Integer square root

Template:Short description In number theory, the integer square root (isqrt) of a non-negative integer Template:Mvar is the non-negative integer Template:Mvar which is the greatest integer less than or equal to the square root of Template:Mvar, $$\mathrm{isqrt}(n)=\lfloor \sqrt{n}\rfloor .$$

For example, $\mathrm{isqrt}(27)=\lfloor \sqrt{27}\rfloor =\lfloor 5.19615242270663...\rfloor =5.$

Introductory remark

Let $y$ and $k$ be non-negative integers.

Algorithms that compute (the decimal representation of) $\sqrt{y}$ run forever on each input $y$ which is not a perfect square.

Algorithms that compute $\lfloor \sqrt{y}\rfloor$ do not run forever. They are nevertheless capable of computing $\sqrt{y}$ up to any desired accuracy $k$.

Choose any $k$ and compute $\lfloor \sqrt{y\times 100^k}\rfloor$.

For example (setting $y=2$): $$\begin{array}{cc}& k=0:\lfloor \sqrt{2\times 100^0}\rfloor =\lfloor \sqrt{2}\rfloor =1\\ & k=1:\lfloor \sqrt{2\times 100^1}\rfloor =\lfloor \sqrt{200}\rfloor =14\\ & k=2:\lfloor \sqrt{2\times 100^2}\rfloor =\lfloor \sqrt{20000}\rfloor =141\\ & k=3:\lfloor \sqrt{2\times 100^3}\rfloor =\lfloor \sqrt{2000000}\rfloor =1414\\ & ⋮\\ & k=8:\lfloor \sqrt{2\times 100^8}\rfloor =\lfloor \sqrt{20000000000000000}\rfloor =141421356\\ & ⋮\\ \end{array}$$

Compare the results with $\sqrt{2}=1.41421356237309504880168872420969807856967187537694...$

It appears that the multiplication of the input by $100^k$ gives an accuracy of Template:Mvar decimal digits.

To compute the (entire) decimal representation of $\sqrt{y}$, one can execute $\mathrm{isqrt}(y)$ an infinite number of times, increasing $y$ by a factor $100$ at each pass.

Assume that in the next program ($\mathrm{sqrtForever}$) the procedure $\mathrm{isqrt}(y)$ is already defined and — for the sake of the argument — that all variables can hold integers of unlimited magnitude.

Then $\mathrm{sqrtForever}(y)$ will print the entire decimal representation of $\sqrt{y}$.

// Print sqrt(y), without halting
void sqrtForever(unsigned int y)
{
	unsigned int result = isqrt(y);
	printf("%d.", result);	// print result, followed by a decimal point

	while (true)	// repeat forever ...
	{
		y = y * 100;	// theoretical example: overflow is ignored
		result = isqrt(y);
		printf("%d", result % 10);	// print last digit of result
	}
}

The conclusion is that algorithms which compute Template:Code are computationally equivalent to [[Methods of computing square roots|algorithms which compute Template:Code]].

Basic algorithms

The integer square root of a non-negative integer $y$ can be defined as $$\lfloor \sqrt{y}\rfloor =x:x^2\le y<(x+1{)}^2,x\in \mathbb{N}$$

For example, $\mathrm{isqrt}(27)=\lfloor \sqrt{27}\rfloor =5$ because $6^2>27\text{ and }{5}^2\ngtr 27$.

Algorithm using linear search

The following C programs are straightforward implementations.

Linear search using addition

In the program above (linear search, ascending) one can replace multiplication by addition, using the equivalence $$(L+1{)}^2=L^2+2L+1=L^2+1+\sum _{i=1}^{L}2.$$

// Integer square root
// (linear search, ascending) using addition
unsigned int isqrt(unsigned int y)
{
	unsigned int L = 0;
	unsigned int a = 1;
	unsigned int d = 3;

	while (a <= y)
	{
		a = a + d;	// (a + 1) ^ 2
		d = d + 2;
		L = L + 1;
	}

	return L;
}

Algorithm using binary search

Linear search sequentially checks every value until it hits the smallest $x$ where $x^2>y$.

A speed-up is achieved by using binary search instead. The following C-program is an implementation.

