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 Proof lines: polynomial eq. over 𝔾 Axiom: vertex v gets color {1,…,k} x v ,1 + x v ,2 + . . . + x v , k = 1 x v , j x v , j 0 = 0 for j ≠ j’ ∈ [k]

Proofs of non- k -colorability, in Polynomial Calculus Proof lines: polynomial eq. over 𝔾 Axiom: vertex v gets color {1,…,k} x v ,1 + x v ,2 + . . . + x v , k = 1 x v , j x v , j 0 = 0 for j ≠ j’ ∈ [k] Axiom: two colors on {u,v} ∈ E(G) x u , j x v , j = 0 for j ∈ [k]

Proofs of non- k -colorability, in Polynomial Calculus Inference rules Proof lines: polynomial eq. over 𝔾 p = 0 q = 0 α , β ∈ F Axiom: vertex v gets color {1,…,k} α p + β q = 0 x v ,1 + x v ,2 + . . . + x v , k = 1 p = 0 x v , j x v , j 0 = 0 for j ≠ j’ ∈ [k] x v , j · p = 0 Axiom: two colors on {u,v} ∈ E(G) x u , j x v , j = 0 for j ∈ [k]

Proofs of non- k -colorability, in Polynomial Calculus Inference rules Proof lines: polynomial eq. over 𝔾 p = 0 q = 0 α , β ∈ F Axiom: vertex v gets color {1,…,k} α p + β q = 0 x v ,1 + x v ,2 + . . . + x v , k = 1 p = 0 x v , j x v , j 0 = 0 for j ≠ j’ ∈ [k] x v , j · p = 0 Refutation Axiom: two colors on {u,v} ∈ E(G) . . . x u , j x v , j = 0 for j ∈ [k] 1 = 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 Easy upper bounds Proof Size ≤ n O( Degree ) Search-Time ≤ n O( Degree ) 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 Easy upper bounds Proof Size ≤ n O( Degree ) Search-Time ≤ n O( Degree ) where n is the number of variables [IPS’99]. Degree Ω (n) implies Proof Size 2 Ω (n)

Our results Main Theorem. Fix k>0. We describe graphs {G n } n such • G n has Θ (k 4 n) vertices of degree O(k 2 ) • G n 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 2 Ω (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 di ff erent encoding for the polynomial equations • finds a special form of PC proof of degree d in time n Θ (d) SUCCESS : non- k -colorable FAILURE : undetermined (maybe d too small?)

De Loera et al. [2008, 2009, 2011, 2015] Algebraic algorithm to decide (non-) k -colorability • uses a di ff erent encoding for the polynomial equations • finds a special form of PC proof of degree d in time n Θ (d) SUCCESS : non- k -colorable FAILURE : undetermined (maybe d too small?) 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. G n require degree Ω (n)

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

i. Construction of the hard graphs

Reduction from FPHP( B n ) — part 1 Bipartite B n Pigeon i picks one hole in {h 1 , …, h k } 1 p i , h 1 + p i , h 2 + . . . + p i , h k = 1 1 2 2 p i , h j · p i , h j 0 = 0 3 for j ≠ j’ ∈ [k] 3 4 4 Pigeons Holes 5 5 [n] [n-1] 6 No two pigeons in same hole 6 7 7 8 8 p i 1 , h · p i 2 , h = 0 9 9 for {i 1 ,h},{i 2 ,h} ∈ E(B n ) 10 10 11 deg ≤ k deg ≤ O(k)

Reduction from FPHP( B n ) — part 2 Theorem [MN’15]. If {B n } n as described here is a family of boundary expanders(*) , then any polynomial calculus refutations of FPHP( B n ) requires degree Ω (n) . (*) boundary expander: for some constant 𝛽 >0 and c>0 every set S of at most 𝛽 n left-vertices has at least c | S | right-vertices that have exactly one neighbor in S each .

