Hardness of Robust Graph Isomorphism, Lasserre Gaps, and Asymmetry of Random Graphs

Hardness of Robust Graph Isomorphism, Lasserre Gaps,and Asymmetry of Random Graphs

Ryan O’Donnell (CMU)John Wright (CMU)

Chenggang Wu (Tsinghua)Yuan Zhou (CMU)

Hardness of Robust Graph Isomorphism, Lasserre Gaps,and Asymmetry of Random Graphs

Ryan O’Donnell (CMU)John Wright (CMU)

Chenggang Wu (Tsinghua)Yuan Zhou (CMU)

Motivating ExampleYesterday’s

Facebook graphYesterday’s

Facebook graph

Graph Isomorphism algorithm A

Facebook graph

Graph Isomorphism algorithm A should:• output “YES, same graph”• unscramble graph #2

Facebook graphToday’s

Facebook graph

Graph Isomorphism algorithm A will:• output “NO, not isomorphic”• terminate

Facebook graph

But these graphs are almost isomorphic!• can we detect this?• can we unscramble graph #2?

Robust Graph Isomorphism

Given two “almost isomorphic” graphs,find the “best almost-isomorphism” between them

(or something close to it)

G = (V(G), E(G)) H = (V(H), E(H))

Isomorphisms

A bijection π:V(G) → V(H) is an isomorphism if

(u, v) ∈ E(G) ⇔ (π(u), π(v)) ∈ E(H)

G = (V(G), E(G)) H = (V(H), E(H))

Isomorphisms

(u, v) ∈ E(G) ⇔ (π(u), π(v)) ∈ E(H)

G = (V(G), E(G)) H = (V(H), E(H))

Isomorphisms

(u, v) ∈ E(G) ⇔ (π(u), π(v)) ∈ E(H)

G = (V(G), E(G)) H = (V(H), E(H))

Isomorphisms

(u, v) ∈ E(G) ⇔ (π(u), π(v)) ∈ E(H)

G = (V(G), E(G)) H = (V(H), E(H))

Isomorphisms

(u, v) ∈ E(G) ⇔ (π(u), π(v)) ∈ E(H)

G = (V(G), E(G)) H = (V(H), E(H))

Isomorphisms

(u, v) ∈ E(G) ⇔ (π(u), π(v)) ∈ E(H)

G = (V(G), E(G)) H = (V(H), E(H))

Isomorphisms

(u, v) ∈ E(G) ⇔ (π(u), π(v)) ∈ E(H)

G = (V(G), E(G)) H = (V(H), E(H))

Isomorphisms

(u, v) ∈ E(G) ⇔ (π(u), π(v)) ∈ E(H)

G = (V(G), E(G)) H = (V(H), E(H))

Isomorphisms

(u, v) ∈ E(G) ⇔ (π(u), π(v)) ∈ E(H)

Isomorphisms

G = (V(G), E(G)) H = (V(H), E(H))

(u, v) ∈ E(G) ⇔ (π(u), π(v)) ∈ E(H)

Isomorphisms, eq.

Pr[(π(u), π(v)) ∈ E(H)] = 1(u, v) E(G)

G = (V(G), E(G)) H = (V(H), E(H))

~ (assuming |E(G)| = |E(H)|)

Isomorphisms, eq.

Pr[(π(u), π(v)) ∈ E(H)] = 1(u, v) E(G)

G = (V(G), E(G)) H = (V(H), E(H))

(uniformly random)

(assuming |E(G)| = |E(H)|)

Isomorphisms, eq.

Pr[(π(u), π(v)) ∈ E(H)] = 1(u, v) E(G)

G = (V(G), E(G)) H = (V(H), E(H))

Approximate Isomorphisms

Pr[(π(u), π(v)) ∈ E(H)] = α(u, v) E(G)

A bijection π:V(G) → V(H) is an α-isomorphism if

Pr[(π(u), π(v)) ∈ E(H)] = α(u, v) E(G)

Fact:π is an isomorphism ⇔ π is a 1-isomorphism

Pr[(π(u), π(v)) ∈ E(H)] = α(u, v) E(G)

Fact:This π is a ½-isomorphism.

