Phan5

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2. Ph thuc hm

Gii thiu

3. nh ngha 4. Biu din PTH bng th 5. Suy din logi cc PTH 6. H tin Amstrong 7. Bao ng 8. Bao ng ca tp thuc tnh 9. Kha Thut ton tm kha 10. Ph ti thiu 11. Gii thiu

L s biu din RBTV di dng hnh thc ton hc.

12. Bo m thng tin khng b tn tht khi phn r hoc kt ni gia cc quan h. 13. nh ngha

Quan h R c nh ngha trn tp thuc tnhU = { A1, A2, ..., An}.

14. A, B U l 2 tp con ca tp thuc tnh U. 15. Nu tn ti mt nh x f: A -> B th ni rng Axc nh hm B, hay B ph thuc hm vo A. 16. K hiu: A -> B. 17. nh ngha

nh ngha hnh thc ca PTH:

18. Quan h Q (A, B, C) c PTHA xc nh B 19. (k hiu l A -> B) nu: 20. q, q Q, sao cho q.A = q.A th q.B = q.B 21. Ngha l:ng vi 1 gi tr ca A th c mt gi tr duy nht ca B 22. A l v tri ca PTH, B l v phi ca PTH 23. PTH A -> A c gi l PTH hin nhin. 24. nh ngha

V d 1: Trong quan h Sinhvien ( Masv , Hoten,Phai, NgSinh, Quequan, Diachi)

C cc PTH sau:

Masv->Quequan, Diachi

25. Masv, Hoten->Ngsinh, Quequan

Khng c cc PTH sau:

Hoten->Ngsinh, Quequan

Tc l xc nh sinh vin th ti thiu phi c m sinh vin nn khng tn ti ph thuc hm suy ra t thuc tnh Hoten.

26. nh ngha

V d 2: Trong QuanH CHITIT_H ( S-han, M-hng , S-lng, n-gi, Tr-gi)

27. C cc PTH sau:

f1: S-ha-n, M-hng-> S-lng

28. f2: S-ha-n, M-hng -> n-gi.

29. f3: S-ha-n, M-hng-> Tr-gi. 30. f4: S-lng, n-gi -> Tr-gi. 31. Biu din PTH bng th(1/3)

PTH c th biu din bng th c hng:

Cc nt trong th chia thnh 2 loi:

Nt thuc tnh: biu din bng tn thuc tnh

32. Nt PTH: biu din bng hnh trn c s th t ca PTH.

Cc cung trong th cng c 2 loi:

Cung n PTH: xut pht t cc thuc tnh v trica cc PTH

33. Cung ri PTH: hng n cc thuc tnh v phica cc PTH

34. Biu din PTH bng th(2/3)

Quan h R (A, B, C, D, E, H)

35. Tp ph thuc hm F = {AB->C, CD->E, EC->A, 36. CD->H, H->B } TH022_11 37. Biu din PTH bng th(3/3)

Quan h R (A, B, C, D, E, G)

38. Tp ph thuc hm F = {A->C; B->DE; AB->G; 39. A->ED; D->E } TH022_11 40. Suy din logic cc PTH(1/2)

Cho lc quan h R vi tp thuc tnh U vtp cc PTH F.

41. X -> Y l 1 PTH; X,Y U. 42. Ta ni rng X -> Y c suy din lgic t F nu

r R, nu r tha tt c cc PTH trong F th r cng tha X -> Y

K hiu l: F = X -> Y.

43. Suy din logic cc PTH(2/2)

Vi F = {X -> Y, X -> Z, Y -> T }

44. Th ta c cc PTH:

X -> YZ

45. X -> T

46. H tin Amstrong(1/4)

Nm 1974, Amstrong a ra h tin (gi l h lut dn Amstrong) : Cho lc quan h Qvi tp thuc tnh U. X, Y, Z, W U. PTH c cctnh cht c bn sau:

A1: Tnh phn x:

Nu Y X th X -> Y

A2: Tnh tng trng:

Nu X -> Y th XZ -> YZ (Z U)

A3: Tnh bc cu:

Nu X -> Y v Y -> Z th X -> Z

V d 1: Cho F = {AB->C, C->A }

CMR: BC->ABC

(1) C->A (gi thit)

48. (2) BC->AB (tng trng 1)

49. (3) AB->C (gi thit) 50. (4) AB-> ABC (tng trng 3) 51. (5) BC->ABC (bc cu 2 & 4) 52. H tin Amstrong(3/4)

Cc tnh cht b sung:

A4: Lut gi bc cu:

Nu X -> Y v YZ -> W th XZ -> W

A5: Lut hp:

Nu X -> Y v X -> Z th X -> YZ

A6: Lut tch:

Nu X -> YZ th X -> Y v X -> Z

V d: Cho R(A,B,C,D,E,G,H). CMR: AB->E viF = {AB->C, B->D, CD->E, CE->GH, G->A }.

