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

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dynamic programming
Dynamic Programming
  • Reading Material: Chapter 7 Sections 1 - 4 and 6.
dynamic programming2
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
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
Computing the Binomial Coefficient
  • Recursive Definition
  • Actual Value
computing the binomial coefficient5
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 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
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 =
straight forward solution
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
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

lcs algorithm
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]


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
Example (Q7.5 pp. 220)
  • Find the length of the longest common subsequence of A=xzyzzyx and B=zxyyzxz
complexity analysis of lcs algorithm
Complexity Analysis of LCS Algorithm
  • What is the time and space complexity of the algorithm?
matrix chain multiplication
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.
straight forward solution15
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
Catalan Numbers
  • Cn=f(n+1)
  • Using Stirling’s Formula, it can be shown that f(n) is approximately
cost of brute force method
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
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
Example (Q7.11 pp. 221-222)
  • Given as input 2 , 3 , 6 , 4 , 2 , 7 compute the minimum number of scalar multiplications:
matchain algorithm
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];

the knapsack problem
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
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

inductive solution
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
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

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];