Pr[(π(u), π(v)) ∈ E(H)] = α(u, v) E(G)

G and H are α-isomorphicif they have a β-isomorphism,

and β ≥ α.

Approximate GISO

(c, s)-approximate GISO

Given G and H, output:• YES if G and H are c-isomorphic • NO if G and H are not s-isomorphic

Fact:(1, s)-approximate GISO is no harder than GISO.

Not so clear for (1-ε, s)-approximate GISO…

c > s,c “close to 1”s “far from 1”

Robust GISO

Given G and H which are (1-ε)-isomorphic,output a (1-r(ε))-isomorphism.

(r(ε)→0 as ε→0+)

• Robust algorithms previously studied for CSPs• a characterization conjectured by [Guruswami and Zhou 2011]• confirmed by [Barto and Kozik 2012]

• Robust GISO introduced in [WYZV 2013]• gives a robust GISO algorithm if G and H are trees• which other classes of graphs have robust GISO algorithms?

Approximate GISO, a brief history

• [Arora et al. 2002] give a PTAS for this problem in the case of dense graphs– Our graphs will be sparse, i.e. m = O(n)

• [Arvind et al. 2012] have shown hardness of approximation results for variants of our problem– e.g., GISO with colored graphs

GISO Hardness

• Famously not known to be in P or NP-complete

• Evidence that it’s not NP-complete

• What about robust GISO?

Our result

Assume Feige’s Random 3XOR Hypothesis.Then there is no poly-time algorithm for Robust GISO.

There exists a constant ε0 such that:For all ε > 0, no poly-time algorithm can distinguish between:• (1-ε)-isomorphic graphs G and H• not (1-ε0)-isomorphic graphs G and H

In other words, no poly-time algorithm solves (1-ε, 1-ε0)-approximate GISO.

(constantly far apart)

Our (newer) result

Assume RP ≠ NP.Then there is no poly-time algorithm for Robust GISO.

There exists a constant ε0 such that:For all ε > 0, no poly-time algorithm can distinguish between:• (1-ε)-isomorphic graphs G and H• not (1-ε0)-isomorphic graphs G and H

In other words, no poly-time algorithm solves (1-ε, 1-ε0)-approximate GISO.

Algorithms for GISO

GISO Algorithms

Algorithm Runtime

Brute force

Weisfeiler-Lehman (WL) algorithm

[Babai Luks 83]

O(n!) ≈ O(2n log n)

exp(O(n log n))

exp(O(n log n)1/2)

WLk algorithm

• Standard heuristic for GISO• Larger k, more powerful. Runs in time nk + O(1).• By [Atserias and Maneva 2013], equivalent to

something familiar:

Level-(k+1) Sherali-Adams LP for GISO

Level-k Sherali-Adams LP for GISOWLk-1

Level-(k-1) Sherali-Adams LP for GISO…

WLk/Level-k Sherali-Adams LP

• “Super LP”• Once speculated that WLk solves GISO with

k= O(log n)• Some graphs require k = Ω(n) rounds

[Cai, Fürer, Immerman 1992]

• How do SDPs do? What about the Lasserre/SOS “Super-Duper SDP”?

Cai, Fürer, Immerman Instance

• Some graphs require k = Ω(n) rounds [Cai, Fürer,

Immerman 1992]• Basically encoded a 3XOR instance as a pair of

graphs.• Our main theorem is similar – we reduce from

random 3XOR.• Known that random 3XOR is as hard as

possible for Lasserre SDP [Schoenebeck 2008]

Our result

There exists a constant ε0 such thatΩ(n) levels of the Lasserre/SOS hierarchy

are needed to distinguish:

• YES: G and H are isomorphic • NO: G and H are (1-ε0)-isomorphic

Our proof

Assume Feige’s Random 3XOR Hypothesis.Then there is no algorithm for Robust GISO.

