Math. 55
Euclid’s GCD Algorithm
May 10, 1999 6:15 pm
Prof. W. Kahan Page 1
Given two positive integers a
≥
b > 0 we seek their
Greatest Common Divisor
( GCD ), which
is the biggest integer d that divides both a and b leaving no remainder. Ordinary long division
computes a positive integer quotient q :=
a/b
and leaves a remainder r := a – q·b that satisfies
0
≤
r < b . Clearly every divisor of both a and b divides r too, and conversely every divisor of
both b and r divides a = q·b + r too; therefore GCD(a, b) = GCD(b, r) . But the pair (b, r) is
smaller
than the pair (a, b) in the sense that b
≤
a and r < b . This leads to an algorithm …
Euclid’s GCD Algorithm
Given integers a
≥
b > 0 , set r
0
:= a and r
1
:= b and perform successive long divisions getting,
for j = 1, 2, 3, …, n in turn until r
n+1
= 0 , quotients q
j
and remainders r
j
that satisfy
r
j–1
= q
j
·r
j
+ r
j+1
with 0
≤
r
j+1
< r
j
.
( Here at step j we divide r
j–1
by r
j
to get quotient q
j
and remainder r
j+1
, stopping when a
remainder r
n+1
= 0 . At that point q
n
> 1 ; can you see why?) The algorithm stops because this
decreasing sequence of n+1 positive integers, r
0
= a
≥
r
1
= b > r
2
> … > r
n–1
> r
n
> r
n+1
= 0 ,
cannot have n > b . Then GCD(a, b) = r
n
because, as explained in the first paragraph,
GCD(a, b) =: GCD(r
0
, r
1
) = GCD(r
1
, r
2
) = … = GCD(r
n–1
, r
n
) = GCD(r
n
, r
n+1
) = r
n
.
The quotients q
j
appear to play no important role in the foregoing algorithm, but appearances
can mislead. By translating the algorithm’s recurrence into matrix language we find uses for q
j
:
Set := first; then for j = 1, 2, 3, …, n in turn confirm that = , with
0
≤
r
j+1
< r
j
and r
n+1
= 0 , so = … .
Now set row := … to obtain two
integers A and B (not both positive) satisfying GCD(a, b) = r
n
= = = B·a + A·b .
We have just found that GCD(a, b) is a linear combination of a and b with integer coefficients,
thus proving the following … ( Cf. text p. 137, and p. 201 ex. 58.)
Theorem 1:
As A and B run independently through all integers the expression B·a + A·b runs
through a set of integers among which the smallest positive integer is GCD(a, b) = B·a + A·b .
Hard Exercise:
Running A and B through
all
integers is unnecessary: Theorem 1 remains true after restrictions
|A| < a and |B|
≤
b
≤
a are imposed; why? Can you prove |A| < a/GCD(a, b) and |B|
≤
b/GCD(a, b) ? See below.
There are two ways to compute A and B . The easiest is to evaluate from-left-to-right the matrix
product defining
after
all the q
j
’s have been computed; this gives rise to a recurrence:
s
n
:= 1 ; s
n–1
:= –q
n–1
; for j = n–2, n–3, …, 2, 1 in turn s
j
:= s
j+2
– q
j
·s
j+1
.
Finally A := s
1
and B := s
2
. Another way to compute them is to evaluate from-right-to-left the
matrix product defining row
simultaneously
with the computation of the q
j
’s :
r0
r1
a
b
rj
rj1+
01
1q
j
–
rj1–
rj
rn
0
01
1q
n
–
01
1q
n1–
–
01
1q
n2–
–
01
1q
2
–
01
1q
1
–
r0
r1
BA 10 01
1q
n
–
01
1q
n1–
–
01
1q
n2–
–
01
1q
2
–
01
1q
1
–
10 rn
0BA a
b
BA
BA
This document was created with FrameMaker404
