Thats a unique idea.You should consider the following idea to enhance your idea so that it would impact uniquely:  prove a complexity dichotomy theorem for counting constraint satisfaction problems (#CSPs) with nonnegative and algebraic ights. This caps a long series of important results on counting problems including counting unighted and ighted graph homomorphisms and the celebrated dichotomy theorem for unighted #CSPs. Our dichotomy theorem gives a succinct criterion for tractability. If a set $\mathcal{F}$ of constraint functions satisfies this criterion, then the problem #CSP$(\mathcal{F})$ defined by $\mathcal{F}$ is solvable in polynomial time; if $\mathcal{F}$ does not satisfy this criterion, then the problem is #P-hard. Furthermore,  show that the question of whether a given $\mathcal{F}$ satisfies the criterion or not is decidable in NP. Surprisingly, our tractability criterion is simpler than the previous criteria for the more restricted classes of counting problems, although when specialized to those classes, they are logically equivalent. Our proof mainly uses linear algebra and represents a ...
['Jin-Yi Cai', 'Xi Chen', 'Pinyan Lu']
I am writing a research paper on a idea Nonnegative Weighted #CSP: An Effective Complexity Dichotomy
[{"from": "human", "value": "I am writing a research paper on a idea Nonnegative Weighted #CSP: An Effective Complexity Dichotomy"}, {"from": "gpt", "value": "Thats a unique idea.You should consider the following idea to enhance your idea so that it would impact uniquely:  prove a complexity dichotomy theorem for counting constraint satisfaction problems (#CSPs) with nonnegative and algebraic ights. This caps a long series of important results on counting problems including counting unighted and ighted graph homomorphisms and the celebrated dichotomy theorem for unighted #CSPs. Our dichotomy theorem gives a succinct criterion for tractability. If a set $\\mathcal{F}$ of constraint functions satisfies this criterion, then the problem #CSP$(\\mathcal{F})$ defined by $\\mathcal{F}$ is solvable in polynomial time; if $\\mathcal{F}$ does not satisfy this criterion, then the problem is #P-hard. Furthermore,  show that the question of whether a given $\\mathcal{F}$ satisfies the criterion or not is decidable in NP. Surprisingly, our tractability criterion is simpler than the previous criteria for the more restricted classes of counting problems, although when specialized to those classes, they are logically equivalent. Our proof mainly uses linear algebra and represents a ..."}]