanh vector

x = utG. (7.1)

Cha’ng ha.n, su.’ du.ng ma tra.n sinh G2 cu’a Vı du. 7.2.2 ta co

x1 = u1 + u3,

x2 = u1 + u3,

x3 = u1 + u2 + u3,

x4 = u2 + u3,

x5 = u3.

Ky hie.u M la so phan tu.’ cu’a ma C (do. dai n). Ta bie’u dien cac phan tu.’ cu’a C bang mo.tma’ng kıch thu.o.c M × n ma cac hang la cac tu. ma.

Gia’ su.’ π la hoan vi. cu’a ta.p {1, 2, . . . , n} va vo.i moi tu. ma x ∈ C ta ap du.ng phep biendo’i, go. i la hoan vi. vi. trı,

π : x 7→ x′

xac di.nh bo.’ ix′

i := xπ(i), i = 1, 2, . . . , n.

Tu.o.ng tu.. , neu π la hoan vi. cu’a cac ky hie.u {0, 1}, ta noi π ca’m sinh mo. t phep hoan vi. kyhie.u neu vo.i chı’ so i nao do, va vo.i moi tu. ma x ∈ C ta ap du.ng phep bien do’i

x 7→ x′,

trong do x′ xac di.nh bo.’ i

x′j :=

{xj neu i 6= j,

π(xi) neu i = j.

Neu ma C ′ co the’ nha.n du.o.. c tu. ma C bang mo.t day cac phep hoan vi. vi. trı hoa. c phephoan vi. ky hie.u thı ta noi hai ma C va C ′ la tu.o.ng du.o.ng.

163

Vı du. 7.2.4. (a) Hai ma sau la tu.o.ng du.o.ng bang cach su.’ du.ng hoan vi. π({1, 2, 3, 4}) ={1, 3, 2, 4} :

0 0 0 00 0 1 11 1 0 01 1 1 1

va

0 0 0 00 1 0 11 0 1 01 1 1 1

.

(b) Ma

C :=

0 0 1 0 00 0 0 1 11 1 1 1 11 1 0 0 0

tu.o.ng du.o.ng vo.i ma

C ′ :=

0 0 0 0 00 1 1 0 11 1 0 1 11 0 1 1 0

qua phep hoan vi.

0 1↓ ↓1 0

cac ky hie.u o.’ vi. trı thu. ba trong C va sau do hoan vi. hai vi. trı thu. 2 va thu. 4.

Bo’ de 7.2.3. Hai ma tra. n Boole cung cap k × n sinh ra hai ma tuyen tınh tu.o.ng du.o.ngneu chung nha. n du.o.. c tu. nhau bang day cac phep toan:

(a) hoan vi. cac hang;

(b) co. ng hai hang; va

(c) hoan vi. cac co. t.

Chu.ng minh. Cac phep toan tren hang (a) va (b) khong thay do’i ha.ng cu’a ma tra.n sinh(chı’ thay do’i cac vector co. so.’ ). Phep toan (c) tu.o.ng du.o.ng vo.i hoan vi. vi. trı cac tu. ma.2

Vı du. 7.2.5. (a) Ma tra.n sinh G1 va G3 co da.ng ba. c thang, tu.c ma tra.n co cac tınh chat:

1. Phan tu.’ khac khong ben trai nhat trong moi hang bang 1.

2. Co. t chu.a phan tu.’ ben trai nhat cu’a mo.t hang bang 1 co tat ca’ cac phan tu.’ khac bang0.

164

3. Neu phan tu.’ bang 1 ben trai nhat trong hang thu. i xuat hie.n o.’ co. t ti thı

t1 < t2 < · · · < tn.

(b) Ma tra.n G2 co the’ du.a ve ma tra.n ba.c thang

G′2 :=

1 1 0 0 10 0 1 0 10 0 0 1 1

.

Su.’ du.ng G′2 cho bo. ma C2, ma hoa (7.1) co da.ng

x1 = u1,

x2 = u1,

x3 = u2,

x4 = u3,

x5 = u1 + u3.

D- ieu nay chı’ ra rang cac ky hie.u ba’n tin u1, u2, u3 xuat hie.n tu.o.ng minh trong cac tu. ma;noi chung, ky hie.u ui se xuat hie.n ta. i vi. trı thu. ti cu’a tu. ma x = utG neu phan tu.’ ben trainhat cu’a hang thu. i cu’a G xuat hie.n trong co.t thu. ti.

Nha.n xet rang, cac ma tra.n ba. c thang cu’a ma C1 va C3 co da.ng G = (Ik A), trong do Ik

la ma tra.n do.n vi. cap k. Ap du.ng phu.o.ng phap cu’a Bo’ de 7.2.3, ma tra.n G′2 co the’ du.a ve

G′′

2 :=

1 0 0 1 10 1 0 0 00 0 1 0 1

.

To’ng quat ta co

D- i.nh ly 7.2.4. Gia’ su.’ C la [n, k]-ma. Khi do ton ta. i ma C ′ tu.o.ng du.o.ng C vo.i ma tra. nsinh da.ng (Ik A).


Theo ket qua’ tren, ta luon co the’ gia’ thiet ma tra.n sinh G co da.ng (Ik | A).

D- i.nh nghıa 7.2.5. Gia’ su.’ C la [n, k]-ma tuyen tınh va H la ma tra.n Boole cap (n−k)×n.H go. i la ma tra.n kie’m tra chan le’ cu’a C neu vo.i mo.i tu. ma x ∈ C ta co

Hx = 0 (mod 2). (7.2)

He. (7.2) du.o.. c go. i la he. phu.o.ng trınh kie’m tra chan le’.

165

Vı du. 7.2.6. [4, 3]-ma C4 bang cach them mo.t bit kie’m tra chan le’ trong Phan 7.1.2: Ba’ntin u1u2u3 du.o.. c ma hoa thanh tu. ma x = x1x2x3x4, trong do

x1 = u1, x2 = u2, x3 = u3,

vax1 + x2 + x3 + x4 = 0.

Do do neu ba’n tin la u = 101 thı tu. ma la x = 1010. Co 23 = 8 tu. ma la

0000 0011 0101 10011010 0110 1100 1111.

Tu.c la tat ca’ cac vector co mo.t so chan so bit bang 1. De dang thu.’ la. i ma tra.n kie’m trachan le’ cu’a C4 la H5 = (1 1 1 1).

Vı du. 7.2.7. Xet [6, 3]-ma la.p C5 : Ba’n tin u1u2u3 du.o.. c ma hoa thanh tu.ma x = x1x2 . . . x6,trong do

x1 = u1, x2 = u2, x3 = u3,

va

x2 + x3 + x4 = 0,

x1 + x3 + x5 = 0,

x1 + x2 + x6 = 0.

Ma C5 co ma tra.n kie’m tra chan le’:

H :=

0 1 1 1 0 01 0 1 0 1 01 1 0 0 0 1

.

D- i.nh ly 7.2.6. Gia’ su.’ G va H la cac ma tra. n vo.i cac hang do. c la. p tuyen tınh co kıchthu.o.c tu.o.ng u.ng k×n va (n− k)×n. Khi do G va H la cac ma tra. n sinh va ma tra. n kie’mtra chan le’ cu’a mo. t ma neu va chı’ neu GH t = 0.

Chu.ng minh. Gia’ su.’ GH t = 0. Khi do moi hang cu’a G la nghie.m cu’a he. phu.o.ng trınh (7.2)va do do khong gian sinh bo.’ i tat ca’ cac to’ ho.. p tuyen tınh cu’a cac hang cu’a G chu.a trongkhong gian cac nghie.m cu’a (7.2). Nhu.ng ca’ hai khong gian nay co chieu bang k nen chungbang nhau. Bang cach suy lua.n tu.o.ng tu.. ta co chieu ngu.o.. c la. i. 2

Vı du. 7.2.8. Cac ma C1, C2, C3 trong cac vı du. tren co cac ma tra.n kie’m tra chan le’ tu.o.ngu.ng la

H1 :=

1 1 0 0 01 0 1 0 01 0 0 1 01 0 0 0 1

, H2 :=

(1 1 0 0 01 0 1 1 1

),

va

H3 :=

0 1 1 1 1 0 01 0 1 1 0 1 01 1 0 1 0 0 1

.

