Dynamic Programming. CIS 110, Fall University of Pennsylvania

Transcription

1 Dynamic Programming CIS 110, Fall 2012 University of Pennsylvania

2 Dynamic Programming Dynamic programming records saves computation for reuse later. Programming: in the optimization sense ( Linear Programming ) Dynamic:... it s impossible to use [it] in a pejorative way. (Richard Bellman) he name was designed to sound cool to RND management and the US Department of Defense more descriptive term is look-up table Richard Bellman 2 CIS 110, Fall 2012

3 Dynamic Programming Dynamic programming records saves computation for reuse later. Programming: in the optimization sense ( Linear Programming ) Dynamic:... it s impossible to use [it] in a pejorative way. (Richard Bellman) he name was designed to sound cool to RND management and the US Department of Defense more descriptive term is look-up table When is dynamic programming useful? Exponential number of solutions Cost of soluion is recursively computed Different solutions recursively compute same values Richard Bellman 2 CIS 110, Fall 2012

4 Dynamic Programming Dynamic programming records saves computation for reuse later. Programming: in the optimization sense ( Linear Programming ) Dynamic:... it s impossible to use [it] in a pejorative way. (Richard Bellman) he name was designed to sound cool to RND management and the US Department of Defense more descriptive term is look-up table When is dynamic programming useful? Exponential number of solutions Cost of soluion is recursively computed Different solutions recursively compute same values he trick is exposing the common subproblems Richard Bellman 2 CIS 110, Fall 2012

5 Rod Cutting Start with a rod of integer length n and cut it into several smaller pieces (of integer length). 3 CIS 110, Fall 2012

6 Rod Cutting Start with a rod of integer length n and cut it into several smaller pieces (of integer length). Now suppose each length has a different value: = 1 = 5 = 8 = = = = = = = 25 3 CIS 110, Fall 2012

7 Rod Cutting Start with a rod of integer length n and cut it into several smaller pieces (of integer length). Now suppose each length has a different value: = 1 = 5 = 8 = = = = = = = 25 How should we cut the rod into pieces? 2 n 1 possibilities for a rod of length n 3 CIS 110, Fall 2012

8 Rod Cutting First rod-cutting strategy (brute-force): For every possible cut, compute the value of the left part plus the value of optimally cutting the right part. ake the best cut. 4 CIS 110, Fall 2012

9 Rod Cutting First rod-cutting strategy (brute-force): For every possible cut, compute the value of the left part plus the value of optimally cutting the right part. ake the best cut. } {{ } Cut recursively } {{ } Cut recursively } {{ } Cut recursively } {{ } Cut recursively }{{} Cut recursively. 4 CIS 110, Fall 2012

10 Rod Cutting First rod-cutting strategy (brute-force): For every possible cut, compute the value of the left part plus the value of optimally cutting the right part. ake the best cut. } {{ } Cut recursively } {{ } Cut recursively Exponential number of recursive calls! 4 CIS 110, Fall 2012 } {{ } Cut recursively } {{ } Cut recursively }{{} Cut recursively.

11 Rod Cutting Second rod-cutting strategy (top-down): For every cut, compute value of left part and store it in a table Find value of optimal cut for right part in table Compute it recursively if it doesn t exist yet 5 CIS 110, Fall 2012

12 Rod Cutting Second rod-cutting strategy (top-down): For every cut, compute value of left part and store it in a table Find value of optimal cut for right part in table Compute it recursively if it doesn t exist yet } {{ } Cut recursively } {{ } Look up cost in table 5 CIS 110, Fall 2012

13 Rod Cutting Second rod-cutting strategy (top-down): For every cut, compute value of left part and store it in a table Find value of optimal cut for right part in table Compute it recursively if it doesn t exist yet } {{ } Cut recursively } {{ } Look up cost in table Reduces computation from O(2 n ) to O(n 2 ) (Why?) Requires an array of length n to store intermediate computations 5 CIS 110, Fall 2012

14 Rod Cutting hrid rod-cutting strategy (bottom-up): Compute the value of a rod of length 1. Store it. Compute the value of a rod of length 2. You can only cut it into rods of length 1. he value of a rod of length 1 is already computed, so there is no recursion. Compute the value of successively longer rods up to length n. he optimal values of shorter rods are always computed first so there is no recursion. 6 CIS 110, Fall 2012

