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Harbin, China

ACM today… ! (2009). DP for NP!. DP for APSP!. Max Flow. Knapsack. Floyd-Warshall. Harbin, China. 2010 finals. Jotto!. A word-guessing game similar to mastermind…. Sophs. Jrs. Srs. Profs. Chalk 1. Chalk 0. Chalk 1. Chalk 1. Quine 1. Quine 1. Quine 2. Quine 2. ?. ?. ?. ?.

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Harbin, China

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  1. ACM today… ! (2009) DP for NP! DP for APSP! Max Flow Knapsack Floyd-Warshall Harbin, China 2010 finals

  2. Jotto! A word-guessing game similar to mastermind… Sophs Jrs Srs Profs Chalk 1 Chalk 0 Chalk 1 Chalk 1 Quine 1 Quine 1 Quine 2 Quine 2 ? ? ? ? next guesses?

  3. Candy! 1800's 1920's 1940's 1950's 1970's

  4. v1 = 100 Maximizing Candy! w1 = 2 i = 1 Suppose you can consume 13 candywt units of candy. v2 = 120 w2 = 3 i = 2 What choice will maximize your candy-value experience? v3 = 230 w3 = 5 i = 3 (1) if you can take fractional parts of candy packages…? v4 = 560 w4 = 7 (2) if you can take any number of whole candy packages…? i = 4 v5 = 675 (3) if you can take 0 or 1 of each whole candy package…? w5 = 9 i = 5

  5. v1 = 100 Maximizing Candy! w1 = 2 i = 1 vpw1 = 50 Suppose you can consume 13 candywt units of candy. v2 = 120 w2 = 3 i = 2 vpw2 = 40 What choice will maximize your candy-value experience? v3 = 230 w3 = 5 GO WITH GREED! i = 3 vpw3 = 46 (1) if you can take fractional parts of candy packages…? v4 = 560 w4 = 7 (2) if you can take any number of whole candy packages…? i = 4 vpw4 = 80 v5 = 675 (3) if you can take 0 or 1 of each whole candy package…? w5 = 9 i = 5 vpw5 = 75

  6. v1 = 100 Knapsack problem w1 = 2 "unbounded" i = 1 vpw1 = 50 Suppose you can consume 13 candywt units of candy. v2 = 120 w2 = 3 (2) if you can take any number of whole candy packages…? i = 2 vpw2 = 40 v3 = 230 w3 = 5 i = 3 vpw3 = 46 IDEA: Consider all possible weights (integers) from 0 up to the weight you can carry - For each one choose the best from all N items v4 = 560 w4 = 7 i = 4 vpw4 = 80 v5 = 675 w5 = 9 i = 5 vpw5 = 75

  7. v1 = 100 Knapsack problem w1 = 2 "unbounded" i = 1 vpw1 = 50 Suppose you can consume 13 candywt units of candy. v2 = 120 w2 = 3 (2) if you can take any number of whole candy packages…? i = 2 vpw2 = 40 v3 = 230 TOTAL WEIGHT w3 = 5 0 1 2 3 4 5 6 7 8 9 10 11 12 13 i = 3 vpw3 = 46 0 0 v4 = 560 100 120 200 230 300 560 560 675 680 775 800 875 w4 = 7 i = 4 vpw4 = 80 max total value v5 = 675 w5 = 9 max V(w-wi) + vi V(w) = i i = 5 vpw5 = 75

  8. v1 = 100 Knapsack problem w1 = 2 "0-1 problem" i = 1 vpw1 = 50 Suppose you can consume 13 candywt units of candy. v2 = 120 w2 = 3 (3) if you can take 0 or 1 of each whole candy package…? i = 2 vpw2 = 40 v3 = 230 w3 = 5 i = 3 vpw3 = 46 IDEA: v4 = 560 Do the same thing as before, but consider sublists of items that grow one-at-a-time w4 = 7 i = 4 vpw4 = 80 v5 = 675 w5 = 9 i = 5 vpw5 = 75

  9. v1 = 100 Knapsack problem w1 = 2 "0-1 problem" i = 1 vpw1 = 50 Suppose you can consume 13 candywt units of candy. (3) if you can take 0 or 1 of each whole candy package…? v2 = 120 V(i-1,w) w2 = 3 V(i,w) = max i = 2 vpw2 = 40 V(i,w-wi) + vi i v3 = 230 TOTAL WEIGHT w3 = 5 0 1 2 3 4 5 6 7 8 9 10 11 12 13 i = 3 vpw3 = 46 0 01 v4 = 560 w4 = 7 012 i = 4 vpw4 = 80 0123 01234 v5 = 675 012345 w5 = 9 i = 5 Total Items vpw5 = 75 max total value

  10. elite exact number of cows we need, indexed from 1 .. 100 Input but the sum-of-squares of their indices must be < 30 3 30 Input the number of different subsets of cows satisfying that constraint. 4 total 14 1 2 3 1 2 4 2 3 4 1 3 4 21 29 26

