siegel s theorem edge coloring and a holant dichotomy
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Siegels theorem, edge coloring, and a holant dichotomy Tyson - PowerPoint PPT Presentation

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Siegels theorem, edge coloring, and a holant dichotomy Tyson Williams (University of Wisconsin-Madison) Joint with: Jin-Yi Cai and Heng Guo (University of Wisconsin-Madison) Appeared at FOCS 2014 1 / 43 Edge Coloring Definition 2 / 43

  1. Hardness of Holant 3 ( − ; AD 3 ) Hardness of Holant 3 ( − ; AD 3 ) proved by the following reduction chain: #P ≤ T Holant 3 ( − ; � 2 , 1 , 0 , 1 , 0 � ) ≤ T Holant 3 ( − ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant 3 ( − ; AD 3 )  w z a if w = x = y = z    b if w = x � = y = z     f f ( w z x y ) = � a , b , c , d , e � = c if w = y � = x = z  d if w = z � = x = y  y  x    e otherwise.  15 / 43

  2. Hardness of Holant 3 ( − ; AD 3 ) Hardness of Holant 3 ( − ; AD 3 ) proved by the following reduction chain: #P ≤ T Holant 3 ( − ; � 2 , 1 , 0 , 1 , 0 � ) ≤ T Holant 3 ( − ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant 3 ( − ; AD 3 )  w z a if w = x = y = z    b if w = x � = y = z     f f ( w z x y ) = � a , b , c , d , e � = c if w = y � = x = z  d if w = z � = x = y  y  x    e otherwise.  15 / 43

  3. Hardness of Holant 3 ( − ; AD 3 ) Hardness of Holant 3 ( − ; AD 3 ) proved by the following reduction chain: #P ≤ T Holant 3 ( − ; � 2 , 1 , 0 , 1 , 0 � ) ≤ T Holant 3 ( − ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant 3 ( − ; AD 3 ) First reduction: From a #P-hard point on the Tutte polynomial. 15 / 43

  4. Hardness of Holant 3 ( − ; AD 3 ) Hardness of Holant 3 ( − ; AD 3 ) proved by the following reduction chain: #P ≤ T Holant 3 ( − ; � 2 , 1 , 0 , 1 , 0 � ) ≤ T Holant 3 ( − ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant 3 ( − ; AD 3 ) First reduction: From a #P-hard point on the Tutte polynomial. Second reduction: Via polynomial interpolation. 15 / 43

  5. Hardness of Holant 3 ( − ; AD 3 ) Hardness of Holant 3 ( − ; AD 3 ) proved by the following reduction chain: #P ≤ T Holant 3 ( − ; � 2 , 1 , 0 , 1 , 0 � ) ≤ T Holant 3 ( − ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant 3 ( − ; AD 3 ) First reduction: From a #P-hard point on the Tutte polynomial. Second reduction: Via polynomial interpolation. Third reduction: Via a gadget construction. 15 / 43

  6. Hardness of Holant 3 ( − ; AD 3 ) Hardness of Holant 3 ( − ; AD 3 ) proved by the following reduction chain: #P ≤ T Holant 3 ( − ; � 2 , 1 , 0 , 1 , 0 � ) ≤ T Holant 3 ( − ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant 3 ( − ; AD 3 ) First reduction: From a #P-hard point on the Tutte polynomial. Second reduction: Via polynomial interpolation. Third reduction: Via a gadget construction. 15 / 43

  7. Reduce From Tutte Polynomial Definition The Tutte polynomial of an undirected graph G is E ( G ) = ∅ ,  1    xT ( G \ e ; x , y ) e ∈ E ( G ) is a bridge,  T ( G ; x , y ) = yT ( G \ e ; x , y ) e ∈ E ( G ) is a loop,     T ( G \ e ; x , y ) + T ( G / e ; x , y ) otherwise, where G \ e is the graph obtained by deleting e and G / e is the graph obtained by contracting e . 16 / 43

  8. Reduce From Tutte Polynomial Definition The Tutte polynomial of an undirected graph G is E ( G ) = ∅ ,  1    xT ( G \ e ; x , y ) e ∈ E ( G ) is a bridge,  T ( G ; x , y ) = yT ( G \ e ; x , y ) e ∈ E ( G ) is a loop,     T ( G \ e ; x , y ) + T ( G / e ; x , y ) otherwise, where G \ e is the graph obtained by deleting e and G / e is the graph obtained by contracting e . The chromatic polynomial is χ ( G ; λ ) = ( − 1) | V |− 1 λ T( G ; 1 − λ, 0) . 16 / 43

  9. Reduction From Tutte Polynomial: Medial Graph Definition (a) (b) (c) A plane graph (a), its medial graph (c), and the two graphs superimposed (b). 17 / 43

  10. Reduction From Tutte Polynomial: Directed Medial Graph Definition (a) (b) (c) A plane graph (a), its directed medial graph (c), and the two graphs superimposed (b). 18 / 43

