ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Tue, 05 Nov 2013 17:46:17 +0100Three-Pass Protocolhttps://ask.sagemath.org/question/10701/three-pass-protocol/I am currently trying to make an example for the three pass protocol.
Maybe someone can tell me my mistake, because it doesn't work out.
There is no error, but the numbers don't fit.
#Prime p = 8885569519.
#Let a = 7 and b = 17.
#Alice knows K = 263785119.
#Over the Ring mod p-1
R = IntegerModRing(8885569518)
K = 263785120
a = 11
b = 17
#Alice calculates K^a:
Ka = R(K)^a
#Bob calculates (K^a)^b:
Kab = R(Ka)^b
#Alice recives Kab and calculates((K^a)^b)^a^(-1).
#first a^(-1)
y = 1/R(a)
#then y*a % p-1 = 1 (p-1 = 8885569518)
Kaby = R(Kab)^y
#Kaby should be the same as K^b:
Kb = R(K)^b
#but it isn't
#The inverse of b:
z = 1/R(b)
KK = R(Kaby)^z
`K^a^b^a^(-1)` should be the same as `K^b`, but it doesn't work out. Does someone see my mistake?
Thank you and best,
Luca
00Luca00Tue, 05 Nov 2013 17:46:17 +0100https://ask.sagemath.org/question/10701/