Dynamic Programming Comp 122, Fall 2004

Transcription

1 Dynamic Programming Comp 122, Fall 2004

2 Review: the previous lecture Principles of dynamic programming: optimization problems, optimal substructure property, overlapping subproblems, trade space for time, implementation via bottom-up/memoization. Example of Match Game. Example of Fibinacci: from Divide and Conquer (exponential time complexity) to Dynamic Programming (O(n) time and space), then to more efficient dynamic programming (constant space, O(logn) time) Example of LCS in O(m n) time and space. Today we will show how to reduce the space complexity of LCS to O(n) and in future lectures we will show how to reduce the time complexity of LCS. Comp 122, Spring 2004

3 Longest Common Subsequence Problem: Given 2 sequences, X = x 1,...,x m and Y = y 1,...,y n, find a common subsequence whose length is maximum. X: springtime Comp 122, Spring 2004 Y: printing LCS(X,Y): printi

4 c[, ] 0 c[ prefix, prefix ] 1 max( c[ prefix, ], c[, prefix ]) if if if empty or empty, end( ) end( ), end( ) end( ). Keep track of c[, ] in a table of nm entries: top/down: increasing row order within each row left-to-right: increasing column order p r i n t i n g s p r i n g t i m e Comp 122, Spring 2004

5 c[, ] 0 c[ prefix, prefix ] 1 max( c[ prefix, ], c[, prefix ]) if if if empty or empty, end( ) end( ), end( ) end( ). Time Complexity: O(nm). Space Complexity: O(nm). Can the space complexity be improved if we just compute the length of the LCS and do not need to recover an actual LCS? In this case we only need to compute the alignment score. Can the space complexity be improved and still allow the recovery of an actual LCS? Yes, but for this we will have to address LCS as a longest path problem. p r i n t i n g s p r i n g t i m e Comp 122, Spring 2004

6 Other sequence questions Edit distance: Given 2 sequences, X = x 1,...,x m and Y = y 1,...,y n, what is the minimum number of deletions, insertions, and changes that you must do to change one to another? Protein sequence alignment: Given a score matrix on amino acid pairs, s(a,b) for a,b { } A, and 2 amino acid sequences, X = x 1,...,x m A m and Y = y 1,...,y n A n, find the alignment with lowest score Comp 122, Spring 2004

7 Outline DNA Sequence Comparison: Biological background Sequence Alignment First Biological Success Stories More Grid Graphs: Manhattan Tourist Problem Longest Paths in Grid Graphs Back to Longest Common Subsequence Problem Review: divide and conquer paradigm Reducing the space requirement of LCS

8 Dynamic Programming: String Editing

9 The Central Dogma of Molecular Biology Gene DNA RNA PROTEIN Function > DNA sequence AATTCATGAAAATCGTATACTGGTCTGGTACCGG CAACACTGAGAAAATGGCAGAGCTCATCGCTAAA GGTATCATCGAATCTGGTAAAGACGTCAACACCA TCAACGTGTCTGACGTTAACATCGATGAACTGCT GAACGAAGATATCCTGATCCTGGGTTGCTCTGCC ATGGGCGATGAAGTTCTCGAGGAAAGCGAATTTG > Protein sequence MKIVYWSGTGNTEKMAELIAKGIIES GKDVNTINVSDVNIDELLNEDILILGC SAMGDEVLEESEFEPFIEEISTKISG KKVALFGSYGWGDGKWMRDFEER MNGYGCVVVETPLIVQNEPDEAEQD CIEFGKKIANI

10 The Central Dogma of Molecular Biology Genome: The digital backbone of molecular biology Transcripts from Gene(DNA subsequence) to Proten sequence: Perform functions encoded in the genome

11 The Sequence Alignment Problem A = c t a c g a g a c B = a a c g a c g a t The Scoring Matrix - a c g t a c g t Compare two strings A and B and measure their similarity by finding the optimal alignment between them. The alignment is classically based on the transformation of one sequence into the other, via operations of substitutions, insertions, and deletions (indels). 13

12 Two Sequence Alignment Problems Global Alignment. A = c t a c g a g a c B = a a c g a c g a t Local Alignment. A = c t a c g a g a c B = a a c g a c g a t 14

13 Two Sequence Alignment Problems Global Alignment. A = c t a c g a g a c B = a a c g a c g a t The Scoring Matrix - a c g t a c g t Local Alignment. A = c t a c g a g a c B = a a c g a c g a t 15

