Division precision issue

[Artemio]

Hello all!

I just stumbled across a weird problem with precision of a division
operation. I am on Mac OS X, GCC 4.0.1.

Say I have two float or double numbers, and I want to divide one by
another, e.g.

float A, B, C;
C = A / B;

The problem is that the result my program gives for two different A
and B values are the same:

1.002323 / 11.025554 = 0.090909090909
1.036174 / 11.397911 = 0.090909090909

But with a calculator application, I get:

1.002323 / 11.025554 = 0.0909090826637 827
1.036174 / 11.397911 = 0.0909091148369 205

Does anyone have any suggestions as what I may be doing wrong? Is
there any special division function I should be using?
Thanks very much,

Artemiy.

[Tim Prince]

Artemio wrote:

Hello all!

I just stumbled across a weird problem with precision of a division
operation. I am on Mac OS X, GCC 4.0.1.

Say I have two float or double numbers, and I want to divide one by
another, e.g.

float A, B, C;
C = A / B;

The problem is that the result my program gives for two different A
and B values are the same:

1.002323 / 11.025554 = 0.090909090909
1.036174 / 11.397911 = 0.090909090909

But with a calculator application, I get:

1.002323 / 11.025554 = 0.0909090826637 827
1.036174 / 11.397911 = 0.0909091148369 205

Does anyone have any suggestions as what I may be doing wrong? Is
there any special division function I should be using?

How can you call this weird? You specify C float data type, defined in
std C as supporting 6 or more equivalent decimal digits precision, then
you compare to a calculator which has 15 decimal digits precision. YOur
comparison should be much closer if you use C double data type. But, if
your data are accurate only to 6 or 7 decimal digits, you may not care
how digits beyond those are generated.

[Artemio]

Well but with double the results are exactly the same...

[Chris Dollin]

Artemio wrote:

Well but with double the results are exactly the same...

Show code.

--
Chris "electric hedgehog" Dollin
"It was the first really clever thing the King had said that day."
/Alice in Wonderland/

[santosh]

Artemio wrote:

Well but with double the results are exactly the same...

Generally calculator programs use platform specific types or
multiprecision libraries to provide a high level of precision. Plain C
has a long double type which might be sufficient for most purposes,
but it makes little sense to compare it with specialised mathematical
programs.

PS. Please quote the message to which you reply. Some people download
new messages and read, and respond, to them offline. This becomes
difficult when messages are not fairly standalone, in terms of context.

[Artemio]

Okay guys, sorry for not being more clear with the issue.

Let's say I have this code:

double A, B;

printf("%0.12f / %0.12f = %0.12f\n", A, B, A / B );
For two different A and B it gives:

1.002323075899 / 11.025553834888 = 0.090909090909

and

1.048065059110 / 11.528715650212 = 0.090909090909

E.g. 1/11 and 1/11.5 are not that precise as you guess but the result
is the same! How in the world can that be? :-(
Artemiy.

On Mar 8, 5:51 pm, "santosh" <santosh....@gm ail.comwrote:

Artemio wrote:

Well but with double the results are exactly the same...

Generally calculator programs use platform specific types or
multiprecision libraries to provide a high level of precision. Plain C
has a long double type which might be sufficient for most purposes,
but it makes little sense to compare it with specialised mathematical
programs.

PS. Please quote the message to which you reply. Some people download
new messages and read, and respond, to them offline. This becomes
difficult when messages are not fairly standalone, in terms of context.

[Yevgen Muntyan]

Artemio wrote:

Okay guys, sorry for not being more clear with the issue.

Let's say I have this code:

double A, B;

printf("%0.12f / %0.12f = %0.12f\n", A, B, A / B );
For two different A and B it gives:

1.002323075899 / 11.025553834888 = 0.090909090909

and

1.048065059110 / 11.528715650212 = 0.090909090909

E.g. 1/11 and 1/11.5 are not that precise as you guess but the result
is the same! How in the world can that be? :-(

muntyan@munt10: ~$ calc
C-style arbitrary precision calculator (version 2.12.1.5)
Calc is open software. For license details type: help copyright
[Type "exit" to exit, or "help" for help.]

; 1.002323075899 / 11.025553834888
~0.090909090909 09915440
; 1.048065059110 / 11.528715650212
~0.090909090909 07513819
; 1.002323075899* 11.528715650212 - 1.048065059110* 11.025553834888
~0.000000000003 05271121

So maybe results are actually good and your estimates aren't?

