The **shifting nth root algorithm** is an algorithm for extracting the

*n*th root of a positive real number which proceeds iteratively by shifting in

*n*digits of the radicand, starting with the most significant, and produces one digit of the root on each iteration, in a manner similar to long division.

## Algorithm

### Notation

Let *B* be the base of the number system you are using, and *n* be the degree of the root to be extracted. Let be the radicand processed thus far, be the root extracted thus far, and be the remainder. Let be the next digits of the radicand, and be the next digit of the root. Let be the new value of for the next iteration, be the new value of for the next iteration, and be the new value of for the next iteration. These are all integers.

### Invariants

At each iteration, the invariant will hold. The invariant will hold. Thus is the largest integer less than or equal to the *n*th root of , and is the remainder.

### Initialization

The initial values of , and should be 0. The value of for the first iteration should be the most significant aligned block of digits of the radicand. An aligned block of digits means a block of digits aligned so that the decimal point falls between blocks. For example, in 123.4 the most significant aligned block of two digits is 01, the next most significant is 23, and the third most significant is 40.

### Main loop

On each iteration we shift in digits of the radicand, so we have and we produce one digit of the root, so we have . The first invariant implies that . We want to choose so that the invariants described above hold. It turns out that there is always exactly one such choice, as will be proved below.

**Proof of existence and uniqueness of**—

By definition of a digit, , and by definition of a block of digits,

The first invariant says that:

or

So, pick the largest integer such that

Such a always exists, since and if then , but since , this is always true for . Thus, there will always be a that satisfies the first invariant

Now consider the second invariant. It says:

or

Now, if is not the largest admissible for the first invariant as described above, then is also admissible, and we have

This violates the second invariant, so to satisfy both invariants we must pick the largest allowed by the first invariant. Thus we have proven the existence and uniqueness of .

To summarize, on each iteration:

- Let be the next aligned block of digits from the radicand
- Let
- Let be the largest such that
- Let
- Let

Now, note that , so the condition

is equivalent to

and

is equivalent to

Thus, we do not actually need , and since and , or , or , so by using instead of we save time and space by a factor of 1/. Also, the we subtract in the new test cancels the one in , so now the highest power of we have to evaluate is rather than .

### Summary

- Initialize and to 0.
- Repeat until desired precision is obtained:
- Let be the next aligned block of digits from the radicand.
- Let be the largest such that
- Let .
- Let
- Assign and

- is the largest integer such that , and , where is the number of digits of the radicand after the decimal point that have been consumed (a negative number if the algorithm has not reached the decimal point yet).

## Paper-and-pencil *n*th roots

As noted above, this algorithm is similar to long division, and it lends itself to the same notation:

1. 4 4 2 2 4 —————————————————————— _ 3/ 3.000 000 000 000 000 \/ 1 = 3(10×0)^{2}×1 +3(10×0)×1^{2}+1^{3}— 2 000 1 744 = 3(10×1)^{2}×4 +3(10×1)×4^{2}+4^{3}————— 256 000 241 984 = 3(10×14)^{2}×4 +3(10×14)×4^{2}+4^{3}——————— 14 016 000 12 458 888 = 3(10×144)^{2}×2 +3(10×144)×2^{2}+2^{3}—————————— 1 557 112 000 1 247 791 448 = 3(10×1442)^{2}×2 +3(10×1442)×2^{2}+2^{3}———���————————— 309 320 552 000 249 599 823 424 = 3(10×14422)^{2}×4 +3(10×14422)×4^{2}+4^{3}���—————————————— 59 720 728 576

Note that after the first iteration or two the leading term dominates the , so we can get an often correct first guess at by dividing by .

## Performance

On each iteration, the most time-consuming task is to select . We know that there are possible values, so we can find using comparisons. Each comparison will require evaluating . In the *k*th iteration, has digits, and the polynomial can be evaluated with multiplications of up to digits and additions of up to digits, once we know the powers of and up through for and for . has a restricted range, so we can get the powers of in constant time. We can get the powers of with multiplications of up to digits. Assuming -digit multiplication takes time and addition takes time , we take time
for each comparison, or time to pick . The remainder of the algorithm is addition and subtraction that takes time , so each iteration takes . For all digits, we need time .

