1. Lecture 2: Dynamic Programming 主講人 : 虞台文

2. Content What is Dynamic Programming? Matrix Chain-Products Sequence Alignments Knapsack Problem All-Pairs Shortest Path Problem Traveling Salesman Problem Conclusion

3. Lecture 2: Dynamic Programming What is Dynamic Programming?

4. Dynamic Programming (DP) tends to break the original problem to sub-problems, i.e., in a smaller size The optimal solution in the bigger sub-problems is found through a retroactive formula which connects the optimal solutions of sub-problems. Used when the solution to a problem may be viewed as the result of a sequence of decisions.

5. Properties for Problems Solved by DP Simple Subproblems – The original problem can be broken into smaller subproblems with the same structure Optimal Substructure of the problems – The solution to the problem must be a composition of subproblem solutions (the principle of optimality) Subproblem Overlap – Optimal subproblems to unrelated problems can contain subproblems in common

6. The Principle of Optimality The basic principle of dynamic programming Developed by Richard Bellman An optimal path has the property that whatever the initial conditions and control variables (choices) over some initial period, the control (or decision variables) chosen over the remaining period must be optimal for the remaining problem, with the state resulting from the early decisions taken to be the initial condition.

7. Example: Shortest Path Problem 10 5 3 Goal Start

8. 25 28 40 Example: Shortest Path Problem 10 5 3 Start Goal

9. 25 28 40 Example: Shortest Path Problem 10 5 3 Start Goal

10. Recall  Greedy Method for Shortest Paths on a Multi-stage Graph Problem – Find a shortest path from v 0 to v 3 Is the greedy solution optimal?

11. Recall  Greedy Method for Shortest Paths on a Multi-stage Graph Problem – Find a shortest path from v 0 to v 3 Is the greedy solution optimal? The optimal path 

12. Example  Dynamic Programming

13. Lecture 2: Dynamic Programming Matrix Chain-Products

14. Matrix Multiplication C = A × B A is d × e and B is e × f O(def ) e B f e j C d f i,j A d i

15. Matrix Chain-Products Given a sequence of matrices, A 1, A 2, …, A n, find the most efficient way to multiply them together. Facts: – A(BC) = (AB)C – Different parenthesizing may need different numbers of operation. Example: A:10 × 30, B: 30 × 5, C : 5 × 60 – (AB)C = (10×30×5) + (10×5×60) = 1500 + 3000 = 4500 ops – A(BC) = (30×5×60) + (10×30×60) = 9000 + 18000 = 27000 ops

16. Matrix Chain-Products Given a sequence of matrices, A 1, A 2, …, A n, find the most efficient way to multiply them together. A Brute-force Approach: – Try all possible ways to parenthesize A=A 1  A 2  …  A n – Calculate number of operations for each one – Pick the best one Time Complexity: – #paranethesizations = #binary trees of n nodes – O(4 n )

17. A Greedy Approach Idea #1: – repeatedly select the product that uses the most operations. Counter-example: – A: 10  5, B: 5  10, C: 10  5, and D: 5  10 – Greedy idea #1 gives (AB)(CD), which takes 500+1000+500 = 2000 ops – A((BC)D) takes 500+250+250 = 1000 ops

18. Another Greedy Approach Idea #2: – repeatedly select the product that uses the least operations. Counter-example: – A: 101  11, B: 11  9, C: 9  100, and D: 100  999 – Greedy idea #2 gives A((BC)D), which takes 109989+9900+108900=228789 ops – (AB)(CD) takes 9999+89991+89100=189090 ops

19. DP  Define Subproblem Original Problem Subproblem ( P ij, i  j ) (P1n)(P1n) Suppose #operations for the optimal solution of P ij is N ij #operations for the optimal solution of the original problem P 1n is N 1n

20. DP  Define Subproblem Original Problem Subproblem ( P ij, i  j ) (P1n)(P1n) Suppose #operations for the optimal solution of P ij is N ij #operations for the optimal solution of the original problem P 1n is N 1n

21. DP  Define Subproblem Original Problem Subproblem ( P ij, i  j ) (P1n)(P1n) Suppose #operations for the optimal solution of P ij is N ij #operations for the optimal solution of the original problem P 1n is N 1n What is the relation btw N ij ( P ij ) and N 1n ( P 1n )?

