Integer square root
Template:Short descriptionTemplate:Use dmy dates 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, <math display="block">\operatorname{isqrt}(n) = \lfloor \sqrt n \rfloor.</math>
For example, <math>\operatorname{isqrt}(27) = \lfloor \sqrt{27} \rfloor = \lfloor 5.19615242270663 ... \rfloor = 5.</math>
Introductory remark
Let <math>y</math> and <math>k</math> be non-negative integers.
Algorithms that compute (the decimal representation of) <math>\sqrt y</math> run forever on each input <math>y</math> which is not a perfect square.<ref group="note">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.</ref>
Algorithms that compute <math>\lfloor \sqrt y \rfloor</math> do not run forever. They are nevertheless capable of computing <math>\sqrt y</math> up to any desired accuracy <math>k</math>.
Choose any <math>k</math> and compute <math display="inline">\lfloor \sqrt {y \times 100^k} \rfloor</math>.
For example (setting <math>y = 2</math>): <math display="block">\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}</math>
Compare the results with <math>\sqrt {2} = 1.41421356237309504880168872420969807856967187537694 ...</math>
It appears that the multiplication of the input by <math>100^k</math> gives an accuracy of Template:Mvar decimal digits.<ref group="note">It is no surprise that the repeated multiplication by Template:Math is a feature in Template:Harvtxt</ref>
To compute the (entire) decimal representation of <math>\sqrt y</math>, one can execute <math>\operatorname{isqrt}(y)</math> an infinite number of times, increasing <math>y</math> by a factor <math>100</math> at each pass.
Assume that in the next program (<math>\operatorname{sqrtForever}</math>) the procedure <math>\operatorname{isqrt}(y)</math> is already defined and — for the sake of the argument — that all variables can hold integers of unlimited magnitude.
Then <math>\operatorname{sqrtForever}(y)</math> will print the entire decimal representation of <math>\sqrt y</math>.<ref group="note">The fractional part of square roots of perfect squares is rendered as Template:Math.</ref>
- <syntaxhighlight lang="python">
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
</syntaxhighlight>
The conclusion is that algorithms which compute <syntaxhighlight lang="text" class="" style="" inline="1">isqrt()</syntaxhighlight> are computationally equivalent to algorithms which compute <syntaxhighlight lang="text" class="" style="" inline="1">sqrt()</syntaxhighlight>.
Another derivation of <math>\sqrt y</math> from <math>\lfloor \sqrt y \rfloor</math> is given in section Continued fraction of √c based on isqrt below.
Basic algorithms
The integer square root of a non-negative integer <math>y</math> can be defined as <math display="block">\lfloor \sqrt y \rfloor = \max \{ x : x^2 \leq y <(x+1)^2, x \in \mathbb{N} \}</math>
For example, <math>\operatorname{isqrt}(27) = \lfloor \sqrt{27} \rfloor = 5</math> because <math>6^2 > 27 \text{ and } 5^2 \ngtr 27</math>.
Algorithm using linear search
The following Python programs are straightforward implementations.
Linear search using addition
In the program above (linear search, ascending) one can replace multiplication by addition, using the equivalence <math display="block">(L+1)^2 = L^2 + 2L + 1 = L^2 + 1 + \sum_{i=1}^L 2.</math>
- <syntaxhighlight lang="python">
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
</syntaxhighlight>
Algorithm using binary search
Linear search sequentially checks every value until it hits the smallest <math>x</math> where <math>x^2 > y</math>.
A speed-up is achieved by using binary search instead.
- <syntaxhighlight lang="python">
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
</syntaxhighlight>
Numerical examples
- <math>\operatorname{isqrt}(0) = 0</math>, <math>\operatorname{isqrt}(1) = 1</math>.
- Using binary search the computation of <math>\operatorname{isqrt}(131072)</math> converges to <math>362</math> in <math>17</math> iteration steps via the <math>[L,R]</math> sequence
- <math>\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}</math>
- The computation of <math>\operatorname{isqrt}(2000000)</math> converges to <math>1414</math> in <math>21</math> steps via the <math>[L,R]</math> sequence
- <math>\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}</math>
- Linear search (ascending, starting from <math>0</math>) needs Template:Val steps.
