What is the difference between Turing-reducibility and m-reducibility?












1












$begingroup$


The set $A$ is many-one reducible (m-reducible) to the set $ B $ if there is a
total computable function $f$ such that $x ∈ A $ iff $f(x) ∈ B$ for all $x$.



We shall write $A ≤_{m} B$



Let be $A, B subset mathbb{N}$



is it true that $A leq_{m} B implies Aleq_{T}B$?



Where $Aleq_{T}B$ is that $A$ is Turing-reducible to the set B.



If it is true, why do the mathematicians use more Turing-reducibility than m-reducibility? Can you give me some counterexample of two sets $A,Bsubset mathbb{N}$ where $Aleq_{T}B$ and $neg (Aleq_{m}B)$?










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    The set $A$ is many-one reducible (m-reducible) to the set $ B $ if there is a
    total computable function $f$ such that $x ∈ A $ iff $f(x) ∈ B$ for all $x$.



    We shall write $A ≤_{m} B$



    Let be $A, B subset mathbb{N}$



    is it true that $A leq_{m} B implies Aleq_{T}B$?



    Where $Aleq_{T}B$ is that $A$ is Turing-reducible to the set B.



    If it is true, why do the mathematicians use more Turing-reducibility than m-reducibility? Can you give me some counterexample of two sets $A,Bsubset mathbb{N}$ where $Aleq_{T}B$ and $neg (Aleq_{m}B)$?










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      The set $A$ is many-one reducible (m-reducible) to the set $ B $ if there is a
      total computable function $f$ such that $x ∈ A $ iff $f(x) ∈ B$ for all $x$.



      We shall write $A ≤_{m} B$



      Let be $A, B subset mathbb{N}$



      is it true that $A leq_{m} B implies Aleq_{T}B$?



      Where $Aleq_{T}B$ is that $A$ is Turing-reducible to the set B.



      If it is true, why do the mathematicians use more Turing-reducibility than m-reducibility? Can you give me some counterexample of two sets $A,Bsubset mathbb{N}$ where $Aleq_{T}B$ and $neg (Aleq_{m}B)$?










      share|cite|improve this question











      $endgroup$




      The set $A$ is many-one reducible (m-reducible) to the set $ B $ if there is a
      total computable function $f$ such that $x ∈ A $ iff $f(x) ∈ B$ for all $x$.



      We shall write $A ≤_{m} B$



      Let be $A, B subset mathbb{N}$



      is it true that $A leq_{m} B implies Aleq_{T}B$?



      Where $Aleq_{T}B$ is that $A$ is Turing-reducible to the set B.



      If it is true, why do the mathematicians use more Turing-reducibility than m-reducibility? Can you give me some counterexample of two sets $A,Bsubset mathbb{N}$ where $Aleq_{T}B$ and $neg (Aleq_{m}B)$?







      logic computability






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 10 '18 at 22:18







      Nicolás Castro

















      asked Dec 10 '18 at 22:11









      Nicolás CastroNicolás Castro

      83




      83






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          It is true that $le_m$ implies $le_T$. The converse, however, is false, so they are truly different things.



          An easy example of this is supplied by the following: any set is Turing equivalent to its complement, but in general a set need not be many-one reducible to its complement. For example, if $C$ is a non-recursive recursively enumerable set, then $C$ is not many-one reducible to its complement.



          This helps motivate $le_T$ as opposed to $le_m$: intuitively, a set has exactly as much information as its complement, just "twisted around" (in a very computable way); $le_T$ reflects this but $le_m$ doesn't. (That said, there are also situations where $le_m$ is more natural.)





          Incidentally, it's worth noting that there are other reducibilities strictly between $le_m$ and $le_T$, such as truth-table reducibility $le_{tt}$ (with respect to which a set is equivalent to its complement).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
            $endgroup$
            – Nicolás Castro
            Dec 10 '18 at 22:34












          • $begingroup$
            @NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
            $endgroup$
            – Noah Schweber
            Dec 10 '18 at 23:23













          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: "69"
          };
          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: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          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
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034560%2fwhat-is-the-difference-between-turing-reducibility-and-m-reducibility%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









          1












          $begingroup$

          It is true that $le_m$ implies $le_T$. The converse, however, is false, so they are truly different things.



          An easy example of this is supplied by the following: any set is Turing equivalent to its complement, but in general a set need not be many-one reducible to its complement. For example, if $C$ is a non-recursive recursively enumerable set, then $C$ is not many-one reducible to its complement.



