Dynamic Programming. Prof. S.J. Soni


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1 Dynamic Programming Prof. S.J. Soni
2 Idea is Very Simple.. Introduction void calculating the same thing twice, usually by keeping a table of known results that fills up as subinstances are solved. Dynamic Programming is a bottomup technique. We usually start with the smallest, and hence the simplest, subinstances. y combining their solutions, we obtain the answers to subinstances of increasing size, until finally we arrive at the solution of the original instance.
3 alculating the binomial coefficient onsider the problem of calculating the binomial coefficient 3
4 alculating the binomial coefficient Function (n,k) if k= or k=n then return 1 else return (n1,k1) + (n1,k) alculate for (5,3) (4,), (4,3) (3,1), (3,) (3,), (3,3) (,), (,1), (,1), (,) (,1), (,), (,), (,3) This algorithm calculates same values again & again, so that it is inefficient algorithm. 4
5 Solution using Dynamic Programming Pascal s Triangle : 1 1: 1 1 : 1 1 3: : : : : :
6 Solution using Dynamic Programming Pascal s Triangle k1 k : 1 1: 1 1 : 1 1 3: : : n1: (n1,k1) (n1,k) n: (n,k) 6
7 Making hange () Problem There are several instances of problem in which greedy algorithm fails to generate answer. For example, There are coins for 1,4 and 6 units. If we have to make change for 8 units, the greedy algorithm will propose doing so using one 6unit coin and two 1unit coins, for a total of three coins. However it is clearly possible to do better than this: we can give the customer his change using just two 4unit coins. lthough the greedy algorithm does not find this solution, it is easily obtained using dynamic programming. 7
8 Solution using Dynamic Programming To solve this problem using DP, we set up a table c [1 n, N], with one row for each available denomination and one column for each amount from to N units. In this table c [i,j] will be the minimum number of coins required to pay an amount of j units, <= j <=N. c [i,j] = min (c[i1,j], 1+c[i, j  di]) 8
9 Solution using Dynamic Programming mount d1= d= d3= c [i,j] = min (c[i1,j], 1+c[i, j  di]) c [,1] = min (c[1,1], 1+c[, 14]) = min(1,inf) =1 c [,] = min (c[1,], 1+c[,  4]) = min(,inf) = c [,4] = min (c[1,4], 1+c[, 44]) = min(4,1) =1 c [3,8] = min (c[,8], 1+c[3, 86]) = min(,3) = 9
10 Solution using Dynamic Programming Which oins? mount d1= d= d3= If c[i,j]=c[i1,j], we move up to c[i1, j] c[3,8] = c[,8], we move on to c[,8] Otherwise check c[i,j]=1+c[i,jdi] so check c[,8]=1+c[,84] = 1+c[,4]= [select coin] gain check c[,4]<>c[1,4] so, c[,4]=1+c[,44]=1 [select coin] 1
11 Knapsack Problem () We are given number of objects and knapsack. This time, however, we suppose that the objects may not be broken into smaller pieces, we may decide either to take an object or to leave it behind, but we may not take a fraction of an object. So it s called Nonfractional knapsack problem or /1 knapsack problem. 11
12 Knapsack Problem () Example for which greedy method cannot work. Object 1 3 Weight Value W=1 Greedy method generates the value 8 which is not optimal. 1
13 Solution using Dynamic Programming To solve this problem using DP, we set up a table V [1 n, W], with one row for each available object and one column for each weight from to W. In this table V [i,j] will be the maximum value of the objects, <= j <=W & 1<= i <=n V [i,j] = max (V[i1,j], V[i1, j  wi]+vi) 13
14 Knapsack Problem () Example Objects Weight Value Vi/Wi W=11 14
15 Solution using Dynamic Programming Weight Limit w1=1, v1= w= v= w3=5, v3= w4=6, v4= w5=7, v5= V [i,j] = max (V[i1,j], V[i1, j  wi]+vi) V [5,11] = max (V[4,11], V[4,117] +8 )= max(4,35)=4 15
