1. 1 Algorithmic Paradigms Greed. 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.

2. 2 Dynamic Programming History Bellman. Pioneered the systematic study of dynamic programming in the 1950s. Etymology. n Dynamic programming = planning over time. n Secretary of Defense was hostile to mathematical research. n Bellman sought an impressive name to avoid confrontation. – "it's impossible to use dynamic in a pejorative sense" – "something not even a Congressman could object to" Reference: Bellman, R. E. Eye of the Hurricane, An Autobiography.

3. 3 Dynamic Programming Applications Areas. n Bioinformatics. n Control theory. n Information theory. n Operations research. n Computer science: theory, graphics, AI, systems, …. Some famous dynamic programming algorithms. n Viterbi for hidden Markov models. n Unix diff for comparing two files. n Smith-Waterman for sequence alignment. n Bellman-Ford for shortest path routing in networks. n Cocke-Kasami-Younger for parsing context free grammars.

4. 6.1 Weighted Interval Scheduling

5. 5 Weighted Interval Scheduling Weighted interval scheduling problem. n Job j starts at s j, finishes at f j, and has weight or value v j. n Two jobs compatible if they don't overlap. n Goal: find maximum weight subset of mutually compatible jobs. Time 01234567891011 f g h e a b c d

6. 6 Unweighted Interval Scheduling Review Recall. Greedy algorithm works if all weights are 1. n Consider jobs in ascending order of finish time. n Add job to subset if it is compatible with previously chosen jobs. Observation. Greedy algorithm can fail spectacularly if arbitrary weights are allowed. Time 01234567891011 b a weight = 999 weight = 1

7. 7 Weighted Interval Scheduling Notation. Label jobs by finishing time: f 1  f 2 ...  f n. Def. p(j) = largest index i < j such that job i is compatible with j. Ex: p(8) = 5, p(7) = 3, p(2) = 0. Time 01234567891011 6 7 8 4 3 1 2 5

8. 8 Dynamic Programming: Binary Choice Notation. OPT(j) = value of optimal solution to the problem consisting of job requests 1, 2,..., j. n Case 1: OPT selects job j. – can't use incompatible jobs { p(j) + 1, p(j) + 2,..., j - 1 } – must include optimal solution to problem consisting of remaining compatible jobs 1, 2,..., p(j) n Case 2: OPT does not select job j. – must include optimal solution to problem consisting of remaining compatible jobs 1, 2,..., j-1 optimal substructure

9. 9 Input: n, s 1,…,s n, f 1,…,f n, v 1,…,v n Sort jobs by finish times so that f 1  f 2 ...  f n. Compute p(1), p(2), …, p(n) Compute-Opt(j) { if (j = 0) return 0 else return max(v j + Compute-Opt(p(j)), Compute-Opt(j-1)) } Weighted Interval Scheduling: Brute Force Brute force algorithm.

10. 10 Weighted Interval Scheduling: Brute Force Observation. Recursive algorithm fails spectacularly because of redundant sub-problems  exponential algorithms. Ex. Number of recursive calls for family of "layered" instances grows like Fibonacci sequence. 3 4 5 1 2 p(1) = 0, p(j) = j-2 5 43 3221 21 10 1010

11. 11 Input: n, s 1,…,s n, f 1,…,f n, v 1,…,v n Sort jobs by finish times so that f 1  f 2 ...  f n. Compute p(1), p(2), …, p(n) for j = 1 to n M[j] = empty M[j] = 0 M-Compute-Opt(j) { if (M[j] is empty) M[j] = max(w j + M-Compute-Opt(p(j)), M-Compute-Opt(j-1)) return M[j] } global array Weighted Interval Scheduling: Memoization Memoization. Store results of each sub-problem in a cache; lookup as needed.

12. 12 Weighted Interval Scheduling: Running Time Claim. Memoized version of algorithm takes O(n log n) time. n Sort by finish time: O(n log n). n Computing p(  ) : O(n) after sorting by start time. M-Compute-Opt(j) : each invocation takes O(1) 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  = # nonempty entries of M[]. – initially  = 0, throughout   n. – (ii) increases  by 1  at most 2n recursive calls. Overall running time of M-Compute-Opt(n) is O(n). ▪ Remark. O(n) if jobs are pre-sorted by start and finish times.

