KISS4691, a potentially top-ranked RNG.

Discussion in 'C Programming' started by geo, Jul 24, 2010.

  1. geo

    geo Guest

    I have been asked to recommend an RNG
    (Random Number Generator) that ranks
    at or near the top in all of the categories:
    performance on tests of randomness,
    length of period, simplicity and speed.
    The most important measure, of course, is
    performance on extensive tests of randomness, and for
    those that perform well, selection may well depend
    on those other measures.

    The following KISS version, perhaps call it KISS4691,
    seems to rank at the top in all of those categories.
    It is my latest, and perhaps my last, as, at age 86,
    I am slowing down.

    Compiling and running the following commented
    C listing should produce, within about four seconds,
    10^9 calls to the principal component MWC(), then
    10^9 calls to the KISS combination in another ~7 seconds.

    Try it; you may like it.

    George Marsaglia


    /*
    The KISS4691 RNG, a Keep-It-Simple-Stupid combination of
    MWC (Multiply-With-Carry), Congruential and Xorshift sequences.
    Expressed as a simple MWC() function and C macros
    #define CNG ( xcng=69069*xcng+123 )
    #define XS ( xs^=(xs<<13), xs^=(xs>>17), xs^=(xs<<5) )
    #define KISS ( MWC()+CNG+XS )
    but easily expressed in other languages, with a slight
    complication for signed integers.

    With its immense period, >10^45000, and speed: MWC()s at
    around 238 million/sec or full KISSes at around 138 million,
    there are few RNGs that do as well as this one
    on tests of randomness and are comparable in even one
    of the categories: period, speed, simplicity---not to
    mention comparable in all of them.

    The MWC4691 sequence x[n]=8193*x[n-4691]+carry mod b=2^32
    is based on p=8193*b^4691-1, period ~ 2^150124 or 10^45192
    For the MWC (Multiply-With-Carry) process, given a current
    x and c, the new x is (8193*x+c) mod b,
    the new c is the integer part of (8193*x+c)/b.

    The routine MWC() produces, in reverse order, the base-b=2^32
    expansion of some j/p with 0<j<p=8193*2^150112-1 with j
    determined by the 64 bits of seeds xcng and xs, or more
    generally, by 150112 random bits in the Q[] array.
    */

    static unsigned long xs=521288629,xcng=362436069,Q[4691];

    unsigned long MWC(void) /*takes about 4.2 nanosecs or 238 million/
    second*/
    {unsigned long t,x,i; static c=0,j=4691;
    j=(j<4690)? j+1:0;
    x=Q[j];
    t=(x<<13)+c+x; c=(t<x)+(x>>19);
    return (Q[j]=t);
    }

    #define CNG ( xcng=69069*xcng+123 )
    #define XS ( xs^=(xs<<13), xs^=(xs>>17), xs^=(xs<<5) )
    #define KISS ( MWC()+CNG+XS ) /*138 million/sec*/

    #include <stdio.h>
    int main()
    {unsigned long i,x;
    for(i=0;i<4691;i++) Q=CNG+XS;
    for(i=0;i<1000000000;i++) x=MWC();
    printf(" MWC result=3740121002 ?\n%22u\n",x);
    for(i=0;i<1000000000;i++) x=KISS;
    printf("KISS result=2224631993 ?\n%22u\n",x);
    }

    /*
    This RNG uses two seeds to fill the Q[4691] array by
    means of CNG+XS mod 2^32. Users requiring more than two
    seeds will need to randomly seed Q[] in main().
    By itself, the MWC() routine passes all tests in the
    Diehard Battery of Tests, but it is probably a good
    idea to include it in the KISS combination.

