Copyright 2000, Kevin Wayne 1

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1 /9/ lgorithmic Paradigms hapter Dynamic Programming reed. Build up a solution incrementally, myopically optimizing some local criterion. Divide-and-conquer. Break up a problem into two sub-problems, solve each sub-problem independently, and combine solution to sub-problems 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 sub-problems. Slides by Kevin Wayne. opyright Pearson-ddison Wesley. ll rights reserved. Dynamic Programming History Dynamic Programming pplications Bellman. Pioneered the systematic study of dynamic programming in the 9s. Etymology. 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" Reference: Bellman, R. E. Eye of the Hurricane, n utobiography. reas. Bioinformatics. ontrol theory. Information theory. Operations research. omputer science: theory, graphics, I, systems,. Some famous dynamic programming algorithms. Viterbi for hidden Markov models. nix diff for comparing two files. Smith-Waterman for sequence alignment. Bellman-Ford for shortest path routing in networks. ocke-kasami-younger for parsing context free grammars. opyright, Kevin Wayne

2 /9/ 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. Two jobs compatible if they don't overlap. oal: find maximum weight subset of mutually compatible jobs. a b c d e f g 8 9 h Time nweighted Interval Scheduling Review Weighted Interval Scheduling Recall. 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. Notation. Label jobs by finishing time: f f... f n. Def. p(j) = largest index i < j such that job i is compatible with j. Ex: p(8) =, p() =, p() =. Observation. reedy algorithm can fail spectacularly if arbitrary weights are allowed. weight = 999 b weight = a 8 9 Time Time 8 opyright, Kevin Wayne

3 /9/ Dynamic Programming: Binary hoice Weighted Interval Scheduling: Brute Force Notation. OPT(j) = value of optimal solution to the problem consisting of job requests,,..., j. Brute force algorithm. ase : OPT selects job j. can't use incompatible jobs { p(j) +, p(j) +,..., j - must include optimal solution to problem consisting of remaining compatible jobs,,..., p(j) ase : OPT does not select job j. optimal substructure must include optimal solution to problem consisting of remaining compatible jobs,,..., j- # if j = OPT( j) = $ % max { v j + OPT( p( j)), OPT( j ) otherwise 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-)) 9 Weighted Interval Scheduling: Brute Force Weighted Interval Scheduling: Memoization Observation. Recursive algorithm fails spectacularly because of redundant sub-problems exponential algorithms. Memoization. Store results of each sub-problem in a cache; lookup as needed. Ex. Number of recursive calls for family of "layered" instances grows like Fibonacci sequence. 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) p() =, p(j) = j- for j = to n M[j] = empty M[j] = 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] opyright, Kevin Wayne

4 /9/ Weighted Interval Scheduling: Running Time laim. Memoized version of algorithm 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 and either (i) returns an existing value M[j] (ii) fills in one new entry M[j] and makes two recursive calls Progress measure Φ = # nonempty entries of M[]. initially Φ =, throughout Φ n. (ii) increases Φ by at most n recursive calls. Overall running time of M-ompute-Opt(n) is O(n). Remark. O(n) if jobs are pre-sorted by start and finish times. omputing the Index p(.) in Linear Time Interval 8 Finish time 8 9 Interval 8 Start time 8 Interval 8 p(.) 8 Time 8 9 utomated Memoization Weighted Interval Scheduling: Finding a Solution utomated memoization. Many functional programming languages (e.g., Lisp) have built-in support for memoization. Q. Why not in imperative languages (e.g., Java)? Q. Dynamic programming algorithms computes optimal value. What if we want the solution itself?. Do some post-processing. (defun F (n) (if (<= n ) n (+ (F (- n )) (F (- n ))))) Lisp (efficient) static int F(int n) { if (n <= ) return n; else return F(n-) + F(n-); F(8) F(9) Java (exponential) F() F() F() F(8) F() 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-) F() F() F() F() F() F() F() F() # of recursive calls n O(n). opyright, Kevin Wayne