13. 13 Automated Memoization Automated memoization. Many functional programming languages (e.g., Lisp) have built-in support for memoization. Q. Why not in imperative languages (e.g., Java)? F(40) F(39)F(38) F(37)F(36) F(37) F(36)F(35) F(36) F(35)F(34) F(37) F(36)F(35) static int F(int n) { if (n <= 1) return n; else return F(n-1) + F(n-2); } (defun F (n) (if (<= n 1) n (+ (F (- n 1)) (F (- n 2))))) Lisp (efficient) Java (exponential)

14. 14 Weighted Interval Scheduling: Finding a Solution Q. Dynamic programming algorithms computes optimal value. What if we want the solution itself? A. Do some post-processing. n # of recursive calls  n  O(n). Run M-Compute-Opt(n) Run Find-Solution(n) Find-Solution(j) { if (j = 0) output nothing else if (v j + M[p(j)] > M[j-1]) print j Find-Solution(p(j)) else Find-Solution(j-1) }

15. 15 Weighted Interval Scheduling: Bottom-Up Bottom-up dynamic programming. Unwind recursion. Input: n, s 1,…,s n, f 1,…,f n, v 1,…,v n Sort jobs by finish times so that f 1  f 2 ...  f n. Compute p(1), p(2), …, p(n) Iterative-Compute-Opt { M[0] = 0 for j = 1 to n M[j] = max(v j + M[p(j)], M[j-1]) }

16. 6.3 Segmented Least Squares

17. 17 Segmented Least Squares Least squares. n Foundational problem in statistic and numerical analysis. n Given n points in the plane: (x 1, y 1 ), (x 2, y 2 ),..., (x n, y n ). n Find a line y = ax + b that minimizes the sum of the squared error: Solution. Calculus  min error is achieved when x y

18. 18 Segmented Least Squares Segmented least squares. n Points lie roughly on a sequence of several line segments. n Given n points in the plane (x 1, y 1 ), (x 2, y 2 ),..., (x n, y n ) with n x 1 < x 2 <... < x n, find a sequence of lines that minimizes f(x). Q. What's a reasonable choice for f(x) to balance accuracy and parsimony? x y goodness of fit number of lines

19. 19 Segmented Least Squares Segmented least squares. n Points lie roughly on a sequence of several line segments. n Given n points in the plane (x 1, y 1 ), (x 2, y 2 ),..., (x n, y n ) with n x 1 < x 2 <... < 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 n Tradeoff function: E + c L, for some constant c > 0. x y

20. 20 Dynamic Programming: Multiway Choice Notation. n OPT(j) = minimum cost for points p 1, p i+1,..., p j. n e(i, j) = minimum sum of squares for points p i, p i+1,..., p j. To compute OPT(j): n Last segment uses points p i, p i+1,..., p j for some i. n Cost = e(i, j) + c + OPT(i-1).

21. 21 Segmented Least Squares: Algorithm Running time. O(n 3 ). n Bottleneck = computing e(i, j) for O(n 2 ) pairs, O(n) per pair using previous formula. INPUT: n, p 1,…,p N, c Segmented-Least-Squares() { M[0] = 0 for j = 1 to n for i = 1 to j compute the least square error e ij for the segment p i,…, p j for j = 1 to n M[j] = min 1  i  j (e ij + c + M[i-1]) return M[n] } can be improved to O(n 2 ) by pre-computing various statistics

22. 6.4 Knapsack Problem

23. 23 Knapsack Problem Knapsack problem. n Given n objects and a "knapsack." n Item i weighs w i > 0 kilograms and has value v i > 0. n Knapsack has capacity of W kilograms. n Goal: fill knapsack so as to maximize total value. Ex: { 3, 4 } has value 40. Greedy: repeatedly add item with maximum ratio v i / w i. Ex: { 5, 2, 1 } achieves only value = 35  greedy not optimal. 1 Value 18 22 28 1 Weight 5 6 62 7 Item 1 3 4 5 2 W = 11

24. 24 Dynamic Programming: False Start Def. OPT(i) = max profit subset of items 1, …, i. n Case 1: OPT does not select item i. – OPT selects best of { 1, 2, …, i-1 } n Case 2: 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 Conclusion. Need more sub-problems!

