Independent sets in sparse
graphs
[Bβ15]: Bansal
[BGG]: Bansal, Anupam Gupta, Guru Guruganesh
Independent set Problem
Given graph G find the largest independent set (size = πΌ(G))
Approx. ratio (Alg) = max πΌ(G)/ Alg(G)
πΊ
Notoriously hard: Ξ© π0.999β¦
Best approx:
π/log 3 π
[Feigeβ 04]
Our Focus: max. degree = d
(avg. degree d suffices)
(d+1) approximation trivial
[Hastad 96]
[Greedy β₯
π
]
π+1
Pick each vertex independently with prob. 1/(2d).
Retain v if no conflicts
v
N(v)
Sparse graphs
In general n/d bound is tight
n/d disjoint copies of πΎπ
IP: max
π π₯π
s. t. π₯π + π₯π β€ 1
if i, j β πΈ
π₯π β 0,1
LP relaxation useless: Ξ©(d) integrality gap (each π₯π = 1/2)
First o(d) guarantee: O( d/ log log d)
[Bopanna, Halldorsson β94] [Ajtai, Erdos, Kolmos, Szemerediβ 81]
Current best: d ( log log d ) / log d using SDPs
[Halperinβ 02 , Alon Kahaleβ 96 + Vishwanathan, Halldorssonβ98 ]
Hardness
Ξ©( π/ log 2 π ) assuming UGC
Ξ©(π/ log 4 π) assuming P β NP.
[Austrin-Khot-Safraβ11]
[Chanβ13]
Hardness only for small d
(For d=n: best approx O(n/log 3 π) [Feigeβ 04])
Right answer: π/ log 2 π or π/ log π?
(i) SDP seems to not help beyond d log log d/log d
(ii) Ramsey theoretic barrier to showing > π/ log 2 π hardness
Generic Approach
To get d/k approximation
0
n/d
(say k = log d)
n/k
If OPT β€ n/k (easy, Alg returns greedy n/d)
OPT > n/k must return something non-trivial
(i) Certify πΌ(G) < n/k
(ii) If cannot certify, find non-trivial solution.
n
Certifying πΌ(G) is small
Clique cover: V = π1 βͺ β― βͺ ππ‘
(each ππ a clique)
πΌ πΊ β€ π πΊ
Theta number: πΌ πΊ β€ π πΊ β€ π πΊ
0
n/d
(SDP captures cliques)
n/k
n
If π πΊ > n/k, find non-trivial independent set.
Ramsey theory: No large clique β large independent sets
Ramsey theory (bounded degree graphs)
[Ajtai, Komlos, Szemerediβ 80]: If πΎ3 -free, πΌ πΊ β₯
π
π
log π
(celebrated result; pioneered nibble method)
Tight: Random graph G(n,d/n)
(girth β log π)
Several other proofs known by now.
Johanssonβ96a: π πΊ β€
π
log π
[unpublished; Molloy-Reedβ02]
(several nice ideas)
But πΎ3 is special: Even πΎ4 -free case, much more challenging
Ramsey theory (bounded degree graphs)
πΎπ β free
[Ajtai, Erdos, Komlos, Szemerediβ 82]: πΌ πΊ β₯
π
ππ
π
log log π
[Shearerβ95]: Beautiful entropy method (non-algorithmic)
πΌ πΊ β₯ ππ
ππ β
1
π
π
log π
π log log π
(major question: remove log log d)
(get trivial bound for r > log d / log log d)
[Johanssonβ96b]: Never published, impossible to find, never verified β¦
Much stronger: π πΊ = π
π π log log π
log π
& Algorithmic!
What does Ramsey give us?
Suppose SDP = n/r
(β each vertex contributes 1/r, say πΎπ -free)
Shearer: πΌ πΊ β₯
Integrality gap =
1 π
log π
π π log log π
ππ·π
πΌ(πΊ)
β€ π
log log π
log π
Same as given by Halperin, Alon-Kahale, β¦ (perhaps not so surprising)
Can we combine Halperin + Ramsey?
Our Results
Thm[Bβ15]: π
log 4 π
levels of SA+ hierarchy, integrality gap β
π
log2 π
(Entropy method: does not give a algorithmic result)
Thm[Bβ15]: ππ quasipoly d time, d/log d approx.
