How many repetitions does the loop












1












$begingroup$


Here is the following algorithm:



for(k=2; k<n; k=k^k)


I understand that I need to check when
$n=k^k$.



But I'm stuck on $k=log_k(n)$



How many repetitions are performed by the loop?






calculus computational-complexity







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have you computed what $k$ is the first few passes through the loop? It grows very rapidly. I doubt there is a nice formula, but for any reasonable $n$ it will be no more than $3$
    $endgroup$
    – Ross Millikan
    Jan 8 at 19:44










  • $begingroup$
    I know it grow very rapidly, but it's depends on "n", i guess that's will be log(log(n)) but i still don't know how to show it.
    $endgroup$
    – shay
    Jan 8 at 20:35
















1












$begingroup$


Here is the following algorithm:



for(k=2; k<n; k=k^k)


I understand that I need to check when
$n=k^k$.



But I'm stuck on $k=log_k(n)$



How many repetitions are performed by the loop?






calculus computational-complexity







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have you computed what $k$ is the first few passes through the loop? It grows very rapidly. I doubt there is a nice formula, but for any reasonable $n$ it will be no more than $3$
    $endgroup$
    – Ross Millikan
    Jan 8 at 19:44










  • $begingroup$
    I know it grow very rapidly, but it's depends on "n", i guess that's will be log(log(n)) but i still don't know how to show it.
    $endgroup$
    – shay
    Jan 8 at 20:35














1












1








1





$begingroup$


Here is the following algorithm:



for(k=2; k<n; k=k^k)


I understand that I need to check when
$n=k^k$.



But I'm stuck on $k=log_k(n)$



How many repetitions are performed by the loop?






calculus computational-complexity







share|cite|improve this question











$endgroup$




Here is the following algorithm:



for(k=2; k<n; k=k^k)


I understand that I need to check when
$n=k^k$.



But I'm stuck on $k=log_k(n)$



How many repetitions are performed by the loop?






calculus computational-complexity




calculus computational-complexity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 21:04









the_candyman

8,94132145




8,94132145










asked Jan 8 at 19:25









shayshay

103




103












  • $begingroup$
    Have you computed what $k$ is the first few passes through the loop? It grows very rapidly. I doubt there is a nice formula, but for any reasonable $n$ it will be no more than $3$
    $endgroup$
    – Ross Millikan
    Jan 8 at 19:44










  • $begingroup$
    I know it grow very rapidly, but it's depends on "n", i guess that's will be log(log(n)) but i still don't know how to show it.
    $endgroup$
    – shay
    Jan 8 at 20:35


















  • $begingroup$
    Have you computed what $k$ is the first few passes through the loop? It grows very rapidly. I doubt there is a nice formula, but for any reasonable $n$ it will be no more than $3$
    $endgroup$
    – Ross Millikan
    Jan 8 at 19:44










  • $begingroup$
    I know it grow very rapidly, but it's depends on "n", i guess that's will be log(log(n)) but i still don't know how to show it.
    $endgroup$
    – shay
    Jan 8 at 20:35
















$begingroup$
Have you computed what $k$ is the first few passes through the loop? It grows very rapidly. I doubt there is a nice formula, but for any reasonable $n$ it will be no more than $3$
$endgroup$
– Ross Millikan
Jan 8 at 19:44




$begingroup$
Have you computed what $k$ is the first few passes through the loop? It grows very rapidly. I doubt there is a nice formula, but for any reasonable $n$ it will be no more than $3$
$endgroup$
– Ross Millikan
Jan 8 at 19:44












$begingroup$
I know it grow very rapidly, but it's depends on "n", i guess that's will be log(log(n)) but i still don't know how to show it.
$endgroup$
– shay
Jan 8 at 20:35




$begingroup$
I know it grow very rapidly, but it's depends on "n", i guess that's will be log(log(n)) but i still don't know how to show it.
$endgroup$
– shay
Jan 8 at 20:35










4 Answers
4






active

oldest

votes


















1












$begingroup$

After the first pass we have $k=2^2=4$ If $n$ is $3$ or $4$ there will be only one pass.

After the second pass we have $k=4^4=256$. If $4 lt n le 256$ there will be two passes.

