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Integer square root

Greatest integer less than or equal to square root

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Source updated September 9, 2026

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

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

01Introductory 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, but are nevertheless capable of computing ${\sqrt {y}}$ up to any desired accuracy $k$ .

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

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

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

It appears that the multiplication of the input by $100^{k}$ gives an accuracy of k decimal digits.

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

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

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

import math # assume isqrt computation as given here def sqrtForever(y: int): """ Print sqrt(y), without halting """ result = math.isqrt(y) print(str(result) + ".", end="") # print result, followed by a decimal point while True: # repeat forever ... y *= 100 # theoretical example: overflow is ignored result = math.isqrt(y) print(str(result % 10), end="") # print last digit of result

The conclusion is that algorithms which compute isqrt() are computationally equivalent to algorithms which compute sqrt().

Another derivation of ${\sqrt {y}}$ from $\lfloor {\sqrt {y}}\rfloor$ is given in section Continued fraction of √c based on isqrt below.

02Basic algorithms

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

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

Algorithm using linear search

The following Python programs are straightforward implementations.

def isqrt(y: int) -> int: """ Integer square root (linear search, ascending) """ # initial underestimate, L <= isqrt(y) L = 0 while (L + 1) * (L + 1) <= y: L += 1 return L def isqrt(y: int) -> int: """ Integer square root (linear search, descending) """ # initial overestimate, isqrt(y) <= R R = y while (R * R > y): R -= 1 return R

Linear search using addition

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

def isqrt(y: int) -> int: """ Integer square root (linear search, ascending) using addition """ L = 0 a = 1 d = 3 while a <= y: a = a + d 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.

def isqrt(y: int) -> int: """ Integer square root (binary search) """ L = 0 # lower bound of the square root R = y + 1 # upper bound of the square root while (L != R - 1): M = (L + R) // 2 # midpoint to test if (M * M <= y): L = M else: R = M return L

Numerical examples

$\operatorname {isqrt} (0)=0$ , $\operatorname {isqrt} (1)=1$ .
Using binary search the computation of $\operatorname {isqrt} (131072)$ converges to $362$ in $17$ iteration steps via the $[L,R]$ sequence

${\begin{aligned}&[0,131073]\to [0,65536]\to [0,32768]\to [0,16384]\to [0,8192]\to [0,4096]\rightarrow [0,2048]\to [0,1024]\to [0,512]\\&\to [256,512]\to [256,384]\to [320,384]\to [352,384]\to [352,368]\to [360,368]\to [360,364]\to [362,364]\to [362,363]\end{aligned}}$

The computation of $\operatorname {isqrt} (2000000)$ converges to $1414$ in $21$ steps via the $[L,R]$ sequence

${\begin{aligned}&[0,2000001]\to [0,1000000]\to [0,500000]\to [0,250000]\to [0,125000]\to [0,62500]\to [0,31250]\to [0,15625]\\&\to [0,7812]\to [0,3906]\to [0,1953]\to [976,1953]\to [976,1464]\to [1220,1464]\to [1342,1464]\to [1403,1464]\\&\to [1403,1433]\to [1403,1418]\to [1410,1418]\to [1414,1418]\to [1414,1416]\to [1414,1415]\end{aligned}}$

Linear search (ascending, starting from $0$ ) needs 1414 steps.

03Algorithm using Newton's method

One way of calculating ${\sqrt {n}}$ and $\operatorname {isqrt} (n)$ 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 $x_{k+1}={\frac {1}{2}}\!\left(x_{k}+{\frac {n}{x_{k}}}\right),\quad k\geq 0,\quad x_{0}>0.$

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

Using only integer division

For computing $\lfloor {\sqrt {n}}\rfloor$ 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 unnecessary.

Let $n>0$ and initial guess $x_{0}>0$ . Define the integer sequence:

$x_{k+1}=\left\lfloor {\frac {x_{k}+\left\lfloor n/x_{k}\right\rfloor }{2}}\right\rfloor ,\quad k=0,1,2,\dots$

Proof of convergence

1. Positivity: All terms are positive integers: $x_{k}>0$ for all $k$ .

2. Monotonicity:

If $x_{k}>{\sqrt {n}}$ , then $\lfloor n/x_{k}\rfloor \leq n/x_{k}$ ;

so $x_{k+1}=\left\lfloor {\frac {x_{k}+\lfloor n/x_{k}\rfloor }{2}}\right\rfloor <{\frac {x_{k}+n/x_{k}}{2}}<x_{k}$ .

