How to tell if a number has decimal places different than 0

[Gaijinco]

Sooner or later everytime I found recreational programming challenges I
stumble with how I test if a number is has decimal places differnt than
0?

For example if I want to know if a number is a square number (i.e. a
number which square root is a positive number as 4, 9, 16 have) I do
something like:

int square = sqrt(number);

if((int)square==square)
// number is a perfect square
else
// number is not a perfect square

Is there a function or a language-specific-way to do this?

 

[Emmanuel Delahaye]

Gaijinco wrote on 26/09/05 :

Sooner or later everytime I found recreational programming challenges I
stumble with how I test if a number is has decimal places differnt than
0?

For example if I want to know if a number is a square number (i.e. a
number which square root is a positive number as 4, 9, 16 have) I do
something like:

int square = sqrt(number);

if((int)square==square)
// number is a perfect square
else
// number is not a perfect square

Is there a function or a language-specific-way to do this?

fmod()

--
Emmanuel
The C-FAQ: http://www.eskimo.com/~scs/C-faq/faq.html
The C-library: http://www.dinkumware.com/refxc.html

"It's specified. But anyone who writes code like that should be
transmogrified into earthworms and fed to ducks." -- Chris Dollin CLC

 

[John Harrison]

Sooner or later everytime I found recreational programming challenges I
stumble with how I test if a number is has decimal places differnt than
0?

For example if I want to know if a number is a square number (i.e. a
number which square root is a positive number as 4, 9, 16 have) I do
something like:

int square = sqrt(number);

if((int)square==square)
// number is a perfect square
else
// number is not a perfect square

Is there a function or a language-specific-way to do this?

Language-specific? What do you mean by that?

Your test is not perfect because it will fail when square is bigger than
the biggest int. This test always works (within the limitations of
floating point accuracy).

#include <math>

double square = sqrt(number);
if (floor(square) == square)
// number is a perfect square
else
// number is not a perfect square

john

 

[Eric Sosman]

Gaijinco wrote On 09/26/05 16:07,:

Sooner or later everytime I found recreational programming challenges I
stumble with how I test if a number is has decimal places differnt than
0?

For example if I want to know if a number is a square number (i.e. a
number which square root is a positive number as 4, 9, 16 have) I do
something like:

int square = sqrt(number);

if((int)square==square)
// number is a perfect square

.... and the test is a tautology.

else
// number is not a perfect square

Is there a function or a language-specific-way to do this?

The code you've shown will (if it doesn't invoke
undefined behavior) declare that every number is a
perfect square: 1, 2, 3.14, and even -42.

To test whether a floating-point number is an
integer with no fractional part, you could try

if ((int)fpn == fpn) ...

This runs into trouble when the magnitude of fpn
is large, so large that its value is outside the range
of numbers representable as `int'.

As an improvement you might try

if (fmod(fpn, 1.0) == 0.0) ...

This is still vulnerable to the "graininess" of
floating-point numbers, which are not mathematical real
numbers with infinite precision.

You cannot usually expect sqrt(fpn) to be the exact square
root of fpn. sqrt(fpn) will be very close to the exact root,
but will (in general) be just a little bit different from the
true value. There could be several different fpn values for
which sqrt(fpn) would deliver exactly the same slightly wrong
answer: both sqrt(4.0) and sqrt(4.0 + tiny_number) might
produce 2.0 as an answer. If you decide that a number is a
perfect square if its computed square root turns out to be an
integer, you will erroneously conclude that 4.0+tiny_number is
a perfect square.

A possibly more thorough test might compute the square
root, test whether it's an integer, and then test whether
its square equals the original number:

double root = sqrt(number);
if (fmod(root, 1.0) == 0.0 && root * root == number)

.... but even this may have some problems. I am always uneasy
when comparing floating-point quantities for exact equality,
mostly because fpn's are usually regarded as approximations
to begin with. You usually need an "approximately equal"
test of some kind, and such a test isn't well suited to the
purely yet/no nature of perfect squaredom.

