Coloring Count Cones of Planar Graphs Zden ek Dvo r ak Bernard - PowerPoint PPT Presentation

Coloring Count Cones of Planar Graphs Zdenˇ ek Dvoˇ r´ ak Bernard Lidick´ y CanaDAM Vancouver, BC May 28, 2019

Motivation Theorem (4CT) Every planar graph is 4 -colorable. Problem Is there a polynomial-time algorithm to decide if a precoloring of a 4 -face extends? (all other faces are triangles) 2

Motivation Theorem (4CT) Every planar graph is 4 -colorable. Problem Is there a polynomial-time algorithm to decide if a precoloring of a 4 -face extends? (all other faces are triangles) Not all precolorings extend. 2

Motivation Theorem (4CT) Every planar graph is 4 -colorable. Problem Is there a polynomial-time algorithm to decide if a precoloring of a 4 -face extends? (all other faces are triangles) Not all precolorings extend. 2

Motivation Theorem (4CT) Every planar graph is 4 -colorable. Problem Is there a polynomial-time algorithm to decide if a precoloring of a 4 -face extends? (all other faces are triangles) Not all precolorings extend. 2

Motivation Theorem (4CT) Every planar graph is 4 -colorable. Problem Is there a polynomial-time algorithm to decide if a precoloring of a 4 -face extends? (all other faces are triangles) Not all precolorings extend. 2

Motivation Theorem (4CT) Every planar graph is 4 -colorable. Problem Is there a polynomial-time algorithm to decide if a precoloring of a 4 -face extends? (all other faces are triangles) Not all precolorings extend. 2

Motivation Theorem (4CT) Every planar graph is 4 -colorable. Problem Is there a polynomial-time algorithm to decide if a precoloring of a 4 -face extends? (all other faces are triangles) Not all precolorings extend. 2

Motivation Theorem (4CT) Every planar graph is 4 -colorable. Problem Is there a polynomial-time algorithm to decide if a precoloring of a 4 -face extends? (all other faces are triangles) Not all precolorings extend. 2

Motivation Theorem (4CT) Every planar graph is 4 -colorable. Problem Is there a polynomial-time algorithm to decide if a precoloring of a 4 -face extends? (all other faces are triangles) Not all precolorings extend. 2

Motivation Theorem (4CT) Every planar graph is 4 -colorable. Problem Is there a polynomial-time algorithm to decide if a precoloring of a 4 -face extends? (all other faces are triangles) Not all precolorings extend. 2

Motivation Theorem (4CT) Every planar graph is 4 -colorable. Problem Is there a polynomial-time algorithm to decide if a precoloring of a 4 -face extends? (all other faces are triangles) Not all precolorings extend. Number of extensions satisfies some constraints. 2

Motivation Theorem (4CT) Every planar graph is 4 -colorable. Problem Is there a polynomial-time algorithm to decide if a precoloring of a 4 -face extends? (all other faces are triangles) # ≤ # + # Not all precolorings extend. Number of extensions satisfies some constraints. 2

Dual 3

Dual 3

Dual 3

Dual 3

Dual 1 2 4 3 3

Dual 1 2 4 3 G is a near cubic plane graph , 3-edge-coloring of G ψ precoloring of half edges of G n G ( ψ ) := # extensions of ψ to G Our goal is to “describe” vectors ( n G ( ψ 1 ) , n G ( ψ 2 ) , n G ( ψ 3 ) , . . . ) 3

n G ( ψ ) := # extensions of ψ to G For G with d half edges if n G ( ψ ) � = 0 then | ψ − 1 ( R ) | ≡ | ψ − 1 ( G ) | ≡ | ψ − 1 ( B ) | ≡ d (mod 2). G R R R R R R B B R B R B R B B R n RRRR = n GGGG = n BBBB Goal: Describe vectors ( n RRRR , n RRBB , n RBRB , n RBBR ) . 4

Reductions with fixed ψ 1 4 2 3 5

Reductions with fixed ψ 1 4 2 3 5

Reductions with fixed ψ 1 4 1 4 2 3 2 3 5

Reductions with fixed ψ 1 4 1 4 2 3 2 3 5

Reductions with fixed ψ 1 4 1 4 2 3 2 3 = 5

Reductions with fixed ψ 1 4 1 4 2 3 2 3 = = 2 · 5

Reductions with fixed ψ 1 4 1 4 2 3 2 3 = = 2 · = + − 5

= + − = + 2 −   = + 2 − + −     = + 6

Representation as a linear subspace Let G 4 be vectors ( n RRRR , n RRBB , n RBRB , n RBBR ) of all graphs with 4 half-edges. 1 4   1 4 1 4 1 4   G       ∈ G 4 ⊂ L , , 2 3 2 3 2 3 2 3      (1 , 0 , 0 , 1) (1 , 1 , 0 , 0) (0 , 1 , 1 , 0)  (? , ? , ? , ?)     G d is in a linear combination of vectors corresponding to forests. 7

Representation as a linear subspace Let G 4 be vectors ( n RRRR , n RRBB , n RBRB , n RBBR ) of all graphs with 4 half-edges. 1 4   1 4 1 4 1 4   G       ∈ G 4 ⊂ L , , 2 3 2 3 2 3 2 3      (1 , 0 , 0 , 1) (1 , 1 , 0 , 0) (0 , 1 , 1 , 0)  (? , ? , ? , ?)     G d is in a linear combination of vectors corresponding to forests. Can one do better and find a cone? (linear combinations with non-negative coefficients preserve positive coordinates ) 7

