| Re: Simple MODULO question. [message #6454] |
Wed, 03 July 1996 00:00 |
peter
Messages: 80
Registered: February 1994
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Member |
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S. Penzes (penzes@dres.dnd.ca) wrote:
: It truly is a simple question and the online help doesn't seem to
: answer it. So it's either a bug or I have a short cicuit in a synapse
: someplace. Anyway here it is:
: Why does print,1.0 mod 0.25 return 0.00000 (as expected)
: generally 1.0 mod (1.0/2^n) returns 0
: while print,1.0 mod 0.2
: or print,1.0 mod (1.0/5.0) return 0.200000
: generally 1.0 mod (1.0 / n)
Probably because, in floating point, 0.25 is represented exactly, so
that 1.0/0.25 = 0.0, while 0.2 is not represented exactly. I'll guess
that 0.2 is actually something like 0.200000001, so that 5*0.2 is
greater than 1.0, while 1.0-4*0.2 is 0.199999996, which prints at 0.2.
Relying on true equality of floating point numbers as a test (which this
is doing, relying on 5.0*0.2=1.0) is, well, unreliable.
Peter
--------------------------------
Peter Webb, HP Labs Medical Dept
E-Mail: peter_webb@hpl.hp.com
Phone: (415) 813-3756
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| Re: Simple MODULO question. [message #6465 is a reply to message #6454] |
Wed, 03 July 1996 00:00  |
Peter Mason
Messages: 145
Registered: June 1996
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Senior Member |
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On 2 Jul 1996, S. Penzes wrote:
> Why does print,1.0 mod 0.25 return 0.00000 (as expected)
> generally 1.0 mod (1.0/2^n) returns 0
> while print,1.0 mod 0.2
> or print,1.0 mod (1.0/5.0) return 0.200000
> generally 1.0 mod (1.0 / n)
>
> In case you're wondering, I am trying to determine if:
> for x mod y whether x is an integer multiple of y.
I guess you're experiencing one of the pitfalls of working with floating-
point numbers - they're not necessarily exact.
I'd have to hazard a guess at exactly what's happening in this example:
. In the first case with 1.0 mod 0.25, both 1.0 and 0.25 (== 1/8) can be
represented exactly in floating-point, and so 0.25 divides an integral
number of times into 1.0 and the mod (remainder) is 0.
. In the second case with 1.0 mod 0.2, although 0.2 looks like a
simple, exact number in base 10, it is a bit of a headache in base 2
and can't be represented exactly - rather like 1/3 in base 10. My
guess is that it's floating-point representation is fractionally
larger than 0.2, and so 1.0 / "0.2" is fractionally less than 5.
So 1.0 mod "0.2" = 5.0 - 4 * "0.2", which is fractionally less than 0.2.
You can get a hint that there's a problem with: PRINT,1.0D/0.2. The
result printed is 4.9999999. (This is a fluke of the print formatting, I
think - both 1.0/0.2 and 1.0D/0.2D return 5!)
I think that you might have to use a steam-driven method for your test:
Instead of testing ((x mod y) eq 0.0), test:
(abs(x - y*round(x/y)) lt some_small_tolerance)
or such.
Peter Mason
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