Our proof

• By a reduction from (a variant of) 3XOR

Instance I

x1 + x2 + x3 = 0 (mod 2)

x10 + x15 + x1 = 1 (mod 2)

x4 + x5 + x12 = 1 (mod 2)

x7 + x8 + x9 = 0 (mod 2)

xi ∈ {0, 1}

Our proof

Instance I

x1 + x2 + x3 = 0 (mod 2)

x10 + x15 + x1 = 1 (mod 2)

x4 + x5 + x12 = 1 (mod 2)

x7 + x8 + x9 = 0 (mod 2)

000 010 100 110001 011 101 111

xi ∈ {0, 1}

Our proof

Instance I

x1 + x2 + x3 = 0 (mod 2)

x10 + x15 + x1 = 1 (mod 2)

x4 + x5 + x12 = 1 (mod 2)

x7 + x8 + x9 = 0 (mod 2)

000 010 100 110001 011 101 111

xi ∈ {0, 1}

Our proof

Instance I

x1 + x2 + x3 = 0 (mod 2)

x10 + x15 + x1 = 1 (mod 2)

x4 + x5 + x12 = 1 (mod 2)

x7 + x8 + x9 = 0 (mod 2)

3XOR easy to solve on satisfiable instances (Gaussian elimination)

Thm: [Håstad 2001]

Given a 3XOR instance I, it is NP-hard to distinguish between:

• YES: I is (1 - ε)-satisfiable• NO: I is (½ + ε)-satisfiable

xi ∈ {0, 1}

Our proof

Instance I

x1 + x2 + x3 = 0 (mod 2)

x10 + x15 + x1 = 1 (mod 2)

x4 + x5 + x12 = 1 (mod 2)

x7 + x8 + x9 = 0 (mod 2)

3XOR easy to solve on satisfiable instances (Gaussian elimination)

Thm: [Håstad 2001]

Given a 3XOR instance I, it is NP-hard to distinguish between:

• YES: I is almost-satisfiable• NO: I is far-from-satisfiable

xi ∈ {0, 1}

Our proof

(reduction)

Almost-satisfiable3XOR instance I

Almost-isomorphicgraphs (G, H)

Far-from-satisfiable3XOR instance I

Far-from-isomorphicgraphs (G, H)

Our proof

(reduction)

✔✗

(only works for most far-from-satisfiable 3XOR instances)

Random 3XOR

• n variables• m equations (m = C * n)

Of thepossible sets of size 3, pick m of them

Random 3XOR

x1 + x2 + x3 = ? (mod 2)

Random 3XOR

x1 + x2 + x3 = ? (mod 2)

x3 + x5 + x7 = ? (mod 2)

Random 3XOR

x1 + x2 + x3 = ? (mod 2)

x3 + x5 + x7 = ? (mod 2)

Random 3XOR

x1 + x2 + x3 = ? (mod 2)

x3 + x5 + x7 = ? (mod 2)

x9 + x10 + xn = ? (mod 2)

Random 3XOR

x1 + x2 + x3 = ? (mod 2)

x3 + x5 + x7 = ? (mod 2)

x9 + x10 + xn = ? (mod 2)

Random 3XOR

x1 + x2 + x3 = ? (mod 2)

x3 + x5 + x7 = ? (mod 2)

x9 + x10 + xn = ? (mod 2)

Random 3XOR

x1 + x2 + x3 = 0 (mod 2)

x3 + x5 + x7 = 1 (mod 2)

x9 + x10 + xn = 1 (mod 2)

Random 3XOR

x1 + x2 + x3 = 0 (mod 2)

x3 + x5 + x7 = 1 (mod 2)

x9 + x10 + xn = 1 (mod 2) For some C > 0,~50%-satisfiable whp.

Feige’s R3XOR Hypothesis

No poly-time algorithm can distinguish between:• an almost-satisfiable 3XOR instance• a random 3XOR instance

• Well-believed and “standard” complexity assumption

• Variants of this hypothesis used as basis for cryptosystems, hardness of approximation results, etc.