    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)
    */
    geo, Jul 24, 2010
    #1
    1. Advertising

  2. geo

    jacob navia Guest

    geo a écrit :
    > I have been asked to recommend an RNG
    > (Random Number Generator) that ranks
    > at or near the top in all of the categories:
    > performance on tests of randomness,
    > length of period, simplicity and speed.
    > The most important measure, of course, is
    > performance on extensive tests of randomness, and for
    > those that perform well, selection may well depend
    > on those other measures.
    >
    > The following KISS version, perhaps call it KISS4691,
    > seems to rank at the top in all of those categories.
    > It is my latest, and perhaps my last, as, at age 86,
    > I am slowing down.
    >
    > Compiling and running the following commented
    > C listing should produce, within about four seconds,
    > 10^9 calls to the principal component MWC(), then
    > 10^9 calls to the KISS combination in another ~7 seconds.
    >
    > Try it; you may like it.
    >
    > George Marsaglia
    >
    >
    > /*
    > The KISS4691 RNG, a Keep-It-Simple-Stupid combination of
    > MWC (Multiply-With-Carry), Congruential and Xorshift sequences.
    > Expressed as a simple MWC() function and C macros
    > #define CNG ( xcng=69069*xcng+123 )
    > #define XS ( xs^=(xs<<13), xs^=(xs>>17), xs^=(xs<<5) )
    > #define KISS ( MWC()+CNG+XS )
    > but easily expressed in other languages, with a slight
    > complication for signed integers.
    >
    > With its immense period, >10^45000, and speed: MWC()s at
    > around 238 million/sec or full KISSes at around 138 million,
    > there are few RNGs that do as well as this one
    > on tests of randomness and are comparable in even one
    > of the categories: period, speed, simplicity---not to
    > mention comparable in all of them.
    >
    > The MWC4691 sequence x[n]=8193*x[n-4691]+carry mod b=2^32
    > is based on p=8193*b^4691-1, period ~ 2^150124 or 10^45192
    > For the MWC (Multiply-With-Carry) process, given a current
    > x and c, the new x is (8193*x+c) mod b,
    > the new c is the integer part of (8193*x+c)/b.
    >
    > The routine MWC() produces, in reverse order, the base-b=2^32
    > expansion of some j/p with 0<j<p=8193*2^150112-1 with j
    > determined by the 64 bits of seeds xcng and xs, or more
    > generally, by 150112 random bits in the Q[] array.
    > */
    >
    > static unsigned long xs=521288629,xcng=362436069,Q[4691];
    >
    > unsigned long MWC(void) /*takes about 4.2 nanosecs or 238 million/
    > second*/
    > {unsigned long t,x,i; static c=0,j=4691;
    > j=(j<4690)? j+1:0;
    > x=Q[j];
    > t=(x<<13)+c+x; c=(t<x)+(x>>19);
    > return (Q[j]=t);
    > }
    >
    > #define CNG ( xcng=69069*xcng+123 )
    > #define XS ( xs^=(xs<<13), xs^=(xs>>17), xs^=(xs<<5) )
    > #define KISS ( MWC()+CNG+XS ) /*138 million/sec*/
    >
    > #include <stdio.h>
    > int main()
    > {unsigned long i,x;
    > for(i=0;i<4691;i++) Q=CNG+XS;
    > for(i=0;i<1000000000;i++) x=MWC();
    > printf(" MWC result=3740121002 ?\n%22u\n",x);
    > for(i=0;i<1000000000;i++) x=KISS;
    > printf("KISS result=2224631993 ?\n%22u\n",x);
    > }
    >
    > /*
    > This RNG uses two seeds to fill the Q[4691] array by
    > means of CNG+XS mod 2^32. Users requiring more than two
    > seeds will need to randomly seed Q[] in main().
    > By itself, the MWC() routine passes all tests in the
    > Diehard Battery of Tests, but it is probably a good
    > idea to include it in the KISS combination.
    >
    > 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)
    > */


    This doesn't work with systems that have unsigned long as a 64 bit quantity.

