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- 1. Dynamic Programming Dr. C.V. Suresh Babu
- 2. Topics What is Dynamic Programming Binomial Coefficient Floyd’s Algorithm Chained Matrix Multiplication Optimal Binary Search Tree Traveling Salesperson 2
- 3. Why Dynamic Programming? Divide-and-Conquer: a top-down approach. Many smaller instances are computed more than once. Dynamic programming: a bottom-up approach. Solutions for smaller instances are stored in a table for later use. 3
- 4. Dynamic Programming An Algorithm Design Technique A framework to solve Optimization problems Elements of Dynamic Programming Dynamic programming version of a recursive algorithm. Developing a Dynamic Programming Algorithm – Example: Multiplying a Sequence of Matrices 4
- 5. Why Dynamic Programming? • It sometimes happens that the natural way of dividing an instance suggested by the structure of the problem leads us to consider several overlapping subinstances. • If we solve each of these independently, they will in turn create a large number of identical subinstances. • If we pay no attention to this duplication, it is likely that we will end up with an inefficient algorithm. • If, on the other hand, we take advantage of the duplication and solve each subinstance only once, saving the solution for later use, then a more efficient algorithm will result. 5
- 6. Why Dynamic Programming? … The underlying idea of dynamic programming is thus quite simple: avoid calculating the same thing twice, usually by keeping a table of known results, which we fill up as subinstances are solved. • Dynamic programming is a bottom-up technique. • Examples: 1) Fibonacci numbers 2) Computing a Binomial coefficient 6
- 7. Dynamic Programming • Dynamic Programming is a general algorithm design technique. • Invented by American mathematician Richard Bellman in the 1950s to solve optimization problems. • “Programming” here means “planning”. • Main idea: • solve several smaller (overlapping) subproblems. • record solutions in a table so that each subproblem is only solved once. • final state of the table will be (or contain) solution. 7
- 8. Dynamic Programming Define a container to store intermediate results Access container versus recomputing results Fibonacci – numbers example (top down) Use vector to store results as calculated so they are not re-calculated 8
- 9. Dynamic Programming Fibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 24 Recurrence Relation of Fibonacci numbers ? 9
- 10. Example: Fibonacci numbers • Recall definition of Fibonacci numbers: f(0) = 0 f(1) = 1 f(n) = f(n-1) + f(n-2) for n ≥ 2 • Computing the nth Fibonacci number recursively (topdown): f(n) f(n-1) f(n-2) + + f(n-3) f(n-2) f(n-3) + f(n-4) 10
- 11. Fib vs. fibDyn int fib(int n) { if (n <= 1) return n; // stopping conditions else return fib(n-1) + fib(n-2); // recursive step } int fibDyn(int n, vector<int>& fibList) { int fibValue; if (fibList[n] >= 0) // check for a previously computed result and return return fibList[n]; // otherwise execute the recursive algorithm to obtain the result if (n <= 1) // stopping conditions fibValue = n; else // recursive step fibValue = fibDyn(n-1, fibList) + fibDyn(n-2, fibList); // store the result and return its value fibList[n] = fibValue; return fibValue; } 11
- 12. Example: Fibonacci numbers Computing the nth fibonacci number using bottom-up iteration: • f(0) = 0 • f(1) = 1 • f(2) = 0+1 = 1 • f(3) = 1+1 = 2 • f(4) = 1+2 = 3 • f(5) = 2+3 = 5 • • • • f(n-2) = f(n-3)+f(n-4) • f(n-1) = f(n-2)+f(n-3) • f(n) = f(n-1) + f(n-2) 12
