Graph Colouring is Hard for Algorithms Based on Hilberts - - PowerPoint PPT Presentation

Graph Colouring is Hard for Algorithms Based on Hilbert’s Nullstellensatz and Gröbner Bases Massimo Lauria Sapienza - Università di Roma CCC 2017 — Riga

joint work with: Jakob Nordström (KTH, Stockholm)

k-coloring of a graph G

Colors = { } Assign a color to each vertex… …while avoiding monochromatic edges

Proofs of non-k-colorability, in Polynomial Calculus

Proof lines: polynomial eq. over 𝔾

Proofs of non-k-colorability, in Polynomial Calculus

Axiom: vertex v gets color {1,…,k}

xv,1 + xv,2 + . . . + xv,k = 1 xv,jxv,j0 = 0

for j≠j’ ∈ [k] Proof lines: polynomial eq. over 𝔾

Proofs of non-k-colorability, in Polynomial Calculus

Axiom: two colors on {u,v} ∈ E(G)

xu,jxv,j = 0

for j ∈ [k] Axiom: vertex v gets color {1,…,k}

xv,1 + xv,2 + . . . + xv,k = 1 xv,jxv,j0 = 0

for j≠j’ ∈ [k] Proof lines: polynomial eq. over 𝔾

Proofs of non-k-colorability, in Polynomial Calculus

Axiom: two colors on {u,v} ∈ E(G)

xu,jxv,j = 0

for j ∈ [k] Axiom: vertex v gets color {1,…,k}

xv,1 + xv,2 + . . . + xv,k = 1 xv,jxv,j0 = 0

for j≠j’ ∈ [k] Proof lines: polynomial eq. over 𝔾

p = 0 q = 0 αp + βq = 0 α, β ∈ F

Inference rules

p = 0 xv,j · p = 0

Proofs of non-k-colorability, in Polynomial Calculus

Axiom: two colors on {u,v} ∈ E(G)

xu,jxv,j = 0

for j ∈ [k] Axiom: vertex v gets color {1,…,k}

xv,1 + xv,2 + . . . + xv,k = 1 xv,jxv,j0 = 0

for j≠j’ ∈ [k]

. . . 1 = 0

Refutation Proof lines: polynomial eq. over 𝔾

p = 0 q = 0 αp + βq = 0 α, β ∈ F

Inference rules

p = 0 xv,j · p = 0

Complexity of proofs of non-k-colorability

Proof Size = cumulative #monomials in proof lines Degree = largest degree among the monomials in the proof

Complexity of proofs of non-k-colorability

Proof Size = cumulative #monomials in proof lines Degree = largest degree among the monomials in the proof

Proof Size ≤ nO(Degree) Search-Time ≤ nO(Degree)

Easy upper bounds where n is the number of variables

Complexity of proofs of non-k-colorability

Proof Size = cumulative #monomials in proof lines Degree = largest degree among the monomials in the proof

Proof Size ≤ nO(Degree) Search-Time ≤ nO(Degree)

Easy upper bounds where n is the number of variables

[IPS’99]. Degree Ω(n) implies Proof Size 2Ω(n)

Our results

Main Theorem. Fix k>0. We describe graphs {Gn}n such

• Gn has Θ(k4n) vertices of degree O(k2)
• Gn is non-k-colorable
• PC proofs of non-k-colorability require degree Ω(n)

Corollary [IPS’99]. PC proofs of non-k-colorability require size 2Ω(n).

Proof theoretic analysis of k-coloring algorithms

• Their trace is a verifiable proof of non-k-colorability
• If NP ≠ coNP then non-k-colorable G with no short proof

Proof theoretic analysis of k-coloring algorithms

• Their trace is a verifiable proof of non-k-colorability
• If NP ≠ coNP then non-k-colorable G with no short proof
• Algebraic algorithms determines (non-)k-colorability
• A. Nullstellensatz certificate; or
• B. a Gröbner basis computation;

Proof theoretic analysis of k-coloring algorithms

• Their trace is a verifiable proof of non-k-colorability
• If NP ≠ coNP then non-k-colorable G with no short proof
• Algebraic algorithms determines (non-)k-colorability
• A. Nullstellensatz certificate; or
• B. a Gröbner basis computation;
• Algebraic algorithms produce polynomial calculus proofs thus

we unconditional exponential lower bounds.

