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CSC 310 – Algorithms and Data Structures Dynamic programming Longest Common Subsequence

CSC 310 – Algorithms and Data Structures Dynamic programming Longest Common Subsequence. Dynamic programming. It is used, when the solution can be recursively described in terms of solutions to subproblems ( optimal substructure )

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CSC 310 – Algorithms and Data Structures Dynamic programming Longest Common Subsequence

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  1. CSC 310 – Algorithms and Data Structures Dynamic programming Longest Common Subsequence

  2. Dynamic programming • It is used, when the solution can be recursively described in terms of solutions to subproblems (optimal substructure) • Algorithm finds solutions to subproblems and stores them in memoryfor later use • More efficient than “brute-force methods”, which solve the same subproblems over and over again

  3. Dynamic Programming • In computer science, a problem which can be broken down recursively is said to have optimal substructure. • Two key attributes for a problem to apply dynamic programming: optimal substructure andoverlapping sub-problems which are only slightly smaller. • When the overlapping problems are half the size of the original problem the strategy is called "divide and conquer" rather than "dynamic programming". (mergesort, quicksort)

  4. Dynamic Programming • Optimal substructure means that the solution to a given optimization problem can be obtained by the combination of optimal solutions to its sub-problems. • Consequently, the first step towards devising a dynamic programming solution is to check whether the problem exhibits such optimal substructure. Such optimal substructures are usually described by means of recursion.

  5. Dynamic Programming • Overlapping sub-problems means that the space of sub-problems must be small, that is, any recursive algorithm solving the problem should solve the same sub-problems over and over, rather than generating new sub-problems. • Dynamic programming takes account of this fact and solves each sub-problem only once.

  6. Dynamic Programming • If sub-problems are overlapping, then one can easily memorize or store the solutions to the sub-problems in a table. • Whenever we attempt to solve a new sub-problem, we first check the table to see if it is already solved. • If a solution has been recorded, we can use it directly; otherwise we solve the sub-problem and add its solution to the table.

  7. Dynamic Programming • Solve the sub-problems first and use their solutions to build-on and arrive at solutions to bigger sub-problems. This is also usually done in a tabular form by • iteratively generating solutions to bigger and bigger sub-problems by using the solutions to small sub-problems. • For example, if we already know the Fibonacci values of F41 and F40, we can directly calculate the value of F42.

  8. Longest Common Subsequence (LCS) Application: comparison of two DNA strings Ex: X= {A B C B D A B }, Y= {B D C A B A} Longest Common Subsequence: X = A BCB D A B Y = B D C A BA Brute force algorithm would compare each subsequence of X with the symbols in Y

  9. LCS Algorithm • if |X| = m, |Y| = n, then there are about 2m subsequences of x; we must compare each with Y (n comparisons) • So the running time of the brute-force algorithm is O(n 2m) • Notice that the LCS problem has optimal substructure: solutions of subproblems are parts of the final solution. • Subproblems: “find LCS of pairs of prefixes of X and Y” Notation: prefix Xi= x1,...,xiis the first iletters of X.

  10. LCS Algorithm Theorem Let Z = z1, . . . , zk be any LCS of X and Y . 1. If xm= yn, then zk= xm= ynand Zk-1 is an LCS of Xm-1 and Yn-1. 2. If xmyn, then either zkxmand Z is an LCS of Xm-1 and Y . 3. or zkynand Z is an LCS of X and Yn-1. Notation: prefix Xi= x1,...,xiis the first i letters of X.

  11. LCS Algorithm Theorem Let Z = z1, . . . , zk be any LCS of X and Y . 1. If xm= yn, then zk= xm= ynand Zk-1 is an LCS of Xm-1 and Yn-1. 2. If xmyn, then either zkxmand Z is an LCS of Xm-1 and Y . 3. or zkynand Z is an LCS of X and Yn-1. Proof: (case 1: xm= yn) Any sequence Z’ that does not end in xm= yncan be made longer by adding xm= ynto the end. Therefore, • longest common subsequence (LCS) Z must end in xm= yn. • Zk-1 is a common subsequence of Xm-1 and Yn-1, and • there is no longer CS of Xm-1 and Yn-1, or Z would not be an LCS.