Best cones we found! 8

Kempe chain relations Kempe chains are paths and cycles. n G ( ψ ) := # extensions of ψ to G R R R R n RRRR = 9

Kempe chain relations Kempe chains are paths and cycles. n G ( ψ ) := # extensions of ψ to G R R R R n RRRR = n + + + 9

Kempe chain relations Kempe chains are paths and cycles. n G ( ψ ) := # extensions of ψ to G R R R R + n RRRR = n + n + + + 9

Kempe chain relations Kempe chains are paths and cycles. n G ( ψ ) := # extensions of ψ to G R B B R + n RRRR = n + n + + + + n RBBR = + n + 9

Kempe chain relations Kempe chains are paths and cycles. n G ( ψ ) := # extensions of ψ to G R B B R + n RRRR = n + n + + + + n RBBR = n − − + n + 9

Resulting system of equations + n RRRR = n + + + n + n RRBB = n + − + + n − n RBRB = n − − − + n − + n RBBR = n − − + n + and all ≥ 0. 10

Resulting system of equations + n RRRR = n + + + n + n RRBB = n + − + + n − n RBRB = n − − − + n − + n RBBR = n − − + n + and all ≥ 0. Solution:   1 4 1 4 1 4 1 4         G 4 ⊆ C one , , 2 3 2 3 2 3 2 3      (1 , 0 , 0 , 1) (1 , 1 , 0 , 0) (0 , 1 , 1 , 0) (0 , 0 , 1 , 1)      10

  G 2 ⊆ C one  1 2  =: K 2     1   G 3 ⊆ C one  =: K 3      2 3  4  1 4 1 4 1 4 1   G 4 ⊆ C one  =: K 4    2 3 2 3 2 3 2 3 11

Rays for G 5 cone 1 1 1 1 2 5 2 5 2 5 2 5 3 4 3 4 3 4 3 4 R 5 , 1 R 5 , 2 R 5 , 3 R 5 , 4 1 1 1 1 2 5 2 5 2 5 2 5 3 4 3 4 3 4 3 4 R 5 , 5 R 5 , 6 R 5 , 7 R 5 , 8 1 1 1 1 2 5 2 5 2 5 2 5 3 4 3 4 3 4 3 4 R 5 , 9 R 5 , 10 R 5 , 11 R 5 , 12 12

Lemma 1 The following claims are equivalent. 2 5 (a) Every planar cubic 2 -edge-connected graph is 3 -edge-colorable. (4CT) 3 4 (b) For every plane near-cubic graph G with 5 R 5 , 12 half-edges, if n G ∈ ray ( R 5 , 12 ) , then n G = 0 . G 5 R 5 , 12 0 13

Lemma 1 The following claims are equivalent. 2 5 (a) Every planar cubic 2 -edge-connected graph is 3 -edge-colorable. (4CT) 3 4 (b) For every plane near-cubic graph G with 5 R 5 , 12 half-edges, if n G ∈ ray ( R 5 , 12 ) , then n G = 0 . G 5 R 5 , 12 0 13

Sketch ( a ) = ⇒ ( b ) Let G have n G ∈ ray ( R 5 , 12 ). Goal n G = 0. G 14

Sketch ( a ) = ⇒ ( b ) Let G have n G ∈ ray ( R 5 , 12 ). Goal n G = 0. G Glue G with C 5 to G as G ⊕ C 5 . 14

Sketch ( a ) = ⇒ ( b ) Let G have n G ∈ ray ( R 5 , 12 ). Goal n G = 0. G Glue G with C 5 to G as G ⊕ C 5 . 14

Sketch ( a ) = ⇒ ( b ) Let G have n G ∈ ray ( R 5 , 12 ). Goal n G = 0. G Glue G with C 5 to G as G ⊕ C 5 . G ⊕ C 5 is not 3-edge-colorable (Petersen graph). 14

Sketch ( a ) = ⇒ ( b ) Let G have n G ∈ ray ( R 5 , 12 ). Goal n G = 0. G Glue G with C 5 to G as G ⊕ C 5 . G ⊕ C 5 is not 3-edge-colorable (Petersen graph). By ( a ), G has a bridge. G no precoloring extends so n G = 0 14

Sketch ( b ) = ⇒ ( a ) Let G be a smallest plane 2-edge-connected graph that is not 3-edge-colorable. Assume ( b ) and show G is 3-edge-colorable. G 15

Sketch ( b ) = ⇒ ( a ) Let G be a smallest plane 2-edge-connected graph that is not 3-edge-colorable. Assume ( b ) and show G is 3-edge-colorable. G Find a 5-face C 5 15

Sketch ( b ) = ⇒ ( a ) Let G be a smallest plane 2-edge-connected graph that is not 3-edge-colorable. Assume ( b ) and show G is 3-edge-colorable. G H Find a 5-face C 5 , replace it by a path 15

Sketch ( b ) = ⇒ ( a ) Let G be a smallest plane 2-edge-connected graph that is not 3-edge-colorable. Assume ( b ) and show G is 3-edge-colorable. G H Find a 5-face C 5 , replace it by a path, now H is 3-edge-colorable. 15

Sketch ( b ) = ⇒ ( a ) Let G be a smallest plane 2-edge-connected graph that is not 3-edge-colorable. Assume ( b ) and show G is 3-edge-colorable. G H Find a 5-face C 5 , replace it by a path, now H is 3-edge-colorable. G − C 5 has a 3-edge-coloring. 15