Proof theoretic analysis of k-coloring algorithms

• Their trace is a verifiable proof of non-k-colorability
• If NP ≠ coNP then non-k-colorable G with no short proof
• Algebraic algorithms determines (non-)k-colorability
• A. Nullstellensatz certificate; or
• B. a Gröbner basis computation;
• Algebraic algorithms produce polynomial calculus proofs thus

we unconditional exponential lower bounds.

• Known for Resolution based algorithms in [Beame et al.’05]

De Loera et al. [2008, 2009, 2011, 2015]

Algebraic algorithm to decide (non-)k-colorability

• uses a different encoding for the polynomial equations
• finds a special form of PC proof of degree d in time nΘ(d)

SUCCESS: non-k-colorable FAILURE: undetermined (maybe d too small?)

De Loera et al. [2008, 2009, 2011, 2015]

Open Problem [De Loera et al ’09, Li et al. ’16]. Is there, for every integer d>0, a graph that requires degree >d?

Our Answer. Gn require degree Ω(n) Algebraic algorithm to decide (non-)k-colorability

• uses a different encoding for the polynomial equations
• finds a special form of PC proof of degree d in time nΘ(d)

SUCCESS: non-k-colorable FAILURE: undetermined (maybe d too small?)

Outline

• i. Construction of the hard graphs
• ii. Sketch of the lower bound
• iii. Conclusions

i.

Construction of the hard graphs

Reduction from FPHP(Bn) — part 1

pi,h1 + pi,h2 + . . . + pi,hk = 1

for j≠j’ ∈ [k]

pi,hj · pi,hj0 = 0

for {i1,h},{i2,h} ∈ E(Bn)

pi1,h · pi2,h = 0

Pigeon i picks one hole in {h1, …, hk} No two pigeons in same hole

Bipartite Bn

[n]

4 5 6 11 7 2 3 10 9 1 8 5 10 4 1 2 3 9 8 7 6

[n-1] Pigeons Holes deg≤k deg≤O(k)

Reduction from FPHP(Bn) — part 2

Theorem [MN’15]. If {Bn}n as described here is a family of boundary expanders(*), then any polynomial calculus refutations of FPHP(Bn) requires degree Ω(n).

(*) boundary expander: for some constant 𝛽>0 and c>0 every set S

• f at most 𝛽n left-vertices has at least c|S| right-vertices that have

exactly one neighbor in S each.

Reduction from FPHP(Bn) — part 3

We build {Gn}n such that k-colorability formula for Gn is an obfuscated version of formula FPHP(Bn) We start with a family {Bn}n

• left-degree ≤ k
• right-degree O(k)
• boundary expanders, i.e. FPHP(Bn) is hard

Construction of Gn — mapping (e.g. k=4)

pi,h1 + pi,h2 + . . . + pi,hk = 1

for j≠j’ ∈ [k]

pi,hj · pi,hj0 = 0

Pigeon i picks one hole in {h1, …, hk}

h1 i h4 h3 h2

Construction of Gn — mapping (e.g. k=4)

pi,h1 + pi,h2 + . . . + pi,hk = 1

for j≠j’ ∈ [k]

pi,hj · pi,hj0 = 0

Pigeon i picks one hole in {h1, …, hk} Vertex i gets one color in {1,…,k} for j≠j’ ∈ [k]

xi,j · xi,j0 = 0 xi,1 + xi,2 + . . . + xi,k = 1

h1 i h4 h3 h2

Special vertex i

i xi,1 xi,1 xi,2 xi,2 xi,3 xi,3 xi,4 xi,4

Construction of Gn — conflicts (e.g. k=4)

for {i1,h},{i2,h} ∈ E(Bn)

pi1,h · pi2,h = 0

No two pigeons in same hole i1 i2 h pi1,h pi1,h pi2,h pi2,h

Construction of Gn — conflicts (e.g. k=4)

for {i1,h},{i2,h} ∈ E(Bn)

pi1,h · pi2,h = 0

No two pigeons in same hole We add a gadget to encode: i1 i2

i1 i2

i1 i2 h pi1,h pi1,h pi2,h pi2,h

(E.g. k=4)

i1 i2

Forbid red to vertex i1 and green to vertex i2

i1 i2 Pre-colored red Pre-colored green

(E.g. k=4)

i1 i2

Forbid red to vertex i1 and green to vertex i2

i1 i2 Pre-colored red Pre-colored green The special vertices are the only one shared among gadgets