16 Solution using Dynamic Programming Weight Limit w1=1, v1= w= v= w3=5, v3= w4=6, v4= w5=7, v5= Which Objects? If V[i,j]=V[i1,j], we move up to V[i1, j] V[5,11] = V[4,11], we move on to V[4,11] Otherwise check V[i,j]=V[i1,jwi]+vi so check V[4,11]=V[3,116]+ = V[3,5]+=4 [select object 4] now check V[3,5]=V[,55]+18=V[,]+18=18 [select object 3] 16
17 Practice Examples Solve following knapsack problem using dynamic programming algorithm with given capacity W=5, Weight and Value are as follows : (,1),(1,1),(3,),(,15). Solve the following Knapsack Problem using Dynamic Programming Method. Write the equation for solving above problem. n = 5, W = 1 Object Weight (w) Value (v)
18 Shortest Paths Let G = <N,> be a directed graph; N is a set of nodes and is a set of edges. We want to calculate the length of the shortest path between each pair of nodes. Suppose the nodes of G are numbered from 1 to n, so N=[1,, n] and suppose matrix L gives the length of each edge, with L[i,j]= for i=1,,..n. L[i,j] >= for all i and j, and L[i,j]= if the edge (i,j) does not exist. 18
19 Shortest Paths The principle of optimality: If k is a node on the shortest path from i to j, then the part of the path from i to k, and the part from k to j, must also be optimal. We construct matrix D that gives the length of the shortest path between each pair of nodes. D k [i,j] = min(d k1 [i,j], D k1 [i,k]+d k1 [k,j]) 19
20 Floyd s lgorithm D k [i,j] = min(d k1 [i,j], D k1 [i,k]+d k1 [k,j]) D 1 = D = L = D 1 [3,] = min(d [3,], D [3,1]+D [1,]) = min(inf, 3+5) = 35
21 Floyd s lgorithm D = L = D 1 = D k [i,j] = min(d k1 [i,j], D k1 [i,k]+d k1 [k,j]) D 1 [1,1] = min(d [1,1], D [1,1]+D [1,1]) = min(, + ) = D 1 [1,] = min(d [1,], D [1,1]+D [1,]) = min(5, + 5) = 5 D 1 [,1] = min(d [,1], D [,1]+D [1,1]) = min(5, 5+ ) = 5 D 1 [3,] = min(d [3,], D [3,1]+D [1,]) = min(inf, 3+5) = 35 1
22 Floyd s lgorithm D = L = D 1 = D = D 3 = D 4 = D k [i,j] = min(d k1 [i,j], D k1 [i,k]+d k1 [k,j])
23 Practice Example Write the equation for finding out shortest path using Floyd s algorithm. Use Floyd s method to find shortest path for below mentions all pairs
24 hained Matrix Multiplication Recall that the product of a p x q matrix and a q x r matrix is the p x r matrix given by q ij = a ik b kj k=1 1<=i<=p, 1<=j<=r It is clear that a total of pqr scalar multiplications are required to calculate the matrix product using this algorithm. e.g. (3X5) (5X4) => (3X4) [3X5X4=6 multiplications] 4
25 hained Matrix Multiplication Suppose now we want to calculate the product of more than two matrices. Matrix multiplication is associative, so we can compute the matrix product M = M 1 M. M n in a number of ways, which all gives the same answer. M = (((M 1 M ) M 3 ). M n ) = ((M 1 M ) (M 3. M n )) = ((M 1 (M M 3 )). M n )) and so on. However matrix multiplication is not commutative, so we are not allowed to change the order of the matrices in these arrangements. 5
26 MM  Example Suppose for example, we want to calculate the product D of four matrices, where is 13X5, is 5X89, is 89X3 and D is 3X34 Instances (())D => =5785, ()=3471, (())D=136 => = 158 scalar multpls ()(D) = 541 (()D) = 455 (())D = 856 ((D)) = 6418 The most efficient method is almost 19 times faster than the slowest. 6
27 MM Example Number of combinations N T(n) The values of T(n) are called the atalan numbers. 7
28 hained Matrix Multiplication Suppose the dimensions of the matrices are given by a vector d[..n] such that the matrix M i, 1<= i<=n, is of dimension d i1 X d i We fill the table m ij using the following rules for s=,1,, n1. s= : m ii = i=1,,.,n s=1 : m i,i+1 = d i1 d i d i+1 i=1,,.,n1 1<s<n: m i,i+s = min (m ik +m k+1,i+s +d i1 d k d i+s ) i<= k < i+s i=1,,,ns 8