166

Nha.n xet rang neu G = (Ik A) thı ma tra.n kie’m tra chan le’ co da.ng H = (At In−k). Khido he. phu.o.ng trınh kie’m tra chan le’ (7.2) cho mo.t phu. thuo.c ham tu.o.ng minh giu.a cac kyhie.u ba’n tin (cac bit thong tin) va cac ky hie.u kie’m tra. Ma tra.n sinh va ma tra.n kie’m trachan le’ cu’a ma tuyen tınh khong chı’ co y nghıa ve ma.t ly thuyet ma no con co nhu.ng u.ngdu.ng chu’ yeu trong vie.c ma hoa va gia’i ma. Tha.t va.y, moi ba’n tin u ∈ Bk du.o.. c ma hoaduy nhat thanh tu. ma x = utG. Vı cac hang cu’a ma tra.n sinh do. c la.p tuyen tınh nen anhxa. u 7→ utG la song anh tu. Bk len C. Vie.c gia’i ma kho khan ho.n se du.o.. c trınh bay trongmu.c tiep theo.

Vı du. 7.2.9. (a) Ma C1 co x1 = u1 la bit thong tin va cac ky hie.u con la. i la cac bit kie’mtra: x2 = x3 = x4 = x5 = x1. Do do C1 co hai tu. ma la 00000 va 11111.

(b) Ma C2 co x1, x3, x4 la cac bit thong tin va cac ky hie.u con la. i la cac bit kie’m tra:x2 = x1, x5 = x1 + x3 + x4. Do do C2 co 23 = 8 tu. ma la

00000 10000 01000 0010011000 10100 01100 11100.

(c) Ma C3 co x1, x2, x3, x4 la cac bit thong tin va cac ky hie.u con la. i la cac bit kie’m tra:

x5 = x2 + x3 + x4,

x6 = x1 + x3 + x4,

x7 = x1 + x2 + x4,

Do do C3 co 24 = 16 tu. ma (hay lie.t ke chung!).

Tren Bn xet tıch vo hu.o.ng cu’a hai vector di.nh nghıa bo.’ i

〈x, y〉 :=n∑

i=1

xiyi (mod 2).

Chu y rang, khac vo.i tıch vo hu.o.ng thong thu.o.ng tren khong gian Euclide, co the’ xa’y ra〈x, x〉 = 0 vo.i vector x 6= 0 nao do.

D- i.nh nghıa 7.2.7. Ma doi ngau hay ma tru.. c giao, ky hie.u C⊥, cu’a ma tuyen tınh C xacdi.nh bo.’ i

C⊥ := {y ∈ Bn | 〈x, y〉 = 0 vo.i mo.i x ∈ C}.

De dang thay rang C⊥ la ma tuyen tınh thoa’ dimC + dimC⊥ = n. Ho.n nu.a

D- i.nh ly 7.2.8. Vo.i mo. i ma tuyen tınh C, ma tra. n kie’m tra chan le’ cu’a C⊥ bang ma tra. nsinh cu’a C va ngu.o.. c la. i.


167

Bai ta.p

1. Gia’ su.’ H la ma tra.n Boole cap r × n. Chu.ng minh ta.p C := {x ∈ Bn|Hx = 0} la matuyen tınh.

2. Chu.ng minh neu C la [n, k]-ma thı

C := {(x, xn+1) ∈ Bn × B1|x = x1x2 . . . xn ∈ C, xn+1 := x1 + x2 + · · · + xn}

cung la ma tuyen tınh (go. i la ma mo.’ ro.ng). Tım moi lien he. giu.a cac ma tra.n kie’mtra chan le’ cu’a C va C.

3. Chu.ng minh rang trong mo.t ma nhi. phan tuyen tınh, hoa. c tat ca’ cac tu. ma bat daubang so 0, hoa. c co chınh xac mo.t nu.’ a bat dau bang so 0, va mo.t nu.’ a bat dau bangso 1.

4. D- u.a cac ma tra.n sinh sau ve da.ng chua’n (Ik A) :

(0 1 11 0 1

),

1 1 1 1 1 1 11 0 0 0 1 0 11 1 0 0 0 1 00 1 1 0 0 0 1

,

0 0 0 1 1 10 1 1 0 1 01 0 0 0 1 1

.

5. Chu.ng minh rang cac ma tra.n sinh

G :=

1 1 0 00 1 1 00 0 1 1

, G′ :=

1 0 0 10 1 0 10 0 1 1

,

sinh ra cac ma tu.o.ng du.o.ng.

6. Chu.ng minh rang cac ma tra.n sinh

G :=

1 1 0 0 0 00 0 1 1 0 00 0 0 0 1 1

, G′ :=

1 1 1 1 1 10 1 1 0 1 10 0 1 0 0 1

,

sinh ra cac ma tu.o.ng du.o.ng.

7. Gia’ su.’ C co ma tra.n sinh

G :=

1 0 1 1 10 1 1 0 11 1 0 0 0

.

Tım ma tra.n A sao cho ma co ma tra.n sinh (I3 A) tu.o.ng du.o.ng vo.i C. Lie.t ke tat ca’

cac tu. ma cu’a C.

8. Gia’ su.’ ma C co ma tra.n sinh da.ng chua’n (Ik A). Chu.ng minh hoan vi. cac hang cu’aA cho ma tra.n sinh cu’a ma tu.o.ng du.o.ng C.

168

9. Chu.ng minh rang quan he. “ma tu.o.ng du.o.ng” la quan he. tu.o.ng du.o.ng.

10. Gia’ su.’ C la [n, k]-ma va a ∈ Bn. Chu.ng minh rang ton ta. i ma C ′ chu.a a va tu.o.ngdu.o.ng vo.i C.

11. Chu.ng minh so cac ma khong tu.o.ng du.o.ng vo.i do. dai n va chu.a hai tu. ma la n.

12. Gia’ su.’ C la [7, 4]-ma tuyen tınh vo.i ma tra.n sinh

G :=

1 0 0 0 1 0 10 1 0 0 1 1 10 0 1 0 1 1 00 0 0 1 0 1 1

.

Ma hoa cac ba’n tin: 0000, 1000 va 1110.

13. Tım ma tra.n sinh va cac tu. ma cu’a [6, 3]−ma co ma tra.n kie’m tra chan le’

H :=

0 1 1 1 0 01 0 1 0 1 01 1 0 0 0 1

.

14. (Ma la.p) Tım ma tra.n sinh va cac tu. ma cu’a [5, 1]−ma co ma tra.n kie’m tra chan le’

H :=

1 1 0 0 01 0 1 0 01 0 0 1 01 0 0 0 1

.

15. (Ma tro.ng lu.o.. ng chan) Cho ma tra.n kie’m tra chan le’

H :=(1 1 1 1 1

).

Tım ma tra.n sinh va cac tu. ma.

16. Tım ma tra.n sinh va cac tu. ma co ma tra.n kie’m tra chan le’

H :=

(1 0 1 01 1 0 1

).

17. Cho ma tra.n kie’m tra chan le’

H :=

0 1 1 1 1 0 01 0 1 1 0 1 01 1 0 1 0 0 1

.

Lie.t ke tat ca’ cac tu. ma.

169

18. Cho ma tra.n kie’m tra chan le’:

H :=

1 0 1 0 01 1 0 1 00 1 0 0 1

.

Tım ma tra.n sinh va cac tu. ma.

19. Tım ma tra.n kie’m tra chan le’ tu.o.ng u.ng vo.i ma du.o.. c thiet la.p bang cach them mo.tbit kie’m tra chan le’ doi vo.i chuoi bit do. dai 4.

20. Tım ma tra.n kie’m tra chan le’ tu.o.ng u.ng vo.i ma la.p ba doi vo.i chuoi bit do. dai 3.

21. Tım ma tra.n kie’m tra chan le’ H neu ma tra.n sinh la

G :=

1 0 0 0 1 1 10 1 0 0 1 0 10 0 1 0 0 1 10 0 0 1 1 1 0

.

