If I'm just trying to show something is NP-hard (as opposed to NP-complete) does my reduction need to be...
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I have a problem that I believe is NP-hard. If I reduce a NP-complete problem to it in exponential time (and not polynomial time), does that prove the problem is NP-hard?
complexity-theory
asked Nov 25 at 22:38
cccompro
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I have a problem that I believe is NP-hard. If I reduce a NP-complete problem to it in exponential time (and not polynomial time), does that prove the problem is NP-hard?
complexity-theory
asked Nov 25 at 22:38
cccompro
82
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cccompro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a problem that I believe is NP-hard. If I reduce a NP-complete problem to it in exponential time (and not polynomial time), does that prove the problem is NP-hard?
complexity-theory
asked Nov 25 at 22:38
cccompro
82
New contributor
cccompro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have a problem that I believe is NP-hard. If I reduce a NP-complete problem to it in exponential time (and not polynomial time), does that prove the problem is NP-hard?
complexity-theory
complexity-theory
asked Nov 25 at 22:38
cccompro
82
New contributor
cccompro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 25 at 22:38
cccompro
82
New contributor
cccompro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 25 at 22:38
cccompro
82
New contributor
cccompro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 25 at 22:38
cccompro
82
asked Nov 25 at 22:38
cccompro
82
82
New contributor
cccompro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
cccompro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
cccompro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
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Hardness/completeness is always defined with respect to a specific kind of reduction.
Using exponential-time reductions doens't work at all, because that means your reduction is even more powerful than the thing you're reducing to.*
Every language except $emptyset$ and $Sigma^*$ is NP-hard under exponential-time reductions. To see this, let $L$ be any language except $emptyset$ and $Sigma^*$, and let $X$ be in NP. We can reduce $X$ to $L$ as follows. Fix two strings $w_{mathrm{yes}}in L$ and $w_{mathrm{no}}notin L$. Now, given a string $x$, we can determine in exponential time if $xin X$. If it is, the reduction maps $xmapsto w_{mathrm{yes}}$; otherwise, it maps $xmapsto w_{mathrm{no}}$.
* OK, technically, we don't know that EXP is more powerful than NP, but it's certainly at least as powerful, and probably more powerful.
answered Nov 25 at 23:05
David Richerby
64.8k1597186
Thanks! That makes a lot of sense. But let's say I found a reduction of time O(n^k), where k is a variable such that k <= n. What complexity class would this be?
– cccompro
Nov 25 at 23:16
The only way you could have $kle n$ for all $n$ would be to have $k=0$ or $k=1$, and I don't think that's what you mean.
– David Richerby
Nov 26 at 1:10
I'm not sure how to explain it. I have a grid with k squares. The upper bound of the size of the grid is n squares.
– cccompro
Nov 26 at 1:22
@cccompro then you have a grid with $O(n)$ squares?? No need for $k$ if we don’t need an tight bound—this is linear.
– D. Ben Knoble
Nov 26 at 4:08
@D.BenKnoble It's not a tight bound because the problem is packing things into the grid (it is similar to bin packing but not exactly) and I have a reduction of time O(n^k)
– cccompro
Nov 26 at 19:26
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hardness/completeness is always defined with respect to a specific kind of reduction.
Using exponential-time reductions doens't work at all, because that means your reduction is even more powerful than the thing you're reducing to.*
Every language except $emptyset$ and $Sigma^*$ is NP-hard under exponential-time reductions. To see this, let $L$ be any language except $emptyset$ and $Sigma^*$, and let $X$ be in NP. We can reduce $X$ to $L$ as follows. Fix two strings $w_{mathrm{yes}}in L$ and $w_{mathrm{no}}notin L$. Now, given a string $x$, we can determine in exponential time if $xin X$. If it is, the reduction maps $xmapsto w_{mathrm{yes}}$; otherwise, it maps $xmapsto w_{mathrm{no}}$.
* OK, technically, we don't know that EXP is more powerful than NP, but it's certainly at least as powerful, and probably more powerful.
answered Nov 25 at 23:05
David Richerby
64.8k1597186
Thanks! That makes a lot of sense. But let's say I found a reduction of time O(n^k), where k is a variable such that k <= n. What complexity class would this be?
