simplifying expressions - How do I invoke the default complexity function?

Documentation on ComplexityFunction says:

With the default setting ComplexityFunction->Automatic, forms are ranked primarily according to their LeafCount, with corrections to treat integers with more digits as more complex.

I need to use this default function on its own, not in Simplify or FullSimplify. Can I invoke it from my code?

If no, could you give me a custom function that behaves as close to the default function as possible?

Thanks!

Answer

The code for the default ComplexityFunction was posted on MathSource a number of years ago by Adam Strzebonski (of Wolfram Research). You will see reference to the original reply from Adam referenced in a MathGroup reply from Andrzej Kozlowski dated 12 Jan 2010 with the subject: "[mg106386] Re : Radicals simplify". I mention all that because I can't get the hyperlink to work (: The code Adam provided is there as well. The implementation from Adam used nested If statements. I can't resist the urge to use Which instead. I give my version below. I don't know for sure that the same function is used, but I have seen no reports indicating that it has changed.

SimplifyCount[p_]:=With[{hd=Head[p]},

  Which[
    hd===Symbol,1,
    hd===Integer,If[p===0,1,Floor[N[Log[2,Abs[p]]/Log[2,10]]]+If[Positive[p],1,2]],
    hd===Rational,SimplifyCount[Numerator[p]]+SimplifyCount[Denominator[p]]+1,
    hd===Complex,SimplifyCount[Re[p]]+SimplifyCount[Im[p]]+1,
    NumberQ[p],2,
    True,SimplifyCount[Head[p]]+If[Length[p]==0,0,Plus@@(SimplifyCount/@(List@@p))]
  ]
]

Comments

plotting - How to draw lines between specified dots on ListPlot?

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calculus and analysis - Integral of the Sinc product

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