Decision problem that can be verified but not run in n^2 time
$begingroup$
A much much weaker idea similar to the P=NP question, is there a decision problem that can be verified in $O(n^2)$ time, but it can be proven that there is no algorithm that decides it $O(n^2)$ time?
time-complexity
asked 4 hours ago
while1forkwhile1fork
183
$endgroup$
add a comment |
$begingroup$
A much much weaker idea similar to the P=NP question, is there a decision problem that can be verified in $O(n^2)$ time, but it can be proven that there is no algorithm that decides it $O(n^2)$ time?
time-complexity
asked 4 hours ago
while1forkwhile1fork
183
$endgroup$
$begingroup$
In what model of computation?
$endgroup$
– Curtis F
3 hours ago
$begingroup$
@CurtisF if it matters then please write that up as an answer. I was under the impression that that sort of thing didn't matter within reason, or else how can one ask "does P=NP"?
$endgroup$
– while1fork
3 hours ago
3
$begingroup$
@while1fork. It does matter if you're giving precise timing estimates, like $O(n^2)$.
$endgroup$
– Rick Decker
3 hours ago
add a comment |
$begingroup$
A much much weaker idea similar to the P=NP question, is there a decision problem that can be verified in $O(n^2)$ time, but it can be proven that there is no algorithm that decides it $O(n^2)$ time?
time-complexity
asked 4 hours ago
while1forkwhile1fork
183
$endgroup$
A much much weaker idea similar to the P=NP question, is there a decision problem that can be verified in $O(n^2)$ time, but it can be proven that there is no algorithm that decides it $O(n^2)$ time?
time-complexity
time-complexity
asked 4 hours ago
while1forkwhile1fork
183
asked 4 hours ago
while1forkwhile1fork
183
edited 4 hours ago
while1fork
asked 4 hours ago
while1forkwhile1fork
183
asked 4 hours ago
while1forkwhile1fork
183
asked 4 hours ago
while1forkwhile1fork
183
183
$begingroup$
In what model of computation?
$endgroup$
– Curtis F
3 hours ago
$begingroup$
@CurtisF if it matters then please write that up as an answer. I was under the impression that that sort of thing didn't matter within reason, or else how can one ask "does P=NP"?
$endgroup$
– while1fork
3 hours ago
3
$begingroup$
@while1fork. It does matter if you're giving precise timing estimates, like $O(n^2)$.
$endgroup$
– Rick Decker
3 hours ago
add a comment |
$begingroup$
In what model of computation?
$endgroup$
– Curtis F
3 hours ago
$begingroup$
@CurtisF if it matters then please write that up as an answer. I was under the impression that that sort of thing didn't matter within reason, or else how can one ask "does P=NP"?
$endgroup$
– while1fork
3 hours ago
3
$begingroup$
@while1fork. It does matter if you're giving precise timing estimates, like $O(n^2)$.
$endgroup$
– Rick Decker
3 hours ago
$begingroup$
In what model of computation?
$endgroup$
– Curtis F
3 hours ago
$begingroup$
In what model of computation?
$endgroup$
– Curtis F
3 hours ago
$begingroup$
@CurtisF if it matters then please write that up as an answer. I was under the impression that that sort of thing didn't matter within reason, or else how can one ask "does P=NP"?
$endgroup$
– while1fork
3 hours ago
$begingroup$
@CurtisF if it matters then please write that up as an answer. I was under the impression that that sort of thing didn't matter within reason, or else how can one ask "does P=NP"?
$endgroup$
– while1fork
3 hours ago
3
3
$begingroup$
@while1fork. It does matter if you're giving precise timing estimates, like $O(n^2)$.
$endgroup$
– Rick Decker
3 hours ago
$begingroup$
@while1fork. It does matter if you're giving precise timing estimates, like $O(n^2)$.
$endgroup$
– Rick Decker
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The answer depends a lot on the model of computation.
