Re: Cryptographie asymétrique

Stefan Tillich a écrit :
remy schrieb:
Stefan Tillich a écrit :

remy schrieb:

bonjour

alice
(a+b)=x
alice -> x bob

Select a = 2, b = 3
x = 2+3 = 5

bob
(x+c)*d+f=y
bob ->y alice

Select c = 5, d = 7, f = 11
y = (5+5)*7+11 = 81

alice
y%a+cle=z
alice -> z a bob

Select key: cle = 13
z = (81 mod 2) + 13 = 14

bob z%d-f=cle

cle = (14 mod 7) - 11 = -11

You might be interested to know that
((b*d) mod a) mod d
is not necessarily 0.

Also
(f mod a) - f
is not generally 0.

algo dans le dommaine public

I guess the public domain is better off without it.

Stefan


alice
(a+b)

bob
(((a+b)+c)*d+f)

alice
a*d+b*d+c*d+f %a=b*d+c*d+f ---> a >(b+c)*d+f

b*d+c*d+f+clee


bob
(b*d+c*d+f+cle) %d=f+cle=x ---> d >f+cle

x-f=cle

Eve:

Knows x, y, z
x = a+b
y = (x+c)*d + f = (a+b+c)*d + f = ad + bd + cd + f
z = y%a + cle = (ad + bd + cd + f)%a + cle = bd + cd + f + cle (a >
(b+c)*d + f)

Does
k = y - z = (ad + bd + cd + f) - (bd + cd + f + cle) = ad - cle
m = y % x

merde
= (a+b)*d + cd + f) % (a+b) = cd + f (because a > (b+c)*d +
f => (a+b) > (b+c)*d + f + b > cd + f)
n = z - m = (bd + cd + f + cle) - (cd + f) = bd + cle
p = k + n = (ad - cle) + (bd + cle) = ad + bd = (a+b)*d
=> p / x = ((a+b)*d) / (a+b) = d
=> m % d = (cd + f) % d = f (because d > f + cle => d > f)
=> z % d - f = (bd + cd + f + cle) % d - f = (f + cle) - f = cle
(because d > f + cle)

Stefan

ok
.

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