4
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Let's say that we have at our disposal

nbfunc = 4;

functions, whose exact expressions are gathered in the list

tabfunc = Table[LegendreP[n,x], {n, 1, nbfunc}];

For the purpose of the example, I gave myself the list of Legendre polynomials, whose evaluation can be compiled. For a given run, the number of functions nbfunc and their expressions tabfunc are fixed forever. But, they may change from one run to another.

Now, I would like to evaluate efficiently the function f[neval_,xeval_] defined as

f[neval_,xeval_] := tabfunc[[neval]] /. {x -> xeval};

where the integer neval is always assumed to satisfy 1 <= neval <= nbfunc. I can compile this evaluation by defining the function fC as

fC = Compile[{{n, _Integer}, {x, _Real}},
tabfunc[[n]],
CompilationTarget -> "C",
CompilationOptions -> {"ExpressionOptimization" -> True, "InlineCompiledFunctions" -> True, "InlineExternalDefinitions" -> True},
RuntimeOptions -> {"CatchMachineOverflow" -> False , "CatchMachineUnderflow" -> False, "CatchMachineIntegerOverflow" -> False, "CompareWithTolerance" -> False, "EvaluateSymbolically" -> False}];

The compilation works correctly (in particular no calls to MainEvaluate), and the timings are improved

f[4, 0.2] == fC[4, 0.2]
Table[f[4, 0.2], {i, 1, 1000}]; // AbsoluteTiming // First
Table[fC[4, 0.2], {i, 1, 1000}]; // AbsoluteTiming // First
(*True
0.005839
0.000169*)

Yet, when inspecting the compiled code of fC,

Needs["CompiledFunctionTools`"];
CompilePrint[fC]

we note that the code contains lines of the form

22  T(R1)0 = {R0, R6, R8, R5}
23  R6 = Part[ T(R1)0, I0]
24  Return

This implies that to compute fC[4,0.2], the compiled function first computes the list {f[1,0.2],f[2,0.2],f[3,0.2],f[4,0.2]}, then returns the fourth element of this list, i.e. the value f[4,0.2]. Unfortunately, this is unsatisfactory, as to compute fC[4,0.2], there should be no need to compute all the others f[i,0.2] for 1<=i<=3.

My question is therefore as follows:

How should one proceed to compile the function fC[n_,x_], so that only the needed expression for the n that is asked is effectively evaluated?

In particular, I face the difficulty that the number nbfunc may change from one run to another (and can be quite large), so that this cannot be done by hand.

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3
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Since Which is compilable, you can try this:

fC = Block[{n},
   With[{

    code = Which @@Join[Riffle[Thread[n == Range[nbfunc]], N[tabfunc]], {True,0.}]

    },
    Compile[{{n, _Integer}, {x, _Real}},
     code,
     CompilationTarget -> "C",
     Parallelization -> True,
     RuntimeAttributes -> {Listable},
     RuntimeOptions -> "Speed"
     ]
    ]
   ];

This way, the function is both listable and parallelized (although parallelization does not help very much):

k = 1000000;
fC[RandomInteger[{1, 4}, k], RandomReal[{-1, 1}, k]]; // 
  AbsoluteTiming // First

Don't forget to give a default case (with True); it might not work otherwise. Notice also how I inlined the actual code with With in order to "precompute" what can be computed in advance. The respective "CompilationOptions" for inlining are infamous for their lack of robustness.

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  • $\begingroup$ Thank you very much for your prompt answer! To be sure, I understand all the details, could you elaborate on: (i) "Notice also how inlined the actual code with With in order to precompute what can be computed in advance.": I am not sure I understand what this means, could you explain it to me? (ii) "The respective CompilationOptions for inlining are infamous for their lack of robustness.": I was unaware of these infamous issues, would you have a relevant link to point me to? $\endgroup$ – jibe Aug 21 '18 at 18:02
  • 1
    $\begingroup$ Right, there is no real computation done. What I meant is that the expression for code is create. But one can performs some simplifications in advance, for example, the conversion to machine floating point numbers of the constants in tabfunc; otherwise, these conversion will be done by type casts at runtime. $\endgroup$ – Henrik Schumacher Aug 21 '18 at 18:06
  • $\begingroup$ Towards (ii): mathematica.stackexchange.com/a/13289 $\endgroup$ – Henrik Schumacher Aug 22 '18 at 10:06

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