// Integer square root (using binary search)
unsigned int isqrt(unsigned int y)
{
	unsigned int L = 0;
	unsigned int M;
	unsigned int R = y + 1;

    while (L != R - 1)
    {
        M = (L + R) / 2;

		if (M * M <= y)
			L = M;
		else
			R = M;
	}

    return L;
}

Numerical example

For example, if one computes $\mathrm{isqrt}(2000000)$ using binary search, one obtains the $[L,R]$ sequence $$\begin{array}{cc}& [0,2000001]\to [0,1000000]\to [0,500000]\to [0,250000]\to [0,125000]\to [0,62500]\to [0,31250]\to [0,15625]\\ & \to [0,7812]\to [0,3906]\to [0,1953]\to [976,1953]\to [976,1464]\to [1220,1464]\to [1342,1464]\to [1403,1464]\\ & \to [1403,1433]\to [1403,1418]\to [1410,1418]\to [1414,1418]\to [1414,1416]\to [1414,1415]\end{array}$$

This computation takes 21 iteration steps, whereas linear search (ascending, starting from $0$) needs Template:Val steps.

Algorithm using Newton's method

One way of calculating $\sqrt{n}$ and $\mathrm{isqrt}(n)$ is to use Heron's method, which is a special case of Newton's method, to find a solution for the equation $x^2-n=0$, giving the iterative formula $$x_{k+1}=\frac{1}{2}(x_k+\frac{n}{x_k}),k\ge 0,x_0>0.$$

The sequence $\{x_k\}$ converges quadratically to $\sqrt{n}$ as $k\to \mathrm{\infty }$.

Stopping criterion

One can proveTemplate:Citation needed that $c=1$ is the largest possible number for which the stopping criterion $$|x_{k+1}-x_k|<c$$ ensures $\lfloor x_{k+1}\rfloor =\lfloor \sqrt{n}\rfloor$ in the algorithm above.

In implementations which use number formats that cannot represent all rational numbers exactly (for example, floating point), a stopping constant less than 1 should be used to protect against round-off errors.

Domain of computation

Although $\sqrt{n}$ is irrational for many $n$, the sequence $\{x_k\}$ contains only rational terms when $x_0$ is rational. Thus, with this method it is unnecessary to exit the field of rational numbers in order to calculate $\mathrm{isqrt}(n)$, a fact which has some theoretical advantages.

Using only integer division

For computing $\lfloor \sqrt{n}\rfloor$ for very large integers n, one can use the quotient of Euclidean division for both of the division operations. This has the advantage of only using integers for each intermediate value, thus making the use of floating point representations of large numbers unnecessary. It is equivalent to using the iterative formula $$x_{k+1}=\lfloor \frac{1}{2}(x_k+\lfloor \frac{n}{x_k}\rfloor )\rfloor ,k\ge 0,x_0>0,x_0\in \mathbb{Z}.$$

By using the fact that $$\lfloor \frac{1}{2}(x_k+\lfloor \frac{n}{x_k}\rfloor )\rfloor =\lfloor \frac{1}{2}(x_k+\frac{n}{x_k})\rfloor ,$$

one can show that this will reach $\lfloor \sqrt{n}\rfloor$ within a finite number of iterations.

In the original version, one has $x_k\ge \sqrt{n}$ for $k\ge 1$, and $x_k>x_{k+1}$ for $x_k>\sqrt{n}$. So in the integer version, one has $\lfloor x_k\rfloor \ge \lfloor \sqrt{n}\rfloor$ and $x_k\ge \lfloor x_k\rfloor >x_{k+1}\ge \lfloor x_{k+1}\rfloor$ until the final solution $x_s$ is reached. For the final solution $x_s$, one has $\lfloor \sqrt{n}\rfloor \le \lfloor x_s\rfloor \le \sqrt{n}$ and $\lfloor x_{s+1}\rfloor \ge \lfloor x_s\rfloor$, so the stopping criterion is $\lfloor x_{k+1}\rfloor \ge \lfloor x_k\rfloor$.

However, $\lfloor \sqrt{n}\rfloor$ is not necessarily a fixed point of the above iterative formula. Indeed, it can be shown that $\lfloor \sqrt{n}\rfloor$ is a fixed point if and only if $n+1$ is not a perfect square. If $n+1$ is a perfect square, the sequence ends up in a period-two cycle between $\lfloor \sqrt{n}\rfloor$ and $\lfloor \sqrt{n}\rfloor +1$ instead of converging.