    I obtain:

    MWC result=3740121002 ?
    4169348530
    KISS result=2224631993 ?
    1421918629

    Compiling with 32 bit machine yields:
    MWC result=3740121002 ?
    3740121002
    KISS result=2224631993 ?
    2224631993
    jacob navia, Jul 24, 2010
    #2
    1. Advertising

  3. geo

    Gib Bogle Guest

    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
    Gib Bogle, Jul 25, 2010
    #3
  4. geo

    geo Guest

    On Jul 24, 11:34 pm, Gib Bogle <> 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
    geo, Jul 25, 2010
    #4
  5. On 7/25/10 8:53 AM, geo wrote:
    [snip]
    >
    > I still like inline functions in Fortan,
    > so would tend to define
    > isign(x)=ishft(x,-31)

    I don't know anything about random numbers, but "sign" is an intrinsic
    function in Fortran and compilers should generate near perfect code
    inline for things like sign(1,x). The generic sign function was added
    to Fortran 90. In FORTRAN 77, you'd need to use the specific ISIGN(1,X)
    function.

    Dick Hendrickson

    [snip]
    >
    > gm
    Dick Hendrickson, Jul 25, 2010
    #5
  6. geo

    Gib Bogle Guest

    geo wrote:
    > 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.


    That produces different c values from those generated by the method based on the
    value of (t<x), therefore it deviates from the C code. This is what I used:

    m = shiftl(x,13) + c
    if (sign(1,m) == sign(1,x)) then
    c = sign(1,x) + shiftr(x,19)
    else
    c = 1 - sign(1,m) + shiftr(x,19)
    endif
    Gib Bogle, Jul 26, 2010
    #6
  7. geo

    robin Guest

    "geo" <> wrote in message news:...
    |I have been asked to recommend an RNG
    | (Random Number Generator) that ranks
    | at or near the top in all of the categories:
    | performance on tests of randomness,
    | length of period, simplicity and speed.
    | The most important measure, of course, is
    | performance on extensive tests of randomness, and for
    | those that perform well, selection may well depend
    | on those other measures.
    |
    | The following KISS version, perhaps call it KISS4691,
    | seems to rank at the top in all of those categories.
    | It is my latest, and perhaps my last, as, at age 86,
    | I am slowing down.
    |
    | Compiling and running the following commented
    | C listing should produce, within about four seconds,
    | 10^9 calls to the principal component MWC(), then
    | 10^9 calls to the KISS combination in another ~7 seconds.
    |
    | Try it; you may like it.
    |
    | George Marsaglia

    Here's the PL/I version:

    (NOSIZE, NOFOFL):
    RNG: PROCEDURE OPTIONS (MAIN, REORDER);

    declare (xs initial (521288629), xcng initial (362436069),
    Q(0:4690) ) static fixed binary (32) unsigned;

    MWC: procedure () returns (fixed binary (32) unsigned);
    /*takes about 4.2 nanosecs or 238 million/second*/
    declare (t,x,i) fixed binary (32) unsigned;
    declare (c initial (0), j initial (4691) ) fixed binary (32) unsigned static;

    if j < 4690 then j = j + 1; else j = 0;
    x = Q(j);
    t = isll(x,13)+c+x; c = (t<x)+isrl(x,19);
    Q(j)=t;
    return (t);
    end MWC;

    CNG: procedure returns (fixed binary (32) unsigned);
    xcng=bin(69069)*xcng+bin(123);
    return (xcng);
    end CNG;

    XXS: procedure returns (fixed binary (32) unsigned);
    xs = ieor (xs, isll(xs, 13) );
    xs = ieor (xs, isrl(xs, 17) );
    xs = ieor (xs, isll(xs, 5) );
    return (xs);
    end XXS;