Construction of Gn — precolored vertices (e.g. k=4)

v2 v1 v3 v4 v5 v6 v7 v8 v9 v10

v1+4m v2+4m v3+4m v4+4m

A long enough chain allows to identify at most one pre-colored vertex to a vertex in the chain. (To keep Gn sparse)

ii.

Sketch of the lower bound

Proof by reduction from FPHP(Bn)

Proof of degree d k−color(Gn)[xv,j] k−color(Gn)[xv,j] 1 = 0 1 = 0

• Thm. k-color(Gn) requires degree Ω(n)

proof.

Proof by reduction from FPHP(Bn)

Proof of degree d k−color(Gn)[xv,j] k−color(Gn)[xv,j] 1 = 0 1 = 0 Proof of degree 2d 1 = 0 1 = 0 k−color(Gn)[ fv,j(pi,h)] k−color(Gn)[ fv,j(pi,h)] xv,j ←

− fv,j(pi,h)

xv,j ←

− fv,j(pi,h) polynomial substitution of degree 2

• Thm. k-color(Gn) requires degree Ω(n)

proof.

Proof by reduction from FPHP(Bn)

Proof of degree d k−color(Gn)[xv,j] k−color(Gn)[xv,j] 1 = 0 1 = 0 Proof of degree 2d 1 = 0 1 = 0 k−color(Gn)[ fv,j(pi,h)] k−color(Gn)[ fv,j(pi,h)] xv,j ←

− fv,j(pi,h)

xv,j ←

− fv,j(pi,h) polynomial substitution of degree 2

FPHP(Bn)[pi,h] FPHP(Bn)[pi,h]

• Pf. of degree O(k)
• Thm. k-color(Gn) requires degree Ω(n)

proof.

Proof by reduction from FPHP(Bn)

Proof of degree d k−color(Gn)[xv,j] k−color(Gn)[xv,j] 1 = 0 1 = 0 Proof of degree 2d 1 = 0 1 = 0 k−color(Gn)[ fv,j(pi,h)] k−color(Gn)[ fv,j(pi,h)] xv,j ←

− fv,j(pi,h)

xv,j ←

− fv,j(pi,h) polynomial substitution of degree 2

FPHP(Bn)[pi,h] FPHP(Bn)[pi,h]

• Pf. of degree O(k)

[MN’15] Degree Ω(n)

• Thm. k-color(Gn) requires degree Ω(n)

proof.

Proof by reduction from FPHP(Bn)

Proof of degree d k−color(Gn)[xv,j] k−color(Gn)[xv,j] 1 = 0 1 = 0 Proof of degree 2d 1 = 0 1 = 0 k−color(Gn)[ fv,j(pi,h)] k−color(Gn)[ fv,j(pi,h)] xv,j ←

− fv,j(pi,h)

xv,j ←

− fv,j(pi,h) polynomial substitution of degree 2

FPHP(Bn)[pi,h] FPHP(Bn)[pi,h]

• Pf. of degree O(k)

Degree Ω(n) [MN’15] Degree Ω(n)

• Thm. k-color(Gn) requires degree Ω(n)

proof.

i1 i2 h pi1,h pi1,h pi2,h pi2,h

xv,j ←

− . . .

Substitution

v i1 i1

i2 i2

v v

pi1,h1 = 1 pi1,h3 = 1 pi1,h4 = 1 pi1,h = 1 pi2,h = 1 pi2,h0

2 = 1

pi2,h0

3 = 1

pi2,h0

4 = 1

i2 i2 i2 i2 i1 i1 i1 i1

i1 i2 h pi1,h pi1,h pi2,h pi2,h

v v

pi1,h1 = 1 pi1,h3 = 1 pi1,h4 = 1 pi1,h = 1 pi2,h = 1 pi2,h0

2 = 1

pi2,h0

3 = 1

pi2,h0

4 = 1

i2 i2 i2 i2 i1 i1 i1 i1

xv,1

pi1,h1 pi2,h0

2 + pi1,hpi2,h0 2 + pi1,hpi2,h0 3 + pi1,hpi2,h0 4 + pi1,h3 pi2,h0 4 + pi1,h4 pi2,h0 2