Example implementation in C

// Square root of integer
unsigned int int_sqrt(unsigned int s)
{
	// Zero yields zero
    // One yields one
	if (s <= 1) 
		return s;

    // Initial estimate (must be too high)
	unsigned int x0 = s / 2;

	// Update
	unsigned int x1 = (x0 + s / x0) / 2;

	while (x1 < x0)	// Bound check
	{
		x0 = x1;
		x1 = (x0 + s / x0) / 2;
	}		
	return x0;
}

Numerical example

For example, if one computes the integer square root of Template:Math using the algorithm above, one obtains the sequence $$\begin{array}{cc}& 1000000\to 500001\to 250002\to 125004\to 62509\to 31270\to 15666\to 7896\\ & \to 4074\to 2282\to 1579\to 1422\to 1414\to 1414\end{array}$$ In total 13 iteration steps are needed. Although Heron's method converges quadratically close to the solution, less than one bit precision per iteration is gained at the beginning. This means that the choice of the initial estimate is critical for the performance of the algorithm.

When a fast computation for the integer part of the binary logarithm or for the bit-length is available (like e.g. std::bit_width in C++20), one should better start at $$x_0=2^{\lfloor ({\log }_{2}n)/2\rfloor +1},$$ which is the least power of two bigger than $\sqrt{n}$. In the example of the integer square root of Template:Math, $\lfloor {\log }_{2}n\rfloor =20$, $x_0=2^{11}=2048$, and the resulting sequence is $$2048\to 1512\to 1417\to 1414\to 1414.$$ In this case only four iteration steps are needed.

Digit-by-digit algorithm

The traditional pen-and-paper algorithm for computing the square root $\sqrt{n}$ is based on working from higher digit places to lower, and as each new digit pick the largest that will still yield a square $\le n$. If stopping after the one's place, the result computed will be the integer square root.

Using bitwise operations

If working in base 2, the choice of digit is simplified to that between 0 (the "small candidate") and 1 (the "large candidate"), and digit manipulations can be expressed in terms of binary shift operations. With * being multiplication, << being left shift, and >> being logical right shift, a recursive algorithm to find the integer square root of any natural number is:

def integer_sqrt(n: int) -> int:
    assert n >= 0, "sqrt works for only non-negative inputs"
    if n < 2:
        return n

    # Recursive call:
    small_cand = integer_sqrt(n >> 2) << 1
    large_cand = small_cand + 1
    if large_cand * large_cand > n:
        return small_cand
    else:
        return large_cand


# equivalently:
def integer_sqrt_iter(n: int) -> int:
    assert n >= 0, "sqrt works for only non-negative inputs"
    if n < 2:
        return n

    # Find the shift amount. See also [[find first set]],
    # shift = ceil(log2(n) * 0.5) * 2 = ceil(ffs(n) * 0.5) * 2
    shift = 2
    while (n >> shift) != 0:
        shift += 2

    # Unroll the bit-setting loop.
    result = 0
    while shift >= 0:
        result = result << 1
        large_cand = (
            result + 1
        )  # Same as result ^ 1 (xor), because the last bit is always 0.
        if large_cand * large_cand <= n >> shift:
            result = large_cand
        shift -= 2

    return result

Traditional pen-and-paper presentations of the digit-by-digit algorithm include various optimizations not present in the code above, in particular the trick of pre-subtracting the square of the previous digits which makes a general multiplication step unnecessary. See Template:Section link for an example.

Karatsuba square root algorithm

The Karatsuba square root algorithm is a combination of two functions: a public function, which returns the integer square root of the input, and a recursive private function, which does the majority of the work.

The public function normalizes the actual input, passes the normalized input to the private function, denormalizes the result of the private function, and returns that.

The private function takes a normalized input, divides the input bits in half, passes the most-significant half of the input recursively to the private function, and performs some integer operations on the output of that recursive call and the least-significant half of the input to get the normalized output, which it returns.

For big-integers of "50 to 1,000,000 digits", Burnikel-Ziegler Karatsuba division and Karatsuba multiplication are recommended by the algorithm's creator.

An example algorithm for 64-bit unsigned integers is below. The algorithm:

1. Normalizes the input inside Template:Mono.
2. Calls Template:Mono, which requires a normalized input.
3. Calls Template:Mono with the most-significant half of the normalized input's bits, which will already be normalized as the most-significant bits remain the same.
4. Continues on recursively until there's an algorithm that's faster when the number of bits is small enough.
5. Template:Mono then takes the returned integer square root and remainder to produce the correct results for the given normalized Template:Mono.
6. Template:Mono then denormalizes the result.