14 Two Sequence Alignment Problems Global Alignment. A = c t a c g a g a c B = a a c g a c g a t Value: 2 The Scoring Matrix - a c g t a c g t Local Alignment. A = c t a c g a g a c B = a a c g a c g a t 16

15 Two Sequence Alignment Problems Global Alignment. A = c t a c g a g a c B = a a c g a c g a t Value: 2 The Scoring Matrix - a c g t a c g t Local Alignment. A = c t a c g a g a c B = a a c g a c g a t Value: 5 17

16 2 The O(n ) time, Classical Dynamic Programming Algorithm A = n c t a c c g 5 a g a The Alignment Graph B = n a a c g a c g a t The Scoring Matrix - a c g t a c g t

17 Computing the Optimal Global Alignment Value A = n c t 2 a 3 c a c 1 4 g 5 a g B = n 0 a a c g a c g a t Classical Dynamic Programming: O(n ) 2 Score of = 1 Score of = -1 19

18 Computing an Optimal Local Alignment Value A = n c t a c c g 5 a g a B = n 0 a a c g a c g a t Classical Dynamic Programming: O(n 2 ) Score of = 1 Score of = -1 20

19 DNA Sequence Comparison: First Success Story Finding sequence similarities with genes of known function is a common approach to infer a newly sequenced gene s function In 1984 Russell Doolittle and colleagues found similarities between cancer-causing gene and normal growth factor (PDGF) gene

20 Cystic Fibrosis Cystic fibrosis (CF) is a chronic and frequently fatal genetic disease of the body's mucus glands (abnormally high level of mucus in glands). CF primarily affects the respiratory systems in children. Mucus is a slimy material that coats many epithelial surfaces and is secreted into fluids such as saliva

21 Cystic Fibrosis: Inheritance In early 1980s biologists hypothesized that CF is an autosomal recessive disorder caused by mutations in a gene that remained unknown till 1989 Heterozygous carriers are asymptomatic Must be homozygously recessive in this gene in order to be diagnosed with CF

22 Cystic Fibrosis: Finding the Gene

23 Cystic Fibrosis and the CFTR Protein CFTR (Cystic Fibrosis Transmembrane conductance Regulator) protein is acting in the cell membrane of epithelial cells that secrete mucus These cells line the airways of the nose, lungs, the stomach wall, etc.

24 Mechanism of Cystic Fibrosis The CFTR protein (1480 amino acids) regulates a chloride ion channel Adjusts the wateriness of fluids secreted by the cell Those with cystic fibrosis are missing one single amino acid in their CFTR Mucus ends up being too thick, affecting many organs

25 Cystic Fibrosis and the CFTR Protein CFTR (Cystic Fibrosis Transmembrane conductance Regulator) protein is acting in the cell membrane of epithelial cells that secrete mucus These cells line the airways of the nose, lungs, the stomach wall, etc.

26 Finding Similarities between the Cystic Fibrosis Gene and ATP binding proteins ATP binding proteins are present on cell membrane and act as transport channel In 1989 biologists found similarity between the cystic fibrosis gene and ATP binding proteins A plausible function for cystic fibrosis gene, given the fact that CF involves sweet secretion with abnormally high sodium level

27 Cystic Fibrosis: Mutation Analysis If a high % of cystic fibrosis (CF) patients have a certain mutation in the gene and the normal patients don t, then that could be an indicator of a mutation that is related to CF A certain mutation was found in 70% of CF patients, convincing evidence that it is a predominant genetic diagnostics marker for CF

28 Outline DNA Sequence Comparison: Biological background The Sequence Alignment problem First Biological Success Stories Grid Graphs: Manhattan Tourist Problem Longest Paths in Grid Graphs Back to Longest Common Subsequence Problem Review: divide and conquer paradigm Reducing the space requirement of LCS

29 Manhattan Tourist Problem (MTP) Imagine seeking a path (from source to sink) to travel (only eastward and southward) with the most number of attractions (*) in the Manhattan grid Source * * * * * * * * * * * * Sink

30 Manhattan Tourist Problem (MTP) Imagine seeking a path (from source to sink) to travel (only eastward and southward) with the most number of attractions (*) in the Manhattan grid Source * * * * * * * * * * * * Sink

31 Manhattan Tourist Problem: Formulation Goal: Find the longest (highest scoring) path in a weighted grid. Input: A weighted grid G with two distinct vertices, one labeled source and the other labeled sink Output: A longest path in G from source to sink

