1. ICS 353: Design and Analysis of Algorithms
King Fahd University of Petroleum & Minerals
Information & Computer Science Department
ICS 353: Design and Analysis of Algorithms
Dynamic Programming

2. Reading Assignment
M. Alsuwaiyel, Introduction to Algorithms: Design Techniques and Analysis, World Scientific Publishing Co., Inc
Chapter 7

3. Dynamic Programming
Dynamic Programming algorithms address problems whose solution is recursive in nature, but has the following property: The direct implementation of the recursive solution results in identical recursive calls that are executed more than once.
Dynamic programming implements such algorithms by evaluating the recurrence in a bottom-up manner, saving intermediate results that are later used in computing the desired solution

4. Fibonacci Numbers
What is the recursive algorithm that computes Fibonacci numbers? What is its time complexity?
Note that it can be shown that

5. Computing the Binomial Coefficient
Recursive Definition
Actual Value

6. Computing the Binomial Coefficient
What is the direct recursive algorithm for computing the binomial coefficient? How much does it cost?
Note that

7. Optimization Problems and Dynamic Programming
Optimization problems with certain properties make another class of problems that can be solved more efficiently using dynamic programming.
Development of a dynamic programming solution to an optimization problem involves four steps
Characterize the structure of an optimal solution
Optimal substructures, where an optimal solution consists of sub-solutions that are optimal.
Overlapping sub-problems where the space of sub-problems is small in the sense that the algorithm solves the same sub-problems over and over rather than generating new sub-problems.
Recursively define the value of an optimal solution.
Compute the value of an optimal solution in a bottom-up manner.
Construct an optimal solution from the computed optimal value.

8. Longest Common Subsequence Problem
Problem Definition: Given two strings A and B over alphabet , determine the length of the longest subsequence that is common in A and B.
A subsequence of A=a1a2…an is a string of the form ai1ai2…aik where 1i1<i2<…<ik n
Example: Let  = { x , y , z }, A = xyxyxxzy, B=yxyyzxy, and C= zzyyxyz
LCS(A,B)=yxyzy Hence the length =
LCS(B,C)= Hence the length =
LCS(A,C)= Hence the length =

9. Straight-Forward Solution
Brute-force search
How many subsequences exist in a string of length n?
How much time needed to check a string whether it is a subsequence of another string of length m?
What is the time complexity of the brute-force search algorithm of finding the length of the longest common subsequence of two strings of sizes n and m?
How many subsequences exist in a string of length n? 2n
How much time needed to check a string whether it is a subsequence of another string of length m? (m)

10. Dynamic Programming Solution
Let L[i,j] denote the length of the longest common subsequence of a1a2…ai and b1b2…bj, which are substrings of A and B of lengths n and m, respectively. Then
L[i,j] = when i = 0 or j = 0
L[i,j] = when i > 0, j > 0, ai=bj
L[i,j] = when i > 0, j > 0, aibj

11. LCS Algorithm Algorithm LCS(A,B)
Input: A and B strings of length n and m respectively
Output: Length of longest common subsequence of A and B
Initialize L[i,0] and L[0,j] to zero;
for i ← 1 to n do
for j ← 1 to m do
if ai = bj then
L[i,j] ← 1 + L[i-1,j-1]
else
L[i,j] ← max(L[i-1,j],L[i,j-1])
end if
end for;
return L[n,m];

12. Example (Q7.5 pp. 220)
Find the length of the longest common subsequence of A=xzyzzyx and B=zxyyzxz

13. Example (Cont.) x z y

14. Example (Cont.) x z y   1 2  3 4

15. Complexity Analysis of LCS Algorithm
What is the time and space complexity of the algorithm?

16. Matrix Chain Multiplication
Assume Matrices A, B, and C have dimensions 210, 102, and 210 respectively. The number of scalar multiplications using the standard Matrix multiplication algorithm for
(A B) C is
A (B C) is
Problem Statement: Find the order of multiplying n matrices in which the number of scalar multiplications is minimum.

17. Straight-Forward Solution
Again, let us consider the brute-force method. We need to compute the number of different ways that we can parenthesize the product of n matrices.
e.g. how many different orderings do we have for the product of four matrices?
Let f(n) denote the number of ways to parenthesize the product M1, M2, …, Mn.
(M1M2…Mk) (M k+1M k+2…Mn)
What is f(2), f(3) and f(1)?

18. Catalan Numbers
Cn=f(n+1)
Using Stirling’s Formula, it can be shown that f(n) is approximately

19. Cost of Brute Force Method
How many possibilities do we have for parenthesizing n matrices?
How much does it cost to find the number of scalar multiplications for one parenthesized expression?
Therefore, the total cost is

20. The Recursive Solution
Since the number of columns of each matrix Mi is equal to the number of rows of Mi+1, we only need to specify the number of rows of all the matrices, plus the number of columns of the last matrix, r1, r2, …, rn+1 respectively.
Let the cost of multiplying the chain Mi…Mj (denoted by Mi,j) be C[i,j]
If k is an index between i+1 and j, what is the cost of multiplying Mi,j considering multiplying Mi,k-1 with Mk,j?
Therefore, C[1,n]=

