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?










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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?










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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?










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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?







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      asked Nov 25 at 22:38









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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.






          share|cite|improve this answer





















          • 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











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          1 Answer
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          1 Answer
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          active

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          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.






          share|cite|improve this answer





















          • 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















          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.






          share|cite|improve this answer





















          • 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













          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • 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










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