  11. Reduction From Tutte Polynomial: Eulerian Graphs and Eulerian Partitions Definition Digraph is Eulerian if “in degree” = “out degree”. 1 19 / 43

  12. Reduction From Tutte Polynomial: Eulerian Graphs and Eulerian Partitions Definition Digraph is Eulerian if “in degree” = “out degree”. 1 Eulerian partition of an Eulerian digraph � G is a partition of the edges 2 of � G such that each part induces an Eulerian digraph. 19 / 43

  13. Reduction From Tutte Polynomial: Eulerian Graphs and Eulerian Partitions Definition Digraph is Eulerian if “in degree” = “out degree”. 1 Eulerian partition of an Eulerian digraph � G is a partition of the edges 2 of � G such that each part induces an Eulerian digraph. Let π κ ( � G ) be the set of Eulerian partitions of � G into at most κ parts. 3 κ ≥ 2 19 / 43

  14. Reduction From Tutte Polynomial: Eulerian Graphs and Eulerian Partitions Definition Digraph is Eulerian if “in degree” = “out degree”. 1 Eulerian partition of an Eulerian digraph � G is a partition of the edges 2 of � G such that each part induces an Eulerian digraph. Let π κ ( � G ) be the set of Eulerian partitions of � G into at most κ parts. 3 Let µ ( c ) be the number of monochromatic vertices in c . 4 κ ≥ 2 µ ( c ) = 1 19 / 43

  15. Reduction From Tutte Polynomial: Crucial Identity Theorem (Ellis-Monaghan) For a plane graph G, � 2 µ ( c ) . κ T( G ; κ + 1 , κ + 1) = c ∈ π κ ( � G m ) 20 / 43

  16. Reduction From Tutte Polynomial: Connection to Holant Then 2 µ ( c ) = Holant κ ( G m ; � 2 , 1 , 0 , 1 , 0 � ) , � c ∈ π κ ( � G m ) where  2 if w = x = y = z E    1 if w = x � = y = z     E ( w z if w = y � = x = z x y ) = 0  1 if w = z � = x = y      0 otherwise,  where E = � 2 , 1 , 0 , 1 , 0 � . 21 / 43

  17. Reduction From Tutte Polynomial: Connection to Holant Then 2 µ ( c ) = Holant κ ( G m ; � 2 , 1 , 0 , 1 , 0 � ) , � c ∈ π κ ( � G m ) where  2 if w = x = y = z E    1 if w = x � = y = z     E ( w z if w = y � = x = z x y ) = 0  1 if w = z � = x = y      0 otherwise,  where E = � 2 , 1 , 0 , 1 , 0 � . 21 / 43

  18. Reduction From Tutte Polynomial: Connection to Holant Then 2 µ ( c ) = Holant κ ( G m ; � 2 , 1 , 0 , 1 , 0 � ) , � c ∈ π κ ( � G m ) where  2 if w = x = y = z E    1 if w = x � = y = z     E ( w z if w = y � = x = z x y ) = 0  1 if w = z � = x = y      0 otherwise,  where E = � 2 , 1 , 0 , 1 , 0 � . 21 / 43

  19. Reduction From Tutte Polynomial: Connection to Holant Then 2 µ ( c ) = Holant κ ( G m ; � 2 , 1 , 0 , 1 , 0 � ) , � c ∈ π κ ( � G m ) where  2 if w = x = y = z E    1 if w = x � = y = z     E ( w z if w = y � = x = z x y ) = 0  1 if w = z � = x = y      0 otherwise,  where E = � 2 , 1 , 0 , 1 , 0 � . 21 / 43

  20. Reduction From Tutte Polynomial: Connection to Holant Then 2 µ ( c ) = Holant κ ( G m ; � 2 , 1 , 0 , 1 , 0 � ) , � c ∈ π κ ( � G m ) where  2 if w = x = y = z E    1 if w = x � = y = z     E ( w z if w = y � = x = z x y ) = 0  1 if w = z � = x = y      0 otherwise,  where E = � 2 , 1 , 0 , 1 , 0 � . 21 / 43

  21. Reduction From Tutte Polynomial: Upshot Corollary For a plane graph G, κ T ( G ; κ + 1 , κ + 1) = Holant κ ( G m ; � 2 , 1 , 0 , 1 , 0 � ) 22 / 43

  22. Reduction From Tutte Polynomial: Upshot Corollary For a plane graph G, κ T ( G ; κ + 1 , κ + 1) = Holant κ ( G m ; � 2 , 1 , 0 , 1 , 0 � ) y Theorem (Vertigan) 4 For any x , y ∈ C , the problem of 3 evaluating the Tutte polynomial at ( x , y ) over planar graphs is #P-hard 2 unless ( x − 1)( y − 1) ∈ { 1 , 2 } or 1 ( x , y ) ∈ { ( ± 1 , ± 1) , ( ω, ω 2 ) , ( ω 2 , ω ) } , where ω = e 2 π i / 3 . In each of these x � 2 � 1 1 2 3 4 exceptional cases, the computation � 1 can be done in polynomial time. � 2 22 / 43