32 i coordinate MTP: An Example source j coordinate sink

33 MTP: Greedy Algorithm Is Not Optimal source promising start, but leads to bad choices! sink

34 MTP: Dynamic Programming source j 0 1 i S 0,1 = S 1,0 = 5 Calculate optimal path score for each vertex in the graph Each vertex s score is the maximum of the prior vertices score plus the weight of the respective edge in between

35 MTP: Dynamic Programming (cont d) source j i S 0,2 = S 1,1 = S 2,0 = 8

36 MTP: Dynamic Programming (cont d) source j i S 3,0 = S 1,2 = S 2,1 = S 3,0 = 8

37 MTP: Dynamic Programming (cont d) source j i S 1,3 = S 2,2 = greedy alg. fails! 9 S 3,1 = 9

38 MTP: Dynamic Programming (cont d) source j i S 2,3 = S 3,2 = 9

39 MTP: Dynamic Programming (cont d) source j i Done! (showing all back-traces) S 3,3 = 16

40 MTP: Recurrence Computing the score for a point (i,j) by the recurrence relation: s i, j = max s i-1, j + weight of the edge between (i-1, j) and (i, j) s i, j-1 + weight of the edge between (i, j-1) and (i, j) The running time is n x m for a n by m grid (n = # of rows, m = # of columns)

41 Adding Diagonal Edges to the Grid A 2 A 3 A 1 B What about diagonals? The score at point B is given by: s B = max of s A1 + weight of the edge (A 1, B) s A2 + weight of the edge (A 2, B) s A3 + weight of the edge (A 3, B)

42 Adding Diagonal Edges to the Grid More generally, computing the score for point x is given by the recurrence relation: s x = max of s y + weight of vertex (y, x) where y є Predecessors(x) Predecessors (x) set of vertices that have edges leading to x

43 Traveling in the Grid The only hitch is that one must decide on the order in which visit the vertices By the time the vertex x is analyzed, the values s y for all its predecessors y should be computed otherwise we are in trouble. We need to traverse the vertices in some order Try to find such order for a directed acyclic grid graph???

44 Traversing the Manhattan Grid 3 different strategies: a) Column by column b) Row by row c) Along diagonals a) b) c)

45 Outline DNA Sequence Comparison: Biological background The Sequence Alignment problem First Biological Success Stories Grid Graphs: Manhattan Tourist Problem Generalizing to Longest Paths in Grid Graphs Back to Longest Common Subsequence Problem Review: divide and conquer paradigm Reducing the space requirement of LCS

46 Generalizing to computing longest paths in DAGs: Directed Acyclic Graphs The Manhattan grid is a DAG Exemplify Topological Ordering: DAG for Dressing in the morning problem

47 Topological Ordering A numbering of vertices of the graph is called topological ordering of the DAG if every edge of the DAG connects a vertex with a smaller label to a vertex with a larger label In other words, if vertices are positioned on a line in an increasing order of labels then all edges go from left to right.

48 Topological ordering 2 different topological orderings of the DAG

49 Longest Path in DAG Problem Goal: Find a longest path between two vertices in a weighted DAG Input: A weighted DAG G with source and sink vertices Output: A longest path in G from source to sink

50 Longest Path in DAG: Dynamic Programming Suppose vertex v has indegree k and predecessors {u 1, u 2 u k} Longest path to v from source is: s v = max of su 1 + weight of edge from u 1 to v su 2 + weight of edge from u 2 to v... su k + weight of edge from u 3 to v In General, compute by increasing Topological Order: s v = max (s u + weight of edge from u to v) u є Predecessors(v)

51 Generalizing to Directed Acylclic Grid Graphs Computing the score for point x is given by the recurrence relation: s v = max of s u + weight of vertex (u, v) where u є Predecessors(v) Predecessors (v) set of vertices that have edges leading to v What is the running time for a DAG G(V, E), where V is the set of all vertices and E is the set of all edges? Answer: O(E) since each edge is evaluated once

52 Outline DNA Sequence Comparison: Biological background The Sequence Alignment problem First Biological Success Stories Grid Graphs: Manhattan Tourist Problem Longest Paths in Grid Graphs Back to Longest Common Subsequence Problem Review: divide and conquer paradigm Reducing the space requirement of LCS

53 Edit Graph for LCS Problem j A T C T G A T C i T Every path is a common subsequence. G C A Every diagonal edge adds an extra element to common subsequence T A C LCS Problem: Find a path with maximum number of diagonal edges

54 Alignment: Dynamic Programming s i,j = s i-1, j-1 +1 if v i = w j max s i-1, j +0 s i, j-1 +0 This recurrence corresponds to the Manhattan Tourist problem (three incoming edges into a vertex) with all horizontal and vertical edges weighted by zero and diagonal edges weighted by one.

