Algorithms in The Real World Fall 2002 Homework Assignment 2 Solutions

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1 Algorithms i The Real Worl Fall 00 Homewor Assigmet Solutios Problem. Suppose that a bipartite graph with oes o the left a oes o the right is costructe by coectig each oe o the left to raomly-selecte oes o the right (each chose with probability /. All raom choices are mae iepeetly, a there is o restrictio o the egree of a oe o the right, i.e., oes o the right may have ay egree from 0 to. Show that for ay fixe β >, a ay fixe > β +, there exists a fixe α such that the graph has (α, β expasio with probability >0. Solutio. (Maria-Floria Balca. We ow that, i geeral, a graph has (α, β expasio if every subset of < α oes o the left has at least β eighbors o the right, where α a β are fixe costats, which satisfy 0 α, β > 0. Let s suppose first that a (, bipartite graph is create i the followig way: for each oe o the left, we pic a raom subset of oes o the right as its eighbors (so each oe o the left has exactly istict eighbors o the right. We have β > fixe, fixe, > β +. We choose α such that: e β++ β α = ( ββ It is clear from above that 0 α. We show that for this choice, our graph has (α, β expasio with probability >0. We have: where: P (graph has (α, β expasio property = = P (graph oes ot have (α, β expasio property = p p = P (graph oes ot have (α, β expasio property = = P (there exist α oes o the left that have at most β eighbors o the right Cosier α; eote A the evet that a subset of vertices o the left has fewer tha β eighbors o the right. It s clear that:

2 p α, N P (A Fix a subset S( of oes o the left, a a subset ( T of β oes o the right. There are ways of choosig S a ways of choosig T. β The probability that all eighbors of S lie isie T is less or equal to : β So, ( P (A ( β β which is the probability that all eges emaatig from some oes o the left fall withi some β oes o the right. But the we have: ( P (A P (A Usig α, we have: [ ( e ( ( β e β β β ( ] βe / ( P (A ( β+ β β β e β++ P (A ( α β β β e β++ Due to the choice of α, it follows that: P (A ( β β e β++ β β e β++ = ( We have thus showe that: p P (A ( = α, N So, p <, a thus the probability that the graph has (α, β expasio property is p > 0. We have just prove that there exists a fixe α such that the graph has the (α, β expasio property with probability >0. =

3 Problem. A. Prove that the bisectio with (i.e. the umber of eges that must be remove to separate a graph ito two equal-size parts, withi of the complete graph with vertices is (/. B. Prove that the bisectio with of the -oe hypercube is /. This shoul be prove from below a above. (Hit: show that the complete graph ca be embee i the hypercube so that vertices map to vertices, eges i the complete graph map to paths i the hypercube, a each hypercube ege es up supportig the same umber of paths. Solutio. (Rya Williams. A. Let G be a complete graph. We will prove that the bisectio with of a complete graph with oes is. For eve this is exactly (/, a for o this is +, which correspos to a separatig the graph i two equal-size parts, withi. Clearly, there exists a bisectio of G, of size, if we place of G s oes i a set S, a the other oes i a set T. Each oe i S is resposible for exactly eges crossig the cut (S, T a S =, so this is a bisectio. Notice the S a T were arbitrary, so this hols for all bisectios of G. B. We ca easily thi of a = oe hypercube as a set of strigs over {0, }, with a ege betwee strig x to y iff h(x, y = (where h is the Hammig istace. So let E = {(x, y : h(x, y = }. Cosier the bisectio (S, T where S = {0b : 0b {0, } }, T = {b : b {0, } }. Give ay s S, ote there is a uique t T such that (s, t E, a viceversa, i.e. a - correspoece betwee members of S a members of T. There are oes i S, hece = / eges i the bisectio. So, the bisectio with is at most /. Now, for the lower bou. We embe a irecte complete graph K, ito the hypercube. First, umber the oes i K (from 0 to, a associate oes i K with oes i the hypercube by associatig with its biary ecoig b(. The associate each irecte ege (u, v i K with a path i the hypercube from b(u to b(v. Note a path i the hypercube of legth from b(u to b(v ca be thought of as a sequece of bit flips that start with b(u a e with b(v. Remember from A that the bisectio with of K is ( / = (/. Let (S, T be a bisectio of the hypercube. By our argumet above, there is a correspoig (S, T bisectio of K with labels o the oes. Cosier ay ege (b(u, b(v crossig the bisectio (S, T. There are / = / pairs of oes {b(s, b(t} i the hypercube, where b(s S a b(t T, that tae oly the ege (b(u, b(v i the path from b(s to b(t (a o other eges crossig the bisectio. Therefore, if the bisectio with of the hypercube were less tha /, the the umber of paths i the hypercube that use those eges i the correspoig (S, T bisectio of K is less tha ( /, which cotraicts the result from A. Problem. I class, a i the Karypis a Kumar reaig, we covere a