54. Ta c:

(1) AB->C (chotrc)

55. (2) AB->AB (phn x)

56. (3) AB->B (lut tch) 57. (4) B->D (cho trc) 58. (5) AB->D (bc cu 3 & 4) 59. (6) AB->CD (hp 1 & 5) 60. (7) CD->E (cho trc) 61. (8) AB->E (bc cu 6 & 7) 62. Bao ng (Closure)(1/3)

Gi F+ l bao ng (Closure) ca F, tc l tp cc PTH c suy din lgic t F.

63. Nu F = F+ th ta ni F l h y (full family)ca cc PTH. 64. Bao ng (Closure)(2/3)

Bi ton thnh vin (MemberShip):

65. Kim tra PTH X -> Y c c suy din lgc t F khng? (tc l X -> Y F+ ? )

y l mt bi ton kh gii.

66. i hi phi c mt h lut dn suy din lgiccc PTH.

67. Bao ng (Closure)(3/3)

Bi ton thnh vin V d: Cho Q(ABCDEG).

F = {AE -> C, CG -> A, BD -> G, GA -> E }

68. CMR: BDC -> Q+ F+ (Q+ = ABCDEG+)

(1) BDC->BDC (phn x)

69. (2) BD->G (gi thit f3)

70. (3) CG->A (gi thit f2) 71. (4) BDC->A (gi bc cu 2,3) 72. (5) BDC->GA (hp 2 & 4) 73. (6) BDC->E (bc cu 5 & f4) 74. (7) BDC->Q+ (hp 1,2,4,6) 75. Bao ng ca tp thuc tnh(1/5)

Thut ton Tm bao ng ca tp thuc tnh

76. Input: Tp U hu hn cc thuc tnh & tp cc 77. PTH F trn U & X U. 78. Output: X + F 79. Phng php: Tnh lin tip X 0 , X 1 , X 2 , theo quy tc nh sau: 80. Bao ng ca tp thuc tnh(2/5)

Thut ton Tm bao ng ca tp thuc tnh

81. Bc 1. X 0= X 82. Bc 2. X i+1= X i A sao cho (Y->Z ) F,m A Z v Y Xi 83. Bc 3. Cho n khi X i+1= Xi(V X= X 0 X 1 X 2 U, m U hu hncho nn s tn ti 1 ch s i no m X i+1= X i ) 84. Khi X + F= X i 85. Bao ng ca tp thuc tnh(3/5)

Tm bao ng ca tp thuc tnh V d 1

86. Cho R(U) vi U=ABCDEG 87. F = {AB -> C, C -> A, BC -> D, ACD -> B, 88. D -> EG, BE -> C, CG -> BD, CE -> AG } 89. Tnh X + F, vi: 90. X= D 91. X= BD 92. Bao ng ca tp thuc tnh(4/5)

Vy D += DEG

93. Bao ng ca tp thuc tnh(5/5)

Vy BD += ABCDEG

94. Kha Thut ton tm kha

R l lc quan h nh ngha trn tp ccthuc tnh U = { A 1 , A 2 , ... , A n}

95. Tp cc PTH F = { f 1 , f 2 , ..., f m} xc nh trn R. 96. K U l kha ca R nu tha mn hai iu kin sau y:

K -> U. ( K l siu kha )

97. K K m K -> U.

98. Bi Ton Tm kha(1/6)

Xc nh tt c cc kha ca 1 lc quan h.

99. Bi ton ny c gii quyt qua 2 giai on:

Giai on 1 : Xy dng tp S cha tt c cc siukha ca R

100. Giai on 2 : Xy dng tp K cha tt c cc khaca R t tp S bng cch loi b khi S nhng siu kha khng ti thiu.

101. Bi Ton Tm kha(2/6)

xc nh tt c cc siu kha ca 1 lc quan h R, ta ln lt xt (2 n -1) tp hp con ca

R +: X 1 , X 2 ,

Nu 1 tp con X ica R +c bao ng bng ngR +th tp con X ichnh l 1 siu kha.