Algorithm using Newton's method
One way of calculating <math>\sqrt{n}</math> and <math>\operatorname{isqrt}(n)</math> is to use Heron's method, which is a special case of Newton's method, to find a solution for the equation <math>x^2 - n = 0</math>, giving the iterative formula <math display="block">x_{k+1} = \frac{1}{2}\!\left(x_k + \frac{n}{x_k}\right), \quad k \ge 0, \quad x_0 > 0.</math>
The sequence <math>\{x_k\}</math> converges quadratically to <math>\sqrt{n}</math> as <math>k\to\infty</math>.<ref>Template:Cite web</ref><ref group="note">See Heron's method.</ref>
Using only integer division
For computing <math>\lfloor \sqrt n \rfloor</math> 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 <math>n > 0</math> and initial guess <math>x_0 > 0</math>. Define the integer sequence:
- <math>x_{k+1} = \left\lfloor \frac{x_k + \left\lfloor n / x_k \right\rfloor}{2} \right\rfloor, \quad k = 0, 1, 2, \dots</math>
Proof of convergence
1. Positivity: All terms are positive integers: <math>x_k > 0</math> for all <math>k</math>.
2. Monotonicity:
- If <math>x_k > \sqrt{n}</math>, then <math>\lfloor n / x_k \rfloor \le n / x_k</math>;
- so <math>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</math>.
- Hence the sequence decreases.
- If <math>x_k < \sqrt{n}</math>, then <math>\lfloor n / x_k \rfloor \ge n / x_k - 1</math>;
- so <math>x_{k+1} \ge \frac{x_k + n / x_k - 1}{2} > x_k - 1</math>.
- Hence the sequence increases or stays the same.
3. Boundedness: The sequence is bounded below by 1 and above by <math>x_0</math>, so it is bounded.
4. Stabilization / Oscillation: A bounded monotone integer sequence either stabilizes or oscillates between two consecutive integers:
- <math>x_{k+1} = x_k</math> or <math>x_{k+1} = x_k \pm 1</math>.
5. Integer "Fixed-point" Condition: At stabilization or oscillation:
- <math>x_{k+1} = \lfloor (x_k + \lfloor n / x_k \rfloor)/2 \rfloor</math>.
- This ensures that the sequence is either at <math>\lfloor \sqrt n \rfloor</math> or oscillating between the two nearest integers around <math>\sqrt n</math>.
6. Conclusion: The sequence eventually stabilizes at <math>\lfloor \sqrt n \rfloor</math> or oscillates between <math>\lfloor \sqrt n \rfloor</math> and <math>\lceil \sqrt n \rceil</math>.
Remark:
- <math>\lfloor \sqrt n \rfloor</math> is a strict fixed point unless <math>n + 1</math> is a perfect square.
- If <math>n + 1</math> is a perfect square, the sequence oscillates between <math>\lfloor \sqrt n \rfloor</math> and <math>\lceil \sqrt n \rceil</math>.
Example implementation
- <syntaxhighlight lang="python">
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
</syntaxhighlight>
Numerical examples
The call isqrt(2000000) converges to <math>1414</math> in 14 passes through while:
- <math>\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}</math>.
One iteration is gained by setting x0 to <math>\lfloor n/2 \rfloor</math> 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.<ref name="Newton2" group="note">Newton's method can be given as follows
(with the initial guess set to <math>s</math>):
<syntaxhighlight lang="python">
def isqrt(s: int) -> int:
"""isqrt via Newton/Heron iteration."""
L, R = 1, s
while L < R:
R = L + ((R - L) // 2)
L = s // R
return R
</syntaxhighlight> Computations
- <math>\operatorname{isqrt}(0) = 0</math>, <math>\operatorname{isqrt}(1) = 1</math>.
