Square Root Distance from Integers












4












$begingroup$


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











$endgroup$

















    4












    $begingroup$


    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











    $endgroup$















      4












      4








      4





      $begingroup$


      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 16 mins ago







      Stephen

















      asked 1 hour ago









      StephenStephen

      7,37823395




      7,37823395






















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          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











          $endgroup$





















            1












            $begingroup$


            Japt, 23 bytes



            _%1©(Z%1<Uª-Z%1Ä<U}a¬²r


            Try it online!






            share|improve this answer











            $endgroup$





















              1












              $begingroup$


              Wolfram Language (Mathematica), 36 bytes



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


              Try it online!



              Explanation



              The result must be of the form $n^2 pm 1$ for some $n in mathbb{N}$. Solving the inequations $sqrt{n^2+1} - n < k$ and $n - sqrt{n^2+1} < k$, we get $n > frac{1-k^2}{2k}$ and $n > frac{1+k^2}{2k}$ 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











              $endgroup$













                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

























                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                2












                $begingroup$

                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












                  $begingroup$

                  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












                    2








                    2





                    $begingroup$

                    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 12 mins ago

























                    answered 46 mins ago









                    ArnauldArnauld

                    76.8k693322




                    76.8k693322























                        1












                        $begingroup$


                        Japt, 23 bytes



                        _%1©(Z%1<Uª-Z%1Ä<U}a¬²r


                        Try it online!






                        share|improve this answer











                        $endgroup$


















                          1












                          $begingroup$


                          Japt, 23 bytes



                          _%1©(Z%1<Uª-Z%1Ä<U}a¬²r


                          Try it online!






                          share|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$


                            Japt, 23 bytes



                            _%1©(Z%1<Uª-Z%1Ä<U}a¬²r


                            Try it online!






                            share|improve this answer











                            $endgroup$




                            Japt, 23 bytes



                            _%1©(Z%1<Uª-Z%1Ä<U}a¬²r


                            Try it online!







                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited 20 mins ago

























                            answered 26 mins ago









                            ASCII-onlyASCII-only

                            3,3721236




                            3,3721236























                                1












                                $begingroup$


                                Wolfram Language (Mathematica), 36 bytes



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


                                Try it online!



                                Explanation



                                The result must be of the form $n^2 pm 1$ for some $n in mathbb{N}$. Solving the inequations $sqrt{n^2+1} - n < k$ and $n - sqrt{n^2+1} < k$, we get $n > frac{1-k^2}{2k}$ and $n > frac{1+k^2}{2k}$ 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











                                $endgroup$


















                                  1












                                  $begingroup$


                                  Wolfram Language (Mathematica), 36 bytes



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


                                  Try it online!



                                  Explanation



                                  The result must be of the form $n^2 pm 1$ for some $n in mathbb{N}$. Solving the inequations $sqrt{n^2+1} - n < k$ and $n - sqrt{n^2+1} < k$, we get $n > frac{1-k^2}{2k}$ and $n > frac{1+k^2}{2k}$ 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











                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$


                                    Wolfram Language (Mathematica), 36 bytes



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


                                    Try it online!



                                    Explanation



                                    The result must be of the form $n^2 pm 1$ for some $n in mathbb{N}$. Solving the inequations $sqrt{n^2+1} - n < k$ and $n - sqrt{n^2+1} < k$, we get $n > frac{1-k^2}{2k}$ and $n > frac{1+k^2}{2k}$ 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), 36 bytes



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


                                    Try it online!



                                    Explanation



                                    The result must be of the form $n^2 pm 1$ for some $n in mathbb{N}$. Solving the inequations $sqrt{n^2+1} - n < k$ and $n - sqrt{n^2+1} < k$, we get $n > frac{1-k^2}{2k}$ and $n > frac{1+k^2}{2k}$ 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 7 secs ago

























                                    answered 18 mins ago









                                    alephalphaalephalpha

                                    21.4k32991




                                    21.4k32991






























                                        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

                                        Сан-Квентин

                                        Image
                                        Сан-Квентин Местоположение Сан-Квентин, Калифорния Координаты 37°56′13″ с. ш. 122°29′20″ з. д. H G Я O L Текущий статус Действует Режим безопасности минимальный и максимальный Количество мест 3 302 Открытие Июнь 1852 Находится в ведомстве California Department of Corrections and Rehabilitation Начальник Mike Martel, начальник  Сан-Квентин на Викискладе Сан-Квентин (англ.  San Quentin ) — тюрьма, располагающаяся на 432 акрах (около 1,7 км²), на мысе Сан-Квентин, в округе Марин, штат Калифорния, США. Тюрьма штата Калифорния Сан-Квентин была открыта в июле 1852 года и является старейшей в штате. Она была построена заключёнными, которые проживали в течение строительства на тюремном судне Вабан. В Сан-Квентине отбывали наказание и мужчины, и женщины до 1934 года, когда построили женскую тюрьму в Техачапи и туда перевели всех заключённых женщин. В Сан-Квентине приводятся в исполнение смертные казни, для этого имеется газовая камера, но...
                                        Read more

                                        8-я гвардейская общевойсковая армия

                                        Image
                                        8-я гвардейская общевойсковая ордена Ленина армия Награды: Войска: РККА → Советская армия → СВ России Род войск: Формирование: 1943 Расформирование (преобразование): 1992 Предшественник: 62-я армия (1942—43) → 8-я гвардейская общевойсковая армия (1943—92) → 8-й гвардейский армейский корпус (1992—98) Боевой путь Изюм-Барвенковская операция Донбасская операция Запорожская операция Битва за Днепр Днепропетровская операция Березнеговато-Снигирёвская операция Одесская операция Люблин-Брестская операция Варшавско-Познанская операция Восточно-Померанская операция Берлинская наступательная операция У этого термина существуют и другие значения, см. 8-я армия. Фотография Гвардейского Красного Знамени (лицевая сторона) 8-й гвардейской армии , образца 1943 года. Фотография Гвардейского Красного Знамени (оборотная сторона) 8-й гвардейской армии , образца 1943 года. 8-я гвардейская общевойсковая ордена Ленина армия  — гвардейское фор...
                                        Read more

                                        Алькесар

                                        Image
                                        Муниципалитет Алькесар Alquézar Герб 42°10′16″ с. ш. 0°01′27″ в. д. H G Я O L Страна Испания   Испания Автономное сообщество Арагон Провинция Уэска Комарка Сомонтано-де-Барбастро Алькальд Хосе Марияно Альтемир Ласкорс (ИСРП) История и география Площадь 32,36 км² Высота НУМ 660 м Часовой пояс UTC+1, летом UTC+2 Население Население 313 человек ( 2010 ) Плотность 9,92 чел./км² Этнохороним alquezarano/-a Цифровые идентификаторы Телефонный код +34 974 Почтовые индексы 22112 Автомобильный код HU alquezar.org   (исп.) Показать/скрыть карты Алькесар Алькесар Алькесар  Аудио, фото и видео на Викискладе Алькесар (исп.  Alquézar ) — муниципалитет в Испании, входит в провинцию Уэска, в составе автономного сообщества Арагон. Муниципалитет находится в составе района (комарки) Сомонтано-де-Барбастро. Занимает площадь 32,36 км². Население — 321 челове...
                                        Read more