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Data Structures and Algorithms Dynamic Programming

Dr. Muhammad Safyan Department of computer Science Government College University, Lahore. Data Structures and Algorithms Dynamic Programming. Today’s Agenda. Problem solving Approaches Dynamic programming. Approaches to Solve a problem. Brute Force

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Data Structures and Algorithms Dynamic Programming

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  1. Dr. Muhammad Safyan Department of computer Science Government College University, Lahore Data Structures andAlgorithmsDynamic Programming

  2. Today’s Agenda Problem solving Approaches Dynamic programming

  3. Approaches to Solve a problem • Brute Force • A brute force algorithm blindly iterates an entire domain of possible solutions in search of one or more solutions which satisfy a condition. Imagine attempting to open one of these: • Divide and Conquer: • Divide a large problem in such a way to become sub-problem of the original ones and again divide sub-problem in to it s sub-problem and subsequently reach to a position where the problem become very simple. • Very simple case become the base case. • Usually Applies to the problem when the sub-problems are disjoint e.gMerge Sort

  4. Approaches to Solve a problem Dynamic Programming: is a special case of Divide and conquer approach applies when the subproblems overlap e.g. Fibonacci problem Greedy Approach: ?

  5. Dynamic Programming is a technique used for Optimizations.Goal: Either get the Maximum or Minimum Results.In Dynamic Programming, Procedure are not well defined.Instead:It find out all possible solution and then pickup the solution i.e. optimal. Dynamic programming may consume more memory than normal --> Choose best from multiple solutions. Dynamic Programming

  6. Dynamic Programming (DP) • Use Recursive approach though not using Recursive Formula. • Recursive approach use Recursion/ Iteration. • DP follows Principle of Optimality-taking the sequence of steps while solving a problem for decision making. • Following Conditions met for the Dynamic problem • Recursive Equation • Optimal Substructure • Overlapping sub problem

  7. Dynamic Programming Recursive Equations: Function Call itself Optimal Sub-Structure Optimal solution to a problem contains optimal solution to sub- Problem. Overlapping Sub-problem: Repeating sub-problem A function call another function that also part of another function of problem. Advantage + Dynamic Programming reduce Time Complexity

  8. Dynamic Programming • Recursion Methodologies • There are two way solve recursive solution • Top-Down: use recursion • Bottom-Up: use Memoization or Tabulation • use recursive Equation and For Loop

  9. Fibonacci Series Fib(n)= 0 if n=0 1 if n=1 fib(n-2)+fib(n-1) Int fib( int n) { if (n<=1) return n; Else return(fib(n-2)+fib(n-1) } Time Complexity=T(n)= O(2n). How can we reduce it?

  10. Top Down: Fibonacci Series What’s the problem?

  11. Top: Down Iterative Mehtod: Fib(n) { if (n == 0) return M[0]; if (n == 1) return M[1]; if (Fib(n-2) is not already calculated) call Fib(n-2); if(Fib(n-1) is already calculated) call Fib(n-1); //Store the ${n}^{th}$ Fibonacci no. in memory & use previous results. M[n] = M[n-1] + M[n-2] Return M[n]; }

  12. already calculated …

  13. Fibonacci Series Store the result into Global array Total call=6 T(n)= n+1 T(n)= O(n) This is called Memoization. Memoization brings big Difference. This is called Bottom-Up approach. We use iterative method -> called Tabulation Method

  14. Dynamic Problem • Dynamic programming is a method of solving optimization problems by combining the solutions of subproblems • Developing these algorithms follows four steps: • Characterize the optimality - formally state what properties an optimal solution exhibits • Recursively define an optimal solution - analyze the problem in a top-down fashion to determine how subproblems relate to the original • Solve the subproblems - start with a base case and solve the sub-problems in a bottom-up manner to find the optimal value • Reconstruct the optimal solution - (optionally) determine the solution that produces the optimal value

  15. Rod Cutting Problem • Assume a company buys long steel rods and cuts them into shorter rods for sale to its customers. If each cut is free and rods of different lengths can be sold for different amounts, we wish to determine how to best cut the original rods to maximize the revenue. • Brute Force Solution: • Let the length of the rod be ninches. There are 2n-1 different ways to cut the rod. • Binary decision of whether or not to make a cut. Number of permutations of lengths is equal to the number of binary patterns of  n-1 bits of which there are 2n-1.