  23. Hardness of Holant 3 ( − ; AD 3 ) Hardness of Holant 3 ( − ; AD 3 ) proved by the following reduction chain: #P ≤ T Holant 3 ( − ; � 2 , 1 , 0 , 1 , 0 � ) ≤ T Holant 3 ( − ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant 3 ( − ; AD 3 ) First reduction: From a #P-hard point on the Tutte polynomial. Second reduction: Via polynomial interpolation. Third reduction: Via a gadget construction. 23 / 43

  24. Hardness of Holant 3 ( − ; AD 3 ) Hardness of Holant 3 ( − ; AD 3 ) proved by the following reduction chain: #P ≤ T Holant 3 ( − ; � 2 , 1 , 0 , 1 , 0 � ) ≤ T Holant 3 ( − ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant 3 ( − ; AD 3 ) First reduction: From a #P-hard point on the Tutte polynomial. Second reduction: Via polynomial interpolation. Third reduction: Via a gadget construction. 23 / 43

  25. Gadget Construction Holant( G ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant( G ′ ; AD 3 ) AD 3 w z AD 3 y x  0 if w = x = y = z    1 if w = x � = y = z     f ( w z x y ) = � 0 , 1 , 1 , 0 , 0 � = if w = y � = x = z 1  0 if w = z � = x = y      0 otherwise.  24 / 43

  26. Gadget Construction Holant( G ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant( G ′ ; AD 3 ) AD 3 w z AD 3 y x  0 if w = x = y = z    1 if w = x � = y = z     f ( w z x y ) = � 0 , 1 , 1 , 0 , 0 � = if w = y � = x = z 1  0 if w = z � = x = y      0 otherwise.  24 / 43

  27. Gadget Construction Holant( G ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant( G ′ ; AD 3 ) AD 3 w z AD 3 y x  0 if w = x = y = z    1 if w = x � = y = z     f ( w z x y ) = � 0 , 1 , 1 , 0 , 0 � = if w = y � = x = z 1  0 if w = z � = x = y      0 otherwise.  24 / 43

  28. Gadget Construction Holant( G ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant( G ′ ; AD 3 ) AD 3 w z AD 3 y x  0 if w = x = y = z    1 if w = x � = y = z     f ( w z x y ) = � 0 , 1 , 1 , 0 , 0 � = if w = y � = x = z 1  0 if w = z � = x = y      0 otherwise.  24 / 43

  29. Gadget Construction Holant( G ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant( G ′ ; AD 3 ) AD 3 w z AD 3 y x  0 if w = x = y = z    1 if w = x � = y = z     f ( w z x y ) = � 0 , 1 , 1 , 0 , 0 � = if w = y � = x = z 1  0 if w = z � = x = y      0 otherwise.  24 / 43

  30. Gadget Construction Holant( G ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant( G ′ ; AD 3 ) AD 3 w z AD 3 y x  0 if w = x = y = z    1 if w = x � = y = z     f ( w z x y ) = � 0 , 1 , 1 , 0 , 0 � = if w = y � = x = z 1  0 if w = z � = x = y      0 otherwise.  24 / 43

  31. Gadget Construction Holant( G ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant( G ′ ; AD 3 ) AD 3 w z AD 3 y x  0 if w = x = y = z    1 if w = x � = y = z     f ( w z x y ) = � 0 , 1 , 1 , 0 , 0 � = if w = y � = x = z 1  0 if w = z � = x = y      0 otherwise.  24 / 43

  32. Hardness of Holant 3 ( − ; AD 3 ) Hardness of Holant 3 ( − ; AD 3 ) proved by the following reduction chain: #P ≤ T Holant 3 ( − ; � 2 , 1 , 0 , 1 , 0 � ) ≤ T Holant 3 ( − ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant 3 ( − ; AD 3 ) First reduction: From a #P-hard point on the Tutte polynomial. Second reduction: Via polynomial interpolation. Third reduction: Via a gadget construction. 25 / 43

  33. Hardness of Holant 3 ( − ; AD 3 ) Hardness of Holant 3 ( − ; AD 3 ) proved by the following reduction chain: #P ≤ T Holant 3 ( − ; � 2 , 1 , 0 , 1 , 0 � ) ≤ T Holant 3 ( − ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant 3 ( − ; AD 3 ) First reduction: From a #P-hard point on the Tutte polynomial. Second reduction: Via polynomial interpolation. Third reduction: Via a gadget construction. 25 / 43

  34. Polynomial Interpolation: Recursive Construction Holant 3 ( G ; � 2 , 1 , 0 , 1 , 0 � ) ≤ T Holant 3 ( G s ; � 0 , 1 , 1 , 0 , 0 � ) N s N 1 N 2 N s +1 Vertices are assigned � 0 , 1 , 1 , 0 , 0 � . 26 / 43