22. Tım cac ma doi ngau cu’a cac ma C2 va C3 trong cac Vı du. 4.2.2 va 4.2.3.

23. Tım cac ma doi ngau cu’a cac ma sau:

C1 :=

0000

1100

0011

1111

, C2 :=

000

110

011

101

.

24. (a) Chu.ng minh rang (C⊥)⊥ = C.

(b) D- a.t C + D := {x + y|x ∈ C, y ∈ D}. Chu.ng minh (C + D)⊥ = C⊥ + D⊥.

25. Ky hie.u En la ta.p tat ca’ cac vector do. dai n co tro.ng lu.o.. ng chan.

(a) Chu.ng minh En la ma tuyen tınh. Tım cac tham so [n, k], ma tra.n kie’m tra chanle’ va ma tra.n sinh cu’a En.

(b) Tım ma E⊥n .

7.3 Khoa’ng cach Hamming

D- i.nh nghıa 7.3.1. Khoa’ng cach Hamming, ky hie.u d(x, y), giu.a hai vector x = x1x2 . . . xn

va y = y1y2 . . . yn la so cac vi. trı i ma xi 6= yi, i = 1, 2, . . . , n.

Nha.n xet rang d(x, y) chınh la so lan thay do’i can thiet tu.ng bit tu. x sang y.

170

Vı du. 7.3.1. d(10111, 00101) = 2, d(0111, 0000) = 3.

D- i.nh ly 7.3.2. Khoa’ng cach Hamming d(x, y) la mo. t metric, tu.c la

(a) d(x, y) ≥ 0 vo.i mo. i x, y ∈ C; dau bang xa’y ra khi va chı’ khi x = y.

(b) d(x, y) = d(y, x).

(c) d(x, y) ≤ d(x, z) + d(z, y) vo.i mo. i x, y, z ∈ C.

Chu.ng minh. Hai kha’ng di.nh dau suy tru.. c tiep tu. di.nh nghıa.

Chu.ng minh (c): Nha.n xet rang

{i | xi 6= yi} ⊂ {i | xi 6= zi} ∪ {i | zi 6= yi},

vı neu xi 6= yi thı hoa. c xi 6= zi hoa. c zi 6= yi. Suy ra

#{i | xi 6= yi} ≤ #{i | xi 6= zi} + #{i | zi 6= yi}.

Ap du.ng nguyen ly bao ham-loa. i tru. va bat da’ng thu.c:

#(A ∪ B) = #A + #B − #(A ∩ B) ≤ #A + #B

ta co dieu can chu.ng minh. 2

Gia’ su.’ rang mo.t ba’n tin du.o.. c ma hoa thanh tu. ma x ∈ C du.o.. c gu.’ i di va nha.n du.o.. cvector y. Co hai tru.o.ng ho.. p xa’y ra

(a) Hoa.c y ∈ C khi do y = x.

(b) Hoa.c y 6∈ C khi do vector loi e := y − x 6= 0.

Trong tru.o.ng ho.. p (b), van de da. t ra la lam sao su.’ a du.o.. c loi sai, phu. c hoi du.o.. c tu. ma xtu. vector nha.n du.o.. c y?

Phu.o.ng phap gia’i ma du.a ra o.’ day, go. i la gia’i ma theo lan ca.n gan nhat, nham tınhkhoa’ng cach Hamming giu.a y vo.i moi tu. ma trong C. D- e’ gia’i ma y, chung ta tım tu. ma xco khoa’ng cach Hamming den y nho’ nhat. Neu

(+) khoa’ng cach giu.a hai tu. ma gan nhat trong C du’ lo.n; va

(+) neu cac loi du’ ıt;

thı x la duy nhat-chınh la tu. ma du.o.. c gu.’ i.

171

Vı du. 7.3.2. Gia’ su.’ C = {0000, 1110, 1011, 1111}. Thı

d(0000, 0110) = 2, d(1110, 0110) = 1, d(1011, 0110) = 3.

Do do neu nha.n du.o.. c y = 0110 /∈ C thı chung ta ket lua.n (gia’i ma theo lan ca.n gan nhat)tu. ma gu.’ i la 1110.

Gia’ su.’ moi bit gu.’ i di co cung xac suat sai p, 0 ≤ p < 1/2. Chung ta go. i kenh nhu. the lakenh doi xu.ng nhi. phan.

Vı du. 7.3.3. Ky hie.u P (X) la xac suat xa’y ra bien co X. Ta co trong kenh doi xu.ng nhi.phan

P ({e = 00000}) = (1 − p)5,P ({e = 01000}) = p(1 − p)4,P ({e = 10010}) = p2(1 − p)3.

Mo.t cach to’ng quat, neu v la vector co a bit bang 1 thı

P ({e = v}) = pa(1 − p)n−a.

Vı p < 1/2 nen 1 − p > p; do do

(1 − p)n > p(1 − p)n−1 > p2(1 − p)n−2 > · · ·

Phu.o.ng phap gia’i ma ho.. p ly nhat nhu. sau: Gia’ su.’ nha.n du.o.. c vector y, chung ta tım tu.

ma x sao cho xac suat P (x|y) cu’a su.. kie.n truyen tu. ma x vo.i dieu kie.n nha.n du.o.. c y la cu.. cda. i. Noi cach khac, tım mo.t tu. ma ho.. p ly nhat trong bo. ma tu.o.ng u.ng vo.i thong bao nha.ndu.o.. c.

D- i.nh ly 7.3.3. Gia’ su.’ tat ca’ cac tu. ma du.o.. c truyen vo.i cung kha’ nang va su.’ du. ng kenhdoi xu.ng nhi. phan. Khi do gia’i ma ho.. p ly nhat trung vo.i gia’i ma theo lan ca. n gan nhat.

Chu.ng minh. Trong kenh doi xu.ng nhi. phan, neu d(x, y) = d thı co d loi khi thay do’i tu.

x sang y; do do xac suat co dieu kie.n P (y|x) cu’a su.. kie.n nha.n du.o.. c y vo.i dieu kie.n tu.

ma x du.o.. c truyen la pd(1 − p)n−d. Ma.t khac, theo gia’ thiet, xac suat truyen tu. ma x laP (x) = 1

#C. Do do

P (x|y) = pd(1 − p)n−d(1/#C)P (nha.n du.o.. c y),

la ham gia’m theo d. Va.y P (x|y) cu.. c da. i khi x la tu. ma gan vo.i y nhat. 2

D- i.nh nghıa 7.3.4. Khoa’ng cach (Hamming) cu’a bo. ma C, ky hie.u d(C), la khoa’ng cachnho’ nhat giu.a hai tu. ma khac nhau, tu.c la

d(C) := min{d(x, y) | x, y ∈ C, x 6= y}.

[n, k]-ma C vo.i khoa’ng cach d du.o.. c ky hie.u la [n, k, d]-ma.

172

Vı du. 7.3.4. (a) Vo.i C = {00000000, 11111000, 01010111, 10101111} thı d(C) = 5.

(b) Vo.i C = {000000, 111111}, thı d(C) = 6.

Khoa’ng cach Hamming xac di.nh kha’ nang phat hie.n va/hoa. c su.’ a sai cac loi.

D- i.nh ly 7.3.5. Ma C co the’ phat hie.n du.o.. c k loi neu va chı’ neu d(C) ≥ k + 1.

Chu.ng minh. ⇒ Bang pha’n chu.ng. Gia’ su.’ C co the’ phat hie.n k loi va d(C) ≤ k. Khi doton ta. i a, b ∈ C sao cho d(a, b) = d(C) ≤ k. Noi cach khac a va b chı’ khac nhau nhieu nhatk vi. trı. Do do se xuat hie.n k loi khi truyen tu. ma a va nha.n du.o.. c tu. ma b. Vı va.y ngu.o.inha.n khong the’ phat hie.n du.o.. c cac loi nay.