/// Performs a Karatsuba square root on a `u64`.
pub fn u64_isqrt(mut n: u64) -> u64 {
    if n <= u32::MAX as u64 {
        // If `n` fits in a `u32`, let the `u32` function handle it.
        return u32_isqrt(n as u32) as u64;
    } else {
        // The normalization shift satisfies the Karatsuba square root
        // algorithm precondition "a₃ ≥ b/4" where a₃ is the most
        // significant quarter of `n`'s bits and b is the number of
        // values that can be represented by that quarter of the bits.
        //
        // b/4 would then be all 0s except the second most significant
        // bit (010...0) in binary. Since a₃ must be at least b/4, a₃'s
        // most significant bit or its neighbor must be a 1. Since a₃'s
        // most significant bits are `n`'s most significant bits, the
        // same applies to `n`.
        //
        // The reason to shift by an even number of bits is because an
        // even number of bits produces the square root shifted to the
        // left by half of the normalization shift:
        //
        // sqrt(n << (2 * p))
        // sqrt(2.pow(2 * p) * n)
        // sqrt(2.pow(2 * p)) * sqrt(n)
        // 2.pow(p) * sqrt(n)
        // sqrt(n) << p
        //
        // Shifting by an odd number of bits leaves an ugly sqrt(2)
        // multiplied in.
        const EVEN_MAKING_BITMASK: u32 = !1;
        let normalization_shift = n.leading_zeros() & EVEN_MAKING_BITMASK;
        n <<= normalization_shift;

        let (s, _) = u64_normalized_isqrt_rem(n);

        let denormalization_shift = normalization_shift / 2;
        return s >> denormalization_shift;
    }
}

/// Performs a Karatsuba square root on a normalized `u64`, returning the square
/// root and remainder.
fn u64_normalized_isqrt_rem(n: u64) -> (u64, u64) {
    const HALF_BITS: u32 = u64::BITS >> 1;
    const QUARTER_BITS: u32 = u64::BITS >> 2;
    const LOWER_HALF_1_BITS: u64 = (1 << HALF_BITS) - 1;

    debug_assert!(
        n.leading_zeros() <= 1,
        "Input is not normalized: {n} has {} leading zero bits, instead of 0 or 1.",
        n.leading_zeros()
    );

    let hi = (n >> HALF_BITS) as u32;
    let lo = n & LOWER_HALF_1_BITS;

    let (s_prime, r_prime) = u32_normalized_isqrt_rem(hi);

    let numerator = ((r_prime as u64) << QUARTER_BITS) | (lo >> QUARTER_BITS);
    let denominator = (s_prime as u64) << 1;

    let q = numerator / denominator;
    let u = numerator % denominator;

    let mut s = (s_prime << QUARTER_BITS) as u64 + q;
    let mut r = (u << QUARTER_BITS) | (lo & ((1 << QUARTER_BITS) - 1));
    let q_squared = q * q;
    if r < q_squared {
        r += 2 * s - 1;
        s -= 1;
    }
    r -= q_squared;

    return (s, r);
}

In programming languages

Some programming languages dedicate an explicit operation to the integer square root calculation in addition to the general case or can be extended by libraries to this end.

Programming language	Example use	Version introduced
Chapel	BigInteger.sqrt(result, n); BigInteger.sqrtRem(result, remainder, n);	Unknown
Common Lisp	(isqrt n)	Unknown
Crystal	Math.isqrt(n)	1.2
Java	n.sqrt() (BigInteger only)	9
Julia	isqrt(n)	0.3
Maple	isqrt(n)	Unknown
PARI/GP	sqrtint(n)	1.35a (as isqrt) or before
Python	math.isqrt(n)	3.8
Racket	(integer-sqrt n) (integer-sqrt/remainder n)	Unknown
Ruby	Integer.sqrt(n)	2.5.0
Rust	n.isqrt() n.checked_isqrt()	1.84.0
SageMath	isqrt(n)	Unknown
Scheme	(exact-integer-sqrt n)	R6RS
Tcl	isqrt($n)	8.5
Zig	std.math.sqrt(n)	Unknown

See also

• Methods of computing square roots

External links

• Template:Cite journal
• Template:Cite book
• Template:Cite web