    KISS: procedure returns (fixed binary (32) unsigned);
    return ( MWC()+CNG+XXS ); /*138 million/sec*/
    end KISS;

    declare (i,x) fixed binary (32) unsigned;
    /* Initialize: */
    do i = 0 to 4691-1; Q(i) = CNG+XXS; end;
    put skip list (q(0), q(4690));
    put skip list ('initialized'); put skip;
    do i = 0 to 1000000000-1; x=MWC(); end;
    put skip edit (" MWC result=3740121002 ",x) (a, f(23));
    do i = 0 to 1000000000-1; x=KISS; end;
    put skip edit ("KISS result=2224631993 ",x) (a, f(23));

    end RNG;
    robin, Jul 26, 2010
    #7
  8. geo

    robin Guest

    "Gib Bogle" <> wrote in message news:i2ij74$kd6$...
    | geo wrote:
    | > 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(x<<13+c)+(x>>19)
    | >
    | > Sorry for that error.
    |
    | That produces different c values from those generated by the method based on the
    | value of (t<x), therefore it deviates from the C code. This is what I used:
    |
    | m = shiftl(x,13) + c
    | if (sign(1,m) == sign(1,x)) then
    | c = sign(1,x) + shiftr(x,19)
    | else
    | c = 1 - sign(1,m) + shiftr(x,19)
    | endif

    Maybe I missed something,
    but isn't this exactly equivalent to what George wrote?
    Just substitute x<<13+c for m in your last two assignments ...
    robin, Jul 26, 2010
    #8
  9. On Jul 24, 3:02 pm, geo <> wrote:
    > This RNG uses two seeds to fill the Q[4691] array by
    > means of CNG+XS mod 2^32. Users requiring more than two
    > seeds will need to randomly seed Q[] in main().
    > By itself, the MWC() routine passes all tests in the
    > Diehard Battery of Tests, but it is probably a good
    > idea to include it in the KISS combination.


    Does this mean that using different seeds will lead to
    streams that are always statistically independent
    (as long as one does not exhaust the RNG's period)?
    Or are there restrictions on the possible combinations
    of seeds?

    I am currently using D.E.Knuth's generator from TAOCP,
    which IIRC allows for 2^30-2 independent streams, and
    which are asserted to be independent, but being able to
    extend the "limit" would be nice.

    Harald
    Harald Anlauf, Jul 26, 2010
    #9
  10. geo

    Gib Bogle Guest

    robin wrote:
    > "Gib Bogle" <> wrote in message news:i2ij74$kd6$...
    > | geo wrote:
    > | > 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(x<<13+c)+(x>>19)
    > | >
    > | > Sorry for that error.
    > |
    > | That produces different c values from those generated by the method based on the
    > | value of (t<x), therefore it deviates from the C code. This is what I used:
    > |
    > | m = shiftl(x,13) + c
    > | if (sign(1,m) == sign(1,x)) then
    > | c = sign(1,x) + shiftr(x,19)
    > | else
    > | c = 1 - sign(1,m) + shiftr(x,19)
    > | endif
    >
    > Maybe I missed something,
    > but isn't this exactly equivalent to what George wrote?
    > Just substitute x<<13+c for m in your last two assignments ...


    I hope so. Maybe I didn't express myself clearly enough. I'll try again.
    Using my implementation of George's corrected code, I get results from the
    Fortran code that differ from those generated by his C code.
    Gib Bogle, Jul 26, 2010
    #10
  11. geo

    Gib Bogle Guest

    Geoff wrote:

    > I am no FORTRAN hacker but I think there's a difference between
    > sign(x) and sign(1,x).


    sign(1,x) returns the sign of x.
    Gib Bogle, Jul 26, 2010
    #11
  12. geo

    Guest

    In article <>,
    Harald Anlauf <> wrote:
    >On Jul 24, 3:02=A0pm, geo <> wrote:
    >> This RNG uses two seeds to fill the Q[4691] array by
    >> means of CNG+XS mod 2^32. Users requiring more than two
    >> seeds will need to randomly seed Q[] in main().
    >> By itself, the MWC() routine passes all tests in the
    >> Diehard Battery of Tests, but it is probably a good
    >> idea to include it in the KISS combination.

    >
    >Does this mean that using different seeds will lead to
    >streams that are always statistically independent
    >(as long as one does not exhaust the RNG's period)?
    >Or are there restrictions on the possible combinations
    >of seeds?


