Square Root Distance from Integers












8












$begingroup$


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.










share|improve this question











$endgroup$

















    8












    $begingroup$


    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.










    share|improve this question











    $endgroup$















      8












      8








      8





      $begingroup$


      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.










      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






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 3 hours ago







      Stephen

















      asked 3 hours ago









      StephenStephen

      7,39823395




      7,39823395






















          6 Answers
          6






          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
                2 hours ago



















              2












              $begingroup$


              J, 39 29 bytes



              [:<./_1 1++:*:@>.@%~1+(,-)@*:


              NB. This shorter version simply uses @alephalpha's formula.



              Try it online!



              39 bytes, original, brute force



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


              Try it online!



              Handles all test cases






              share|improve this answer











              $endgroup$





















                1












                $begingroup$


                C# (Visual C# Interactive Compiler), 89 bytes





                k=>{double n=1,p;for(;Math.Abs(Math.Round(p=Math.Sqrt(0d+n++))-p)>k|p%1==0;);return n-1;}


                Try it online!






                share|improve this answer









                $endgroup$





















                  1












                  $begingroup$


                  Python, 42 bytes





                  lambda k:((k-1/k)//2)**2+1-2*(k<1/k%2<2-k)


                  Try it online!



                  Based on alephalpha's formula, explicitly checking if we're in the $m^2-1$ or $m^2+1$ case via the condition k<1/k%2<2-k.



                  Python 3.8 can save a byte with an inline assignment.



                  Python 3.8, 41 bytes





                  lambda k:((a:=k-1/k)//2)**2-1+2*(a/2%1<k)


                  Try it online!



                  These beat my recursive solution:



                  50 bytes





                  f=lambda k,x=1:k>.5-abs(x**.5%1-.5)>0 or-~f(k,x+1)


                  Try it online!






                  share|improve this answer











                  $endgroup$













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                    6 Answers
                    6






                    active

                    oldest

                    votes








                    6 Answers
                    6






                    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 2 hours ago

























                        answered 3 hours 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 3 hours ago

























                                answered 3 hours 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
                                      2 hours 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
                                      2 hours 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 2 hours ago

























                                    answered 3 hours ago









                                    ASCII-onlyASCII-only

                                    3,3821236




                                    3,3821236












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


















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
















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




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











                                    2












                                    $begingroup$


                                    J, 39 29 bytes



                                    [:<./_1 1++:*:@>.@%~1+(,-)@*:


                                    NB. This shorter version simply uses @alephalpha's formula.



                                    Try it online!



                                    39 bytes, original, brute force



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


                                    Try it online!



                                    Handles all test cases






                                    share|improve this answer











                                    $endgroup$


















                                      2












                                      $begingroup$


                                      J, 39 29 bytes



                                      [:<./_1 1++:*:@>.@%~1+(,-)@*:


                                      NB. This shorter version simply uses @alephalpha's formula.



                                      Try it online!



                                      39 bytes, original, brute force



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


                                      Try it online!



                                      Handles all test cases






                                      share|improve this answer











                                      $endgroup$
















                                        2












                                        2








                                        2





                                        $begingroup$


                                        J, 39 29 bytes



                                        [:<./_1 1++:*:@>.@%~1+(,-)@*:


                                        NB. This shorter version simply uses @alephalpha's formula.



                                        Try it online!



                                        39 bytes, original, brute force



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


                                        Try it online!



                                        Handles all test cases






                                        share|improve this answer











                                        $endgroup$




                                        J, 39 29 bytes



                                        [:<./_1 1++:*:@>.@%~1+(,-)@*:


                                        NB. This shorter version simply uses @alephalpha's formula.



                                        Try it online!



                                        39 bytes, original, brute force



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


                                        Try it online!



                                        Handles all test cases







                                        share|improve this answer














                                        share|improve this answer



                                        share|improve this answer








                                        edited 1 hour ago

























                                        answered 2 hours ago









                                        JonahJonah

                                        2,351916




                                        2,351916























                                            1












                                            $begingroup$


                                            C# (Visual C# Interactive Compiler), 89 bytes





                                            k=>{double n=1,p;for(;Math.Abs(Math.Round(p=Math.Sqrt(0d+n++))-p)>k|p%1==0;);return n-1;}


                                            Try it online!






                                            share|improve this answer









                                            $endgroup$


















                                              1












                                              $begingroup$


                                              C# (Visual C# Interactive Compiler), 89 bytes





                                              k=>{double n=1,p;for(;Math.Abs(Math.Round(p=Math.Sqrt(0d+n++))-p)>k|p%1==0;);return n-1;}


                                              Try it online!