Reduction from FPHP( B n ) — part 3 We start with a family {B n } n • left-degree ≤ k • right-degree O(k) • boundary expanders, i.e. FPHP( B n ) is hard We build {G n } n such that k -colorability formula for G n is an obfuscated version of formula FPHP( B n )

Construction of G n — mapping (e.g. k=4 ) h 1 Pigeon i picks one hole in {h 1 , …, h k } h 2 p i , h 1 + p i , h 2 + . . . + p i , h k = 1 i h 3 p i , h j · p i , h j 0 = 0 for j ≠ j’ ∈ [k] h 4

Construction of G n — mapping (e.g. k=4 ) h 1 Pigeon i picks one hole in {h 1 , …, h k } h 2 p i , h 1 + p i , h 2 + . . . + p i , h k = 1 i h 3 p i , h j · p i , h j 0 = 0 for j ≠ j’ ∈ [k] h 4 Vertex i gets one color in {1,…,k} Special vertex i x i ,1 + x i ,2 + . . . + x i , k = 1 i for j ≠ j’ ∈ [k] x i , j · x i , j 0 = 0 x i ,1 x i ,1 x i ,2 x i ,2 x i ,3 x i ,3 x i ,4 x i ,4

Construction of G n — conflicts (e.g. k=4 ) No two pigeons in same hole i 1 p i 1 , h p i 1 , h h p i 1 , h · p i 2 , h = 0 for {i 1 ,h},{i 2 ,h} ∈ E(B n ) p i 2 , h p i 2 , h i 2

Construction of G n — conflicts (e.g. k=4 ) No two pigeons in same hole i 1 p i 1 , h p i 1 , h h p i 1 , h · p i 2 , h = 0 for {i 1 ,h},{i 2 ,h} ∈ E(B n ) p i 2 , h p i 2 , h i 2 We add a gadget to encode: i 1 i 2 i 1 i 2

Forbid red to vertex i 1 and green to vertex i 2 i 1 i 2 (E.g. k=4 ) Pre-colored green Pre-colored red i 1 i 2

Forbid red to vertex i 1 and green to vertex i 2 i 1 i 2 (E.g. k=4 ) Pre-colored green Pre-colored red i 1 i 2 The special vertices are the only one shared among gadgets

Construction of G n — precolored vertices (e.g. k=4 ) v 10 v 2 v 4 v 6 v 8 v 1 v 3 v 5 v 7 v 9 v 1+4m A long enough chain allows to identify at v 2+4m most one pre-colored vertex to a vertex in the chain. (To keep G n sparse) v 3+4m v 4+4m

ii. Sketch of the lower bound

Proof by reduction from FPHP( B n ) Thm. k -color( G n ) requires degree Ω (n) proof. k − color ( G n )[ x v , j ] k − color ( G n )[ x v , j ] Proof of degree d 1 = 0 1 = 0

Proof by reduction from FPHP( B n ) Thm. k -color( G n ) requires degree Ω (n) proof. k − color ( G n )[ x v , j ] k − color ( G n )[ x v , j ] k − color ( G n )[ f v , j ( p i , h )] k − color ( G n )[ f v , j ( p i , h )] Proof of degree d Proof of degree 2 d 1 = 0 1 = 0 1 = 0 1 = 0 − f v , j ( p i , h ) polynomial substitution of degree 2 − f v , j ( p i , h ) x v , j ← x v , j ←

Proof by reduction from FPHP( B n ) Thm. k -color( G n ) requires degree Ω (n) proof. FPHP ( B n )[ p i , h ] FPHP ( B n )[ p i , h ] Pf. of degree O(k) k − color ( G n )[ x v , j ] k − color ( G n )[ x v , j ] k − color ( G n )[ f v , j ( p i , h )] k − color ( G n )[ f v , j ( p i , h )] Proof of degree d Proof of degree 2 d 1 = 0 1 = 0 1 = 0 1 = 0 − f v , j ( p i , h ) polynomial substitution of degree 2 − f v , j ( p i , h ) x v , j ← x v , j ←

Proof by reduction from FPHP( B n ) Thm. k -color( G n ) requires degree Ω (n) proof. FPHP ( B n )[ p i , h ] FPHP ( B n )[ p i , h ] Pf. of degree O(k) k − color ( G n )[ x v , j ] k − color ( G n )[ x v , j ] k − color ( G n )[ f v , j ( p i , h )] k − color ( G n )[ f v , j ( p i , h )] Proof of degree d Proof of degree 2 d [MN’15] Degree Ω (n) 1 = 0 1 = 0 1 = 0 1 = 0 − f v , j ( p i , h ) polynomial substitution of degree 2 − f v , j ( p i , h ) x v , j ← x v , j ←