Contradiction proof for inequality of P and NP?
$begingroup$
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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
inverted_indexinverted_index
1384
asked 1 hour ago
inverted_indexinverted_index
1384
asked 1 hour ago
inverted_indexinverted_index
1384
asked 1 hour ago
inverted_indexinverted_index
1384
asked 1 hour 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
6,1251826
$endgroup$
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
6,1251826
$endgroup$
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
6,1251826
$endgroup$
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
6,1251826
$endgroup$
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
6,1251826
edited 10 mins ago
answered 1 hour ago
orlporlp
6,1251826
answered 1 hour ago
orlporlp
6,1251826
answered 1 hour ago
orlporlp
6,1251826
6,1251826
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