- The computation of <math>\operatorname{isqrt}(2000000)</math> consists in <math>13 \; [L,R]</math> steps:
- <math>\begin{aligned} &[1,2000000] \to [2,1000000] \to [3,500001] \\& \to [7,250002] \to [15,125004] \to [31,62509] \\& \to [63,31270] \to [127,15666] \to [253,7896] \\& \to [490,4074] \to [876,2282] \to [1266,1579] \\& \to [1406,1422] \to [1414,1414] \end{aligned}</math>
- One sees that the performance gain of Newton's method over binary search is due to the fact that <math>\lfloor \sqrt s \rfloor</math> is approached simultaneously from Left and Right, whereas binary search adjusts only one side at each iteration.</ref> 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 <math display="block">x_0 = 2^{\lfloor (\log_2 n) / 2 \rfloor + 1},</math> which is the least power of two bigger than <math>\sqrt n</math>. In the example of the integer square root of Template:Math, <math>\lfloor \log_2 n \rfloor = 20</math>, <math>x_0 = 2^{11} = 2048</math>, and the resulting sequence is
<math display="block">2048 \rightarrow 1512 \rightarrow 1417 \rightarrow 1414 \rightarrow 1414.</math> In this case only four iteration steps are needed. This corresponds to the call isqrt(2000000, 2048).
Digit-by-digit algorithm
The traditional pen-and-paper algorithm for computing the square root <math>\sqrt{n}</math> is based on working from higher digit places to lower, and as each new digit pick the largest that will still yield a square <math>\leq n</math>. 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:
- <syntaxhighlight lang="python">
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 = integer_sqrt(n >> 2) << 1
large_cand = small_cand + 1
if large_cand * large_cand > n:
return small_cand
else:
return large_cand
</syntaxhighlight> Equivalent non-recursive program:<ref name="Guy">Template:Cite web</ref><ref group="note">The algorithm is explained in Square_root_algorithms#Binary numeral system (base 2)</ref>
- <syntaxhighlight lang="python">
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
# "one" starts at the highest power of four <= x
one = 1
while one <= op:
one <<= 2
one >>= 2
while one != 0:
if op >= res + one:
op -= res + one
res += 2 * one
res //= 2
one //= 4
return res
</syntaxhighlight>
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.<ref name="Guy"/>
Karatsuba 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).<ref name="Zimmermann">Template:Cite web</ref> 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.<ref name="Zimmermann"/>
- <math>\text{Algorithm } \text{SqrtRem}(n = a_3 b^3 + a_2 b^2 + a_1 b + a_0)</math>
- <math>\text{Input: } 0 \le a_i < b, \text{ with } a_3 \ge b/4</math>
- <math>\text{Output: } (s,r) \text{ such that } s^2 \le n = s^2 + r < (s + 1)^2</math>
- <math>\qquad (s', r') \gets \text{SqrtRem}(a_3 b + a_2)</math>
- <math>\qquad (q, u) \gets \text{DivRem}(r' b + a_1, 2 s')</math>
- <math>\qquad s \gets s' b + q</math>
- <math>\qquad r \gets u b + a_0 - q^2</math>
- <math>\qquad \text{if } r < 0 \text{ then }</math>
- <math>\qquad \qquad r \gets r + 2s - 1</math>
- <math>\qquad \qquad s \gets s - 1</math>
- <math>\qquad \text{return } (s,r)</math>
Because only one recursive call is made per level, the total complexity remains <math>O(n)</math> 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 | <math>T(n)=3T(n/2)+O(n)</math> | <math>T(n)=T(n/2)+O(n)</math> |
| Asymptotic complexity | <math>O(n^{\log_2 3})\!\approx\!O(n^{1.585})</math> | <math>O(n)</math> |
| 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 Template:Ill and Karatsuba multiplication. It was first analyzed formally by Paul Zimmermann (1999).<ref name="Zimmermann"/> Earlier practical work includes Martin Guy (1985),<ref name="Guy"/> and recursive versions appear in Donald Knuth (1998).<ref>Template:Cite book</ref> Modern GMP and MPIR libraries implement similar recursive techniques.
