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58 CHAPTER 4. THE EUCLIDEAN ALGORITHM
The individual computations are not difficult, but the continual swapping of
gcd entries, as well as the rewriting of “gcd” is a bit much. In addition while
we have kept the remainder information, we have lost the quotients that we will
soon need. So we are going to introduce a more compact way of presenting this
computation, perhaps due to [Glasby].
To see how this is built, consider gcd(69, 27) = gcd(27, 69%27). The remainder
69%27 is 15, and the quotient of 69 ÷ 27 is 2. So we write
q
r 69
2
27 15
where q and r remind us which line contains the quotient and which the remainder.
Similarly, for gcd(27, 15) = gcd(15, 27%15) we write
q
r 17
1
15 12
since the quotient of 27 ÷ 15 is 1 and the remainder is 12.
The advantage of this representation is that
,
q
r 69
2
27 15
and
q
r 17
1
15 12
may be combined as
q
r 69
Adding the 12 ÷ 3 results then gives
2
27
1
15 12
.
q
r 69
2
27
1
15
1
12
4
3 0
.
Examples:
(1) Compute gcd(15, 85).
For this first example we work one division step at a time and write in
bold the numbers added at each step.
Enter the two numbers, larger one first:
q
r 85 15
from 85 ÷ 15:
from 15 ÷ 10:
from 10 ÷ 5:
q
r 85
q
r 85
q
r 85
5
15
5
15
1
10
5
15 10
1
10 5
2
5 0
The gcd is the final non-zero remainder. So gcd(15, 85) = 5.