  35. Polynomial Interpolation: Recursive Construction Holant 3 ( G ; � 2 , 1 , 0 , 1 , 0 � ) ≤ T Holant 3 ( G s ; � 0 , 1 , 1 , 0 , 0 � ) N s N 1 N 2 N s +1 Vertices are assigned � 0 , 1 , 1 , 0 , 0 � . Let f s be the function corresponding to N s . Then f s = M s f 0 , where  0 2 0 0 0   1  1 1 0 0 0 0         M = 0 0 0 1 0 and f 0 = 0 .         0 0 1 0 0 1     0 0 0 0 1 0 Obviously f 1 = � 0 , 1 , 1 , 0 , 0 � . 26 / 43

  36. Polynomial Interpolation: Eigenvectors and Eigenvalues Spectral decomposition M = P Λ P − 1 , where  1 − 2 0 0 0   2 0 0 0 0  1 1 0 0 0 0 − 1 0 0 0         P = 0 0 1 1 0 and Λ = 0 0 1 0 0 .         0 0 1 − 1 0 0 0 0 − 1 0     0 0 0 0 1 0 0 0 0 1 27 / 43

  37. Polynomial Interpolation: Eigenvectors and Eigenvalues Spectral decomposition M = P Λ P − 1 , where  1 − 2 0 0 0   2 0 0 0 0  1 1 0 0 0 0 − 1 0 0 0         P = 0 0 1 1 0 and Λ = 0 0 1 0 0 .         0 0 1 − 1 0 0 0 0 − 1 0     0 0 0 0 1 0 0 0 0 1 Let x = 2 2 s . Then x − 1  x 0 0 0 0   + 1  3 x − 1 0 1 0 0 0    3  f 2 s = P Λ 2 s P − 1 f 0 = P   P − 1 f 0 =   0 0 1 0 0 0 .         0 0 0 1 0 1     0 0 0 0 1 0 27 / 43

  38. Polynomial Interpolation: Eigenvectors and Eigenvalues Spectral decomposition M = P Λ P − 1 , where  1 − 2 0 0 0   2 0 0 0 0  1 1 0 0 0 0 − 1 0 0 0         P = 0 0 1 1 0 and Λ = 0 0 1 0 0 .         0 0 1 − 1 0 0 0 0 − 1 0     0 0 0 0 1 0 0 0 0 1 Let x = 2 2 s . Then x − 1  x 0 0 0 0   + 1  3 x − 1 0 1 0 0 0    3  f ( x ) = f 2 s = P Λ 2 s P − 1 f 0 = P   P − 1 f 0 =   0 0 1 0 0 0 .         0 0 0 1 0 1     0 0 0 0 1 0 27 / 43

  39. Polynomial Interpolation: Eigenvectors and Eigenvalues Spectral decomposition M = P Λ P − 1 , where  1 − 2 0 0 0   2 0 0 0 0  1 1 0 0 0 0 − 1 0 0 0         P = 0 0 1 1 0 and Λ = 0 0 1 0 0 .         0 0 1 − 1 0 0 0 0 − 1 0     0 0 0 0 1 0 0 0 0 1 Let x = 2 2 s . Then x − 1  x 0 0 0 0   + 1  3 x − 1 0 1 0 0 0    3  f ( x ) = f 2 s = P Λ 2 s P − 1 f 0 = P   P − 1 f 0 =   0 0 1 0 0 0 .         0 0 0 1 0 1     0 0 0 0 1 0 Note f (4) = � 2 , 1 , 0 , 1 , 0 � . 27 / 43

  40. Polynomial Interpolation: Eigenvectors and Eigenvalues Spectral decomposition M = P Λ P − 1 , where  1 − 2 0 0 0   2 0 0 0 0  1 1 0 0 0 0 − 1 0 0 0         P = 0 0 1 1 0 and Λ = 0 0 1 0 0 .         0 0 1 − 1 0 0 0 0 − 1 0     0 0 0 0 1 0 0 0 0 1 Let x = 2 2 s . Then x − 1  x 0 0 0 0   + 1  3 x − 1 0 1 0 0 0    3  f ( x ) = f 2 s = P Λ 2 s P − 1 f 0 = P   P − 1 f 0 =   0 0 1 0 0 0 .         0 0 0 1 0 1     0 0 0 0 1 0 Note f (4) = � 2 , 1 , 0 , 1 , 0 � . (Side note: picking s = 1 so that x = 4 only works when κ = 3.) 27 / 43

  41. Polynomial Interpolation: The Interpolation Holant 3 ( − ; � 2 , 1 , 0 , 1 , 0 � ) ≤ T Holant 3 ( − ; � 0 , 1 , 1 , 0 , 0 � ) 28 / 43

