# Evaluating Jacobi's symbol using Eisenstein's algorithm

The algorithm is described as follow,

Eisenstein proposed the following algorithm for computing the Jacobi symbol $(a|n)$ where $a$ and $n$ are odd numbers. Write $n = aq + er$ where $q = \lfloor n/a \rfloor$ if this is even, or $q = \lfloor n/a \rfloor + 1$ otherwise, and $e = \pm 1$ accordingly. Then $r$ is odd, and you can use the quadratic reciprocity law and the formula for $(-1|r)$ to continue by replacing the pair $n,a$ with the pair $a, r$.

I understand the Euler's algorithm (division), and got it to work correctly. Unfortunately, I have no idea how this algorithm works. Furthermore, what's the end condition of this algorithm? My guess was $r \neq 1$ is the end condition, but the result was completely wrong. An example to illustrate this algorithm would be greatly appreciated.

Edit This is my solution, written in C#

public int eisensteinAlgo( int a, int n ) {
int parity = 1;
int q = 0;
int epsilon;
int r = 0;
while ( ( n % a ) == 0 ) {
q = n / a;
if ( ( q % 2 ) == 1 ) {
q += 1;
epsilon = -1;
}
else {
epsilon = 1;
}
r = n - ( a * q );
r *= epsilon;
parity *= ( int )Math.Pow( -1.0, ( r - 1 ) / 2 );
n = a;
a = r;
}
return parity;
}


Thank you,

-
Notes on your code: a. It might be better to use (q & 1) == 1 instead of ( q % 2 ) == 1 b. n - ( a * q ) is just n % a all over again; why are you computing it again? c. Exponentiation is slow; figure out a conditional to replace your computation for parity (which I'll leave as an exercise). – J. M. Apr 7 '11 at 5:13
@J.M.: Many thanks for your comments. However, my rule of thumb is to get the right code first, then optimize it later. Because I think writing the right code in the first place is often much harder than optimize it. – Chan Apr 7 '11 at 5:29
"get the right code first, then optimize it later." - of course. :) – J. M. Apr 7 '11 at 5:30
Thanks for the agreement ;) – Chan Apr 7 '11 at 5:37

The reduction step is simply to apply quadratic reciprocity in order to rewrite $(a|n)$ to $\pm (n|a)\:,\:$ then reduce $n$ modulo $a\:$ in a way that the remainder stays odd (by forcing the quotient $q\:$ to be even). The algorithm terminates when $a$ divides $n\:,$ with the result being either the accumulated sign (if $a=1)\:,$ or else $\:0\:$ (if $a>1)\:.\:$ The choice of odd remainder eliminates the need to pull out factors of $2\:.$ But one pays a price for this algorithmic simplification since it makes the algorithm less efficient.
Sorry for late reply. I was trying to solve other problems. First of all, thanks a lot for your answer. But I don't understand how the result were calculated. Do we calculate $(-1|r)$ each time using the formula $(-1)^{\frac{n-1}{2}}$ and then multiply it with the result? – Chan Apr 7 '11 at 4:03
Yes, you can keep a $\:\pm 1\:$ sign "accumulator" multiplying the symbol during the calculation. If you're stuck then show your work so we can see precisely where you're stuck. – Bill Dubuque Apr 7 '11 at 4:19