5 /9/ Weighted Interval Scheduling: Bottom-p Bottom-up dynamic programming. nwind recursion.. Segmented Least Squares 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-]) Segmented Least Squares Segmented Least Squares Least squares. Foundational problem in statistic and numerical analysis. iven n points in the plane: (x, y ), (x, y ),..., (x n, y n ). Find a line y = ax + b that minimizes the sum of the squared error: Segmented least squares. Points lie roughly on a sequence of several line segments. iven n points in the plane (x, y ), (x, y ),..., (x n, y n ) with x < x <... < x n, find a sequence of lines that minimizes f(x). n SSE = ( y i ax i b) i= y Q. What's a reasonable choice for f(x) to balance accuracy and parsimony? number of lines goodness of fit x y Solution. alculus min error is achieved when a = n x i i y i ( i x i ) ( i y i ), b = n x i ( x i ) i i i y i a i x i n x 9 opyright, Kevin Wayne

6 /9/ Segmented Least Squares Dynamic Programming: Multiway hoice Segmented least squares. Points lie roughly on a sequence of several line segments. iven n points in the plane (x, y ), (x, y ),..., (x n, y n ) with x < x <... < x n, find a sequence of lines that minimizes: the sum of the sums of the squared errors E in each segment the number of lines L Tradeoff function: E + c L, for some constant c >. Notation. OPT(j) = minimum cost for points p, p i+,..., p j. e(i, j) = minimum sum of squares for points p i, p i+,..., p j. To compute OPT(j): Last segment uses points p i, p i+,..., p j for some i. ost = e(i, j) + c + OPT(i-). y $ & if j = OPT( j) = % min { e(i, j) + c + OPT(i ) otherwise '& i j x Segmented Least Squares: lgorithm INPT: n, p,,p N, c. Knapsack Problem Segmented-Least-Squares() { M[] = for j = to n for i = to j compute the least square error e ij for the segment p i,, p j for j = to n M[j] = min i j (e ij + c + M[i-]) return M[n] Running time. O(n ). can be improved to O(n ) by pre-computing various statistics Bottleneck = computing e(i, j) for O(n ) pairs, O(n) per pair using previous formula. opyright, Kevin Wayne

7 /9/ Knapsack Problem Dynamic Programming: False Start 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. Ex: {, has value. W = Item Value reedy: repeatedly add item with maximum ratio v i / w i. Ex: {,, achieves only value = greedy not optimal. 8 8 Weight Def. OPT(i) = max profit subset of items,, i. ase : OPT does not select item i. OPT selects best of {,,, i- ase : OPT 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 onclusion. Need more sub-problems! Dynamic Programming: dding a New Variable Knapsack Problem: Bottom-p Def. OPT(i, w) = max profit subset of items,, i with weight limit w. Knapsack. Fill up an n-by-w array. ase : OPT does not select item i. OPT selects best of {,,, i- using weight limit w ase : OPT selects item i. new weight limit = w w i OPT selects best of {,,, i using this new weight limit # if i = % OPT(i, w) = $ OPT(i, w) if w i > w % & max{ OPT(i, w), v i + OPT(i, w w i ) otherwise 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] 8 opyright, Kevin Wayne

8 /9/ Knapsack lgorithm Knapsack Problem: Running Time φ W Running time. Θ(n W). Not polynomial in input size! "Pseudo-polynomial." Decision version of Knapsack is NP-complete. [hapter 8] n + { {, {,, {,,, Knapsack approximation algorithm. There exists a polynomial algorithm that produces a feasible solution that has value within.% of optimum. [Section.8] {,,,, Item Value Weight OPT: {, value = + 8 = W = 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 opyright, Kevin Wayne 8

