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Re: Radicals simplify

Andrzej Kozlowski wrote:
> Yes, but Adam Strzebonski has already posted here the default 
> CoplexityFunction in the past, and knowing it really wont help you much:
> SimplifyCount[p_] :=
> If[Head[p]===Symbol, 1,
>    If[IntegerQ[p],
>       If[p==0, 1, Floor[N[Log[2, Abs[p]]/Log[2, 10]]]+If[p>0, 1, 
> 2]],
>       If[Head[p]===Rational,
>          SimplifyCount[Numerator[p]]+SimplifyCount[Denominator[p]]+1,
>          If[Head[p]===Complex,
>             SimplifyCount[Re[p]]+SimplifyCount[Im[p]]+1,
>             If[NumberQ[p], 2,
> 	       SimplifyCount[Head[p]]+If[Length[p]==0, 0,
> 	       Plus@@(SimplifyCount/@(List@@p))]]]]]]
> Would it be helpful to have this in the Documentation?

It would be more helpful if it were more readable, that's for sure.

I think Adam's function is equivalent to

SimplifyCount[p_Integer]:= Floor[N[Log[2, Abs[p]]/Log[2, 10]]]+If[p>0, 1, 2]
SimplifyCount[p_Complex]:= SimplifyCount[Re[p]]+SimplifyCount[Im[p]]+1,
SimplifyCount[p_?NumberQ]:= 2
SimplifyCount[p_] :=  Plus@@(SimplifyCount/@(Append[List@@p,Head[p]])) 
(*everything else*)

not tested, but you get the idea.  Adam's program may be faster in some 
way, or he may have written because he thinks in terms of If[Head[]==] 
  and it did not occur to him to use pattern matching. Or maybe my program
doesn't work because the interactions of the multiple patterns cause 
problems -- I require that the patterns be applied in the right order 
and Mathematica may be randomizing them in some way. (That is, 0 is an 
Integer, but we want to use the first line, not the 3rd).

One way to make documentation readable is to have a system that is 
logically consistent --- the semantics are simpler. By this measure, it 
would be much easier to understand the system and read this 
documentation if the pattern matching for p/q and r+s*I did not need 
special cases when p and q , or s were numbers, but just
looked like the same patterns as the cases when p,q, s are symbols.
Then the complexity function would look like this

SimplifyCount[p_Integer]:= Floor[N[Log[2,Abs[p]]/Log[2,10]]]+If[p>0,1,2]
SimplifyCount[p_?NumberQ]:= 2
SimplifyCount[p_] :=  Plus@@(SimplifyCount/@(Append[List@@p,Head[p]]))

  5 lines, not 11. [In fact, it may be that this documentation would be 
worth providing if it were noted that this is a "simplification" of how 
it really works.]

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