Weirdness: (-0.666667 + 0.333333)

[Corne' Cornelius]

Hi,

I'm experiencing some weirdness in a program, when subtracting 2
(double)'s which should result in 0, but instead it returns
-1.11022e-16. It looks to me that changing the double x_step decleration
to unsigned type, might help, but the compiler complains when i try that.

Any ideas ?

#include <iostream>
using namespace std;

int main(int argc, char *argv[]) {

double x1 = -1;
double x2 = 1;
double nx = 6;
int pos = 0;
double x = 0.0;
double x_step = 0.0;

x_step = (x2 - x1) / nx;

x = x1;
for (pos = 0; pos < nx; pos++) {
cout << x << " +\t" << x_step << " =\t " << (x +
x_step) << endl;
x = x + x_step;
}

return 0;
}


Output:

-1 + 0.333333 = -0.666667
-0.666667 + 0.333333 = -0.333333
-0.333333 + 0.333333 = -1.11022e-16
-1.11022e-16 + 0.333333 = 0.333333
0.333333 + 0.333333 = 0.666667
0.666667 + 0.333333 = 1

 

[bartek]

Hi,

I'm experiencing some weirdness in a program, when subtracting 2
(double)'s which should result in 0, but instead it returns
-1.11022e-16. It looks to me that changing the double x_step decleration
to unsigned type, might help, but the compiler complains when i try that.

Any ideas ?

That's the nature of floating point arithmetic. You shouldn't compare
against exact values, beacause of rounding errors (which are unavoidable,
though usually very small).

You should check against some small epsilon range instead of comparing
exact numbers.

 

[Bill Seurer]

I'm experiencing some weirdness in a program, when subtracting 2
(double)'s which should result in 0, but instead it returns
-1.11022e-16.

Doubles do not always exactly represent the value of a number but are an
approximation of the value. This is especially true for computed values.

See http://docs.sun.com/source/806-3568/ncg_goldberg.html

 

[Corne' Cornelius]

I've rounded the x_step var to a couple of places after the decimal
point which seem to help, i need a number closer to 0 then what the
overflow results in.

Thanks !

 

[bartek]

I've rounded the x_step var to a couple of places after the decimal
point which seem to help, i need a number closer to 0 then what the
overflow results in.

first...

#include <limits>

....then...

std::numeric_limits<double>::epsilon()

....will give you the smallest possible increment from 1.
Unfortunately rounding errors can be bigger than that.
It all depends on what you're actually doing. You should choose the
threshold based on that.

 

[bartek]

first...

#include <limits>

...then...

std::numeric_limits<double>::epsilon()

...will give you the smallest possible increment from 1.
Unfortunately rounding errors can be bigger than that.
It all depends on what you're actually doing. You should choose the
threshold based on that.

I mean -- based on what you're doing, not on numeric_limits::epsilon()
that is.

 

[Chris Dams]

Hi!

I'm experiencing some weirdness in a program, when subtracting 2
(double)'s which should result in 0, but instead it returns
-1.11022e-16. It looks to me that changing the double x_step decleration
to unsigned type, might help, but the compiler complains when i try that.

The problem is that real numbers on a computer have limited accuracy.
Results of substracting two real numbers that are (almost) equal, should
not be trusted. Solutions to this problem generally depend on what
exactly you want to do.

Bye,
Chris Dams

 

[Pete Becker]

Doubles do not always exactly represent the value of a number but are an
approximation of the value. This is especially true for computed values.

Doubles exactly represent the value that they hold. They sometimes don't
hold the result that you would get from some computation with infinite
precision arithemetic. But, then, neither do ints. In their first week
future programmers learn to accept that (1/3)*3 isn't 1. For some reason
<g> it's harder for them to accept that with floating point values.

 

[Gert Van den Eynde]

Hi,

I'm experiencing some weirdness in a program, when subtracting 2
(double)'s which should result in 0, but instead it returns
-1.11022e-16. It looks to me that changing the double x_step decleration
to unsigned type, might help, but the compiler complains when i try that.