⇐ Gia’ su.’ d(C) ≥ k+1, va khi truyen tu. ma x ta nha.n du.o.. c y vo.i d(x, y) ≤ k. Do khoa’ngcach giu.a hai tu. ma ıt nhat la k + 1, thı tu. ma truyen pha’i la x. Vı va.y ngu.o.i nha.n co the’

phat hie.n du.o.. c cac loi nay. 2

Gia’ su.’ k ∈ N. Ta noi C co the’ su.’ a k loi neu vo.i mo.i thong bao nha.n du.o.. c y ∈ Bn ton ta. inhieu nhat mo.t tu. ma x sao cho d(x, y) ≤ k. D- ieu nay co nghıa rang, neu mo.t tu. ma du.o.. ctruyen va co nhieu nhat k loi thı gia’i ma theo lan ca.n gan nhat se thu du.o.. c dung mo.t tu.

ma du.o.. c truyen.

D- i.nh ly 7.3.6. Ma C co the’ su.’ a k loi neu va chı’ neu d(C) ≥ 2k + 1.

Chu.ng minh. ⇒ Gia’ su.’ C co the’ su.’ a du.o.. c k loi. Neu d(C) ≤ 2k thı ton ta. i hai tu. ma ava b khac nhau l vi. trı, vo.i l ≤ 2k. Thay do’i [l/2] bit trong a sao cho co vector c chı’ khacvector b dung [l/2] vi. trı. Khi do

d(a, c) = d(b, c) = [l/2].

Do do khong the’ su.’ a du.o.. c [l/2] ≤ k loi khi nha.n du.o.. c c, mau thuan!

⇐ Ngu.o.. c la. i gia’ su.’ d(C) ≥ 2k + 1. Gia’ su.’ tu. ma x du.o.. c truyen va nha.n du.o.. c vector zvo.i d(x, z) ≤ k. De thay neu y la tu. ma khac x thı d(z, y) ≥ k + 1, vı neu d(z, y) ≤ k ta seco

d(x, y) ≤ d(x, z) + d(z, y) ≤ k + k = 2k.

Mau thuan vo.i d(C) ≥ 2k + 1. D- ieu pha’i chu.ng minh. 2

Vı du. 7.3.5. D- a.t

C := {00000000, 11111000, 01010111, 10101111}.

Ta co d(C) = 5 va do do co the’phat hie.n du.o.. c 5−1 = 4 loi va co the’ su.’ a du.o.. c [(5−1)/2] = 2loi.

173

Co mo.t cach de dang de’ tım khoa’ng cach toi thie’u cu’a bo. ma. Tru.o.c het ta co khai nie.msau:

D- i.nh nghıa 7.3.7. Tro.ng lu.o.. ng Hamming, ky hie.u wt(x), cu’a vector x = x1x2 . . . xn la socac chı’ so i sao cho xi 6= 0.

Vı du. 7.3.6. wt(00000) = 0, wt(10111) = 4, wt(11111) = 5.

Bo’ de 7.3.8. Gia’ su.’ x, y la cac tu. ma cu’a ma tuyen tınh C. Khi do d(x, y) = wt(x− y).

Chu.ng minh. Cac vi. trı bang 1 trong vector x− y chınh la nhu.ng vi. trı ma hai vector x vay khac nhau. Do do d(x, y) = wt(x− y). 2

D- i.nh ly 7.3.9. Khoa’ng cach cu’a ma C bang tro. ng lu.o.. ng toi thie’u cu’a tu. ma khac khongtrong C.

Chu.ng minh. Gia’ su.’ d(C) = d thı ton ta. i x, y ∈ C, x 6= y, sao cho d(x, y) = d. Do do

wt(x− y) = d.

Nhu.ng C la ma tuyen tınh nen x − y ∈ C.

Ngu.o.. c la. i gia’ su.’ x ∈ C la tu. ma khac khong vo.i tro.ng lu.o.. ng toi thie’u. Do C la tuyentınh nen 0 ∈ C. Va.y

wt(x) = wt(x− 0) = d(x, 0) ≥ d(C).

2

Bai ta.p

1. Tım khoa’ng cach Hamming cu’a cac ca.p chuoi bit sau:

(a) 00000, 11111;

(b) 1010101, 0011100;

(c) 000000001, 111000000;

(d) 1111111111, 0100100011.

2. Co bao nhieu loi co the’ phat hie.n va bao nhieu loi co the’ su.’ a sai trong cac ma sau:

(a) {0000000, 1111111}.(b) {00000, 00111, 10101, 10010}.(c) {00000000, 11111000, 01100111, 10011111}.

3. Chu.ng minh rang neu khoa’ng cach toi thie’u giu.a cac tu. ma la bon, thı co the’ su.’ a saidung mo.t loi va phat hie.n sai ba loi.

174

4. Chu.ng minh rang mo.t ma co the’ su.’ a sai dong tho.i ≤ a loi va phat hie.n a + 1, . . . , bloi neu va chı’ neu no co khoa’ng cach toi thie’u ıt nhat a + b + 1.

5. Chu.ng minh neu mo.t ma co khoa’ng cach toi thie’u la d, tu. ma x du.o.. c truyen, khongco qua (d − 1)/2 loi xuat hie.n va y nha.n du.o.. c, thı

d(x, y) < d(y, z)

vo.i tat ca’ cac tu. ma z 6= x.

6. Chu.ng minh rang:wt(x + y) ≥ wt(x) −wt(y).

Dau bang xa’y ra neu va chı’ neu xi = 1 khi yi = 1.

7. Gia’ su.’ rang x va y la cac chuoi bit co do. dai n, va m la so cac vi. trı ma o.’ do ca’ x vay bang 1. Chu.ng minh rang

wt(x + y) = wt(x) + wt(y)− 2m.

8. Cho cac ma tra.n sinh

G1 :=

(1 1 1 1 00 0 1 1 1

), G2 :=

1 0 0 1 1 0 10 1 0 1 0 1 10 0 1 0 1 1 1

.

(a) Lie.t ke cac tu. ma tu.o.ng u.ng cac ma tra.n sinh tren.

(b) Tım khoa’ng cach toi thie’u cu’a cac bo. ma.

9. Tıch cu’a hai vector nhi. phan x va y la vector, ky hie.u x ∗ y, xac di.nh bo.’ i

x ∗ y = (x1y1, . . . , xnyn),

ma bang 1 ta. i vi. trı thu. i neu va chı’ neu xi = yi = 1. Chu.ng minh rang

(a) wt(x + y) = wt(x) + wt(y)− 2wt(x ∗ y).

(b) wt(x + z) + wt(y + z) + wt(x + y + z) ≥ 2wt(x + y + x ∗ y)−wt(z). Dau bang xa’yra neu va chı’ neu khong xa’y ra dong tho.i xi = 0, yi = 0, zi = 1.

10. Chu.ng minh rang trong mo.t ma nhi. phan tuyen tınh, hoa. c tat ca’ cac tu. ma co tro.nglu.o.. ng chan, hoa. c co chınh xac mo.t nu.’ a tro.ng lu.o.. ng chan va mo.t nu.’ a tro.ng lu.o.. ng le’.

11. Tınh khoa’ng cach cu’a ma En (gom tat ca’ vector do. dai n co tro.ng lu.o.. ng chan).

12. Chu.ng minh vo.i mo.i x, y ∈ Bn ta co:

[n∑

i=1

(xi − yi)2

]1/2

=√

d(x, y).

175

13. Gia’ su.’ x va y la cac vector nhi. phan vo.i d(x, y) = d. Chu.ng minh rang so cac vector zsao cho d(x, z) = r va d(y, z) = s la C(d, i)C(n− d, r − i), trong do i = (d + r − s)/2.Neu d + r − s le’ thı so nay bang 0, trong khi neu r + s = d, no bang C(d, r).


〈x, y〉 :=

n∑

i=1

xiyi = 0 (mod 2)

neu va chı’ neu wt(x ∗ y) chan va bang 1 neu va chı’ neu wt(x ∗ y) le’. Suy ra 〈x, x〉 = 0neu va chı’ neu wt(x) chan.

15. Gia’ su.’ u, v,w, x la bon vector doi mo.t co khoa’ng cach d (d pha’i la so chan).

(a) Chu.ng minh rang ton ta. i chınh xac mo.t vector ma khoa’ng cach den cac vectoru, v,w bang d/2.