25. 25 Dynamic Programming: Adding a New Variable Def. OPT(i, w) = max profit subset of items 1, …, i with weight limit w. n Case 1: OPT does not select item i. – OPT selects best of { 1, 2, …, i-1 } using weight limit w n Case 2: OPT selects item i. – new weight limit = w – w i – OPT selects best of { 1, 2, …, i–1 } using this new weight limit

26. 26 Input: n, w 1,…,w N, v 1,…,v N for w = 0 to W M[0, w] = 0 for i = 1 to n for w = 1 to W if (w i > w) M[i, w] = M[i-1, w] else M[i, w] = max {M[i-1, w], v i + M[i-1, w-w i ]} return M[n, W] Knapsack Problem: Bottom-Up Knapsack. Fill up an n-by-W array.

27. 27 Knapsack Algorithm n + 1 1 Value 18 22 28 1 Weight 5 6 62 7 Item 1 3 4 5 2  { 1, 2 } { 1, 2, 3 } { 1, 2, 3, 4 } { 1 } { 1, 2, 3, 4, 5 } 0 0 0 0 0 0 0 1 0 1 1 1 1 1 2 0 6 6 6 1 6 3 0 7 7 7 1 7 4 0 7 7 7 1 7 5 0 7 18 1 6 0 7 19 22 1 7 0 7 24 1 28 8 0 7 25 28 1 29 9 0 7 25 29 1 34 10 0 7 25 29 1 34 11 0 7 25 40 1 W + 1 W = 11 OPT: { 4, 3 } value = 22 + 18 = 40

28. 28 Knapsack Problem: Running Time Running time.  (n W). n Not polynomial in input size! n "Pseudo-polynomial." n Decision version of Knapsack is NP-complete. [Chapter 8] Knapsack approximation algorithm. There exists a polynomial algorithm that produces a feasible solution that has value within 0.01% of optimum. [Section 11.8]

29. 6.5 RNA Secondary Structure

30. 30 RNA Secondary Structure RNA. String B = b 1 b 2  b n over alphabet { A, C, G, U }. Secondary structure. RNA is single-stranded so it tends to loop back and form base pairs with itself. This structure is essential for understanding behavior of molecule. G U C A GA A G CG A U G A U U A G A CA A C U G A G U C A U C G G G C C G Ex: GUCGAUUGAGCGAAUGUAACAACGUGGCUACGGCGAGA complementary base pairs: A-U, C-G

31. 31 RNA Secondary Structure Secondary structure. A set of pairs S = { (b i, b j ) } that satisfy: n [Watson-Crick.] S is a matching and each pair in S is a Watson- Crick complement: A-U, U-A, C-G, or G-C. n [No sharp turns.] The ends of each pair are separated by at least 4 intervening bases. If (b i, b j )  S, then i < j - 4. n [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. Usual hypothesis is that an RNA molecule will form the secondary structure with the optimum total free energy. Goal. Given an RNA molecule B = b 1 b 2  b n, find a secondary structure S that maximizes the number of base pairs. approximate by number of base pairs

32. 32 RNA Secondary Structure: Examples Examples. C GG C A G U U UA A UGUGGCCAU GG C A G U UA A UGGGCAU C GG C A U G U UA A GUUGGCCAU sharp turncrossingok G G 44 base pair

33. 33 RNA Secondary Structure: Subproblems First attempt. OPT(j) = maximum number of base pairs in a secondary structure of the substring b 1 b 2  b j. Difficulty. Results in two sub-problems. n Finding secondary structure in: b 1 b 2  b t-1. n Finding secondary structure in: b t+1 b t+2  b n-1. 1 tn match b t and b n OPT(t-1) need more sub-problems

34. 34 Dynamic Programming Over Intervals Notation. OPT(i, j) = maximum number of base pairs in a secondary structure of the substring b i b i+1  b j. n Case 1. If i  j - 4. – OPT(i, j) = 0 by no-sharp turns condition. n Case 2. Base b j is not involved in a pair. – OPT(i, j) = OPT(i, j-1) n Case 3. Base b j pairs with b t for some i  t < j - 4. – non-crossing constraint decouples resulting sub-problems – OPT(i, j) = 1 + max t { OPT(i, t-1) + OPT(t+1, j-1) } Remark. Same core idea in CKY algorithm to parse context-free grammars. take max over t such that i  t < j-4 and b t and b j are Watson-Crick complements