(beats Halperin by log log d)
Use O(log d) levels of SA+ to simulate AEKSβ82 + Halperin
Thm[BGG] : Can get β d/ log 2 π approximation in ππ exp(d) time.
(Use Johansson instead of entropy method, but need d levels of SA+)
(Johansson described in BGG)
Our Results
Thm [BGG] : Standard SDP, integrality gap β π/ log 3/2 π
Non-algorithmic. Extend Shearerβs result.
[Shearerβ95]: πΌ πΊ β₯
Thm[BGG]: For any r,
π
log π
π π log log π
πΌ πΊ β₯
π
π
log π/ log π
Eg. can set r = log100 π
Thm [BGG]: LP based β d/log d approximation (SA d-levels)
Rest of the talk
β’
β’
β’
β’
SA+ -> better guarantees
Shearerβs entropy method
Sketch of our πΎπ -free result
Nibble Methods
See main idea behind each
Standard SDP formulation
SDP (vector program): vector π£π for vertex i.
Intended solution: π£π = π£0 if i chosen, 0 otherwise.
Max Ξ£π π£π β
π£π
π£π β
π£π = 0
if π, π β πΈ
π£π β
π£0 = π£π β
π£π
π£0 β
π£0 = 1
π£0
π£0
π£0 β
π£0 1
π£π β
β
π£π β
=
=
2
2
4
4
Exercise: Total contribution of a clique β€ 1
0
π£0
Halperinβs Rounding
Halperin: If SDP value = c n log log d/ log d,
gives independent set of size (n/d) log π 2πβ0.5
Sketch: Consider subspace π£0β₯
Let π€π = Proj π£π on π£0β₯
π€π = π£π β ππ π£0
0
If π£π and π£π orthogonal, then π€π and π€π
are negatively correlated
KMS rounding: Get advantage over random
0
n/d
n/k
n
Worst case: SDP sits here
π£0
k= log d/log log d
π
log2 π
-approx with d levels of SA+
hierarchy
Hierarchies
Automatic way to strengthen LP or SDP relaxation by adding
constraints (Lovasz Schrijver, Sherali Adams, Lasserre, β¦)
k-levels: β ππ variables
max π₯π,1
π₯π,1 + π₯π,1 β€ 1
π₯π,0 + π₯π,1 = 1
n-levels = exact integer program
i, j β πΈ
Distributional view: π₯π,1 = Pr[vertex i is picked] (marginals)
For an edge (i,j), cannot pick simultaneously
All-half solution: For edge (i,j) have (1,0) or (0,1) with prob. 0.5
Hierarchies
Automatic way to strengthen LP or SDP relaxation by adding
constraints (Lovasz Schrijver, Sherali Adams, Lasserre, β¦)
k-levels: β ππ variables
max
n-levels = exact integer program
π₯π,1
π₯π,1 + π₯π,1 β€ 1
π₯π,0 + π₯π,1 = 1
i, j β πΈ
Variables: π₯ π,πΌ
πΌ β 0,1
|π|
Various constraints
For any subset of β€ k vertices S,
1) A distribution over valid 0-1 assignments to S.
2) For S and T, distributions are consistent over π β© π.
S
Strengthen SDP with d-levels of SA
Let π₯π = π£π β
π£π
(contribution of i to objective)
If Val β€ π/log 2 π, done.
Assume each π₯π β₯ 1/(2 log 2 π) (delete i if not)
Claim: For any v, neighborhood N(v) is O(log 3 π) colorable.
Proof: π π£ β€ π, SA solution defines distribution over
independent sets of N(v), s.t. each vertex lies in
β₯ 1/(2 log 2 π) fraction of sets.
Alonβ96: For locally-r colorable graphs
π
log π
πΌ πΊ β₯π
π log π
Make algorithmic via Johansson
v
N(v)
Recall SDP β n/log d
Locally-r-colorable graphs [Alon]
Goal: πΌ πΊ β₯ π
π
π
log π
log π
Proof: Pick set S uniformly from all independent sets.