15 Sequence Matching Human genes are coded by four bases: denine (), hymine (), uanine (), Cytosine (C) DN undergoes mutations with each copy: Substitutions: replace one base with another Deletions: some bases are dropped Suppose we isolate a gene in a new organism: C C C Predict function by comparing to genes in know, organism: e.g. C How similar are C C C and C? 7 CIS 110, Fall 2012

16 Sequence Matching How similar are C C C and C? How many mutations to change first sequence into second? How (un)likely is each mutation Edit Distance: minimum cost to convert one string into another. Each change (mutation) has an associated cost: ap 2 Mismatch 1 Match 0 Example matchings: C C C C C C C - - C CIS 110, Fall 2012

17 Edit Distance Brute-force recursive solution: Start at end of sequence and work backwards C C C C 1 C C - 2 C C C C C - C 2 C C C 0 1 C C C C 2 1 C C C C C C - C 0 2 C C 2 2 Recurse until we have all possible matches, then find minimum hree recursive calls per node O(3 n ) matching cost We need to do better! 9 CIS 110, Fall 2012

18 Edit Distance Dynamic Programming recursive solution Consider a pair of characters in the middle: C C C C What is the cost of matching from this pair of s to the end? Cost of matching s (0) + lowest cost of matching C C to C. Brute force solution computes all possible costs Idea: For each pair of characters, keep track of best match up to end 10 CIS 110, Fall 2012

19 Dynamic Programming Idea: For each pair of characters, keep track of best match to end C C C C Initialization: Cost of zero-length match (lower right) is zero Inserting a gap (move right or down in table) costs two 11 CIS 110, Fall 2012

20 Dynamic Programming Idea: For each pair of characters, keep track of best match to end C C C C Iteration: Work back from lower right Cost of cell (i, j) is C(i, j) = min(c(i + i, j) + 2, C(i, j + 1) + 2, C(i + 1, j + 1)) + δ where δ = 1 if the i th character of string and j th character of string B are identical. 11 CIS 110, Fall 2012

21 Dynamic Programming Idea: For each pair of characters, keep track of best match to end C C C C Iteration: Work back from lower right Cost of cell (i, j) is C(i, j) = min(c(i + i, j) + 2, C(i, j + 1) + 2, C(i + 1, j + 1)) + δ where δ = 1 if the i th character of string and j th character of string B are identical. 11 CIS 110, Fall 2012

22 Dynamic Programming Idea: For each pair of characters, keep track of best match to end C C C C Iteration: Work back from lower right Cost of cell (i, j) is C(i, j) = min(c(i + i, j) + 2, C(i, j + 1) + 2, C(i + 1, j + 1)) + δ where δ = 1 if the i th character of string and j th character of string B are identical. 11 CIS 110, Fall 2012

23 Dynamic Programming Idea: For each pair of characters, keep track of best match to end C C C C Iteration: Work back from lower right Cost of cell (i, j) is C(i, j) = min(c(i + i, j) + 2, C(i, j + 1) + 2, C(i + 1, j + 1)) + δ where δ = 1 if the i th character of string and j th character of string B are identical. 11 CIS 110, Fall 2012

24 Dynamic Programming Idea: For each pair of characters, keep track of best match to end C C C C Iteration: Work back from lower right Cost of cell (i, j) is C(i, j) = min(c(i + i, j) + 2, C(i, j + 1) + 2, C(i + 1, j + 1)) + δ where δ = 1 if the i th character of string and j th character of string B are identical. 11 CIS 110, Fall 2012

25 Dynamic Programming Idea: For each pair of characters, keep track of best match to end C C C C 4 1O Iteration: Work back from lower right Cost of cell (i, j) is C(i, j) = min(c(i + i, j) + 2, C(i, j + 1) + 2, C(i + 1, j + 1)) + δ where δ = 1 if the i th character of string and j th character of string B are identical. 11 CIS 110, Fall 2012

