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geo
Posts:
38
Registered:
6/19/08
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Re: KISS4691, a potentially top-ranked RNG.
Posted:
Jul 25, 2010 9:53 AM
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On Jul 24, 11:34 pm, Gib Bogle <g.bo...@auckland.no.spam.ac.nz> wrote:
> geo wrote:
>
> > Languages requiring signed integers should use equivalents
> > of the same operations, except that the C statement:
> > c=(t<x)+(x>>19);
> > can be replaced by that language's version of
> > if sign(x<<13+c)=sign(x) then c=sign(x)+(x>>19)
> > else c=1-sign(t)+(x>>19)
> > */
>
> Hi George,
>
> I translated this into Fortran, and found that I get different results than with
> C. I've tracked the difference into MWC(). The following Fortran code, with my
> laborious comparison of two signed integers treating them as unsigned, gives
> correct results. If I comment out the line
> c = tLTx + SHIFTR(x,19)
> and uncomment the following lines that implement your suggestion above to
> compute c, I get different results.
>
> integer function MWC()
> integer :: t, x, i
> integer, save :: c = 0, j = 4691
> integer :: tLTx
>
> if (j < 4690) then
> j = j + 1
> else
> j = 0
> endif
> x = Q(j)
> t = SHIFTL(x,13) + c + x
> if ((t >= 0 .and. x >= 0) .or. (t < 0 .and. x < 0)) then
> if (t < x) then
> tLTx = 1
> else
> tLTx = 0
> endif
> elseif (x < 0) then
> tLTx = 1
> elseif (t < 0) then
> tLTx = 0
> endif
>
> c = tLTx + SHIFTR(x,19)
>
> !if (sign(1,SHIFTL(x,13)+c) == sign(1,x)) then
> ! c = sign(1,x) + SHIFTR(x,19)
> !else
> ! c = 1 - sign(1,t) + SHIFTR(x,19)
> !endif
> Q(j) = t
> MWC = t
> end function
>
> Is it the case that although your suggested workaround gives different results
> from the C code in this case, it is still equivalent as a RNG?
>
> Cheers
> Gib
Thanks very much for the Fortran version.
I made a mistake in the comment on versions
for signed integers. This:
Languages requiring signed integers should use equivalents
of the same operations, except that the C statement:
c=(t<x)+(x>>19);
can be replaced by that language's version of
if sign(x<<13+c)=sign(x) then c=sign(x)+(x>>19)
else c=1-sign(t)+(x>>19)
should have been:
Languages requiring signed integers should use equivalents
of the same operations, except that the C statement:
c=(t<x)+(x>>19);
can be replaced by that language's version of
if sign(x<<13+c)=sign(x) then c=sign(x)+(x>>19)
else c=1-sign(x<<13+c)+(x>>19)
Sorry for that error.
I still like inline functions in Fortan,
so would tend to define
isign(x)=ishft(x,-31)
and
m=ishft(x,13)+c
if(isign(m).eq.isign(x)) then c=isign(x)+ishft(x,-19)
else c=1-isign(m)+ishft(x,-19)
and
Q[j]=m+x
If calculating the carry c of the MWC operation fails
to fix that extra increment properly, then rather than
a systematic expansion, in reverse order, 32 bits at a time,
of the binary representation of j/(1893*2^150112-1) for some
j determined by the random seeds, we will be jumping around
in that expansion, and we can't be sure that the period will
still be the order of b=2^32 for the prime p=1893*b^4196-1.
gm
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