  11. elite exact number of cows we need, indexed from 1 .. 100 Input but the sum-of-squares of their indices must be < 30 3 30 "use it or lose it" Input analysis the number of different subsets of cows satisfying that constraint. 4 total 14 1 2 3 1 2 4 2 3 4 1 3 4 21 You're either going to USE the first cow - or you're not. Sum up both cases! 29 26

  12. elite exact number of cows we need, indexed from 1 .. 100 Input but the sum-of-squares of their indices must be < 30 3 30 # of len-3 subsets, totaling under 30, starting from 1 N(3,30,1) = N(2,29,2) + N(3,30,2) Input # of len-2 subsets, totaling under 29, starting from 2 # of len-3 subsets, totaling under 30, starting from 2 the number of different subsets of cows satisfying that constraint. 4 "use it or lose it" total 14 1 2 3 1 2 4 2 3 4 1 3 4 21 29 26

  13. Thanks to Allie Russell for this…

  14. elite but it's much too slow!

  15. Pythonfunction decorators plain

  16. Pythonfunction decorators @dec42 decorated!

  17. Pythonfunction decorators fib = dec42(fib) implemented

  18. DP for APSP… all-pairs-shortest-paths Directed graph as adjacency matrix: dst Directed graph: "to" 1 2 3 4 100 inf 100 0 14 1 2 14 1 14 inf 0 50 2 50 src "from" inf inf 0 14 3 10 14 inf 10 inf 0 4 4 3 14 0 intermediate nodes

  19. Floyd-Warshall! dst dst "to" "to" 1 2 3 4 1 2 3 4 inf 100 inf 100 0 14 0 14 1 1 14 14 inf 0 50 inf 0 50 2 2 src src "from" "from" inf inf inf 0 14 inf 0 14 3 3 24 inf 10 inf 0 10 inf 0 4 4 1 intermediate node(s)! 0 intermediate nodes 1

  20. Floyd-Warshall! dst dst "to" "to" 1 2 3 4 1 2 3 4 inf 64 inf 100 0 14 0 14 1 1 14 14 inf 0 50 inf 0 50 2 2 src src "from" "from" inf inf inf 0 14 inf 0 14 3 3 24 24 10 inf 0 10 inf 0 4 4 1 intermediate node(s) 2 intermediate node(s)! 1 1 2

  21. Floyd-Warshall! All-pairs shortest paths: dst dst "to" "to" 1 2 3 4 1 2 3 4 28 42 28 42 0 14 0 14 1 1 14 14 38 0 28 inf 0 28 2 2 src src "from" "from" 24 inf 38 0 14 inf 0 14 3 3 24 24 10 38 0 10 38 0 4 4 3 intermediate node(s) 4 intermediate node(s) ~ done! 1 2 3 1 2 3 4

  22. Floyd-Warshall! dst T[src][dst][k] "to" 1 2 3 4 minimum distance from src to dst using intermediate nodes 1..k 28 42 0 14 1 T[src][dst][k-1] inf 14 inf 0 28 2 min = src T[src][k][k-1] + T[k][dst][k-1] 14 + 10 "from" inf inf 0 14 3 24 10 38 0 4 24 4 intermediate node(s) 1 2 3 4

  23. Tracking the paths… Floyd Warshall… not a huge amount of extra work to keep an extra table (pred) of the first intermediate vertex on the path from i to j

  24. This week's problems… These two are available if you haven't submitted them successfully before…

  25. This week's problems… !

  26. number of nodes hurdles Input number of edges number of "tasks" 5 6 3 1 2 12 3 2 8 1 3 5 2 5 3 3 4 4 2 4 8 3 4 1 2 5 1 from node 1 to node 2 the cost ("height") is 12 12 2 1 Output 3 8 5 What is the minimum cost along any one of the edges required to get from start to end ? 5 3 8 4 4 8 -1 4 not possible!

  27. number of cows dining Input total # of foods total # of drinks 4 3 3 2 2 1 2 3 1 2 2 2 3 1 2 2 2 1 3 1 2 2 1 1 3 3 Likes 1 2 2 3 1 3 1 3 3 1 1 2 1 2 3 0 # of foods cow[i] likes foods drinks 1 # of drinks cow[i] likes 2 Output 3 What is a cow-satisfying assignment here? 3 # of cows that can receive both a food and a drink they like… foods drinks each can be used only once

  28. Max Flow Ford-Fulkerson algorithm 12 B D 20 16 sink F A 10 4 9 7 source 4 13 C E 14 capacity The problem how much traffic (flow) can get from the source to the sink ?

