알고리즘 설계 및 분석 Foundations of Algorithm 유관우. Digital Media Lab. 2 Chap. 5...

알고리즘 설계 및 분석Foundations of Algorithm유관우

Digital Media Lab. 2

Chap. 5 Backtracking(Eg) 큰 성 (castle)- 수 백 개의 방 (rooms) 그리고 “ Sleeping Beauty” Systematic search - BFS, DFS 갈필요 없는 길 – 미리 알 수 있다면… (Bounding function or promising function) DFS with Bounding function (promising function) Worst-case : exhaustive search But, 대부분의 겨우 : 몹시 효율적 NP-Complete problems (Eg) 0-1 knapsack problem D.P. : O(min(2n, nW)) Backtracking : can be very efficient if good bounding function (promising function). Recursion


Backtracking : DFS of a tree except that nodes are visited if promising

Promising function DFS (Depth First Search)

Procedure d_f_s_tree( v: node)var u : node{ visit v; // some action. // for each child u of v do d_f_s_tree(u);}

12 3

4 5 6

7 8

9

DFS : 1 2 4 7 9 8 6 3 5

BFS : 1 2 3 4 5 6 7 8 9

1

2

3 5 7 114

6

8 9 10


n-Queens Problem nⅹn Chess board n Queens (e.g.) 8-Queens problem

Promising function( Bounding fcn) (i) same row ⅹ (ii) same column ⅹ (col(i) ≠ col(k)) (iii) same diagonal ⅹ (|col(i)-col(k)|≠|i-k|)

, ,)(2

2

nn

n n

,! , nnn

QQ

QQ

1 2 3 4 5 6 7 8

1 2 3 4 5 6 7 8

Q (i, col

(i)) Q (k, col

(k))

Q (i,col(i))

Q (k,col(k))


#leaf nodes : n!=4! DFS : “Backtrack “ at dead ends – 4! Leaf nodes (n!) Backtracking : “Backtrack “ if non-promising (prom. fcn.) (pruning)

X1=1

X2=2 3 4

43

4 3 4 2 3 2 3 2

2 4 32 3 4 1 4 1 3 2 4

1 3 4 1 2 4 321

2 3 4

4-Queen 문제의 state-space tree

(x1, x2, x3, x4)=(2, 4, 1, 3)


procedure queens ( i: index);var j : index;{ if promising(i) then if I==n then write (col[1]~col[n]); else for j=1 to n do { col[i+1]=j; queens(i+1); }function promising ( i : index) : boolean;var k : index;{ k=1; promising=true; while k<i and promising do { if (col[i]==col[k]) or abs(col[i]-col[k])==abs(i-k)) promising=false; k++;}} main : queens(0); Print all solutions. (Need to Exit)


Better (faster) AlgorithmMonte Carlo Algorithm“place almost all queens randomly, then do remaining queens using backtracking.”

#Nodes DFS (same row X)

(nn)

#Nodes DFS(same col. X)

(n!)

#NodesBacktracking

#Nodes Promising

Backtracking341

19,173,9619.73ⅹ1012

1.20ⅹ1016

2440,320

4.79ⅹ108

8.72ⅹ1010

6115,721

1.01ⅹ107

3.78ⅹ108

48

1214

n

172,057

8.56ⅹ105

2.74ⅹ107


Sum-of-subsets problem A special case of 0/1-knapsack S = {item1, item2,…, itemn} n items w[1..n] = (w1, w2, …, wn) weights W. Find a subset A⊆S s.t. ∑wi=W(Eg) n=5, W=21 w[1..5]=(5,6,10,11,16) Solutions : {w1,w2,w3}, {w1,w5}, {w3,w4}

0/1-knapsack 과의 관계 ? NP-Complete State space tree : 2n leaves

A

w1

w2

w3

w2

w3 w3 w3

O

O O

OOOO

{w1,w2} {w3}


Assumption : weights are in sorted order Promising function

ⅹ weight + w i+1 > W at i-th level ⅹ weight + total < W(Eg) n=4, W=13 (w1,w2,w3,w4)=(3,4,5,6)

function promising (i: index) : boolean;{ promising=(weight + total ≥ W) and (weight=W or weight+w[i+1]≤ W)}

