2006/7/16, Henrik Sundberg <[EMAIL PROTECTED]>:
2006/7/16, Andy Pepperdine <[EMAIL PROTECTED]>:
> On Sunday 16 July 2006 06:48, Andrius Aštrauskas wrote:
> > On Fri, 14 Jul 2006 13:32:45 +0200
> >
> > Jurgen Stigter <[EMAIL PROTECTED]> wrote:
> > > /I am using version 2.0 of OpenOffice.org. My problem is :/
> > >
> > > The MOD function is returning -512 instead of 0 (though each of the
> > > factors is divisable by 23):
> > > MOD (9951585559 *370469809;23) results in -512, but when I copy the
> > > formula and put it in another cell, the result is -320
> > > while
> > > MOD (9951585559;23) results in 0
> > > MOD (370469809;23) results 0
>
> This is almost certainly due to an integer overflow. There will be a limit to
> the size of exact integer values. I don't know how big they can be; perhaps
> someone else can say.
log2 (9951585559) => 33.21 bits
log2(370469809) => 28.46 bits
The product needs 61.68 bits, less than 63 (one taken by the sign).
I.e. 64 bit arithmetics ought to work.
both MOD (9951585559 *370469809;23) and MOD ((9951585559
*370469809)-1;23) gives the result 0 for me (XP, OOO 2.03), implying
overflow. But why?
MOD (9951585559-1;23) gives 22, so it works for more than 32 bits.
Are the integers handled as float numbers when they are big enough?
9951585559 *370469809 is displayed as 3,69E+018, implying this and
making sense to the result.
I tried a little more, by making a table in calc.
X 2^X =MOD(2^X;23)-MOD(2^X-1;23) =(2^X+1)-(2^X)
47 140737488355328 1 1
48 281474976710656 1 0
49 562949953421312 1 0
50 1,13E+015 1 0
51 2,25E+015 1 0
52 4,50E+015 0 0
53 9,01E+015 0 0
The third column indicates that 64-bit float values are used. I.e.
With a 52-bit mantissa according to:
http://en.wikipedia.org/wiki/IEEE_floating-point_standard
The fourth column indicates that something is wrong.
I think that numeral values always are represented as double precision
float values and therefore that integer values above ~2^52 are
meaningless.
My table indicates that integer values above 2^47 does not work. This
might be a bug.
> In this particular case, you can rewrite the formula because:
> a*b (mod p) = a (mod p) * b (mod p)
Not really. Max(a*b (mod p))=p-1; Max(a (mod p) * b (mod p))=(p-1)**2.
(a (mod p) * b (mod p)) (mod p) might work.
/$
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