15-859E: Advanced Algorithms CMU, Spring 2015 Lecture #2: Randomized MST and MST Verification January 14, 2015

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1 15-859E: Advaced Algorithms CMU, Sprig 2015 Lecture #2: Radomized MST ad MST Verificatio Jauary 14, 2015 Lecturer: Aupam Gupta Scribe: Yu Zhao 1 Prelimiaries I this lecture we are talkig about two cotets: Karger-Klai-Tarja algorithm ad MST Verificatio. Our settig is the same as last lecture. Let G = (V, E to be a simple graph with vertex set V ad edge set E, where V = ad E = m. We also assume that all the weights of edges are distict. I Lecture 1 we saw some determiistic algorithms fidig MST i time O(m log, O(m + log ad O(m log. Our goal i this lecture is to fid MST i the graph i O(m + expected time. Remember the two rules we metioed i Lecture 1: cut rule ad cycle rule. 1.1 Heavy & light Figure 2.1: Example of heavy ad light edges Take ay tree 1 i a graph. For example, i Figure 2.1 we pick the tree with black edges. The tree might ot be a MST, or eve ot be coected (a forest. Now look at ay other edge i the graph. The blue edge o the right side is heavier tha ay other edge i the cycle. The by cycle rule this edge will ot be i the MST of the graph. What about the red edge o the left side? This edge is ot the heaviest edgei the cycle, ad i fact this edge appears i the MST of the graph. Therefore we ca discard some edges which will ever appear i the MST based o ay give tree of a graph. Defiitio 2.1. Let T be a forest of graph G ad e E(G. If e create a cycle whe addig to T ad e is heaviest edge i the cycle, the we call edge e is T -heavy. Otherwise, edge e is T -light. Every edge should be either T -heavy or T -light. Notice that if a edge e does ot make a cycle, the e is T -light. If edge e is i the tree T, edge e is also T -light. I Figure 2.1, the red edge is T -light, the blue edge is T -heavy, ad all black edges are T -light. Fact 2.2. Edge e is T -light iff e MST(T {e}. 1 Here whe we talk about trees, i fact we mea trees or forests. Ad MST meas the miimum spaig forest if the graph is ot coected. 1

2 Fact 2.3. If T is a MST of G the edge e E(G is T -light iff e T. Therefore our idea is that pick a good tree/forest T, fid all the T -heavy edges ad get rid of them. Hopefully the umber of edges remaied is small. The we will fid the MST i remaiig edges. We wat to fid some tree T such that there are a lot of T -heavy edges. 1.2 Magic Blackbox How should we fid all the T -heavy edges? Here we first assume a magic blackbox: MST Verificatio. We ll show how to implemet MST Verificatio later. Theorem 2.4 (MST Verificatio. Give a tree T G, we ca output all the T -light edges i E(G i time O( V + E. This woderful blackbox ca output all the T -light edges i liear time. 2 Karger-Klai-Tarja Algorithm Suppose a graph G = (V, E has vertices ad m edges. KKT(G as followig. Here we will preset the algorithm 1. Ru 3 rouds of Borůvka s Algorithm o G to get a graph G = (V, E with /8 vertices ad m m edges. 2. E 1 a radom sample of edges E of G where each edge is picked idepedetly with probability 1/2. 3. T 1 KKT(G 1 = (V, E E 2 all the T 1 -light edges i E. 5. T 2 KKT(G 2 = (V, E Retur T 2 (combie with the edges chose i Step 1. Here the idea is that we radomly choose half of the edges ad fid the MST o those edges. We hope this tree T will have a lot of T -heavy edges therefore we ca discard these edges ad fid the MST o the remaiig graph. Theorem 2.5. KKT algorithm will retur MST(G. Proof. We ca prove it by iductio. For the basic case it is trivial, because we will fid the MST i Step 1. Otherwise i Step 3 we fid a MST T 1 of graph G 1, ad i E 2 just discard all the T 1 -heavy edges from the etire edges E (ot oly from E 1!! which are ot possible to be i the MST of G. Or i other words, MST(G E 2. Therefore what Step 5 returs is the MST of G 2, which is also the MST of G. Combie it with the edges we choose i Step 1, we get the MST of graph G. The key idea of this proof is that discardig heavy edges of ay tree i a graph will remai MST the same. Now we eed to deal with the complexity. 2