29 hained Matrix Multiplication Suppose for example, we want to calculate the product D of four matrices, where is 13X5, is 5X89, is 89X3 and D is 3X34 do d1 d d3 d4 The vector d is therefore (13, 5, 89, 3, 34). s=1 : m i,i+1 = d i1 d i d i+1 i=1,,.,n1 For s=1, we find m1 = d d1 d = 13X5X89 = 5785 m3 = d1 d d3 = 5X89X3 = 1335 m34 = d d3 d4 = 89X3X34 = 978 9
30 hained Matrix Multiplication 1<s<n: m i,i+s = min (m ik +m k+1,i+s +d i1 d k d i+s ) i<= k < i+s For s=, we obtain, [where 1<s<n=1<<4 and i = 1,..,4, so i=1,] [ i<=k<i+s = 1<=k<3, so k=1 & k=] m13 = min (m11+m3+13x5x3, // for k=1 m1+m33+13x89x3) // for k= = min (153,956) = 153 [ i<=k<i+s = <=k<4, so k= & k=3] m4 = min (m+m34+5x89x34, // for k= m3+m44+5x3x34) // for k=3 = min (48,1845) = 1845 i=1,,,ns 3
31 hained Matrix Multiplication 1<s<n: m i,i+s = min (m ik +m k+1,i+s +d i1 d k d i+s ) i<= k < i+s Finally for s=3, we obtain, [where 1<s<n=1<3<4 and i = 1,..,43, so i=1 only] [ i<=k<i+s = 1<=k<4, so k=1, k=, k=3] m14 = min (m11+m4+13x5x34, // for k=1 m1+m34+13x89x34, // for k= m13+m14+13x3x34) // for k=3 = min (455,541,856) = 856 i=1,,,ns 31
32 hained Matrix Multiplication j=1 3 4 i= s=3 s= s=1 s= Solution: (( ( )) D) is 13X5, is 5X89, is 89X3 and D is 3X34 [()= ()=195 + (())D=136] = 856 3
33 GTU Practice Examples Given the four matrix find out optimal sequence for multiplication D=<5,4,6,,7> Using algorithm find an optimal parenthesization of a matrix chain product whose sequence of dimension is 5,1,3,1,5,5,6 (use dynamic programming). 33
34 Longest ommon Subsequence Problem using Dynamic Programming 34
35 Dynamic programming It is used, when the solution can be recursively described in terms of solutions to subproblems (optimal substructure) lgorithm finds solutions to subproblems and stores them in memory for later use More efficient than bruteforce methods, which solve the same subproblems over and over again 35
36 Longest ommon Subsequence (LS) pplication: comparison of two DN strings Ex: X= { D }, Y= { D } Longest ommon Subsequence: X = D Y = D rute force algorithm would compare each subsequence of X with the symbols in Y 36
37 LS lgorithm if X = m, Y = n, then there are m subsequences of x; we must compare each with Y (n comparisons) So the running time of the bruteforce algorithm is O(n m ) Notice that the LS problem has optimal substructure: solutions of subproblems are parts of the final solution. Subproblems: find LS of pairs of prefixes of X and Y 37
38 LS lgorithm First we ll find the length of LS. Later we ll modify the algorithm to find LS itself. Define X i, Y j to be the prefixes of X and Y of length i and j respectively Define c[i,j] to be the length of LS of X i and Y j Then the length of LS of X and Y will be c[m,n] c[ i, j] c[ i 1, j 1] 1 max( c[ i, j 1], c[ i 1, j]) if x[ i] y[ otherwise j], 38
39 LS recursive solution c[ i, j] c[ i 1, j 1] 1 max( c[ i, j 1], c[ i 1, j]) if x[ i] y[ otherwise j], We start with i = j = (empty substrings of x and y) Since X and Y are empty strings, their LS is always empty (i.e. c[,] = ) LS of empty string and any other string is empty, so for every i and j: c[, j] = c[i,] = 39
40 LS recursive solution c[ i, j] c[ i 1, j 1] 1 max( c[ i, j 1], c[ i 1, j]) if x[ i] y[ otherwise j], When we calculate c[i,j], we consider two cases: First case: x[i]=y[j]: one more symbol in strings X and Y matches, so the length of LS X i and Y j equals to the length of LS of smaller strings X i1 and Y i1, plus 1 4
