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The Strong Exponential Time Hypothesis Daniel Lokshtnaov

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Tight lower bounds Have seen that ETH can give tight lower bounds How tight? ETH «ignores» constants in exponent How to distinguish 1.85 n from 1.0001 n ?

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SAT Fastest algorithm for SAT: 2 n poly(m)

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d-SAT

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Strong ETH

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Showing Lower Bounds under SETH Your Problem Too fast algorithm? d-SAT The number of 9’s MUST be independent of d

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Dominating Set Input: n vertices, integer k Question: Is there a set S of at most k vertices such that N[S] = V(G)? Naive: n k+1 Smarter: n k+o(1) Assuming ETH: no f(k)n o(k) n k/10 ? n k-1 ?

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SAT k-Dominating Set Variables SAT-formula k groups, each on n/k variables. One vertex for each of the 2 n/k assignments to the variables in the group.

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xyxyxyxyxy Variables SAT-formula k groups, each on n/k variables. Selecting one vertex from each cloud corrsponds to selecting an assignment to the variables. Cliques

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xyxyxyxyxy Variables SAT-formula k groups, each on n/k variables. One vertex per clause in the formula Edge if the partial assignment satisfies the clause

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SAT k-Dominating Set analysis Too fast algorithm for k-Dominating Set: n k-0.01 For any fixed k (like k=3)

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Dominating Set, wrapping up A O(n 2.99 ) algorithm for 3-Dominating Set, or a O(n 3.99 ) algorithm for 4-Dominating Set, or a a O(n 4.99 ) algorithm for 5-Dominating Set, or a … … would violate SETH.

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Independent Set / Treewidth DP: O(2 t n) time algorithm Can we do it in 1.99 t poly(n) time? Next: If yes, then SETH fails!

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Independent Set / Treewidth

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Independent Sets on an Even Path tftftftftf True False In independent set: Not in solution: first True then False

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tftftf tftftf tftftf tftftf a b c d ac b ad c bd c

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tftftf tftftf tftftf tftftf a b c d ac b ad c bd c True False But what about the first true then false independent sets?

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Dealing with true false a b c d Clause gadgets 123123123123123 Every variable flips true false at most once!

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Treewidth Bound by picture tftftf tftftf tftftf tftftf a b c d ac b ad c bd c … … … … n d Formal proof - exercise

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Independent Set / Treewidth wrap up Thus, no 1.99 t algorithm for Independent Set assuming SETH

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3 t lower bound for Dominating Set? Need to reduce k-SAT formulas on n-variables to Dominating Set in graphs of treewidth t, where

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Hitting Set / n

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Budget = 4

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d-SAT vs Hitting Set A c n algorithm for Hitting Set makes a c 2n algorithm for d-SAT. Since 1.41 2n < 1.9999 n, a 1.41 n algorithm for Hitting Set violates the SETH. Have a 2 n algorithm and a 1.41 n lower bound. Next: 2 n lower bound

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Hitting Set

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Some deep math Why is this relevant?

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Variables: g g gg g t t t t t Elements:

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Analyzing a group Group of g variables Group of t elements 2 g assignments to variables Injection

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Lets call these sets guards

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Analyzing a group Group of g variables Group of t elements assignments to variables Injection

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Variables: g g gg g t t t t t Elements: potential solutions assignments

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Forbidding partial assignments

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Variables … … Bad assignment Set added to F to forbid the bad assignment

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Forbidding partial assignments For each bad assignment to at most d groups, forbid it by adding a «bad assignment guard» This adds O(n d 2 gd ) = O(n d ) sets to F.

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Hitting Set wrap up

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Conclusions SETH can be used to give very tight running time bounds. SETH recently has been used to give lower bounds for polynomial time solvable problems, and for running time of approximation algorithms.

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Important Open Problems Can we show a 2 n lower bound for Set Cover assuming SETH? Can we show a 1.00001 lower bound for 3-SAT assuming SETH?

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Exercises Show that SETH implies ETH (Hint: use the sparsification lemma) Exercise 4.16, 4.17, 4.18, 4.19.

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Postdoc in Bergen? Much better than Budapest. Really.

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Minicourse on parameterized algorithms and complexity Part 8: The Strong Exponential-Time Hypothesis Dániel Marx (slides by Daniel Lokshtanov) Jagiellonian.

Minicourse on parameterized algorithms and complexity Part 8: The Strong Exponential-Time Hypothesis Dániel Marx (slides by Daniel Lokshtanov) Jagiellonian.

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