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Re: count zeros in a number

  • To: mathgroup at smc.vnet.net
  • Subject: [mg121862] Re: count zeros in a number
  • From: DrMajorBob <btreat1 at austin.rr.com>
  • Date: Wed, 5 Oct 2011 04:01:39 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201110020636.CAA28027@smc.vnet.net> <j6brvm$8om$1@smc.vnet.net>
  • Reply-to: drmajorbob at yahoo.com

A person just introduced to Mathematica is NOT, as yet, a Mathematica  
programmer.

Bobby

On Tue, 04 Oct 2011 15:57:30 -0500, Richard Fateman  
<fateman at eecs.berkeley.edu> wrote:

> 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..
> Do[.....,{1000}]//Timing
>
> 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...
>
> RJF
>
>
>


-- 
DrMajorBob at yahoo.com



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