55 Alignment as a Path in the Edit Graph Every path in the edit graph corresponds to an alignment:

56 Alignment as a Path in the Edit Graph Old Alignment v= AT_GTTAT_ w= ATCGT_A_C New Alignment v= AT_GTTAT_ w= ATCG_TA_C

57 Alignment as a Path in the Edit Graph v= AT_GTTAT_ w= ATCGT_A_C (0,0), (1,1), (2,2), (2,3), (3,4), (4,5), (5,5), (6,6), (7,6), (7,7)

58 Alignment: Dynamic Programming s i,j = s i-1, j-1 +1 if v i = w j max s i-1, j s i, j-1

59 Dynamic Programming Example Initialize 1 st row and 1 st column to be all zeroes. Or, to be more precise, initialize 0 th row and 0 th column to be all zeroes.

60 Dynamic Programming Example S i,j = S i-1, j-1 max S i-1, j S i, j-1 value from NW +1, if v i = w j value from North (top) value from West (left)

61 Dynamic Programming Example A T C G T A C A T G T T A T S i,j = S i-1, j-1 max S i-1, j S i, j-1 value from NW +1, if v i = w j value from North (top) value from West (left)

62 Alignment: Backtracking Arrows from. show where the score originated if from the top if from the left if v i = w j

63 Alignment: Dynamic Programming s i,j = s i-1, j-1 +1 if v i = w j max s i-1, j s i, j-1

64 LCS Algorithm 1. LCS(v,w) 2. for i 1 to n 3. s i, for j 1 to m 5. s 0,j 0 6. for i 1 to n 7. for j 1 to m 8. s i-1,j 9. s i,j max s i,j s i-1,j-1 + 1, if v i = w j 11. if s i,j = s i-1,j b i,j if s i,j = s i,j-1 if s i,j = s i-1,j return (s n,m, b)

65 LCS(v,w) created the alignment grid Now we need a way to read the best alignment of v and w Follow the arrows backwards from sink Now What?

66 Printing LCS: Backtracking 1. PrintLCS(b,v,i,j) 2. if i = 0 or j = 0 3. return 4. if b i,j = 5. PrintLCS(b,v,i-1,j-1) 6. print v i 7. else 8. if b i,j = 9. PrintLCS(b,v,i-1,j) 10. else 11. PrintLCS(b,v,i,j-1)

67 LCS Runtime It takes O(nm) time to fill in the nxm dynamic programming matrix. Why O(nm)? The pseudocode consists of a nested for loop inside of another for loop to set up a nxm matrix. Memory? Naively O(nm) we next show how to reduce it to O(n) by combining divide and conquer with dynamic programming.

68 Outline DNA Sequence Comparison: Biological background The Sequence Alignment problem First Biological Success Stories Grid Graphs: Manhattan Tourist Problem Longest Paths in Grid Graphs Back to Longest Common Subsequence Problem Review: divide and conquer paradigm Reducing the space requirement of LCS

69 Divide & Conquer Algorithms

70 Divide and Conquer Algorithms Divide problem into sub-problems Conquer by solving sub-problems recursively. If the sub-problems are small enough, solve them in brute force fashion Combine the solutions of sub-problems into a solution of the original problem (tricky part)

71 Sorting Problem Revisited Given: an unsorted array Goal: sort it

72 Mergesort: Divide Step Step 1 Divide log(n) divisions to split an array of size n into single elements

73 Mergesort: Conquer Step Step 2 Conquer O(n) O(n) O(n) O(n) logn iterations, each iteration takes O(n) time. Total Time: O(n logn)

74 Mergesort: Combine Step Step 3 Combine arrays of size 1 can be easily merged to form a sorted array of size 2 2 sorted arrays of size n and m can be merged in O(n+m) time to form a sorted array of size n+m

75 Mergesort: Combine Step Combining 2 arrays of size Etcetera

76 Merge Algorithm 1. Merge(a,b) 2. n1 size of array a 3. n2 size of array b 4. a n a n i 1 7. j 1 8. for k 1 to n1 + n2 9. if a i < b j 10. c k a i 11. i i else 13. c k b j 14. j j return c

