[Rd] Error in FrF2 example on Mac OS

[Ulrike Grömping]

---------- Original Message ----------- 
 From: Simon Urbanek <simon.urbanek at r-project.org> 
 To: Ulrike Grömping <groemp at tfh-berlin.de> 
 Cc: r-devel at r-project.org 
 Sent: Wed, 25 Mar 2009 10:32:59 -0400 
 Subject: Re: [Rd] Error in FrF2 example on Mac OS

> On Mar 24, 2009, at 10:41 , Ulrike Grömping wrote: 
> 
> > 
> > 
> > Petr Savicky wrote: 
> >> 
> >> On Tue, Mar 24, 2009 at 02:45:57PM +0100, Uwe Ligges wrote: 
> >>>> gives the custom error message "nruns must be a power of 2.",   
> >>>> which is 
> >>>> generated in the first check within function FrF2: 
> >>>> 
> >>>>   if (!is.null(nruns)){ 
> >>>>      k <- floor(log2(nruns)) 
> >>>>      if (!2^k==nruns) stop("nruns must be a power of 2.")} 
> >>> 
> >>> 
> >>> Probably a rounding issue on different platforms? 
> >>> I guess the test should be something like: 
> >>> 
> >>> if (!is.null(nruns)){ 
> >>>  if(!isTRUE(all.equal(log2(nruns) %% 1, 0))) 
> >>>    stop("nruns must be a power of 2.") 
> >>> } 
> >> 
> >> Probably, k is needed also later. Assumig that 2^k works correctly, 
> >> the following could be sufficient 
> >> 
> >>   if (!is.null(nruns)){ 
> >>      k <- round(log2(nruns)) 
> >>      if (!2^k==nruns) stop("nruns must be a power of 2.")} 
> >> 
> >> In order to test the assumption, one can use 
> >> 
> >>  x <- 2^(0:100 + 0) # use double exponent to be sure 
> >>  all(x == floor(x)) 
> >> 
> >> Powers of two are represented exactly, since they have only one 
> >> significant bit. 
> >> 
> >> Petr. 
> >> 
> > 
> > Yes, round instead of floor should also do the job, if rounding is the 
> > issue. But then, with powers of 2 indeed being represented exactly   
> > (I would 
> > expect even on Macs), maybe rounding is not the issue? I have no   
> > possibility 
> > to check this, since I do not have access to a Mac with R installed.   
> > On my 
> > windows machine, 
> > all(log2(x)==floor(log2(x))) 
> > with x as defined above yields TRUE. 
> > 
> 
> What you're missing is that you cannot rely on log2 to give you an   
> integer. The test above bears no relevance to your problem - this is   
> not about representing 2^x - this is about log2 which you cannot   
> expect to satisfy log2(2^b) == b numerically since it could as well be   
> computed log(x)/log(2) which is not exactly representable. Use round   
> and all is well :). 
> 
> > which(floor(log2(2^x))!=x) 
>  [1]  4  7  8 13 14 15 25 27 29 49 53 57 64 97 
> > which(round(log2(2^x))!=x) 
> integer(0) 
> 
> Cheers, 
> Simon 
> 

Yes, round did indeed solve the problem, it just surprises me that the Mac is
so different from the other (binary) animals.

Regards,
Ulrike