Best regards,
Yevgen

[Mike Wahler]

"Artemio" <ar*****@kdemai l.netwrote in message
news:11******** **************@ n33g2000cwc.goo glegroups.com.. .

Hello all!

I just stumbled across a weird problem with precision of a division
operation. I am on Mac OS X, GCC 4.0.1.

Say I have two float or double numbers, and I want to divide one by
another, e.g.

float A, B, C;
C = A / B;

The problem is that the result my program gives for two different A
and B values are the same:

1.002323 / 11.025554 = 0.090909090909
1.036174 / 11.397911 = 0.090909090909

But with a calculator application, I get:

1.002323 / 11.025554 = 0.0909090826637 827
1.036174 / 11.397911 = 0.0909091148369 205

Does anyone have any suggestions as what I may be doing wrong? Is
there any special division function I should be using?

#include <stdio.h>

int main()
{
double a = 1.002323075899;
double b = 11.025553834888 ;
double c = a / b;
double x = 1.048065059110;
double y = 11.528715650212 ;
double z = x / y;

printf("a == %.20f\n"
"b == %.20f\n"
"a / b == %.20f\n\n"
"x == %.20f\n"
"y == %.20f\n"
"x / y == %.20f\n",
a, b, c, x, y, z);

return 0;
}

Output (MSVC 2005 Express, Pentium 4):

a == 1.0023230758990 0000000
b == 11.025553834887 99900000
a / b == 0.0909090909090 9916900

x == 1.0480650591100 0000000
y == 11.528715650212 00000000
x / y == 0.0909090909090 7514600
"What Every Computer Scientist Should
Know About Floating-Point Arithmetic":
http://docs.sun.com/source/806-3568/ncg_goldberg.html
-Mike

[Artemio]

Thanks very much Mike, I'll give that one a thorough read!

Artemiy.

On Mar 8, 7:46 pm, "Mike Wahler" <mkwah...@mkwah ler.netwrote:

"Artemio" <arte...@kdemai l.netwrote in message

news:11******** **************@ n33g2000cwc.goo glegroups.com.. .

Hello all!

I just stumbled across a weird problem with precision of a division
operation. I am on Mac OS X, GCC 4.0.1.

Say I have two float or double numbers, and I want to divide one by
another, e.g.

float A, B, C;
C = A / B;

The problem is that the result my program gives for two different A
and B values are the same:

1.002323 / 11.025554 = 0.090909090909
1.036174 / 11.397911 = 0.090909090909

But with a calculator application, I get:

1.002323 / 11.025554 = 0.0909090826637 827
1.036174 / 11.397911 = 0.0909091148369 205

Does anyone have any suggestions as what I may be doing wrong? Is
there any special division function I should be using?

#include <stdio.h>

int main()
{
double a = 1.002323075899;
double b = 11.025553834888 ;
double c = a / b;
double x = 1.048065059110;
double y = 11.528715650212 ;
double z = x / y;

printf("a == %.20f\n"
"b == %.20f\n"
"a / b == %.20f\n\n"
"x == %.20f\n"
"y == %.20f\n"
"x / y == %.20f\n",
a, b, c, x, y, z);

return 0;

}

Output (MSVC 2005 Express, Pentium 4):

a == 1.0023230758990 0000000
b == 11.025553834887 99900000
a / b == 0.0909090909090 9916900

x == 1.0480650591100 0000000
y == 11.528715650212 00000000
x / y == 0.0909090909090 7514600

"What Every Computer Scientist Should
Know About Floating-Point Arithmetic":http://docs.sun.com/source/806-3568/ncg_goldberg.html

-Mike

[Thad Smith]

Artemio wrote:

Say I have two float or double numbers, and I want to divide one by
another, e.g.

float A, B, C;
C = A / B;

The problem is that the result my program gives for two different A
and B values are the same:

1.002323 / 11.025554 = 0.090909090909
1.036174 / 11.397911 = 0.090909090909

But with a calculator application, I get:

1.002323 / 11.025554 = 0.0909090826637 827
1.036174 / 11.397911 = 0.0909091148369 205

Does anyone have any suggestions as what I may be doing wrong?

The problem is loss of precision, but the important question is where.
The actual results are 6 and 5 significant digits, worse than expected
for most single precision floating point implementations .

You get much closer to the results indicated earlier with

1.002323 / 11.025553
1.036174 / 11.397914

I suspect this is due to rounding error with limited precision of
conversion on the denominators. Try increasing the precision of the
conversion and storing all values in double, or long double, if the
results you get above is not accurate enough for your application.

--
Thad