1 / 41

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. ?. ?. ?. ?.

art
Télécharger la présentation

Harbin, China

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related