Square Root Distance from Integers












6












$begingroup$


Given a decimal number k, find the smallest integer n such that the square root of n is within k of an integer. However, the distance should be nonzero - n cannot be a perfect square.



Given k, a decimal number or a fraction (whichever is easier for you), such that 0 < k < 1, output the smallest positive integer n such that the difference between the square root of n and the closest integer to the square root of n is less than or equal to k but nonzero.



If i is the closest integer to the square root of n, you are looking for the first n where 0 < |i - sqrt(n)| <= k.



Rules




  • You cannot use a language's insufficient implementation of non-integer numbers to trivialize the problem.

  • Otherwise, you can assume that k will not cause problems with, for example, floating point rounding.


Test Cases



.9         > 2

.5         > 2

.4         > 3

.3         > 3

.25        > 5

.2         > 8

.1         > 26

.05        > 101

.03        > 288

.01        > 2501

.005       > 10001

.003       > 27888

.001       > 250001

.0005      > 1000001

.0003      > 2778888

.0001      > 25000001

.0314159   > 255

.00314159  > 25599

.000314159 > 2534463


Comma separated test case inputs: 



0.9, 0.5, 0.4, 0.3, 0.25, 0.2, 0.1, 0.05, 0.03, 0.01, 0.005, 0.003, 0.001, 0.0005, 0.0003, 0.0001, 0.0314159, 0.00314159, 0.000314159


This is code-golf, so shortest answer in bytes wins.






code-golf number integer







share|improve this question











$endgroup$

















    6












    $begingroup$


    Given a decimal number k, find the smallest integer n such that the square root of n is within k of an integer. However, the distance should be nonzero - n cannot be a perfect square.



    Given k, a decimal number or a fraction (whichever is easier for you), such that 0 < k < 1, output the smallest positive integer n such that the difference between the square root of n and the closest integer to the square root of n is less than or equal to k but nonzero.



    If i is the closest integer to the square root of n, you are looking for the first n where 0 < |i - sqrt(n)| <= k.



    Rules




    • You cannot use a language's insufficient implementation of non-integer numbers to trivialize the problem.

    • Otherwise, you can assume that k will not cause problems with, for example, floating point rounding.


    Test Cases



    .9         > 2
    
.5         > 2
    
.4         > 3
    
.3         > 3
    
.25        > 5
    
.2         > 8
    
.1         > 26
    
.05        > 101
    
.03        > 288
    
.01        > 2501
    
.005       > 10001
    
.003       > 27888
    
.001       > 250001
    
.0005      > 1000001
    
.0003      > 2778888
    
.0001      > 25000001
    
.0314159   > 255
    
.00314159  > 25599
    
.000314159 > 2534463


    Comma separated test case inputs: 



    0.9, 0.5, 0.4, 0.3, 0.25, 0.2, 0.1, 0.05, 0.03, 0.01, 0.005, 0.003, 0.001, 0.0005, 0.0003, 0.0001, 0.0314159, 0.00314159, 0.000314159


    This is code-golf, so shortest answer in bytes wins.






    code-golf number integer







    share|improve this question











    $endgroup$















      6












      6








      6





      $begingroup$


      Given a decimal number k, find the smallest integer n such that the square root of n is within k of an integer. However, the distance should be nonzero - n cannot be a perfect square.



      Given k, a decimal number or a fraction (whichever is easier for you), such that 0 < k < 1, output the smallest positive integer n such that the difference between the square root of n and the closest integer to the square root of n is less than or equal to k but nonzero.



      If i is the closest integer to the square root of n, you are looking for the first n where 0 < |i - sqrt(n)| <= k.



      Rules




      • You cannot use a language's insufficient implementation of non-integer numbers to trivialize the problem.

      • Otherwise, you can assume that k will not cause problems with, for example, floating point rounding.