- 13. Recursive calls for fib(5) f ib ( 5 ) f ib ( 3 ) f ib ( 4 ) f ib ( 3 ) f ib ( 2 ) f ib ( 1 ) f ib ( 2 ) f ib ( 1 ) f ib ( 1 ) f ib ( 1 ) f ib ( 2 ) f ib ( 0 ) f ib ( 1 ) f ib ( 0 ) f ib ( 0 ) 13
- 14. fib(5) Using Dynamic Programming f ib ( 5 ) 6 f ib ( 4 ) f ib ( 3 ) 5 f ib ( 3 ) f ib ( 2 ) f ib ( 1 ) f ib ( 2 ) 4 f ib ( 2 ) f ib ( 1 ) f ib ( 1 ) f ib ( 0 ) f ib ( 1 ) f ib ( 0 ) 3 f ib ( 1 ) f ib ( 0 ) 1 2 14
- 15. Statistics (function calls) fib fibDyn N = 20 21,891 39 N = 40 331,160,281 79 15
- 16. Top down vs. Bottom up Top down dynamic programming moves through recursive process and stores results as algorithm computes Bottom up dynamic programming evaluates by computing all function values in order, starting at lowest and using previously computed values. 16
- 17. Examples of Dynamic Programming Algorithms • Computing binomial coefficients • Optimal chain matrix multiplication • Floyd’s algorithms for all-pairs shortest paths • Constructing an optimal binary search tree • Some instances of difficult discrete optimization problems: • travelling salesman • knapsack 17
- 18. A framework to solve Optimization problems For – – – each current choice: Determine what subproblem(s) would remain if this choice were made. Recursively find the optimal costs of those subproblems. Combine those costs with the cost of the current choice itself to obtain an overall cost for this choice Select a current choice that produced the minimum overall cost. 18
- 19. Elements of Dynamic Programming Constructing solution to a problem by building it up dynamically from solutions to smaller (or simpler) subproblems – – sub-instances are combined to obtain sub-instances of increasing size, until finally arriving at the solution of the original instance. make a choice at each step, but the choice may depend on the solutions to sub-problems. 19
- 20. Elements of Dynamic Programming … Principle of optimality – the optimal solution to any nontrivial instance of a problem is a combination of optimal solutions to some of its sub-instances. Memorization (for overlapping sub-problems) – – avoid calculating the same thing twice, usually by keeping a table of know results that fills up as subinstances are solved. 20
- 21. Development of a dynamic programming algorithm Characterize the structure of an optimal solution – Breaking a problem into sub-problem – whether principle of optimality apply Recursively define the value of an optimal solution – define the value of an optimal solution based on value of solutions to sub-problems Compute the value of an optimal solution in a bottom-up fashion – compute in a bottom-up fashion and save the values along the way – later steps use the save values of pervious steps Construct an optimal solution from computed information 21
- 22. Binomial Coefficient Binomial coefficient: n n! = k k!( n −k )! for 0 ≤ k ≤ n Cannot compute using this formula because of n! Instead, use the following formula: 22
- 23. Binomial Using Divide & Conquer Binomial formula: n − 1 n − 1 C k − 1 + C k 0 < k < n n C = k n n 1 k = 0 or k = n (C or C ) 0 n 23
- 24. Binomial using Dynamic Programming Just like Fibonacci, that formula is very inefficient Instead, we can use the following: (a + b) n = C (n,0)a n + ... + C (n, i )a n −i b i + ... + C (n, n)b n 24
