Contradiction proof for inequality of P and NP? Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Proof for P-complete is not closed under intersectionProof of sum of powerset?Contradiction between best-case running time of insertion sort and $nlog n$ lower bound?bounded length CoNP proofLogarithmic Randomness is Necessary for PCP TheoremTrouble seeing the contradiction in diagonalization proofIs it always possible to have one part of the reduction?Is this language NP Hard?Testing algorithm for a modified sieve of EratosthenesFinding a complexity by solving inequality
Multiple fireplaces in an apartment building?
Expansion//Explosion and Siren Stormtamer
A Dictionary or Encyclopedia of Fantasy or Fairy Tales from the 1960s
Does Feeblemind produce an ongoing magical effect that can be dispelled?
Multiple options vs single option UI
Is there any hidden 'W' sound after 'comment' in : Comment est-elle?
What is the ongoing value of the Kanban board to the developers as opposed to management
Is a 5 watt UHF/VHF handheld considered QRP?
What's the difference between using dependency injection with a container and using a service locator?
Mistake in years of experience in resume?
How long after the last departure shall the airport stay open for an emergency return?
My admission is revoked after accepting the admission offer
Trumpet valves, lengths, and pitch
Do I need to protect SFP ports and optics from dust/contaminants? If so, how?
Can I criticise the more senior developers around me for not writing clean code?
What ability score does a Hexblade's Pact Weapon use for attack and damage when wielded by another character?
Co-worker works way more than he should
What is it called when you ride around on your front wheel?
Holes in ElementMesh with ToElementMesh of ImplicitRegion
Are there moral objections to a life motivated purely by money? How to sway a person from this lifestyle?
Will I lose my paid in full property
What is this word supposed to be?
Flattening the sub-lists
Additive group of local rings
Contradiction proof for inequality of P and NP?
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Proof for P-complete is not closed under intersectionProof of sum of powerset?Contradiction between best-case running time of insertion sort and $nlog n$ lower bound?bounded length CoNP proofLogarithmic Randomness is Necessary for PCP TheoremTrouble seeing the contradiction in diagonalization proofIs it always possible to have one part of the reduction?Is this language NP Hard?Testing algorithm for a modified sieve of EratosthenesFinding a complexity by solving inequality
$begingroup$
I'm trying to argue that N is not equal NP using hierarchy theorems. This is my argument, but when I showed it to our teacher and after deduction, he said that this is problematic where I can't find a compelling reason to accept.
We start off by assuming that $P=NP$. Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$. As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$. On the contrary, the time hierarchy theorem states that there should be a language $A in TIME(n^k+1)$, that's not in $TIME(n^k)$. This would lead us to conclude that $A$ is in $P$, while not in $NP$, which is a contradiction to our first assumption. So, we came to the conclusion that $P neq NP$.
Is there something wrong with my proof? I was struggling for hours before asking this, though!
complexity-theory time-complexity
asked 2 hours ago
inverted_indexinverted_index
1384
$endgroup$
add a comment |
$begingroup$
I'm trying to argue that N is not equal NP using hierarchy theorems. This is my argument, but when I showed it to our teacher and after deduction, he said that this is problematic where I can't find a compelling reason to accept.
We start off by assuming that $P=NP$. Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$. As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$. On the contrary, the time hierarchy theorem states that there should be a language $A in TIME(n^k+1)$, that's not in $TIME(n^k)$. This would lead us to conclude that $A$ is in $P$, while not in $NP$, which is a contradiction to our first assumption. So, we came to the conclusion that $P neq NP$.
Is there something wrong with my proof? I was struggling for hours before asking this, though!
complexity-theory time-complexity
asked 2 hours ago
inverted_indexinverted_index
1384
$endgroup$
add a comment |
$begingroup$
I'm trying to argue that N is not equal NP using hierarchy theorems. This is my argument, but when I showed it to our teacher and after deduction, he said that this is problematic where I can't find a compelling reason to accept.
We start off by assuming that $P=NP$. Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$. As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$. On the contrary, the time hierarchy theorem states that there should be a language $A in TIME(n^k+1)$, that's not in $TIME(n^k)$. This would lead us to conclude that $A$ is in $P$, while not in $NP$, which is a contradiction to our first assumption. So, we came to the conclusion that $P neq NP$.
Is there something wrong with my proof? I was struggling for hours before asking this, though!
complexity-theory time-complexity
asked 2 hours ago
inverted_indexinverted_index
1384
$endgroup$
I'm trying to argue that N is not equal NP using hierarchy theorems. This is my argument, but when I showed it to our teacher and after deduction, he said that this is problematic where I can't find a compelling reason to accept.