(b) Chu.ng minh rang ton ta. i nhieu nhat mo.t vector ma khoa’ng cach den cac vectoru, v,w, x bang d/2.

16. Gia’ su.’ C la [n, k]-ma vo.i ma tra.n kie’m tra chan le’ H = (A In−k) va 1 ≤ t ≤ k. MaCt tu.o.ng u.ng ma tra.n kie’m tra chan le’ Ht = (At In−k) trong do At la ma tra.n cap(n − k) × (k − t) nha.n du.o.. c tu. A bang cach xoa di t co.t dau tien.

(a) Chu.ng minh Ct gom tat ca’ cac tu. ma cu’a C vo.i t to.a do. dau tien bang 0 bi. xoa.

(b) Chu.ng minh Ct la [n − t, k − t]-ma.

(c) Chu.ng minh d(Ct) ≥ d(C).

17. Vo.i moi n ∈ N, mieu ta’ ma C vo.i he. so k/n lo.n nhat va d(C) = 2. Ton ta. i duy nhatC?

18. Chu.ng minh rang hai ma tu.o.ng du.o.ng co cung khoa’ng cach.

19. Ky hie.u [n, k, d]-ma co nghıa [n, k]-ma vo.i do. dai d. Chu.ng minh rang neu ton ta. i[n, k, 2d]-ma thı ton ta. i ma vo.i cung tham so nhu.ng tat ca’ cac tu. ma co do. dai chan.

20. Chu.ng minh rang neu H la ma tra.n kie’m tra chan le’ cu’a ma C co do. dai n thı C cokhoa’ng cach toi thie’u d neu va chı’ neu mo.i ta.p gom d − 1 co.t cu’a H do. c la.p tuyentınh, nhu.ng ton ta. i ta.p gom d co.t phu. thuo.c tuyen tınh. Tu. do suy ra:

(a) Neu C la [n, k, d]-ma thı d ≤ n − k + 1.

(b) Khoa’ng cach toi thie’u cu’a ma co ma tra.n sinh:

I7

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1 1 0 01 0 1 00 1 1 01 1 1 11 1 0 10 1 0 11 0 0 1

.

176

21. (a) Chu.ng minh rang ton ta. i ma tuyen tınh gom M phan tu.’ , co do. dai n, nhieu nhatr bit kie’m tra chan le’, va khoa’ng cach toi thie’u d, neu

d−2∑

i=0

(M − 1)iC(n − 1, i) < M r.

(b) Chu.ng minh rang neu

2k

d−2∑

i=0

C(n − 1, i) < 2n.

thı ton ta. i ma tuyen tınh [n, k] vo.i khoa’ng cach toi thie’u d.

22. Gia’ su.’ C la [n, k, d]-ma C vo.i n < 2d. Chu.ng minh

2k(2k − 1)d ≤∑

x,y∈C

d(x, y) ≤ n22k−1.

23. Neu cach xay du.. ng [30, 11, 6]-ma? Bo. ma nay co bao nhieu tu. ma va kha’ nang phathie.n loi la bao nhieu?

24. Ky hie.u (n,M, d)-ma nghıa la [n, k, d]-ma, trong do M := 2k la so cac tu. ma. Xaydu.. ng, neu ton ta. i, cac (n,M, d)-ma vo.i cac tham so sau:

(6, 2, 6), (3, 8, 1), (4, 8, 2), (5, 3, 4), (8, 4, 5), (8, 30, 3).

(Neu khong ton ta. i, gia’i thıch ta. i sao).

25. (a) Gia’ su.’ d le’. Chu.ng minh ton ta. i (n,M, d)-ma neu va chı’neu ton ta. i (n+1,M, d+1)-ma.

(b) Chu.ng minh neu ton ta. i (n,M, d)-ma thı ton ta. i (n− 1,M ′, d)-ma vo.i M ′ ≥ M/2.

26. (To’ ho.. p hai ma) Gia’ su.’ G1, G2 la hai ma tra.n sinh tu.o.ng u.ng cac ma [n1, k, d1] va[n2, k, d2]. Chu.ng minh rang cac ma tra.n(

G1 00 G2

)

va (G1|G2) la cac ma tra.n sinh cu’a cac [n1+n2, 2k,min(d1, d2)]-ma va [n1+n2, 2k, d]-ma(d ≥ d1 + d2).

27. Vo.i x = x1x2 . . . xm ∈ Bm, y = y1y2 . . . yn ∈ Bn ta ky hie.u

(x, y) := x1x2 . . . xmy1y2 . . . yn ∈ Bm+n.

Gia’ su.’ C1 la (n,M1, d1)-ma va C2 la (n,M2, d2)-ma. D- a.t

C3 := {(x, x + y)|x ∈ C1, y ∈ C2}.Chu.ng minh C3 la (2n,M1M2, d)-ma tuyen tınh vo.i d = min{2d1, d2}.

28. Gia’ su.’ C := {x = x1x2 . . . xn ∈ Bn | x1 = x2 = · · · = xn}.(a) Chu.ng minh C la [n, 1, n]-ma.

(b) Chu.ng minh C⊥ la [n, n − 1, 2]-ma.

177

7.4 Ho. i chu.ng

D- i.nh nghıa 7.4.1. Gia’ su.’ C la [n, k]−ma tuyen tınh. Vo.i moi vector a ∈ Bn ta.p ho.. p

Ca := a + C = {a + x | x ∈ C}

du.o.. c go. i la coset (modulo hay ti.nh tien) cu’a C.

Nha.n xet 18. (a) Mo.i vector b ∈ Bn thuo. c mo. t coset nao do.

(b) Hai vector a va b thuo.c cung mo.t coset neu va chı’ neu (a − b) ∈ C.

(c) Moi coset chu.a 2k vector.

Me.nh de 7.4.2. Hai coset hoa. c ro.i nhau hoa. c trung nhau.

Chu.ng minh. Gia’ su.’ v ∈ (a + C) ∩ (b + C). Khi do ton ta. i x, y ∈ C sao cho

v = a + x = b + y.

Va.yb = a + x − y = a + x′,

trong do x′ = x − y ∈ C.

Suy rab + C ⊂ a + C.

Tu.o.ng tu..a + C ⊂ b + C.

2

Tu. Me.nh de 7.4.2 ta co the’ phan tıch Bn thanh ho.. p cac coset ro.i nhau cu’a C :

Bn = C ∪ (a1 + C) ∪ · · · ∪ (at + C), (7.3)

trong do t = 2n−k − 1, ai ∈ C, i = 1, 2, . . . , t, (ai + C) ∩ (aj + C) = ∅, i 6= j.

Gia’ su.’ ngu.o.i gia’i ma nha.n du.o.. c vector y. Khi do ton ta. i i sao cho

y = ai + x, x ∈ C.

Neu x′ la tu. ma truyen thı vector loi

e = y − x′ = ai + x− x′ = ai + x′′ ∈ ai + C,

trong do x′′ := x − x′ ∈ C. Noi cach khac vector loi chınh la vector trong coset chu.a y.

178

Do do quyet di.nh cu’a ngu.o.i gia’i ma la, neu nha.n du.o.. c vector y thı cho.n mo.t vector cotro.ng lu.o.. ng nho’ nhat e trong coset chu.a y va gia’i ma y la x = y − e. Vector tro.ng lu.o.. ngnho’ nhat trong coset du.o.. c go. i la coset leader (neu co ho.n mo.t vector vo.i tro.ng lu.o.. ng nho’

nhat, thı cho.n ngau nhien mo.t va go. i la coset leader).

Gia’ su.’ rang ai trong (7.3) la coset leader. Cach thong thu.o.ng de’ gia’i ma la su.’ du. ng ba’ngchua’n du.o.. c di.nh nghıa nhu. sau. Hang dau tien gom chınh bo. ma, vo.i tu. ma khong da. t bentrai:

x(1) = 0, x(2), . . . , x(s), s = 2k;

cac hang tiep theo la cac coset ai + C du.o.. c sap xep theo cung thu. tu.. vo.i coset leader da. tben trai:

ai + x(1), ai + x(2), . . . , ai + x(s).

