Algorithmic Paradigms. Chapter 6 Dynamic Programming. Steps in Dynamic Programming. Dynamic Programming. Dynamic Programming Applications

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1 lgorithmic Paradigms reed. Build up a solution incrementally, only optimizing some local criterion. hapter Dynamic Programming Divide-and-conquer. Break up a problem into two sub-problems, solve each sub-problem independently, and combine solution to subproblems to form solution to original problem. Dynamic programming. Break up a problem into a series of overlapping sub-problems, and build up solutions to larger and larger subproblems. Dynamic Programming History Richard Bellman. Pioneered the systematic study of dynamic programming in the 9s. HOIE OF HE NME DYNMI PRORMMIN. Dynamic programming = planning over time. Secretary of Defense was hostile to mathematical research. Bellman sought an impressive name to avoid confrontation. "it's impossible to use dynamic in a pejorative sense" "something not even a ongressman could object to" Dynamic Programming pplications reas. Bioinformatics. ontrol theory. Information theory. Operations research. omputer science: theory, graphics, I, systems,. Some famous dynamic programming algorithms. Bellman-Ford for shortest path routing in networks. Smith-Waterman for sequence alignment. Viterbi for hidden Markov models. Unix diff for comparing two files. Reference: Bellman, R. E. Eye of the Hurricane, n utobiography. Dynamic Programming Dynamic Programming is an algorithm design technique for optimization problems: often minimizing or maximizing. Like divide and conquer, DP solves problems by combining solutions to subproblems. Unlike divide and conquer, subproblems are not independent. Subproblems may share subsubproblems, However, solution to one subproblem may not affect the solutions to other subproblems of the same problem. DP reduces computation by Solving subproblems in a bottom-up fashion. Storing solution to a subproblem the first time it is solved. Looking up the solution when subproblem is encountered again. Key: determine structure of optimal solutions Steps in Dynamic Programming. haracterize the structure of an optimal solution.. Define value of optimal solution recursively.. ompute optimal solution values either top-down with caching or bottom-up in a table.. onstruct an optimal solution from computed values.

2 Examples Longest increasing subsequence Problem: iven a sequence of numbers a, a, a,, a n Find max value of k such that a i < a i < a i < a ik and i < i < < i k. Example: 8 9 Naïve lgorithm For every subsequence of X, check whether it s increasing. ime:? Θ(n n ). n subsequences of X to check. Each subsequence takes Θ(n) time to check: scan Y for first letter, for second, and so on. Longest increasing subsequence Longest increasing subsequence { } {, } 8 9 Length of the subsequence ended with,, 8,,,, 9, n + {,, 8 } {,, 8, } {,, 8,, } {,, 8,,, } {,, 8,,,, 9 } {,, 8,,,, 9, } Longest ommon Subsequence Problem: iven sequences, X = x,...,x m and Y = y,...,y n, find a common subsequence whose length is maximum. springtime ncaa tournament basketball printing akron krzyzewski Subsequence need not be consecutive, but must be in order. Naïve lgorithm For every subsequence of X, check whether it s a subsequence of Y. ime:? Θ(n m ). m subsequences of X to check. Each subsequence takes Θ(n) time to check: scan Y for first letter, for second, and so on.

3 Longest ommon Subsequence Problem: iven sequences, X = x,...,x m and Y = y,...,y n, find a common subsequence whose length is maximum. springtime printing Optimal Substructure heorem: Let Z = z,..., z k be any LS of X and Y.. If x m = y n, then z k = x m = y n and Z k- is an LS of X m- and Y n-.. If x m y n, then either z k x m and Z is an LS of X m- and Y.. or z k y n and Z is an LS of X and Y n-. Notation: prefix X i = x,...,x i is the first i letters of X. Optimal Substructure heorem: Let Z = z,..., z k be any LS of X and Y.. If x m = y n, then z k = x m = y n and Z k- is an LS of X m- and Y n-.. If x m y n, then either z k x m and Z is an LS of X m- and Y.. or z k y n and Z is an LS of X and Y n-. Proof: (case : x m = y n ) ny sequence Z that does not end in x m = y n can be made longer by adding x m = y n to the end. herefore, () longest common subsequence (LS) Z must end in x m = y n. () Z k- is a common subsequence of X m- and Y n-, and () there is no longer S of X m- and Y n-, or Z would not be an LS. Optimal Substructure heorem: Let Z = z,..., z k be any LS of X and Y.. If x m = y n, then z k = x m = y n and Z k- is an LS of X m- and Y n-.. If x m y n, then either z k x m and Z is an LS of X m- and Y.. or z k y n and Z is an LS of X and Y n-. Proof: (case : x m y n, and z k x m ) Since Z does not end in x m, () Z is a common subsequence of X m- and Y, and () there is no longer S of X m- and Y, or Z would not be an LS. Recursive Solution Recursive Solution Define c[i, j] = length of LS of X i and Y j, where X i = x,...,x i is the first i letters of X.. We want c[m,n]. c[springtime, printing] c[springtim, printing] c[springtime, printin] [springti, printing] [springtim, printin] [springtim, printin] [springtime, printi] his gives a recursive algorithm and solves the problem. But does it solve it well? [springt, printing] [springti, printin] [springtim, printi] [springtime, print] ost: O(?)