22. DP  Principle of Optimality N ik N k +1, n d i  d k +1 d k  d j +1

23. DP  Implementation 1 2 i n 12jn N ij

24. DP  Implementation 1 2 i n 12jn N ij

25. DP  Implementation 1 2 i n 12jn N ij ?

26. DP  Implementation 1 2 i n 12jn N ij ?

27. DP  Implementation 1 2 i n 12jn N ij ?

28. DP  Implementation 1 2 i n 12jn N ij

29. DP  Implementation 1 2 i n 12jn N ij ?

30. DP  Implementation 1 2 i n 12jn N ij ?

31. DP  Implementation 1 2 i n 12jn N ij

32. DP  Implementation 1 2 i n 12jn N ij ?

33. DP  Implementation 1 2 i n 12jn N ij ?

34. DP  Implementation 1 2 i n 12jn N ij

35. DP for Matrix Chain-Products Algorithm matrixChain(S): Input: sequence S of n matrices to be multiplied Output: number of operations in an optimal parenthesization of S for i  1 to n // main diagonal terms are all zero N i,i  0 for d  2 to n // each diagonal do following for i  1 to n  d+1 // do from top to bottom for each diagonal j  i+d  1 N i,j  infinity for k  i to j  1 // counting minimum N i,j  min(N i,j, N i,k +N k+1,j +d i d k+1 d j+1 ) Algorithm matrixChain(S): Input: sequence S of n matrices to be multiplied Output: number of operations in an optimal parenthesization of S for i  1 to n // main diagonal terms are all zero N i,i  0 for d  2 to n // each diagonal do following for i  1 to n  d+1 // do from top to bottom for each diagonal j  i+d  1 N i,j  infinity for k  i to j  1 // counting minimum N i,j  min(N i,j, N i,k +N k+1,j +d i d k+1 d j+1 )

36. Time Complexity Algorithm matrixChain(S): Input: sequence S of n matrices to be multiplied Output: number of operations in an optimal parenthesization of S for i  1 to n // main diagonal terms are all zero N i,i  0 for d  2 to n // each diagonal do following for i  1 to n  d+1 // do from top to bottom for each diagonal j  i+d  1 N i,j  infinity for k  i to j  1 // counting minimum N i,j  min(N i,j, N i,k +N k+1,j +d i d k+1 d j+1 ) Algorithm matrixChain(S): Input: sequence S of n matrices to be multiplied Output: number of operations in an optimal parenthesization of S for i  1 to n // main diagonal terms are all zero N i,i  0 for d  2 to n // each diagonal do following for i  1 to n  d+1 // do from top to bottom for each diagonal j  i+d  1 N i,j  infinity for k  i to j  1 // counting minimum N i,j  min(N i,j, N i,k +N k+1,j +d i d k+1 d j+1 ) O(n3)O(n3) O(n3)O(n3)

37. Exercises 1. The matrixChain algorithm only computes #operations of an optimal parenthesization. But, it doesn’t report the optimal parenthesization scheme. Please modify the algorithm so that it can do so. 2. Given an example with 5 matrices to illustrate your idea using a table.

38. Lecture 2: Dynamic Programming Sequence Alignment

39. Question Given two strings are they similar? what is their distance? and

40. Example Y:Y: applicable plausibly X:X: How similar they are? Can you give them a score?

41. Example Y’:Y’: applica---ble -p-l--ausibly X’:X’: Match Mismatch Indel Three cases: Matches Mismatches Insertions & deletions (indel)

42. Example Y’:Y’: applica---ble -p-l--ausibly X’:X’: Match Mismatch Indel Three cases: Matches Mismatches Insertions & deletions (indel) (+1) (  1) The values depends on applications. It can be described using a so-called substitution matrix, to be discussed shortly.

43. Example Y’:Y’: applica---ble -p-l--ausibly X’:X’: Match Mismatch Indel Three cases: Matches Mismatches Insertions & deletions (indel) (+1) (  1) Score = 5  (+1) + 1  (  1) + 7  (  1) =  3

44. Example Y’:Y’: applica---ble -p-l--ausibly X’:X’: Match Mismatch Indel Three cases: Matches Mismatches Insertions & deletions (indel) (+1) (  1) Score = 5  (+1) + 1  (  1) + 7  (  1) =  3 Is the alignment optimal?

45. Sequence Alignment In bioinformatics, a sequence alignment is a way of arranging the primary sequences of DNA, RNA, or protein to identify regions of similarity that may be a consequence of functional, structural, or evolutionary relationships between the sequences.

