Séance no 1 : Autour de l`algorithme d`Euclide 1 Rappels sur l

Eric.Wegrzynowski (a) univ-lille1.fr
a= 33 b= 18
33 −18 = 15
18 −15 = 3
15 −3 = 12
12 −3=9
9−3=6
6−3=3
pgcd(33,18) = 3.
33 = 18 ×1 + 15
18 = 15 ×1+3
15 = 3 ×5+0
a= 33 b= 18 q
irem

a b
d= pgcd(a, b)
a1=a b1=b
b16= 0
r a1b1
a1b1
b1r
a1
i q r a1b1
pgcd(33,18)
a1a2r
pgcd(a,0) = a,
a≥0b > 0b a (mod b)
pgcd(a, b) = pgcd(b, a (mod b)).
b

euclide1 (a,b) := {
local a1, b1, r;
a1 := a ;
b1 := b ;
tantque b1 <> 0 faire
r := irem(a1,b1) ;
a1 := b1 ;
b1 := r ;
ftantque ;
retourne a1 ;
}
pgcd(33,18)
pgcd(33,18) = pgcd(18,15)
= pgcd(15,3)
= pgcd(3,0)
= 3
a b
d= pgcd(a, b)
b= 0
a
pgcd(b, a (mod b))
euclide2

cout(a, b) pgcd(a, b)
N×N N
a= 33 b= 18
cout(33,18) = 3.
cout
(a, b)a≥b
(a, b)a≥b
F0= 1
F1= 1
n≥0
Fn+1 =Fn+1 +Fn
(a, b)a≥b
b
(a, b)a≤Fn+1 b≤Fn
cout(a, b)≤n
Fnn

F5=F4+F3
= (F3+F2)+(F2+F1)
= ((F2+F1)+(F1+F0)) + ((F1+F0) + F1) = (((F1+F0) + F1)+(F1+F0)) + ((F1+F0) + F1)
= (((1 + 1) + 1) + (1 + 1)) + ((1 + 1) + 1)
= 8
F5F3F2
Fn
Fn+1
Fn
evalf
rsolve Fnn
Fn
Fn∼5 + √5
10 Φn,
Φ = (1+√5)
2
a≥b
b