  12. LCS Algorithm Theorem Let Z = z1, . . . , zk be any LCS of X and Y . 1. If xm= yn, then zk= xm= ynand Zk-1 is an LCS of Xm-1 and Yn-1. 2. If xmyn, then either zkxmand Z is an LCS of Xm-1 and Y . 3. or zkynand Z is an LCS of X and Yn-1. Proof: (case 2: xmyn, andzkxm) Since Z does not end in xm, • Z is a common subsequence of Xm-1 and Y, and • there is no longer CS of Xm-1 and Y, or Z would not be an LCS.

  13. LCS Algorithm • First we’ll find the length of LCS. Later we’ll modify the algorithm to find LCS itself. • Define Xi, Yj to be the prefixes of X and Y of length i and j respectively • Define c[i,j] to be the length of LCS of Xi and Yj • Then the length of LCS of X and Y will be c[m,n]

  14. LCS recursive solution • We start with i = j = 0 (empty substrings of x and y) • Since X0 and Y0 are empty strings, their LCS is always empty (i.e. c[0,0] = 0) • LCS of empty string and any other string is empty, so for every i and j: c[0, j] = c[i,0] = 0

  15. LCS recursive solution • 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 LCS Xi and Yjequals to the length of LCS of smaller strings Xi-1 and Yi-1 , plus 1

  16. LCS recursive solution • Second case:x[i] != y[j] • As symbols don’t match, our solution is not improved, and the length of LCS(Xi , Yj) is the same as before (i.e. maximum of LCS(Xi, Yj-1) and LCS(Xi-1,Yj)

  17. LCS Length Algorithm LCS-Length(X, Y) 1. m = length(X) // get the # of symbols in X 2. n = length(Y) // get the # of symbols in Y 3. for i = 1 to m c[i,0] = 0 // special case: Y0 4. for j = 1 to n c[0,j] = 0 // special case: X0 5. for i = 1 to m // for all Xi 6. for j = 1 to n // for all Yj 7. if ( Xi == Yj ) 8. c[i,j] = c[i-1,j-1] + 1 9. else c[i,j] = max( c[i-1,j], c[i,j-1] ) 10. return c

  18. LCS Example We’ll see how LCS algorithm works on the following example: • X = ABCB • Y = BDCAB What is the Longest Common Subsequence of X and Y? LCS(X, Y) = BCB X = A BCB Y = B D C A B

  19. ABCB BDCAB LCS Example (0) j 0 1 2 3 4 5 i Yj B D C A B Xi 0 A 1 B 2 3 C 4 B X = ABCB; m = |X| = 4 Y = BDCAB; n = |Y| = 5 Allocate array c[5,4]

  20. ABCB BDCAB LCS Example (1) j 0 1 2 3 4 5 i Yj B D C A B Xi 0 0 0 0 0 0 0 A 1 0 B 2 0 3 C 0 4 B 0 for i = 1 to m c[i,0] = 0 for j = 1 to n c[0,j] = 0

  21. ABCB BDCAB LCS Example (2) j 0 1 2 3 4 5 i Yj B D C A B Xi 0 0 0 0 0 0 0 A 1 0 0 B 2 0 3 C 0 4 B 0 if ( Xi == Yj ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] )

  22. ABCB BDCAB LCS Example (3) j 0 1 2 3 4 5 i Yj B D C A B Xi 0 0 0 0 0 0 0 A 1 0 0 0 0 B 2 0 3 C 0 4 B 0 if ( Xi == Yj ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] )

  23. ABCB BDCAB LCS Example (4) j 0 1 2 3 4 5 i Yj B D C A B Xi 0 0 0 0 0 0 0 A 1 0 0 0 0 1 B 2 0 3 C 0 4 B 0 if ( Xi == Yj ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] )

  24. ABCB BDCAB LCS Example (5) j 0 1 2 3 4 5 i Yj B D C A B Xi 0 0 0 0 0 0 0 A 1 0 0 0 0 1 1 B 2 0 3 C 0 4 B 0 if ( Xi == Yj ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] )

  25. ABCB BDCAB LCS Example (6) j 0 1 2 3 4 5 i Yj B D C A B Xi 0 0 0 0 0 0 0 A 1 0 0 0 0 1 1 B 2 0 1 3 C 0 4 B 0 if ( Xi == Yj ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] )

  26. ABCB BDCAB LCS Example (7) j 0 1 2 3 4 5 i Yj B D C A B Xi 0 0 0 0 0 0 0 A 1 0 0 0 0 1 1 B 2 0 1 1 1 1 3 C 0 4 B 0 if ( Xi == Yj ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] )

