the strong exponential time hypothesis

The Strong Exponential Time Hypothesis

Daniel Lokshtnaov

Tight lower bounds

Have seen that ETH can give tight lower bounds

How tight? ETH «ignores» constants in exponent

How to distinguish 1.85n from 1.0001n?

SAT

Input: Formula with m clauses over n boolean variables.Question: Does there exist an assignment to the variables that satisfies all clauses?

Note: Input can have size superpolynomial in n!

Fastest algorithm for SAT: 2npoly(m)

d-SAT

Here all clauses have size dInput size nd

Fastest algorithm for 2-SAT: n+mFastest algorithm for 3-SAT: 1.31n

Fastest algorithm for 4-SAT: 1.47n

…Fastest algorithm for d-SAT:Fastest algorithm for SAT:

Strong ETH

Let d-SAT has a algorithm

Know: 0 Sd 1

ETH: s3 0 SETH: 1

Let

Showing Lower Bounds under SETH

Your ProblemToo fast algorithm?

d-SAT1.99999𝑛

The number of 9’s MUST be independent of d

Dominating Set

Input: n vertices, integer kQuestion: Is there a set S of at most k vertices

such that N[S] = V(G)?

Naive: nk+1

Smarter: nk+o(1)

Assuming ETH: no f(k)no(k)

nk/10?

nk-1?

SAT k-Dominating SetVariables

SAT-formula

k groups, each onn/k variables.

One vertex for each of the 2n/k assignments to the variablesin the group.

x yx y x y x y x y

Variables

SAT-formula


Selecting one vertex from eachcloud corrsponds to selectingan assignment to the variables.

Cliques

x yx y x y x y x y

Variables

SAT-formula


One vertex per clause in the formula

Edge if the partial assignmentsatisfies the clause

SAT k-Dominating Setanalysis

Too fast algorithm for k-Dominating Set: nk-0.01

For any fixed k (like k=3)

The output graph has k2n/k + m 2k 2n/k vertices

If m 2n/k then 2n is at most mk, which is polynomial!

So m

≤(2𝑘)𝑘⋅ 2𝑛 𝑘−0.01

𝑘

¿𝑂 (1.999𝑛)

Dominating Set, wrapping up

A O(n2.99) algorithm for 3-Dominating Set, or a O(n3.99) algorithm for 4-Dominating Set, or aa O(n4.99) algorithm for 5-Dominating Set, or a …… would violate SETH.

Independent Set / Treewidth

Input: Graph G, integer k, tree-decomposition of G of width t.Question: Does G have an independent set of size at least k?

DP: O(2tn) time algorithmCan we do it in 1.99t poly(n) time?

Next: If yes, then SETH fails!

Independent Set / Treewidth

Will reduce n-variable d-SAT to Independent Set in graphs of treewidth t, where t n+d.

So a 1.99tpoly(N) algorithm for Independent Set gives a 1.99n+dpoly(n) O(1.999n) time algorithm for d-SAT.

Independent Sets on an Even Path

t f t f t f t f t f

True

False

In independent set:

Not in solution:

first

Truethen

False

d-SAT Independent Setproof by example

= (a b c) (a c d) (b c d)

t f t f t f

t f t f t f

t f t f t f

t f t f t f

a

b

c

d

a c

b

a d

c

b d

c

Independent Sets Assignments

= (a b c) (a c d) (b c d)

t f t f t f

t f t f t f

t f t f t f

t f t f t f

a

b

c

d

a c

b

a d

c

b d

c

True

True

False

False

But what about the first true then false independent sets?

Dealing with truefalse

a

bc

d

Clausegadgets

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

Every variable flips truefalse at most once!

Treewidth Boundby picture

t f t f t f

t f t f t f

t f t f t f

t f t f t f

a

b

c

d

a c

b

a d

c

b d

c

………

…

n

d

Formal proof - exercise

Independent Set / Treewidthwrap up

Reduced n-variable d-SAT to Independent Set in graphs of treewidth t, where t n+d.

A 1.99tpoly(N) algorithm for Independent Set gives a 1.99n+dpoly(n) O(1.999n) time algorithm for d-SAT.