After the third we have $k=256^{256}$, which is enormous-about $600$ digits. Only if $n$ is larger than this will there be a fourth pass.



I don't know of a nice function that can take $n$ and give back the number of passes.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Notice that



    $$2^2=4,4^4=256,256^{256}approx3.2317cdot10^{616}.$$



    Needless to say, the next term is astronomical.



    So you take little risk by saying zero to three iterations. (If $n$ is a 32 bits integer uniformly drawn, say three and you'll be right with probability $0.999999940$)






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      We start with $k=2$. Since it is the first, let's call it $k_0$. We observe that:



      $$k_0 = 2 = 2^1 = 2^{2^0}.$$



      The next element of the sequence is:



      $$k_1 = k_0^{k_0} = 2^2 = 2^{2^1}.$$



      By continuing...



      $$k_2 = k_3^{k_3} = (2^2)^{2^2} = 2^{2^3}.$$
      $$k_3 = k_4^{k_4} = (2^{2^3})^{2^{2^3}} = 2^{2^{11}}.$$
      $$k_4 = k_5^{k_5} = (2^{2^{11}})^{2^{2^{11}}} = 2^{2^{2059}}.$$



      Let's introduce
      $s_i = log_2 log_2 k_i$ (i.e. $k_i = 2^{2^{s_i}})$. In our case, $s_0 = 0$, $s_1 = 1$, $s_2 = 3$, $s_3=11$, $s_4 = 2059$, ...



      After a deep analysis of the showed calculation, it turns out that the $s$ sequence satisfies the following:



      $$s_{i+1} = s_i + 2^{s_i},$$



      with $s_0 = 0.$






      share|cite|improve this answer











      $endgroup$





















        0












        $begingroup$

        I looked for the number of repetitions, here is the answer I thought of:
        $$
        2^(2^k) = n
        $$
        $$
        log(2^(2^k)) = log(n)
        $$
        $$
        2^klog(2)=log(n)
        $$
        $$
        2^k=log(n)
        $$



        $$
        k=log(log(n))
        $$






        share|cite|improve this answer









        $endgroup$













          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%2f3066612%2fhow-many-repetitions-does-the-loop%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          After the first pass we have $k=2^2=4$ If $n$ is $3$ or $4$ there will be only one pass.

          After the second pass we have $k=4^4=256$. If $4 lt n le 256$ there will be two passes.

          After the third we have $k=256^{256}$, which is enormous-about $600$ digits. Only if $n$ is larger than this will there be a fourth pass.



          I don't know of a nice function that can take $n$ and give back the number of passes.






          share|cite|improve this answer









          $endgroup$


















            1












            $begingroup$

            After the first pass we have $k=2^2=4$ If $n$ is $3$ or $4$ there will be only one pass.

            After the second pass we have $k=4^4=256$. If $4 lt n le 256$ there will be two passes.

            After the third we have $k=256^{256}$, which is enormous-about $600$ digits. Only if $n$ is larger than this will there be a fourth pass.



            I don't know of a nice function that can take $n$ and give back the number of passes.






            share|cite|improve this answer









            $endgroup$
















              1












              1








              1





              $begingroup$

              After the first pass we have $k=2^2=4$ If $n$ is $3$ or $4$ there will be only one pass.

              After the second pass we have $k=4^4=256$. If $4 lt n le 256$ there will be two passes.

              After the third we have $k=256^{256}$, which is enormous-about $600$ digits. Only if $n$ is larger than this will there be a fourth pass.



              I don't know of a nice function that can take $n$ and give back the number of passes.






              share|cite|improve this answer









              $endgroup$



              After the first pass we have $k=2^2=4$ If $n$ is $3$ or $4$ there will be only one pass.

              After the second pass we have $k=4^4=256$. If $4 lt n le 256$ there will be two passes.

              After the third we have $k=256^{256}$, which is enormous-about $600$ digits. Only if $n$ is larger than this will there be a fourth pass.



              I don't know of a nice function that can take $n$ and give back the number of passes.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 8 at 21:02









              Ross MillikanRoss Millikan

              297k23198371




              297k23198371























                  1












                  $begingroup$

                  Notice that



                  $$2^2=4,4^4=256,256^{256}approx3.2317cdot10^{616}.$$



                  Needless to say, the next term is astronomical.