Hence the sequence decreases.

If $x_{k}<{\sqrt {n}}$ , then $\lfloor n/x_{k}\rfloor \geq n/x_{k}-1$ ;

so $x_{k+1}\geq {\frac {x_{k}+n/x_{k}-1}{2}}>x_{k}-1$ .

Hence the sequence increases or stays the same.

3. Boundedness: The sequence is bounded below by 1 and above by $x_{0}$ , so it is bounded.

4. Stabilization / Oscillation: A bounded monotone integer sequence either stabilizes or oscillates between two consecutive integers:

$x_{k+1}=x_{k}$ or $x_{k+1}=x_{k}\pm 1$ .

5. Integer "Fixed-point" Condition: At stabilization or oscillation:

$x_{k+1}=\lfloor (x_{k}+\lfloor n/x_{k}\rfloor )/2\rfloor$ .

This ensures that the sequence is either at $\lfloor {\sqrt {n}}\rfloor$ or oscillating between the two nearest integers around ${\sqrt {n}}$ .

6. Conclusion: The sequence eventually stabilizes at $\lfloor {\sqrt {n}}\rfloor$ or oscillates between $\lfloor {\sqrt {n}}\rfloor$ and $\lceil {\sqrt {n}}\rceil$ .

Remark:

$\lfloor {\sqrt {n}}\rfloor$ is a strict fixed point unless $n+1$ is a perfect square.
If $n+1$ is a perfect square, the sequence oscillates between $\lfloor {\sqrt {n}}\rfloor$ and $\lceil {\sqrt {n}}\rceil$ .

Example implementation

def isqrt(n: int, x0: int = 1) -> int: """ isqrt via Newton-Heron iteration with specified initial guess. Uses 2-cycle oscillation detection. Preconditions: n >= 0 # isqrt(0) = 0 x0 > 0, defaults to 1 # initial guess Output: isqrt(n) """ assert n >= 0 and x0 > 0, "Invalid input" # isqrt(0) = 0; isqrt(1) = 1 if n < 2: return n prev2 = -1 # x_{i-2} prev1 = x0 # x_{i-1} while True: x1 = (prev1 + n // prev1) // 2 # Case 1: converged (steady value) if x1 == prev1: return x1 # Case 2: oscillation (2-cycle) if x1 == prev2 and x1 != prev1: # We’re flipping between prev1 and prev2 # Choose the smaller one (the true integer sqrt) return min(prev1, x1) # Move forward prev2, prev1 = prev1, x1

Numerical examples

The call isqrt(2000000) converges to $1414$ in 14 passes through while:

${\begin{aligned}&1000000\to 500001\to 250002\to 125004\to 62509\to 31270\to 15666\to 7896\to 4074\to 2282\\&\to 1579\to 1422\to 1414\rightarrow 1414\end{aligned}}$ .

One iteration is gained by setting x0 to $\lfloor n/2\rfloor$ with the call isqrt(2000000, 1000000). 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. n.bit_length() in Python), 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 2000000, $\lfloor \log _{2}n\rfloor =20$ , $x_{0}=2^{11}=2048$ , and the resulting sequence is $2048\rightarrow 1512\rightarrow 1417\rightarrow 1414\rightarrow 1414.$ In this case only four iteration steps are needed. This corresponds to the call isqrt(2000000, 2048).

04Digit-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 $\leq n$ . 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 isqrt_recursive(n: int) -> int: assert n >= 0, "n must be a non-negative integer" if n < 2: return n # Recursive call: small_cand = isqrt_recursive(n >> 2) << 1 # same as 2 * isqrt(n // 4) large_cand = small_cand + 1 if large_cand * large_cand > n: return small_cand else: return large_cand

Equivalent non-recursive program:

def isqrt_iterative(x: int) -> int: """ Guy, Martin (1985). "Fast integer square root by Mr. Woo's abacus algorithm" """ assert x >= 0, "x must be a non-negative integer" op = x; res = 0 # note: i << 1 is 2 * i, i << 2 is 4 * i, # i >> 1 is i // 2, and i >> 2 is i // 4... # "one" starts at the highest power of four <= x one = 1 while one <= op: one <<= 2 one >>= 2 while one != 0: dltasqr = res + one if op >= dltasqr: op -= dltasqr res += one << 1 res >>= 1 one >>= 2 return res

See Methods of computing square roots § Binary numeral system (base 2) for an example.