      Test Cases



      .9         > 2
      
.5         > 2
      
.4         > 3
      
.3         > 3
      
.25        > 5
      
.2         > 8
      
.1         > 26
      
.05        > 101
      
.03        > 288
      
.01        > 2501
      
.005       > 10001
      
.003       > 27888
      
.001       > 250001
      
.0005      > 1000001
      
.0003      > 2778888
      
.0001      > 25000001
      
.0314159   > 255
      
.00314159  > 25599
      
.000314159 > 2534463


      Comma separated test case inputs: 



      0.9, 0.5, 0.4, 0.3, 0.25, 0.2, 0.1, 0.05, 0.03, 0.01, 0.005, 0.003, 0.001, 0.0005, 0.0003, 0.0001, 0.0314159, 0.00314159, 0.000314159


      This is code-golf, so shortest answer in bytes wins.






      code-golf number integer







      share|improve this question











      $endgroup$




      Given a decimal number k, find the smallest integer n such that the square root of n is within k of an integer. However, the distance should be nonzero - n cannot be a perfect square.



      Given k, a decimal number or a fraction (whichever is easier for you), such that 0 < k < 1, output the smallest positive integer n such that the difference between the square root of n and the closest integer to the square root of n is less than or equal to k but nonzero.



      If i is the closest integer to the square root of n, you are looking for the first n where 0 < |i - sqrt(n)| <= k.



      Rules




      • You cannot use a language's insufficient implementation of non-integer numbers to trivialize the problem.

      • Otherwise, you can assume that k will not cause problems with, for example, floating point rounding.


      Test Cases



      .9         > 2
      
.5         > 2
      
.4         > 3
      
.3         > 3
      
.25        > 5
      
.2         > 8
      
.1         > 26
      
.05        > 101
      
.03        > 288
      
.01        > 2501
      
.005       > 10001
      
.003       > 27888
      
.001       > 250001
      
.0005      > 1000001
      
.0003      > 2778888
      
.0001      > 25000001
      
.0314159   > 255
      
.00314159  > 25599
      
.000314159 > 2534463


      Comma separated test case inputs: 



      0.9, 0.5, 0.4, 0.3, 0.25, 0.2, 0.1, 0.05, 0.03, 0.01, 0.005, 0.003, 0.001, 0.0005, 0.0003, 0.0001, 0.0314159, 0.00314159, 0.000314159


      This is code-golf, so shortest answer in bytes wins.






      code-golf number integer




      code-golf number integer






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 1 hour ago







      Stephen

















      asked 1 hour ago









      StephenStephen

      7,38823395




      7,38823395






















          4 Answers
          4






          active

          oldest

          votes


















          4












          $begingroup$


          Wolfram Language (Mathematica), 34 bytes



          Min[⌈.5/#+{-#,#}/2⌉^2+{1,-1}]&


          Try it online!



          Explanation



          The result must be of the form $m^2 pm 1$ for some $m in mathbb{N}$. Solving the inequations $sqrt{m^2+1} - m le k$ and $m - sqrt{m^2-1} le k$, we get $m ge frac{1-k^2}{2k}$ and $m ge frac{1+k^2}{2k}$ respectively. So the result is $operatorname{min}left({leftlceil frac{1-k^2}{2k} rightrceil}^2+1, {leftlceil frac{1+k^2}{2k} rightrceil}^2-1right)$.






          share|improve this answer











          $endgroup$





















            3












            $begingroup$

            JavaScript (ES7),  51  50 bytes





            f=(k,n)=>!(d=(s=n**.5)+~(s-.5))|d*d>k*k?f(k,-~n):n


            Try it online!



            (fails for the test cases that require too much recursion)




            Non-recursive version,  57  56 bytes





            k=>{for(n=1;!(d=(s=++n**.5)+~(s-.5))|d*d>k*k;);return n}


            Try it online!



            Or for 55 bytes:



            k=>eval(`for(n=1;!(d=(s=++n**.5)+~(s-.5))|d*d>k*k;);n`)


            Try it online!



            (but this one is significantly slower)






            share|improve this answer











            $endgroup$





















              2












              $begingroup$


              Japt, 18 bytes



              _¬%1©U>½-Z¬u1 a½}a


              Try it online!






              share|improve this answer











              $endgroup$













              • $begingroup$
                Might be shorter using Arnauld's solution
                $endgroup$
                – ASCII-only
                33 mins ago



















              2












              $begingroup$


              J, 39 bytes



              2(>:@])^:((<+.0=])(<.-.)@(-<.)@%:)^:_~]


              Try it online!



