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

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          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









              $endgroup$













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                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






























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