3 Claim 2.6. E[#E 1 ] = 1 2 m. This claim is trivial sice we pick each edge with probability 1/2. Claim 2.7. E[#E 2 ] 2. This meas graph G 2 will be hopefully very sparse. Figure 2.2: Illustratio of aother order of coi tossig How should we aalysis E[#E 2 ]? I Step 2 of KKG algorithm, we first flip a coi o each edge, the take all the edges tossig head ad fially i Step 3 we use KKG algorithm o this ew graph to get the MST of G 1. Let s thik of this process i aother order. First of all, Step 3 is to calculate the MST of the ew graph. It does ot matter if we use KKG algorithm or some other method, ad we will still get the same MST. The istead of first flippig cois for all edges the calculatig the MST. We ca do these two thigs simultaeously: We flip the cois for the edges i a icreasig order by weight, ad do Kruskal immediately after ay edge tossig head. We defie a coi toss is useful, if Kruskal algorithm would like to add this edge i the MST, ad defie a coi toss is useless if ot. For example, Figure 2.2(a gives a graph with coi toss result. The solid edges are the edges with tossig head, ad the dashed edges are the edges with tossig tail. Figure 2.2(b is the MST o all edges tossig head. Now let s check the usefuless of all these coi flips, as i Figure 2.2(c. I the begiig the MST is a empty set. The the coi flip of the first edge is useful, sice it should be added ito the MST. The tossig result is head, so we add this edge ito MST. The coi flip of the secod edge is useful, sice it should be added ito the MST. However the tossig result is tail, so sadly we ca ot add this edge ito MST. The coi flip of the third, forth, fifth edges are all useful, ad the tossig result are all head, so we add all of them ito the MST. The coi toss of sixth edge is useless, because we will ot add this edge ito MST sice it will creat a cycle, so is the seveth edge. Claim 2.8. T 1 = MST(G 1. e E is T 1 -light if ad oly if the coi flip of e is useful. 3

4 Proof. If the coi flip of e is useful, the at the momet we flip the coi by addig e ito the MST there is o cycle, which meas either e T 1 (coi tossig head, or there is o cycle i T 1 {e}, or edge e is ot the heaviest edge i the cycle of T 1 {e}. I ay of these cases, edge e is T 1 -light. O the other had, if the coi flip of e is useless, the at the momet we flip the coi there would be a cycle if we add e ito the MST, therefore edge e is T 1 -heavy. The we ca prove Claim 2.7. Proof of Claim 2.7. E[#E 2 ] = E[#useful coi flips] 1 1/2 2 The first iequality holds sice by each useful coi flip we may add a edge ito the MST of G 1 with probability 1/2, ad the fial T 1 = MST (G 1 has at most 1 edges. Therefore the expectatio of useful coi flips is at most 2( 1 2. Theorem 2.9. KKT(G = (V, E ca retur the MST i time O(m +. Proof. Let X G be the expect ruig time o graph G, ad X m, := max {X G} G=(V,E, V =, E =m I KKG algorithm, Step 1, 2, 4 ad 6 will be doe i liear time, so we assume the time cost of Step 1, 2, 4 ad 6 is at most cm. Step 3 will sped time X G1, ad Step 5 will sped time X G2. The we have X G cm + X G1 + X G2 cm + X m1, + X m 2, Here we assume that X m, c(2m +, the X G = cm + E[c(2m 1 + ] + E[c(2m 2 + ] c(m + m + 6 c(2m + The first iequality holds because E[m 1 ] 1 2 m ad E[m 2 ] 2. The secod iequality holds because /8 ad m m. 3 MST Verificatio Now we come back the implemet of the magic blackbox. Here we oly cosider about the trees (ot forests. Here we refie Theorem 2.4 as followig. Theorem 2.10 (MST Verificatio. Give T = (V, E where V = ad m pairs of vertices (u i, v i, we ca fid the heaviest edge o the path i T from u i to v i for all i i O(m + time. Tarja fid a algorithm i Time O(mα(m, usig Uio-Fid. Komlos fid how to do it with O(m + comparisos. Dixo-Rauch-Tarja o RAM machie with O(m. Here we will give the idea of Komlos. 4