05Karatsuba square root algorithm

The Karatsuba square root algorithm applies the same divide-and-conquer principle as the Karatsuba multiplication algorithm to compute integer square roots. The method was formally analyzed by Paul Zimmermann (1999). It recursively splits the input number into high and low halves, computes the square root of the higher half, and then determines the lower half algebraically.

Algorithm

Paul Zimmermann (1999) gives the following algorithm.

${\text{Algorithm }}{\text{SqrtRem}}(n=a_{3}b^{3}+a_{2}b^{2}+a_{1}b+a_{0})$

${\text{Input: }}0\leq a_{i}<b,{\text{ with }}a_{3}\geq b/4$

${\text{Output: }}(s,r){\text{ such that }}s^{2}\leq n=s^{2}+r<(s+1)^{2}$

$\qquad (s',r')\gets {\text{SqrtRem}}(a_{3}b+a_{2})$

$\qquad (q,u)\gets {\text{DivRem}}(r'b+a_{1},2s')$

$\qquad s\gets s'b+q$

$\qquad r\gets ub+a_{0}-q^{2}$

$\qquad {\text{if }}r<0{\text{ then }}$

$\qquad \qquad r\gets r+2s-1$

$\qquad \qquad s\gets s-1$

$\qquad {\text{return }}(s,r)$

Because only one recursive call is made per level, the total complexity remains $O(n)$ in the number of bits. Each level performs only linear-time arithmetic on half-size numbers.

Comparison with Karatsuba multiplication

Property	Karatsuba multiplication	Karatsuba-style square root
Recursive calls per level	3	1
Recurrence	$T(n)=3T(n/2)+O(n)$	$T(n)=T(n/2)+O(n)$
Asymptotic complexity	$O(n^{\log _{2}3})\!\approx \!O(n^{1.585})$	$O(n)$
Key operation	Three partial multiplications and recombination	One recursive square root and algebraic correction

Use and history

The Karatsuba-style square root is mainly used for arbitrary-precision arithmetic on very large integers, where it combines efficiently with Burnikel-Ziegler division and Karatsuba multiplication. It was first analyzed formally by Paul Zimmermann (1999). Earlier practical work includes Martin Guy (1985), and recursive versions appear in Donald Knuth (1998). Modern GMP and MPIR libraries implement similar recursive techniques.

Implementation in Python

The Python program below implements Zimmermann’s algorithm. Given an integer $n\geq 0$ , SqrtRem computes simultaneously its integer square root $s=\lfloor {\sqrt {n}}\rfloor$ and the corresponding remainder $r=n-s^{2}$ . The choice of isqrt() is ad libitum.

