Bsp:
3 = gcd(15, 21), p = ..., q = ...
Beweis: konstruktiv, (p, q)
egcd :: N -> N -> (Z,Z)
egcd a b =
if b == 0 then ( ... , ... )
else let (d , m) = divMod a b
(p' , q') = egcd b m
in ( ... , ... )
Beweis für Algorithms:
Johannes Waldmann
2015-12-11