Integer square root

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

Algorithms that compute (the decimal representation of) $\backslash sqrt\; y$ run forever on each input $y$ which is not a perfect square.The square roots of the perfect squares (e.g., 0, 1, 4, 9, 16) are integers. In all other cases, the square roots of positive integers are irrational numbers.

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

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

For example (setting $y\; =\; 2$):

$$\backslash begin\{align\}$$

& 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 \\

& \vdots \\

& k = 8: \lfloor \sqrt {2 \times 100^{8}} \rfloor = \lfloor \sqrt {20000000000000000} \rfloor = 141421356 \\

& \vdots \\

\end{align}

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

It appears that the multiplication of the input by $100^k$ gives an accuracy of {{mvar|k}} decimal digits.It is no surprise that the repeated multiplication by {{math|100}} is a feature in {{harvtxt|Jarvis|2006}}

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

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

Then $\backslash operatorname\{sqrtForever\}(y)$ will print the entire decimal representation of $\backslash sqrt\; y$.The fractional part of square roots of perfect squares is rendered as {{math|000...}}.

// 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 {{code|isqrt()}} are computationally equivalent to Methods of computing square roots.

The integer square root of a non-negative integer $y$ can be defined as

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

The following C programs are straightforward implementations.

{{flex columns

|1=

// Integer square root

// (using linear search, ascending)

unsigned int isqrt(unsigned int y)

{

// initial underestimate, L <= isqrt(y)

unsigned int L = 0;

while ((L + 1) * (L + 1) <= y)

L = L + 1;

return L;

}

|2=

// Integer square root

// (using linear search, descending)

unsigned int isqrt(unsigned int y)

{

// initial overestimate, isqrt(y) <= R

unsigned int R = y;

while (R * R > y)

R = R - 1;

return R;

}

}}

In the program above (linear search, ascending) one can replace multiplication by addition, using the equivalence

$$(L+1)^2\; =\; L^2\; +\; 2L\; +\; 1\; =\; L^2\; +\; 1\; +\; \backslash sum\_\{i=1\}^L\; 2.$$

// 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;

}

{{Disputed-section|date=May 2025}}

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 $\backslash operatorname\{isqrt\}(2000000)$ using binary search, one obtains the $[L,R]$ sequence

$$\backslash begin\{align\}$$

& [0,2000001] \rightarrow [0,1000000] \rightarrow [0,500000] \rightarrow [0,250000] \rightarrow [0,125000] \rightarrow [0,62500] \rightarrow [0,31250] \rightarrow [0,15625] \\

& \rightarrow [0,7812] \rightarrow [0,3906] \rightarrow [0,1953] \rightarrow [976,1953] \rightarrow [976,1464] \rightarrow [1220,1464] \rightarrow [1342,1464] \rightarrow [1403,1464] \\

& \rightarrow [1403,1433] \rightarrow [1403,1418] \rightarrow [1410,1418] \rightarrow [1414,1418] \rightarrow [1414,1416] \rightarrow [1414,1415]

\end{align}

This computation takes 21 iteration steps, whereas linear search (ascending, starting from $0$) needs {{val|1414}} steps.

One way of calculating $\backslash sqrt\{n\}$ and $\backslash operatorname\{isqrt\}(n)$ 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

$$x\_\{k+1\}\; =\; \backslash frac\{1\}\{2\}\backslash !\backslash left(x\_k\; +\; \backslash frac\{n\}\{x\_k\}\backslash right),\; \backslash quad\; k\; \backslash ge\; 0,\; \backslash quad\; x\_0\; >\; 0.$$

The sequence $\backslash \{x\_k\backslash \}$ converges quadratically to $\backslash sqrt\{n\}$ as $k\backslash to\backslash infty$.

One can prove{{Citation needed|date=August 2021}} that $c=1$ is the largest possible number for which the stopping criterion

ensures $\backslash lfloor\; x\_\{k+1\}\; \backslash rfloor=\backslash lfloor\; \backslash sqrt\; n\; \backslash rfloor$ 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.

Although $\backslash sqrt\{n\}$ is irrational for many $n$, the sequence $\backslash \{x\_k\backslash \}$ 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 $\backslash operatorname\{isqrt\}(n)$, a fact which has some theoretical advantages.

