Thats a unique idea.You should consider the following idea to enhance your idea so that it would impact uniquely:  show that the quantum query complexity of evaluating nand-tree instances with average choice complexity at most W is O(W), where average choice complexity is a measure of the difficulty of winning the associated two-player game. This generalizes a superpolynomial speedup over classical query complexity due to Zhan et al.  further show that the player with a winning strategy for the two-player game associated with the nand-tree can win the game with an expected O(N^(1/4) √C(x)) quantum queries against a random opponent, where C(x) is the average choice complexity of the instance. This gives an improvement over the query#R##N#complexity of the naive strategy, which costs O(√N) queries.#R##N#The results rely on a connection beten nand-tree evaluation and st-connectivity problems on certain graphs, and span programs for st-connectivity problems. Our results follow from relating average choice complexity to the effective resistance of these graphs, which itself#R##N#corresponds to the span program witness size.
['Stacey Jeffery', 'Shelby Kimmel']
I am writing a research paper on a idea NAND-Trees, Average Choice Complexity, and Effective Resistance
[{"from": "human", "value": "I am writing a research paper on a idea NAND-Trees, Average Choice Complexity, and Effective Resistance"}, {"from": "gpt", "value": "Thats a unique idea.You should consider the following idea to enhance your idea so that it would impact uniquely:  show that the quantum query complexity of evaluating nand-tree instances with average choice complexity at most W is O(W), where average choice complexity is a measure of the difficulty of winning the associated two-player game. This generalizes a superpolynomial speedup over classical query complexity due to Zhan et al.  further show that the player with a winning strategy for the two-player game associated with the nand-tree can win the game with an expected O(N^(1/4) \u221aC(x)) quantum queries against a random opponent, where C(x) is the average choice complexity of the instance. This gives an improvement over the query#R##N#complexity of the naive strategy, which costs O(\u221aN) queries.#R##N#The results rely on a connection beten nand-tree evaluation and st-connectivity problems on certain graphs, and span programs for st-connectivity problems. Our results follow from relating average choice complexity to the effective resistance of these graphs, which itself#R##N#corresponds to the span program witness size."}]