1. Dynamic Programming
Reporter : 林明慧

2. Outline
The Longest Common Subsequence Problem
0/1 Knapsack Problem

3. The Longest Common Subsequence Problem
The longest common subsequence problem is to find a longest common subsequence between two strings.
A straightforward method is to find all common subsequences by an exhaustive search.

4. The Longest Common Subsequence Problem
Example
A = a b a a d e c
B = b a f c
Try subsequences chosen from B with length three :
a f c
b f c
b a c
b a f

5. The Longest Common Subsequence Problem
Disadvantage
Exceedingly time-consuming
New method
By tracing back the procedure finding the length of the longest common subsequence.

6. The Longest Common Subsequence Problem
A = x1 x2 … xm , B = y1 y2 … yn
Two possibilities :
xm = yn. xm is contained in the longest common subsequence.
x1 x2 … xm-1
y1 y2 … yn-1
xm yn.
x1 x2 … xm with y1 y2 … yn-1
x1 x2 … xm-1 with y1 y2 … yn
Example :
A = a b a a d e c
B = b a f c e c f c

7. Recursive formula x1 x2 …
xi-1 xi xm y1 y2
yj-1 yj yn 1 1

8. Example A = a b a a d e c , B = c a a c e d c xi yj a b d e c 1 2 3 41 2 3 4 5 6 7
L(i,j) 1 2 1 2 1 2 3 1 2 3 1 2 3 4 4 1 2 3 1 1 2 1 1 1 1 1 2 1 2 1 1 2

9. Experiment_1 A = a b a a d e c , B = c a a c e d c xi yj a b d e c 1 21 2 3 4 5 6 7
L(i,j) 1 2 3 1 2 3 1 2 3 4 1 1 1 2 1 2 3

10. Experiment_2 A = a b a a d e c , B = c a a c e d c xi yj a b d e c 1 21 2 3 4 5 6 7
L(i,j) 1 2 3 1 2 3 1 2 3 4 1 1 1 2 1 2

11. Result_1 A = a b a a d e c , B = c a a c e d c
A longest common subsequence is a a e c.
A = a b a a d e c , B = c a a c e d c xi yj a b d e c 1 2 3 4 5 6 7
L(i,j)

12. Result_2 A = a b a a d e c , B = c a a c e d c
A longest common subsequence is a a d c.
A = a b a a d e c , B = c a a c e d c xi yj a b d e c 1 2 3 4 5 6 7
L(i,j)

13. 0/1 Knapsack Problem
The 0/1 knapsack problem is to maximize the total profit under the constraint that the total weight of all chosen objects is at most capacity.

14. 0/1 Knapsack Problem It is defined as follows:
Given n objects and a knapsack.
Object i has a weight Wi.
Knapsack has a capacity M.
If object i is placed into the knapsack, we will obtain a profit Pi.

15. Example Capacity M = 10 i Wi Pi 1 10 40 2 3 20 5 30 x2=0 x3=0 A C F40
10-10=0 G S
x3=1 30 D
7-5=2
x2=1 20 H T
10-3=7
x3=0
x1=0 B
7-0=7 I
10-0=10
x3=1 E 30
x2=0
10-5=5
10-0=10 J
x3=0
10-0=10

16. Example d(S,T) = max{40+d(A,T) , 0+d(B,T)} d(A,T) d(B,T) A 40 S T B SC D E F G H I J T
d(S,T) = max{40+d(A,T) , 0+d(B,T)} A 40
d(A,T) S T B
d(B,T)

17. Example d(A,T) = 0+0+0 = 0 d(B,T) = max{20+d(D,T) , 0+d(E,T)} d(D,T)S A B C D E F G H I J T
d(A,T) = = 0
d(B,T) = max{20+d(D,T) , 0+d(E,T)} A C F T D
d(D,T) 20 B T
d(E,T) E

18. Example S A B C D E F G H I J T
d(D,T) = max{30+d(G,T) , 0+d(H,T)} = max{30+0 , 0+0} = max{30 , 0} = 30
d(E,T) = max{30+d(I,T) , 0+d(J,T)} = max{30+0 , 0+0} = max{30 , 0} = 30 G 30 D T H E I J T 30

19. Example S A B C D E F G H I J T
d(B,T) = max{20+d(D,T) , 0+d(E,T)} = max{20+30 , 0+30} = max{50 , 30} = 50
d(S,T) = max{40+d(A,T) , 0+d(B,T)}
= max{40+0 , 0+50}
= max{40 , 50}
= 50 B D E T 20
d(D,T)
d(E,T) S A B T 40
d(A,T)
d(B,T)

20. Example Capacity M = 10 0 + 20 + 30 + 0 = 50 i Wi Pi 1 10 40 2 3 20 5
= 50
x2=0
x3=0 A C F
x1=1 40 G S
x3=1 30 D
x2=1 20 H T
x3=0
x1=0 B
x3=1 I E 30
x2=0 J
x3=0

21. Another Expression
Recursive formula

22. Example Capacity M = 10 Capacity j i 1 2 3 4 5 6 7 8 9 10 W1=10, P1=401 2 3 4 5 6 7 8 9 10
W1=10, P1=40 40
W2=3, P2=20 20
W3=5, P3=30 30 50
P[1,10] = max{P[1-1,10],40+P[1-1,10-10]} = 40
P[2,10] = max{P1,P2} = 40
P[2,3] = max{P[2-1,3],20+P[2-1,3-3]} = 20
P[3,3] = P2 = 20
P[3,5] = max{P[3-1,5],30+P[3-1,5-5]} = 30
P[3,8] = P2+P3 = 50
P[3,10] = max{P1,(P2+P3)} = 50

23. The End Have A Nice Holiday!