Steve,

> cycles = 597.65625ms.  Find the smallest integer constants A and B
> such that (A+B)*600 = A*601.56125 + B*597.65625.  If A and/or B are too
> large, pick different loop times.  Ideally, one of A or B should be
> 1 since this simplifies the loop logic.
>
> [Anyone got a good algorithm for finding A and B?]

The smallest integer solution to the Diophantine equation ax-by=0 (a,b>0)
is (x,y) = (b/gcd(a,b), a/gcd(a,b)). I your case,

(A+B)*600 = A*601.56125 + B*597.65625

<=>

1.56125 A = 2.34375 B

<=>

249 A = 375 B

The smallest solution is (A,B) = (375,249), since
gcd(249,375)=1. Close, but not exact, is (A,B) = (3,2), since 249 is
near 250.

(A good way to find approximate solutions with small integers is to
truncate the continued fractions expansion of a/b. In some reasonable
sense this gives the "best" possible approximation, but I guess you
don't need this here.)

 -- Martin

Martin Nilsson                           http://www.sics.se/~mn/
Swedish Institute of Computer Science    E-mail: mn@sics.se
Box 1263, S-164 28 Kista                 Fax: +46-8-751-7230
Sweden                                   Tel: +46-8-752-1574