          This helps motivate $le_T$ as opposed to $le_m$: intuitively, a set has exactly as much information as its complement, just "twisted around" (in a very computable way); $le_T$ reflects this but $le_m$ doesn't. (That said, there are also situations where $le_m$ is more natural.)





          Incidentally, it's worth noting that there are other reducibilities strictly between $le_m$ and $le_T$, such as truth-table reducibility $le_{tt}$ (with respect to which a set is equivalent to its complement).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
            $endgroup$
            – Nicolás Castro
            Dec 10 '18 at 22:34












          • $begingroup$
            @NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
            $endgroup$
            – Noah Schweber
            Dec 10 '18 at 23:23


















          1












          $begingroup$

          It is true that $le_m$ implies $le_T$. The converse, however, is false, so they are truly different things.



          An easy example of this is supplied by the following: any set is Turing equivalent to its complement, but in general a set need not be many-one reducible to its complement. For example, if $C$ is a non-recursive recursively enumerable set, then $C$ is not many-one reducible to its complement.



          This helps motivate $le_T$ as opposed to $le_m$: intuitively, a set has exactly as much information as its complement, just "twisted around" (in a very computable way); $le_T$ reflects this but $le_m$ doesn't. (That said, there are also situations where $le_m$ is more natural.)





          Incidentally, it's worth noting that there are other reducibilities strictly between $le_m$ and $le_T$, such as truth-table reducibility $le_{tt}$ (with respect to which a set is equivalent to its complement).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
            $endgroup$
            – Nicolás Castro
            Dec 10 '18 at 22:34












          • $begingroup$
            @NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
            $endgroup$
            – Noah Schweber
            Dec 10 '18 at 23:23
















          1












          1








          1





          $begingroup$

          It is true that $le_m$ implies $le_T$. The converse, however, is false, so they are truly different things.



          An easy example of this is supplied by the following: any set is Turing equivalent to its complement, but in general a set need not be many-one reducible to its complement. For example, if $C$ is a non-recursive recursively enumerable set, then $C$ is not many-one reducible to its complement.



          This helps motivate $le_T$ as opposed to $le_m$: intuitively, a set has exactly as much information as its complement, just "twisted around" (in a very computable way); $le_T$ reflects this but $le_m$ doesn't. (That said, there are also situations where $le_m$ is more natural.)





          Incidentally, it's worth noting that there are other reducibilities strictly between $le_m$ and $le_T$, such as truth-table reducibility $le_{tt}$ (with respect to which a set is equivalent to its complement).






          share|cite|improve this answer









          $endgroup$



          It is true that $le_m$ implies $le_T$. The converse, however, is false, so they are truly different things.



          An easy example of this is supplied by the following: any set is Turing equivalent to its complement, but in general a set need not be many-one reducible to its complement. For example, if $C$ is a non-recursive recursively enumerable set, then $C$ is not many-one reducible to its complement.



          This helps motivate $le_T$ as opposed to $le_m$: intuitively, a set has exactly as much information as its complement, just "twisted around" (in a very computable way); $le_T$ reflects this but $le_m$ doesn't. (That said, there are also situations where $le_m$ is more natural.)





          Incidentally, it's worth noting that there are other reducibilities strictly between $le_m$ and $le_T$, such as truth-table reducibility $le_{tt}$ (with respect to which a set is equivalent to its complement).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 22:18









          Noah SchweberNoah Schweber

          123k10150286




          123k10150286












          • $begingroup$
            Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
            $endgroup$
            – Nicolás Castro
            Dec 10 '18 at 22:34












          • $begingroup$
            @NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
            $endgroup$
            – Noah Schweber
            Dec 10 '18 at 23:23




















          • $begingroup$
            Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
            $endgroup$
            – Nicolás Castro
            Dec 10 '18 at 22:34












          • $begingroup$
            @NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
            $endgroup$
            – Noah Schweber
            Dec 10 '18 at 23:23


















          $begingroup$
          Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
          $endgroup$
          – Nicolás Castro
          Dec 10 '18 at 22:34






          $begingroup$
          Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
          $endgroup$
          – Nicolás Castro
          Dec 10 '18 at 22:34














          $begingroup$
          @NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
          $endgroup$
          – Noah Schweber
          Dec 10 '18 at 23:23






          $begingroup$
          @NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
          $endgroup$
          – Noah Schweber
          Dec 10 '18 at 23:23




















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics 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.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034560%2fwhat-is-the-difference-between-turing-reducibility-and-m-reducibility%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

          Quarter-circle Tiles

          build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

          Mont Emei