On one extreme: in the decision tree model, many lower bounds can be proven. For instance, consider the 3SUM problem. You can verify an alleged solution to the 3SUM problem in $O(n)$ time, but it's conjectured that no algorithm can solve the 3SUM problem in $O(n^{2-epsilon})$ time for any $epsilon>0$. One can prove this holds in some version of the algebraic decision tree model; one can prove that any algorithm to solve it in this model must take $Omega(n^2)$ time. This provides a gap of $O(n^2)$ vs $O(n)$ for solving vs verifying solutions, if the verifier is limited to the algebraic decision tree model. This is a pretty limited result; in this case, the model means that we're restricting attention to algorithms of a particular form. So, it doesn't rule out the possibility of some other algorithm (that does something weird) being faster.
On the other extreme: if we allow arbitrary (non-uniform) boolean circuits, then there is no explicitly stated function $f$ on $n$ bits where we can currently prove that every circuit for computing $f$ needs $ge 3.1n$ gates. In other words, this is saying that we have no clue how to prove lower bounds for circuits. Roughly speaking, we have no known result of a problem where we can prove that solving it requires circuits of size $omega(n)$. See, e.g., https://cstheory.stackexchange.com/a/8005/5038 and https://cstheory.stackexchange.com/q/21400/5038. So, given our current state of knowledge, we have no hope of proving a result like that if the model of computation is unrestricted boolean circuits.
answered 2 hours ago
D.W.♦D.W.
99.8k12122286
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f104894%2fdecision-problem-that-can-be-verified-but-not-run-in-n2-time%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
$begingroup$
The answer depends a lot on the model of computation.
On one extreme: in the decision tree model, many lower bounds can be proven. For instance, consider the 3SUM problem. You can verify an alleged solution to the 3SUM problem in $O(n)$ time, but it's conjectured that no algorithm can solve the 3SUM problem in $O(n^{2-epsilon})$ time for any $epsilon>0$. One can prove this holds in some version of the algebraic decision tree model; one can prove that any algorithm to solve it in this model must take $Omega(n^2)$ time. This provides a gap of $O(n^2)$ vs $O(n)$ for solving vs verifying solutions, if the verifier is limited to the algebraic decision tree model. This is a pretty limited result; in this case, the model means that we're restricting attention to algorithms of a particular form. So, it doesn't rule out the possibility of some other algorithm (that does something weird) being faster.
On the other extreme: if we allow arbitrary (non-uniform) boolean circuits, then there is no explicitly stated function $f$ on $n$ bits where we can currently prove that every circuit for computing $f$ needs $ge 3.1n$ gates. In other words, this is saying that we have no clue how to prove lower bounds for circuits. Roughly speaking, we have no known result of a problem where we can prove that solving it requires circuits of size $omega(n)$. See, e.g., https://cstheory.stackexchange.com/a/8005/5038 and https://cstheory.stackexchange.com/q/21400/5038. So, given our current state of knowledge, we have no hope of proving a result like that if the model of computation is unrestricted boolean circuits.
answered 2 hours ago
D.W.♦D.W.
99.8k12122286
$endgroup$
add a comment |
$begingroup$
The answer depends a lot on the model of computation.
On one extreme: in the decision tree model, many lower bounds can be proven. For instance, consider the 3SUM problem. You can verify an alleged solution to the 3SUM problem in $O(n)$ time, but it's conjectured that no algorithm can solve the 3SUM problem in $O(n^{2-epsilon})$ time for any $epsilon>0$. One can prove this holds in some version of the algebraic decision tree model; one can prove that any algorithm to solve it in this model must take $Omega(n^2)$ time. This provides a gap of $O(n^2)$ vs $O(n)$ for solving vs verifying solutions, if the verifier is limited to the algebraic decision tree model. This is a pretty limited result; in this case, the model means that we're restricting attention to algorithms of a particular form. So, it doesn't rule out the possibility of some other algorithm (that does something weird) being faster.
On the other extreme: if we allow arbitrary (non-uniform) boolean circuits, then there is no explicitly stated function $f$ on $n$ bits where we can currently prove that every circuit for computing $f$ needs $ge 3.1n$ gates. In other words, this is saying that we have no clue how to prove lower bounds for circuits. Roughly speaking, we have no known result of a problem where we can prove that solving it requires circuits of size $omega(n)$. See, e.g., https://cstheory.stackexchange.com/a/8005/5038 and https://cstheory.stackexchange.com/q/21400/5038. So, given our current state of knowledge, we have no hope of proving a result like that if the model of computation is unrestricted boolean circuits.
answered 2 hours ago
D.W.♦D.W.
