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

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          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











              $endgroup$













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                3 Answers
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                3 Answers
                3






                active

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                active

                oldest

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                active

                oldest

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                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






























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