  16. Rod Cutting Problem Eight possible ways to cut a rod of length 4

  17. Rod Cutting Problem To find the optimal value we simply add up the prices for all the pieces of each permutation and select the highest value. Dynamic Programming Solution: Formalize the problem by assuming that a piece of length i has price pi. Optimal solution cuts the rod into k pieces of lengths i1, i2, ... , ik. such that n = i1 + i2 + ... + ik, then the revenue for a rod of length nis

  18. Rod Cutting Problem Optimal Substructure:

  19. Rod Cutting Problem Recursive Equation: Complexity where T(j) is the number of times the recursion occurs for each iteration of the for loop with j = n-i. The solution to this recursion can be shown to be T(n) = 2n which is still exponential behavior.  The problem with the top-down naive solution is that we recomputed all possible cuts thus producing the same run time as brute-force (only in a recursive fashion).

  20. Rod Cutting Problem: Bottom-Up we can store the solutions to the smaller problems in a bottom-up manner rather than recomputed them. the run time can be drastically improved (at the cost of additional memory usage).  To implement this approach we simply solve the problems starting for smaller lengths and store these optimal revenues in an array (of size n+1). Then when evaluating longer lengths we simply look-up these values to determine the optimal revenue for the larger piece. We can formulate this recursively as follows

  21. Rod Cutting Problem: Bottom-Up Total Length Profit ↓ 0 0 0 0 0 0 0 → 2 8 0 2 4 6 10 ↓ 5 0 10 2 5 7 12 ↓ ↓ 11 0 2 5 9 14 9 ↓ ↓ ↓ ↓ ↓ 11 0 2 5 9 14 6 Max(Profit by excluding new piece, Profit by including new piece)

  22. Rod Cutting Problem: Bottom-Up Note that to compute any rj we only need the values r0 to rj-1 which we store in an array. Hence we will compute the new element using only previously computed values. The implementation of this approach is

  23. Rod Cutting Problem: Bottom-Up

  24. Rod Cutting Problem: Extended Bottom-Up

  25. Matrix Multiplication: Dynamic Programming Recalling Matrix Multiplication

  26. Matrix Multiplication: Dynamic Programming Recalling Matrix Multiplication

  27. Matrix Multiplication: Dynamic Programming Recalling Matrix Multiplication

  28. Matrix-Chain multiplication (cont.) Cost of the matrix multiplication: An example:

  29. Matrix-Chain multiplication (cont.)

  30. Matrix-Chain multiplication (cont.) • The problem: Given a chain of nmatrices, where matrix Ai has dimension pi-1xpi, fully paranthesize the product in a way that minimizes the number of scalar multiplications.

  31. Elements of dynamic programming (cont.) Overlapping subproblems: (cont.) 1..4 1..1 2..4 1..2 3..4 1..3 4..4 2..2 3..4 2..3 4..4 1..1 2..2 3..3 4..4 1..1 2..3 1..2 3..3 2..2 3..3 1..1 2..2 3..3 4..4 2..2 3..3 The recursion tree of RECURSIVE-MATRIX-CHAIN( p, 1, 4). The computations performed in a shaded subtree are replaced by a single table lookup in MEMOIZED-MATRIX-CHAIN( p, 1, 4).

