The equation ax + by = gcd(a,b) is particularly useful when a and b are relatively prime: x is then the multiplicative inverse of a modulo b.
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2 Computing a multiplicative inverse in a finite field 3 External links |
Consider as an example the computation of gcd(120,23) with Euclid's algorithm:
In this case, the remainder in the second-to-last line indicates that the gcd is 1; that is, 120 and 23 are coprime. Now do a little algebra on each of the above lines:
Now observe that the first line contains multiples of 120 and 23. Also, the rightmost values are in each case the remainders listed on the previous line, and the left side of the differences are the residues from two lines up. We can thus progressively calculate each successive remainder as sums of products of our two original values.
Here we rewrite the second equations in the above table:
Notice that the last line says that 1 = -9*120 + 47*23, which is what we wanted: x = -9 and y = 47.
This means that -9 is the multiplicative inverse of 120 modulo 23, because 1 = -9 * 120 (mod 23).
Here is a JavaScript implementation of the Extended Euclidean algorithm which should work in most browsers:
// Save original values.
a0 = a;
b0 = b;
// Initializations. We maintain the invariant p*a0 + q*b0 = a and r*a0 + s*b0 = b.
p = 1; q = 0;
r = 0; s = 1;
// The algorithm:
while (b != 0) {
// Show result.
alert("gcd(" + a0 + "," + b0 + ")=" + p + "*" + a0 +
The extended Euclidian algorithm can also be used to calculate the multiplicative inverse in a finite field.
Given the irreducible polynomial f(x) used to define the finite field, and the element a(x) whose inverse is desired, then a form of the algorithm suitable for determining the inverse is given by the following pseudocode:
For example, if the polynomial used to define the finite field GF(28) is f(x) = x8 + x4 + x3 + x + 1, and x6 + x4 + x + 1 = {53} in big-endian hexadecimal notation, is the element whose inverse is desired, then performing the algorithm results in the following:
Calculating a GCD (and multiplicative inverse)
120 / 23 = 5 r 5
23 / 5 = 4 r 3
5 / 3 = 1 r 2
3 / 2 = 1 r 1
2 / 1 = 2 r 0
120 / 23 = 5 r 5 => 5 = 120 - 5*23
23 / 5 = 4 r 3 => 3 = 23 - 4*5
5 / 3 = 1 r 2 => 2 = 5 - 1*3
3 / 2 = 1 r 1 => 1 = 3 - 1*2
2 / 1 = 2 r 0 => 0 = 2 - 2*1
5 = 120 - 5*23 = 1*120 - 5*23
3 = 23 - 4*5 = 1*23 - 4*(1*120 - 5*23) = -4*120 + 21*23
2 = 5 - 1*3 = (1*120 - 5*23) - 1*(-4*120 + 21*23) = 5*120 - 26*23
1 = 3 - 1*2 = (-4*120 + 21*23) - 1*(5*120 - 26*23) = -9*120 + 47*23
Javascript implementation
// This program works only for non-negative inputs.
// Get data from user and convert strings to integers.
var a = parseInt(prompt("Enter non-negative value for a",0))
var b = parseInt(prompt("Enter non-negative value for b",0))
c = a % b;
quot = Math.floor(a/b); //Javascript doesn't have an integer division operator
a = b;
b = c;
new_r = p - quot * r; new_s = q - quot * s;
p = r; q = s;
r = new_r; s = new_s;
} "+(" + q + ")*" + b0 + "=" + a)
Computing a multiplicative inverse in a finite field
Pseudocode
remainder[1] = f(x)
remainder[2] = a(x)
auxiliary[1] = 0
auxiliary[2] = 1
i = 2
do while remainder[i] <> 1
i = i + 1
remainder[i] = remainder(remainder[i-2] / remainder[i-1])
quotient[i] = quotient(remainder[i-2] / remainder[i-1])
auxiliary[i] = quotient[i] * auxiliary[i-1] + auxiliary[i-2]
inverse = auxiliary[i]
Example
i | remainder[i] | quotient[i] | auxiliary[i] |
---|---|---|---|
1 | x8 + x4 + x3 + x + 1 | 0 | |
2 | x6 + x4 + x + 1 | 1 | |
3 | x2 | x2 + 1 | x2 + 1 |
4 | x + 1 | x4 + x2 | x6 + |
5 | 1 | x + 1 | x7 + x6 + x3 + |
Thus, the inverse is x7 + x6 + x3 + x = {CA}