Vı du. 7.4.1. [4, 2]−Ma vo.i ma tra.n sinh G =

(1 0 1 10 1 0 1

)co ba’ng chua’n

Ba’n tin 00 10 01 11 Ho. i chu.ng

Bo. ma C 0000 1011 0101 1110

(00

)

Coset a1 + C 1000 0011 1101 0110

(11

)

Coset a2 + C 0100 1111 0001 1010

(01

)

Coset a3 + C 0010 1001 0111 1100

(10

)

cosetleader

7.4.1 Gia’i ma dung ba’ng chua’n

Neu nha.n du.o.. c vector y, gia’ su.’ 1111, ta se tım du.o.. c vi. trı cu’a no trong ba’ng. Khi dovector loi e la coset leader nam o.’ vi. trı ben trai nhat cung hang vo.i y, trong tru.o.ng ho.. pnay e = 0100, va tu. ma du.o.. c truyen la:

x = y − e = 1011

nam tren dı’nh cu’a co.t chu.a y, ba’n tin tu.o.ng u.ng la 10.

Nha.n xet 19. (a) Gia’i ma dung ba’ng chua’n la gia’i ma ho.. p ly cu.. c da. i. D- e’ tım coset chu.ay, chung ta tım vector s := Hy ∈ Bn−k, du.o.. c go. i la ho. i chu.ng (syndrome) cu’a y.

(b) Neu y la tu. ma thı s = 0. Tha. t va.y neu y = x + e, s ∈ C, thı

s = Hy = Hx + He = He (7.4)

179

(c) Neu cac loi xuat hie.n ta. i cac vi. trı a, b, c . . . tu.c la

e = 00 . . . 01a00 . . . 00 . . . 01

b00 . . . 00 . . . 01

c00 . . .

thı tu. (7.4) ta co

s =∑

ejHj = Ha + Hb + Hc + · · ·

trong do Hj la vector tu.o.ng u.ng co.t thu. j cu’a ma tra.n H.Va.y

D- i.nh ly 7.4.3. Ho. i chu.ng cu’a vector y bang to’ng cac vector co. t Hj cu’a ma tra. n H, trongdo chı’ so j tu.o.ng u.ng vi. trı xuat hie.n loi .

Ho.n nu.a, hai vector cung mo.t coset cu’a C neu va chı’ neu chung co cung ho. i chu.ng.

Tha.t va.y u va v cung coset neu va chı’ neu (u − v) ∈ C; tu.c la H(u − v) = 0; hay tu.o.ngdu.o.ng Hu = Hv. Do do

D- i.nh ly 7.4.4. Ton ta. i tu.o.ng u.ng mo. t-mo. t len giu.a coset va ho. i chu.ng trong ma C.

Vı du. 7.4.2. Su.’ du.ng ma tra.n kie’m tra chan le’ trong Vı du. 7.2.7 de’ xac di.nh tu. ma du.o.. cgu.’ i neu nha.n du.o.. c thong bao 001111 (gia’ thiet co nhieu nhat mo.t loi xuat hie.n). Ta co

Hy =

0 1 1 1 0 01 0 1 0 1 01 1 0 0 0 1

001111

=

001

.

D- ay la co.t thu. sau cu’a H. Do do bit thu. sau cu’a 001111 la sai. Va.y tu. ma du.o.. c truyen la001110.

Bai ta.p

1. Gia’ su.’ C la [n, k]-ma va a ∈ Bn. Chu.ng minh coset Ca = C neu va chı’ neu a ∈ C. Tu.

do suy ra

(a) So phan tu.’ cu’a ta.p {x ∈ C|d(x, a) = i} bang Ai-so cac tu. ma co tro.ng lu.o.. ng i.

(b) So cac ca.p tu. ma (x, y) sao cho d(x, y) = i bang 2kAi.

2. Chu.ng minh rang neu C la ma tuyen tınh va a 6∈ C, thı C ∪Ca cung la ma tuyen tınh.

3. (a) Xay du.. ng ma’ng chua’n doi vo.i cac ma co cac ma tra.n sinh

G1 :=

(1 00 1

), G2 :=

(1 0 10 1 1

), G3 :=

(1 0 1 1 00 1 0 1 1

).

(b) Su.’ du. ng ma’ng chua’n thu. ba de’ gia’i ma cac vector 11111 va 01011.

(c) Cho cac vı du. : Hai loi xuat hie.n trong tu. ma va su.’ a dung; hai loi xuat hie.n trongtu. ma va su.’ a khong dung.

180

4. Gia’ su.’ C la [4, 2]-ma vo.i ma tra.n sinh

G :=

(1 0 0 10 1 0 1

).

(a) Tım cac tu. ma.

(b) Tım cac coset, coset leader cu’a C.

(c) Xay du.. ng ma’ng chua’n. Tu. do gia’i ma khi nha.n du.o.. c cac vector 0011, 0001, 0100.

5. Xay du.. ng ma’ng chua’n doi vo.i ma co ma tra.n kie’m tra chan le’

H :=

0 1 1 1 0 01 0 1 0 1 01 1 0 0 0 1

.

Su.’ du.ng ma’ng nay de’ gia’i ma cac vector 110100 va 111111.

6. (a) Xay du.. ng ma’ng chua’n doi vo.i ma co ma tra.n sinh

G :=

(1 0 1 00 1 1 1

).

(b) Tım vector ho. i chu.ng cu’a y = 1111. Tu. do gia’i ma.

7. Gia’ su.’ [7, 4]-ma co ma tra.n kie’m tra chan le’

H :=

1 1 0 0 1 0 00 0 1 1 0 1 01 1 1 1 0 0 1

.

Xay du.. ng ma’ng chua’n. Tu. do gia’i ma cac vector nha.n du.o.. c: 1111111, 1101011, 0110111va 0111000.

8. Neu C ⊂ C⊥, ta noi rang C la tu.. doi ngau yeu, viet tat w.s.d (weakly self dual). C latu.. doi ngau neu C = C⊥. Vı du. ma la.p [n, 1, n] la w.s.d neu va chı’ neu n chan. Khin = 2, ma la.p {00, 11} la tu.. doi ngau. Chu.ng minh rang

(a) C la w.s.d neu 〈x, y〉 = 0, vo.i mo.i x, y ∈ C.

(b) C tu.. doi ngau neu no la w.s.d va co chieu k = n/2 (do do n chan).

9. Xay du.. ng cac ma tu.. doi ngau co do. dai 4 va 8.

10. Gia’ su.’ n chan va C la [n, (n− 1)/2] ma w.s.d. Chu.ng minh rang C⊥ = C ∪Ca, trongdo a la vector co tat ca’ cac to.a do. bang 1.

11. Chu.ng minh rang ma vo.i ma tra.n kie’m tra chan le’ H = (A I) tu.. doi ngau neu va chı’

neu A la ma tra.n vuong sao cho AAt = I.

181

12. Gia’ su.’ C la ma w.s.d. Chu.ng minh rang mo.i tu. ma co tro.ng lu.o.. ng chan. Ho.n nu.a,neu moi hang cu’a ma tra.n sinh cu’a C co tro.ng lu.o.. ng chia het cho 4 thı mo.i tu. macung co tro.ng lu.o.. ng chia het cho 4.

13. Gia’ su.’ [8, 4, 4]-ma C co ma tra.n kie’m tra chan le’

H :=

1 1 1 1 1 1 1 10 0 0 1 1 1 1 00 1 1 0 0 1 1 01 0 1 0 1 0 1 0

.

Chu.ng minh C la ma tu.. doi ngau.

7.5 Ma hoan ha’o

D- e’ co the’ su.’ a cac loi xuat hie.n khi truyen du. lie.u, chung ta can xay du.. ng bo. ma C cokhoa’ng cach d(C) lo.n. Nhu.ng dieu do se lam gio.i ha.n so lu.o.. ng tu. ma trong bo. ma. Phannay se chı’ ra moi lien he. giu.a d(C) va so phan tu.’ cu’a ta.p ho.. p C.