35. 35 Bottom Up Dynamic Programming Over Intervals Q. What order to solve the sub-problems? A. Do shortest intervals first. Running time. O(n 3 ). RNA(b 1,…,b n ) { for k = 5, 6, …, n-1 for i = 1, 2, …, n-k j = i + k Compute M[i, j] return M[1, n] } using recurrence 000 00 0 2 3 4 1 i 6789 j

36. 36 Dynamic Programming Summary Recipe. n Characterize structure of problem. n Recursively define value of optimal solution. n Compute value of optimal solution. n Construct optimal solution from computed information. Dynamic programming techniques. n Binary choice: weighted interval scheduling. n Multi-way choice: segmented least squares. n Adding a new variable: knapsack. n Dynamic programming over intervals: RNA secondary structure. Top-down vs. bottom-up: different people have different intuitions. Viterbi algorithm for HMM also uses DP to optimize a maximum likelihood tradeoff between parsimony and accuracy CKY parsing algorithm for context-free grammar has similar structure

37. 6.6 Sequence Alignment

38. 38 String Similarity How similar are two strings? n ocurrance n occurrence ocurrance ccurrenceo - ocurrnce ccurrnceo --a e- ocurrance ccurrenceo - 5 mismatches, 1 gap 1 mismatch, 1 gap 0 mismatches, 3 gaps

39. 39 Applications. n Basis for Unix diff. n Speech recognition. n Computational biology. Edit distance. [Levenshtein 1966, Needleman-Wunsch 1970] n Gap penalty  ; mismatch penalty  pq. n Cost = sum of gap and mismatch penalties. 2  +  CA CGACCTACCT CTGACTACAT TGACCTACCT CTGACTACAT - T C C C  TC +  GT +  AG + 2  CA - Edit Distance

40. 40 Goal: Given two strings X = x 1 x 2... x m and Y = y 1 y 2... y n find alignment of minimum cost. Def. An 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 j'. Ex: CTACCG vs. TACATG. Sol: M = x 2 -y 1, x 3 -y 2, x 4 -y 3, x 5 -y 4, x 6 -y 6. Sequence Alignment CTACC- TACAT- G G y1y1 y2y2 y3y3 y4y4 y5y5 y6y6 x2x2 x3x3 x4x4 x5x5 x1x1 x6x6

41. 41 Sequence Alignment: Problem Structure Def. OPT(i, j) = min cost of aligning strings x 1 x 2... x i and y 1 y 2... y j. n Case 1: OPT matches x i -y j. – pay mismatch for x i -y j + min cost of aligning two strings x 1 x 2... x i-1 and y 1 y 2... y j-1 n Case 2a: OPT leaves x i unmatched. – pay gap for x i and min cost of aligning x 1 x 2... x i-1 and y 1 y 2... y j n Case 2b: OPT leaves y j unmatched. – pay gap for y j and min cost of aligning x 1 x 2... x i and y 1 y 2... y j-1

42. 42 Sequence Alignment: Algorithm Analysis.  (mn) time and space. English words or sentences: m, n  10. Computational biology: m = n = 100,000. 10 billions ops OK, but 10GB array? Sequence-Alignment(m, n, x 1 x 2...x m, y 1 y 2...y n, ,  ) { for i = 0 to m M[0, i] = i  for j = 0 to n M[j, 0] = j  for i = 1 to m for j = 1 to n M[i, j] = min(  [x i, y j ] + M[i-1, j-1],  + M[i-1, j],  + M[i, j-1]) return M[m, n] }

43. 6.7 Sequence Alignment in Linear Space

44. 44 Sequence Alignment: Linear Space Q. Can we avoid using quadratic space? Easy. Optimal value in O(m + n) space and O(mn) time. n Compute OPT(i, ) from OPT(i-1, ). n No longer a simple way to recover alignment itself. Theorem. [Hirschberg 1975] Optimal alignment in O(m + n) space and O(mn) time. n Clever combination of divide-and-conquer and dynamic programming. n Inspired by idea of Savitch from complexity theory.

