Some Complete and Intermediate Polynomials in Algebraic Complexity Theory

Abstract

We provide a list of new natural V N P-intermediate polynomial families, based on basic (combinatorial) N P-complete problems that are complete under parsimonious reductions. Over finite fields, these families are in V N P, and under the plausible hypothesis M o d _p P ⫅̸ P / p o l y, are neither V N P-hard (even under oracle-circuit reductions) nor in V P. Prior to this, only the Cut Enumerator polynomial was known to be V N P-intermediate, as shown by Bürgisser in 2000. We show next that over rationals and reals, the clique polynomial cannot be obtained as a monotone p-projection of the permanent polynomial, thus ruling out the possibility of transferring monotone clique lower bounds to the permanent. We also show that two of our intermediate polynomials, based on satisfiability and Hamiltonian cycle, are not monotone affine polynomial-size projections of the permanent. These results augment recent results along this line due to Grochow. Finally, we describe a (somewhat natural) polynomial defined independent of a computation model, and show that it is V P-complete under polynomial-size projections. This complements a recent result of Durand et al. (2014) which established V P-completeness of a related polynomial but under constant-depth oracle circuit reductions. Both polynomials are based on graph homomorphisms. A simple restriction yields a family similarly complete for V B P.

Appendix

In this appendix we prove that the graphs G _i , i ∈ {1,2,3}, from Fig. 1, are rigid and pairwise incomparable. We briefly recall the construction of these graphs. For the graph G, in Fig. 1, there is an edge between i and j if 1⩽|i − j|⩽4. Further add an edge between 1 and 16. The G _i ’s are obtained, as shown in Fig. 1, by adding an extra edge between 1 and 7 + i. We state some definitions that will be useful to us in the proof.

Definition 3

A graph H is asymmetric if the only automorphism (isomorphism from H to itself) is the identity.

Definition 4

A graph H is a core if every endomorphism (homomorphism from H to itself) is an isomorphism (and hence an automorphism).

Recall a graph H is rigid if the only endomorphism is the identity. Thus, H is rigid if and only if it is an asymmetric core.

Let χ _H denote the chromatic number of H, that is, the least k such that some map from V(H) to the set of colours [k] gives all adjacent vertices distinct colours. We say that H is χ(H)-chromatic. A graph H is said to be vertex-critical if for every u ∈ V(H), χ _H∖{u}<χ _H . If there is a homomorphism from G to H, then the definition of homomorphism implies that χ(G)⩽χ(H). It follows that every vertex-critical graph is a core.

Claim 1

Each graph in {G, G ₁, G ₂, G ₃} is a core.

Claim 2

Each graph in {G ₁, G ₂, G ₃} is asymmetric.

Hence, each G _i is rigid.

Claim 3

The graphs in {G ₁, G ₂, G ₃} are pairwise incomparable; for i ≠ j, there is no homomorphism from G _i to G _j .

Proof of Claim 1

We show that G (and hence also each G _i ) is not 5-colourable, while for every u ∈ [16], each G _i ∖{u} is 5-colourable. Hence all four graphs are 6-chromatic vertex-critical.

Non-5-colourability : The vertices 1 to 5 form a clique and must get distinct colours, say vertex i gets the colour c _i for i ∈ [5]. Now there is a unique way of extending the colouring sequentially to vertices 6,7,8,…, if we use only five colours. But this assigns colour c ₁ to 16, and vertices 1 and 16 are neighbours. So no 5-colouring is possible.

5-colourability : Consider G _i ∖{u}. Colour node j with colour c _{j mod 5} if j<u, with colour c _{(j−1) mod 5} if j > u. This satisfies all edge constraints: For a black edge (j, k), 1⩽|j − k|⩽4, so if both j and k are present, then their colours are distinct even if j<u<k. If the blue-red edge is present, note that the red vertex gets colour c ₂, c ₃, c ₄, or c ₅, while vertex 1 always gets colour c ₁.

Proof of Claim 2

Since isomorphisms must preserve degrees vertex-wise, consider the degrees of vertices in the graphs. First, group the vertices of G by degree. degree 5:{1,2,15,16} degree 6:{3,14} degree 7:{4,13} degree 8:{5,6,7,8,9,10, 11,12}.

Similarly, group the vertices of G _i by degree. degree 5:{2,15,16} degree 6:{1,3,14} degree 7:{4,13} degree 8:{5,6,7,8,9,10, 11,12}∖{the red node 7+i} degree 9: the red node 7 + i

Consider an automorphism f on G ₁. Since only vertex 8 has degree 9, f must map 8 to 8. Vertex 1 is the only neighbour of 8 with degree 6, so f must map 1 to 1. Vertex 1 has two degree-5 neighbours, 2 and 16, but 16 has another degree-5 neighbour 15 while 2 does not have any degree-5 neighbour, so f cannot swap these degree-5 neighbours of 1. So f maps 2 to 2 and 16 to 16. Proceeding this way based on degree, we see that f must in fact fix every vertex.

An identical argument works for G ₂. For G ₃, one additional twist: The red vertex 10 gets mapped to 10. Now 10 has two degree-6 neighbours, 1 and 14. Can f map 1 to 14? No, since 1 has a degree-6 neighbour 3, while 14 has no degree-6 neighbour. Thus f cannot swap 1 and 14.

Proof of Claim 3

Suppose to the contrary that f:V ₁→V ₂ is a homomorphism from G ₁ to G ₂ (the argument is similar for other pairs). If f is not surjective, then by vertex-criticality, G ₁ has a homomorphism to a 5-colourable graph, but χ(G ₁) = 6, a contradiction. So f must be surjective.

Furthermore, f must induce a bijection between the edges of G ₁ and G ₂. If it didn’t, then two edges of G ₁ are mapped to the same edge of G ₂. This implies that two vertices of G ₁ are mapped to the same vertex of G ₂, violating surjectivity.

Thus the vertex degrees must be preserved exactly: for each u ∈ V ₁, the degree of u in G ₁ is the same as the degree of f(u) in G ₂.

Since the red vertices are the only vertices with degree 9, f must map the red vertex of G ₁, vertex 8, to the red vertex of G ₂, vertex 9. Now use the argument as used in Claim 2 to extend this mapping. f must map 1 to 1, 2 to 2, and so on. We thus reach the conclusion that f must map 8 to 8, contradicting f(8)=9. Hence no such map f is possible.