The only internal storage needed is , which is digits on the *k*th iteration. That this algorithm does not have bounded memory usage puts an upper bound on the number of digits which can be computed mentally, unlike the more elementary algorithms of arithmetic. Unfortunately, any bounded memory state machine with periodic inputs can only produce periodic outputs, so there are no such algorithms which can compute irrational numbers from rational ones, and thus no bounded memory root extraction algorithms.

Note that increasing the base increases the time needed to pick by a factor of , but decreases the number of digits needed to achieve a given precision by the same factor, and since the algorithm is cubic time in the number of digits, increasing the base gives an overall speedup of . When the base is larger than the radicand, the algorithm degenerates to binary search, so it follows that this algorithm is not useful for computing roots with a computer, as it is always outperformed by much simpler binary search, and has the same memory complexity.

## Examples

### Square root of 2 in binary

1. 0 1 1 0 1 ------------------ _ / 10.00 00 00 00 00 1 \/ 1 + 1 ----- ---- 1 00 100 0 + 0 -------- ----- 1 00 00 1001 10 01 + 1 ----------- ------ 1 11 00 10101 1 01 01 + 1 ---------- ------- 1 11 00 101100 0 + 0 ---------- -------- 1 11 00 00 1011001 1 01 10 01 1 ---------- 1 01 11 remainder

### Square root of 3

1. 7 3 2 0 5 ---------------------- _ / 3.00 00 00 00 00 \/ 1 = 20×0×1+1^2 - 2 00 1 89 = 20×1×7+7^2 (27 x 7) ---- 11 00 10 29 = 20×17×3+3^2 (343 x 3) ----- 71 00 69 24 = 20×173×2+2^2 (3462 x 2) ----- 1 76 00 0 = 20×1732×0+0^2 (34640 x 0) ------- 1 76 00 00 1 73 20 25 = 20×17320×5+5^2 (346405 x 5) ---------- 2 79 75

### Cube root of 5

1. 7 0 9 9 7 ---------------------- _ 3/ 5. 000 000 000 000 000 \/ 1 = 300×(0^2)×1+30×0×(1^2)+1^3 - 4 000 3 913 = 300×(1^2)×7+30×1×(7^2)+7^3 ----- 87 000 0 = 300×(17^2×0+30×17×(0^2)+0^3 ------- 87 000 000 78 443 829 = 300×(170^2)×9+30×170×(9^2)+9^3 ---------- 8 556 171 000 7 889 992 299 = 300×(1709^2)×9+30×1709×(9^2)+9^3 ------------- 666 178 701 000 614 014 317 973 = 300×(17099^2)×7+30×17099×(7^2)+7^3 --------------- 52 164 383 027

### Fourth root of 7

1. 6 2 6 5 7 --------------------------- _ 4/ 7.0000 0000 0000 0000 0000 \/ 1 = 4000×(0^3)×1+600×(0^2)×(1^2)+40×0×(1^3)+1^4 - 6 0000 5 5536 = 4000×(1^3)×6+600×(1^2)×(6^2)+40×1×(6^3)+6^4 ------ 4464 0000 3338 7536 = 4000×(16^3)×2+600×(16^2)×(2^2)+40×16×(2^3)+2^4 --------- 1125 2464 0000 1026 0494 3376 = 4000×(162^3)×6+600×(162^2)×(6^2)+40×162×(6^3)+6^4 -------------- 99 1969 6624 0000 86 0185 1379 0625 = 4000×(1626^3)×5+600×(1626^2)×(5^2)+ ----------------- 40×1626×(5^3)+5^4 13 1784 5244 9375 0000 12 0489 2414 6927 3201 = 4000×(16265^3)×7+600×(16265^2)×(7^2)+ ---------------------- 40×16265×(7^3)+7^4 1 1295 2830 2447 6799

## See also

## External links

- Why the square root algorithm works "Home School Math". Also related pages giving examples of the long-division-like pencil and paper method for square roots.
- Reflections on The Square Root of Two "Medium". With an example of a C++ implementation.