Almost-polynomial Ratio ETH-Hardness of ApproxIMATIng Densest k-subgraph. Pasin Manurangsi UC Berkeley

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1 Almost-polynomial Ratio ETH-Hardness of ApproxIMATIng Densest k-subgraph Pasin Manurangsi UC Berkeley

2 Densest k-subgraph (DkS) G S Input An undirected graph G = (V, E) A positive integer k Goal Find a subset of vertices S V of size k that maximizes the number of edges within S

3 Densest k-subgraph: What is Known? Hardness Results [Feige02] Random 3SAT Hypothesis Some constant [Khot04] NP ԑ>0 ځ BPTIME(2 nԑ ) Some constant [Raghavendra-Steurer10] Small Set Expansion Hypothesis Every constant [Alon-Arora-Manokaran-Moshkovitz-Weinstein11] Planted Cliue Hypothesis [Braverman-Ko-Rubinstein-Weinstein17] ETH Some constant Conjecture [BCCFV10, AAMMW11, ] NP-hard n c for some c > 0 2 Ω(log2/3 n) Approximation Algorithms (Note: n = V ) [Kortsarz-Peleg93] O(n ) [Feige-Kortsarz-Peleg01] O(n ) [Bhaskara-Charikar-Chlamtac-Feige-Vijayaraghavan10] O(n 0.25+ԑ )

4 Densest k-subgraph: Our Results Hardness Results [Feige02] Random 3SAT Hypothesis Some constant [Khot04] NP ԑ>0 ځ BPTIME(2 nԑ ) Some constant [Raghavendra-Steurer10] Small Set Expansion Hypothesis Every constant [Alon-Arora-Manokaran-Moshkovitz-Weinstein11] Planted Cliue Hypothesis [Braverman-Ko-Rubinstein-Weinstein17] ETH Some constant Conjecture [BCCFV10, AAMMW11, ] NP-hard n c for some c > 0 ETH Gap-ETH OUR Results n 1/polyloglog n n o(1) 2 Ω(log2/3 n)

5 Densest k-subgraph: Our Results Holds even for perfect completeness Theorem Assuming ETH, no polynomial time algorithm can, on every input graph G that contains k-cliue, find a n -1/polyloglog n -dense k-subgraph. Note: Den(S) = E(S) / k 2 This regime is easy! [Feige-Seltser97] There is an (1 + ԑ)-approximation algorithm with running time no(log n/ԑ) There is an n ԑ -approximation algorithm with running time n O(1/ԑ) ETH Gap-ETH OUR Results n 1/polyloglog n n o(1) Theorem Assuming Gap- ETH, no n O (1/ԑ 1/3 ) -time algorithm can, on every G that contains k-cliue, find a n -ԑ -dense k-subgraph.

6 The Hypotheses: ETH & Gap-ETH Exponential Time Hypothesis (ETH) [Impagliazzo-Paturi99] No 2 o(n) -time algorithm can solve 3SAT. (n is the number of variables.) Gap Exponential Time Hypothesis (Gap-ETH) [Dinur16, M-Raghavendra16] No 2 o(n) -time algorithm can distinguish between (i) a satisfiable 3CNF formula, and, (ii) a 3CNF formula which is not even 0.99-satisfiable.

7 ETH/Gap-ETH & Hardness of Approximation The Birthday Repetition Framework [Aaronson-Impagliazzo-Moshkovitz14] 3SAT Instance Φ Size: n YES: val(φ) = 1 NO: val(φ) < 1 Lower Bound: 2 Ω(n) PCP Thm [Dinur07] 3SAT Instance Φ Size: n = n polylog n YES: val(φ ) = 1 NO: val(φ ) < 0.99 Lower Bound: 2Ω(n /polylog n ) DKS Instance (G, k) Size: N = 2 n /polylog n = 2 o(n) YES: k-cliue NO: Every k-subgraph has density < N -1/polyloglog N Lower Bound: N ω(1) Used in other works before and since: Not necessary, assuming Gap-ETH Densest k-subgraph [BKRW17] ԑ-best ԑ-nash [Braverman-Ko-Weinstein15] Community Detection [Rubinstein17], etc.

8 Parameter: (1 n) 3SAT Φ Size: n YES: val(φ) = 1 NO: val(φ) < 0.99 (x 1 x 3 x 5 ) (x 1 x 2 x n ) The Reduction DKS Instance (G = (V, E), k) Size: N = n 2 Just pick all partial assignments YES: k-cliue to a satisfying assignment! NO: Every k-subgraph has density < exp(-ω( 4 /n 3 )) {x 1 0, x 2 0,, x 0} inconsistent Representative Case: = n / polylog n N = exp(n/polylog n) Gap: exp(n/polylog n) = N 1/polyloglog N V = {all partial assignments to variables} E = {(u, v) that are consistent and do not violate any clauses} {x 1 0, x 2 0,, x 1} {x n +1 1, x n +2 1,, x n 0} violate a clause k = n inconsistent {x n +1 1, x n +2 1,, x n 1}

9 Soundness Proof Goal val(φ) < 0.99 every k-subgraph of G has density < exp(-ω( 4 /n 3 )) A Less Ambitious Goal val(φ) < 0.99 G does not contain a bicliue of size larger than k exp(-ω( 4 /n 3 ))

10 Soundness Proof Goal val(φ) < 0.99 every k-subgraph of G has density < exp(-ω( 4 /n 3 )) A Less Ambitious Goal val(φ) < 0.99 G does not contain a bicliue of size larger than k exp(-ω( 2 /n)) An Even Less Ambitious Goal G does not contain a bicliue of size larger than k