45. 45 Edit distance graph. n Let f(i, j) be shortest path from (0,0) to (i, j). n Observation: f(i, j) = OPT(i, j). Sequence Alignment: Linear Space i-j m-n x1x1 x2x2 y1y1 x3x3 y2y2 y3y3 y4y4 y5y5 y6y6   0-0  

46. 46 Edit distance graph. n Let f(i, j) be shortest path from (0,0) to (i, j). n Can compute f (, j) for any j in O(mn) time and O(m + n) space. Sequence Alignment: Linear Space i-j m-n x1x1 x2x2 y1y1 x3x3 y2y2 y3y3 y4y4 y5y5 y6y6   0-0 j

47. 47 Edit distance graph. n Let g(i, j) be shortest path from (i, j) to (m, n). n Can compute by reversing the edge orientations and inverting the roles of (0, 0) and (m, n) Sequence Alignment: Linear Space i-j m-n x1x1 x2x2 y1y1 x3x3 y2y2 y3y3 y4y4 y5y5 y6y6   0-0  

48. 48 Edit distance graph. n Let g(i, j) be shortest path from (i, j) to (m, n). n Can compute g(, j) for any j in O(mn) time and O(m + n) space. Sequence Alignment: Linear Space i-j m-n x1x1 x2x2 y1y1 x3x3 y2y2 y3y3 y4y4 y5y5 y6y6   0-0 j

49. 49 Observation 1. The cost of the shortest path that uses (i, j) is f(i, j) + g(i, j). Sequence Alignment: Linear Space i-j m-n x1x1 x2x2 y1y1 x3x3 y2y2 y3y3 y4y4 y5y5 y6y6   0-0

50. 50 Observation 2. let q be an index that minimizes f(q, n/2) + g(q, n/2). Then, the shortest path from (0, 0) to (m, n) uses (q, n/2). Sequence Alignment: Linear Space i-j m-n x1x1 x2x2 y1y1 x3x3 y2y2 y3y3 y4y4 y5y5 y6y6   0-0 n / 2 q

51. 51 Divide: find index q that minimizes f(q, n/2) + g(q, n/2) using DP. n Align x q and y n/2. Conquer: recursively compute optimal alignment in each piece. Sequence Alignment: Linear Space i-j x1x1 x2x2 y1y1 x3x3 y2y2 y3y3 y4y4 y5y5 y6y6   0-0 q n / 2 m-n

52. 52 Theorem. Let T(m, n) = max running time of algorithm on strings of length at most m and n. T(m, n) = O(mn log n). Remark. Analysis is not tight because two sub-problems are of size (q, n/2) and (m - q, n/2). In next slide, we save log n factor. Sequence Alignment: Running Time Analysis Warmup

53. 53 Theorem. Let T(m, n) = max running time of algorithm on strings of length m and n. T(m, n) = O(mn). Pf. (by induction on n) n O(mn) time to compute f(, n/2) and g (, n/2) and find index q. n T(q, n/2) + T(m - q, n/2) time for two recursive calls. n Choose constant c so that: n Base cases: m = 2 or n = 2. n Inductive hypothesis: T(m, n)  2cmn. Sequence Alignment: Running Time Analysis

54. 6.8 Shortest Paths

55. 55 Shortest Paths Shortest path problem. Given a directed graph G = (V, E), with edge weights c vw, find shortest path from node s to node t. Ex. Nodes represent agents in a financial setting and c vw is cost of transaction in which we buy from agent v and sell immediately to w. s 3 t 2 6 7 4 5 10 18 -16 9 6 15 -8 30 20 44 16 11 6 19 6 allow negative weights

56. 56 Shortest Paths: Failed Attempts Dijkstra. Can fail if negative edge costs. Re-weighting. Adding a constant to every edge weight can fail. u t sv 2 1 3 -6 st 2 3 2 -3 3 5 5 6 6 0

57. 57 Shortest Paths: Negative Cost Cycles Negative cost cycle. Observation. If some path from s to t contains a negative cost cycle, there does not exist a shortest s-t path; otherwise, there exists one that is simple. st W c(W) < 0 -6 7 -4

58. 58 Shortest Paths: Dynamic Programming Def. OPT(i, v) = length of shortest v-t path P using at most i edges. n Case 1: P uses at most i-1 edges. – OPT(i, v) = OPT(i-1, v) n Case 2: P uses exactly i edges. – if (v, w) is first edge, then OPT uses (v, w), and then selects best w-t path using at most i-1 edges Remark. By previous observation, if no negative cycles, then OPT(n-1, v) = length of shortest v-t path.