    No. Without checking it more carefully, I can't say definitely,
    but it looks as if you would be very unlikely to notice a PRACTICAL
    association with two different seeds, provided that you throw
    away the first 10,000 or so numbers - and even that qualification
    may be unnecessary. But, unless I have some missed some subtlety,
    the sequences cannot be guaranteed to be pseudo-independent.

    The only two methods I know of of guaranteeing pseudo-independence
    are using coprime sequences and by choosing them using the spectral
    test or equivalent. Even then, there are some qualifications on
    what is meant by pseudo-independence. However, in practice, it's
    rare to have trouble with high-quality generators.

    >I am currently using D.E.Knuth's generator from TAOCP,
    >which IIRC allows for 2^30-2 independent streams, and
    >which are asserted to be independent, but being able to
    >extend the "limit" would be nice.


    Eh? Which version? There was assuredly nothing with those
    properties in edition 2. The first papers on the topic were later.

    You need to be very careful to distinguish separate (i.e. disjoint)
    sequences from pseudo-independent ones, and FAR too many papers
    written by people who ought to know better confuse the two. Doing
    that is a common cause of seriously wrong answers in some types of
    calculation.


    Regards,
    Nick Maclaren.
    , Jul 26, 2010
    #12
  13. On Jul 26, 10:09 pm, wrote:
    > In article <>,
    > Harald Anlauf  <> wrote:
    >
    > >Does this mean that using different seeds will lead to
    > >streams that are always statistically independent
    > >(as long as one does not exhaust the RNG's period)?
    > >Or are there restrictions on the possible combinations
    > >of seeds?

    >
    > No.  Without checking it more carefully, I can't say definitely,
    > but it looks as if you would be very unlikely to notice a PRACTICAL
    > association with two different seeds, provided that you throw
    > away the first 10,000 or so numbers - and even that qualification
    > may be unnecessary.  But, unless I have some missed some subtlety,
    > the sequences cannot be guaranteed to be pseudo-independent.
    >
    > The only two methods I know of of guaranteeing pseudo-independence
    > are using coprime sequences and by choosing them using the spectral
    > test or equivalent.  Even then, there are some qualifications on
    > what is meant by pseudo-independence.  However, in practice, it's
    > rare to have trouble with high-quality generators.


    I was shown funny experiences with simple generators... =:-o

    > >I am currently using D.E.Knuth's generator from TAOCP,
    > >which IIRC allows for 2^30-2 independent streams, and
    > >which are asserted to be independent, but being able to
    > >extend the "limit" would be nice.

    >
    > Eh?  Which version?  There was assuredly nothing with those
    > properties in edition 2.  The first papers on the topic were later.


    http://www-cs-faculty.stanford.edu/~uno/news02.html#rng

    > You need to be very careful to distinguish separate (i.e. disjoint)
    > sequences from pseudo-independent ones, and FAR too many papers
    > written by people who ought to know better confuse the two.  Doing
    > that is a common cause of seriously wrong answers in some types of
    > calculation.


    For an application which needs to distribute a large problem over many
    processors it is very useful to have many (pseudo-)independent
    streams,
    at least they should be practically indistinguishable from independent
    ones. An accidental correlation is likely worse than a short period.

    Harald
    Harald Anlauf, Jul 26, 2010
    #13
  14. In comp.lang.fortran wrote:
    > In article <>,
    > Harald Anlauf <> wrote:

    (snip)

    >>Does this mean that using different seeds will lead to
    >>streams that are always statistically independent
    >>(as long as one does not exhaust the RNG's period)?
    >>Or are there restrictions on the possible combinations
    >>of seeds?


    > No. Without checking it more carefully, I can't say definitely,
    > but it looks as if you would be very unlikely to notice a PRACTICAL
    > association with two different seeds, provided that you throw
    > away the first 10,000 or so numbers - and even that qualification
    > may be unnecessary. But, unless I have some missed some subtlety,
    > the sequences cannot be guaranteed to be pseudo-independent.