              Handles all test cases






              share|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.ifUsing("editor", function () {
                StackExchange.using("externalEditor", function () {
                StackExchange.using("snippets", function () {
                StackExchange.snippets.init();
                });
                });
                }, "code-snippets");

                StackExchange.ready(function() {
                var channelOptions = {
                tags: "".split(" "),
                id: "200"
                };
                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: 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
                });


                }
                });














                draft saved

                draft discarded


















                StackExchange.ready(
                function () {
                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f180412%2fsquare-root-distance-from-integers%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









                4












                $begingroup$


                Wolfram Language (Mathematica), 34 bytes



                Min[⌈.5/#+{-#,#}/2⌉^2+{1,-1}]&


                Try it online!



                Explanation



                The result must be of the form $m^2 pm 1$ for some $m in mathbb{N}$. Solving the inequations $sqrt{m^2+1} - m le k$ and $m - sqrt{m^2-1} le k$, we get $m ge frac{1-k^2}{2k}$ and $m ge frac{1+k^2}{2k}$ respectively. So the result is $operatorname{min}left({leftlceil frac{1-k^2}{2k} rightrceil}^2+1, {leftlceil frac{1+k^2}{2k} rightrceil}^2-1right)$.






                share|improve this answer











                $endgroup$


















                  4












                  $begingroup$


                  Wolfram Language (Mathematica), 34 bytes



                  Min[⌈.5/#+{-#,#}/2⌉^2+{1,-1}]&


                  Try it online!



                  Explanation



                  The result must be of the form $m^2 pm 1$ for some $m in mathbb{N}$. Solving the inequations $sqrt{m^2+1} - m le k$ and $m - sqrt{m^2-1} le k$, we get $m ge frac{1-k^2}{2k}$ and $m ge frac{1+k^2}{2k}$ respectively. So the result is $operatorname{min}left({leftlceil frac{1-k^2}{2k} rightrceil}^2+1, {leftlceil frac{1+k^2}{2k} rightrceil}^2-1right)$.






                  share|improve this answer











                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$


                    Wolfram Language (Mathematica), 34 bytes



                    Min[⌈.5/#+{-#,#}/2⌉^2+{1,-1}]&


                    Try it online!



                    Explanation



                    The result must be of the form $m^2 pm 1$ for some $m in mathbb{N}$. Solving the inequations $sqrt{m^2+1} - m le k$ and $m - sqrt{m^2-1} le k$, we get $m ge frac{1-k^2}{2k}$ and $m ge frac{1+k^2}{2k}$ respectively. So the result is $operatorname{min}left({leftlceil frac{1-k^2}{2k} rightrceil}^2+1, {leftlceil frac{1+k^2}{2k} rightrceil}^2-1right)$.






                    share|improve this answer











                    $endgroup$




                    Wolfram Language (Mathematica), 34 bytes



                    Min[⌈.5/#+{-#,#}/2⌉^2+{1,-1}]&


                    Try it online!



                    Explanation



                    The result must be of the form $m^2 pm 1$ for some $m in mathbb{N}$. Solving the inequations $sqrt{m^2+1} - m le k$ and $m - sqrt{m^2-1} le k$, we get $m ge frac{1-k^2}{2k}$ and $m ge frac{1+k^2}{2k}$ respectively. So the result is $operatorname{min}left({leftlceil frac{1-k^2}{2k} rightrceil}^2+1, {leftlceil frac{1+k^2}{2k} rightrceil}^2-1right)$.







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited 32 mins ago

























                    answered 1 hour ago









                    alephalphaalephalpha

                    21.4k32991




                    21.4k32991























                        3












                        $begingroup$

                        JavaScript (ES7),  51  50 bytes





                        f=(k,n)=>!(d=(s=n**.5)+~(s-.5))|d*d>k*k?f(k,-~n):n


                        Try it online!



                        (fails for the test cases that require too much recursion)




                        Non-recursive version,  57  56 bytes





                        k=>{for(n=1;!(d=(s=++n**.5)+~(s-.5))|d*d>k*k;);return n}


                        Try it online!