ⅹ

12

7

3 97 8

13

4

0437

0

033 0

4

5

0 04

55

6

0 00

0ⅹ

ⅹ

ⅹⅹⅹⅹ


Sum_of_subsets(0,0,total); Initially, total =

procedure sum_of_subsets ( i : index; weight, total : integer);{ if promising(i) then

if weight= W thenwrite(include[1]~include[i]);

else{ include[i+1] = ‘yes’; sum_of_subsets(i+1 , weight+w[i+1], total-w[i+1]);

include[i+1] = ‘no’; sum_of_subsets(i+1 , weight, total-w[i+1]); }}

#nodes in state space tree: For almost all instances : Only small portion But

122 1

0

nn

k

k

nodesWwWw nn

n

ii )12(:, 1

1

1

n

iiw

1


Graph Coloring(p199) M-coloring problem;

Color undirected graph with colors.2 adjacent vertices : diff.color(Eg)

m

V1 V2

V4 V3

2-coloring X3-coloring O

State-space tree : leaves

Promising function : check adjacent vertices for the same color.

시작1 2 3

21 3

21 3

21 3X

X

X

X X

nmnodes

11...1

121

mmmmmn

n

m


Hamiltonian Circuits Problem TSP problem in chapter 3.

Brute-force: n!, (n-1)! D.P. : When n=20, B.F 3800 years D.P. 45 sec. More efficient solution? See chapter 6

Given an undirected or directed graph, find a tour(HC). (need not be S.P.)

32)2)(1( nnn


State-space tree(n-1)! Leaves worst-case.

Promising function:1. i-th node (i+1)-th node.2.(n-1)-th node 0-th node3. i-th node ≠ 0 ~ (i-1)-th node

V1 V2

V6V5

V3

V7

V4

V8

V1 V2

V5

V3

V4HC : X

(Eg)


0/1 Knapsack problem w[1..n] = (w1, w2,…, wn) p[1..n] = (p1, p2,…, pn) W: Knapsack size Determine A S maximizing⊆

State-space tree

2n leaves

x1=1x2=

1x3=1x4=

1

00

0

00000

00000

01

11 1 1 01

1 11

1

x2=1

WwtspA

iA

i ..


StrategyProcedure checknode(v : node);{ if value(v) > best then best = value(v);

if promising(v) then for each child u of v, checknode(u); } Obvious promising fcn

weight ≥W (need not expand)

Less obvious promising function.Note: 0/1 Knapsack 의 해≤ fractional Knapsack 의 해At level i,

maxprofit: best solution so far.Then, nonpromising if bound ≤ maxprofit.(Note: Sorted according to Pi/Wi)지금까지 이익 + 남은 자루용량으로 최대이익 ≤ 지금까지 최고이익 nonpromising

k

kk

ij

k

ij

wpWp

Ww

)totweight()profit(bound

)(,weighttotweight

1

1j

1

1j


(Eg) n=4, W=16(p1, p2, p3, p4)=(40, 30, 50,10)(w1, w2, w3, w4)=(2, 5,10, 5)(Pi/Wi)=(20, 6, 5, 2)

3

2

Item 1

X

X X X X

X

X

00

115

402

115

00

82

707

115

402

98

12017

707

80801280

707

70

901298

402

5010017

901290


Procedure knapsack(i :index) ; profit, weight : integer);{ if weight ≤ W and profit >maxprofit then

maxprofit = profit; numbest = i ; bestset = include;} // array of size n //if promising(i) then {include[i +1]=‘yes’;knapsack(i+1, profit+p[i +1], weight+w[i +1]);include[i +1]=‘no’;knapsack(i +1, profit, weight);

}}Numbest = maxprofit =0;knapsack(0,0,0);write(maxprofit ,bestset[1]~bestset[i]);

Very bad instance :pi= wi =1 (1 ≤i ≤n-1)pn= wn =W=n(2n+1-1)nodes in total

알고리즘 설계 및 분석 Foundations of Algorithm 유관우. Digital Media Lab. 2 Chap. 5...

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