    My biggest complaint about the current standard RANDOM_SEED
    is that it doens't provide a way to get a reliably good
    seed from a default (likely 32 bit) integer.

    There are many generators with extrememly long periods,
    and correspondingly long state. As the designers of the RNG
    are the ones likely to know how to choose a good seed, it
    would seem they would be the best ones to supply a good
    seed generator.

    > The only two methods I know of of guaranteeing pseudo-independence
    > are using coprime sequences and by choosing them using the spectral
    > test or equivalent. Even then, there are some qualifications on
    > what is meant by pseudo-independence. However, in practice, it's
    > rare to have trouble with high-quality generators.


    Now, one can supply an array of the appropriate length to
    RANDOM_SEED(PUT=...), but how to generate such an array
    from a smaller seed? There is no way to know.

    (snip)

    > You need to be very careful to distinguish separate (i.e. disjoint)
    > sequences from pseudo-independent ones, and FAR too many papers
    > written by people who ought to know better confuse the two. Doing
    > that is a common cause of seriously wrong answers in some types of
    > calculation.


    -- glen
    glen herrmannsfeldt, Jul 26, 2010
    #14
  15. geo

    Guest

    In article <>,
    Harald Anlauf <> wrote:
    >>
    >> The only two methods I know of of guaranteeing pseudo-independence
    >> are using coprime sequences and by choosing them using the spectral
    >> test or equivalent. =A0Even then, there are some qualifications on
    >> what is meant by pseudo-independence. =A0However, in practice, it's
    >> rare to have trouble with high-quality generators.

    >
    >I was shown funny experiences with simple generators... =3D:-o


    Yup. Very common.

    >> >I am currently using D.E.Knuth's generator from TAOCP,
    >> >which IIRC allows for 2^30-2 independent streams, and
    >> >which are asserted to be independent, but being able to
    >> >extend the "limit" would be nice.

    >>
    >> Eh? =A0Which version? =A0There was assuredly nothing with those
    >> properties in edition 2. =A0The first papers on the topic were later.

    >
    >http://www-cs-faculty.stanford.edu/~uno/news02.html#rng


    Ah. Thanks. It's new, then. I haven't been active in this area
    ina couple of decades, so haven't been keeping track.

    >> You need to be very careful to distinguish separate (i.e. disjoint)
    >> sequences from pseudo-independent ones, and FAR too many papers
    >> written by people who ought to know better confuse the two. =A0Doing
    >> that is a common cause of seriously wrong answers in some types of
    >> calculation.

    >
    >For an application which needs to distribute a large problem over many
    >processors it is very useful to have many (pseudo-)independent
    >streams,
    >at least they should be practically indistinguishable from independent
    >ones. An accidental correlation is likely worse than a short period.


    Right. That's precisely why I got involved.


    Regards,
    Nick Maclaren.
    , Jul 26, 2010
    #15
  16. geo

    robin Guest

    "Geoff" <> wrote in message news:...
    | On Mon, 26 Jul 2010 23:32:27 +1000, "robin" <>
    | wrote:
    |
    | >"Gib Bogle" <> wrote in message news:i2ij74$kd6$...
    | >| geo wrote:
    | >| > 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(x<<13+c)+(x>>19)
    | >| >
    | >| > Sorry for that error.
    | >|
    | >| That produces different c values from those generated by the method based on the
    | >| value of (t<x), therefore it deviates from the C code. This is what I used:
    | >|
    | >| m = shiftl(x,13) + c
    | >| if (sign(1,m) == sign(1,x)) then
    | >| c = sign(1,x) + shiftr(x,19)
    | >| else
    | >| c = 1 - sign(1,m) + shiftr(x,19)
    | >| endif
    | >
    | >Maybe I missed something,
    | >but isn't this exactly equivalent to what George wrote?
    | >Just substitute x<<13+c for m in your last two assignments ...
    |
    | I am no FORTRAN hacker but I think there's a difference between
    | sign(x) and sign(1,x).