77 Mergesort: Example Divide Conquer

78 1. MergeSort(c) MergeSort Algorithm 2. n size of array c 3. if n = 1 4. return c 5. left list of first n/2 elements of c 6. right list of last n-n/2 elements of c 7. sortedleft MergeSort(left) 8. sortedright MergeSort(right) 9. sortedlist Merge(sortedLeft,sortedRight) 10. return sortedlist

79 MergeSort: Running Time The problem is simplified to baby steps for the i th merging iteration, the complexity of the problem is O(n) number of iterations is O(log n) running time: O(n logn)

80 Divide and Conquer Approach to LCS Path(source, sink) if(source & sink are in consecutive columns) output the longest path from source to sink else middle middle vertex between source & sink Path(source, middle) Path(middle, sink)

81 Divide and Conquer Approach to LCS Path(source, sink) if(source & sink are in consecutive columns) output the longest path from source to sink else middle middle vertex between source & sink Path(source, middle) Path(middle, sink) The only problem left is how to find this middle vertex!

82 Computing Alignment Path Requires Quadratic Memory Alignment Path Space complexity for computing alignment path for sequences of length n and m is O(nm) We need to keep all backtracking references in memory to reconstruct the path (backtracking) Note that the longest path itself is linear in size. n m

83 Crossing the Middle Line m/2 m We want to calculate the longest path from (0,0) to (n,m) that passes through (i,m/2) where i ranges from 0 to n and represents the i-th row n Prefix(i) Suffix(i) Define length(i) as the length of the longest path from (0,0) to (n,m) that passes through vertex (i, m/2)

84 Crossing the Middle Line m/2 m Prefix(i) n Suffix(i) Define (mid,m/2) as the vertex where the longest path crosses the middle column. length(mid, m/2) = optimal length = max 0 i n length(i)

85 Computing Prefix(i) prefix(i) is the length of the longest path from (0,0) to (i,m/2) Compute prefix(i) by dynamic programming in the left half of the matrix store prefix(i) column 0 m/2 m

86 Computing Suffix(i) suffix(i) is the length of the longest path from (i,m/2) to (n,m) suffix(i) is the length of the longest path from (n,m) to (i,m/2) with all edges reversed Compute suffix(i) by dynamic programming in the right half of the reversed matrix store suffix(i) column 0 m/2 m

87 Length(i) = Prefix(i) + Suffix(i) Add prefix(i) and suffix(i) to compute length(i): length(i)=prefix(i) + suffix(i) You now have a middle vertex of the maximum path (i,m/2) as maximum of length(i) 0 middle point found i 0 m/2 m

88 Finding the Middle Point 0 m/4 m/2 3m/4 m

89 Finding the Middle Point again 0 m/4 m/2 3m/4 m

90 And Again 0 m/8 m/4 3m/8 m/2 5m/8 3m/4 7m/8 m

91 Time = Area: First Pass On first pass, the algorithm covers the entire area Area = n m

92 Time = Area: First Pass On first pass, the algorithm covers the entire area Area = n m But according to this figure this computation seems to require O(m n) space! Computing prefix(i) Computing suffix(i) However, note that in each such DP step we only compute the Scores, note the paths

93 Computing Alignment Score with Linear Memory Alignment Score Space complexity of computing just the score itself is O(n) We only need the previous column to calculate the current column, and we can then throw away that previous column once we re done using it n n 2

94 Computing Alignment Score: Recycling Columns Only two columns of scores are saved at any given time memory for column 1 is used to calculate column 3 memory for column 2 is used to calculate column 4

95 Time = Area: First Pass On first pass, the algorithm covers the entire area Area = n m Space requirement: O(n)? (In addition to the two recycled columns used for the rectangle DP, we maintain an O(n) sized vector to store the accumulated mid points). Computing prefix(i) Computing suffix(i)

96 Time = Area: Second Pass On second pass, the algorithm covers only 1/2 of the area Area/2

97 Time = Area: Third Pass On third pass, only 1/4th is covered. Area/4

98 Geometric Reduction At Each Iteration 1 + ½ + ¼ (½) k 2 Runtime: O(Area) = O(nm) 5 th pass: 1/16 3 rd pass: 1/4 first pass: 1 4 th pass: 1/8 2 nd pass: 1/2