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hw6solns - 15-750 Graduate Algorithms Spring 2009 Miller...

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15-750 — Graduate Algorithms — Spring 2009 Miller and Dinitz and Tangwongsan Assignment 6 Solutions 1 Separators in outer planar graphs (20 pts.) (a) We say that a graph is outer planar if it is 1) planar, and 2) there exists a planar embedding such that all of the vertices are on the boundary of the outer face. Prove that all outer-planar graphs have 2-vertex-separators. Do not give a proof by algo- rithm. 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 is k -outer planar if it can be drawn in the plane so that the vertices can be partitioned into k nested “rings” where all all edges stay within one ring or go between adjacent rings. Prove that all k -outer planar graphs have 2 k -vertex separators. (c) Given an embedded triangulated k -outer planar give an O ( n ) time algorithm to find this 2 k -separator. Solution: (a) Here is one clever approach. Also, the result for the second part applies to this question for k = 1 . Start with a planar drawing of G . 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 suffices to find a 1 -edge separator in the dual. We did this in class. The two ends of the edge are vertices of the desired separator.
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15-750 HW 6 Solutions 2 (b) For k -outer planar graphs, augment the graph to have a vertex in the outer face with edge to every vertex in the outer ring. Add edges so there is at least one edge from every vertex to another vertex in a smaller ring. Now, the resulting graph has a spanning tree of diameter 2 k . We showed in class that such graphs have 2 k -vertex separators. It remains to show that these augmentations did not violate the planarity assumption. (c) See Kozen for how to to the whole planar separator theorem constructively in linear time. 2 More NP-Completeness (20 pts.) Let φ be a 3CNF formula. A 6 = -assignment to the variables of φ is one where each clause contains two literals with unequal truth values. In other words a 6 =-assignment satisfies φ without assigning three true literals in any clause. (a) Show that the negation of an 6 =-assignment to φ is also a 6 =-assignment. (b) Let 6 =SAT be the collection of 3CNF formulas that have a 6 =-assignment. Prove that 6 =SAT is NP-complete. Hint: consider replacing each clause c i = ( y 1 y 2 y 3 ) by the two clauses ( y 1 y 2 z i ) and ( z i y 3 b ), where z i is a new variable for each clause c i and b is a single additional new variable. (c) A cut in an undirected graph is a partition of the vertices into two disjoint subsets S and T . The size of the cut is the number of edges that have one endpoint in S and the other in T . Let MAXCUT = {h G, k i | G has a cut of size k or more } . Prove that MAXCUT is NP-complete.
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