99.8k12122286
$endgroup$
add a comment |
$begingroup$
The answer depends a lot on the model of computation.
On one extreme: in the decision tree model, many lower bounds can be proven. For instance, consider the 3SUM problem. You can verify an alleged solution to the 3SUM problem in $O(n)$ time, but it's conjectured that no algorithm can solve the 3SUM problem in $O(n^{2-epsilon})$ time for any $epsilon>0$. One can prove this holds in some version of the algebraic decision tree model; one can prove that any algorithm to solve it in this model must take $Omega(n^2)$ time. This provides a gap of $O(n^2)$ vs $O(n)$ for solving vs verifying solutions, if the verifier is limited to the algebraic decision tree model. This is a pretty limited result; in this case, the model means that we're restricting attention to algorithms of a particular form. So, it doesn't rule out the possibility of some other algorithm (that does something weird) being faster.
On the other extreme: if we allow arbitrary (non-uniform) boolean circuits, then there is no explicitly stated function $f$ on $n$ bits where we can currently prove that every circuit for computing $f$ needs $ge 3.1n$ gates. In other words, this is saying that we have no clue how to prove lower bounds for circuits. Roughly speaking, we have no known result of a problem where we can prove that solving it requires circuits of size $omega(n)$. See, e.g., https://cstheory.stackexchange.com/a/8005/5038 and https://cstheory.stackexchange.com/q/21400/5038. So, given our current state of knowledge, we have no hope of proving a result like that if the model of computation is unrestricted boolean circuits.
answered 2 hours ago
D.W.♦D.W.
99.8k12122286
$endgroup$
The answer depends a lot on the model of computation.
On one extreme: in the decision tree model, many lower bounds can be proven. For instance, consider the 3SUM problem. You can verify an alleged solution to the 3SUM problem in $O(n)$ time, but it's conjectured that no algorithm can solve the 3SUM problem in $O(n^{2-epsilon})$ time for any $epsilon>0$. One can prove this holds in some version of the algebraic decision tree model; one can prove that any algorithm to solve it in this model must take $Omega(n^2)$ time. This provides a gap of $O(n^2)$ vs $O(n)$ for solving vs verifying solutions, if the verifier is limited to the algebraic decision tree model. This is a pretty limited result; in this case, the model means that we're restricting attention to algorithms of a particular form. So, it doesn't rule out the possibility of some other algorithm (that does something weird) being faster.
On the other extreme: if we allow arbitrary (non-uniform) boolean circuits, then there is no explicitly stated function $f$ on $n$ bits where we can currently prove that every circuit for computing $f$ needs $ge 3.1n$ gates. In other words, this is saying that we have no clue how to prove lower bounds for circuits. Roughly speaking, we have no known result of a problem where we can prove that solving it requires circuits of size $omega(n)$. See, e.g., https://cstheory.stackexchange.com/a/8005/5038 and https://cstheory.stackexchange.com/q/21400/5038. So, given our current state of knowledge, we have no hope of proving a result like that if the model of computation is unrestricted boolean circuits.
answered 2 hours ago
D.W.♦D.W.
99.8k12122286
answered 2 hours ago
D.W.♦D.W.
99.8k12122286
answered 2 hours ago
D.W.♦D.W.
99.8k12122286
answered 2 hours ago
D.W.♦D.W.
99.8k12122286
99.8k12122286
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%2f104894%2fdecision-problem-that-can-be-verified-but-not-run-in-n2-time%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
$begingroup$
In what model of computation?
$endgroup$
– Curtis F
3 hours ago
$begingroup$
@CurtisF if it matters then please write that up as an answer. I was under the impression that that sort of thing didn't matter within reason, or else how can one ask "does P=NP"?
$endgroup$
– while1fork
3 hours ago
3
$begingroup$
@while1fork. It does matter if you're giving precise timing estimates, like $O(n^2)$.
$endgroup$
– Rick Decker
3 hours ago