• Well-known complexity assumption

• Variants of this hypothesis used as basis for cryptosystems, hardness of approximation results, etc.

• Solvable in time 2O(n/log(n)) [Blum, Kalai, Wasserman 2003]

Our proof

• Assume Feige’s R3XOR Hypothesis

(reduction)

Our proof

(reduction)

Random3XOR instance I

Our proof

(reduction)

Far-from-isomorphicgraphs (G, H)(w.h.p.)

The reduction• Assume graph:3XOR instances → graphs

x1 + x2 + x3 = 0 (mod 2)

x2 + x4 + xn = 1 (mod 2)

x1 + x3 + x8 = 1 (mod 2)

3XOR instance I

x1 + x2 + x3 = 0 (mod 2)

x2 + x4 + xn = 1 (mod 2)

x1 + x3 + x8 = 1 (mod 2)

3XOR instance I

x1 + x2 + x3 = 0 (mod 2)

x2 + x4 + xn = 0 (mod 2)

x1 + x3 + x8 = 0 (mod 2)

instance sat(I)

x1 + x2 + x3 = 0 (mod 2)

x2 + x4 + xn = 1 (mod 2)

x1 + x3 + x8 = 1 (mod 2)

3XOR instance I

x1 + x2 + x3 = 0 (mod 2)

x2 + x4 + xn = 0 (mod 2)

x1 + x3 + x8 = 0 (mod 2)

instance sat(I)

Fact:sat(I) is satisfiable (just set xi’s to 0)

x1 + x2 + x3 = 0 (mod 2)

x2 + x4 + xn = 1 (mod 2)

x1 + x3 + x8 = 1 (mod 2)

3XOR instance I

x1 + x2 + x3 = 0 (mod 2)

x2 + x4 + xn = 0 (mod 2)

x1 + x3 + x8 = 0 (mod 2)

instance sat(I)

x1 + x2 + x3 = 0 (mod 2)

x2 + x4 + xn = 1 (mod 2)

x1 + x3 + x8 = 1 (mod 2)

3XOR instance I

x1 + x2 + x3 = 0 (mod 2)

x2 + x4 + xn = 0 (mod 2)

x1 + x3 + x8 = 0 (mod 2)

instance sat(I)

• G := graph(I)• H := graph(sat(I))Output (G, H)

Equation 0-Gadgetx + y + z = 0 (mod 2)

Equation 0-Gadgetx + y + z = 0 (mod 2) 000 010 100 110

001 011 101 111

good assignments

Variable vertices:

000 010 100 110001 011 101 111

good assignments

Variable vertices:

x → 0y → 0z → 0

x → 0y → 1z → 1

x → 1y → 0z → 1

x → 1y → 1z → 0

Equation vertices:

000 010 100 110001 011 101 111

good assignments

Variable vertices:

x → 0y → 0z → 0

x → 0y → 1z → 1

x → 1y → 0z → 1

x → 1y → 1z → 0

Equation vertices:

000 010 100 110001 011 101 111

good assignments

Variable vertices:

x → 0y → 0z → 0

x → 0y → 1z → 1

x → 1y → 0z → 1

x → 1y → 1z → 0

Equation vertices:

000 010 100 110001 011 101 111

good assignments

Variable vertices:

x → 0y → 0z → 0

x → 0y → 1z → 1

x → 1y → 0z → 1

x → 1y → 1z → 0

Equation vertices:

000 010 100 110001 011 101 111

good assignments

Variable vertices:

x → 0y → 0z → 0

x → 0y → 1z → 1

x → 1y → 0z → 1

x → 1y → 1z → 0

Equation vertices:

000 010 100 110001 011 101 111

good assignments

Variable vertices:

x → 0y → 0z → 0

x → 0y → 1z → 1

x → 1y → 0z → 1

x → 1y → 1z → 0

Equation vertices:

000 010 100 110001 011 101 111

good assignments

Variable vertices:

x → 0y → 0z → 0

x → 0y → 1z → 1

x → 1y → 0z → 1

x → 1y → 1z → 0

Equation vertices:

000 010 100 110001 011 101 111

good assignments

Equation 0-gadget, zoomed out

x → 0y → 0z → 0

x → 0y → 1z → 1

x → 1y → 0z → 1

x → 1y → 1z → 0

x + y + z = 0 (mod 2)

x → 0y → 0z → 0

x → 0y → 1z → 1

x → 1y → 0z → 1

x → 1y → 1z → 0

x + y + z = 0 (mod 2)

001 011 101 111

good assignments

001 011 101 111

good assignments

Variable vertices:

001 011 101 111

good assignments

Variable vertices:

x → 0y → 0z → 1

x → 0y → 1z → 0

x → 1y → 0z → 0

x → 1y → 1z → 1

Equation vertices:

001 011 101 111

good assignments

Variable vertices:

x → 0y → 0z → 1

x → 0y → 1z → 0

x → 1y → 0z → 0

x → 1y → 1z → 1

Equation vertices:

x → 0y → 0z → 1

x → 0y → 1z → 0

x → 1y → 0z → 0

x → 1y → 1z → 1

x + y + z = 1 (mod 2)

x → 0y → 0z → 1

x → 0y → 1z → 0

x → 1y → 0z → 0

x → 1y → 1z → 1

x + y + z = 1 (mod 2)

graph: 3XOR instance → Graph

3XOR instance

Eq 1: x1 + x2 + x3 = 0 (mod 2)

Eq 2: x1 + x4 + xn = 1 (mod 2)

Eq m: x7 + x11 + xn = 1 (mod 2)

Eq1 Eq2 Eqm

x2 x3 xn

3XOR instance

Eq 1: x1 + x2 + x3 = 0 (mod 2)

Eq 2: x1 + x4 + xn = 1 (mod 2)

Eq m: x7 + x11 + xn = 1 (mod 2)

Eq1 Eq2 Eqm

x2 x3 xn

3XOR instance

Eq 1: x1 + x2 + x3 = 0 (mod 2)

Eq 2: x1 + x4 + xn = 1 (mod 2)

Eq m: x7 + x11 + xn = 1 (mod 2)

Place the equation gadget over equation and variable blobs

Eq1 Eq2 Eqm

x2 x3 xn

3XOR instance

Eq 1: x1 + x2 + x3 = 0 (mod 2)

Eq 2: x1 + x4 + xn = 1 (mod 2)

Eq m: x7 + x11 + xn = 1 (mod 2)

Eq1 Eq2 Eqm

x2 x3 xn

3XOR instance

Eq 1: x1 + x2 + x3 = 0 (mod 2)

Eq 2: x1 + x4 + xn = 1 (mod 2)

Eq m: x7 + x11 + xn = 1 (mod 2)

Eq1 Eq2 Eqm

x2 x3 xn

3XOR instance

Eq 1: x1 + x2 + x3 = 0 (mod 2)

Eq 2: x1 + x4 + xn = 1 (mod 2)

Eq m: x7 + x11 + xn = 1 (mod 2)

Eq1 Eq2 Eqm

x2 x3 xn

3XOR instance

Eq 1: x1 + x2 + x3 = 0 (mod 2)

Eq 2: x1 + x4 + xn = 1 (mod 2)

Eq m: x7 + x11 + xn = 1 (mod 2)

Eq1 Eq2 Eqm

x2 x3 xn

3XOR instance

Eq 1: x1 + x2 + x3 = 0 (mod 2)

Eq 2: x1 + x4 + xn = 1 (mod 2)

Eq m: x7 + x11 + xn = 1 (mod 2)

Eq1 Eq2 Eqm

x2 x3 xn

3XOR instance

Eq 1: x1 + x2 + x3 = 0 (mod 2)

Eq 2: x1 + x4 + xn = 1 (mod 2)

