# on arithmetic derivations of square roots

**R – Xi'an's Og**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

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**An** intriguing question made a short-lived appearance on the CodeGolf section of Stack Exchange, before being removed, namely the (most concise possible) coding of an arithmetic derivation of the square root of an integer, S, with a 30 digit precision and using only arithmetic operators. I was not aware of the myriad of solutions available, as demonstrated on the dedicated WIkipedia page. And ended playing with three of them during a sleepless pre-election night!

The first solution for finding √S is based on a continued fraction representation of the root,

with a²≤S and r=S-a². It is straightforward to code-golf:

while((r<-S-T*T)^2>1e-9)T=(F<-2*T+r/(2*T+F))-T;F

but I found it impossible to reach the 30 digit precision (even when decreasing the error bound from 10⁻⁹). Given the strict rules of the game, this would have been considered to be a failure.

The second solution is Goldschmidt’s algorithm

b=S T=1/sum((1:S)^2~~1e-9){ b=b*T[1]^2 T=c((3-b)/2,T)} S*prod(T)~~

which is longer for code-golfing but produces both √S and 1/√S (and is faster than the Babylonian method and Newton-Raphson). Again no luck with high precision and almost surely unacceptable for the game.

The third solution is the most interesting [imho] as it mimicks long division, working two digits at a time (and connection with Napier’s bones)

`~`=length D=~S S=c(S,0*(1:30)) p=d=0 a=1:9 while(~S){ F=c(F,x<-sum(a*(20*p+a)<=(g<-100*d+10*S[1]+S[2]))) d=g-x*(20*p+x) p=x+10*p S=S[-1:-2]} sum(10^{1+D/2-1:~F}*F)

plus providing an arbitrary number of digits with no error. This code requires S to be entered as a sequence of digits (with a possible extra top digit 0 to make the integer D even). Returning one digit at a time, it would further have satisfied the constraints of the question (if in a poorly condensed manner).

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**R – Xi'an's Og**.

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