def SqrtRem(n: int, word_bits: int = 32) -> tuple[int, int]: """ Implementation based on Zimmermann's Karatsuba-style integer square root algorithm [Zimmermann, 1999]. It recursively splits the input n into "limbs" of size `word_bits` and combines partial results to compute the integer square root. Args: n (int): Non-negative integer to compute the square root of. word_bits (int, optional): Number of bits per "limb" or chunk used when recursively splitting n. Default is 32. Each limb represents a fixed-size part of n for the algorithm. Returns: tuple[int, int]: s = integer square root of n, r = remainder (n - s*s). Notes: The limb size controls recursion granularity. Larger word_bits reduces recursion depth but increases the size of subproblems; smaller word_bits increases recursion depth but works on smaller chunks. Reference: Zimmermann, P. (1999). "Karatsuba Square Root", Research report #3805, Inria. Archived: https://inria.hal.science/inria-00072854v1/file/RR-3805.pdf """ if n < 0: raise ValueError("n must be non-negative") if n == 0: return 0, 0 # trivial case # Determine number of word-sized limbs (mimics “limb” splitting in Zimmermann) limblen = (n.bit_length() + word_bits - 1) // word_bits # Base case: single limb, compute directly if limblen <= 1: s = isqrt(n) # any isqrt, e.g., math.isqrt or custom r = n - s*s return s, r # --- Step 1: Split n into high and low parts --- half_limbs = limblen // 2 shift = half_limbs * word_bits hi = n >> shift # high half, corresponds to a3*b + a2 lo = n & ((1 << shift) - 1) # low half, corresponds to a1*b + a0 # --- Step 2: Recursive call on the high part --- s_high, r_high = SqrtRem(hi, word_bits) # approximate sqrt of high half # --- Step 3: Recombine to approximate full sqrt --- quarter = shift // 2 numerator = (r_high << quarter) | (lo >> quarter) # simulate Zimmermann’s DivRem step denominator = s_high << 1 # 2*s' term q = numerator // denominator if denominator else 0 # integer division s_candidate = (s_high << quarter) + q # recombine high and low # --- Step 4: Verification and correction --- # Ensure remainder is non-negative and s*s <= n < (s+1)*(s+1) s = s_candidate r = n - s*s while r < 0: # overestimate correction s -= 1 r = n - s*s while (s + 1)*(s + 1) <= n: # underestimate correction s += 1 r = n - s*s return s, r

Example usage

for n in [(2**32) + 5, 12345678901234567890, (1 << 1512) - 1]: s, r = SqrtRem(n) print(f"SqrtRem({n}) = {s}, remainder = {r}")

Computation
SqrtRem(4294967301) = 65536, remainder = 5
SqrtRem(12345678901234567890) = 3513641828, remainder = 5763386306
SqrtRem(143665816004337822710282600285310394341474369045835074863414468709543787931907367746179403311452034095731304066234112071267510472464260955153084575408147254672957261763907982395337943906645864229014250227057207826232751957053220218983971305018634078800548055251973907806245884614087189937340865371691338441989956445051526543084039211962387469415699218979531585795574920384684004258007709014706216763392717018544247025174258411677231986785008489302218244095) = 379032737378102767370356320425415662904513187772631008578870126471203845870697482014374611530431269030880793627229265919475483409207718357286202948008100864063587640630090308972232735749901964068667724412528434753635948938919935, remainder = 758065474756205534740712640850831325809026375545262017157740252942407691741394964028749223060862538061761587254458531838950966818415436714572405896016201728127175281260180617944465471499803928137335448825056869507271897877839870

06In 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
PHP	sqrt($num)	4
Python	math.isqrt(n)	3.8
Racket	(integer-sqrt n) (integer-sqrt/remainder n)	Unknown
Ruby	Integer.sqrt(n)	2.5.0
Rust	n.isqrt() n.checked_isqrt()	1.84.0
SageMath	isqrt(n)	Unknown
Scheme	(exact-integer-sqrt n)	R⁶RS
Tcl	isqrt($n)	8.5
Zig	std.math.sqrt(n)	Unknown

07Continued fraction of √c based on isqrt

The computation of the simple continued fraction of ${\sqrt {c}}$ can be carried out using only integer operations, with $\operatorname {isqrt} (c)$ serving as the initial term. The algorithm generates continued fraction expansion in canonical form.

Let $a_{0}=\lfloor {\sqrt {c}}\rfloor$ be the integer square root of $c$ .

If $c$ is a perfect square, the continued fraction terminates immediately:

${\sqrt {c}}=[a_{0}].$

Otherwise, the continued fraction is periodic:

${\sqrt {c}}=[a_{0};{\overline {a_{1},a_{2},\dots ,a_{m}}}]$ ,

where the overline indicates the repeating part.

The continued fraction can be obtained by the following recurrence, which uses only integer arithmetic:

$m_{0}=0,\quad d_{0}=1,\quad a_{0}=\lfloor {\sqrt {c}}\rfloor .$

For $k\geq 0$ ,

$m_{k+1}=d_{k}a_{k}-m_{k},\quad d_{k+1}={\frac {c-m_{k+1}^{2}}{d_{k}}},\quad a_{k+1}=\left\lfloor {\frac {a_{0}+m_{k+1}}{d_{k+1}}}\right\rfloor .$

Since there are only finitely many possible triples $(m_{k},d_{k},a_{k})$ , eventually one repeats, and from that point onward the continued fraction becomes periodic.