For computing $\backslash lfloor\; \backslash sqrt\; n\; \backslash rfloor$ 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

$$x\_\{k+1\}\; =\; \backslash left\backslash lfloor\; \backslash frac\{1\}\{2\}\backslash !\backslash left(x\_k\; +\; \backslash left\backslash lfloor\; \backslash frac\{n\}\{x\_k\}\; \backslash right\backslash rfloor\; \backslash right)\; \backslash right\backslash rfloor,\; \backslash quad\; k\; \backslash ge\; 0,\; \backslash quad\; x\_0\; >\; 0,\; \backslash quad\; x\_0\; \backslash in\; \backslash mathbb\{Z\}.$$

By using the fact that

$$\backslash left\backslash lfloor\; \backslash frac\{1\}\{2\}\backslash !\backslash left(x\_k\; +\; \backslash left\backslash lfloor\; \backslash frac\{n\}\{x\_k\}\; \backslash right\backslash rfloor\; \backslash right)\; \backslash right\backslash rfloor\; =\; \backslash left\backslash lfloor\; \backslash frac\{1\}\{2\}\backslash !\backslash left(x\_k\; +\; \backslash frac\{n\}\{x\_k\}\; \backslash right)\; \backslash right\backslash rfloor,$$

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

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

However, $\backslash lfloor\; \backslash sqrt\; n\; \backslash rfloor$ is not necessarily a fixed point of the above iterative formula. Indeed, it can be shown that $\backslash lfloor\; \backslash sqrt\; n\; \backslash rfloor$ 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 $\backslash lfloor\; \backslash sqrt\; n\; \backslash rfloor$ and $\backslash lfloor\; \backslash sqrt\; n\; \backslash rfloor\; +\; 1$ instead of converging.

// 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;

}

For example, if one computes the integer square root of {{math|2000000}} using the algorithm above, one obtains the sequence

$$\backslash begin\{align\}$$

& 1000000 \rightarrow 500001 \rightarrow 250002 \rightarrow 125004 \rightarrow 62509 \rightarrow 31270 \rightarrow 15666 \rightarrow 7896 \\

& \rightarrow 4074 \rightarrow 2282 \rightarrow 1579 \rightarrow 1422 \rightarrow 1414 \rightarrow 1414

\end{align}

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

$$x\_0\; =\; 2^\{\backslash lfloor\; (\backslash log\_2\; n)\; /2\; \backslash rfloor+1\},$$

which is the least power of two bigger than $\backslash sqrt\; n$. In the example of the integer square root of {{math|2000000}}, $\backslash lfloor\; \backslash log\_2\; n\; \backslash rfloor\; =\; 20$, $x\_0\; =\; 2^\{11\}\; =\; 2048$, and the resulting sequence is

$$2048\; \backslash rightarrow\; 1512\; \backslash rightarrow\; 1417\; \backslash rightarrow\; 1414\; \backslash rightarrow\; 1414.$$

In this case only four iteration steps are needed.

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

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

1. 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 {{section link|Methods of computing square roots|Binary numeral system (base 2)}} for an example.{{cite web | last= Woo |first= C |title= Square root by abacus algorithm (archived) |date = June 1985 | url = http://medialab.freaknet.org/martin/src/sqrt/sqrt.c | archive-url = https://web.archive.org/web/20120306040058/http://medialab.freaknet.org/martin/src/sqrt/sqrt.c | archive-date = 2012-03-06 }}

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.{{cite web | last = Zimmermann | first = Paul | title = Karatsuba Square Root |date = 1999 | publisher = Inria | publication-date = 2006-05-24 | series = Research report #3805 | url = https://inria.hal.science/inria-00072854v1/file/RR-3805.pdf | archive-url = https://web.archive.org/web/20230511212802/https://inria.hal.science/inria-00072854v1/file/RR-3805.pdf | archive-date = 2023-05-11 }}

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

1. Normalizes the input inside {{mono|u64_isqrt}}.
2. Calls {{mono|u64_normalized_isqrt_rem}}, which requires a normalized input.
3. Calls {{mono|u32_normalized_isqrt_rem}} 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. {{mono|u64_normalized_isqrt_rem}} then takes the returned integer square root and remainder to produce the correct results for the given normalized {{mono|u64}}.
6. {{mono|u64_isqrt}} 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);

}

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.

• Methods of computing square roots

{{Reflist|group=note}}

{{Reflist}}

{{Refbegin}}

• {{cite journal

|last= Jarvis

|first= Ashley Frazer

|journal= Mathematical Spectrum

|title= Square roots by subtraction

|year= 2006

|volume= 37

|pages= 119–122

|url= http://www.afjarvis.staff.shef.ac.uk/maths/jarvisspec02.pdf

}}

• {{cite book

|last= Minsky

|first= Marvin

|author-link= Marvin Minsky

|title= Computation: Finite and Infinite Machines

|chapter= 9. The Computable Real Numbers

|url= https://archive.org/details/computationfinit0000mins

|url-access= registration

|publisher= Prentice-Hall

|year= 1967

|isbn= 0-13-165563-9

|oclc= 0131655639

}}

• {{cite web

|url= http://mathcentral.uregina.ca/RR/database/RR.09.95/grzesina1.html

|title= A geometric view of the square root algorithm

}}

{{Refend}}

{{number theoretic algorithms}}

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