Randomized Algorithms III Min Cut

Transcription

1 Chapter 11 Randomized Algorithms III Min Cut CS 57: Algorithms, Fall 01 October 1, Min Cut Problem Definition 11. Min cut Min cut G = V, E): undirected graph, n vertices, m edges. Interested in cuts in G. S V \ S Definition cut in G: a partition of V : S and V \ S. Edges of the cut: S, V \ S) is size of the cut S, V \ S) = { uv u S, v V \ S, and uv E }, minimum cut / mincut: cut in graph with min size Some definitions A) conditional probability of X given Y is Pr [ X = x Y = y ] [ = Pr Pr [ X = x) Y = y) ] = Pr [ X = x Y = y ] Pr[Y = y]. 1 ] X=x) Y =y) [ ]. Pr Y =y

2 B) X, Y events are independent, if Pr [ X = x Y = y ] = Pr [ X = x ] Pr [ Y = y ]. = Pr [ X = x ] [ ] Y = y = Pr X = x Some more probability Lemma E 1,..., E n : n events not necessarily independent). Then, Pr [ ] [ ] [ ] [ ] n i=1 E i = Pr E1 Pr E E 1 Pr E E1 E... Pr [ ] E n E1... E n The Algorithm Edge contraction... y x G: A) edge contraction: e = xy in G. B)... merge x, y into a single vertex. C)...remove self loops. D)... parallel edges multi-graph. E)... weights/ multiplicities on the edges Min cut in weighted graph G/xy: {x, y} Edge contraction implemented in On) time: A) Graph represented using adjacency lists. B) Merging the adjacency lists of the two vertices being contracted. C) Using hashing to do fix-ups. i.e., fix adjacency list of vertices connected to x, y.) D) Include edge weight in computing cut weight Cuts under contractions Observation A) A cut in G/xy is a valid cut in G. B) There cuts in G are not in G/xy. C) The cut S = {x} is not in G/xy. D) = size mincut in G/xy mincut in G. A) Idea: Repeatedly perform edge contractions benefits: shrink graph)... B) Every vertex in contracted graph is a connected component in the original graph.)

3 Contraction x y {x, y} ) ) ) 5) 7) 5 5 8) 9) 10) 11) 1) 9 1) 1) Contraction - all together now x y a) b) c) d) 5 5 e) f) g) h) But... i) j) A) Not min cut!

4 B) Contracted wrong edge somewhere... C) If never contract an edge in the cut... D)...get min cut in the end! E) We might still get min cut even if we contract edge min cut. Why??? The resulting algorithm The algorithm... Algorithm MinCutG) G 0 G i = 0 while G i has more than two vertices do e i random edge from EG i ) G i+1 G i /e i i i + 1 Let S, V \ S) be the cut in the original graph corresponding to the single edge in G i return S, V \ S) How to pick a random edge? Lemma X = {x 1,..., x n }: elements, ωx i ): integer positive weight. Pick randomly, in On) time, an element X, with prob picking x i being ωx i ) /W, where W = ni=1 ωx i ). Proof : Randomly choose r [0, W ]. Precompute β i = i k=1 ωx k ) = β i 1 + ωx i ). Find first index i, β i 1 < r β i. Return x i. A) Edges have weight... B)...compute total weight of each vertex adjacent edges). C) Pick randomly a vertex by weight. D) Pick random edge adjacent to this vertex Analysis The probability of success Lemma... Lemma G: mincut of size k and n vertices, then EG) kn. Proof : Each vertex degree is at least k, otherwise the vertex itself would form a minimum cut of size smaller than k. As such, there are at least v V degreev)/ nk/ edges in the graph Lemma... Lemma If we pick in random an edge e from a graph G, then with probability at most n belong to the minimum cut. it Proof : There are at least nk/ edges in the graph and exactly k edges in the minimum cut. Thus, the probability of picking an edge from the minimum cut is smaller then k/nk/) = /n.

5 11... Lemma Lemma MinCut outputs the mincut with prob. nn 1). Proof A) E i : event that e i is not in the minimum cut of G i. B) MinCut outputs mincut if all the events E 0,..., E n happen. C) Pr [ ] E i E0 E 1... E i 1 1 V G i ) = 1 n i. = = Pr[E 0... E n ] = Pr[E 0 ] Pr [ [ [ ] E 1 E0 ] Pr E E0 E 1 ]... Pr En E0... E n Proof continued... As such, we have n i=0 = n n = 1 ) n n i = n i i=0 n i n n 1 n n... 1 n n 1) Running time analysis Running time analysis... Observation MinCut runs in On ) time. Observation The algorithm always outputs a cut, and the cut is not smaller than the minimum cut. Definition Amplification: running an experiment again and again till the things we want to happen, with good probability, do happen Getting a good probability MinCutRep: algorithm runs MinCut nn 1) times and return the minimum cut computed. Lemma probability MinCutRep fails to return the minimum cut is < 0.1. Proof : MinCut fails to output the mincut in each execution is at most 1. nn 1) MinCutRep fails, only if all nn 1) executions of MinCut fail. 1 nn 1) Result ) nn 1) exp nn 1) nn 1)) = exp ) < 0.1, since 1 x e x for 0 x 1. Theorem One can compute mincut in On ) time with constant probability to get a correct result. In On log n) time the minimum cut is returned with high probability. 5