i1 i2 h pi1,h pi1,h pi2,h pi2,h

v v

pi1,h1 = 1 pi1,h3 = 1 pi1,h4 = 1 pi1,h = 1 pi2,h = 1 pi2,h0

2 = 1

pi2,h0

3 = 1

pi2,h0

4 = 1

i2 i2 i2 i2 i1 i1 i1 i1

xv,1

pi1,h1 pi2,h0

2 + pi1,hpi2,h0 2 + pi1,hpi2,h0 3 + pi1,hpi2,h0 4 + pi1,h3 pi2,h0 4 + pi1,h4 pi2,h0 2

xv,2 ←

− 0

i1 i2 h pi1,h pi1,h pi2,h pi2,h

v v

pi1,h1 = 1 pi1,h3 = 1 pi1,h4 = 1 pi1,h = 1 pi2,h = 1 pi2,h0

2 = 1

pi2,h0

3 = 1

pi2,h0

4 = 1

i2 i2 i2 i2 i1 i1 i1 i1

xv,3

pi1,h1 pi2,h + pi1,h1 pi2,h0

3 + pi1,h3 pi2,h0 2 + pi1,h4 pi2,h0 3 + pi1,h4 pi2,h0 4

xv,1

pi1,h1 pi2,h0

2 + pi1,hpi2,h0 2 + pi1,hpi2,h0 3 + pi1,hpi2,h0 4 + pi1,h3 pi2,h0 4 + pi1,h4 pi2,h0 2

xv,2 ←

− 0

i1 i2 h pi1,h pi1,h pi2,h pi2,h

v v

pi1,h1 = 1 pi1,h3 = 1 pi1,h4 = 1 pi1,h = 1 pi2,h = 1 pi2,h0

2 = 1

pi2,h0

3 = 1

pi2,h0

4 = 1

i2 i2 i2 i2 i1 i1 i1 i1

xv,3

pi1,h1 pi2,h + pi1,h1 pi2,h0

3 + pi1,h3 pi2,h0 2 + pi1,h4 pi2,h0 3 + pi1,h4 pi2,h0 4

xv,1

pi1,h1 pi2,h0

2 + pi1,hpi2,h0 2 + pi1,hpi2,h0 3 + pi1,hpi2,h0 4 + pi1,h3 pi2,h0 4 + pi1,h4 pi2,h0 2

xv,2 ←

− 0

xv,4

pi1,h1 pi2,h0

4 + pi1,h3 pi2,h + pi1,h3 pi2,h0 3 + pi1,h4 pi2,h

i1 i2 h pi1,h pi1,h pi2,h pi2,h

v v

Proof of degree d k−color(Gn)[xv,j] k−color(Gn)[xv,j] 1 = 0 1 = 0 Proof of degree 2d 1 = 0 1 = 0 k−color(Gn)[ fv,j(pi,h)] k−color(Gn)[ fv,j(pi,h)] FPHP(Bn)[pi,h] FPHP(Bn)[pi,h]

• Pf. of degree O(k)

FPHP(Bn) degree Ω(n) ⟶ k-color(Gn) degree Ω(n)

iii.

Conclusions

We have built hard graphs for k-coloring

Open 1 — Average case complexity

Random G~𝒣(n,2∆/(n − 1)) with ∆ > k ln(k) + O(k ln ln(k)) (known for Resolution [Beame et al.’05])

• Graphs that require high complexity polynomial calculus proofs
• … therefore hard for algebraic algorithms (e.g. De Loera et al.)

Our graphs are easy for Cutting Planes (CP)

Open 2 — stronger proof systems

find graphs hard for stronger proof systems (e.g. cutting planes)

Open 3 — pseudo boolean solvers

improve Pseudo Boolean sat solver to search for CP proofs

• CP is a proof language for Linear Integer Programming
• Our graphs have short CP proofs of rank O(k) [our paper]
• Still hard for Pseudo Boolean SAT solvers [SAT’16]

Thank you