41 LS recursive solution c[ i, j] c[ i 1, j 1] 1 max( c[ i, j 1], c[ i 1, j]) if x[ i] y[ otherwise j], Second case: x[i]!= y[j] s symbols don t match, our solution is not improved, and the length of LS(X i, Y j ) is the same as before (i.e. maximum of LS(X i, Y j1 ) and LS(X i1,y j ) 41
42 LS Example We ll see how LS algorithm works on the following example: X = Y = D What is the Longest ommon Subsequence of X and Y? LS(X, Y) = X = Y = D 4
43 LS Example () j D i Yj D Xi X = ; m = X = 4 Y = D; n = Y = 5 llocate array c[4,5] 43
44 LS Example (1) j D i Yj D Xi for i = 1 to m c[i,] = for j = 1 to n c[,j] = 44
45 LS Example () j D i Yj D Xi if ( X i == Y j ) c[i,j] = c[i1,j1] + 1 else c[i,j] = max( c[i1,j], c[i,j1] ) 45
46 LS Example (3) j D i Yj D Xi if ( X i == Y j ) c[i,j] = c[i1,j1] + 1 else c[i,j] = max( c[i1,j], c[i,j1] ) 46
47 LS Example (4) j D i Yj D Xi if ( X i == Y j ) c[i,j] = c[i1,j1] + 1 else c[i,j] = max( c[i1,j], c[i,j1] ) 47
48 LS Example (5) j D i Yj D Xi if ( X i == Y j ) c[i,j] = c[i1,j1] + 1 else c[i,j] = max( c[i1,j], c[i,j1] ) 48
49 LS Example (6) j D i Yj D Xi if ( X i == Y j ) c[i,j] = c[i1,j1] + 1 else c[i,j] = max( c[i1,j], c[i,j1] ) 49
50 LS Example (7) j D i Yj D Xi if ( X i == Y j ) c[i,j] = c[i1,j1] + 1 else c[i,j] = max( c[i1,j], c[i,j1] ) 5
51 LS Example (8) j D i Yj D Xi if ( X i == Y j ) c[i,j] = c[i1,j1] + 1 else c[i,j] = max( c[i1,j], c[i,j1] ) 51
52 LS Example (1) j D i Yj D Xi if ( X i == Y j ) c[i,j] = c[i1,j1] + 1 else c[i,j] = max( c[i1,j], c[i,j1] ) 5
53 LS Example (11) j D i Yj D Xi if ( X i == Y j ) c[i,j] = c[i1,j1] + 1 else c[i,j] = max( c[i1,j], c[i,j1] ) 53
54 LS Example (1) j D i Yj D Xi if ( X i == Y j ) c[i,j] = c[i1,j1] + 1 else c[i,j] = max( c[i1,j], c[i,j1] ) 54
55 LS Example (13) j D i Yj D Xi if ( X i == Y j ) c[i,j] = c[i1,j1] + 1 else c[i,j] = max( c[i1,j], c[i,j1] ) 55
56 LS Example (14) j D i Yj D Xi if ( X i == Y j ) c[i,j] = c[i1,j1] + 1 else c[i,j] = max( c[i1,j], c[i,j1] ) 56
57 LS Example (15) j D i Yj D Xi if ( X i == Y j ) c[i,j] = c[i1,j1] + 1 else c[i,j] = max( c[i1,j], c[i,j1] ) 57
58 LS lgorithm Running Time LS algorithm calculates the values of each entry of the array c[m,n] So what is the running time? O(m*n) since each c[i,j] is calculated in constant time, and there are m*n elements in the array 58
59 How to find actual LS So far, we have just found the length of LS, but not LS itself. We want to modify this algorithm to make it output Longest ommon Subsequence of X and Y Each c[i,j] depends on c[i1,j] and c[i,j1] or c[i1, j1] For each c[i,j] we can say how it was acquired: 3 For example, here c[i,j] = c[i1,j1] +1 = +1=3 59
60 How to find actual LS  continued Remember that c[ i, j] c[ i 1, j 1] 1 max( c[ i, j 1], c[ i 1, j]) if x[ i] y[ otherwise j], So we can start from c[m,n] and go backwards Whenever c[i,j] = c[i1, j1]+1, remember x[i] (because x[i] is a part of LS) When i= or j= (i.e. we reached the beginning), output remembered letters in reverse order 6
61 Finding LS j i Yj D Xi
62 Finding LS () j i Yj D Xi LS (reversed order): LS (straight order): (this string turned Prof. S.J. Soni out  SPE, to Visnagar be a palindrome) 6
63 GTU Practice Examples Given two sequences of characters, P=<MLNOM> Q=<MNOM> Obtain the longest common subsequence. Find Longest ommon Subsequence using Dynamic Programming Technique with illustration X={,,,,D,,} Y={,D,,,,} Using algorithm determine an Longest ommon Sequence of,,,d,,,,d,f and,,,f (use dynamic programming). Explain how to find out Longest ommon Subsequence of two strings using Dynamic Programming method.find any one Longest ommon Subsequence of given two strings using Dynamic Programming. S1=abbacdcba S=bcdbbcaac 63
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