59. 59 Shortest Paths: Implementation Analysis.  (mn) time,  (n 2 ) space. Finding the shortest paths. Maintain a "successor" for each table entry. Shortest-Path(G, t) { foreach node v  V M[0, v]   M[0, t]  0 for i = 1 to n-1 foreach node v  V M[i, v]  M[i-1, v] foreach edge (v, w)  E M[i, v]  min { M[i, v], M[i-1, w] + c vw } }

60. 60 Shortest Paths: Practical Improvements Practical improvements. n Maintain only one array M[v] = shortest v-t path that we have found so far. n No need to check edges of the form (v, w) unless M[w] changed in previous iteration. Theorem. Throughout the algorithm, M[v] is length of some v-t path, and after i rounds of updates, the value M[v] is no larger than the length of shortest v-t path using  i edges. Overall impact. n Memory: O(m + n). n Running time: O(mn) worst case, but substantially faster in practice.

61. 61 Bellman-Ford: Efficient Implementation Push-Based-Shortest-Path(G, s, t) { foreach node v  V { M[v]   successor[v]   } M[t] = 0 for i = 1 to n-1 { foreach node w  V { if (M[w] has been updated in previous iteration) { foreach node v such that (v, w)  E { if (M[v] > M[w] + c vw ) { M[v]  M[w] + c vw successor[v]  w } If no M[w] value changed in iteration i, stop. }

62. 6.9 Distance Vector Protocol

63. 63 Distance Vector Protocol Communication network. n Nodes  routers. n Edges  direct communication link. n Cost of edge  delay on link. Dijkstra's algorithm. Requires global information of network. Bellman-Ford. Uses only local knowledge of neighboring nodes. Synchronization. We don't expect routers to run in lockstep. The order in which each foreach loop executes in not important. Moreover, algorithm still converges even if updates are asynchronous. naturally nonnegative, but Bellman-Ford used anyway!

64. 64 Distance Vector Protocol Distance vector protocol. n Each router maintains a vector of shortest path lengths to every other node (distances) and the first hop on each path (directions). n Algorithm: each router performs n separate computations, one for each potential destination node. n "Routing by rumor." Ex. RIP, Xerox XNS RIP, Novell's IPX RIP, Cisco's IGRP, DEC's DNA Phase IV, AppleTalk's RTMP. Caveat. Edge costs may change during algorithm (or fail completely). t v 1 s 1 1 deleted "counting to infinity" 2 1

65. 65 Path Vector Protocols Link state routing. n Each router also stores the entire path. n Based on Dijkstra's algorithm. n Avoids "counting-to-infinity" problem and related difficulties. n Requires significantly more storage. Ex. Border Gateway Protocol (BGP), Open Shortest Path First (OSPF). not just the distance and first hop

66. 6.10 Negative Cycles in a Graph

67. 67 Detecting Negative Cycles Lemma. If OPT(n,v) = OPT(n-1,v) for all v, then no negative cycles. Pf. Bellman-Ford algorithm. Lemma. If OPT(n,v) < OPT(n-1,v) for some node v, then (any) shortest path from v to t contains a cycle W. Moreover W has negative cost. Pf. (by contradiction) n Since OPT(n,v) < OPT(n-1,v), we know P has exactly n edges. n By pigeonhole principle, P must contain a directed cycle W. n Deleting W yields a v-t path with < n edges  W has negative cost. v t W c(W) < 0

68. 68 Detecting Negative Cycles Theorem. Can detect negative cost cycle in O(mn) time. n Add new node t and connect all nodes to t with 0-cost edge. n Check if OPT(n, v) = OPT(n-1, v) for all nodes v. – if yes, then no negative cycles – if no, then extract cycle from shortest path from v to t v 18 2 5 -23 -15 -11 6 t 0 0 0 0 0

69. 69 Detecting Negative Cycles: Application Currency conversion. Given n currencies and exchange rates between pairs of currencies, is there an arbitrage opportunity? Remark. Fastest algorithm very valuable! F$ £ ¥ DM 1/7 3/10 2/3 2 17056 3/50 4/3 8 IBM 1/10000 800

70. 70 Detecting Negative Cycles: Summary Bellman-Ford. O(mn) time, O(m + n) space. n Run Bellman-Ford for n iterations (instead of n-1). n Upon termination, Bellman-Ford successor variables trace a negative cycle if one exists. n See p. 288 for improved version and early termination rule.