Bo’ de 7.5.1. Gia’ su.’ x ∈ Bn, 0 ≤ k ≤ n. Khi do

#{y ∈ Bn | d(x, y) ≤ k} = C(n, 0) + C(n, 1) + · · · + C(n, k).

Chu.ng minh. Vo.i moi i ∈ {0, 1, . . . , n} co di.nh, ta co

#{y ∈ Bn | d(x, y) = i}

bang so cac cach cho.n i vi. trı sao cho x va y khac nhau ta. i cac vi. trı do va bo.’ i va.y bangC(n, i). 2

Bo’ de 7.5.2. Gia’ su.’ C la bo. ma gom cac tu. ma co do. dai n va

d(C) ≥ 2k + 1.

Khi do vo.i moi y ∈ Bn, ton ta. i nhieu nhat mo. t phan tu.’ x ∈ C, sao cho

y ∈ B(x, k) := {z ∈ Bn | d(x, z) ≤ k}.

Chu.ng minh. Gia’ su.’ y ∈ B(x, k) ∩ B(x′, k), x, x′ ∈ C (x 6= x′). Khi do

d(x, x′) ≤ d(x, y) + d(x′, y) ≤ 2k.

Mau thuan vo.i gia’ thiet. 2

182

D- i.nh ly 7.5.3. Gia’ su.’ C la bo. ma gom cac tu. ma do. dai n va d(C) ≥ 2k + 1. Thı

#C ≤ 2n

[C(n, 0) + C(n, 1) + · · · + C(n, k)].

Chu.ng minh. Du.. a tren cac nha.n xet sau

+ Co 2n vector do. dai n (do #Bn = 2n).

+ Moi qua’ cau B(x, k), x ∈ C, chu.a

C(n, 0) + C(n, 1) + · · · + C(n, k)

vector (xem Bo’ de 7.5.1).

+ Vo.i moi x, x′ ∈ C, x 6= x′, thı B(x, k) ∩ B(x′, k) = ∅. 2

Vı du. 7.5.1. Neu C co do. dai 7 va d(C) = 3 thı

#C ≤ 27

[C(7, 0) + C(7, 1)]= 128/8 = 16.

D- i.nh nghıa 7.5.4. Ma hoan ha’o (perfect code) la bo. ma C sao cho d(C) = 2k + 1 va

#C =2n

[C(n, 0) + C(n, 1) + · · · + C(n, k)].

Noi cach khac, ma hoan ha’o la ma co so phan tu.’ nhieu nhat trong tat ca’ cac ma co khoa’ngcach 2k + 1 cho tru.o.c.

Vı du. 7.5.2. Ma gom hai tu. ma 00000 va 11111 la ma hoan ha’o.

Bai ta.p

1. Tım so cu.. c da. i cac tu. ma trong mo.t bo. ma ma cac tu. ma la chuoi cac bit co do. daichın va khoa’ng cach toi thie’u giu.a cac tu. ma la nam.

2. Chu.ng minh rang neu n la so tu.. nhien le’, thı ma gom hai tu. ma co do. dai n gom toanso 0 va 1 la mo.t ma hoan ha’o.

3. Chu.ng minh rang neu C la ma hoan ha’o khong tam thu.o.ng vo.i khoa’ng cach toi thie’u7 thı do. dai tu. ma la 23.

183

4. Gia’ su.’ G24 co ma tra.n sinh G = (I12 A) trong do

A :=

0 1 1 1 1 1 1 1 1 1 1 11 1 1 0 1 1 1 0 0 0 1 01 1 0 1 1 1 0 0 0 1 0 11 0 1 1 1 0 0 0 1 0 1 11 1 1 1 0 0 0 1 0 1 1 01 1 1 0 0 0 1 0 1 1 0 11 1 0 0 0 1 0 1 1 0 1 11 0 0 0 1 0 1 1 0 1 1 11 0 0 1 0 1 1 0 1 1 1 01 0 1 0 1 1 0 1 1 1 0 01 1 0 1 1 0 1 1 1 0 0 01 0 1 1 0 1 1 1 0 0 0 1

.

(a) Chu.ng minh G24 tu.. doi ngau, tu.c la: G⊥24 = G24.

(b) Chu.ng minh (A I12) cung la ma tra.n sinh cu’a G24.

(c) Chu.ng minh mo.i tu. ma cu’a G24 co tro.ng lu.o.. ng chia het cho 4.

(d) Chu.ng minh G24 khong co tu. ma vo.i tro.ng lu.o.. ng 4.

(e) Chu.ng minh G24 la [24, 12, 8]-ma (go. i la ma Golay).

(f) Gia’ su.’ G23 nha.n du.o.. c tu. G24 bang cach bo’ tat ca’ cac to.a do. cuoi trong cac tu. ma.Suy ra cac tham so cu’a ma G23 va do do G23 la ma hoan ha’o.

5. Gia’ su.’ x, y ∈ Bn. Ta noi vector x phu’ vector y neu x ∗ y = y. Cha’ng ha.n, 111001 phu’

101000.

(a) Chu.ng minh rang neu vector y ∈ B23 tro.ng lu.o.. ng 4 thı ton ta. i duy nhat tu. max ∈ G23 phu’ y.

(b) Suy ra so cac tu. ma tro.ng lu.o.. ng 7 trong G23 la 253.

6. Chu.ng minh khong ton ta. i [90, 278, 5]-ma tuyen tınh.

7. Chu.ng minh khong ton ta. i [13, 64, 5]-ma tuyen tınh. (HD. Gia’ su.’ C la [13, 6, 5]-matu.o.ng u.ng ma tra.n sinh

[1 1 1 1 1

G1

∣∣∣∣0 0 0 0 0 0 0 0

G1

].

Chu.ng minh G2 sinh ra [8, 5, 3]-ma, mau thuan vo.i D- i.nh ly 7.5.3).

7.6 Ma Hamming

Phan nay nghien cu.u cac ma Hamming la mo.t trong nhu.ng bo. ma co the’ de dang ma hoava gia’i ma. D- ay la bo. ma co the’ su.’ a sai mo.t loi. Theo D- i.nh ly 7.4.3, ho. i chu.ng cu’a vector

184

nha.n du.o.. c bang to’ng cac co.t cu’a ma tra.n kie’m tra chan le’ H u.ng vo.i loi xuat hie.n. Do dode’ xay du.. ng bo. ma su.’ a sai mo.t loi, chung ta pha’i co (ta. i sao?) cac co.t cu’a H khac khongva doi mo.t khac nhau. Neu H co r hang thı se co 2r − 1 vector co.t co do. dai r tho’a gia’

thiet tren.

Vı du. 7.6.1. Neu r = 3 thı co 23 − 1 = 7 co.t

1 2 3 4 5 6 7

0 0 0 1 1 1 10 1 1 0 0 1 11 0 1 0 1 0 1

la bie’u dien nhi. phan cu’a cac so tu. 1 den 7.

D- i.nh nghıa 7.6.1. Ma Hamming ba. c r la ma co ma tra.n kie’m tra chan le’ H cap r×(2r−1)sao cho cac co.t cu’a H khac khong va doi mo.t khac nhau.

Vı du. 7.6.2. Ma tra.n H cu’a ma Hamming ba.c 2 co da.ng

H =

(1 1 01 0 1

).

Bo. ma nay co hai tu. ma la 000 va 111. D- ay la ma la.p tuyen tınh ba.c 3.

Vı du. 7.6.3. Ma tra.n H cu’a ma Hamming ba.c 3 co da.ng

H =

0 1 1 1 1 0 01 0 1 1 0 1 01 1 0 1 0 0 1

.

Bo. ma nay co 16 tu. ma.

Bo’ de 7.6.2. Ma Hamming ba. c r chu.a 2n−r tu. ma vo.i n = 2r − 1.

Chu.ng minh. Hie’n nhien theo di.nh nghıa. 2

Bo’ de 7.6.3. Khoa’ng cach toi thie’u cu’a ma Hamming ba. c r bang 3.