26 Dynamic Programming Idea: For each pair of characters, keep track of best match to end C C C C O Iteration: Work back from lower right Cost of cell (i, j) is C(i, j) = min(c(i + i, j) + 2, C(i, j + 1) + 2, C(i + 1, j + 1)) + δ where δ = 1 if the i th character of string and j th character of string B are identical. 11 CIS 110, Fall 2012

27 Dynamic Programming Idea: For each pair of characters, keep track of best match to end C C C C O Iteration: Work back from lower right Cost of cell (i, j) is C(i, j) = min(c(i + i, j) + 2, C(i, j + 1) + 2, C(i + 1, j + 1)) + δ where δ = 1 if the i th character of string and j th character of string B are identical. 11 CIS 110, Fall 2012

28 Dynamic Programming Idea: For each pair of characters, keep track of best match to end C C C C O Iteration: Work back from lower right Cost of cell (i, j) is C(i, j) = min(c(i + i, j) + 2, C(i, j + 1) + 2, C(i + 1, j + 1)) + δ where δ = 1 if the i th character of string and j th character of string B are identical. 11 CIS 110, Fall 2012

29 Dynamic Programming Idea: For each pair of characters, keep track of best match to end C C C C O Iteration: Work back from lower right Cost of cell (i, j) is C(i, j) = min(c(i + i, j) + 2, C(i, j + 1) + 2, C(i + 1, j + 1)) + δ where δ = 1 if the i th character of string and j th character of string B are identical. 11 CIS 110, Fall 2012

30 Dynamic Programming Idea: For each pair of characters, keep track of best match to end C C C C O Iteration: Work back from lower right Cost of cell (i, j) is C(i, j) = min(c(i + i, j) + 2, C(i, j + 1) + 2, C(i + 1, j + 1)) + δ where δ = 1 if the i th character of string and j th character of string B are identical. 11 CIS 110, Fall 2012

31 Dynamic Programming Idea: For each pair of characters, keep track of best match to end C C C C O Recovering the best alignment: Final cost is in cell (0, 0) Follow arrows to reconstruct string aligns letter in the current column with a gap aligns letter in the current row with a gap matches letters in current row and column with each other otal running time: O(mn)! 11 CIS 110, Fall 2012

32 Some More Examples Image Compositing iven a set of overlapping images, what is the best way to stitch them? Cut the images along an invisible seam, and splice them together. he optimal seam can be found through dynamic programming. Even better: shortest path Davis98 12 CIS 110, Fall 2012

33 Some More Examples Matrix parenthesization Need to multiply a sequence of rectangular matrices Which matrices should be multiplied first to minimize the number of operations ( ( )) b1 ( a 1 a 2 a 3 ) b 2 ( c 1 c 2 c 3 ) = ( a 1 b 1 +a2b 2 +a 3 b 3 ) ( c 1 c 2 c 3 ) b 3 ( a 1 a 2 a 3 ) (( ) ) ( ) b1 b1 c 1 b 1 c 2 b 1 c 3 b 2 ( c 1 c 2 c 3 ) = ( a 1 a 2 a 3 ) b 2 c 1 b 2 c 2 b 2 c 3 b 3 b 3 c 1 b 3 c 2 b 3 c 3 13 CIS 110, Fall 2012

34 Some More Examples Matrix parenthesization Need to multiply a sequence of rectangular matrices Which matrices should be multiplied first to minimize the number of operations ( ( )) b1 ( a 1 a 2 a 3 ) b 2 ( c 1 c 2 c 3 ) = ( a 1 b 1 +a2b 2 +a 3 b 3 ) ( c 1 c 2 c 3 ) b 3 ( a 1 a 2 a 3 ) (( ) ) ( ) b1 b1 c 1 b 1 c 2 b 1 c 3 b 2 ( c 1 c 2 c 3 ) = ( a 1 a 2 a 3 ) b 2 c 1 b 2 c 2 b 2 c 3 b 3 b 3 c 1 b 3 c 2 b 3 c 3 Seam carving (see demo) Shrink an image by finding one row or column of pixels to remove he seam doesn t have to be straight it can wiggle Use dynamic programming to find the best set of pixels to remove 13 CIS 110, Fall 2012