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The graph CLIQUE problem

- 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

CLIQUE is NP hard for the “special” kind of graphs

- CLIQUE is “harder” for general kind of graph
- CLIQUE is NP-hard for general kind of graph.
- CLIQUE IS NP-COMPLETE.

VERTEX COVER

- 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

- 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:

- 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.

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