21. The Dynamic Programming Algorithm

22. Example (Q7.11 pp )
Given as input 2 , 3 , 6 , 4 , 2 , 7 compute the minimum number of scalar multiplications:

23. Example (Q7.11 pp ) M1 M2 M3 M4 M5

24. Example (Q7.11 pp. 221-222) M1 36 M1..M2 84 (M1..M2) M3 96 M1 (M2..M4)M1 36
M1..M2 84
(M1..M2) M3 96
M1 (M2..M4) M2 72
M2..M3
M2(M3..M4)
126
(M2..M4)M5 M3 48
M3..M4
132
(M3..M4) M5 M4 56
M4..M5 M5

25. Another Example (Activity Sheet)
Given as input 5 , 2 , 3 , 6 , 4 , 2, 4
Compute the minimum number of scalar multiplications.
Find the optimal parenthesization

26. MatChain Algorithm Algorithm MatChain
Input: r[1..n+1] of +ve integers corresponding to the dimensions of a chain of matrices
Output: Least number of scalar multiplications required to multiply the n matrices
for i := 1 to n do
C[i,i] := 0; // diagonal d0
for d := 1 to n-1 do // for diagonals d1 to dn-1
for i := 1 to n-d do
j := i+d;
C[i,j] := ;
for k := i+1 to j do
C[i,j] := min{C[i,j],C[i,k-1]+C[k,j]+r[i]r[k]r[j+1];
end for;
return C[1,n];

27. Time and Space Complexity of MatChain Algorithm
Time Complexity
Space Complexity

28. All-Pairs Shortest Paths
Problem: For every vertex u, v  V, calculate (u, v).
Possible Solution 1:
Cost of Possible Solution 1:

29. Dynamic Programming Solution
Define a k-path from u to v, where
u,v  {1 , 2 , … , n}
to be any path whose intermediate vertices all have indices less than or equal to k.
What is a 0-path?
What is a 1-path? …
What is an n-path?

30. Floyd’s Algorithm Algorithm Floyd
Input: An n  n matrix length[1..n, 1..n] such that length[i,j] is the weight of the edge (i,j) in a directed graph G = ({1,2,…,n}, E)
Output: A matrix D with D[i,j] = [i,j]
1 D = length; //copy the input matrix length into D
2 for k = 1 to n do
3 for i = 1 to n do
4 for j = 1 to n do
D[i,j] = min{D[i,j] , D[i,k] + D[k,j]}

31. Example 2 11 5 12 15 4 8 2 4 1 11 1 3 5

32. Example (Cont.)
0-p 1 2 3 4
1-p 1 2 3 4
2-p 1 2 3 4
3-p 1 2 3 4

33. Example (Cont.)
0-p 1 2 3 4 12 5  15 8 11
1-p 1 2 3 4 12 5  15 8 11 6
2-p 1 2 3 4 12 5 23 15 8 11 19 6
3-p 1 2 3 4 9 5 7 15 8 10 19 6

34. Example (Cont.)
3-p 1 2 3 4 9 5 7 15 8 10 19 6
4-p 1 2 3 4 9 5 7 11 8 10 6

35. Time and Space Complexity
Time Complexity:
Space Complexity:

36. The Knapsack Problem
Let U = {u1, u2, …, un} be a set of n items to be packed in a knapsack of size C.
Let sj and vj be the size and value of the jth item, where sj, vj , 1  j  n.
The objective is to fill the knapsack with some items from U whose total size does not exceed C and whose total value is maximum.
Assume that the size of each item does not exceed C.

37. The Knapsack Problem Formulation
Given n +ve integers in U, we want to find a subset SU s.t.
is maximized subject to the constraint
Sometimes referred to as the 0-1 knapsack problem, since no more than one item is allowed from each type.

38. = max {V[i-1,j], V[ , ]+vi} if
Inductive Solution
Let V[i,j] denote the value obtained by filling a knapsack of size j with items taken from the first i items {u1, u2, …, ui} in an optimal way:
The range of i is
The range of j is
The objective is to find V[ , ]
V[i,0] = V[0,j] =
V[i,j] = V[i-1,j] if
= max {V[i-1,j], V[ , ]+vi} if

39. Example (pp. 223 Question 7.22)
There are five items of sizes 3, 5, 7, 8, and 9 with values 4, 6, 7, 9, and 10 respectively. The size of the knapsack is 22. 4

40. Example (pp. 223 Question 7.22)
There are five items of sizes 3, 5, 7, 8, and 9 with values 4, 6, 7, 9, and 10 respectively. The size of the knapsack is 22. 4 6 10 7 11 13 17 15 19 20 22 14 16 21 23 25

42. Algorithm Knapsack Algorithm Knapsack
Input: A set of items U = {u1,u2,…,un} with sizes s1,s2,…,sn and values v1,v2,…,vn, respectively and knapsack capacity C.
Output: the maximum value of subject to
for i := 0 to n do
V[i,0] := 0;
for j := 0 to C do
V[0,j] := 0;
for i := 1 to n do
for j := 1 to C do
V[i,j] := V[i-1,j];
if si  j then
V[i,j] := max{V[i,j], V[i-1,j-si]+vi}
end for;
return V[n,C];

43. Time and Space Complexity of the Knapsack Algorithm