4 multilevel ege-separator algorithm. I this problem you ee to geeralize this techique to wor for vertex separators irectly (o ot use a postprocessig stage. I particular:. Argue why coarseig usig a maximal matchig is or is ot still appropriate.. Describe what we shoul eep trac of whe coarseig (cotractig the graph (e.g. o the ege separator versio each ege ept a weight represetig the umber of origial eges betwee two multivertices.. Describe how we project the solutio of the coarsee versio bac oto the origial graph (the recursive solutio must retur a vertex separator. 4. Describe a variat of Kerigha-Li or (preferably the Fiuccia-Mattheyses heuristic for vertex separators. Be explicit about what the gai metric is. Solutio. (Cha Zhag. Coarseig usig a maximal matchig is still appropriate. Because if at a coarser level G i we have a vertex separator that has vertices, the after we expa the graph bac to G i, we have a vertex separator that has maximum vertices. There wo t be eges ewly geerate betwee the two subsets that are beig separate. Hece this provies still a relatively goo separator..whe coarseig the graph, we ee to eep a weight for each oe, represetig how may oes were merge to this oe.. As state i, whe projectig the coarsee versio bac to the origial graph, the result is still a reasoably goo vertex separator. We just expa the separator a the two other members of the partitio separately. The umber of eges betwee A a C, or betwee C a B may icrease, because each e of a ege may expa to two oes i the ext level, a thus for each ege i the coarser graph we may obtai 4 eges i the fier versio. However, we oly care about the umber of vertices, a as alreay state i, this umber (i ay of the partitio members ca icrease at most two times. 4. Variat of the Fiuccia-Mattheyses heuristic: For each oe v, we efie gai G(v as follows: if v belogs to A, the G(v is the reuctio i the size of the vertex separator if we move v ito the separator, a the move v eighbors from the separator ito B, if these oes have o ege to A. Similarly, we ca efie the gai if v belogs to B. Our algorithm woul be: F M(G, A, B, C For every u i A, Put u i a priority queue Q A, base o priority G(u For every v i B, Put u i a priority queue Q B, base o priority G(v While possible, Fi the maximum from Q A a Q B a o the correspoig vertex-movig operatio (as i the efiitio of G. Upate the gais. 4

5 Note: We ca also cout the reuctio i size of the separator, whe movig vertex v out of it a ito A (or B a brig its eighbors from B (or A, respectively ito the separator. Problem 4. Prove that give a class of graphs satisfyig a O( ( / ege-separator theorem, all members must have boue egree. or (more challegig Prove that give a class of graphs satisfyig a O( ( / vertex-separator theorem, the eges of ay member ca be irecte so that the out-egree is boue. You ca use the fact metioe i the class that such graphs have boue esity (i.e. the average egree is boue. Solutio. (Doru-Cristia Balca I We have to prove that, give S a class of graphs satisfyig a ( / ege separator theorem, all the members must have boue egree. Let s suppose the opposite: t N, G t S such that eg(g t t N, G t S, v t V t such that eg(v t t (* I orer for S to satisfy a ( / ege separator theorem, it is ecessary (by efiitio that S be close with respect to the subgraph operatio. That is, if G t S, the all its subgraphs are i S, satisfyig the ( / coitio. From (* it follows that, for every t N, there exists a graph cotaiig a vertex of egree larger tha t. If from G t we extract the star graph forme by this vertex v t a its eighbors, it follows that for every t N, S cotais a member (a star graph with at least t + oes. From the course, we ow that the class of the star graphs oes ot satisfy a O( ( /, but a O( ege separator theorem. But this simply cotraicts our suppositio. I coclusio, the egrees of all S s members must be boue by a positive costat. II Let s tae a class of graphs satisfyig a O( ( / vertex separator theorem. From the course, we ow that the average egree of such graphs is boue. Let be such a bou. The, for a graph G with vertices, we have: eg(v i eg(v i We stuy the possibility of givig a orietatio for the eges of G, such that the out-egree of all vertices is boue (basically, we mea that the maximum out-egree bou oes ot epe o. For a irecte graph, we ow that: eg (v i = eg + (v i = E, a eg (v i + eg + (v i = eg(v i 5

