Best way to test whether a number is an integer

[rpmu...@gmail.com]

I'm using sage as my secret weapon in solving some of the problems on
Project Euler. One of the things I need to do is figure out whether a
number is an integer or not.

If I have a rational number, I can do something like:

sage: (3/4).is_integral()
False

But this doesn't work if I have square roots:

sage: sqrt(9).is_integral()
---------------------------------------------------------------------------
<type 'exceptions.AttributeError'> Traceback (most recent call
last)

/Users/rmuller/Python/Pistol/ProjectEuler/<ipython console> in
<module>()

<type 'exceptions.AttributeError'>: 'SymbolicComposition' object has
no attribute 'is_integral'

Is there some way to do this intelligently? The python routine I use:

def isint(n): return int(n) == n

works until the python floats start getting big an inaccurate.

[David Joyner]

This doesn't answer your general question, but as a workaround in the
example you cited, you can use is_square:

sage: a = 9
sage: a.is_square()
True

[rpmu...@gmail.com]

Thanks, that does indeed get me past my block. I'd be glad to hear
other solutions, as well, since it will help me learn more about sage.

On Feb 7, 9:12 am, "David Joyner" <wdjoy...@gmail.com> wrote:
> This doesn't answer your general question, but as a workaround in the
> example you cited, you can use is_square:
>
> sage: a = 9
> sage: a.is_square()
> True
>

[William Stein]

On Feb 7, 2008 8:20 AM, rpmu...@gmail.com <rpmu...@gmail.com> wrote:
>
> Thanks, that does indeed get me past my block. I'd be glad to hear
> other solutions, as well, since it will help me learn more about sage.
>
> On Feb 7, 9:12 am, "David Joyner" <wdjoy...@gmail.com> wrote:
> > This doesn't answer your general question, but as a workaround in the
> > example you cited, you can use is_square:
> >
> > sage: a = 9
> > sage: a.is_square()
> > True

Coerce to Integer and use try/except:

sage: try:
Integer(sqrt(7))
except TypeError:
print "sqrt(7) is not an integer."
....:
sqrt(7) is not an integer.

sage: try:
Integer(sqrt(9))
except TypeError:
print "sqrt(9) is not an integer."
....:
3

[rpmu...@gmail.com]

Thanks for all of the suggestions. I have a follow-on question:

I can see by looking at the source code to "is_square??" that the Sage
function just wraps Pari's is_square function. Can anyone tell me a
simple way to figure out whether a number is square, other than:

>>> squares = Set(i*i for i in range(1,very_big_number))
>>> def is_square(n): return n in squares

which gets memory intensive for really large values of
very_big_number.

On Feb 7, 11:49 am, "William Stein" <wst...@gmail.com> wrote:

[rpmu...@gmail.com]

I was curious enough about this to answer my own question. I coded up
the following for the integer square root:
def isqrt(n):
xn = 1
xn1 = (xn + n/xn)/2
while abs(xn1 - xn) > 1:
xn = xn1
xn1 = (xn + n/xn)/2
while xn1*xn1 > n:
xn1 -= 1
return xn1
which I got from:
http://www.codecodex.com/wiki/index.php?title=Calculate_an_integer_square_root

I can then use:

def is_square(n): return isqrt(n)**2 == n

which works quite nicely.

[William Stein]

On Feb 11, 2008 10:38 AM, rpmu...@gmail.com <rpmu...@gmail.com> wrote:
>
> Thanks for all of the suggestions. I have a follow-on question:
>
> I can see by looking at the source code to "is_square??" that the Sage
> function just wraps Pari's is_square function. Can anyone tell me a
> simple way to figure out whether a number is square, other than:
>
> >>> squares = Set(i*i for i in range(1,very_big_number))
> >>> def is_square(n): return n in squares
>
> which gets memory intensive for really large values of
> very_big_number.

Use the is_square *method* of integers:

sage: n = 100
sage: n.is_square()
True
sage: n = 100919080981232
sage: n.is_square()
False

The above is implemented by a single call into
the GMP library:

sage: n.is_square??
...
def is_square(self):
r"""
Returns \code{True} if self is a perfect square

EXAMPLES:
sage: Integer(4).is_square()
True
sage: Integer(41).is_square()
False
"""
return mpz_perfect_square_p(self.value)

This is going to be *vastly* faster than anything
you are likely to write:

sage: n = 100919080981232
sage: time for _ in xrange(10^6): a = n.is_square()
CPU times: user 0.29 s, sys: 0.00 s, total: 0.29 s
sage: time v = [n.is_square() for n in [1..10^6]]
CPU times: user 0.53 s, sys: 0.14 s, total: 0.68 s
Wall time: 2.50
sage: len(v)
1000000