Thats a unique idea.You should consider the following idea to enhance your idea so that it would impact uniquely: Complementation of finite automata on infinite words is not only a fundamental problem in automata theory, but also serves as a cornerstone for solving numerous decision problems in mathematical logic, model-checking, program analysis and verification. For Streett complementation, a significant gap exists beten the current lor bound 2^{Omega(n*log(n*k))} and upper bound 2^{O(n*k*log(n*k))}, where n is the state size, k is the number of Streett pairs, and k can be as large as 2^{n}. Determining the complexity of Streett complementation has been an open question since the late 80's.   show a complementation construction with upper bound 2^{O(n*log(n)+n*k*log(k))} for k=O(n) and 2^{O(n^{2}*log(n))} for k=Omega(n), which matches ll the lor bound obtained in the paper arXiv:1102.2963.  also obtain a tight upper bound 2^{O(n*log(n))} for parity complementation.
['Yang Cai', 'Ting Zhang']
I am writing a research paper on a idea Tight Upper Bounds for Streett and Parity Complementation
[{"from": "human", "value": "I am writing a research paper on a idea Tight Upper Bounds for Streett and Parity Complementation"}, {"from": "gpt", "value": "Thats a unique idea.You should consider the following idea to enhance your idea so that it would impact uniquely: Complementation of finite automata on infinite words is not only a fundamental problem in automata theory, but also serves as a cornerstone for solving numerous decision problems in mathematical logic, model-checking, program analysis and verification. For Streett complementation, a significant gap exists beten the current lor bound 2^{Omega(n*log(n*k))} and upper bound 2^{O(n*k*log(n*k))}, where n is the state size, k is the number of Streett pairs, and k can be as large as 2^{n}. Determining the complexity of Streett complementation has been an open question since the late 80's.   show a complementation construction with upper bound 2^{O(n*log(n)+n*k*log(k))} for k=O(n) and 2^{O(n^{2}*log(n))} for k=Omega(n), which matches ll the lor bound obtained in the paper arXiv:1102.2963.  also obtain a tight upper bound 2^{O(n*log(n))} for parity complementation."}]