    George gave general advice on how to do it.
    That advice wasn't specific to Fortran.
    It's necessary to use the appropriate Fortran construct --
    and sign(1,x) is the only way to do that.
    robin, Jul 27, 2010
    #16
  17. geo

    robin Guest

    "geo" <> wrote in message news:...
    |I have been asked to recommend an RNG
    | (Random Number Generator) that ranks
    | at or near the top in all of the categories:
    | performance on tests of randomness,
    | length of period, simplicity and speed.
    | The most important measure, of course, is
    | performance on extensive tests of randomness, and for
    | those that perform well, selection may well depend
    | on those other measures.

    I have already posted a PL/I version using unsigned arithmetic.

    Here is another version, this time using signed arithmetic :--

    (NOSIZE, NOFOFL):
    RNG: PROCEDURE OPTIONS (MAIN, REORDER);

    declare (xs initial (521288629), xcng initial (362436069),
    Q(0:4690) ) static fixed binary (31);

    MWC: procedure () returns (fixed binary (31));
    declare (t,x,i) fixed binary (31);
    declare (c initial (0), j initial (4691) ) fixed binary (31) static;
    declare (t1, t2, t3) fixed binary (31);

    if j < hbound(Q,1) then j = j + 1; else j = 0;
    x = Q(j);
    t = isll(x,13)+c+x;
    t1 = iand(x, 3) - iand(t, 3);
    t2 = isrl(x, 2) - isrl(t, 2);
    if t2 = 0 then t2 = t1;
    if t2 > 0 then t3 = 1; else t3 = 0;
    c = t3 + isrl(x, 19);
    Q(j)=t;
    return (t);
    end MWC;

    CNG: procedure returns (fixed binary (31));
    xcng=bin(69069)*xcng+bin(123);
    return (xcng);
    end CNG;

    XXS: procedure returns (fixed binary (31));
    xs = ieor (xs, isll(xs, 13) );
    xs = ieor (xs, isrl(xs, 17) );
    xs = ieor (xs, isll(xs, 5) );
    return (xs);
    end XXS;

    KISS: procedure returns (fixed binary (31));
    return ( MWC()+CNG+XXS );
    end KISS;

    declare (i,x) fixed binary (31);
    declare y fixed decimal (11);

    Q = CNG+XXS; /* Initialize. */
    do i = 1 to 1000000000; x=MWC(); end;
    put skip edit (" Expected MWC result = 3740121002", 'computed =', x)
    (a, skip, x(12), a, f(11));
    y = iand(x, 2147483647);
    if x < 0 then y = y + 2147483648;
    put skip edit (y) (x(11), f(22)); put skip;
    do i = 1 to 1000000000; x=KISS; end;
    put skip edit ("Expected KISS result = 2224631993", 'computed =', x)
    (a, skip, x(12), a, f(11));
    y = iand(x, 2147483647);
    if x < 0 then y = y + 2147483648;
    put skip edit (y) (x(11), f(22));

    end RNG;
    robin, Jul 27, 2010
    #17
  18. geo

    Guest

    In article <>,
    Geoff <> wrote:
    >>|
    >>| I am no FORTRAN hacker but I think there's a difference between
    >>| sign(x) and sign(1,x).
    >>
    >>George gave general advice on how to do it.
    >>That advice wasn't specific to Fortran.
    >>It's necessary to use the appropriate Fortran construct --
    >>and sign(1,x) is the only way to do that.
    >>

    >Thanks for that clarification. I have not done any Fortran code since
    >1975 and it looked a whole lot different than it does today.


    Actually, that aspect hasn't changed since :) SIGN(X) always was
    an implementation-dependent extension (now called processor-dependent).