Chu.ng minh. Vı ma tra.n kie’m tra chan le’ H co cac co.t khac 0 va khong co hai co.tnao giong nhau nen ma Hamming ba. c r co the’ su.’ a sai mo.t loi. Theo D- i.nh ly 7.3.6 ta cod(C) ≥ 3. Trong so cac co.t cu’a ma tra.n H co ba co.t sau

C1 =

1100...0

, C2 =

1000...0

, C3 =

0100...0

.

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Chu y rang

C1 + C2 + C3 = 0 (mod 2).

D- a.t x la vector bang 1 o.’ vi. trı cu’a cac co.t nay va bang 0 neu ngu.o.. c la. i. Khi do Hx = 0.Noi cach khac, x la tu. ma. Nhu.ng wt(x) = 3. Do do, theo D- i.nh ly 7.3.9 thı

d(C) ≤ wt(x) = 3.

2

D- i.nh ly 7.6.4. Ma Hamming ba. c r la ma hoan ha’o.

Chu.ng minh. D- a.t n := 2r − 1. Theo Bo’ de 7.6.2 thı #C = 2n−r. Theo Bo’ de 4.6.3 thıd(C) = 3. Va.y

2n−r[C(n, 0) + C(n, 1)] = 2n−r(1 + n) = 2n−r(1 + 2r − 1) = 2n.

2

D- i.nh ly 7.6.4 chı’ ra rang ma Hamming la ma hoan ha’o. Nghien cu.u ma hoan ha’o la mo.ttrong nhu.ng lınh vu.. c quan tro.ng nhat cu’a ly thuyet ma va da co nhu.ng ket qua’ nhat di.nh.

Bai ta.p

1. [7, 4]−ma C co ma tra.n kie’m tra chan le’

H :=

1 1 1 0 1 0 01 1 0 1 0 1 01 0 1 1 0 0 1

.

(a) Ma hoa thong bao gom hai ba’n tin u = 0000 1101.

(b) Gia’i ma khi nha.n du.o.. c chuoi bit 0000111 0001110 (gia’ su.’ co nhieu nhat mo.t loisai).

(c) Tım cac tham so n, k, d cu’a C.

2. Gia’ su.’ [7, 4, 3]-ma Hamming co ma tra.n kie’m tra chan le’

H :=

0 1 1 1 1 0 01 0 1 1 0 1 01 1 0 1 0 0 1

.

(a) Tım ma tra.n sinh da.ng chua’n.

(b) Gia’i ma vector nha.n du.o.. c y = 1010110 (gia’ thiet co nhieu nhat mo.t loi sai).

186

3. Tım mo.t ma tra.n kie’m tra chan le’ cu’a ma Hamming ba. c 4. Gia’i ma cac vector nha.ndu.o.. c (gia’ thiet co nhieu nhat mo.t loi sai):

(a) 100 000 000 000 000.

(b) 111 111 111 111 111.

4. Chu.ng minh rang cac ma tra.n

H :=

0 0 0 1 1 1 10 1 1 0 0 1 11 0 1 0 1 0 1

, H ′ :=

0 1 1 1 1 0 01 0 1 1 0 1 01 1 0 1 0 0 1

sinh ra cung mo.t ma Hamming ba. c 3.

5. Lie.t ke ba da.ng cu’a ma tra.n kie’m tra chan le’ H cu’a ma Hamming ba.c 4. VietH du.o.i da.ng (A Ir). Tu. do ma hoa ba’n tin u = 11111100000, va gia’i ma vector111000111000111.

6. Chu.ng minh rang ma Hamming ba. c r du.o.. c xac di.nh duy nhat theo nghıa: Bat ky mavo.i cac tham so [2r − 1, 2r − 1 − r, 3] tu.o.ng du.o.ng vo.i ma Hamming ba. c r.

7. Tım ma tra.n sinh G cu’a ma Hamming ba. c r va su.’ du.ng no de’ chu.ng minh mo.i tu.

ma khac khong co tro.ng lu.o.. ng ≥ 3. (HD. Neu ton ta. i tu. ma co tro.ng lu.o.. ng ≤ 2 thıno pha’i la to’ng cu’a nhieu nhat hai hang cu’a G).

8. Tım ma tra.n kie’m tra chan le’ cu’a [15, 11, 3]-ma Hamming. Gia’i thıch cach gia’i maneu co dung mo.t loi xuat hie.n. Neu co ho.n hai loi thı sao?

9. Chu.ng minh so cac ma Hamming khac nhau co do. dai n = 2r − 1 la

(2r − 1)!∏m−1i=0 (2m − 2i)

.

187

Tai lie.u tham kha’o

[1] S. Arlinghaus, W. Arlinghaus, J. Nystuen, The Hedetniemi matrix sum: an algorithmfor shortest path and shortest distance, Geographical Analysis, Vol. 22, No. 4, Oct.,351-360 (1990).

[2] C. Berge, Ly thuyet do thi. va u.ng du. ng, NXB Khoa ho.c va ky thua.t Ha No. i, 1971.

[3] A. Cayley, Collected papers, Quart. Jl. of Mathematics, 13 Cambridge, 26 (1897).

[4] N. Biggs, Discrete mathematic, Clarendon Press Oxford, 1989.

[5] N. Christofides, Graph theory an algorithmic approach, Academic Press INC. (1975).

[6] E. W. Dijkstra, A note on two problems in connection with graphs, Numerische Math-ematik, 1, 269 (1959).

[7] P. J. Cameron, Combinatorics: topics, techniques, algorithms, Cambridge UniversityPress, 1994.

[8] N. Deo, Graph theory with applications to engineering and computer science, Prentice-Hall Inc., 1974.

[9] R. J. MC Eliece, M. Kac, The theory of information and coding, Addison-Wesley, 1977.

[10] R. W. Floyd, Algorithm 97-Shortest path, Comm. of ACM, 5, 345 (1962).

[11] C. M. Goldie, R. G. E. Pinch, Communication theory, Cambridge University Press,1991.

[12] M. Gondran, M. Minoux, S. Vajda, Graphs and algorithms, John Wiley & Sons (1990).

[13] R. W. Hamming, Coding and information theory, Prentice Hall, 1980.

[14] R. Hill, A first course in coding theory, Clarendon Press Oxford, 1985.

[15] T. C. Hu, Integer programming and network flows, Addison-Wesley, Reading, Mas-sachusetts (1969).

[16] R. Johnsonbaugh, An introduction to discrete mathematic, Macmillan Publishing Com-pany, 1992.

189

[17] A. R. Kenneth, C. R. B. Wright, Discrete mathematics, Prentice-Hall InternationalEditions, 1978.

[18] V. Kevin, M. Whitney, Algorithm 422-Minimum spanning tree, Comm. of ACM, 15,273 (1972).

[19] G. Kirchhoff, in “Annalen der Physik and Chemie” 72, 497 (1847).

[20] S. Lipschutz, Essential computer mathematic, McGraw-Hill, 1992.

[21] S. Lipschutz, M. L. Lipson, 2000 sloved problems in discrete mathematics, McGraw-Hill,1992.

[22] C. L. Liu, Introduction to combinationnal mathematic, McGraw-Hill, 1985.

[23] F. J. MacWilliams, N. J. A. Soane, The theory of error-correcting codes, North-Holland,1981.

[24] A. A. Michael, A. J. Kfoury, R. N. Moll, D. Gries, A basis for theoretical computerscience, Springer-Verlag NewYork Inc., 1981.

[25] J. G. Michaels, K. H. Rosen, Applications of discrete mathematics, McGraw-Hill, 1991.

[26] J. D. Murchland, A new method for finding all elementary paths in a complete directedgraph, London School of Economics, Report LSE-TNT-22 (1965).

[27] R. C. Prim, Shortest connection networks and some generalizations, Bell Syst. Tech.Jl., 36, 1389 (1957).

[28] S. Roman, An introduction to discrete mathematic, Saunders College, 1982.

[29] K. H. Rosen, Discrete mathematics and its applications, McGraw-Hill, 1995.

[30] B. M. Stephen, A. Ralston, Discrete algorithmic mathematics, Addision-Wesley Pub-lishing Company, 1991.

[31] D. Welsh, Codes and cryptography, Clarendon Press Oxford, 1987.

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