  42. Polynomial Interpolation: The Interpolation Holant 3 ( − ; � 2 , 1 , 0 , 1 , 0 � ) = T Holant 3 ( − ; f (4)) ≤ T Holant 3 ( − ; f ( x )) ≤ T Holant 3 ( − ; � 0 , 1 , 1 , 0 , 0 � ) 28 / 43

  43. Polynomial Interpolation: The Interpolation Holant 3 ( − ; � 2 , 1 , 0 , 1 , 0 � ) = T Holant 3 ( − ; f (4)) ≤ T Holant 3 ( − ; f ( x )) ≤ T Holant 3 ( − ; � 0 , 1 , 1 , 0 , 0 � ) If G has n vertices, then p ( G , x ) = Holant 3 ( G ; f ( x )) ∈ Z [ x ] has degree n . 28 / 43

  44. Polynomial Interpolation: The Interpolation Holant 3 ( − ; � 2 , 1 , 0 , 1 , 0 � ) = T Holant 3 ( − ; f (4)) ≤ T Holant 3 ( − ; f ( x )) ≤ T Holant 3 ( − ; � 0 , 1 , 1 , 0 , 0 � ) If G has n vertices, then p ( G , x ) = Holant 3 ( G ; f ( x )) ∈ Z [ x ] has degree n . Let G 2 s be the graph obtained by replacing every vertex in G with N 2 s . Then Holant 3 ( G 2 s ; � 0 , 1 , 1 , 0 , 0 � ) = p ( G , 2 2 s ). 28 / 43

  45. Polynomial Interpolation: The Interpolation Holant 3 ( − ; � 2 , 1 , 0 , 1 , 0 � ) = T Holant 3 ( − ; f (4)) ≤ T Holant 3 ( − ; f ( x )) ≤ T Holant 3 ( − ; � 0 , 1 , 1 , 0 , 0 � ) If G has n vertices, then p ( G , x ) = Holant 3 ( G ; f ( x )) ∈ Z [ x ] has degree n . Let G 2 s be the graph obtained by replacing every vertex in G with N 2 s . Then Holant 3 ( G 2 s ; � 0 , 1 , 1 , 0 , 0 � ) = p ( G , 2 2 s ). Using oracle for Holant 3 ( − ; � 0 , 1 , 1 , 0 , 0 � ), evaluate p ( G , x ) at n + 1 distinct points x = 2 2 s for 0 ≤ s ≤ n . By polynomial interpolation, efficiently compute the coefficients of p ( G , x ). QED. 28 / 43

  46. Proof Outline for Dichotomy of Holant( − ; � a , b , c � ) For all a , b , c ∈ C , want to show that Holant( − ; � a , b , c � ) is in P or #P-hard . 29 / 43

  47. Proof Outline for Dichotomy of Holant( − ; � a , b , c � ) For all a , b , c ∈ C , want to show that Holant( − ; � a , b , c � ) is in P or #P-hard . Using � a , b , c � : Attempt to construct a special unary constraint. 1 Attempt to interpolate all binary constraints of a special form, 2 assuming we have the special unary constraint. Construct a special ternary constraint that we show is #P-hard , 3 assuming we have the special unary and binary constraints. 29 / 43

  48. Proof Outline for Dichotomy of Holant( − ; � a , b , c � ) For all a , b , c ∈ C , want to show that Holant( − ; � a , b , c � ) is in P or #P-hard . Using � a , b , c � : Attempt to construct a special unary constraint. 1 Attempt to interpolate all binary constraints of a special form, 2 assuming we have the special unary constraint. Construct a special ternary constraint that we show is #P-hard , 3 assuming we have the special unary and binary constraints. For some a , b , c ∈ C , our attempts fail. In those cases, we either show the problem is in P or 1 prove #P-hardness without the help of additional signatures. 2 29 / 43

  49. Holant( � a, b, c � ) Lemma 8.2 Attempts 1 and 2 Construct � 1 � Fail Lemma 8.1 Lemma 8.3 Succeed Attempt 1 Lemma 9.4 Counting Attempt 2 Interpolate Weighted Fail Cases 1, 2, 3, 4, 5 all � x, y � Corollary 8.4 Eulerian Lemmas 9.5, 9.6, 9.7, 9.11, 9.12 B = 0 Corollary 9.13 Partitions Corollary 7.13 Attempts 3 and 4 All Cases Lemma B.1 Succeed Construct Construct � a, b, b � Fail Attempt 1 Lemmas 7.14 � 3( κ − 1) , κ − 3 , − 3 � with a � = b Lemma 7.1 and 7.15 A = 0 Lemma 7.3 Succeed Bobby Fischer Gadget Lemma 4.18 Counting Vertex κ -Colorings Corollary 4.19 30 / 43