    Regards,
    Nick Maclaren.
    , Jul 27, 2010
    #18
  19. geo

    robin Guest

    "jacob navia" <> wrote in message news:i2fir2$op4$...

    | This doesn't work with systems that have unsigned long as a 64 bit quantity.
    |
    | I obtain:
    |
    | MWC result=3740121002 ?
    | 4169348530
    | KISS result=2224631993 ?
    | 1421918629

    For a 64-bit machine (using 64-bit integer arithmetic),
    you'd need to truncate each result to 32 bits. That not
    only applies to the multiplication, it also applies to addition, etc.
    On a 32-bit machine, these extra bits are discarded,
    but in 64-bit arithmetic, they are retained,
    and unless they are similarly discarded,
    you won't get the same results.
    I suggest using IAND(k, 2*2147483647+1)
    for the truncation.

    With such modifications in the program,
    it should then produce the same results on both 32-bit and
    64-bit machines.

    P.S. the product 2*2... is best obtained using ISHFT.

    | Compiling with 32 bit machine yields:
    | MWC result=3740121002 ?
    | 3740121002
    | KISS result=2224631993 ?
    | 2224631993
    robin, Jul 27, 2010
    #19
  20. geo

    Uno Guest

    glen herrmannsfeldt wrote:
    > In comp.lang.fortran wrote:
    >> In article <>,
    >> Harald Anlauf <> wrote:

    > (snip)
    >
    >>> Does this mean that using different seeds will lead to
    >>> streams that are always statistically independent
    >>> (as long as one does not exhaust the RNG's period)?
    >>> Or are there restrictions on the possible combinations
    >>> of seeds?

    >
    >> No. Without checking it more carefully, I can't say definitely,
    >> but it looks as if you would be very unlikely to notice a PRACTICAL
    >> association with two different seeds, provided that you throw
    >> away the first 10,000 or so numbers - and even that qualification
    >> may be unnecessary. But, unless I have some missed some subtlety,
    >> the sequences cannot be guaranteed to be pseudo-independent.

    >
    > My biggest complaint about the current standard RANDOM_SEED
    > is that it doens't provide a way to get a reliably good
    > seed from a default (likely 32 bit) integer.
    >
    > There are many generators with extrememly long periods,
    > and correspondingly long state. As the designers of the RNG
    > are the ones likely to know how to choose a good seed, it
    > would seem they would be the best ones to supply a good
    > seed generator.
    >
    >> The only two methods I know of of guaranteeing pseudo-independence
    >> are using coprime sequences and by choosing them using the spectral
    >> test or equivalent. Even then, there are some qualifications on
    >> what is meant by pseudo-independence. However, in practice, it's
    >> rare to have trouble with high-quality generators.

    >
    > Now, one can supply an array of the appropriate length to
    > RANDOM_SEED(PUT=...), but how to generate such an array
    > from a smaller seed? There is no way to know.


    So for the put= values in fortran, you need a vector of pseudorandom
    integers, which is as good as it gets without truly random devices,
    making--one hopes-a period that is large with respect to the interval
    you're interested in.

    It doesn't seem like a problem with epistemology as much a mathematical
    ceiling on how much randomness you can create by a handful of values.
    --
    Uno
    Uno, Jul 29, 2010
    #20
    1. Advertising

Want to reply to this thread or ask your own question?

It takes just 2 minutes to sign up (and it's free!). Just click the sign up button to choose a username and then you can ask your own questions on the forum.
Similar Threads
  1. Francois Grieu
    Replies:
    7
    Views:
    437
    Ben Bacarisse
    Apr 8, 2009
  2. geo
    Replies:
    5
    Views:
    355
    Tim Little
    Aug 30, 2010
  3. geo

    Implementing the KISS4691 RNG

    geo, Sep 5, 2010, in forum: C Programming
    Replies:
    22
    Views:
    773
    David Thompson
    Sep 17, 2010
  4. Dann Corbit

    Re: Implementing the KISS4691 RNG

    Dann Corbit, Sep 8, 2010, in forum: C++
    Replies:
    2
    Views:
    328
    Victor Bazarov
    Sep 9, 2010
  5. robin
    Replies:
    79
    Views:
    1,257
    Frederick Williams
    Jul 1, 2011
Loading...

Share This Page