102. Nu R ch c 1 siu kha S th siu kha cng l kha ca lc quan h R 103. Bi Ton Tm kha(3/6)

Trong trng hp R c nhiu hn 1 siu kha (hu hn), xc nh tt c cc kha ch nh, ta so snh 1 cp siu kha S iv S j . Nu S i S j , ta loi S jv gi li S i .

104. Ln lt so snh tng cp siu kha loi b tp ln, cui cng thu c tp cc kha ch nhca R. 105. Suy raThut ton khng kh thi khi n ln. 106. Bi ton tm kha(4/6)

Bi Ton Tm kha Thut ton ci tin:

Chng ta s ci tin thut ton da trn vic phn loi tp thuc tnh R +

A gi l thuc tnh ngun nu A khng xut hin v phi ca bt k PTH khng hin nhin no ca F.

107. Tp cc thuc tnh ngun k hiu l N

108. A gi l thuc tnh ch nu A khng phi thuctnh ngun v A khng xut hin v tri ca bt k PTH khng hin nhin no ca F.

109. Tp cc thuc tnh ch k hiu l D. 110. Bi ton tm kha(5/6)

Tp hp cc thuc tnh khng phi ngun vkhng phi ch gi l tp trung gian. K hiu l L

111. Cc tp hp N, D, L ri nhau tng i mt vN D L = R +

Nhn xt

Nu K l kha ca R th K cha tt c cc thuctnh ngun v khng cha bt k thuc tnh chno.

112. Bi ton tm kha(6/6)

B1: Xy dng 2v tp con ca L: L1, L2, bngphng php ng chy nh phn.

113. B2: Xy dng tp K cha cc siu kha

114. L i , X i= N L i

115. Tnh X i + F. Nu X i + F= R +th K = K X i

B3: Loi b dn cc siu kha ln

116. V d thut ton ci tin(1/3)

V d: Cho R(ABCDEG) vi tp PTH F = { AE ->C, CG ->A, BD ->G, GA ->E }Xc nh tt c cc kha ca R

117. Ta c:

N = { B, D }

118. D =

119. L = { A, C, E, G } 120. Xy dng tp thuc tnh L ibngphng phpng chynh phn. 121. V d thut ton ci tin(2/3) 122. V d thut ton ci tin(3/3)

Vy c 2 siu kha:

123. BDA

V 2 kha:

124. BDA

125. Ph ti thiu(1/4)

PTH tng ng

Gi F v G l cc tp PTH. Ta ni rng F v Gl tng ng nu F += G + .

126. Nu F v G tng ng, i khi cn niF ph G (hay G ph F).

Ph ti thiu: Gi F l ti thiu nu

Mi v phi ca 1 PTH F ch c 1 thuc tnh

128. Khng tn ti 1 PTH X -> A thuc F, m:

F += (F - {X -> A} ) +

Khng tn ti 1 PTH X -> A thuc F, v 1 tp con Z ca X m:

F += (F - {X -> A} {Z -> A} ) +

129. V d - Ph ti thiu(3/4)

V d 1:

Cho F = { AB ->C, C->A, BC ->D, ACD -> B,D ->E, D ->G, BE ->C, CG ->B, CG ->D, CE->A, CE->G }

130. Tm Ph ti thiu .

131. Ta c:

CE->A l d tha (v C->A )

132. CG->B l d tha (v CG->D, C->A v ACD->B)

133. ACD ->B thay bng CD ->B v C ->A

Suy ra PTT = { AB ->C, C->A, BC ->D, CD -> B, D ->E, D ->G, BE ->C, CG ->D, CE ->G }

V d 2:

Cho F = { AB -> C, C->A, BC ->D, ACD ->B,D -> E, D -> G, BE ->C, CG ->B, CG -> D, CE->A, CE->G }

Tm Ph ti thiu .

Loi b CE ->A, CG ->D v ACD ->B s c tp ti thiu:

135. Suy ra PTT = { AB ->C, C->A, BC ->D, D ->E, D ->G, BE ->C, CG ->B, CE ->G }

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