Decision problem that can be verified but not run in n^2 timeA Problem on Time Complexity of AlgorithmsCan we...
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Decision problem that can be verified but not run in n^2 time
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Decision problem that can be verified but not run in n^2 time
A Problem on Time Complexity of AlgorithmsCan we construct problems that can be solved in $Theta(n^c)$ time, and tested in $O(n)$ timeProblems that provably require quadratic timeIs there a decision algorithm with time complexity of Ө(n²)?Lower bounds and $P$ vs $NP$Is P the set of all algorithms whose run-time is $Oleft( n^{ O left( 1 right) }right)$?Does there exist a Turing-machine that runs in time $o(nlog n)$, but not $O(n)$?reducing $CLIQUE$ from decision to search problemHow is the Time Complexity of a Function Problem different than its corresponding Decision Problem?Show that NP is not equal to SPACE(n)
$begingroup$
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
4
$begingroup$
@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
4
$begingroup$
@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
4
$begingroup$
@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
4
4
$begingroup$
@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
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$begingroup$
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 3 hours ago
D.W.♦D.W.
99.8k12122286
$endgroup$
add a comment |
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$begingroup$
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 3 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 3 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 3 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 3 hours ago
D.W.♦D.W.
99.8k12122286
answered 3 hours ago
D.W.♦D.W.
99.8k12122286
answered 3 hours ago
D.W.♦D.W.
99.8k12122286
answered 3 hours ago
D.W.♦D.W.
99.8k12122286
99.8k12122286
add a comment |
add a comment |
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$begingroup$
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
4
$begingroup$
@while1fork. It does matter if you're giving precise timing estimates, like $O(n^2)$.
$endgroup$
– Rick Decker
3 hours ago