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Unformatted text preview: 15750 Graduate Algorithms Spring 2009Miller and Dinitz and TangwongsanAssignment 6 Solutions1Separators in outer planar graphs(20 pts.)(a)We say that a graph isouter planarif it is1) planar, and2) there exists a planar embedding such that all of the vertices are on the boundary of theouter face.Prove that all outerplanar graphs have 2vertexseparators. Do not give a proof by algorithm. Use the tools developed in class.HINT:Recall that augmenting the graph does not change the size of the separators.(b)There is a natural extension of the definition of outer planar graphs. Say that a graph iskouter planarif it can be drawn in the plane so that the vertices can be partitioned intoknested rings where all all edges stay within one ring or go between adjacent rings. Provethat allkouter planar graphs have 2kvertex separators.(c)Given an embedded triangulatedkouter planar give anO(n) time algorithm to find this2kseparator.Solution:(a) Here is one clever approach. Also, the result for the second part applies to this question fork= 1.Start with a planar drawing ofG. There is some outer face containing all of the vertices.Triangulate the other faces. Observe that the dual of the triangulation is a binary tree. It sufficesto find a1edge separator in the dual. We did this in class. The two ends of the edge are verticesof the desired separator.15750HW 6 Solutions2(b) Forkouter planar graphs, augment the graph to have a vertex in the outer face with edge toevery vertex in the outer ring. Add edges so there is at least one edge from every vertex toanother vertex in a smaller ring. Now, the resulting graph has a spanning tree of diameter2k.We showed in class that such graphs have2kvertex separators. It remains to show that theseaugmentations did not violate the planarity assumption.(c) See Kozen for how to to the whole planar separator theorem constructively in linear time.2More NPCompleteness(20 pts.)Letbe a 3CNF formula. A6=assignmentto the variables ofis one where each clause containstwo literals with unequal truth values. In other words a6=assignment satisfieswithout assigningthree true literals in any clause.(a) Show that the negation of an6=assignment tois also a6=assignment.(b) Let6=SAT be the collection of 3CNF formulas that have a6=assignment. Prove that6=SAT is NPcomplete. Hint: consider replacing each clauseci= (y1y2y3) by thetwo clauses (y1y2zi) and (ziy3b), whereziis a new variable for each clauseciandbis a single additional new variable.(c) Acutin an undirected graph is a partition of the vertices into two disjoint subsetsSandT. The size of the cut is the number of edges that have one endpoint inSandthe other inT. LetMAXCUT={hG,ki Ghas a cut of sizekor more}. Prove thatMAXCUTis NPcomplete....
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This document was uploaded on 11/03/2009.
 Spring '09
 Algorithms

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