Implementation in Python
The Python program below implements Zimmermann’s algorithm. Given an integer <math>n \ge 0</math>, SqrtRem computes simultaneously its integer square root <math>s = \lfloor \sqrt{n} \rfloor</math> and the corresponding remainder <math>r = n - s^2</math>. The choice of isqrt() is ad libitum.<ref name="Newton2" group="note"/>
- <syntaxhighlight lang="python">
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
</syntaxhighlight>
Example usage
- <syntaxhighlight lang="python">
for n in [(2**32) + 5, 12345678901234567890, (1 << 1512) - 1]:
s, r = SqrtRem(n)
print(f"SqrtRem({n}) = {s}, remainder = {r}")
</syntaxhighlight>
| Computation |
|---|
SqrtRem(4294967301) = 65536, remainder = 5 |
SqrtRem(12345678901234567890) = 3513641828, remainder = 5763386306 |
SqrtRem(143665816004337822710282600285310394341474369045835074863414468709543787931907367746179403311452034095731304066234112071267510472464260955153084575408147254672957261763907982395337943906645864229014250227057207826232751957053220218983971305018634078800548055251973907806245884614087189937340865371691338441989956445051526543084039211962387469415699218979531585795574920384684004258007709014706216763392717018544247025174258411677231986785008489302218244095) = 379032737378102767370356320425415662904513187772631008578870126471203845870697482014374611530431269030880793627229265919475483409207718357286202948008100864063587640630090308972232735749901964068667724412528434753635948938919935, remainder = 758065474756205534740712640850831325809026375545262017157740252942407691741394964028749223060862538061761587254458531838950966818415436714572405896016201728127175281260180617944465471499803928137335448825056869507271897877839870 |
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);<ref>Template:Cite web</ref>BigInteger.sqrtRem(result, remainder, n); |
Unknown |
| Common Lisp | (isqrt n)<ref>Template:Cite web</ref> |
Unknown |
| Crystal | Math.isqrt(n)<ref>Template:Cite web</ref> |
1.2 |
| Java | n.sqrt()<ref>Template:Cite web</ref> (BigInteger only) |
9 |
| Julia | isqrt(n)<ref>Template:Cite web</ref> |
0.3 |
| Maple | isqrt(n)<ref>Template:Cite web</ref> |
Unknown |
| PARI/GP | sqrtint(n)<ref>Template:Cite web</ref> |
1.35a<ref>Template:Cite web</ref> (as isqrt) or before
|
| PHP | sqrt($num)<ref>Template:Cite web</ref> |
4 |
| Python | math.isqrt(n)<ref>Template:Cite web</ref> |
3.8 |
| Racket | (integer-sqrt n)<ref>Template:Cite web</ref>(integer-sqrt/remainder n) |
Unknown |
| Ruby | Integer.sqrt(n)<ref>Template:Cite web</ref> |
2.5.0 |
| Rust | n.isqrt()<ref>Template:Cite web</ref>n.checked_isqrt()<ref>Template:Cite web</ref> |
1.84.0 |
| SageMath | isqrt(n)<ref>Template:Cite web</ref> |
Unknown |
| Scheme | (exact-integer-sqrt n)<ref>Template:Cite web</ref> |
R6RS |
| Tcl | isqrt($n)<ref>Template:Cite web</ref> |
8.5 |
| Zig | std.math.sqrt(n)<ref>Template:Cite web</ref> |
Unknown |
Continued fraction of √c based on isqrt
The computation of the simple continued fraction of <math>\sqrt{c}</math> can be carried out using only integer operations, with <math>\operatorname{isqrt}(c)</math> serving as the initial term. The algorithm<ref>Template:Cite web</ref> generates continued fraction expansion in canonical form.
Let <math>a_0 = \lfloor \sqrt{c} \rfloor</math> be the integer square root of <math>c</math>.