  29. - 16 13 - - - - - 10 12 - - - 4 - - 14 - - - 9 - - 20 - - - 7 - 4 - - - - - - Max Flow The problem how much traffic can get from the source to the sink ? 12 B D 20 16 sink F A 10 4 9 7 source 4 13 “Capacity Graph” C E 14 TO C capacity A B C D E F A B C FROM D E F

  30. - 16 13 - - - - - 10 12 - - - 4 - - 14 - - - 9 - - 20 - - - 7 - 4 - - - - - - Find a path in C via BFS The problem how much traffic can get from the source to the sink ? Need to subtract this flow from C ! 12 B D 20 16 sink F A 10 4 9 7 source 4 13 C E 14 TO C capacity A B C D E F A B C FROM D E F

  31. - 12 0 - - - -12 - 0 12 - - - 0 - - 0 - - -12 0 - - 12 - - - 0 - 0 - - - -12 - - Create F Create a FLOW GRAPH with the minimum weight from that path 12 B D 12 20 16 sink 12 12 F A 10 4 9 7 source 4 13 “Flow Graph” C E 14 TO F capacity A B C D E F A B flow in one direction is negative flow in the other direction C FROM D E F

  32. - 4 13 - - - 12 - 10 0 - - - 4 - - 14 - - 12 9 - - 8 - - - 7 - 4 - - - 12 - - R = C - F Use the RESIDUAL GRAPH with the rest of the capacity after that flow 0 B D 12 8 4 sink 12 12 F A 10 4 9 7 source 4 13 “Residual Graph” C E 14 TO R capacity A B C D E F A B reverse edges allow the "undoing" of previous flow! C FROM D E F

  33. Keep going! Continue running this on the residual capacity until BFS fails… 12/12 B D 19/20 11/16 sink F A 0/10 1/4 0/9 7/7 source 4/4 12/13 C E 11/14 max flow:23 Continue until there is no longer a path from A to F !

  34. 1. Set F to all 0 The max flow algorithm 2. Compute R = C - F 3. BFS in R from source to sink. 3a. If no path, you’re done. 3b. If  a path, add the available capacity to F; goto (2). Floyd-Fulkerson 12 B D 3 20 16 sink 2 4 1 0 F A 10 4 2 9 1 7 2 source 4 4 13 C 5 E 14 capacity flow residual “Residual Graph” 3 B D 12 18 9 Don't need to keep R around explicity: Keep only the current flow (F) and the original capacity (C). 4 2 A 1 9 2 5 6 F 7 2 5 C E 4 11 9

  35. Get into the flow! A little bit of name contention… edmonds_karp This is the algorithm. defmax_flow(C, source, sink): n = len(C) # C is the capacity matrix F = [[0] * n for i in range(n)] # F is the flow matrix # residual capacity from u to v is C[u][v] - F[u][v] whileTrue: path = BFS(C, F, source, sink) if not path: break # no path - we're done! # find the path's flow, that is, the "bottleneck" edges = [C[u][v]-F[u][v] for u,v in path] path_flow = min( edges ) print"Augmenting by", path_flow for u,v in path: # traverse path to update flow F[u][v] += path_flow # forward edge up F[v][u] -= path_flow # backward edge down return sum([F[source][i] for i inrange(n)]) # out from source

  36. Useful alone, too A brief BFS algorithm using the Capacity matrix defBFS(C, F, source, sink): queue = [source] # the BFS queue paths = {source: []} # stores 1 path per graph node while queue: u = queue.pop(0) # next node to explore (expand) for v inrange(len(C)): # for each possible next node # path from u to v? and not yet at v? if C[u][v] - F[u][v] > 0 and v notin paths: paths[v] = paths[u] + [(u,v)] if v == sink: return paths[v] queue.append(v) # go from v in the future returnNone

  37. Setting up… And the code needed to run it… if __name__ == "__main__": # make a capacity graph # node A B C D E F C = [ [ 00, 16, 13, 00, 00, 00 ], # A [ 00, 00, 10, 12, 00, 00 ], # B [ 00, 04, 00, 00, 14, 00 ], # C [ 00, 00, 9, 00, 00, 20 ], # D [ 00, 00, 00, 7, 00, 4 ], # E [ 00, 00, 00, 00, 00, 00 ] ] # F print"C is", C source = 0 # A sink = 5 # F max_flow_value = max_flow( C, source, sink ) print"max_flow_value is", max_flow_value Linked at the ACM website by the slides…

  38. Jotto! A word-guessing game similar to mastermind… Sophs Jrs Srs Profs Chalk 1 Chalk 0 Chalk 1 Chalk 1 Quine 1 Quine 1 Quine 2 Quine 2 ? ? ? ? next guesses?

  39. elite eating problem how does the solution depend on subproblems? use it or lose it!

  40. v1 = 100 Knapsack problem w1 = 2 "0-1 problem" i = 1 vpw1 = 50 Suppose you can consume 13 candywt units of candy. (3) if you can take 0 or 1 of each whole candy package…? v2 = 120 w2 = 3 V(i,w) = i = 2 vpw2 = 40 v3 = 230 TOTAL WEIGHT w3 = 5 0 1 2 3 4 5 6 7 8 9 10 11 12 13 i = 3 vpw3 = 46 0 01 v4 = 560 w4 = 7 012 i = 4 vpw4 = 80 0123 01234 v5 = 675 012345 w5 = 9 i = 5 Total Items vpw5 = 75 max total value

  41. Candy! 1800's 1920's 1940's 1950's 1970's

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