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Tag Archives: Clique
Protected: Partially Ordered Knapsack
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Posted in Appendix Mathematical Programming
Tagged Clique, Difficulty 5, Knapsack, Partially Ordered Knapsack, uncited reduction
Protected: Conjunctive Boolean Query
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Posted in Appendix: Storage and Retrieval
Tagged Clique, Conjunctive Boolean Query, Conjunctive Query Foldability, Difficulty 3, SR31, uncited reduction
Protected: KClosure
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Posted in AppendixGraph Theory
Tagged Clique, Difficulty 8, GT58, KClosure, Reductions I needed help with, Weighted KClosure
Protected: Maximum Subgraph Matching
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Posted in AppendixGraph Theory
Tagged Clique, Difficulty 2, Difficulty 5, GT48, GT50, Largest Common Subgraph, Maximum Subgraph Matching, No G&J reference, Subgraph Isomorphism, uncited reduction
Protected: Induced Subgraph with Property π, Induced Connected Subgraph with Property π
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Posted in AppendixGraph Theory
Tagged 3sat, Clique, Difficulty 10, GT21, GT22, Independent Set, Induced Subgraph With Property Pi, Vertex Cover
Protected: Balanced Complete Bipartite Subgraph
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Posted in AppendixGraph Theory
Tagged Balanced Complete Bipartite Subgraph, Clique, Difficulty 7, GT24
Protected: Largest Common Subgraph
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Posted in Chapter 3 Exercises
Tagged Clique, Difficulty 3, GT49, Largest Common Subgraph, No G&J reference, reductions, uncited reduction
Clique
G&J don’t even bother to prove Clique is NPComplete, just stating that it (and Independent Set) is a “different version” of Vertex Cover. But the problem comes up often enough that it’s worth seeing in its own right.
The problem: Clique (I’ve also seen “Max Clique” or “Clique Decision Problem” (CDP))
The definition: Given a graph G=(V,E), and a positive integer J (G&J say J needs to be ≤ V, but I think you can allow large values and make your algorithm just say “no”). Does J contain a subset V’ of V, or size at least J, where every two vertices in V’ are joined by an edge in E?
Example: Using the graph we had for VC:
{a,c,d} form a clique of size 3. There are no cliques of size 4, but if we add the edge (d,g) and the edge (c,h), then the vertices {c,d,g,h} would be one.
The reduction: From Vertex Cover. Given a graph G and an integer K that is an instance of Vertex Cover, take the complement of G (the complement of a graph has the same vertices, but the set of edges is “inverted”: (u,v) is an edge in the complement if and only if it was not an edge in G), and use it as your instance of CDP. Set J= VK
Difficulty: 3. It’s not a 2 because it takes a little thought to see why the complement works.
Note: As stated perviously, I teach this course using the Cormen book, so in class I show Clique is NPHard by reducing from 3SAT. It’s a good example of a hard reduction, probably about as hard as the VC reduction G&J do.
Posted in Core Problems
Tagged Clique, core problems, Difficulty 3, reductions, Vertex Cover
Vertex Cover
Back to the “core 6” with a classic problem from graph theory. There are lots of similar graph problems, so it’s important to keep them all straight.
The Problem: Vertex Cover (VC)
The Definition: Given a graph G=(V,E) and a positive integer K (G&J say that K ≤ V, but really, if it’s bigger than that, you can have the algorithm just return “yes”). Can I find a subset V’ of V, V’ ≤ K, where every edge in E has at least one endpoint in V’?
Example: Hmm, I wonder if I can insert an image..
Ok, looks like that worked. In the graph above, {c,d,g} is a vertex cover, because all edges have at least one endpoint in the set {c,d,g}. As we’ll see later, a graph containing a clique (complete subgraph) of N nodes will need at have at least N1 of those vertices in the cover.
The reduction: From 3SAT. G&J on pages 5456 show the construction. We build a graph with vertices for all variables and their negation, with an edge between them. We have three variables for each clause, connected in a triangle. (The three variables correspond to “positions” in the clause). Then there is an edge from each vertex corresponding to a literal (or its negation) to its corresponding “clause vertex” the vertex that holds the position of that variable in the original formula. K= # of variables + 2* number of clauses.
Difficulty: 7 , assuming it’s the first time a student has seen this sort of reduction. If they have seen something similar (for example, a reduction from 3SAT to Clique), then maybe a 5. The ideas are pretty similar.
Note: I teach my algorithms class using the Cormen book. In it, they reduce 3SAT to Clique, proving Clique is NPComplete, and then reduce Clique to VC. This works in the exact same way as the reduction from VC to Clique that I’ll be doing here next.
Posted in Core Problems
Tagged 3sat, Clique, core problems, Difficulty 7, reductions, Vertex Cover