4 Steps in Dynamic Programming Recursive Solution. haracterize structure of an optimal solution.. Define value of optimal solution recursively.. ompute optimal solution values either top-down with caching or bottom-up in a table.. onstruct an optimal solution from computed values. Keep track of c[ ] in a table of nm entries: top/down bottom/up ost: O(?) s p r i n g t i m e p r i n t i n g omputing the length of an LS LS-LENH (X, Y). m length[x]. n length[y]. for i to m. do c[i, ]. for j to n. do c[, j ]. for i to m 8. do for j to n 9. do if x i = y j. then c[i, j ] c[i, j ] +. b[i, j ]. else if c[i, j ] c[i, j ]. then c[i, j ] c[i, j ]. b[i, j ]. else c[i, j ] c[i, j ]. b[i, j ]. return c and b b[i, j ] points to table entry whose subproblem we used in solving LS of X i and Y j. c[m,n] contains the length of an LS of X and Y. ime: O(mn) onstructing an LS PRIN-LS (b, X, i, j). if i = or j =. then return. if b[i, j ] =. then PRIN-LS(b, X, i, j ). print x i. elseif b[i, j ] =. then PRIN-LS(b, X, i, j) 8. else PRIN-LS(b, X, i, j ) Initial call is PRIN-LS (b, X,m, n). When b[i, j ] =, we have extended LS by one character. So LS = entries with in them. ime: O(m+n) Weighted Interval Scheduling Weighted Interval Scheduling Weighted interval scheduling problem. Job j starts at s j, finishes at f j, and has weight or value v j. wo jobs compatible if they don't overlap. oal: find maximum weight subset of mutually compatible jobs. a b c d e f g h 8 9 ime

5 Unweighted Interval Scheduling reedy algorithm works if all weights are. onsider jobs in ascending order of finish time. dd job to subset if it is compatible with previously chosen jobs. Weighted Interval Scheduling Notation. Label jobs by finishing time: f f... f n. Let p(j)=largest index i (<j) s.t. job i is compatible with j. Ex: p(8) =?, p() =?, p() =?. weight = 999 weight = a b 8 9 ime Observation. reedy algorithm can fail spectacularly if arbitrary weights are allowed ime Dynamic Programming: Binary hoice Problem: Find mutually compatible jobs that maximize total value. Let OP(j) = value of optimal solution to the problem consisting of job requests,,..., j. ase : OP selects job j. can't use incompatible jobs { p(j) +, p(j) +,... } must include optimal solution to problem consisting of remaining compatible jobs,,..., p(j) optimal substructure ase : OP does not select job j. must include optimal solution to problem consisting of remaining compatible jobs,,..., j- ost? Weighted Interval Scheduling: Brute Force Brute force algorithm. Input: n, s,,s n, f,,f n, v,,v n Sort jobs by finish times so that f f... f n. ompute p(), p(),, p(n) ompute-opt(j) { if (j = ) return else return max(v j + ompute-opt(p(j)), ompute-opt(j-)) } Weighted Interval Scheduling: Brute Force p() =, p(j) = j- (+ )/ Ex. Number of recursive calls for family of "layered" instances grows like Fibonacci sequence. Observation. Recursive algorithm fails spectacularly because of redundant sub-problems exponential algorithms.

6 Weighted Interval Scheduling: Memoization Memoization. Store results of each sub-problem in a cache; lookup as needed. Input: n, s,,s n, f,,f n, v,,v n Sort jobs by finish times so that f f... f n. ompute p(), p(),, p(n) for j = to n M[j] = empty global array M-ompute-Opt(j) { if (M[j] is empty) M[j] = max(w j + M-ompute-Opt(p(j)), M-ompute-Opt(j-)) return M[j] } Weighted Interval Scheduling: Running ime laim. Memorized version of alg takes O(n log n) time. Sort by finish time: O(n log n). omputing p( ) : O(n) after sorting by start time. M-ompute-Opt(j): each invocation takes O() time & either (i) returns an existing value M[j] (ii) fills in one new entry M[j] and makes two recursive calls Progress measure = # nonempty entries of M[]. initially =, throughout n. (ii) increases by at most n recursive calls. Overall running time of M-ompute-Opt(n) is O(n). Note: O(n) if jobs are presorted by start & finish times. Weighted Interval Scheduling: Finding a Solution Q. Dynamic programming algorithms computes optimal value. What if we want the solution itself?. Do some post-processing. Run M-ompute-Opt(n) Run Find-Solution(n) Find-Solution(j) { if (j = ) output nothing else if (v j + M[p(j)] > M[j-]) print j Find-Solution(p(j)) else Find-Solution(j-) } # of recursive calls n O(n). Weighted Interval Scheduling: Bottom-Up Bottom-up dynamic programming. Unwind recursion. Input: n, s,,s n, f,,f n, v,,v n Sort jobs by finish times so that f f... f n. ompute p(), p(),, p(n) Iterative-ompute-Opt { M[] = for j = to n M[j] = max(v j + M[p(j)], M[j-]) } exercise ost of calculating p(i) Problem: Find mutually compatible jobs that maximize total value. Start (sorted) Finish (sorted)

