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Re: count zeros in a number

  • To: mathgroup at
  • Subject: [mg121859] Re: count zeros in a number
  • From: Richard Fateman <fateman at>
  • Date: Wed, 5 Oct 2011 04:01:07 -0400 (EDT)
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  • References: <> <j6brvm$8om$> <> <op.v2t544pxtgfoz2@bobbys-imac.local>

On 10/4/2011 9:44 AM, DrMajorBob wrote:
> There's nothing "esoteric" about IntegerDigits, Replace, or Repeated, 
> as in:
> x = 24^24*55^55;
> Replace[IntegerDigits@x, {__, zeroes : 0 ..} :>
>    Length@{zeroes}] // Timing
> {0.000185, 55}

My feeling is that these ARE esoteric.  A person just introduced to 
Mathematica would probably not encounter IntegerDigits, Repeated, or 
patterns other than the trivial one used in pseudo function definitions as
f[x_]:=   x+1.

> If we wanted brute-force arithmetic, we might use Fortran.
> Your suggested solutions are NOT greatly hindered by "the slow 
> implementation of looping constructs in Mathematica", on the other hand:
> (n = 0; While[GCD[x, 10^n] == 10^n, n++]; n - 1) // Timing
> {0.000029, 0}

huh?   I got  {0., 55}.   Perhaps my computer is faster or has a lower 
resolution timer, but we should both get 55.

> (n = 0; While[Mod[x, 10] == 0, (x = x/10; n++)]; n) // Timing
> {0.000018, 0}

I mention the slow implementation of looping constructs simply to warn 
people that if you can resolve your computation by simple composition of 
built-in functions e.g. F[G[H[x]]]   and don't rely on Mathematica
to efficiently execute loops written as While[] For[]  etc, you are 
probably going to run faster.  Let Mathematica's operation on compound 
objects like lists do the job.  Length[] is a lot faster than counting 
with a While etc.

In my timings, doing the operations 1000 times..

I found that my solutions were about 10X faster than the IntegerDigits 
one, on this particular value of x.

Fortran can't be used unless you manage arbitrary precision integers...


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