What does a proof that Co-NP =P entail for the NP versus Co-NP question
What I wonder is what exactly would it entail. Would it,for instance imply that P=NP or would there be different consequences,I haven't found any assorted consequences so far in my research.
Thank You,
Akash
complexity-theory
asked 5 hours ago
AKASH VETRIVEL
111
add a comment |
What I wonder is what exactly would it entail. Would it,for instance imply that P=NP or would there be different consequences,I haven't found any assorted consequences so far in my research.
Thank You,
Akash
complexity-theory
asked 5 hours ago
AKASH VETRIVEL
111
add a comment |
What I wonder is what exactly would it entail. Would it,for instance imply that P=NP or would there be different consequences,I haven't found any assorted consequences so far in my research.
Thank You,
Akash
complexity-theory
asked 5 hours ago
AKASH VETRIVEL
111
What I wonder is what exactly would it entail. Would it,for instance imply that P=NP or would there be different consequences,I haven't found any assorted consequences so far in my research.
Thank You,
Akash
complexity-theory
complexity-theory
asked 5 hours ago
AKASH VETRIVEL
111
asked 5 hours ago
AKASH VETRIVEL
111
asked 5 hours ago
AKASH VETRIVEL
111
asked 5 hours ago
AKASH VETRIVEL
111
asked 5 hours ago
AKASH VETRIVEL
111
111
add a comment |
add a comment |
1 Answer
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$text{P}=text{co-NP}$ implies that $text{co-P}=text{co-(co-NP)}=text{NP}$. But
$text{co-P}=text{P}$: you can just swap the accept and reject states of a deterministic Turing machine to complement the language that it decides.
answered 4 hours ago
David Richerby
65.9k15100190
add a comment |
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1 Answer
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1 Answer
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$text{P}=text{co-NP}$ implies that $text{co-P}=text{co-(co-NP)}=text{NP}$. But
$text{co-P}=text{P}$: you can just swap the accept and reject states of a deterministic Turing machine to complement the language that it decides.
answered 4 hours ago
David Richerby
65.9k15100190
add a comment |
$text{P}=text{co-NP}$ implies that $text{co-P}=text{co-(co-NP)}=text{NP}$. But
$text{co-P}=text{P}$: you can just swap the accept and reject states of a deterministic Turing machine to complement the language that it decides.
answered 4 hours ago
David Richerby
65.9k15100190
add a comment |
$text{P}=text{co-NP}$ implies that $text{co-P}=text{co-(co-NP)}=text{NP}$. But
$text{co-P}=text{P}$: you can just swap the accept and reject states of a deterministic Turing machine to complement the language that it decides.
answered 4 hours ago
David Richerby
65.9k15100190
$text{P}=text{co-NP}$ implies that $text{co-P}=text{co-(co-NP)}=text{NP}$. But
$text{co-P}=text{P}$: you can just swap the accept and reject states of a deterministic Turing machine to complement the language that it decides.
answered 4 hours ago
David Richerby
65.9k15100190
answered 4 hours ago
David Richerby
65.9k15100190
answered 4 hours ago
David Richerby
65.9k15100190
answered 4 hours ago
David Richerby
65.9k15100190
65.9k15100190
add a comment |
add a comment |
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