                  So you take little risk by saying zero to three iterations. (If $n$ is a 32 bits integer uniformly drawn, say three and you'll be right with probability $0.999999940$)






                  share|cite|improve this answer











                  $endgroup$


















                    1












                    $begingroup$

                    Notice that



                    $$2^2=4,4^4=256,256^{256}approx3.2317cdot10^{616}.$$



                    Needless to say, the next term is astronomical.



                    So you take little risk by saying zero to three iterations. (If $n$ is a 32 bits integer uniformly drawn, say three and you'll be right with probability $0.999999940$)






                    share|cite|improve this answer











                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      Notice that



                      $$2^2=4,4^4=256,256^{256}approx3.2317cdot10^{616}.$$



                      Needless to say, the next term is astronomical.



                      So you take little risk by saying zero to three iterations. (If $n$ is a 32 bits integer uniformly drawn, say three and you'll be right with probability $0.999999940$)






                      share|cite|improve this answer











                      $endgroup$



                      Notice that



                      $$2^2=4,4^4=256,256^{256}approx3.2317cdot10^{616}.$$



                      Needless to say, the next term is astronomical.



                      So you take little risk by saying zero to three iterations. (If $n$ is a 32 bits integer uniformly drawn, say three and you'll be right with probability $0.999999940$)







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 8 at 22:02

























                      answered Jan 8 at 21:57









                      Yves DaoustYves Daoust

                      128k675227




                      128k675227























                          0












                          $begingroup$

                          We start with $k=2$. Since it is the first, let's call it $k_0$. We observe that:



                          $$k_0 = 2 = 2^1 = 2^{2^0}.$$



                          The next element of the sequence is:



                          $$k_1 = k_0^{k_0} = 2^2 = 2^{2^1}.$$



                          By continuing...



                          $$k_2 = k_3^{k_3} = (2^2)^{2^2} = 2^{2^3}.$$
                          $$k_3 = k_4^{k_4} = (2^{2^3})^{2^{2^3}} = 2^{2^{11}}.$$
                          $$k_4 = k_5^{k_5} = (2^{2^{11}})^{2^{2^{11}}} = 2^{2^{2059}}.$$



                          Let's introduce
                          $s_i = log_2 log_2 k_i$ (i.e. $k_i = 2^{2^{s_i}})$. In our case, $s_0 = 0$, $s_1 = 1$, $s_2 = 3$, $s_3=11$, $s_4 = 2059$, ...



                          After a deep analysis of the showed calculation, it turns out that the $s$ sequence satisfies the following:



                          $$s_{i+1} = s_i + 2^{s_i},$$



                          with $s_0 = 0.$






                          share|cite|improve this answer











                          $endgroup$


















                            0












                            $begingroup$

                            We start with $k=2$. Since it is the first, let's call it $k_0$. We observe that:



                            $$k_0 = 2 = 2^1 = 2^{2^0}.$$



                            The next element of the sequence is:



                            $$k_1 = k_0^{k_0} = 2^2 = 2^{2^1}.$$



                            By continuing...



                            $$k_2 = k_3^{k_3} = (2^2)^{2^2} = 2^{2^3}.$$
                            $$k_3 = k_4^{k_4} = (2^{2^3})^{2^{2^3}} = 2^{2^{11}}.$$
                            $$k_4 = k_5^{k_5} = (2^{2^{11}})^{2^{2^{11}}} = 2^{2^{2059}}.$$



                            Let's introduce
                            $s_i = log_2 log_2 k_i$ (i.e. $k_i = 2^{2^{s_i}})$. In our case, $s_0 = 0$, $s_1 = 1$, $s_2 = 3$, $s_3=11$, $s_4 = 2059$, ...



                            After a deep analysis of the showed calculation, it turns out that the $s$ sequence satisfies the following:



                            $$s_{i+1} = s_i + 2^{s_i},$$



                            with $s_0 = 0.$






                            share|cite|improve this answer











                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              We start with $k=2$. Since it is the first, let's call it $k_0$. We observe that:



                              $$k_0 = 2 = 2^1 = 2^{2^0}.$$



                              The next element of the sequence is:



                              $$k_1 = k_0^{k_0} = 2^2 = 2^{2^1}.$$



                              By continuing...



