1 / 21

Non-Approximability Results

Non-Approximability Results. Summary. Gap technique Examples: MINIMUM GRAPH COLORING, MINIMUM TSP, MINIMUM BIN PACKING The PCP theorem Application: Non-approximability of MAXIMUM 3-SAT. The Gap Technique. P 1 : NPO minimization problem (same for maximization) P 2 : NP-hard decision problem

kmorehouse
Download Presentation

Non-Approximability Results

An Image/Link below is provided (as is) to download presentation 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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Non-Approximability Results

  2. Summary • Gap technique • Examples: MINIMUM GRAPH COLORING, MINIMUM TSP, MINIMUM BIN PACKING • The PCP theorem • Application: Non-approximability of MAXIMUM 3-SAT

  3. The Gap Technique • P1: NPO minimization problem (same for maximization) • P2: NP-hard decision problem • Function f that maps instances x of P2 into instances f(x)of P1 such that: • If x is a YES-instance, then m*(f(x))=c(x) • If x is a NO-instance, then m*(f(x)) c(x)(1+g) • Theorem: No r-approximation algorithm for P1 exists with r<(1+g) (unless P=NP)

  4. Proof • A: r-approximation algorithm with r<(1+g) • If x is a YES-instance, then m*(f(x))=c(x). Hence, m(f(x),A(f(x))) rm*(f(x))=rc(x)<c(x)(1+g) • If x is a NO-instance, then m*(f(x)) c(x)(1+g). Hence, m(f(x),A(f(x))) c(x)(1+g) • A allows to decide P2 in polynomial time

  5. Inapproximability of graph coloring • NP-hard to decide whether a planar graph can be colored with 3 colors • Any planar graph is 4-colorable • f(G)=G where G is a planar graph • If G is 3-colorable, then m*(f(G))=3 • If G is not 3-colorable, then m*(f(G))=4=3(1+1/3) • Gap: g=1/3 • Theorem: MINIMUM GRAPH COLORING has no r-approximation algorithm with r<4/3 (unless P=NP)

  6. Inapproximability of bin packing • NP-hard to decide whether a set of integers I can be partitioned into two equal sets • f(I)=(I,B) where B is equal to half the total sum • If I is a YES-instance, then m*(f(I))=2 • If G is a NO-instance, then m*(f(G))  3=2(1+1/2) • Gap: g=1/2 • Theorem: MINIMUM BIN PACKING has no r-approximation algorithm with r<3/2 (unless P=NP)

  7. MINIMUM TSP • INSTANCE: Complete graph G=(V,E), weight function on E • SOLUTION: A tour of all vertices, that is, a permutation π of V • MEASURE: Cost of the tour, i.e., 1k |V|-1w(vπ[k], vπ[k+1])+w(vπ[|V|], vπ[1])

  8. Inapproximability of TSP • NP-hard to decide whether a graph contains an Hamiltonian circuit • For any g>0, f(G=(V,E))=(G’=(V,V2),w) where w(u,v)=1 if (u,v) is in E, otherwise w(u,v)=1+|V|g • If G has an Hamiltonian circuit, then m*(f(G))=|V| • If G has no Hamiltonian circuit, then m*(f(G))  |V|-1+1+|V|g=|V|(1+g) • Gap: any g>0 • Theorem: MINIMUMTSP has no r-approximation algorithm with r>1 (unless P=NP)

  9. The NPO world (unless P=NP) NPO MINIMUM TSP APX MINIMUM BIN PACKING MAXIMUM SAT ( ?) MINIMUM VERTEX COVER( ?) MAXIMUM CUT( ?) PTAS MINIMUM PARTITION PO MINIMUM PATH MINIMUM GRAPH COLORING? Certainly not in PTAS

  10. Deterministically checkable proofs input es. Boolean formula proof es. truth assignment Verifier Must read the entire proof Yes/No

  11. Probabilistically checkable proofs input es. Boolean formula proof es. truth assignment random bits Verifier Trade-off between the number of random bits and the number of bits read? Yes/No

