# The Extended Euclid Algorithm

Richard Chang, CSEE Dept, University of Maryland Baltimore County
Updated: March 3, 2009

Consider the following theorem in number theory:

Theorem. For all positive integers a and b there exist integers s and t such that

as + bt = gcd(a, b).

The proof of this theorem can be found in an exercise in the Rosen textbook. However, the proof merely shows that s and t exist. It doesn't show how to find s and t. The Extended Euclid algorithm can be used to find s and t.

Finding s and t is especially useful when we want to compute multiplicative inverses. Suppose that gcd(a, n) = 1. (That is, a and n are relatively prime.) We have seen that in this situation a has a multiplicative inverse modulo n. That is, there exists an integer, which we call a-1, such that

aa-1 ≡ 1    (mod n).
Using the theorem above, we can find s and t so that
as + nt = 1.
Since gcd(a, n) = 1. Now, notice that
as = 1 - nt.
Now, nt ≡ 0 modulo n, so
as ≡ 1    (mod n).
Therefore, sa-1 modulo n and we have found the multiplicative inverse of a.

### The Algorithm

Here's the pseudo-code for the Extended Euclid algorithm: ExtEuclid (a,b) { // returns a triple (d,s,t) such that d = gcd(a,b) and // d == a*s + b*t if (b == 0) return (a,1,0) ; (d1, s1, t1) = ExtEuclid(b,a%b) ; d = d1 ; s = t1 ; t = s1 - (a div b) * t1 ; // note: div = integer division return (d,s,t) ; } Note that when the triple (d,s,t) is returned and assigned to the triple (d1,s1,t1) then d is assigned to d1, s is assigned to s1 and t is assigned to t1. Also, we use div to indicate integer division. For example, 5 div 2 evaluates to 2.

Now, let's do an example. We let a = 99 and b = 78 and call ExtEuclid(99,78). We use the following table to keep track of recursive calls to ExtEuclid.

a b a div b d s t
199781
2
3
4
5
6

Since 78 is not 0, we call ExtEuclid() recursively with parameters 78 and 21 because 99 % 78 is 21. Our table becomes:

a b a div b d s t
199781
278213
3
4
5
6

The recursive calls to ExtEuclid continues until we have b == 0. Then ExtEuclid returns the triple (3,1,0) and our table looks like:

a b a div b d s t
199781
278213
321151
415 62
5 6 32
6 3 0-310

The 5th call to ExtEuclid now receives the return values from the 6th call and has d1 = 3, s1 = 1 and t1 = 0. Still within the 5th call, we calculate that d = d1 = 3, s = t1 = 0 and

t = s1 - (a div b) * t1 = 1 - 2 * 0 = 1 So the table becomes:

a b a div b d s t
199781
278213
321151
415 62
5 6 32301
6 3 0-310

The 4th receives the values d1 = 3, s1 = 0 and t1 = 1 from the 5th call, then computes s = t1 = 1 and

t = s1 - (a div b) * t1 = 0 - 2 * 1 = -2 Note that the value of d will remain the same. Our table becomes:

a b a div b d s t
199781
278213
321151
415 6231-2
5 6 32301
6 3 0-310

In the 3rd call, s = -2 and

t = s1 - (a div b) * t1 = 1 - 1 * (-2) = 3

a b a div b d s t
199781
278213
3211513-23
415 6231-2
5 6 32301
6 3 0-310

In the 2nd call, s = 3 and

t = s1 - (a div b) * t1 = -2 - 3 * (3) = -11

a b a div b d s t
199781
27821333-11
3211513-23
415 6231-2
5 6 32301
6 3 0-310

Finally, in the 1st call, s = -11 and

t = s1 - (a div b) * t1 = 3 - 1 * (-11) = 14

a b a div b d s t
1997813-1114
27821333-11
3211513-23
415 6231-2
5 6 32301
6 3 0-310

Now, we can verify that 99 ⋅ (-11) + 78 ⋅ 14 = 3. So, we've found the s and t such that a s + b t = gcd(a,b).

### An Example for Multiplicative Inverses

Suppose that we want to find the multiplicative inverse of 60 modulo 13. We know this inverse exists because gcd(60,13) = 1. Running ExtEuclid(60,13) will give us this table. (We won't produce the table step by step, but you should confirm the calculations yourself.)

a b a div b d s t
16013415-23
213 811-35
3 8 5112-3
4 5 311-12
5 3 2111-1
6 2 12101
7 1 0-110

We can check the final return value is correct: 60 ⋅ 5 + 13 ⋅ (-23) = 300 − 299 = 1. This also shows that 5 is the multiplicative inverse of 60, because (60 * 5) % 13 = 300 % 13 = 1. to Spring 2009 CMSC 203 Homepage