5 Figure 2.3: Illustratio of balacig a tree 3.1 Balace the tree Suppose T = (V, E is a tree o vertices, ad we ru Boruvka s algorithm o T. (Let V 1 = V be the origial vertices at the begiig of roud 1, V i be the vertices i roud i, ad say there are L rouds so that V L+1 = 1. We build a tree T as follows: the vertices are the uio of all the V i. There is a edge from v V i to w V i+1 if the vertex v belogs to a compoet i roud i ad that is cotracted to form w V i+1 ; the weight of this edge (v, w is the mi-weight edge out of v i roud i. (Note that all vertices i V are ow leaves i T. Figure 2.3 shows a example of balacig a tree. Fact For odes u, v i a tree T, let maxwt T (u, v be the maximum weight of a edge o the (uique path betwee u, v i the tree T. For all u, v V, we have maxwt T (u, v = maxwt T (u, v This is a problem i Homework 1. I Figure 2.3, we have maxwt T (v 1, v 7 is 7 which is the weight of edge (v 4, v 6. We ca check that maxwt T (v 1, v 7 is also 7. Fact T has at most log 2 layers sice each vertex has at least 2 childre ad all leaves of T are at the bottom (the same level.. Here we ca make a balaced tree T based o tree T with some good properties. The we ca just query the heaviest edges of vertex pairs i tree T istead of T. Aother trick is that we ca assume that all queries are acestor-descet queries if we ca fid the least commo acestor quickly. Theorem 2.13 (Harel-Tarja. Give a tree T, we ca preprocess i O( time, ad aswer all LCA queries i O(1 time. 5

6 3.2 Get the aswer Now we have reduced our questio to how to aswer the acestor-descet queries efficietly. For each edge e = (u, v where v is the paret of u, we will look at all queries startig i subtree T u ad edig above vertex v. Say those queries go to w 1, w 2,..., w k. The the query strig is Q e = (w 1, w 2,..., w k. The we eed to calculate the aswer strig A e = (a 1, a 2,, a k where a i is the maximal weight amog the edges betwee w i ad u. Figure 2.4: Illustratio of queries ad aswers Figure 2.4 gives us a example. Suppose Q (b,a = (w 1, w 3, w 4, which meas there are three queries startig from some vertices i the subtree of b ad edig to w 1, w 3, w 4. The we get the aswer A (b,a = (a 1, a 3, a 4 = (6, 4, 4 sice the maximal weight of a edge o the path from w 1 to b is the weight of edge (w 1, w 2, ad the maximal weight of a edge o the path from either w 3, w 4 to b is from the weight of edge (w 3, w 4. Give the aswer of A (b,a, how should we fid the aswer of A (c,b efficietly? Say Q (c,b = (w 1, w 4, b. Comparig with Q (b,a, we may lose some queries sice they are from some other childre of vertex b, ad we may have some other queries edig at b which will ot cotai i Q (b,a. The easiest way is to update every query. Suppose the weight of edge (c, b is t. The we ca calculate A (c,b = (max{a 1, t}, max{a 4, t}, t = (max{6, 5}, max{4, 5}, 5 The trick is that if i Q e = (w 1, w 2,..., w k, w 1, w 2,..., w k is sorted from the top to the bottom. tha the aswers A e = (a 1, a 2,, a k should be o-icreasig, a 1 a 2 a k. Therefore we ca do biary search to reduce the umber of comparisos. Claim Give the questio strig Q e for e = (u, v where v is the paret of u, the aswer strig A e for e, we ca compute aswers A e for e = (w, u where w is a child of u, withi time log( A e

7 Theorem The total umber of comparisos for all queries t e log ( Q e + 1 O(m + Proof. Assume the umber of edges i level i is i. Here the level is couted from the bottom to the top, the edges coected to leaves are at level 0. log 2 (1 + Q e = i avg (log 2 (1 + Q e i log 2 (1 + avg = i log 2 (1 + i log 2 ( 1 + m i ( Q e Q e i ( m + = i log 2 + log 2 i The first iequality holds by Jese s iequality ad covexess of fuctio log 2 (1+x. The secod iequality holds sice the umber of all queries is m ad each query will oly appear o at most oe edge o ay particular level. Therefore we have t = e = i i log 2 (1 + Q e log 2 (1 + Q e ( m + i log 2 + log 2 i = log 2 m + + i i log 2 i m + log 2 + c = O( log m + + = O(m + 7