46. Global and Local Alignments L G P S S K Q T G K G S - S R I W D N Global alignment L N - I T K S A G K G A I M R L G D A - - - - - - - T G K G - - - - - - - - Local alignment - - - - - - - A G K G - - - - - - - -

47. Global and Local Alignments L G P S S K Q T G K G S - S R I W D N Global alignment L N - I T K S A G K G A I M R L G D A - - - - - - - T G K G - - - - - - - - Local alignment - - - - - - - A G K G - - - - - - - -

48. Global and Local Alignments Global Alignment – attempts to align the entire sequence – most useful when the sequences in the query set are similar and of roughly equal size. – Needleman–Wunsch algorithm (1971). Local Alignment – Attempts to align partial regions of sequences with high level of similarity. – Smith-Waterman algorithm (1981)

49. Needleman–Wunsch Algorithm Find the best global alignment of any two sequences under a given substitution matrix. Maximize a similarity score, to give maximum match Maximum match = largest number of residues of one sequence that can be matched with another allowing for all possible gaps Based on dynamic programming Involves an iterative matrix method of calculation

50. Substitution Matrix In bioinformatics, a substitution matrix estimates the rate at which each possible residue in a sequence changes to each other residue over time. Substitution matrices are usually seen in the context of amino acid or DNA sequence alignment, where the similarity between sequences depends on the mutation rates as represented in the matrix.

51. Substitution Matrix (DNA) w/o Gap Cost ACGT A2 11 1 11 C 11 2 11 1 G1 11 2 11 T 11 1 11 2

52. Substitution Matrix (DNA) w/ Gap Cost ACGT A2 11 1 11 C 11 2 11 1 G1 11 2 11 T 11 1 11 2 ACGT  A2 11 1 11 22 C 11 2 11 1 22 G1 11 2 11 22 T 11 1 11 2 22 22 22 22 22 0

53. Substitution Matrix ( 3D-BLAST)

54. DP  Define Subproblem Consider two strings, s of length n and t of length m. Let S be the substitution matrix. Subproblem: Let P ij is defined to be the optimal aligning for the two substrings: t[1..i] and s[1..j], and let M ij be the matching score. Original Problem: P mn (matching score M mn )

55. DP  Principle of Optimality j j1j1 i i1i1   ?

56. Example Step 1. Create a scoring matrix Step 2. Make an empty table for M ij Step 3. Initialize base conditions Step 4. Fill table by Step 5. Trace back s : t : ACGGTAG CCTAAG M ij   A A C C G G G G T T A A G G   C C C C T T A A A A G G

57. Example Step 1. Create a scoring matrix Step 2. Make an empty table for M ij Step 3. Initialize base conditions Step 4. Fill table by Step 5. Trace back s : t : ACGGTAG CCTAAG M ij   A A C C G G G G T T A A G G   C C C C T T A A A A G G 0 22 44 66 88  10  12 22 44 66 88  10  12  14

58. Example Step 1. Create a scoring matrix Step 2. Make an empty table for M ij Step 3. Initialize base conditions Step 4. Fill table by Step 5. Trace back s : t : ACGGTAG CCTAAG M ij   A A C C G G G G T T A A G G   C C C C T T A A A A G G 0 22 44 66 88  10  12  14 22 44 66 88  10  12 11 0 22 44 66 88  10 33 1 11 33 33 55 77 55 11 0 22 11 33 55 44 33 01 11 1 11 66 55 22 1012 88 77 33 0013

59. Example Step 1. Create a scoring matrix Step 2. Make an empty table for M ij Step 3. Initialize base conditions Step 4. Fill table by Step 5. Trace back s : t : ACGGTAG CCTAAG M ij   A A C C G G G G T T A A G G   C C C C T T A A A A G G 0 22 44 66 88  10  12 22 44 66 88  10  12  14 11 33 55 44 66 88 0 1 11 33 55 77 22 11 0 0 22 33 44 33 22 1 1 0 66 33 11 11 0 0 88 55 33 1 1 1  10 77 55 11 2 33 1 11 0 11 1 0 1 s’ :t’ :s’ :t’ : GGGG AAAA T-T- GAGA GTGT CCCC ACAC

60. Needleman–Wunsch Algorithm Step 1. Create a scoring matrix Step 2. Make an empty table for M ij Step 3. Initialize base conditions Step 4. Fill table by Step 5. Trace back for i = 2 to m+1 do for j from 2 to n +1 do

61. s ’  ””, t ’  ”” while i < 1 and j < 1 do if s ’  s j + s ’ t ’  t i + t ’ else if s ’  s j + s ’ t ’  gap+ t ’ else s ’  gap+ s ’ t ’  t i + t ’ while i > 1 do t ’  gap+ t ’ while j > 1 do s ’  gap+ s ’ Needleman–Wunsch Algorithm Step 1. Create a scoring matrix Step 2. Make an empty table for M ij Step 3. Initialize base conditions Step 4. Fill table by Step 5. Trace back

62. Local Alignment Problem Given two strings s = s 1 ……s n, t = t 1 …….t m Find substrings s ’, t ’ whose similarity (optimal global alignment value) is maximum.

