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I want to calculate a loop. For speed, I compiled my code. Even after the compilation, it is still very slow in getting the results.

Are there any further improvements that can be made to my code?

Needs["CCompilerDriver`"]
c = 
  Compile[{{x0, _Integer}}, 
    With[{x = x0}, 
      pover = 0.0; t = 1.0;  
      While[t <= x - 1, 
        pover = 
          pover + 
            Total[
              Table[
                1.0/Sqrt[a^2 + t^2 + (100000.0 - 0.7)^2] - 
                1.0/Sqrt[a^2 + t^2 + (100000.0 + 0.7)^2], 
                {a, t + 1, 40000.0,1}]]; 
        t++];
      pover], 
    CompilationTarget -> "C", 
    Parallelization -> True, 
    RuntimeOptions -> "Speed"]`

Timing[c[2500.0]]

The result is

{126.42, 0.0125525}
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2 Answers 2

8
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Runs in a quarter-second by reformulating as a single two-index sum:

c = Compile[{{x, _Integer}},
      Sum[1/Sqrt[a^2 + t^2 + (100000 - 0.7)^2] - 
          1/Sqrt[a^2 + t^2 + (100000 + 0.7)^2],
          {t, 1, x - 1}, {a, t + 1, 40000}],
      CompilationTarget -> "C", Parallelization -> True, RuntimeOptions -> "Speed"];

c[2500] // AbsoluteTiming
(*    {0.262834, 0.0125525}    *)

If an approximation is enough for you, then we can replace the sum by an integral:

F[q_, y_, x_] = 
  Assuming[q > 0 && 1 <= x < y,
    Integrate[1/Sqrt[a^2 + t^2 + q],
              {t, 1 - 1/2, x - 1 + 1/2},
              {a, t + 1 - 1/2, y + 1/2}]];

c[x_] = F[(100000 - 0.7)^2, 40000, x] - F[(100000 + 0.7)^2, 40000, x];

c[2500] // Re // RepeatedTiming
(*    {0.0000615271, 0.0125525}    *)

evaluates in 60 microseconds.

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2
  • $\begingroup$ would it possible to make it faster? $\endgroup$
    – Q. Wang
    Feb 23, 2021 at 13:31
  • $\begingroup$ Yes, if approximations are allowed. See update. $\endgroup$
    – Roman
    Feb 23, 2021 at 15:39
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This runs in about a second:

Needs["CCompilerDriver`"]
c = Compile[{{x, _Integer}}, Block[{pover,t}, pover = 0.0; t = 1.0;  
   While[t <= x - 1, 
    pover = pover + 
      Total[Table[
        1.0/Sqrt[a^2 + t^2 + (100000.0 - 0.7)^2] - 
         1.0/Sqrt[a^2 + t^2 + (100000.0 + 0.7)^2], 
         {a, t + 1, 40000.0,
          1}]]; t++];
   pover], CompilationTarget -> "C", Parallelization -> True, 
  RuntimeOptions -> "Speed"]
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