9 /9/ Edit Distance Sequence lignment pplications. Basis for nix diff. Speech recognition. omputational biology. Edit distance. [Levenshtein 9, Needleman-Wunsch 9] ap penalty δ; mismatch penalty α pq. ost = sum of gap and mismatch penalties. T T T - T T T 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. Def. The pair x i -y j and x i' -y j' cross if i < i', but j > j'. cost( M ) = α xi y j + δ + δ (x i, y j ) M i : x!#" # $ i unmatched j : y j unmatched!### #" ##### $ mismatch gap T T T α T + α T + α + α T - T T δ + α Ex: T vs. TT. Sol: M = x -y, x -y, x -y, x -y, x -y. x x x x x x T - - T T y y y y y y Sequence lignment: Problem Structure Sequence lignment: lgorithm Def. OPT(i, j) = min cost of aligning strings x x... x i and y y... y j. ase : OPT 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: OPT 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: OPT leaves y j unmatched. pay gap for y j and min cost of aligning x x... x i and y y... y j- " jδ if i = $ " α xi y j + OPT(i, j ) $ $ OPT(i, j) = # min # δ + OPT(i, j) otherwise $ $ δ + OPT(i, j ) % % $ iδ if j = Sequence-lignment(m, n, x x...x m, y y...y n, δ, α) { for i = to m M[, i] = iδ for j = to n M[j, ] = jδ for i = to m for j = to n M[i, j] = min(α[x i, y j ] + M[i-, j-], δ + M[i-, j], δ + M[i, j-]) return M[m, n] nalysis. Θ(mn) time and space. English words or sentences: m, n. omputational biology: m = n =,. billions ops OK, but B array? opyright, Kevin Wayne 9

10 /9/ Sequence lignment: lgorithm y n +. RN Secondary Structure φ T T x m + φ T ap penalty = Mismatch penalty = x x x x x x T - - T T y y y y y y RN Secondary Structure RN Secondary Structure RN. String B = b b b n over alphabet {,,,. Secondary structure. RN is single-stranded so it tends to loop back and form base pairs with itself. This structure is essential for understanding behavior of molecule. Ex: Secondary structure. set of pairs S = { (b i, b j ) that satisfy: [Watson-rick.] S is a matching and each pair in S is a Watson- rick complement: -, -, -, or -. [No sharp turns.] The ends of each pair are separated by at least intervening bases. If (b i, b j ) S, then i < j -. [Non-crossing.] If (b i, b j ) and (b k, b l ) are two pairs in S, then we cannot have i < k < j < l. Free energy. sual hypothesis is that an RN molecule will form the secondary structure with the optimum total free energy. approximate by number of base pairs oal. iven an RN molecule B = b b b n, find a secondary structure S that maximizes the number of base pairs. complementary base pairs: -, - 9 opyright, Kevin Wayne

11 /9/ RN Secondary Structure: Examples RN Secondary Structure: Subproblems Examples. First attempt. OPT(j) = maximum number of base pairs in a secondary structure of the substring b b b j. match b t and b n base pair t n Difficulty. Results in two sub-problems. Finding secondary structure in: b b b t-. Finding secondary structure in: b t+ b t+ b n-. OPT(t-) need more sub-problems ok sharp turn crossing Dynamic Programming Over Intervals Bottom p Dynamic Programming Over Intervals Notation. OPT(i, j) = maximum number of base pairs in a secondary structure of the substring b i b i+ b j. Q. What order to solve the sub-problems?. Do shortest intervals first. ase. If i j -. OPT(i, j) = by no-sharp turns condition. ase. Base b j is not involved in a pair. OPT(i, j) = OPT(i, j-) RN(b,,b n ) { for k =,,, n- for i =,,, n-k j = i + k ompute M[i, j] i ase. Base b j pairs with b t for some i t < j -. non-crossing constraint decouples resulting sub-problems OPT(i, j) = + max t { OPT(i, t-) + OPT(t+, j-) return M[, n] using recurrence 8 9 j take max over t such that i t < j- and b t and b j are Watson-rick complements Running time. O(n ). Remark. Same core idea in KY algorithm to parse context-free grammars. opyright, Kevin Wayne

12 /9/ Dynamic Programming Summary Recipe. haracterize structure of problem. Recursively define value of optimal solution. ompute value of optimal solution. onstruct optimal solution from computed information. Dynamic programming techniques. Binary choice: weighted interval scheduling. Multi-way choice: segmented least squares. dding a new variable: knapsack. Viterbi algorithm for HMM also uses DP to optimize a maximum likelihood tradeoff between parsimony and accuracy Dynamic programming over intervals: RN secondary structure. KY parsing algorithm for context-free grammar has similar structure Top-down vs. bottom-up: different people have different intuitions. opyright, Kevin Wayne