6 It follows that eg + (v i = eg(v i (so, we might have a iea of this bou s magitue. We will prove the out-egree boueess by iuctio over the graph size. We use maily the fact that the class of graphs cosiere here is close with respect to the subgraph operatio. Let v be the vertex with the smallest egree. We shall oriet its iciet eges outwars. Obviously, the, eg + (v = eg(v. Sice eg(v i a eg(v = mi v V eg(v, it follows that eg(v. The problem woul be solve if we fou a orietatio of all the other eges of the graph, such that the out-egree of all the vertices (ifferet from v! is boue. But we ow this, from the iuctio hypothesis: by taig out v a its iciet eges, we o ot iterfere with other vertices future out-egree, a we basically reuce the problem to a ietical oe, but for a smaller graph size. So, we ca give a orietatio of the remaiig eges (those o-iciet with v such that the out-egree of the vertices ifferet from v is boue by a costat b 0. If we choose b = max(, b 0, the it follows that we ca oriet the eges of ay graph i the class, such that the maximum out-egree is smaller tha a costat b. With this, the problem is solve completely. Problem 5. Cosier applyig ivie-a-coquer to graphs a let s say that mergig two recursive solutios tae f(s time, where s is the umber of eges separatig the two graphs. For each of the followig f(s, a assumig you are give a ege-separator tree for which all separators for the subgraphs of size are / / balace a boue by /, what is the ruig time of such a approach.. s. s log s. s 4. s 4 Solutio. (Hubert Cha Let T ( be the ruig time i the case of a graph of size. The recursive formula is: T ( = T ( + T ( + f(, for > 0 T ( = O(, for 0 where 0 is the threshol to stop the recursio.. f(s = s Assume for all 0, we have A > 0 large eough such that: T ( A B ( for > 0, where B = +. 6

7 The iuctive step: T ( = T ( + T ( + f( A B + A B + = A ( + + = A B. f(s = s log(s (log is base Usig a similar approach, we show: T ( A B log C for all, for some costats A, B, C. We first figure out what B a C ca be, by looig at the iuctive step. Deote P ( = A B log C. We ee to have: T ( = T ( + T ( + log ( P ( + P ( + log + (log So, we ee: P ( + P ( + log + log P (. ( We ca easily see that: (: P ( + P ( + log + log = A B log C + A B log = A C C = A log [ B log B log ] + log + log [ B + B ] log C + log + [ B log B [ B log B ] log ] log + C + C log It is quite easy to see that the last member of the above equatio is lower or equal to P (, if we choose: { } B = max, ( / + / { log +B C = max, log +B log / + / A = max{c +, }. f(s = s } Suppose T ( H( log + A for large eough A, such that the iequality hols for 0 a H(p = p log p ( p log ( p. 7

8 We write H = H( for the rest of the solutio. Iuctive step: T ( = T ( + T ( + H log ( + H = H log ( + + A + A ( log + log + H ( log + log + + A = H log + A 4. f(s = s 4 Agai, let A > 0 be large eough, so that for 0, we have: T ( A Iuctive step: T ( = T ( + T ( + 4 ( ( A ( ( + A = A. 8