- 25. Bottom-Up Recursive – B[i] [j] = property: B[i – 1] [j – 1] + B[i – 1][j] 0 < j < i 1 j = 0 or j = i 25
- 26. Pascal’s Triangle 0 1 0 1 1 1 1 2 1 2 1 3 1 3 3 1 4 1 4 6 4 … i n 2 3 4 … j k 1 B[i-1][j-1]+ B[i-1][j] B[i][j] 26
- 27. Binomial Coefficient Record the values in a table of n+1 rows and k+1 columns 0 1 2 0 1 1 2 1 3 3 k 1 3 k-1 1 2 … 1 1 3 1 ... k 1 1 … n-1 1 n 1 n − 1 C k − 1 n − 1 C k n C k 27
- 28. Binomial Coefficient ALGORITHM Binomial(n,k) //Computes C(n, k) by the dynamic programming algorithm //Input: A pair of nonnegative integers n ≥ k ≥ 0 //Output: The value of C(n ,k) for i 0 to n do for j 0 to min (i ,k) do if j = 0 or j = k C [i , j] 1 else C [i , j] C[i-1, j-1] + C[i-1, j] return C [n, k] k i −1 A( n, k ) = ∑∑1 + i =1 j =1 = n k k n i =1 i =K +1 ∑ ∑1 = ∑(i −1) + ∑k i =k +1 j =1 ( k −1) k + k ( n − k ) ∈Θ( nk ) 2 28
- 29. Floyd’s Algorithm: All pairs shortest paths •Find shortest path when direct path doesn’t exist •In a weighted graph, find shortest paths between every pair of vertices • Same idea: construct solution through series of matrices D(0), D(1), … using an initial subset of the vertices as 4 3 intermediaries. 1 • Example: 1 6 1 5 2 3 4 29
- 30. Shortest Path Optimization problem – more than one candidate for the solution Solution is the candidate with optimal value Solution 1 – brute force – – Find all possible paths, compute minimum Efficiency? Worse than O(n2) Solution 2 – dynamic programming – – Algorithm that determines only lengths of shortest paths Modify to produce shortest paths as well 30
- 31. Example 1 2 3 4 5 1 2 3 4 5 1 0 1 ∞ 1 5 1 0 1 3 1 4 2 9 0 3 2 ∞ 2 8 0 3 2 5 3 ∞ ∞ 0 4 ∞ 3 10 11 0 4 7 4 ∞ ∞ 2 0 3 4 6 7 2 0 3 5 3 ∞ ∞ ∞ 0 5 3 4 6 4 0 W - Graph in adjacency matrix D - Floyd’s algorithm 31
- 32. Meanings D(0)[2][5] = lenth[v2, v5]= ∞ D(1)[2][5] = minimum(length[v2,v5], length[v2,v1,v5]) = minimum (∞, 14) = 14 D(2)[2][5] = D(1)[2][5] = 14 D(3)[2][5] = D(2)[2][5] = 14 D(4)[2][5] = minimum(length[v2,v1,v5], length[v2,v4,v5]), length[v2,v1,v5], length[v2, v3,v4,v5]), = minimum (14, 5, 13, 10) = 5 D(5)[2][5] = D(4)[2][5] = 5 32
- 33. Floyd’s Algorithm 2 a 7 6 3 b c d 1 ( ( ( ( ( d ijk ) = min{d ijk −1) , d ikk −1) + d kjk −1) } for k ≥ 1, d ij0 ) = wij 33
- 34. Computing D D(0) =W Now compute D(1) Then D(2) Etc. 34
- 35. Floyd’s Algorithm: All pairs shortest paths • ALGORITHM Floyd (W[1 … n, 1… n]) •For k ← 1 to n do •For i ← 1 to n do •For j ← 1 to n do •W[i, j] ← min{W[i,j], W{i, k] + W[k, j]} •Return W •Efficiency = ? Θ(n) 35
- 36. Example: All-pairs shortest-path problem Example: Apply Floyd’s algorithm to find the t Allpairs shortest-path problem of the digraph defined by the following weight matrix 0 6 ∞ ∞ 3 2 0 ∞ ∞ ∞ ∞ 3 0 2 ∞ 1 2 4 0 ∞ 8 ∞ ∞ 3 0 36
- 37. Visualizations http://www.ifors.ms.unimelb.edu.au/tutorial/path/#list http://www1.math.luc.edu/~dhardy/java/alg/floyd.html http://students.ceid.upatras.gr/%7Epapagel/project/kef5_7_2.htm 37
- 38. Chained Matrix Multiplication Problem: Matrix-chain multiplication – a chain of <A1, A2, …, An> of n matrices – find a way that minimizes the number of scalar multiplications to compute the product A1A2…An Strategy: Breaking a problem into sub-problem – A1A2...Ak, Ak+1Ak+2…An Recursively define the value of an optimal solution – m[i,j] = 0 if i = j – m[i,j]= min{i<=k<j} (m[i,k]+m[k+1,j]+pi-1pkpj) – for 1 <= i <= j <= n 38
- 39. Example Suppose we want to multiply a 2x2 matrix with a 3x4 matrix Result is a 2x4 matrix In general, an i x j matrix times a j x k matrix requires i x j x k elementary multiplications 39
