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Re: count zeros in a number
*To*: mathgroup at smc.vnet.net
*Subject*: [mg121868] Re: count zeros in a number
*From*: Richard Fateman <fateman at eecs.berkeley.edu>
*Date*: Wed, 5 Oct 2011 04:02:44 -0400 (EDT)
*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com
*References*: <201110020636.CAA28027@smc.vnet.net> <j6brvm$8om$1@smc.vnet.net> <201110040530.BAA20698@smc.vnet.net> <op.v2t544pxtgfoz2@bobbys-imac.local> <4E8B733A.8050506@eecs.berkeley.edu> <op.v2ujibmqtgfoz2@bobbys-imac.local>
On 10/4/2011 2:33 PM, DrMajorBob wrote:
> I'm guessing that I got zero because I ran some of the code with an
> undefined x. My bad!
I think that you (me too) might not have noticed that one of my methods
changed x. Here is a safer version.
savedx= 24^24*55^55;
Well, here's another method..
(xs = ToString[savedx];
StringLength[xs] -
StringLength[ToString[ToExpression[StringReverse[xs]]]])
Let's time it, running 1000 times.
Do[ (xs = ToString[savedx];
StringLength[xs] -
StringLength[
ToString[ToExpression[StringReverse[xs]]]]), {1000}] // Timing
{0.063,Null}
This bizarre hack is reasonably fast, indeed twice as fast as your
method, on my system.
Do[Replace[
IntegerDigits@x, {__, zeroes : 0 ..} :>
Length@{zeroes}], {1000}] // Timing
{0.141, Null}
and seven times as fast as my older simple arithmetic method which uses
iteration,
Do[(n = 0; x = savedx; While[Mod[x, 10] == 0, (x = x/10; n++)];
n), {1000}] // Timing
{0.438, Null}
and which (as I contended) was inefficient in Mathematica. While you
seem to think that this method is ancient and outmoded, I think it is
not the method, but the implementation of iteration. Or perhaps of
arithmetic?
Another implementation of this arithmetical method on the same computer
takes 0.063 seconds.
RJF
>
> Correcting that, I find the esoteric method faster than your ancient,
> outmoded ones:
>
> x = 24^24*55^55;
> Replace[IntegerDigits@x, {__, zeroes : 0 ..} :>
> Length@{zeroes}] // Timing
>
> {0.000198, 55}
>
> (n = 0; While[GCD[x, 10^n] == 10^n, n++]; n - 1) // Timing
>
> {0.000417, 55}
>
> (n = 0; While[Mod[x, 10] == 0, (x = x/10; n++)]; n) // Timing
>
> {0.000228, 55}
>
> 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
>>
>>
>>
>
>
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