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    If TWO PLAYER SAT were complete for PH, then TWO PLAYER S... — Carmelics
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    Home/Modality & Possibility
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    Supports→It is widely believed that PH is a proper subset of PSPACE

    If TWO PLAYER SAT were complete for PH, then TWO PLAYER SAT would be in Sigma^P_k for some k

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    If PH = PSPACE, then TWO PLAYER SAT would be complete for PHIf TWO PLAYER SAT were in Sigma^P_k, determining winning strategies for n-round ...It is widely believed that PH is a proper subset of PSPACEThe conclusion that n-round game difficulty collapses to k-round game difficulty...

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    If PH = PSPACE, then TWO PLAYER SAT would be complete for PH (since it...85%If PH = PSPACE, then TWO PLAYER SAT would be complete for PH84%TWO PLAYER SAT_n is complete for Σ^P_n in the Polynomial Hierarchy80%TWO PLAYER SAT_n is complete for Sigma^P_n in the Polynomial Hierarchy80%

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    SEP: computational-complexity
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    \(\sc{TWO}\ \sc{PLAYER}\ \sc{SAT}_n\) may be shown to be complete for the class \(\Sigma^P_n\) in the Polynomial Hierarchy. Note, however, as the value of \(n\) increases, we expect that it should become more difficult to decide membership in \(\sc{TWO}\ \sc{PLAYER}\ \sc{SAT}_n\) in much the same way that it appears to become more difficult to determine whether a given player has a winning strategy for increasingly long games of Go or chess. This observation provides part of the reason why it is

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