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, For example,

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 do not run forever. They are nevertheless capable of computing \sqrt y up to any desired accuracy k. Choose any k and compute. For example (setting y = 2): Compare the results with 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 an infinite number of times, increasing y by a factor 100 at each pass. Assume that in the next program the procedure is already defined and — for the sake of the argument — that all variables can hold integers of unlimited magnitude. Then will print the entire decimal representation of \sqrt y. The conclusion is that algorithms which compute isqrt are computationally equivalent to algorithms which compute sqrt.

Basic algorithms

The integer square root of a non-negative integer y can be defined as For example, because.

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

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. Numerical example For example, if one computes using binary search, one obtains the [L,R] sequence This computation takes 21 iteration steps, whereas linear search (ascending, starting from 0) needs 1,414 steps.

Algorithm using Newton's method

One way of calculating \sqrt{n} and 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 The sequence {x_k} converges quadratically to \sqrt{n} as k\to\infty.

Stopping criterion

One can prove that c=1 is the largest possible number for which the stopping criterion ensures 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, a fact which has some theoretical advantages.

Using only integer division

For computing 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 By using the fact that one can show that this will reach within a finite number of iterations. In the original version, one has for k \ge 1, and for. So in the integer version, one has and until the final solution x_s is reached. For the final solution x_s, one has and, so the stopping criterion is. However, is not necessarily a fixed point of the above iterative formula. Indeed, it can be shown that 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 and instead of converging.

Example implementation in C

Numerical example

For example, if one computes the integer square root of 2000000 using the algorithm above, one obtains the sequence 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. in C++20), one should better start at which is the least power of two bigger than \sqrt n. In the example of the integer square root of 2000000 ,, , and the resulting sequence is 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 \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: 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 for an example.

Karatsuba square root algorithm

The Karatsuba square root algorithm is a fast algorithm for big-integers of "50 to 1,000,000 digits" if Burnikel-Ziegler Karatsuba division and Karatsuba multiplication are used. An example algorithm for 64-bit unsigned integers is below. The algorithm:

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.