  50. Logical Dependencies in Dichotomy of Holant κ ( − ; � a , b , c � ) edge coloring k=r Eulerian partition directed medial graph parity condition major interpolate results ternary and quaternary resutls Tutte diagonal as state sum tau_color: f(P_0) = 0 check orthogonality condition lattice condition (LC) Puiseux series planar T utte dichotomy only 5 solutions in Z for p(x,y) state sum as Holant problem any arity interpolation LC characterization for cubic polys LC satisfied by Sn or An Galois Gps 3R & 2C roots in x for p(x,y) condition for Sn Galois gp Dedkind's Theorem condition from same norm roots edge coloring k>r hard generic binary interpolation Bobby Fischer gadget planar Eulerian partition hard (tau_color) reducible p(x,y) satisfies LC for y>3 irreducible p(x,y) satisfies LC for y>3 p(x,3) satisfies LC local holographic transformation interpolate all binary edge coloring k=r hard eigenvalue shifted triple (EST) special binary interpolation binary interpolation eigenvalues planar pairing def chomatic in Tutte obtain =_4 planar Eulerian partition hard (tau_4) p(x,y) satisfies LC for y=>3 Triangle gadget obtain any a+(k-3)b-(k-2)c=0 construct unary 2nd distinct norms 1st distinct norms extra special cases EST distinct norms interpolate all binaries find planar pairing reduction to vertex coloring construct <1> in two cases generalized edge coloring hard <3(k-1),k-3,-3> hard for k>3 <6,0,-3> hard obtain <a',b',b'> assuming a+(k-3)b-(k-2)c!=0 obtain <3(k-1),k-3,-3> generic generalized anti-gadget interpolation 4th special case arity reduction local holographic transformation <(k-1)(k-2),2-k,2> hard ternary construction summary 2nd special case typical case 3rd special case 1st special case 5th special case edge coloring k>r hard a+(k-3)b-(k-2)c=0 dichotomy binary interpolation summary <a,b,c> dichotomy 31 / 43

  51. Polynomial Interpolation Evaluate x ∈{ 1 , 2 , 3 , 4 } 30 20 p ( x ) = 2 x 3 − 3 x 2 − 17 x + 10 10 � 2 � 1 1 2 3 4 � 10 � 20 Interpolate p (1) = 2 · 1 3 − 3 · 1 2 − 17 · 1 + 10 = − 8 p (2) = 2 · 2 3 − 3 · 2 2 − 17 · 2 + 10 = − 20 p (3) = 2 · 3 3 − 3 · 3 2 − 17 · 3 + 10 = − 14 p (4) = 2 · 4 3 − 3 · 4 2 − 17 · 4 + 10 = 22 32 / 43

  52. Polynomial Interpolation Evaluate x ∈{ 1 , 2 , 3 , 4 } 30 20 p ( x ) = 2 x 3 − 3 x 2 − 17 x + 10 10 � 2 � 1 1 2 3 4 � 10 � 20 Interpolate 1 3 1 2 1 1 1 0       2 − 8 2 3 2 2 2 1 2 0 − 3 − 20        =  3 3 3 2 3 1 3 0      − 17 − 14      4 3 4 2 4 1 4 0 10 22 Vandermonde system 32 / 43

  53. Polynomial Interpolation Evaluate x ∈{ 1 , 2 , 3 , 4 } 30 20 p ( x ) = 2 x 3 − 3 x 2 − 17 x + 10 10 � 2 � 1 1 2 3 4 � 10 � 20 Interpolate 1 3 1 2 1 1 1 0       ? − 8 2 3 2 2 2 1 2 0 ? − 20       =  3 3 3 2 3 1 3 0      ? − 14       4 3 4 2 4 1 4 0 ? 22 Vandermonde system 32 / 43

  54. Polynomial Interpolation Evaluate x ∈{ 1 , 2 , 3 , 4 } 30 20 p ( x ) = 2 x 3 − 3 x 2 − 17 x + 10 10 � 2 � 1 1 2 3 4 � 10 � 20 Interpolate − 1  1 3 1 2 1 1 1 0      ? − 8 2 3 2 2 2 1 2 0 ? − 20        =    3 3 3 2 3 1 3 0    ? − 14      4 3 4 2 4 1 4 0 ? 22 Vandermonde system 32 / 43

  55. Polynomial Interpolation Evaluate x ∈{ 1 , 2 , 3 , 4 } 30 20 p ( x ) = 2 x 3 − 3 x 2 − 17 x + 10 10 � 2 � 1 1 2 3 4 � 10 � 20 Interpolate − 1  1 3 1 2 1 1 1 0      2 − 8 2 3 2 2 2 1 2 0 − 3 − 20        =    3 3 3 2 3 1 3 0    − 17 − 14      4 3 4 2 4 1 4 0 10 22 Vandermonde system 32 / 43

  56. Interpolating Univariate Polynomials Let p d ( X ) = c 0 + c 1 X + · · · + c d X d ∈ Z [ X ]. Can interpolate p d ( X ) from p d ( x 0 ) , p d ( x 1 ) , . . . , p d ( x d ) � x 0 , x 1 , . . . , x d are distinct ( x 0 ) 0 ( x 0 ) 1  ( x 0 ) d      · · · c 0 p d ( x 0 ) ( x 1 ) 0 ( x 1 ) 1 ( x 1 ) d · · · c 1 p d ( x 1 )        =  . . .   .   .  ... . . . . .       . . . . .      ( x d ) 0 ( x d ) 1 ( x d ) d · · · c d p d ( x d ) Vandermonde system 33 / 43

