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# Code Puzzler Response - Find the Integer Square Root of a Positive Integer

12.13.2012
| 3661 views |
This Code Puzzler comes from the community contributor - Erik Colban
Thanks, Eric for today's extra brain-teaser!

Write a function that calculates the integer square root of any positive integer, but only uses add, subtract, left and right shifts, but no division or multiplication. The algorithm must run in O(log n) time. By integer square root I mean the largest integer whose square is less than or equal to a given number.

A similar problem was given before, and all submitted solutions were based on the iteration: x_n+1 = (x_n + x/x_n) / 2. That formula is great for finding good rational approximations of the square root of a number, but it's trivial once you know the formula. This problem is a little more challenging.

Below is an answer. The first while loop finds the largest power of 4 that is less than or equal to value. The next while loop has the following invariants:

1. q * root * root + remainder == value
2. 0 <= remainder < q * (2 * root + 1)
3. d == q * root

It's relatively easy to verify these invariants. When exiting the while loop, q == 1, and the invariants can be used to show that:

1. root * root <= value && value < (root + 1) * (root + 1)

```public static int intSqrt(int value) {
int q = 1;
while (q <= value) {
q <<= 2;
}
q >>= 2;
int d = q;
int remainder = value - q;
int root = 1;
while (q > 1) {
root <<= 1;
q >>= 2;
int s = q + d;
d >>= 1;
if (s <= remainder) {
remainder -= s;
root++;
d += q;
}
}
return root;
}```

Are there any other answers you have that don't use  x_n+1 = (x_n + x/x_n) / 2?

"Starting from scratch" is seductive but disease ridden