  12. PCP[r,q] • A decision problem P belongs to PCP[r,q] if it admits a polynomial-time verifier A such that: • For any input of length n, A uses r(n) bits casuali • For any input of length n, A queries q(n) bits of the proof • For any YES-instance x, there exists a proof such that A answers Yes with probability 1 • For any NO-instance x, for any proof A answers Yes with probability less than 1/2 • Theorem: NP=PCP[log,O(1)]

  13. The PCP theorem • Given a class F of functions, PCP(r, F) is the union of PCP[r,q], for all qF • By definition, NP=PCP(0,poly) where poly is the set of polynomials • Theorem: NP=PCP(log,O(1)) • Proving that NP includes PCP(log, O(1)) is easy • Proving that NP is included in PCP(log, O(1)) is hard (complete proof is more than 50 pages: we will see only a part)

  14. Inapproximability of satisfiability • Gap technique • Intuitive motivation: gap in acceptance probability corresponds to gap in measure • Reduction f from SAT such that • If x is satisfiable, m*(f(x))=c(x) where c(x) is the number of clauses in f(x) • If x is not satisfiable, m*(f(x))<c(x)/(1+g) with g>0 • Gap: g not explicitily computed • Theorem: MAXIMUM SAT is not r-approximable for r<1+g

  15. The reduction • For any random string R of length O(logn), we construct a Boolean formula CRof constant size which is satisfiable if and only if there exists a sequence of answers to the queries that make the verifier answer Yes • CRcan be written in CNF • The final formula f(x) is the union of all these formulas CR

  16. The reduction (continued) • Consider the decision tree of the verifier corresponding to R and let CR encode the accepting paths q1 0 1 q2 q2 0 1 0 q3 q3 q3 q3 1 1 0 0 1 0 0 1 no yes yes no no no no yes (notq1andnotq2andq3) or (not q1andq2andnot q3) or (q1andq2andq3)

  17. Proof • Let k be the number of clauses in CR (we may assume it is the same for each R) • We have p(n)=2hlogn=nh formulas CR, that is, the final formula has c(x)=kp(n) clauses • If x is satisfiable, the verifier accept for any R. Any formula CR is satisfiable: hence, m*(f(x))=c(x) • If x is not satisfiable, less than 1/2 of the formulas CR are satisfiable. Hence, m*(f(x))<c(x)/2+(k-1)p(n)/2=c(x)(1-1/(2k))=c(x)/(1+g) • g depends on the number of queries

  18. MAXIMUM CLIQUE • INSTANCE: Graph G=(V,E) • SOLUTION: A subset U of V such that, for any two vertices u and v in U,(u,v) is in E • MEASURE: Cardinality of U

  19. Product graphs • Given a graph G=(V,E), define G2(V2,E2) asV2={(u,v) : u,vV} andE2={((u,v),(x,y)) : u=x and (v,y) is in E or (u,x) is in E} • Theorem:G has a clique of k nodes iff G2 has a clique of k2 nodes

  20. Self-improvability of clique • Theorem: If MAXIMUM CLIQUE belongs to APX, then MAXIMUM CLIQUE belongs to PTAS • Let A be an r-approximation algorithm for MAXIMUM CLIQUE • Compute G2, then compute U2=A(G2) and, finally, compute the corresponding U in G • From theorem above, m*(G)=sqrt(m*(G2)) and |U|=sqrt(|U2|) • Hence, performance ratio of U is at most sqrt(r) • Iterating we can obtain any performance ratio

  21. The NPO world if PNP NPO MINIMUM TSP APX MINIMUM BIN PACKING MAXIMUM SAT MINIMUM VERTEX COVER( ?) MAXIMUM CUT( ?) PTAS MINIMUM PARTITION PO MINIMUM PATH MINIMUM GRAPH COLORING? Certainly not in PTAS MAXIMUM CLIQUE? Either in PTAS or not in APX

More Related