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Square Root Distance from Integers


Calculate the square root only using ++Sorted Lexical Partition of a NumberReverse and squareThe fastest square root calculatorRobbers - square times square rootCops - square times square rootFermat's factorization helperMiller-Rabin Strong PseudoprimesExact change in fewest bills and coinsApproximate My Squares













5












$begingroup$


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











$endgroup$

















    5












    $begingroup$


    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











    $endgroup$















      5












      5








      5





      $begingroup$


      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 21 mins ago







      Stephen

















      asked 1 hour ago









      StephenStephen

      7,38323395




      7,38323395






















          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











          $endgroup$





















            2












            $begingroup$


            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 $sqrt{n^2+1} - n le k$ and $n - sqrt{n^2+1} le k$, we get $n ge frac{1-k^2}{2k}$ and $n ge frac{1+k^2}{2k}$ 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











            $endgroup$





















              1












              $begingroup$


              Japt, 20 bytes



              _¬%1©½-(Z¬%1 a½ <U}a


              Try it online!






              share|improve this answer











              $endgroup$













              • $begingroup$
                21 but faster: _%1©½-(Z%1 a½ <U}a¬²r
                $endgroup$
                – ASCII-only
                2 mins ago













              Your Answer





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              3 Answers
              3






              active

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              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 17 mins ago

























                  answered 51 mins ago









                  ArnauldArnauld

                  76.8k693322




                  76.8k693322























                      2












                      $begingroup$


                      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 $sqrt{n^2+1} - n le k$ and $n - sqrt{n^2+1} le k$, we get $n ge frac{1-k^2}{2k}$ and $n ge frac{1+k^2}{2k}$ 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











                      $endgroup$


















                        2












                        $begingroup$


                        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 $sqrt{n^2+1} - n le k$ and $n - sqrt{n^2+1} le k$, we get $n ge frac{1-k^2}{2k}$ and $n ge frac{1+k^2}{2k}$ 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











                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$


                          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 $sqrt{n^2+1} - n le k$ and $n - sqrt{n^2+1} le k$, we get $n ge frac{1-k^2}{2k}$ and $n ge frac{1+k^2}{2k}$ 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), 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 $sqrt{n^2+1} - n le k$ and $n - sqrt{n^2+1} le k$, we get $n ge frac{1-k^2}{2k}$ and $n ge frac{1+k^2}{2k}$ 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 5 mins ago

























                          answered 23 mins ago









                          alephalphaalephalpha

                          21.4k32991




                          21.4k32991























                              1












                              $begingroup$


                              Japt, 20 bytes



                              _¬%1©½-(Z¬%1 a½ <U}a


                              Try it online!






                              share|improve this answer











                              $endgroup$













                              • $begingroup$
                                21 but faster: _%1©½-(Z%1 a½ <U}a¬²r
                                $endgroup$
                                – ASCII-only
                                2 mins ago


















                              1












                              $begingroup$


                              Japt, 20 bytes



                              _¬%1©½-(Z¬%1 a½ <U}a


                              Try it online!






                              share|improve this answer











                              $endgroup$













                              • $begingroup$
                                21 but faster: _%1©½-(Z%1 a½ <U}a¬²r
                                $endgroup$
                                – ASCII-only
                                2 mins ago
















                              1












                              1








                              1





                              $begingroup$


                              Japt, 20 bytes



                              _¬%1©½-(Z¬%1 a½ <U}a


                              Try it online!






                              share|improve this answer











                              $endgroup$




                              Japt, 20 bytes



                              _¬%1©½-(Z¬%1 a½ <U}a


                              Try it online!







                              share|improve this answer














                              share|improve this answer



                              share|improve this answer








                              edited 3 mins ago

























                              answered 31 mins ago









                              ASCII-onlyASCII-only

                              3,3721236




                              3,3721236












                              • $begingroup$
                                21 but faster: _%1©½-(Z%1 a½ <U}a¬²r
                                $endgroup$
                                – ASCII-only
                                2 mins ago




















                              • $begingroup$
                                21 but faster: _%1©½-(Z%1 a½ <U}a¬²r
                                $endgroup$
                                – ASCII-only
                                2 mins ago


















                              $begingroup$
                              21 but faster: _%1©½-(Z%1 a½ <U}a¬²r
                              $endgroup$
                              – ASCII-only
                              2 mins ago






                              $begingroup$
                              21 but faster: _%1©½-(Z%1 a½ <U}a¬²r
                              $endgroup$
                              – ASCII-only
                              2 mins ago




















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