1 / 8

# An NP-completeness Example - PowerPoint PPT Presentation

An NP-completeness Example. The graph CLIQUE problem. Undirected Graph G = (V, E) a clique is a complete subgraph of G Subset V’ of vertices that are all connected to each other CLIQUE = {(G,k) : G is a graph with a clique of size k}

Related searches for An NP-completeness Example

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'An NP-completeness Example' - oshin

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### An NP-completeness Example

• Undirected Graph G = (V, E)

• a cliqueis a complete subgraph of G

• Subset V’ of vertices that are all connected to each other

• CLIQUE = {(G,k) : G is a graph with a clique of size k}

• For a general graph, and large-enough |V|, k, straightforward (naïve) algorithm for CLIQUE takes factorial time

• OMEGA(k2 * C(|V|, k) )

• 1. CLIQUE is in NP

• Exercise: Given encoding of G, and a certificate V’, come up with a poly-time verification algorithm to check that V’ is a clique of size k for G

• 2. CLIQUE is NP-hard

• 3-CNF-SAT P CLIQUE

• 1 + 2 implies CLIQUE is NP-complete

CLIQUE is NP-hard 3-CNF-SAT P CLIQUE

• From instance I_3_CNF_SAT of 3-CNF-SAT, generate an instance I_CLIQUE of CLIQUE in polynomial time

• I_CLIQUE returns true exactly whenever I_3_CNF_SAT returns true

• I_CLIQUE includes a graph G = (V, E) and a vertex subset V’

• Build the graph G to do the mapping

• I_CLIQUE

•  = C1 AND C2 AND … Ck

• C1 = lit_11 OR lit_12 OR lit_13.

• Lit_11 = x or NOT x

• Similarly for C2, C3, … Ck

• Build Graph G = (V, E) as follows

• Take the r-th clause in I_3_CNF_SAT

• For lit_1r OR lit_2r OR lit_3r, place vertices vert_1r, vert_2r, vert_3r in V

• put an edge from vertices vert_ir to vert_js (i = 1, 2, 3; j = 1, 2, 3) if

• r != s (I.e., the vertices correspond to different triples in the formula) AND

• lit_ir != NOT lit_js

• don’t put an edge from x to (NOT x) even if they are in different triples

• CAN DO THIS IN POLY TIME!

•  = (x1 OR (NOT x2) OR (NOT x3)) AND ((NOT x1) OR x2 OR x3) AND (x1 OR x2 OR x3) true when x2 =0 and x3 = 1

• GRAPH (Fig 34.14) HERE

• Suppose  is TRUE for some values of the variables.

• Then each Cr is true => at least one of lit_r1, lit_r2, lit_r3 is true.

• Select one such “true” vertex from each Cr (e.g., if lit_r2 and lit_r3 are both true, select vert_r2 at random from vert_r2 and vert_r3)

• Call this a subset V’

• V’ has k elements (one from each Cr) and is a clique

• Take any two vertices in V’, and there is an edge between them by construction.

• By the selection method, they are in different triples

• They are both 1’s by selection method, so they are not negations of each other.

• We mapped from a satisfying assignment for  to a graph with a clique of size k (vert_12, vert_23, vert_33)

• Suppose G has a clique V’ of size k (vert_12, vert_23, vert_33)

• Each vertex in V’ belongs to a different triple and cover all triples between them

• there cannot be an edge in G if two vertices belong to the same triple.

• If vert_ri is in V’, assign lit_ri = 1

• there cannot be an assignment like { xi = 1; (NOT xi) =1 }

• there cannot be an edge in G if two vertices are negations of each other

• Each clause has one TRUE literal by the mapping

• We mapped from a graph with clique of size k to a satisfying assignment for 

VERTEX COVER x3) AND (x1 OR x2 OR x3)

• Vertex cover of an undirected graph G = (V, E) is a subset V’ of V such that

• if (u, v) is in E, then either u or v or both is in V’

• The vertices in V’ together ‘cover’ all the edges of E

• |V’| is the size of a vertex cover.

• VERTEX_COVER = { (G, k) : graph G has a vertex cover of size k}.

• VERTEX_COVER is NP-complete

• VERTEX_COVER is in NP

• exercise given (G, k) and certificate c = V’, find a poly algorithm to verify c

• VERTEX_COVER is NP-hard

• CLIQUE p VERTEX_COVER

• Complement of a graph x3) AND (x1 OR x2 OR x3)

• COMP_G = (V, COMP_E)

• COMP_E = { (u, v) : u, v are in V, u != v, and (u, v) not in E }

• edges that are not in E

• FIG 34.15 HERE

• Reduction algorithm from CLIQUE to VERTEX_COVER

• from I_CLIQUE = (G, k) Compute COMP_G (poly)

• Claim: we now have I_VERTEX_COVER = (COMP_G, |V| - k)

• G has a clique of size k if and only if COMP_G has a vertex cover of size |V| - k

• If part: x3) AND (x1 OR x2 OR x3)

• Suppose G has a clique V’ of size k.

• Claim: V-V’ is a vertex cover in COMP_G

• proof: suppose (u, v) is an edge in COMP_E.

• Then (u, v) is not in E

• So, either u or v is not in V’ since if both u and v are in V’, there must be an edge between them.

• That is, either u or v is in V-V’

• if (u, v) is an edge in COMP_E, then V-V’ covers that edge, hence V-V’ is a vertex cover in COMP_G. and V-V’ has size |V| -k

• Only-if part:

• Suppose COMP_G has a vertex cover V’ of size |V| -k

• claim: V-V’ is a clique in G

• proof: for all u, v in V, if (u, v) is in COMP_E, then u is in V’ or v is in V’ or both

• by contrapositive, for all u, v in V, if u is not in V’ and v is not in V’, then (u, v) is in E.

• I.e., V-V’ is a clique.