  57. Interpolating Univariate Polynomials Let p d ( X ) = c 0 + c 1 X + · · · + c d X d ∈ Z [ X ]. ∀ d ∈ N , Can interpolate p d ( X ) from p d ( x 0 ) , p d ( x 1 ) , . . . , p d ( x d ) � x 0 , x 1 , . . . are distinct ( x 0 ) 0 ( x 0 ) 1  ( x 0 ) d      · · · c 0 p d ( x 0 ) ( x 1 ) 0 ( x 1 ) 1 ( x 1 ) d · · · c 1 p d ( x 1 )        =  . . .   .   .  ... . . . . .       . . . . .      ( x d ) 0 ( x d ) 1 ( x d ) d · · · c d p d ( x d ) Vandermonde system 33 / 43

  58. Interpolating Univariate Polynomials Let p d ( X ) = c 0 + c 1 X + · · · + c d X d ∈ Z [ X ]. ∀ d ∈ N , Can interpolate p d ( X ) from p d ( x 0 ) , p d ( x 1 ) , . . . , p d ( x d ) � x 0 , x 1 , . . . are distinct ( x 0 ) 0 ( x 0 ) 1 ( x 0 ) d p d ( x 0 )       · · · c 0 ( x 1 ) 0 ( x 1 ) 1 ( x 1 ) d p d ( x 1 ) · · · c 1        =  . . .   .   .  ... . . . . .       . . . . .      ( x d ) 0 ( x d ) 1 ( x d ) d p d ( x d ) · · · c d Vandermonde system 33 / 43

  59. Interpolating Univariate Polynomials Let p d ( X ) = c 0 + c 1 X + · · · + c d X d ∈ Z [ X ]. ∀ d ∈ N , Can interpolate p d ( X ) from p d ( x 0 ) , p d ( x 1 ) , . . . , p d ( x d ) � x 0 , x 1 , . . . are distinct � x is not a root of unity ( x 0 ) 0 ( x 0 ) 1 ( x 0 ) d p d ( x 0 )       · · · c 0 ( x 1 ) 0 ( x 1 ) 1 ( x 1 ) d p d ( x 1 ) · · · c 1        =  . . .   .   .  ... . . . . .       . . . . .      ( x d ) 0 ( x d ) 1 ( x d ) d p d ( x d ) · · · c d Vandermonde system 33 / 43

  60. Interpolating Multivariate Polynomials Let p d ( X , Y , Z ) = c 0 , 0 , d X 0 Y 0 Z d + · · · + c d , 0 , 0 X d Y 0 Z 0 ∈ Z [ X , Y , Z ] be a homogeneous multivariate polynomial of degree d . ∀ d ∈ N , Can interpolate p d ( X , Y , Z ) from p d ( x 0 , y 0 , z 0 ) , p d ( x 1 , y 1 , z 1 ) , . . . � ? ( x 0 ) 0 ( y 0 ) 0 ( z 0 ) d ( x 0 ) d ( y 0 ) 0 ( z 0 ) 0       · · · c 0 , 0 , d p d ( x 0 , y 0 , z 0 ) . ( x 1 ) 0 ( y 1 ) 0 ( z 1 ) d ( x 1 ) d ( y 1 ) 0 ( z 1 ) 0 · · · p d ( x 1 , y 1 , z 1 ) .  =       . . . . .      . . . . . . . c d , 0 , 0 . 34 / 43

  61. Interpolating Multivariate Polynomials Let p d ( X , Y , Z ) = c 0 , 0 , d X 0 Y 0 Z d + · · · + c d , 0 , 0 X d Y 0 Z 0 ∈ Z [ X , Y , Z ] be a homogeneous multivariate polynomial of degree d . ∀ d ∈ N , Can interpolate p d ( X , Y , Z ) from p d ( x 0 , y 0 , z 0 ) , p d ( x 1 , y 1 , z 1 ) , . . . � ( x 0 ) 0 ( y 0 ) 0 ( z 0 ) d ( x 0 ) d ( y 0 ) 0 ( z 0 ) 0 p d ( x 0 , y 0 , z 0 )       · · · c 0 , 0 , d . ( x 1 ) 0 ( y 1 ) 0 ( z 1 ) d ( x 1 ) d ( y 1 ) 0 ( z 1 ) 0 p d ( x 1 , y 1 , z 1 ) · · · .  =       . . . . .      . . . . . . . c d , 0 , 0 . Vandermonde system 34 / 43

