Can I rely on this to judge a square number in C++?

Question

Can I rely on

sqrt((float)a)*sqrt((float)a)==a

or

(int)sqrt((float)a)*(int)sqrt((float)a)==a

to check whether a number is a perfect square? Why or why not?
int a is the number to be judged. I'm using Visual Studio 2005.

Edit: Thanks for all these rapid answers. I see that I can't rely on float type comparison. (If I wrote as above, will the last a be cast to float implicitly?) If I do it like

(int)sqrt((float)a)*(int)sqrt((float)a) - a < e

How small should I take that e value?

Edit2: Hey, why don't we leave the comparison part aside, and decide whether the (int) is necessary? As I see, with it, the difference might be great for squares; but without it, the difference might be small for non-squares. Perhaps neither will do. :-(

Answer 1

basics first:

if you (int) a number in a calculation it will remove ALL post-comma data. If I remember my C correctly, if you have an (int) in any calculation (+/-*) it will automatically presume int for all other numbers.

So in your case you want float on every number involved, otherwise you will loose data:

sqrt((float)a)*sqrt((float)a)==(float)a

is the way you want to go

Answer 2

Not to do the same calculation twice I would do it with a temporary number:

 int b = (int)sqrt((float)a);
 if((b*b) == a)
 {
     //perfect square
 }

edit: dav made a good point. instead of relying on the cast you'll need to round off the float first

so it should be:

 int b = (int) (sqrt((float)a) + 0.5f);
 if((b*b) == a)
 {
     //perfect square
 }

Answer 3

Actually, this is not a C++, but a math question.

1. With floating point numbers, you should never rely on equality. Where you would test a == b, just test against abs(a - b) < eps, where eps is a small number (e.g. 1E-6) that you would treat as a good enough approximation.
2. If the number you are testing is an integer, you might be interested in the Wikipedia article about Integer square root

EDIT:

As Krugar said, the article I linked does not answer anything. Sure, there is no direct answer to your question there, phoenie. I just thought that the underlying problem you have is floating point precision and maybe you wanted some math background to your problem.

For the impatient, there is a link in the article to a lengthy discussion about implementing isqrt. It boils down to the code karx11erx posted in his answer.

If you have integers which do not fit into an unsigned long, you can modify the algorithm yourself.

Answer 4

Floating point math is inaccurate by nature.

So consider this code:

int a=35;
float conv = (float)a;
float sqrt_a = sqrt(conv);
if( sqrt_a*sqrt_a == conv )
    printf("perfect square");

this is what will happen:

a = 35
conv = 35.000000
sqrt_a = 5.916079
sqrt_a*sqrt_a = 34.999990734

this is amply clear that sqrt_a^2 is not equal to a.

Answer 5

I didn't follow the formula, I apologize. But you can easily check if a floating point number is an integer by casting it to an integer type and compare the result against the floating point number. So,

bool isSquare(long val) {
    double root = sqrt(val);
    if (root == (long) root)
        return true;
    else return false;
}

Naturally this is only doable if you are working with values that you know will fit within the integer type range. But being that the case, you can solve the problem this way, saving you the inherent complexity of a mathematical formula.

Answer 6

If you don't want to relay on float precition then you can use the following code that uses interger math.

The Isqrt is taken from here and is O(log n)

// Finds the integer square root of a positive number
static int Isqrt(int num)
{
    if (0 == num) { return 0; }  // Avoid zero divide
    int n = (num / 2) + 1;       // Initial estimate, never low
    int n1 = (n + (num / n)) / 2;
    while (n1 < n)
    {
        n = n1;
        n1 = (n + (num / n)) / 2;
    } // end while
    return n;
} // end Isqrt()

static bool IsPerfectSquare(int num)
{
    return Isqrt(num) * Isqrt(num) == num;
}

Answer 7

Your question has already been answered, but here is a working solution.