If <math>c</math> is a perfect square, the continued fraction terminates immediately:
- <math>\sqrt{c} = [a_0].</math>
Otherwise, the continued fraction is periodic:
- <math>\sqrt{c} = [a_0; \overline{a_1, a_2, \dots, a_m}]</math>,
where the overline indicates the repeating part.
The continued fraction can be obtained by the following recurrence, which uses only integer arithmetic:
- <math>m_{0} = 0, \quad d_{0} = 1, \quad a_{0} = \lfloor \sqrt{c} \rfloor.</math>
- For <math>k \geq 0</math>,
- <math>
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. </math>
Since there are only finitely many possible triples <math>(m_{k}, d_{k}, a_{k})</math>, eventually one repeats, and from that point onward the continued fraction becomes periodic.
Implementation in Python
Template:Anchor On input <math>c</math>, a non-negative integer, the following program computes the simple continued fraction of <math>\sqrt{c}</math>. The integer square root <math>\lfloor \sqrt{c} \rfloor</math> is computed once.<ref name="Newton2" group="note"/> Only integer arithmetic is used. The program outputs <math>[a0, (a1, a2, ..., am)]</math>, where the second element is the periodic part.
- <syntaxhighlight lang="python" start="0" highlight="7">
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))
</syntaxhighlight>
Example usage
- <syntaxhighlight lang="python" start="0" highlight="4">
for c in list(range(0, 18)) + [114] + [4097280036]:
cf = continued_fraction_sqrt(c)
print(f"sqrt({c}): {cf}")
</syntaxhighlight> Output
Input (c) Output (cf) Continued fraction 0 [0] <math>\sqrt{0} = 0</math> 1 [1] <math>\sqrt{1} = 1</math> 2 [1; (2,)] <math>\sqrt{2} = [1; \overline{2}]</math> 3 [1; (1, 2)] <math>\sqrt{3} = [1; \overline{1,2}]</math> 4 [2] <math>\sqrt{4} = 2</math> 5 [2; (4,)] <math>\sqrt{5} = [2; \overline{4}]</math> 6 [2; (2, 4)] <math>\sqrt{6} = [2; \overline{2,4}]</math> 7 [2; (1, 1, 1, 4)] <math>\sqrt{7} = [2; \overline{1,1,1,4}]</math> 8 [2; (1, 4)] <math>\sqrt{8} = [2; \overline{1,4}]</math> 9 [3] <math>\sqrt{9} = 3</math> 10 [3; (6,)] <math>\sqrt{10} = [3; \overline{6}]</math> 11 [3; (3, 6)] <math>\sqrt{11} = [3; \overline{3,6}]</math> 12 [3; (2, 6)] <math>\sqrt{12} = [3; \overline{2,6}]</math> 13 [3; (1, 1, 1, 1, 6)] <math>\sqrt{13} = [3; \overline{1, 1, 1, 1, 6}]</math> 14 [3; (1, 2, 1, 6)] <math>\sqrt{14} = [3; \overline{1, 2, 1, 6)}]</math> 15 [3; (1, 6)] <math>\sqrt{15} = [3; \overline{1,6}]</math> 16 [4] <math>\sqrt{16} = 4</math> 17 [4; (8,)] <math>\sqrt{17} = [4; \overline{8}]</math> 114 [10; (1, 2, 10, 2, 1, 20)] <math>\sqrt{114} = [10; \overline{1,2,10,2,1,20}]</math><ref group="note">See the example in the article Periodic continued fraction.</ref> 4097280036 [64009; (1, 1999, 3, 4, 1, 499, 3, 1, 3, 3, 1, 124, ...
Template:Spaces..., 3, 1, 3, 499, 1, 4, 3, 1999, 1, 128018)]
period: 13,032 terms<ref group="note">The continued fraction expansion of <math>\sqrt{4097280036}</math> has a period of 13,032 terms. While Python is unable to display the entire sequence on screen due to its length, writing the output to a file completes successfully.</ref>