ππ£ = πΈπ π β© π£
E[S] =
and
ππ£ = πΈπ [ π β© π π£ ]
and
πΈπ β₯
π£ ππ£
Suffices to show : ππ£ +
ππ£
π
β₯ 2π
1
π
log π
log π
1
π
π£ ππ£
for all v.
In fact will show this holds for every conditioning of π\(π£ βͺ π(π£))
ππ£ +
ππ£
π
β₯ 2π
1
π
log π
log π
v
X
Condition on π β© πΊ\v\N π£
Let X be vertices still available in N(v), and x = |X|
N(v)
G\v\N(v)
To extend independent set to G
1)
Either pick v (only)
2)
Pick an independent set in X (and not v)
Let 2ππ₯ be the number of independent sets in X.
1
2ππ₯ +1
+
1
π
π
2ππ₯
2ππ₯ +1
s = avg. size of independent set in X
ππ£ +
ππ£
π
β₯ 2π
1
π
log π
log π
v
X
Condition on π β© πΊ\v\N π£
Let X be vertices still available in N(v), and x = |X|
N(v)
G\v\N(v)
Let 2ππ₯ be the number of independent sets in X.
Fact: 2ππ₯ subsets in a universe of size x have size β₯ π ππ₯/(log 1/π)
1
2ππ₯ +1
+
1
π
ππ₯
2ππ₯
π
log 1/π 2ππ₯ +1
If 2ππ₯ β€ π done.
Otherwise,
ππ₯
log 1/π
β₯
½ log π
log π
(as π β₯
1
)
π
Reducing the number of levels
Donβt really need locally-r-colorable
Suffices if every subset S of N(v)
with π β€ π log 2 π is fractionally r-colorable
For us, π = log 2 π . So π log 4 π levels of SA+ suffice
Gives an existential proof.
Open: Can Johanssonβs work with this weaker condition?
Shearerβs πΎπ -free proof
πΎπ - free graph has an independent set of size β π₯ 1/π
Off-diagonal Ramsey: R(s,t) β π +π‘
π
So β₯ exp(π₯ 1/π ) independent sets in X.
Apply to entropy method (with π π₯ β₯ π₯ 1/π )
v
X
N(v)
Some algebra: πΌ πΊ β₯
1 π
log π
π π log log π
G\v\N(v)
Our idea to extend to higher r: Do a more refined counting of
independent sets.
Puzzle
π
π
For πΎ3 -free graphs πΌ πΊ β₯ log π
Number of independent sets β₯ 2πΌ
Where is the truth?
πΊ
, at most
π
πΌ πΊ
Thought experiment: Sample each vertex with π =
Get Gβ: πβ β ππ
So πΌ πΊβ β₯
πβ β ππ
log πβ²
πβ
πβ²
so
πβ²
πβ²
β
π
π
β π
π
log π
π
1
.
π
= (n/d) (½ log d)
Key: Now, G must contain many such sets
G
Gβ
A word about nibble methods
Nibble Methods (Johansson)
Coloring with s << d colors.
Each vertex maintains list of available colors.
Initially L(v) = {1,β¦,s}
Each round:
Some vertices activate, pick a random color from list.
Neighbors update their lists.
Thought experiment: If s = 2 d / log d
Each neighbor of v picks independently
Pr [color i left for v] = 1 β
Enough colors will be left for v.
1 π
π
=
1
π
v
N(v)
Nibble Methods (Johansson)
Coloring with s << d colors.
Each vertex maintains list of available colors.
Initially L(v) = {1,β¦,s}
Each round:
Some vertices activate, pick a random color from list.
Neighbors update their lists.
Kimβ95: Girth 5, lists of neighbors of v β independent
Triangle-free: Not true (new ideas: entropy, energy)
Johanssonβ96a: Never published; Molloy Reed book
πΎπ free, locally r-colorable
(new ideas: modifies the process) Johanssonβ96b
Open questions
Is SDP integrality gap d/ log 2 π ?
Best upper bound on πΌ πΊ for πΎπ -free graphs: πΌ πΊ β€
π
π
log π
log π
Matching tight (lower) bound will imply the result
Obtain an algorithmic π /log 3/2 π approximation for SDPs
(extend Johanssonβs approach)?
Using SA+, d/log 2 π approximation in ππ quasi-poly(d) time.
Questions!
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