63. Example: Local Alignment GTAGT CATCAT ATG TGACTGAC G TC CATDOGCAT CC TGACTGAC A GTAGTCATCATATGCCTGACTGACG TCCATDOGCATCCTACTACTGACTGACA difference block Best aligned subsequeces

64. Global Alignment (Needleman–Wunsch Algorithm) Local Alignment (Smith-Waterman Algorithm) Recursive Formulation

65. Exercises 3. Find the best local aligned substrings for the following two DNA strings: GAATTCAGTTA GGATCGA You have to give the detail. Hint: start from the left table.

66. Exercises 4. What is longest common sequence (LCS) problem? How to solve LCS using dynamic programming technique?

67. Lecture 2: Dynamic Programming Knapsack Problem

68. Knapsack Problems Given some items, pack the knapsack to get the maximum total value. Each item has some weight and some benefit. Total weight that we can carry is no more than some fixed capacity. Fractional knapsack problem – Items are divisible: you can take any fraction of an item. – Solved with a greedy algorithm. 0-1 knapsack problem – Items are indivisible; you either take an item or not. – Solved with dynamic programming.

69. Given a knapsack with maximum capacity W, and a set S consisting of n items Each item i has some weight w i and benefit value b i (all w i and W are integer values) Problem: How to pack the knapsack to achieve maximum total value of packed items? 0-1 Knapsack Problem Why it is called a 0-1 Knapsack Problem?

70. Example: 0-1 Knapsack Problem Which boxes should be chosen to maximize the amount of money while still keeping the overall weight under 15 kg ?

71. Example: 0-1 Knapsack Problem Objective Function Unknowns or Variables Constraints

72. Formulation: 0-1 Knapsack Problem

73. 0-1 Knapsack Problem: Brute-Force Approach Since there are n items, there are 2 n possible combinations of items. We go through all combinations and find the one with maximum value and with total weight less or equal to W Running time will be O(2 n )

74. DP  Define Subproblem Suppose that items are labeled 1,..., n. Define a subproblem, say, P k as to finding an optimal solution for items in S k = {1, 2,..., k}.  original problem is P n. Is such a scheme workable? Is the principle of optimality held?

75. A Counterexample 1. 2kgs, 3$ 2. 3kgs, 4$ 3. 4kgs, 5$ 4. 5kgs, 8$ 5. 9kgs, 10$ 20 kgs P 1 P 2 P 3 P 4 P 5

76. A Counterexample Sub- problem OptimumValue P1P1 13$ P2P2 1, 27$ P3P3 1, 2, 312$ P4P4 1, 2, 3, 420$ P5P5 1, 3, 4, 526$ 1. 2kgs, 3$ 2. 3kgs, 4$ 3. 4kgs, 5$ 4. 5kgs, 8$ 5. 9kgs, 10$ 20 kgs

77. A Counterexample Sub- problem OptimumValue P1P1 13$ P2P2 1, 27$ P3P3 1, 2, 312$ P4P4 1, 2, 3, 420$ P5P5 1, 3, 4, 526$ 1. 2kgs, 3$ 2. 3kgs, 4$ 3. 4kgs, 5$ 4. 5kgs, 8$ 5. 9kgs, 10$ 20 kgs Solution for P 4 is not part of the solution for P 5 !!!

78. DP  Define Subproblem Suppose that items are labeled 1,..., n. Define a subproblem, say, P k as to finding an optimal solution for items in S k = {1, 2,..., k}.  original problem is P n. Is such a scheme workable? Is the principle of optimality held?  

79. DP  Define Subproblem Suppose that items are labeled 1,..., n. Define a subproblem, say, P k,w as to finding an optimal solution for items in S k = {1, 2,..., k} and with total weight no more than w.  original problem is P n,W. Is such a scheme workable? Is the principle of optimality held? New version

80. DP  Principle of Optimality Denote the benefit for the optimal solution of P k,w as B k,w.

81. DP  Principle of Optimality Denote the benefit for the optimal solution of P k,w as B k,w. In this case, it is impossible to include the k th object. Not include the k th object include the k th object There are two possible choices.