                        Or for 55 bytes:



                        k=>eval(`for(n=1;!(d=(s=++n**.5)+~(s-.5))|d*d>k*k;);n`)


                        Try it online!



                        (but this one is significantly slower)






                        share|improve this answer











                        $endgroup$


















                          3












                          $begingroup$

                          JavaScript (ES7),  51  50 bytes





                          f=(k,n)=>!(d=(s=n**.5)+~(s-.5))|d*d>k*k?f(k,-~n):n


                          Try it online!



                          (fails for the test cases that require too much recursion)




                          Non-recursive version,  57  56 bytes





                          k=>{for(n=1;!(d=(s=++n**.5)+~(s-.5))|d*d>k*k;);return n}


                          Try it online!



                          Or for 55 bytes:



                          k=>eval(`for(n=1;!(d=(s=++n**.5)+~(s-.5))|d*d>k*k;);n`)


                          Try it online!



                          (but this one is significantly slower)






                          share|improve this answer











                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$

                            JavaScript (ES7),  51  50 bytes





                            f=(k,n)=>!(d=(s=n**.5)+~(s-.5))|d*d>k*k?f(k,-~n):n


                            Try it online!



                            (fails for the test cases that require too much recursion)




                            Non-recursive version,  57  56 bytes





                            k=>{for(n=1;!(d=(s=++n**.5)+~(s-.5))|d*d>k*k;);return n}


                            Try it online!



                            Or for 55 bytes:



                            k=>eval(`for(n=1;!(d=(s=++n**.5)+~(s-.5))|d*d>k*k;);n`)


                            Try it online!



                            (but this one is significantly slower)






                            share|improve this answer











                            $endgroup$



                            JavaScript (ES7),  51  50 bytes





                            f=(k,n)=>!(d=(s=n**.5)+~(s-.5))|d*d>k*k?f(k,-~n):n


                            Try it online!



                            (fails for the test cases that require too much recursion)




                            Non-recursive version,  57  56 bytes





                            k=>{for(n=1;!(d=(s=++n**.5)+~(s-.5))|d*d>k*k;);return n}


                            Try it online!



                            Or for 55 bytes:



                            k=>eval(`for(n=1;!(d=(s=++n**.5)+~(s-.5))|d*d>k*k;);n`)


                            Try it online!



                            (but this one is significantly slower)







                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited 1 hour ago

























                            answered 1 hour ago









                            ArnauldArnauld

                            76.8k693322




                            76.8k693322























                                2












                                $begingroup$


                                Japt, 18 bytes



                                _¬%1©U>½-Z¬u1 a½}a


                                Try it online!






                                share|improve this answer











                                $endgroup$













                                • $begingroup$
                                  Might be shorter using Arnauld's solution
                                  $endgroup$
                                  – ASCII-only
                                  33 mins ago
















                                2












                                $begingroup$


                                Japt, 18 bytes



                                _¬%1©U>½-Z¬u1 a½}a


                                Try it online!






                                share|improve this answer











                                $endgroup$













                                • $begingroup$
                                  Might be shorter using Arnauld's solution
                                  $endgroup$
                                  – ASCII-only
                                  33 mins ago














                                2












                                2








                                2





                                $begingroup$


                                Japt, 18 bytes



                                _¬%1©U>½-Z¬u1 a½}a


                                Try it online!






                                share|improve this answer











                                $endgroup$




                                Japt, 18 bytes



                                _¬%1©U>½-Z¬u1 a½}a


                                Try it online!







                                share|improve this answer














                                share|improve this answer



                                share|improve this answer








                                edited 40 mins ago

























                                answered 1 hour ago









                                ASCII-onlyASCII-only

                                3,3821236




                                3,3821236












                                • $begingroup$
                                  Might be shorter using Arnauld's solution
                                  $endgroup$
                                  – ASCII-only
                                  33 mins ago


















                                • $begingroup$
                                  Might be shorter using Arnauld's solution
                                  $endgroup$
                                  – ASCII-only
                                  33 mins ago
















                                $begingroup$
                                Might be shorter using Arnauld's solution
                                $endgroup$
                                – ASCII-only
                                33 mins ago




                                $begingroup$
                                Might be shorter using Arnauld's solution
                                $endgroup$
                                – ASCII-only
                                33 mins ago











                                2












                                $begingroup$


                                J, 39 bytes



                                2(>:@])^:((<+.0=])(<.-.)@(-<.)@%:)^:_~]


                                Try it online!