                                              share|improve this answer









                                              $endgroup$
















                                                1












                                                1








                                                1





                                                $begingroup$


                                                C# (Visual C# Interactive Compiler), 89 bytes





                                                k=>{double n=1,p;for(;Math.Abs(Math.Round(p=Math.Sqrt(0d+n++))-p)>k|p%1==0;);return n-1;}


                                                Try it online!






                                                share|improve this answer









                                                $endgroup$




                                                C# (Visual C# Interactive Compiler), 89 bytes





                                                k=>{double n=1,p;for(;Math.Abs(Math.Round(p=Math.Sqrt(0d+n++))-p)>k|p%1==0;);return n-1;}


                                                Try it online!







                                                share|improve this answer












                                                share|improve this answer



                                                share|improve this answer










                                                answered 2 hours ago









                                                Embodiment of IgnoranceEmbodiment of Ignorance

                                                1,170119




                                                1,170119























                                                    1












                                                    $begingroup$


                                                    Python, 42 bytes





                                                    lambda k:((k-1/k)//2)**2+1-2*(k<1/k%2<2-k)


                                                    Try it online!



                                                    Based on alephalpha's formula, explicitly checking if we're in the $m^2-1$ or $m^2+1$ case via the condition k<1/k%2<2-k.



                                                    Python 3.8 can save a byte with an inline assignment.



                                                    Python 3.8, 41 bytes





                                                    lambda k:((a:=k-1/k)//2)**2-1+2*(a/2%1<k)


                                                    Try it online!



                                                    These beat my recursive solution:



                                                    50 bytes





                                                    f=lambda k,x=1:k>.5-abs(x**.5%1-.5)>0 or-~f(k,x+1)


                                                    Try it online!






                                                    share|improve this answer











                                                    $endgroup$


















                                                      1












                                                      $begingroup$


                                                      Python, 42 bytes





                                                      lambda k:((k-1/k)//2)**2+1-2*(k<1/k%2<2-k)


                                                      Try it online!



                                                      Based on alephalpha's formula, explicitly checking if we're in the $m^2-1$ or $m^2+1$ case via the condition k<1/k%2<2-k.



                                                      Python 3.8 can save a byte with an inline assignment.



                                                      Python 3.8, 41 bytes





                                                      lambda k:((a:=k-1/k)//2)**2-1+2*(a/2%1<k)


                                                      Try it online!



                                                      These beat my recursive solution:



                                                      50 bytes





                                                      f=lambda k,x=1:k>.5-abs(x**.5%1-.5)>0 or-~f(k,x+1)


                                                      Try it online!






                                                      share|improve this answer











                                                      $endgroup$
















                                                        1












                                                        1








                                                        1





                                                        $begingroup$


                                                        Python, 42 bytes





                                                        lambda k:((k-1/k)//2)**2+1-2*(k<1/k%2<2-k)


                                                        Try it online!



                                                        Based on alephalpha's formula, explicitly checking if we're in the $m^2-1$ or $m^2+1$ case via the condition k<1/k%2<2-k.



                                                        Python 3.8 can save a byte with an inline assignment.



                                                        Python 3.8, 41 bytes





                                                        lambda k:((a:=k-1/k)//2)**2-1+2*(a/2%1<k)


                                                        Try it online!



                                                        These beat my recursive solution:



                                                        50 bytes





                                                        f=lambda k,x=1:k>.5-abs(x**.5%1-.5)>0 or-~f(k,x+1)


                                                        Try it online!






                                                        share|improve this answer











                                                        $endgroup$




                                                        Python, 42 bytes





                                                        lambda k:((k-1/k)//2)**2+1-2*(k<1/k%2<2-k)


                                                        Try it online!



                                                        Based on alephalpha's formula, explicitly checking if we're in the $m^2-1$ or $m^2+1$ case via the condition k<1/k%2<2-k.



                                                        Python 3.8 can save a byte with an inline assignment.



                                                        Python 3.8, 41 bytes





                                                        lambda k:((a:=k-1/k)//2)**2-1+2*(a/2%1<k)


                                                        Try it online!



                                                        These beat my recursive solution:



                                                        50 bytes





                                                        f=lambda k,x=1:k>.5-abs(x**.5%1-.5)>0 or-~f(k,x+1)


                                                        Try it online!







                                                        share|improve this answer














                                                        share|improve this answer



                                                        share|improve this answer








                                                        edited 47 mins ago

























                                                        answered 1 hour ago









                                                        xnorxnor

                                                        91.1k18186442




                                                        91.1k18186442






























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