If I'm just trying to show something is NP-hard (as opposed to NP-complete) does my reduction need to be...











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?










share|cite|improve this question







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.
























    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?










    share|cite|improve this question







    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.






















      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?










      share|cite|improve this question







      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






      share|cite|improve this question







      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.











      share|cite|improve this question







      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.









      share|cite|improve this question




      share|cite|improve this question






      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 5 hours ago









      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.






















          1 Answer
          1






          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.






          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
            4 hours ago










          • 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
            2 hours ago










          • 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
            2 hours ago











          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',
          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
          });


          }
          });






          cccompro is a new contributor. Be nice, and check out our Code of Conduct.










           

          draft saved


          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f100544%2fif-im-just-trying-to-show-something-is-np-hard-as-opposed-to-np-complete-does%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








          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
            4 hours ago










          • 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
            2 hours ago










          • 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
            2 hours ago















          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
            4 hours ago










          • 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
            2 hours ago










          • 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
            2 hours ago













          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 4 hours ago









          David Richerby

          64.6k1597186




          64.6k1597186












          • 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
            4 hours ago










          • 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
            2 hours ago










          • 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
            2 hours ago


















          • 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
            4 hours ago










          • 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
            2 hours ago










          • 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
            2 hours ago
















          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
          4 hours ago




          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
          4 hours ago












          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
          2 hours ago




          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
          2 hours ago












          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
          2 hours ago




          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
          2 hours ago










          cccompro is a new contributor. Be nice, and check out our Code of Conduct.










           

          draft saved


          draft discarded


















          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.















           


          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f100544%2fif-im-just-trying-to-show-something-is-np-hard-as-opposed-to-np-complete-does%23new-answer', 'question_page');
          }
          );

          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







          Popular posts from this blog

          Bundesstraße 106

          Verónica Boquete

          Ida-Boy-Ed-Garten