82. Example 1. 2kgs, 3$ 2. 3kgs, 4$ 3. 4kgs, 5$ 4. 5kgs, 6$ 5 kgs w kw k 01234 0 1 2 3 4 5

83. Example 1. 2kgs, 3$ 2. 3kgs, 4$ 3. 4kgs, 5$ 4. 5kgs, 6$ 5 kgs w kw k 01234 0 1 2 3 4 5 0 0 0 0 0 0 0 0 0 0 Step 1. Setup table and initialize base conditions.

84. Example 1. 2kgs, 3$ 2. 3kgs, 4$ 3. 4kgs, 5$ 4. 5kgs, 6$ 5 kgs w kw k 01234 0 1 2 3 4 5 0 0 0 0 0 0 0 0 0 0 0 3 3 3 3 Step 2. Fill all table entries progressively.

85. Example 1. 2kgs, 3$ 2. 3kgs, 4$ 3. 4kgs, 5$ 4. 5kgs, 6$ 5 kgs w kw k 01234 0 1 2 3 4 5 0 0 0 0 0 0 0 0 0 0 0 3 3 3 3 0 3 4 4 7 Step 2. Fill all table entries progressively.

86. Example 1. 2kgs, 3$ 2. 3kgs, 4$ 3. 4kgs, 5$ 4. 5kgs, 6$ 5 kgs w kw k 01234 0 1 2 3 4 5 0 0 0 0 0 0 0 0 0 0 0 3 3 3 3 0 3 4 4 7 0 3 4 5 7 Step 2. Fill all table entries progressively.

87. Example 1. 2kgs, 3$ 2. 3kgs, 4$ 3. 4kgs, 5$ 4. 5kgs, 6$ 5 kgs w kw k 01234 0 1 2 3 4 5 0 0 0 0 0 0 0 0 0 0 0 3 3 3 3 0 3 4 4 7 0 3 4 5 7 0 3 4 5 7 Step 2. Fill all table entries progressively.

88. Example 1. 2kgs, 3$ 2. 3kgs, 4$ 3. 4kgs, 5$ 4. 5kgs, 6$ 5 kgs w kw k 01234 0 1 2 3 4 5 0 0 0 0 0 0 0 0 0 0 0 3 3 3 3 0 3 4 4 7 0 3 4 5 7 0 3 4 5 7 Step 3. Trace back

89. Pseudo-Polynomial Time Algorithm The time complexity for 0-1 knapsack using DP is O(Wn). Not a polynomial-time algorithm if W is large. This is a pseudo-polynomial time algorithm.

90. Lecture 2: Dynamic Programming All-Pairs Shortest Path Problem

91. All-Pairs Shortest Path Problem Given weighted graph G(V,E), we want to determine the cost d ij of the shortest path between each pair of nodes in V.

92. VkVk Floyd's Algorithm Let be the minimum cost of a path from node i to node j, using only nodes in V k ={v 1,…,v k }. k k Vk1Vk1 i i j j The all-pairs shortest path problem is to find all paths with costs

93. Floyd's Algorithm Input Parameter: D Output Parameter: D, next all_paths(D, next) n = D.NumberOfRows; // initialize next: if no intermediate // vertices are allowed next[i][j] = j for i = 1 to n for j = 1 to n next[i][j] = j; for k = 1 to n // compute D(k) for i = 1 to n for j = 1 to n if (D[i][k] + D[k][j] < D[i][j]) D[i][j] = D[i][k] + D[k][j]; next[i][j] = next[i][k]; Input Parameter: D Output Parameter: D, next all_paths(D, next) n = D.NumberOfRows; // initialize next: if no intermediate // vertices are allowed next[i][j] = j for i = 1 to n for j = 1 to n next[i][j] = j; for k = 1 to n // compute D(k) for i = 1 to n for j = 1 to n if (D[i][k] + D[k][j] < D[i][j]) D[i][j] = D[i][k] + D[k][j]; next[i][j] = next[i][k]; O(n3)O(n3) O(n3)O(n3)