For "number-theoretic" calculations you'll usually be much
better off using integers of some flavor. If the numbers grow
large you may need to resort to a "bignum" package; several
are available.

 

[Mark McIntyre]

On Mon, 26 Sep 2005 21:05:09 GMT, in comp.lang.c , John Harrison

(of testing to see if a float is a perfect square)

This test always works (within the limitations of
floating point accuracy).

I'd not be too sure of that. Remember floating point is not an exact
representation.

 

[John Harrison]

On Mon, 26 Sep 2005 21:05:09 GMT, in comp.lang.c , John Harrison

(of testing to see if a float is a perfect square)




I'd not be too sure of that. Remember floating point is not an exact
representation.

That's why I said 'within the limitations of floating point accuracy'.
My test always checks if a floating point number is integral. Obviously
an integral return from sqrt does not necessarily mean the sqrt
parameter was a perfect square.

john

 

[Eric Sosman]

John Harrison wrote On 09/26/05 17:20,:

That's why I said 'within the limitations of floating point accuracy'.
My test always checks if a floating point number is integral. Obviously
an integral return from sqrt does not necessarily mean the sqrt
parameter was a perfect square.

On the machine in front of me right now, sqrt(1.0)
and sqrt(1.0000000000000002) both give 1.0 as the root.

 

[Martin Ambuhl]

Sooner or later everytime I found recreational programming challenges I
stumble with how I test if a number is has decimal places differnt than
0?

For example if I want to know if a number is a square number (i.e. a
number which square root is a positive number as 4, 9, 16 have) I do
something like:

int square = sqrt(number);

if((int)square==square)
// number is a perfect square
else
// number is not a perfect square

Is there a function or a language-specific-way to do this?

using modf, you can test either the integral part or the fractional
part. Look askance at any solution that involves ints.

#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <math.h>

int main(void)
{
double x, fp, ip;
int loop, cnt;
srand(time(0));
for (loop = cnt = 0; cnt < 10; loop++) {
x = (int) (100. * rand() / (1. + RAND_MAX)) / 10.;
fp = modf(x, &ip);
if (ip != x || fp)
continue;
printf("%4d: x = %g, fractional part (fp) = %g,"
"integer part (ip) = %g\n"
" (ip %s x, fp %s 0)\n", loop, x, fp, ip,
(ip == x) ? "==" : "!=", (fp == 0) ? "==" : "!=");
cnt++;
}
return 0;
}


1: x = 8, fractional part (fp) = 0,integer part (ip) = 8
(ip == x, fp == 0)
36: x = 0, fractional part (fp) = 0,integer part (ip) = 0
(ip == x, fp == 0)
63: x = 6, fractional part (fp) = 0,integer part (ip) = 6
(ip == x, fp == 0)
70: x = 1, fractional part (fp) = 0,integer part (ip) = 1
(ip == x, fp == 0)
79: x = 5, fractional part (fp) = 0,integer part (ip) = 5
(ip == x, fp == 0)
87: x = 5, fractional part (fp) = 0,integer part (ip) = 5
(ip == x, fp == 0)
128: x = 4, fractional part (fp) = 0,integer part (ip) = 4
(ip == x, fp == 0)
147: x = 1, fractional part (fp) = 0,integer part (ip) = 1
(ip == x, fp == 0)
148: x = 8, fractional part (fp) = 0,integer part (ip) = 8
(ip == x, fp == 0)
162: x = 1, fractional part (fp) = 0,integer part (ip) = 1
(ip == x, fp == 0)

 

[John Harrison]

On the machine in front of me right now, sqrt(1.0)
and sqrt(1.0000000000000002) both give 1.0 as the root.

Am I missing something? What point are you making?

john

 

[Keith Thompson]

#include <math>

double square = sqrt(number);
if (floor(square) == square)
// number is a perfect square
else
// number is not a perfect square

I suggest that "square" is a really bad name for a variable that hold
the result of a call to sqrt().