  32. Matrix-Chain multiplication (Contd.) RECURSIVE-MATRIX-CHAIN (p, i, j) 1 if i = j 2 then return 0 3 m[i,j] ← -1 4fork←ito j-1 5 do q←RECURSIVE-MATRIX-CHAIN (p, i, k) + RECURSIVE-MATRIX-CHAIN (p, k+1, j)+ pi-1pkpj 6 if q < m[i,j] 7then m[i,j] ←q 8 return m[i,j]

  33. Elements of dynamic programming (cont.) Overlapping subproblems: (cont.) WE guess that Using the substitution method with

  34. Matrix-Chain multiplication (cont.) • Counting the number of alternative paranthesization : bn

  35. Matrix-Chain multiplication (cont.) Step 1: The structure of an optimal paranthesization(op) • Find the optimal substructure and then use it to construct an optimal solution to the problem from optimal solutions to subproblems. • Let Ai...jwhere i ≤ j, denote the matrix product • Ai Ai+1 ... Aj • Anyparenthesization of Ai Ai+1 ... Aj must split the product between Ak andAk+1 fori ≤ k < j.

  36. Matrix-Chain multiplication (cont.) The optimal substructure of the problem: • Suppose that an op of Ai Ai+1 ... Aj splits the productbetween Akand Ak+1 then the paranthesization of the subchain Ai Ai+1 ... Ak within this parantesization of Ai Ai+1 ... Aj must be an op of Ai Ai+1 ... Ak

  37. Matrix-Chain multiplication (cont.) Step 2: A recursive solution: • Let m[i,j] be the minimum number of scalar multiplications needed to compute the matrix Ai...jwhere 1≤ i ≤ j ≤ n. • Thus, the cost of a cheapest way to compute A1...nwould be m[1,n]. • Assume that the op splits the product Ai...jbetween Akand Ak+1.wherei ≤ k <j. • Then m[i,j] =Theminimum cost for computing Ai...k and Ak+1...j+ the cost of multiplying these two matrices.

  38. Matrix-Chain multiplication (cont.) Recursive defination for the minimum cost of paranthesization:

  39. Matrix-Chain multiplication (cont.) To help us keep track of how to constrct an optimal solution we define s[ i,j] to be a value of kat which we can splitthe product Ai...jto obtain an optimal paranthesization. That is s[ i,j] equals a value k such that

  40. Matrix-Chain multiplication (cont.) Step 3: Computing the optimal costs It is easy to write a recursive algorithm based on recurrence for computing m[i,j]. But the running time will be exponential!...

  41. Matrix-Chain multiplication (cont.) Step 3: Computing the optimal costs We compute the optimal cost by using a tabular, bottom-up approach.

  42. Matrix-Chain multiplication (Contd.) MATRIX-CHAIN-ORDER(p) n←length[p]-1 for i←1 to n do m[i,i]←0 for l←2to n do for i←1 to n-l+1 do j←i+l-1 m[i,j]←∞ for k←ito j-1 do q←m[i,k] + m[k+1,j]+pi-1pkpj if q < m[i,j] then m[i,j] ←qs[i,j] ←k return m and s

  43. Matrix-Chain multiplication (cont.) An example:matrix dimension A1 30 x 35 A2 35 x 15 A3 15 x 5 A4 5 x 10 A5 10 x 20 A6 20 x 25

  44. Matrix-Chain multiplication (cont.) s m 6 1 2 3 6 1 5 i j 2 15125 4 3 5 3 3 i j 10500 3 3 4 11875 3 3 4 3 4 5 575 7125 5 9375 3 3 2 1 3 5 3500 2500 4375 2500 4 5 7875 1 2 3 2 5000 1000 6 2625 750 1000 15750 1 0 0 0 0 0 0 A1 A2 A3 A4 A5 A6

  45. Matrix-Chain multiplication (cont.) Step 4: Constructing an optimal solution An optimal solution can be constructed from the computed information stored in the table s[1...n, 1...n]. We know that the final matrix multiplication is The earlier matrix multiplication can be computed recursively.

  46. Matrix-Chain multiplication (Contd.) • PRINT-OPTIMAL-PARENS (s, i, j) • 1 if i=j • then print “Ai” • else print “ ( “ • PRINT-OPTIMAL-PARENS (s, i, s[i,j]) • PRINT-OPTIMAL-PARENS (s, s[i,j]+1, j) • 6 Print “ ) ”

  47. Matrix-Chain multiplication (Contd.) RUNNING TIME: Recursive solution takes exponential time. Matrix-chain order yields a running time of O(n3)

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