– cccompro
Nov 25 at 23:16
The only way you could have $kle n$ for all $n$ would be to have $k=0$ or $k=1$, and I don't think that's what you mean.
– David Richerby
Nov 26 at 1:10
I'm not sure how to explain it. I have a grid with k squares. The upper bound of the size of the grid is n squares.
– cccompro
Nov 26 at 1:22
@cccompro then you have a grid with $O(n)$ squares?? No need for $k$ if we don’t need an tight bound—this is linear.
– D. Ben Knoble
Nov 26 at 4:08
@D.BenKnoble It's not a tight bound because the problem is packing things into the grid (it is similar to bin packing but not exactly) and I have a reduction of time O(n^k)
– cccompro
Nov 26 at 19:26
|
show 1 more comment
up vote
2
down vote
accepted
Hardness/completeness is always defined with respect to a specific kind of reduction.
Using exponential-time reductions doens't work at all, because that means your reduction is even more powerful than the thing you're reducing to.*
Every language except $emptyset$ and $Sigma^*$ is NP-hard under exponential-time reductions. To see this, let $L$ be any language except $emptyset$ and $Sigma^*$, and let $X$ be in NP. We can reduce $X$ to $L$ as follows. Fix two strings $w_{mathrm{yes}}in L$ and $w_{mathrm{no}}notin L$. Now, given a string $x$, we can determine in exponential time if $xin X$. If it is, the reduction maps $xmapsto w_{mathrm{yes}}$; otherwise, it maps $xmapsto w_{mathrm{no}}$.
* OK, technically, we don't know that EXP is more powerful than NP, but it's certainly at least as powerful, and probably more powerful.
answered Nov 25 at 23:05
David Richerby
64.8k1597186
Thanks! That makes a lot of sense. But let's say I found a reduction of time O(n^k), where k is a variable such that k <= n. What complexity class would this be?
– cccompro
Nov 25 at 23:16
The only way you could have $kle n$ for all $n$ would be to have $k=0$ or $k=1$, and I don't think that's what you mean.
– David Richerby
Nov 26 at 1:10
I'm not sure how to explain it. I have a grid with k squares. The upper bound of the size of the grid is n squares.
– cccompro
Nov 26 at 1:22
@cccompro then you have a grid with $O(n)$ squares?? No need for $k$ if we don’t need an tight bound—this is linear.
– D. Ben Knoble
Nov 26 at 4:08
@D.BenKnoble It's not a tight bound because the problem is packing things into the grid (it is similar to bin packing but not exactly) and I have a reduction of time O(n^k)
– cccompro
Nov 26 at 19:26
|
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hardness/completeness is always defined with respect to a specific kind of reduction.
Using exponential-time reductions doens't work at all, because that means your reduction is even more powerful than the thing you're reducing to.*
Every language except $emptyset$ and $Sigma^*$ is NP-hard under exponential-time reductions. To see this, let $L$ be any language except $emptyset$ and $Sigma^*$, and let $X$ be in NP. We can reduce $X$ to $L$ as follows. Fix two strings $w_{mathrm{yes}}in L$ and $w_{mathrm{no}}notin L$. Now, given a string $x$, we can determine in exponential time if $xin X$. If it is, the reduction maps $xmapsto w_{mathrm{yes}}$; otherwise, it maps $xmapsto w_{mathrm{no}}$.
* OK, technically, we don't know that EXP is more powerful than NP, but it's certainly at least as powerful, and probably more powerful.
answered Nov 25 at 23:05
David Richerby
64.8k1597186
Hardness/completeness is always defined with respect to a specific kind of reduction.
Using exponential-time reductions doens't work at all, because that means your reduction is even more powerful than the thing you're reducing to.*
Every language except $emptyset$ and $Sigma^*$ is NP-hard under exponential-time reductions. To see this, let $L$ be any language except $emptyset$ and $Sigma^*$, and let $X$ be in NP. We can reduce $X$ to $L$ as follows. Fix two strings $w_{mathrm{yes}}in L$ and $w_{mathrm{no}}notin L$. Now, given a string $x$, we can determine in exponential time if $xin X$. If it is, the reduction maps $xmapsto w_{mathrm{yes}}$; otherwise, it maps $xmapsto w_{mathrm{no}}$.