                                Handles all test cases






                                share|improve this answer









                                $endgroup$


















                                  2












                                  $begingroup$


                                  J, 39 bytes



                                  2(>:@])^:((<+.0=])(<.-.)@(-<.)@%:)^:_~]


                                  Try it online!



                                  Handles all test cases






                                  share|improve this answer









                                  $endgroup$
















                                    2












                                    2








                                    2





                                    $begingroup$


                                    J, 39 bytes



                                    2(>:@])^:((<+.0=])(<.-.)@(-<.)@%:)^:_~]


                                    Try it online!



                                    Handles all test cases






                                    share|improve this answer









                                    $endgroup$




                                    J, 39 bytes



                                    2(>:@])^:((<+.0=])(<.-.)@(-<.)@%:)^:_~]


                                    Try it online!



                                    Handles all test cases







                                    share|improve this answer












                                    share|improve this answer



                                    share|improve this answer










                                    answered 27 mins ago









                                    JonahJonah

                                    2,351916




                                    2,351916






























                                        draft saved

                                        draft discarded




















































                                        If this is an answer to a challenge…




                                        • …Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.


                                        • …Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
                                          Explanations of your answer make it more interesting to read and are very much encouraged.


                                        • …Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.



                                        More generally…




                                        • …Please make sure to answer the question and provide sufficient detail.


                                        • …Avoid asking for help, clarification or responding to other answers (use comments instead).





                                        draft saved


                                        draft discarded














                                        StackExchange.ready(
                                        function () {
                                        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f180412%2fsquare-root-distance-from-integers%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

                                        Contact image not getting when fetch all contact list from iPhone by CNContact

                                        Image
                                        up vote 0 down vote favorite I know this question already asked but not getting solution. From this code I will get all the information from the contact but image not found when open vcf files on mac os, also not getting when share this file. I use this stackoverflow link here but It's not help full. var contacts = [CNContact]() let keys = [CNContactVCardSerialization.descriptorForRequiredKeys() ] as [Any] let request = CNContactFetchRequest(keysToFetch: keys as! [CNKeyDescriptor]) do { try self.contactStore.enumerateContacts(with: request) { (contact, stop) in // Array containing all unified contacts from everywhere contacts.append(contact) } } catch { print("unable to fetch contacts") } do { let data = try
                                        Read more

                                        count number of partitions of a set with n elements into k subsets

                                        Image
                                        up vote 6 down vote favorite 1 This program is for count number of partitions of a set with n elements into k subsets I am confusing here return k*countP(n-1, k) + countP(n-1, k-1); can some one explain what is happening here? why we are multiplying with k? NOTE->I know this is not the best way to calculate number of partitions that would be DP // A C++ program to count number of partitions // of a set with n elements into k subsets #include<iostream> using namespace std; // Returns count of different partitions of n // elements in k subsets int countP(int n, int k) { // Base cases if (n == 0 || k == 0 || k > n) return 0; if (k == 1 || k == n) return 1; // S(n+1, k) = k*S(n, k) + S(n, k-1) return k*countP(n-1, k) + countP(n-1, k-1); } // Driver program int ma
                                        Read more

                                        A CLEAN and SIMPLE way to add appendices to Table of Contents and bookmarks

                                        Image
                                        0 I found a lot of questions abount appendices and ToC. Many users want appendices to be grouped in an Appendix part, however some problems arise with ToC, hyperref, PDF viewer bookmarks, and so on. There are different solutions which require extra packages, command patching and other extra code, however none of them satisfies me. I almost found an easy way to accomplish a good result, where appendices are added to bookmarks in the right way and hyperref links point to the right page. However, the number of the "Appendix" part page is wrong (it's the number of appendix A). Is there any EASY way to fix that? This is a MWE: documentclass{book} usepackage[nottoc,notlot,notlof]{tocbibind} usepackage{hyperref} begin{document} frontmatter tableofcontents mainmatter part{First} chapter{
                                        Read more