What does a proof that Co-NP =P entail for the NP versus Co-NP question
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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 Dec 29 '18 at 4:47
AKASH VETRIVELAKASH VETRIVEL
193
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 Dec 29 '18 at 4:47
AKASH VETRIVELAKASH VETRIVEL
193
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 Dec 29 '18 at 4:47
AKASH VETRIVELAKASH VETRIVEL
193
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 Dec 29 '18 at 4:47
AKASH VETRIVELAKASH VETRIVEL
193
asked Dec 29 '18 at 4:47
AKASH VETRIVELAKASH VETRIVEL
193
asked Dec 29 '18 at 4:47
AKASH VETRIVELAKASH VETRIVEL
193
asked Dec 29 '18 at 4:47
AKASH VETRIVELAKASH VETRIVEL
193
asked Dec 29 '18 at 4:47
AKASH VETRIVELAKASH VETRIVEL
193
193
add a comment |
add a comment |
1 Answer
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$textP=textco-NP$ implies that $textco-P=textco-(co-NP)=textNP$. But
$textco-P=textP$: you can just swap the accept and reject states of a deterministic Turing machine to complement the language that it decides.
answered Dec 29 '18 at 5:42
David RicherbyDavid Richerby
66.2k15100190
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1 Answer
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1 Answer
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active
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votes
$textP=textco-NP$ implies that $textco-P=textco-(co-NP)=textNP$. But
$textco-P=textP$: you can just swap the accept and reject states of a deterministic Turing machine to complement the language that it decides.
answered Dec 29 '18 at 5:42
David RicherbyDavid Richerby
66.2k15100190
add a comment |
$textP=textco-NP$ implies that $textco-P=textco-(co-NP)=textNP$. But
$textco-P=textP$: you can just swap the accept and reject states of a deterministic Turing machine to complement the language that it decides.
answered Dec 29 '18 at 5:42
David RicherbyDavid Richerby
66.2k15100190
add a comment |
$textP=textco-NP$ implies that $textco-P=textco-(co-NP)=textNP$. But
$textco-P=textP$: you can just swap the accept and reject states of a deterministic Turing machine to complement the language that it decides.
answered Dec 29 '18 at 5:42
David RicherbyDavid Richerby
66.2k15100190
$textP=textco-NP$ implies that $textco-P=textco-(co-NP)=textNP$. But
$textco-P=textP$: you can just swap the accept and reject states of a deterministic Turing machine to complement the language that it decides.
answered Dec 29 '18 at 5:42
David RicherbyDavid Richerby
66.2k15100190
answered Dec 29 '18 at 5:42
David RicherbyDavid Richerby
66.2k15100190
answered Dec 29 '18 at 5:42
David RicherbyDavid Richerby
66.2k15100190
answered Dec 29 '18 at 5:42
David RicherbyDavid Richerby
66.2k15100190
66.2k15100190
add a comment |
add a comment |
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