94. Floyd's Algorithm Input Parameters: next, i, j Output Parameters: None print_path(next, i, j) { // if no intermediate vertices, just // print i and j and return if (j == next[i][j]) { print(i + “ ” + j); return; } // output i and then the path from the vertex // after i (next[i][j]) to j print(i + “ ”); print_path(next,next[i][j], j); } Input Parameters: next, i, j Output Parameters: None print_path(next, i, j) { // if no intermediate vertices, just // print i and j and return if (j == next[i][j]) { print(i + “ ” + j); return; } // output i and then the path from the vertex // after i (next[i][j]) to j print(i + “ ”); print_path(next,next[i][j], j); }

95. Lecture 2: Dynamic Programming Traveling Salesman Problem

96. Traveling Salesman Problem (TSP)

97. How many feasible paths? n cities

98. Example (TSP) 1 1 2 2 3 3 4 4 2 6 4 7 6 5 (1234) = 18 (1243) = 19 (1324) = 23 (1342) = 19 (1423) = 23 (1432) = 18

99. Subproblem Formulation for TSP 1 1 S i i g(i, S) length of the shortest path from i to 1 visiting each city in S exactly once. g(1, V  {1}) length of the optimal TSP tour.

100. Subproblem Formulation for TSP 1 1 S i i g(i, S) length of the shortest path from i to 1 visiting each city in S exactly once. S{j}S{j} j j d ij Goal: g(1, V  {1})

101. Example Goal: g(1, V  {1}) d 12 d 13 d 14 d 23 d 24 d 32 d 34 d 42 d 43 d 34 d 43 d 24 d 42 d 23 d 32

102. Example Goal: g(1, V  {1}) d 12 d 13 d 14 d 23 d 24 d 32 d 34 d 42 d 43 d 34 d 43 d 24 d 42 d 23 d 32 464 2 62 55 66 77 9 11 10 8 13 9 767565 161314 2 64 18

103. DP  TSP Algorithm Input Parameter: D Output Parameter: P // path TSP(D) n = Dim(D); for i = 1 to n g[i,  ] = D[i, 1]; for k = 1 to n  2 // compute g for subproblems for all S  V  {1} with |S|=k for all i  S  {1} g[i, S] = min j  S {D[i, j], g[j, S  {j}] }; P[i, S] = arg min j  S {D[i, j], g[j, S  {j}] }; // compute the TSP tour g[1, V  {1}] = min j  V  {1} {D[1, j], g[j, V  {1, j}]}; P[1, V  {1}] = arg min j  V  {1} {D[1, j], g[j, V  {1, j}]}; Input Parameter: D Output Parameter: P // path TSP(D) n = Dim(D); for i = 1 to n g[i,  ] = D[i, 1]; for k = 1 to n  2 // compute g for subproblems for all S  V  {1} with |S|=k for all i  S  {1} g[i, S] = min j  S {D[i, j], g[j, S  {j}] }; P[i, S] = arg min j  S {D[i, j], g[j, S  {j}] }; // compute the TSP tour g[1, V  {1}] = min j  V  {1} {D[1, j], g[j, V  {1, j}]}; P[1, V  {1}] = arg min j  V  {1} {D[1, j], g[j, V  {1, j}]}; Goal: g(1, V  {1})

104. DP  TSP Algorithm Input Parameter: D Output Parameter: P // path TSP(D) n = Dim(D); for i = 1 to n g[i,  ] = D[i, 1]; for k = 1 to n  2 // compute g for subproblems for all S  V  {1} with |S|=k for all i  S  {1} g[i, S] = min j  S {D[i, j], g[j, S  {j}] }; P[i, S] = arg min j  S {D[i, j], g[j, S  {j}] }; // compute the TSP tour g[1, V  {1}] = min j  V  {1} {D[1, j], g[j, V  {1, j}]}; P[1, V  {1}] = arg min j  V  {1} {D[1, j], g[j, V  {1, j}]}; Input Parameter: D Output Parameter: P // path TSP(D) n = Dim(D); for i = 1 to n g[i,  ] = D[i, 1]; for k = 1 to n  2 // compute g for subproblems for all S  V  {1} with |S|=k for all i  S  {1} g[i, S] = min j  S {D[i, j], g[j, S  {j}] }; P[i, S] = arg min j  S {D[i, j], g[j, S  {j}] }; // compute the TSP tour g[1, V  {1}] = min j  V  {1} {D[1, j], g[j, V  {1, j}]}; P[1, V  {1}] = arg min j  V  {1} {D[1, j], g[j, V  {1, j}]}; O(2 n ) Goal: g(1, V  {1})