Implementation in Python

On input $c$ , a non-negative integer, the following program computes the simple continued fraction of ${\sqrt {c}}$ . The integer square root $\lfloor {\sqrt {c}}\rfloor$ is computed once. Only integer arithmetic is used. The program outputs $[a0,(a1,a2,...,am)]$ , where the second element is the periodic part.

def continued_fraction_sqrt(c: int) -> tuple[int, tuple[int, ...]]: """ Compute the continued fraction of sqrt(c) using integer arithmetic. Returns [a0, (a1, a2, ..., am)] where the second element is the periodic part. For perfect squares, the period is empty. """ a0 = isqrt(c) # Perfect square: return period empty if a0 * a0 == c: return (a0, ()) m, d, a = 0, 1, a0 period = [] seen = set() while True: m_next = d * a - m d_next = (c - m_next * m_next) // d a_next = (a0 + m_next) // d_next if (m_next, d_next, a_next) in seen: break seen.add((m_next, d_next, a_next)) period.append(a_next) m, d, a = m_next, d_next, a_next return (a0, tuple(period))

Example usage

for c in list(range(0, 18)) + [114] + [4097280036]: cf = continued_fraction_sqrt(c) print(f"sqrt({c}): {cf}")

Output

Input (c)	Output (cf)	Continued fraction
0	[0]	${\sqrt {0}}=0$
1	[1]	${\sqrt {1}}=1$
2	[1; (2,)]	${\sqrt {2}}=[1;{\overline {2}}]$
3	[1; (1, 2)]	${\sqrt {3}}=[1;{\overline {1,2}}]$
4	[2]	${\sqrt {4}}=2$
5	[2; (4,)]	${\sqrt {5}}=[2;{\overline {4}}]$
6	[2; (2, 4)]	${\sqrt {6}}=[2;{\overline {2,4}}]$
7	[2; (1, 1, 1, 4)]	${\sqrt {7}}=[2;{\overline {1,1,1,4}}]$
8	[2; (1, 4)]	${\sqrt {8}}=[2;{\overline {1,4}}]$
9	[3]	${\sqrt {9}}=3$
10	[3; (6,)]	${\sqrt {10}}=[3;{\overline {6}}]$
11	[3; (3, 6)]	${\sqrt {11}}=[3;{\overline {3,6}}]$
12	[3; (2, 6)]	${\sqrt {12}}=[3;{\overline {2,6}}]$
13	[3; (1, 1, 1, 1, 6)]	${\sqrt {13}}=[3;{\overline {1,1,1,1,6}}]$
14	[3; (1, 2, 1, 6)]	${\sqrt {14}}=[3;{\overline {1,2,1,6}}]$
15	[3; (1, 6)]	${\sqrt {15}}=[3;{\overline {1,6}}]$
16	[4]	${\sqrt {16}}=4$
17	[4; (8,)]	${\sqrt {17}}=[4;{\overline {8}}]$
114	[10; (1, 2, 10, 2, 1, 20)]	${\sqrt {114}}=[10;{\overline {1,2,10,2,1,20}}]$
4097280036	[64009; (1, 1999, 3, 4, 1, 499, 3, 1, 3, 3, 1, 124, ... ..., 3, 1, 3, 499, 1, 4, 3, 1999, 1, 128018)] period: 13,032 terms

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Sources and credits

This article is adapted from the Wikipedia article “Integer square root”, written by its contributors and licensed under CC BY-SA 4.0. Fathomly has changed the layout, removed citation markers, navigation and maintenance notices, and adjusted punctuation. This adapted version is shared under the same license. For references, see the original article.

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Integer square root

01Introductory remark

02Basic algorithms

Algorithm using linear search

Linear search using addition

Algorithm using binary search

03Algorithm using Newton's method

Using only integer division

Proof of convergence

Example implementation

Numerical examples

04Digit-by-digit algorithm

Using bitwise operations

05Karatsuba square root algorithm

Algorithm

Comparison with Karatsuba multiplication

Use and history

Implementation in Python

06In programming languages

07Continued fraction of √c based on isqrt

Implementation in Python

Sources and credits

Related topics

Square root algorithms

Periodic continued fraction