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 >> >> >> > >

**References**:**count zeros in a number***From:*dimitris <dimmechan@yahoo.com>

**Re: count zeros in a number***From:*Richard Fateman <fateman@cs.berkeley.edu>

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