Thus, no 1.99t algorithm for

Independent Set assuming SETH

Assuming SETH, the following algorithms are optimal:– 2t poly(n) for Independent Set– 3t poly(n) for Dominating Set– ct poly(n) for c-Coloring– 3t poly(n) for Odd Cycle Transversal– 2t poly(n) for Partition Into Triangles– 2t poly(n) for Max Cut– …

3t lower bound for Dominating Set?

Need to reduce k-SAT formulas on n-variables to Dominating Set in graphs of treewidth t, where

3𝑡≈2𝑛

So

Hitting Set / n

Input: Family F = {S1,…,Sm} of sets over universe U = {v1, …, vn}, integer k.

Question: Does there exist a set X U of size at most k such that for every Si F, Si X ?

Naive algorithm runs in time.

Next: implies that SETH fails

d-SAT Hitting Set

a

a

= (a b c) (a c d) (b c d)

b

b

c

c

d

dBudget = 4

d-SAT vs Hitting Set

A cn algorithm for Hitting Set makes a c2n algorithm for d-SAT.

Since 1.412n < 1.9999n, a 1.41n algorithm for Hitting Set violates the SETH.

Have a 2n algorithm and a 1.41n lower bound.

Next: 2n lower bound

Hitting Set

For any fixed , will reduce k-SAT with n variables to Hitting Set with universe with at most (1+)n elements.

So a 1.99n algorithm for Hitting Set gives a 1.99n(1+) 1.999n time algorithm for k-SAT

Some deep math

For every there exists a natural number g such that, for t = we have:

( 𝑡

⌈ 𝑡2⌉ )≥2𝑔

Why is this relevant?

d-SAT Hitting SetGroup the variables into groups of size g, and set t =.

Variables: g g g g g

tt t t t

Elements (1 + ) variables

Elements:

Solution budget from each group

Will force from each group

Exactly from each group

Analyzing a group

Group of g variables

Group of t elements

2g assignments to variables

subsets of elements of sizeexactly .

Injection

Forcing solution vertices in a group?

Add all subsets of the group of size to the family F.

Any set that picks less than elements the group misses a guard.

Any set that picks at least elements from each group hits all the guards

Lets call these sets guards

Analyzing a groupGroup of g variables

Group of t elements

assignments to variables

subsets of elements of size exactly .

Injection

What about the element subsets of size thatdo not correspond to assignments?

Sets of size

Adding a set of size to the family F ensures that the «group complement» set is not picked.

All other sets of size in the group may still be picked in solution.

Forbid sets of size that do not correspond to assignments.

d-SAT Hitting Set

Variables: g g g g g

tt t t t

Elements:

potentialsolutions

assignments

Want:Solutions Satisfying assignments

Forbidding partial assignments

Pick any d groups of variables, and consider some assignment to these variables.

If this assignment falsifies we want to forbid the corresponding set in the Hitting Set instance from being selected.


Variables …

…

Bad assignment

Set added to F to forbid the bad assignment


For each bad assignment to at most d groups, forbid it by adding a «bad assignment guard»

This adds O(nd2gd) = O(nd) sets to F.

Satisfying Assignments Hitting Sets

A satisfying assignment has no bad sub-assignments corresponds to a hitting set.

A hitting set corresponds to an assignment.

If this assignment falsified a clause C, the assignment would be bad for the d groups C lives in, and miss a bad assignment guard.

Hitting Set wrap up

Can reduce n variable d-SAT to n(1+) element Hitting Set.

So a cn algorithm for Hitting Set yields a (c+)n algorithm for d-SAT.

A 1.99n algorithm for Hitting Set would violate SETH.

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.

Important Open Problems

Can we show a 2n lower bound for Set Cover assuming SETH?

Can we show a 1.00001 lower bound for 3-SAT assuming SETH?

Exercises

Show that SETH implies ETH (Hint: use the sparsification lemma)

Exercise 4.16, 4.17, 4.18, 4.19.

Postdoc in Bergen?Much better than Budapest. Really.

Thank You

the strong exponential time hypothesis

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