- 40. Example Consider multiplication of four matrices: A (20 x 2) x B (2 x 30) x C (30 x 12) x D (12 x 8) Matrix multiplication is associative A(B (CD)) = (AB) (CD) Five different orders for multiplying 4 matrices 1. 2. 3. 4. 5. A(B (CD)) = 30*12*8 + 2*30*8 + 20*2*3 = 3,680 (AB) (CD) = 20*2*30 + 30*12*8 + 20*30*8 = 8,880 A ((BC) D) = 2*30*12 + 2*12*3 + 20*2*8 = 1,232 ((AB) C) D = 20*2*30 + 20*30*12 + 20*12*8 = 10,320 (A (BC)) D = 2*30*12 + 20*2*12 + 20*12*8 = 3,120 40
- 41. Algorithm int minmult (int n, const ind d[], index P[ ] [ ]) { index i, j, k, diagonal; int M[1..n][1..n]; for (i = 1; i <= n; i++) M[i][i] = 0; for (diagonal = 1; diagonal <= n-1; diagonal++) for (i = 1; i <= n-diagonal; i++) { j = i + diagonal; M[i] [j] = minimum(M[i][k] + M[k+1][j] + d[i-1]*d[k]*d[j]); // minimun (i <= k <= j-1) P[i] [j] = a value of k that gave the minimum; } return M[1][n]; } 41
- 42. Optimal Binary Trees Optimal way of constructing a binary search tree Minimum depth, balanced (if all keys have same probability of being the search key) What if probability is not all the same? Multiply probability of accessing that key by number of links to get to that key 42
- 43. Example If p1 = 0.7 Key3 p2 = 0.2 3(0.7) + 2(0.2) + 1(0.1) = 2.6 p3 = 0.1 key2 Θ (n3) Efficiency key1 43
- 44. Traveling Salesperson The Traveling Salesman Problem (TSP) is a deceptively simple combinatorial problem. It can be stated very simply: A salesman spends his time visiting n cities (or nodes) cyclically. In one tour he visits each city just once, and finishes up where he started. In what order should he visit them to minimize the distance traveled? 44
- 45. Why study? The problem has some direct importance, since quite a lot of practical applications can be put in this form. It also has a theoretical importance in complexity theory, since the TSP is one of the class of "NP Complete" combinatorial problems. NP Complete problems are intractable in the sense that no one has found any really efficient way of solving them for large n. – They are also known to be more or less equivalent to each other; if you knew how to solve one kind of NP Complete problem you could solve the lot. 45
- 46. Efficiency The holy grail is to find a solution algorithm that gives an optimal solution in a time that has a polynomial variation with the size n of the problem. The best that people have been able to do, however, is to solve it in a time that varies exponentially with n. 46
- 47. Later… We’ll get back to the traveling salesperson problem in the next chapter…. 47
- 48. Animations http://www.pcug.org.au/~dakin/tsp.htm http://www.ing.unlp.edu.ar/cetad/mos/TSPBIB_hom 48
- 49. Chapter Summary • Dynamic programming is similar to divide-andconquer. • Dynamic programming is a bottom-up approach. • Dynamic programming stores the results (small instances) in the table and reuses it instead of recomputing it. • Two steps in development of a dynamic programming algorithm: • Establish a recursive property • Solve an instance of the problem in a bottom-up 49
- 50. Exercise: Sudoku puzzle 50
- 51. Rules of Sudoku • Place a number (1-9) in each blank cell. • Each row (nine lines from left to right), column (also nine lines from top to bottom) and 3x3 block bounded by bold line (nine blocks) contains number from 1 through 9. 51

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