6 11. A faster algorithm Faster algorithm Why MinCutRep needs so many executions? Probability of failure in first ν iterations is Pr [ ] ν 1 E 0... E ν 1 1 ) ν 1 n i = i=0 n i i=0 n i = n n n n 1 n n... n ν)n ν 1) =. n n 1) = ν = n/: Prob of success 1/. = ν = n n: Prob of success 1/n Faster algorithm... Insight A) As the graph get smaller probability for bad choice increases. B) Currently do the amplification from the outside of the algorithm. C) Put amplification directly into the algorithm Contract ContractG, t) shrinks G till it has only t vertices. FastCut computes the minimum cut using Contract. Contract G, t ) while G) > t do Pick a random edge e in G. G G/e return G FastCutG = V, E)) G -- multi-graph begin n V G) if n 6 then Compute minimum cut of G and return cut. t 1 + n/ H 1 ContractG, t) H ContractG, t) /* Contract is randomized!!! */ X 1 FastCutH 1 ), X FastCutH ) return mincut of X 1 and X. end Lemma... Lemma The running time of FastCutG) is On log n), where n = V G). Proof : Well, we perform two calls to ContractG, t) which takes On ) time. And then we perform two recursive calls on the resulting graphs. We have: T n) = On ) + T n ) The solution to this recurrence is On log n) as one can easily and should) verify. 6

7 Success at each step Lemma Probability that mincut in contracted graph is original mincut is at least 1/. Proof : Plug in ν = n t = n 1 + n/ into success probability: ] Pr [E 0... E n t = tt 1) n n 1) ) 1 + n/ 1 + n/ 1 nn 1) Probability of success... Lemma FastCut finds the minimum cut with probability larger than Ω 1/ log n). See class notes for a formal proof. We provide a more elegant direct argument shortly Amplification Lemma Running FastCut repeatedly c log n times, guarantee that the algorithm outputs mincut with probability 1 1/n. c is a constant large enough. Proof : A) FastCut succeeds with prob c / log n, c is a constant. B)...fails with prob. 1 c / log n. C)...fails in m reps with prob. 1 c / log n) m. But then 1 c / log n) m e c / log n ) m e mc / log n 1 n, for m = log n) /c Theorem Theorem One can compute the minimum cut in a graph G with n vertices in On log n) time. The algorithm succeeds with probability 1 1/n. Proof : We do amplification on FastCut by running it Olog n) times. The running time bound follows from lemma On coloring trees and min-cut Trees and coloring edges... A) T h be a complete binary tree of height h. B) Randomly color its edges by black and white. C) E h : there exists a black path from root T h to one of its leafs. D) ρ h = Pr[E h ]. E) ρ 0 = 1 and ρ 1 = / see below). 7

8 Bounding ρ h A) u root of T h : children u l and u r. B) ρ h 1 : Probability for black path u l children C) Prob of black path from u through u 1 is: Pr [ uu l is black ] ρ h 1 = ρ h 1 / D) Prob. no black path through u l is 1 ρ h 1 /. E) Prob no black path is: 1 ρ h 1 /) F) We have ρ h = 1 1 ρ h 1 ) ρ h 1 = ρ h 1 ) = ρ h 1 ρ h Lemma... Lemma We have that ρ h 1/h + 1). Proof : A) By induction. For h = 1: ρ 1 = / 1/1 + 1). B) ρ h = ρ h 1 ρ h 1 = fρ h 1 ), for fx) = x x /. C) f x) = 1 x/. = f x) > 0 for x [0, 1]. D) fx) is increasing in the range [0, 1] ) 1 E) By induction: ρ h = f ρ h 1 ) f = 1 h 1) + 1 h 1 h. F) hh + 1) h + 1) h h + h h 1 h h 1, h h h Back to FastCut... A) Recursion tree for FastCut corresponds to such a coloring. B) Every call performs two recursive calls. C) Contraction in recursion succeeds with prob 1/. Draw recursion edge in black if successful. D) algorithm succeeds there black path from root of recursion tree to leaf. E) Since depth of tree H + log n. F) by above... probability of success is 1/h + 1) 1/ + log n) Galton-Watson processes A) Start with a single node. B) Each node has two children. C) Each child survives with probability half independently). D) If a child survives then it is going to have two children, and so on. E) A single node give a rise to a random tree. F) Q: Probability that the original node has descendants h generations in the future. G) Prove this probability is at least 1/h + 1). 8

9 Galton-Watson process A) Victorians worried: aristocratic surnames were disappearing. B) Family names passed on only through the male children. C) Family with no male children had its family name disappear. D) # male children of a person is an independent random variable X {0, 1,,...}. E) Starting with a single person, its family as far as male children are concerned) is a random tree with the degree of a node being distributed according to X. F).. A family disappears if E[X] 1, and it has a constant probability of surviving if E[X] > Galton-Watson process A)... Infant mortality is dramatically down. No longer a problem. B) Countries with family names that were introduced long time ago... C)...have very few surnames. Koreans have 50 surnames, and three surnames form 5% of the population). D) Countries introduced surnames recently have more surnames. Dutch have surnames only for the last 00 years, and there are 68, 000 different family names). 9