 

[Pete Becker]

I'd not be too sure of that. Remember floating point is not an exact
representation.

Floating point is exact. Unlike real numbers, it is not continuous.

 

[E. Robert Tisdale]

Remember floating point is not an exact representation.

Floating point is exact. Unlike real numbers, it is not continuous.

In general, *finite precision* floating-point arithmetic is *inexact*.
The set real numbers includes irrational numbers which have *no*
finite precision digital representation in this universe --
they exist only in the minds of mathematicians.

 

[Keith Thompson]

Floating point is exact. Unlike real numbers, it is not continuous.

It depends on how you look at it. A given floating-point
representation can be viewed either as an exact value, or as an
inexact approximation of any of the infinitely many real numbers close
to the exact represented value. The latter, though arguably
incorrect, is probably the more common interpretation, especially
given things like:

double one_third = 1.0/3.0;

Integers are usually considered to be exact because they typically
aren't thought of as being approximations of nearby numbers; 42 really
is 42, not an approximation of 42.0625.

 

[Dik T. Winter]

>
> In general, *finite precision* floating-point arithmetic is *inexact*.

I think you misunderstand Pete Beckers meaning. Assuming IEEE, given
two operands and an operation, it is precisely predictable what the
result is. But I understand you are pretty good at misunderstanding.

 

[Eric Sosman]

Am I missing something? What point are you making?

Just giving a concrete example of the point you made
(and snipped):

>>> [...] Obviously
>>> an integral return from sqrt does not necessarily mean the sqrt
>>> parameter was a perfect square.

 

[Howard]

That's why I said 'within the limitations of floating point accuracy'. My
test always checks if a floating point number is integral. Obviously an
integral return from sqrt does not necessarily mean the sqrt parameter was
a perfect square.

john

I think a lot of the discussion here is assuming the the "number" variable
being passed to the sqrt function is a random floating point variable. But,
I think he was actually asking about testing _integers_, since no value
containing a fractional part could _ever_ be the square of an integer. If
we assume that "number" is itself an integer (even if it's stored in a
float), then testing if the floor of the result of the sqrt call is equal to
the result itself is a perfectly valid choice. The range of integers is
sufficently small that there will never be a case where the result of the
square root will appear to be exactly an integral number when in fact it is
not. And, if we're dealing with a "number" variable that is actually
floating point, then it makes sense to first check if that number is itself
an integer (using floor or fmod) before testing if it's also the square of
an integer.

-Howard

 

[Gaijinco]

The issue at hand was that I would normally do this for testing if an
integrer was a square number:

bool square_number(int number){
if(sqrt(number)==floor(sqrt(number)))
return true;
else
return false;
}

However "sqrt(number)==floor(sqrt(number))" seemed an awkward way to
test it, so I wasn't sure if there was a "clenear" way to do it.

"fmod(sqrt(number),1.0)==0.0" seems to be an alternate but it isn't
very clean to me.

The facts about integers and operations were very intresting! Thank you.

 

[Mark McIntyre]

Mark McIntyre wrote:

(of comparing an int and a float)

That's why I said 'within the limitations of floating point accuracy'.

my point exactly.

My test always checks if a floating point number is integral.

Well, it checks if the floating point number is integral given your
first constraint.

Its possible for (int)square <> square, even though 'square' /is/ a
square number.

As a general rule any expression involving comparison operators and
floats should be regarded as /highly/ suspect.

 

[Mark McIntyre]

Floating point is exact. Unlike real numbers, it is not continuous.

Please exactly represent pi in a float.

 

[Kai-Uwe Bux]

Please exactly represent pi in a float.

He said, it is not continuous. The point is:

a) every float represents a unique real number and does so exactly.

b) not every real number has a representation as a float.


Of course you can *regard* a real number as an approximation to nearby real
numbers. Thus, despite (a) being true, one can regard any double as an
approximation to nearby real nubers. [Depending on context, we do the same
using intergers: every time I encounter $250, it actually happens to mean
$249.98+tax.]


Best

Kai-Uwe Bux