                              $$k_2 = k_3^{k_3} = (2^2)^{2^2} = 2^{2^3}.$$
                              $$k_3 = k_4^{k_4} = (2^{2^3})^{2^{2^3}} = 2^{2^{11}}.$$
                              $$k_4 = k_5^{k_5} = (2^{2^{11}})^{2^{2^{11}}} = 2^{2^{2059}}.$$



                              Let's introduce
                              $s_i = log_2 log_2 k_i$ (i.e. $k_i = 2^{2^{s_i}})$. In our case, $s_0 = 0$, $s_1 = 1$, $s_2 = 3$, $s_3=11$, $s_4 = 2059$, ...



                              After a deep analysis of the showed calculation, it turns out that the $s$ sequence satisfies the following:



                              $$s_{i+1} = s_i + 2^{s_i},$$



                              with $s_0 = 0.$






                              share|cite|improve this answer











                              $endgroup$



                              We start with $k=2$. Since it is the first, let's call it $k_0$. We observe that:



                              $$k_0 = 2 = 2^1 = 2^{2^0}.$$



                              The next element of the sequence is:



                              $$k_1 = k_0^{k_0} = 2^2 = 2^{2^1}.$$



                              By continuing...



                              $$k_2 = k_3^{k_3} = (2^2)^{2^2} = 2^{2^3}.$$
                              $$k_3 = k_4^{k_4} = (2^{2^3})^{2^{2^3}} = 2^{2^{11}}.$$
                              $$k_4 = k_5^{k_5} = (2^{2^{11}})^{2^{2^{11}}} = 2^{2^{2059}}.$$



                              Let's introduce
                              $s_i = log_2 log_2 k_i$ (i.e. $k_i = 2^{2^{s_i}})$. In our case, $s_0 = 0$, $s_1 = 1$, $s_2 = 3$, $s_3=11$, $s_4 = 2059$, ...



                              After a deep analysis of the showed calculation, it turns out that the $s$ sequence satisfies the following:



                              $$s_{i+1} = s_i + 2^{s_i},$$



                              with $s_0 = 0.$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Jan 8 at 21:47

























                              answered Jan 8 at 21:31









                              the_candymanthe_candyman

                              8,94132145




                              8,94132145























                                  0












                                  $begingroup$

                                  I looked for the number of repetitions, here is the answer I thought of:
                                  $$
                                  2^(2^k) = n
                                  $$
                                  $$
                                  log(2^(2^k)) = log(n)
                                  $$
                                  $$
                                  2^klog(2)=log(n)
                                  $$
                                  $$
                                  2^k=log(n)
                                  $$



                                  $$
                                  k=log(log(n))
                                  $$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    I looked for the number of repetitions, here is the answer I thought of:
                                    $$
                                    2^(2^k) = n
                                    $$
                                    $$
                                    log(2^(2^k)) = log(n)
                                    $$
                                    $$
                                    2^klog(2)=log(n)
                                    $$
                                    $$
                                    2^k=log(n)
                                    $$



                                    $$
                                    k=log(log(n))
                                    $$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      I looked for the number of repetitions, here is the answer I thought of:
                                      $$
                                      2^(2^k) = n
                                      $$
                                      $$
                                      log(2^(2^k)) = log(n)
                                      $$
                                      $$
                                      2^klog(2)=log(n)
                                      $$
                                      $$
                                      2^k=log(n)
                                      $$



                                      $$
                                      k=log(log(n))
                                      $$






                                      share|cite|improve this answer









                                      $endgroup$



                                      I looked for the number of repetitions, here is the answer I thought of:
                                      $$
                                      2^(2^k) = n
                                      $$
                                      $$
                                      log(2^(2^k)) = log(n)
                                      $$
                                      $$
                                      2^klog(2)=log(n)
                                      $$
                                      $$
                                      2^k=log(n)
                                      $$



                                      $$
                                      k=log(log(n))
                                      $$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 13 at 16:11









                                      shayshay

                                      103




                                      103






























                                          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%2f3066612%2fhow-many-repetitions-does-the-loop%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

                                          Human spaceflight

                                          Image
                                          Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
                                          Read more

                                          Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                                          Image
                                          0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
                                          Read more

                                          張江高科駅

                                          Image
                                          張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
                                          Read more