Any ideas ?

Don't compare two floating point numbers as such. Use the trick of an
epsilon interval by Knuth (implemented in the fcmp package available at
sourceforge.net) described in TAOCP Vol 2 Seminumerical algorithms.


bye,
gert

 

[Corne' Cornelius]

i should've made it more clear what i actually want to do.

I'm not really trying to compare 2 double which could be in accurate
becuase of rounding errors.

I'm writing a program that draws graphs on a cartesian plane, so in
order to go on the x-axis from say -2, to +2, in 20 intervals, i need
one of these point to be zero or very close to it.

how could i make sure i don't get rounding problems ?

 

[Gert Van den Eynde]

i should've made it more clear what i actually want to do.

I'm not really trying to compare 2 double which could be in accurate
becuase of rounding errors.

I'm writing a program that draws graphs on a cartesian plane, so in
order to go on the x-axis from say -2, to +2, in 20 intervals, i need
one of these point to be zero or very close to it.

how could i make sure i don't get rounding problems ?

You can't. The only thing you can do is to compare your value of x to zero
(using fcmp) and if it's close enough force it to be zero. Btw, Is the
difference between 1e-16 and 0.0 so significant that it will show up on a
graph in the range [-1,1] ?

bye,
gert

 

[Corne' Cornelius]

heh, i also thaught of not worrying about it being 1e-16, but then
there's a pixel missing where x is 0.

Looks like i will have to make 2 loops and 1 statement.

for when:
x - epsilon() < 0
x = 0
x + epsilon() > 0

Corne' Cornelius wrote:

i should've made it more clear what i actually want to do.

I'm not really trying to compare 2 double which could be in accurate
becuase of rounding errors.

I'm writing a program that draws graphs on a cartesian plane, so in
order to go on the x-axis from say -2, to +2, in 20 intervals, i need
one of these point to be zero or very close to it.

how could i make sure i don't get rounding problems ?

You can't. The only thing you can do is to compare your value of x to zero
(using fcmp) and if it's close enough force it to be zero. Btw, Is the
difference between 1e-16 and 0.0 so significant that it will show up on a
graph in the range [-1,1] ?

bye,
gert

 

[Karl Heinz Buchegger]

heh, i also thaught of not worrying about it being 1e-16, but then
there's a pixel missing where x is 0.

Then your transformation from world coordinates to
pixel coordinates is in error, if such a small deviation
from 0 makes that effect.

 

[Karl Heinz Buchegger]

Then your transformation from world coordinates to
pixel coordinates is in error, if such a small deviation
from 0 makes that effect.

The usual error is to truncate instead of round when chopping
of the fractional part in pixel space.

 

[Chris Dams]

Hi!

heh, i also thaught of not worrying about it being 1e-16, but then
there's a pixel missing where x is 0.

If you want to draw a point for every pixels you should loop over pixels
and not over doubles. On the other hand, making small steps and drawing
a straight line between these, is likely to be good enough. Furthermore,
if you draw points for every pixel in the x-direction you should also
worry about what happens if the slope of your graph is bigger than 1.

Bye,
Chris Dams

 

[Richard Herring]

i should've made it more clear what i actually want to do.

I'm not really trying to compare 2 double which could be in accurate
becuase of rounding errors.

I'm writing a program that draws graphs on a cartesian plane, so in
order to go on the x-axis from say -2, to +2, in 20 intervals, i need
one of these point to be zero or very close to it.

how could i make sure i don't get rounding problems ?

You can often improve matters by using an integer loop counter:

instead of: for (double x=xstart; x<xend; x+=step) {...}

use something like: for (int i=istart; i<iend; ++i) { double x=i*xstep;
....}

It doesn't get rid of the rounding errors, but it does ensure that they
don't accumulate as you progress through the loop.