Eq m: x7 + x11 + xn = 1 (mod 2)

Eq1 Eq2 Eqm

x2 x3 xn

3XOR instance

Eq 1: x1 + x2 + x3 = 0 (mod 2)

Eq 2: x1 + x4 + xn = 1 (mod 2)

Eq m: x7 + x11 + xn = 1 (mod 2)

Eq1 Eq2 Eqm

x2 x3 xn

3XOR instance

Eq 1: x1 + x2 + x3 = 0 (mod 2)

Eq 2: x1 + x4 + xn = 1 (mod 2)

Eq m: x7 + x11 + xn = 1 (mod 2)

Eq1 Eq2 Eqm

x2 x3 xn

3XOR instance

Eq 1: x1 + x2 + x3 = 0 (mod 2)

Eq 2: x1 + x4 + xn = 1 (mod 2)

Eq m: x7 + x11 + xn = 1 (mod 2)

Eq1 Eq2 Eqm

x2 x3 xn

3XOR instance

Eq 1: x1 + x2 + x3 = 0 (mod 2)

Eq 2: x1 + x4 + xn = 1 (mod 2)

Eq m: x7 + x11 + xn = 1 (mod 2)

x1 + x2 + x3 = 0 (mod 2)

x2 + x4 + xn = 1 (mod 2)

x1 + x3 + x8 = 1 (mod 2)

3XOR instance I

x1 + x2 + x3 = 0 (mod 2)

x2 + x4 + xn = 0 (mod 2)

x1 + x3 + x8 = 0 (mod 2)

instance sat(I)

• G := graph(I)• H := graph(sat(I))Output (G, H)

Zoomed-out picture

Eq1 Eq2 Eqm

x2 x3 xn

Eq1 Eq2 Eqm

x2 x3 xn

Need to show

(reduction)

Need to show

(reduction)

Completeness:

Completeness

• Almost-satisfiable 3XOR instance I

Completeness

• Satisfiable 3XOR instance I

Completeness

• Satisfiable 3XOR instance I• Let f:{xi} → {0, 1} be a satisfying assignment

Completeness

• Satisfiable 3XOR instance I• Let f:{xi} → {0, 1} be a satisfying assignment• We’ll use f to construct an isomorphism π

between G and H

Completeness

• Satisfiable 3XOR instance I• Let f:{xi} → {0, 1} be a satisfying assignment• We’ll use f to construct an isomorphism π

between G and H• What should π look like?

Constructing π

Eq1 Eq2 Eqm

x2 x3 xn x1…

Eq1 Eq2 Eqm

x2 x3 xn

f:{xi} → {0, 1} a satisfying assignment

Constructing π

Eq1 Eq2 Eqm

x2 x3 xn x1…

Eq1 Eq2 Eqm

x2 x3 xn

If f(x1) = 0, π maps 0 vertex to 0 and 1

vertex to 1

Constructing π

Eq1 Eq2 Eqm

x2 x3 xn x1…

Eq1 Eq2 Eqm

x2 x3 xn

If f(x2) = 1, π swaps the 0and the 1 vertices

Constructing π

Eq1 Eq2 Eqm

x2 x3 xn x1…

Eq1 Eq2 Eqm

x2 x3 xn

If f(x3) = 1, π swaps the 0and the 1 vertices

Constructing π

Eq1 Eq2 Eqm

x2 x3 xn x1…

Eq1 Eq2 Eqm

x2 x3 xn

Constructing π

Eq1 Eq2 Eqm

x2 x3 xn x1…

Eq1 Eq2 Eqm

x2 x3 xn

?Fact:

• For every good even assignment, the 0-gadget has an isomorphism with the 0-gadget• For every good odd assignment, the 1-gadget has an isomorphism with the 0-gadget

Gadget Isomorphism

(0,0,1) (0,1,0) (1,0,0) (1,1,1) (0,0,0) (0,1,1) (1,0,1) (1,1,0)

000 010 100 110001 011 101 111

good assignments