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Posted 29 Aug 2001

Extended Euclidean Alogo

, 29 Aug 2001
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To detemine multiplicative inverse.


The statement of the Extended Euclidean Algorithm Theorem. (Extended Euclidean Algorithm.) Let m and n be positive integers. Define:

a[0] = m,  a[1] = n,

q[k] = Floor (a[k-1]/a[k])  for  k > 0,

a[k] = a[k-2] - a[k-1] q[k-1]  for  k > 1,

x[0] = 1,  x[1] = 0,  y[0] = 0,  y[1] = 1,

x[k] = x[k-2] - q[k-1] x[k-1]  for  k > 1,

y[k] = y[k-2] - q[k-1] y[k-1]  for  k > 1.

If a[p] is the last nonzero a[k], then:

a[p] = (m,n) = x[p] m + y[p] n.

(The Floor of a real number x is the greatest integer less than or equal to x. So Floor(1.5) = 1, Floor(-2.1) = -3, and Floor(0) = 0.)

Don't be alarmed by all the subscripts! Believe it or not, the algorithm isn't bad for hand computation if you're careful to arrange the numbers in a table.

Before I give the proof, here's a quick overview. The qs compute the greatest common divisor by successive division, i.e., using the standard Euclidean algorithm. Nothing new there! The xs and ys are for bookkeeping; they eventually yield the coefficients in the linear combination.

Proof. I'll take for granted the statement and proof of the standard Euclidean algorithm.

Ignoring the xs and ys, the computation of the as and qs is the same as the computation of the standard Euclidean algorithm: the as are the remainders, and the qs are the quotients. Thus, the algorithm terminates, and when it does, the last nonzero a is the greatest common divisor (m,n).

I claim that:

x[k]a[0] + y[k]a[1] = a[k]  for  k > 0.

I'll prove this using induction.

For k = 1, I have:

x[1]a[0] + y[1]a[1] = 0[1] a[0] + 1[1] a[1] = a[1].

Now, take k > 1, and assume that the result is true for all indices less than or equal to k. I want to prove the result for k + 1. Compute:

x[k+1] a[0] + y[k+1] a[1] =

   (x[k-1] - q[k] x[k]) a[0] + (y[k-1] - q[k] y[k]) a[1] =

   (x[k-1] a[0] + y[k-1] a[1]) - q[k] (x[k] a[0] + y[k] a[1]) =

   a[k-1] - q[k] a[k] =


The first equality comes from the definition of the xs and the ys. I used the induction hypothesis to get the third equality. The last equality uses the definition of the as.

The result is true for k + 1, so it's true for all k > 0, by induction. In particular, if a[p] = (m,n) is the last nonzero a, then:

x[p] a[0] + y[p] a[1] = a[p] = (m,n).

The Extended Euclidean Algorithm

If m and n are integers (not both 0), the greatest common divisor (m,n) of m and n is the largest integer which divides both m and n. The Euclidean algorithm uses repeated division to compute the greatest common divisor.

The greatest common divisor of m and n can be expressed as an integer linear combination of m and n. That is, there are integers c and d such that,

(m,n) = c m + d n.

There are infinitely many pairs of numbers c, d that work; sometimes, you can find a pair by trial and error. For example, (10,7) = 1, and by juggling numbers in my head, I see that:

1 = (-2)(10) + (3)(7).

On the other hand, (367,221) = 1, but it's not likely that you'd figure out that:

1 = (-56)(367) + (93)(221)

by juggling numbers in your head!

The Extended Euclidean Algorithm computes the greatest common divisor (m,n) of integers m and n, as well as numbers c and d such that:

(m,n) = c m + d n.


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Junaid Majeed
United States United States
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