11 Soundness Proof (Cont.) Goal: G does not contain a bicliue of size larger than n Immediate from the observations! {x 1 0, x 2 0,, x 0} {x n +1 1, x n +2 1,, x n 0} L {x 1 0, x 2 1,, x 0} {x n +1 1, x n +2 1,, x n 1} R Flattening A(L) = {x 1 0, x 2 0, x 2 1, } A(R) = {, x n 0, x n 1} Observation 1 L A(L) R A(R) and x i 0, x i 1 - x i 0 x i 0 x i Observation 2 A(L) + A(R) 2n

12 Soundness Proof: the good, the bad and the Ugly Goal: val(φ) < 0.99 G does not contain a bicliue of size larger than exp(-ω( 2 /n)) n {x 1 0, x 2 0,, x 0} {x n +1 1, x n +2 1,, x n 0} L {x 1 0, x 2 1,, x 0} {x n +1 1, x n +2 1,, x n 1} R Flattening A(L) = {x 1 0, x 2 0, x 2 1, } A(R) = {, x n 0, x n 1} Observation 1 L A(L) R A(R) and x i 0, x i 1 - x i 0 x i 0 x i Observation 2 A(L) + A(R) 2n bad variable good variable ugly variable

13 Soundness ProoF: The Ugly Goal: val(φ) < 0.99 G does not contain a bicliue of size larger than exp(-ω( 2 /n)) n Extreme Case 1 every variable is ugly A(L) + A(R) n Observation 1 L A(L) and R A(R) A(L) x i 0, x i 1 - x i 0 x i 0 x i Observation 2 A(L) + A(R) 2n A(R) bad variable good variable ugly variable

14 Soundness Proof: The Bad Goal: val(φ) < 0.99 G does not contain a bicliue of size larger than exp(-ω( 2 /n)) n Extreme Case 2 every variable is bad Not valid vertices! Claim Most elements of A(L) contains both x i 0 and x i 1 for some i! Proof of Claim Birthday Paradox In expectation, a random element of A(L) contains Ω( 2 /n) such i s! Only exp(-ω( 2 /n)) fraction of A(L) are actual vertices Observation 1 L A(L) and R A(R) A(L) x i 0, x i 1 - x i 0 x i 0 Observation 2 A(L) + A(R) 2n A(R) bad variable good variable

15 Soundness Proof: The Good Goal: val(φ) < 0.99 G does not contain a bicliue of size larger than exp(-ω( 2 /n)) Extreme Case 3 every variable is good good variable A(L) x i 0 A(R) x i 0 A(L) = A(R) is an actual assignment! 0.01m clauses unsatisfied Suppose a clause (x i x j x k ) is unsatisfied. Claim: x i, x j cannot appear together in any vertex in L L {, x i 0, x j 1, } (x i, x j ) is a prohibited pair that can t appear together in any vertex of L violate clause There are at least 0.01m prohibited pairs {, x k 0, } n R Birthday Paradox Only exp(-ω( 2 /n)) fraction of A(L) can be vertices of L

16 Soundness Proof: From Bicliue to Dense Subgraphs What we have proved: val(φ) < 0.99 G does not contain a bicliue of size n Why Bicliue? Different from cliue: t-cliue free graph can be (1-1/(t - 1))- dense! exp(-ω(l2 /n)) Theorem [Kővári-Sós-Turán54] Every t-bicliue free graph on N vertices is O(N -1/t )-dense. Need t log N to even get constant bound Theorem [Alon02] Every graph on N vertices that contains < ε t2 N 2t copies of t-bicliue is O(ԑ)-dense. Only need to prove that G does not contain many bicliues.? What we actually want: val(φ) < 0.99 every n -subgraph of G has density < exp(-ω( 4 /n 3 )) Aviad Rubinstein told me to do so Our bound is not (nearly) enough

17 Soundness Proof: Bounding Number of Bicliues What we actually showed: For each flattening (A(L), A(R)), L is a subset of a small subset of R is a subset of a small subset of A(L) A(R) and Fix (A(L), A(R)). Only a small number of choices for each vertex in L. Same for R. Bounding number of t-bicliues Sum over all choices of (A(L), A(R)) With appropriate t, Alon s Theorem gives the desired bound for density of n -subgraph!

18 A different Perspective: Parameterized Inapprox. of Bicliue Theorem [Kővári-Sós-Turán54] Every t-bicliue free graph on n vertices is O(n -1/t )-dense. Theorem [Alon02] Every graph on N vertices that contains < ε 2t2 N 2t copies of t-bicliue is O(ԑ)-dense. 3SAT Instance Φ Size: n YES: val(φ ) = 1 NO: val(φ ) < 0.99 Subsample each vertex w.p. p = 0. 1/(ε t N) t-bicliue free graph on O(1/ε t ) vertices OUR Reduction DKS Instance (G, k) Size: N = 2 o(n) YES: k-cliue NO: G contains k 2t copies of t-bicliue for t k Subsample each vertex w.p. p Apply KST Theorem The graph must be O(ε)-dense Parameterized Hardness of Approximating Bicliue! DKS Instance (G, pk) Size: N = 2 o(n) YES: (pk)-cliue NO: G is t-bicliue free (where t pk)

19 Conclusion Sub-exponential time reduction from gap 3SAT to Densest k-subgraph Assuming ETH, gives n 1/polyloglog n -ratio inapproximability for DkS Assuming Gap-ETH, gives n o(1) -ratio inapproximability for DkS Assuming Gap-ETH, gives parameterized inapproximability for bicliue Open Questions NP-hardness of approximation of Densest k-subgraph? Polynomial ratio hardness of approximation of Densest k-subgraph (assuming ETH/Gap-ETH)? THANK YOU! Questions?