* OK, technically, we don't know that EXP is more powerful than NP, but it's certainly at least as powerful, and probably more powerful.
answered Nov 25 at 23:05
David Richerby
64.8k1597186
answered Nov 25 at 23:05
David Richerby
64.8k1597186
answered Nov 25 at 23:05
David Richerby
64.8k1597186
answered Nov 25 at 23:05
David Richerby
64.8k1597186
64.8k1597186
Thanks! That makes a lot of sense. But let's say I found a reduction of time O(n^k), where k is a variable such that k <= n. What complexity class would this be?
– cccompro
Nov 25 at 23:16
The only way you could have $kle n$ for all $n$ would be to have $k=0$ or $k=1$, and I don't think that's what you mean.
– David Richerby
Nov 26 at 1:10
I'm not sure how to explain it. I have a grid with k squares. The upper bound of the size of the grid is n squares.
– cccompro
Nov 26 at 1:22
@cccompro then you have a grid with $O(n)$ squares?? No need for $k$ if we don’t need an tight bound—this is linear.
– D. Ben Knoble
Nov 26 at 4:08
@D.BenKnoble It's not a tight bound because the problem is packing things into the grid (it is similar to bin packing but not exactly) and I have a reduction of time O(n^k)
– cccompro
Nov 26 at 19:26
|
show 1 more comment
Thanks! That makes a lot of sense. But let's say I found a reduction of time O(n^k), where k is a variable such that k <= n. What complexity class would this be?
– cccompro
Nov 25 at 23:16
The only way you could have $kle n$ for all $n$ would be to have $k=0$ or $k=1$, and I don't think that's what you mean.
– David Richerby
Nov 26 at 1:10
I'm not sure how to explain it. I have a grid with k squares. The upper bound of the size of the grid is n squares.
– cccompro
Nov 26 at 1:22
@cccompro then you have a grid with $O(n)$ squares?? No need for $k$ if we don’t need an tight bound—this is linear.
– D. Ben Knoble
Nov 26 at 4:08
@D.BenKnoble It's not a tight bound because the problem is packing things into the grid (it is similar to bin packing but not exactly) and I have a reduction of time O(n^k)
– cccompro
Nov 26 at 19:26
Thanks! That makes a lot of sense. But let's say I found a reduction of time O(n^k), where k is a variable such that k <= n. What complexity class would this be?
– cccompro
Nov 25 at 23:16
Thanks! That makes a lot of sense. But let's say I found a reduction of time O(n^k), where k is a variable such that k <= n. What complexity class would this be?
– cccompro
Nov 25 at 23:16
The only way you could have $kle n$ for all $n$ would be to have $k=0$ or $k=1$, and I don't think that's what you mean.
– David Richerby
Nov 26 at 1:10
The only way you could have $kle n$ for all $n$ would be to have $k=0$ or $k=1$, and I don't think that's what you mean.
– David Richerby
Nov 26 at 1:10
I'm not sure how to explain it. I have a grid with k squares. The upper bound of the size of the grid is n squares.
– cccompro
Nov 26 at 1:22
I'm not sure how to explain it. I have a grid with k squares. The upper bound of the size of the grid is n squares.
– cccompro
Nov 26 at 1:22
@cccompro then you have a grid with $O(n)$ squares?? No need for $k$ if we don’t need an tight bound—this is linear.
– D. Ben Knoble
Nov 26 at 4:08
@cccompro then you have a grid with $O(n)$ squares?? No need for $k$ if we don’t need an tight bound—this is linear.
– D. Ben Knoble
Nov 26 at 4:08
@D.BenKnoble It's not a tight bound because the problem is packing things into the grid (it is similar to bin packing but not exactly) and I have a reduction of time O(n^k)
– cccompro
Nov 26 at 19:26
@D.BenKnoble It's not a tight bound because the problem is packing things into the grid (it is similar to bin packing but not exactly) and I have a reduction of time O(n^k)
– cccompro
Nov 26 at 19:26
|
show 1 more comment
cccompro is a new contributor. Be nice, and check out our Code of Conduct.
cccompro is a new contributor. Be nice, and check out our Code of Conduct.
cccompro is a new contributor. Be nice, and check out our Code of Conduct.
cccompro is a new contributor. Be nice, and check out our Code of Conduct.
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