  27. ABCB BDCAB LCS Example (8) j 0 1 2 3 4 5 i Yj B D C A B Xi 0 0 0 0 0 0 0 A 1 0 0 0 0 1 1 B 2 0 1 1 1 1 2 3 C 0 4 B 0 if ( Xi == Yj ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] )

  28. ABCB BDCAB LCS Example (10) j 0 1 2 3 4 5 i Yj B D C A B Xi 0 0 0 0 0 0 0 A 1 0 0 0 0 1 1 B 2 0 1 1 1 1 2 3 C 0 1 1 4 B 0 if ( Xi == Yj ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] )

  29. ABCB BDCAB LCS Example (11) j 0 1 2 3 4 5 i Yj B D C A B Xi 0 0 0 0 0 0 0 A 1 0 0 0 0 1 1 B 2 0 1 1 1 1 2 3 C 0 1 1 2 4 B 0 if ( Xi == Yj ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] )

  30. ABCB BDCAB LCS Example (12) j 0 1 2 3 4 5 i Yj B D C A B Xi 0 0 0 0 0 0 0 A 1 0 0 0 0 1 1 B 2 0 1 1 1 1 2 3 C 0 1 1 2 2 2 4 B 0 if ( Xi == Yj ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] )

  31. ABCB BDCAB LCS Example (13) j 0 1 2 3 4 5 i Yj B D C A B Xi 0 0 0 0 0 0 0 A 1 0 0 0 0 1 1 B 2 0 1 1 1 1 2 3 C 0 1 1 2 2 2 4 B 0 1 if ( Xi == Yj ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] )

  32. ABCB BDCAB LCS Example (14) j 0 1 2 34 5 i Yj B D C A B Xi 0 0 0 0 0 0 0 A 1 0 0 0 0 1 1 B 2 0 1 1 1 1 2 3 C 0 1 1 2 2 2 4 B 0 1 1 2 2 if ( Xi == Yj ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] )

  33. ABCB BDCAB LCS Example (15) j 0 1 2 3 4 5 i Yj B D C A B Xi 0 0 0 0 0 0 0 A 1 0 0 0 0 1 1 B 2 0 1 1 1 1 2 3 C 0 1 1 2 2 2 3 4 B 0 1 1 2 2 if ( Xi == Yj ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] )

  34. LCS Algorithm Running Time • LCS 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

  35. How to find actual LCS • So far, we have just found the length of LCS, but not LCS itself. • We want to modify this algorithm to make it output Longest Common Subsequence of X and Y Each c[i,j] depends on c[i-1,j] and c[i,j-1] or c[i-1, j-1] For each c[i,j] we can say how it was acquired: For example, here c[i,j] = c[i-1,j-1] +1 = 2+1=3 2 2 2 3

  36. How to find actual LCS - continued • Remember that • So we can start from c[m,n] and go backwards • Whenever c[i,j] = c[i-1, j-1]+1, remember x[i] (because x[i] is a part of LCS) • When i=0 or j=0 (i.e. we reached the beginning), output remembered letters in reverse order

  37. Finding LCS j 0 1 2 3 4 5 i Yj B D C A B Xi 0 0 0 0 0 0 0 A 1 0 0 0 0 1 1 B 2 0 1 1 1 1 2 3 C 0 1 1 2 2 2 3 4 B 0 1 1 2 2

  38. Finding LCS (2) j 0 1 2 3 4 5 i Yj B D C A B Xi 0 0 0 0 0 0 0 A 1 0 0 0 0 1 1 B 2 0 1 1 1 1 2 3 C 0 1 1 2 2 2 3 4 B 0 1 1 2 2 B C B LCS (reversed order): B C B (this string turned out to be a palindrome) LCS (straight order):

  39. Knapsack problem • Given some items, pack the knapsack to get • the maximum total value. Each item has some • weight and some value. Total weight that we can • carry is no more than some fixed number W. • So we must consider weights of items as well as • their value. • Item # Weight Value • 1 1 8 • 2 3 6 • 3 5 5

  40. Knapsack problem There are two versions of the problem: (1) “0-1 knapsack problem” and (2) “Fractional knapsack problem” (1) Items are indivisible; you either take an item or not. Solved with dynamic programming (2) Items are divisible: you can take any fraction of an item. Solved with a greedy algorithm.

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