LinuxQuestions.org - Doubt about numerical precision

- Programming (https://www.linuxquestions.org/questions/programming-9/)

- - Doubt about numerical precision (https://www.linuxquestions.org/questions/programming-9/doubt-about-numerical-precision-692633/)

Bassy

12-23-2008 09:20 PM

Doubt about numerical precision

Hi! I'm writing a graphical editor for curves (homework) and I've come with an instance of this algorithm:

Code:

int main(){



  float a;



  for(a = 0.0; a <= 1.0; a = a + 0.05)

    printf("%f\n", a);



  return 0;

}

Now the thing is that it doesn't print 1.0 XD Now I'd like some opinions about the implications of numerical precision here because its just additions and besides a is an increasing number.

I tried it with "double a;" but it just the same jejeje. I'm using gcc 4.1.0.

I'm pretty curious about this :D

paulsm4

12-23-2008 09:54 PM

Hi -

Here's a minor change to your program:

Code:

#include <stdio.h>



int

main ()

{

  float a;



  for (a=0.0; a <= 1.0; a += 0.05)

    printf ("%f\n", a);

  if (a == 1.0)

    printf ("%f == 1.0\n", a);

  else

    printf ("%f != 1.0\n", a);



  return 0;

}

And here's the sample output:

Quote:

0.000000
0.050000
0.100000
0.150000
0.200000
0.250000
0.300000
0.350000
0.400000
0.450000
0.500000
0.550000
0.600000
0.650000
0.700000
0.750000
0.800000
0.850000
0.900000
0.950000
1.000000 != 1.0

Hint: this is pure "C" (and floating point precision). It has nothing to do with you, your program, or GCC (in fact, I compiled/ran this under Visual Studio. You'll get similar results with ANY C compiler).

Why?

It's an important question - Google for "C" and "floating point" - I'm sure you'll find the answer.

Good luck!

Bassy

12-23-2008 10:00 PM

Nice but...

Thanks! I know it's not my code... and it's not my compiler neither because the original code is in C++ being compiled with g++.

I'm just curious becaise it defies intuition XD

Thanks for the interest ;) I'll google that :D

jschiwal

12-23-2008 10:41 PM

The value you use as an increment may not be exactly representable as a floating point number. In the 80's I read an article in a C-Language magazine about floating point numbers. There were so many caveats, I couldn't believe how complicated it was. A floating point library may even use approximation routines for sin() for very small angles. The routine to calculate the value depends on the input value, and isn't always a series. ( The same thing may be be happening in a floating point processor )

If you evaluate some series, you may need to process the series so that the smaller values are added first. This may mean a for loop that counts down for some series.

Never test for equality in floating point numbers. Test if the value is within a small delta from the value you want to test. Using a for loop, you might want to add a small delta value to the top value in your loop.

Bassy

12-23-2008 11:20 PM

Jajajaja I was actually thinking that.

When I took Numerical Analysis for CS at university I remember I had to evaluate series from the smallest value to the biggest I actually have those proof somewhere in my notes.

The funny thing is that if you re write it like this:

Code:

#include <stdio.h>



int main(){



  float a;



  for(a = 0.0; a <= 1.0;){

    printf("%f\n", a);

    a+=0.05;

  }

  return 0;

}

So you can change the position of the a+= 0.05. If it is after the print (like here) you loose the 1.0 but if it is before you loose the 0.0

Jajajaja nice ah?!?!?

ErV	12-24-2008 03:34 AM

Hi.

Quote:

Originally Posted by Bassy (Post 3385665)

Code:

int main(){



  float a;



  for(a = 0.0; a <= 1.0; a = a + 0.05)

    printf("%f\n", a);



  return 0;

}

Try:

Code:

int main(){



  float a;



  for(a = 0.0; a <= 1.000001; a = a + 0.05)

    printf("%f\n", a);



  return 0;

}

Code:

#include <stdio.h>

int main(){



  int i;



  for(i = 0; i<= 20; i++)

    printf("%f\n", (float)i/20.0f);



  return 0;

}

Sergei Steshenko

12-24-2008 11:38 AM

Quote:

Originally Posted by Bassy (Post 3385665)

Hi! I'm writing a graphical editor for curves (homework) and I've come with an instance of this algorithm:

Code:

int main(){



  float a;



  for(a = 0.0; a <= 1.0; a = a + 0.05)

    printf("%f\n", a);



  return 0;

}

Absolutely never write such loops:

Code:

for(a = 0.0; a <= 1.0; a = a + 0.05)

in any programming language.

Unless you are sure internally numbers are represented in decimal.

Bassy

12-24-2008 01:19 PM

Jejeje

Hey ErV... nice approach with the casts... and defining the epsilon... cool. I think I'm gonna' steal you the one with the casts jejejeje because is pretty elegant :)

What I did was, to let the for stop by its own imprecision and then by taking advantage of a math property of the curves I'm viewing, to make a final assignment and problem solved... it's almost like if my epsilon is defined so I can stop exactly one number earlier :D

And Sergei Steshenko...

Quote:

Unless you are sure internally numbers are represented in decimal.

You mean... that I have to be sure that the real number can actually be represented using the given precision?

Sergei Steshenko

12-24-2008 01:41 PM

Quote:

Originally Posted by Bassy (Post 3386275)

Not given precision, but infinite precision for the fraction you are using as step and for loop limits.

Such loops are a disaster in general case because you can't be sure how many times they are executed.

Bassy

12-24-2008 02:11 PM

Cool

Yep I agree! ;)

All times are GMT -5. The time now is 01:54 PM.