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Dynamic Programming. Reading Material: Chapter 7 Sections 1 - 4 and 6. Dynamic Programming.

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Dynamic Programming

- Reading Material: Chapter 7 Sections 1 - 4 and 6.

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

Fibonacci Numbers

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

Computing the Binomial Coefficient

- Recursive Definition
- Actual Value

Computing the Binomial Coefficient

- What is the direct recursive algorithm for computing the binomial coefficient? How much does it cost?
- Note that

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.

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 1i1<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 =

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?

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, aibj

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;

end for;

return L[n,m];

Example (Q7.5 pp. 220)

- Find the length of the longest common subsequence of A=xzyzzyx and B=zxyyzxz

Complexity Analysis of LCS Algorithm

- What is the time and space complexity of the algorithm?

Matrix Chain Multiplication

- Assume Matrices A, B, and C have dimensions 210, 102, and 210 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.

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)?

Catalan Numbers

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

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

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]=

Example (Q7.11 pp. 221-222)

- Given as input 2 , 3 , 6 , 4 , 2 , 7 compute the minimum number of scalar multiplications:

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;

end for;

return C[1,n];

Time and Space Complexity of MatChain Algorithm

- Time Complexity
- Space Complexity

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.

The Knapsack Problem Formulation

- Given n +ve integers in U, we want to find a subset SU s.t.

is maximized subject to the constraint

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

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.

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;

end for;

return V[n,C];

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