7 String Similarity Sequence lignment How similar are two strings? ocurrance occurrence o c u r r a n c e - o c c u r r e n c e mismatches, gap o c - u r r a n c e o c c u r r e n c e mismatch, gap o c - u r r - a n c e o c c u r r e - n c e mismatches, gaps pplications. Basis for Unix diff. Speech recognition. omputational biology. Edit Distance Edit distance. [Levenshtein 9, Needleman-Wunsch 9] ap penalty ; mismatch penalty pq. ost = sum of gap and mismatch penalties. Scoring Matrices for ligning DN Sequences ransition-ransversion matrix Sequence lignment oal: iven two strings X = x x... x m and Y = y y... y n find alignment of minimum cost. Def. n alignment M is a set of ordered pairs x i -y j such that each item occurs in at most one pair and no crossings. Sequence lignment: Problem Structure Let OP(i, j) = min cost of aligning strings x x... x i and y y... y j. ase : OP matches x i -y j. pay mismatch for x i -y j + min cost of aligning two strings x x... x i- and y y... y j- ase a: OP leaves x i unmatched. pay gap for x i and min cost of aligning x x... x i- and y y... y j ase b: OP leaves y j unmatched. pay gap for y j and min cost of aligning x x... x i and y y... y j- Ex: vs.. Sol: M = x -y, x -y, x -y, x -y, x -y. x x x x x x - - y y y y y y

8 Sequence lignment: lgorithm Exercise - Let gap = Sequence-lignment(m, n, x x...x m, y y...y n,, ) { for i = to m M[, i] = i for j = to n M[j, ] = j empty - ransition-ransversion matrix } for i = to m for j = to n M[i, j] = min( [x i, y j ] + M[i-, j-], + M[i-, j], + M[i, j-]) return M[m, n] empty Let gap = -8 Exercise Elements of Dynamic Programming - empty empty ransition-ransversion matrix Optimal substructure Overlapping subproblems Bottom-up approach Knapsack Problem Knapsack Problem Knapsack problem. iven n objects and a "knapsack." Item i weighs w i > kilograms and has value v i >. Knapsack has capacity of W kilograms. oal: fill knapsack so as to maximize total value. Item Value Let W= Weight 8 8 reedy method: repeatedly add item with max ratio v i /w i. Ex: {,, } achieves only value = optimal? Ex: {, } has value. 8

9 Dynamic Programming: False Start Dynamic Programming Let OP(i)=max profit for subset of items,, i. ase : OP does not select item i. OP selects best of {,,, i- } ase : OP selects item i. accepting item i does not immediately imply that we will have to reject other items without knowing what other items were selected before i, we don't even know if we have enough room for i Let OP(i, w) = max profit subset of items,, i with weight limit w. ase : OP does not select item i. OP selects best of {,,, i- } using weight limit w ase : OP selects item i. new weight limit = w w i OP selects best of {,, i } using the new weight limit onclusion: Need more info about sub-problems! Knapsack Problem: Bottom-Up Knapsack lgorithm Knapsack. Fill up an n-by-w array. Input: n, w,,w N, v,,v N for w = to W M[, w] = for i = to n for w = to W if (w i > w) M[i, w] = M[i-, w] else M[i, w] = max {M[i-, w], v i + M[i-, w-w i ]} return M[n, W] { } n + {, } {,, } {,,, } {,,,, } OP: {, } value = + 8 = M[i, w] = max {M[i-, w], v i +M[i-, w-w i ]} W + 8? W = =max{, 8+M[, ]} Item Value Weight 8 8 Knapsack Problem: Running ime Running time. (n W). Not polynomial in input size! "Pseudo-polynomial." Decision version of Knapsack is NP-complete. Knapsack approximation algorithm. here exists a polynomial algorithm that produces a feasible solution that has value within.% of optimum. oin-hanging: Dynamic programming $ postage reedy:,,,,. Optimal:,. reedy algorithm failed!? 9

10 Shortest path Seam carving Energy function he Operator iven an energy function e, we can define the cost of a seam as look for the optimal seam s* that minimizes this seam cost : he optimal seam can be found using dynamic programming.

11 Bonus p/squared-pics/id99 and a nice review