Your 'perfect squares' are implicitly integer values, so you could easily solve floating point format related accuracy problems by using some integer square root function to determine the integer square root of the value you want to test. That function will return the biggest number r for a value v where r * r <= v. Once you have r, you simply need to test whether r * r == v.

unsigned short isqrt (unsigned long a)
{
    unsigned long rem = 0;
    unsigned long root = 0;

for (int i = 16; i; i--) {
    root <<= 1;
    rem = ((rem << 2) + (a >> 30));
    a <<= 2;
    if (root < rem)
        rem -= ++root;
    }
return (unsigned short) (root >> 1);
}

bool PerfectSquare (unsigned long a)
{
    unsigned short r = isqrt (a);

return r * r == a;
}

Answer 8

As reinier says, you need to add 0.5 to make sure it rounds to the nearest integer, so you get

int b = (int) (sqrt((float)a) + 0.5f);
if((b*b) == a) /* perfect square */

For this to work, b has to be (exactly) equal to the square root of a if a is a perfect square. However, I don't think you can guarantee this. Suppose that int is 64 bits and float is 32 bits (I think that's allowed). Then a can be of the order 2^60, so its square root is of order 2^30. However, a float only stores 24 bits in the significand, so the rounding error is of order 2^(30-24) = 2^6. This is larger to 1, so b may contain the wrong integer. For instance, I think that the above code does not identify a = (2^30+1)^2 as a perfect square.

Answer 9

The more obvious, if slower -- O(sqrt(n)) -- way:

bool is_perfect_square(int i) {
    int d = 1;
    for (int x = 0; x <= i; x += d, d += 2) {
     if (x == i) return true;
    }
    return false; 
}

Answer 10

I would do.

// sqrt always returns positive value. So casting to int is equivalent to floor()
int down =  static_cast<int>(sqrt(value));
int up   = down+1;                           // This is the ceil(sqrt(value))

// Because of rounding problems I would test the floor() and ceil()
// of the value returned from sqrt().
if (((down*down) == value) || ((up*up) == value))
{
     // We have a winner.
}

Answer 11

While others have noted that you should not test for equality with floats, I think you are missing out on chances to take advantage of the properties of perfect squares. First there is no point in re-squaring the calculated root. If a is a perfect square then sqrt(a) is an integer and you should check:

b = sqrt((float)a)
b - floor(b) < e

where e is set sufficiently small. There are also a number of integers that you can cross of as non-square before taking the square root. Checking Wikipedia you can see some necessary conditions for a to be square:

A square number can only end with digits 00,1,4,6,9, or 25 in base 10

Another simple check would be to see that a % 4 == 1 or 0 before taking the root since:

Squares of even numbers are even, since (2n)^2 = 4n^2.
Squares of odd numbers are odd, since (2n + 1)^2 = 4(n^2 + n) + 1.

These would essentially eliminate half of the integers before taking any roots.

Answer 12

The cleanest solution is to use an integer sqrt routine, then do:

bool isSquare( unsigned int a ) {

  unsigned int s = isqrt( a );
  return s * s == a;

}

This will work in the full int range and with perfect precision. A few cases:

a = 0, s = 0, s * s = 0 (add an exception if you don't want to treat 0 as square)
a = 1, s = 1, s * s = 1
a = 2, s = 1, s * s = 1
a = 3, s = 1, s * s = 1
a = 4, s = 2, s * s = 4
a = 5, s = 2, s * s = 4

Won't fail either as you approach the maximum value for your int size. E.g. for 32-bit ints:

a = 0x40000000, s = 0x00008000, s * s = 0x40000000
a = 0xFFFFFFFF, s = 0x0000FFFF, s * s = 0xFFFE0001

Using floats you run into a number of issues. You may find that sqrt( 4 ) = 1.999999..., and similar problems, although you can round-to-nearest instead of using floor().

Worse though, a float has only 24 significant bits which means you can't cast any int larger than 2^24-1 to a float without losing precision, which introduces false positives/negatives. Using doubles for testing 32-bit ints, you should be fine, though.

But remember to cast the result of the floating-point sqrt back to an int and compare the result to the original int. Comparisons between floats are never a good idea; even for square values of x in a limited range, there is no guarantee that sqrt( x ) * sqrt( x ) == x, or that sqrt( x * x) = x.