We start off by assuming that $P=NP$. Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$. As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$. On the contrary, the time hierarchy theorem states that there should be a language $A in TIME(n^k+1)$, that's not in $TIME(n^k)$. This would lead us to conclude that $A$ is in $P$, while not in $NP$, which is a contradiction to our first assumption. So, we came to the conclusion that $P neq NP$.
Is there something wrong with my proof? I was struggling for hours before asking this, though!
complexity-theory time-complexity
complexity-theory time-complexity
asked 2 hours ago
inverted_indexinverted_index
1384
asked 2 hours ago
inverted_indexinverted_index
1384
asked 2 hours ago
inverted_indexinverted_index
1384
asked 2 hours ago
inverted_indexinverted_index
1384
asked 2 hours ago
inverted_indexinverted_index
1384
1384
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$.
Sure.
As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$.
No. Polynomial time reductions aren't free. We can say it takes $O(n^r(L))$ time to reduce language $L$ to $SAT$, where $r(L)$ is the exponent in the polynomial time reduction used. This is where your argument falls apart. There is no finite $k$ such that for all $L in NP$ we have $r(L) < k$. At least this does not follow from $P = NP$ and would be a much stronger statement.
And this stronger statement does indeed conflict with the time hierarchy theorem, which tells us that $P$ can not collapse into $TIME(n^k)$, let alone all of $NP$.
answered 1 hour ago
orlporlp
6,1351826
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "419"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f108496%2fcontradiction-proof-for-inequality-of-p-and-np%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$.
Sure.
As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$.
No. Polynomial time reductions aren't free. We can say it takes $O(n^r(L))$ time to reduce language $L$ to $SAT$, where $r(L)$ is the exponent in the polynomial time reduction used. This is where your argument falls apart. There is no finite $k$ such that for all $L in NP$ we have $r(L) < k$. At least this does not follow from $P = NP$ and would be a much stronger statement.
And this stronger statement does indeed conflict with the time hierarchy theorem, which tells us that $P$ can not collapse into $TIME(n^k)$, let alone all of $NP$.
answered 1 hour ago
orlporlp
6,1351826
$endgroup$
add a comment |
$begingroup$
Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$.
Sure.
As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$.
No. Polynomial time reductions aren't free. We can say it takes $O(n^r(L))$ time to reduce language $L$ to $SAT$, where $r(L)$ is the exponent in the polynomial time reduction used. This is where your argument falls apart. There is no finite $k$ such that for all $L in NP$ we have $r(L) < k$. At least this does not follow from $P = NP$ and would be a much stronger statement.
And this stronger statement does indeed conflict with the time hierarchy theorem, which tells us that $P$ can not collapse into $TIME(n^k)$, let alone all of $NP$.
answered 1 hour ago
orlporlp
6,1351826
$endgroup$
add a comment |
$begingroup$
Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$.
Sure.
As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$.
No. Polynomial time reductions aren't free. We can say it takes $O(n^r(L))$ time to reduce language $L$ to $SAT$, where $r(L)$ is the exponent in the polynomial time reduction used. This is where your argument falls apart. There is no finite $k$ such that for all $L in NP$ we have $r(L) < k$. At least this does not follow from $P = NP$ and would be a much stronger statement.
And this stronger statement does indeed conflict with the time hierarchy theorem, which tells us that $P$ can not collapse into $TIME(n^k)$, let alone all of $NP$.
answered 1 hour ago
orlporlp
6,1351826
$endgroup$
Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$.
Sure.
As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$.
No. Polynomial time reductions aren't free. We can say it takes $O(n^r(L))$ time to reduce language $L$ to $SAT$, where $r(L)$ is the exponent in the polynomial time reduction used. This is where your argument falls apart. There is no finite $k$ such that for all $L in NP$ we have $r(L) < k$. At least this does not follow from $P = NP$ and would be a much stronger statement.
And this stronger statement does indeed conflict with the time hierarchy theorem, which tells us that $P$ can not collapse into $TIME(n^k)$, let alone all of $NP$.
answered 1 hour ago
orlporlp
6,1351826
edited 47 mins ago
answered 1 hour ago
orlporlp
6,1351826
answered 1 hour ago
orlporlp
6,1351826
answered 1 hour ago
orlporlp
6,1351826
6,1351826
add a comment |
add a comment |
Thanks for contributing an answer to Computer Science Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f108496%2fcontradiction-proof-for-inequality-of-p-and-np%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