  62. Interpolating Multivariate Polynomials Let p d ( X , Y , Z ) = c 0 , 0 , d X 0 Y 0 Z d + · · · + c d , 0 , 0 X d Y 0 Z 0 ∈ Z [ X , Y , Z ] be a homogeneous multivariate polynomial of degree d . ∀ d ∈ N , Can interpolate p d ( X , Y , Z ) from p d ( x 0 , y 0 , z 0 ) , p d ( x 1 , y 1 , z 1 ) , . . . � ( x 0 y 0 z d ) 0 ( x d y 0 z 0 ) 0 p d ( x 0 , y 0 , z 0 )       · · · c 0 , 0 , d . ( x 0 y 0 z d ) 1 ( x d y 0 z 0 ) 1 p d ( x 1 , y 1 , z 1 ) · · · .  =       . . . . .      . . . . . . . c d , 0 , 0 . Vandermonde system 34 / 43

  63. Interpolating Multivariate Polynomials Let p d ( X , Y , Z ) = c 0 , 0 , d X 0 Y 0 Z d + · · · + c d , 0 , 0 X d Y 0 Z 0 ∈ Z [ X , Y , Z ] be a homogeneous multivariate polynomial of degree d . ∀ d ∈ N , Can interpolate p d ( X , Y , Z ) from p d ( x 0 , y 0 , z 0 ) , p d ( x 1 , y 1 , z 1 ) , . . . � lattice condition ( x 0 y 0 z d ) 0 ( x d y 0 z 0 ) 0 p d ( x 0 , y 0 , z 0 )       · · · c 0 , 0 , d . ( x 0 y 0 z d ) 1 ( x d y 0 z 0 ) 1 p d ( x 1 , y 1 , z 1 ) · · · .  =       . . . . .      . . . . . . . c d , 0 , 0 . Vandermonde system 34 / 43

  64. Lattice Condition Definition We say that λ 1 , λ 2 , . . . , λ ℓ ∈ C − { 0 } satisfy the lattice condition if ℓ ∀ x ∈ Z ℓ − { 0 } � with x i = 0 , i =1 ℓ we have � λ x i i � = 1 . i =1 35 / 43

  65. Lattice Condition Definition We say that λ 1 , λ 2 , . . . , λ ℓ ∈ C − { 0 } satisfy the lattice condition if ℓ ∀ x ∈ Z ℓ − { 0 } � with x i = 0 , i =1 ℓ we have � λ x i i � = 1 . i =1 Example (Easy) For any i , j , k ∈ Z such that i + j + k = 0 and ( i , j , k ) � = (0 , 0 , 0), it follows that 2 i 3 j 5 k � = 1 . 35 / 43

  66. Lattice Condition: Another Example Example (Medium) For any i , j , k ∈ Z such that i + j + k = 0 and ( i , j , k ) � = (0 , 0 , 0), it follows that √ √ � j � � k 1 i � 3 − 3 + 2 2 � = 1 . 36 / 43

  67. Lattice Condition: Another Example Example (Medium) For any i , j , k ∈ Z such that i + j + k = 0 and ( i , j , k ) � = (0 , 0 , 0), it follows that √ √ √ � j − k � j � � k 1 i � 7 k = 1 i � 3 − 3 + 2 3 + 2 2 � = 1 . 36 / 43

  68. Lattice Condition: Another Example Example (Medium) For any i , j , k ∈ Z such that i + j + k = 0 and ( i , j , k ) � = (0 , 0 , 0), it follows that √ √ √ � j − k � j � � k 1 i � 7 k = 1 i � 3 − 3 + 2 3 + 2 2 � = 1 . Suppose √ � j − k 1 i � 7 k = 1 . 3 + 2 36 / 43

  69. Lattice Condition: Another Example Example (Medium) For any i , j , k ∈ Z such that i + j + k = 0 and ( i , j , k ) � = (0 , 0 , 0), it follows that √ √ √ � j − k � j � � k 1 i � 7 k = 1 i � 3 − 3 + 2 3 + 2 2 � = 1 . Suppose √ � j − k 1 i � 7 k = 1 . 3 + 2 Then j − k = 0 k = 0 j = 0 i = 0 . Contradiction! 36 / 43

  70. “Hard” Lattice Condition Example Want to prove: For all integers y ≥ 4, the roots of p ( x , y ) = x 5 − (2 y + 1) x 3 − ( y 2 + 2) x 2 + ( y − 1) yx + y 3 . satisfy the lattice condition. 37 / 43

  71. “Hard” Lattice Condition Example Want to prove: For all integers y ≥ 4, the roots of p ( x , y ) = x 5 − (2 y + 1) x 3 − ( y 2 + 2) x 2 + ( y − 1) yx + y 3 . satisfy the lattice condition. Lemma Let p ( x ) ∈ Q [ x ] be a polynomial of degree n ≥ 2 . If the Galois group of p over Q is S n or A n and 1 the roots of p do not all have the same complex norm, 2 then the roots of p satisfy the lattice condition. 37 / 43

  72. Factorizations and Roots Galois group of p over Q is S n or A n 38 / 43

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