Well, you can take it this way:
If then it must be that both and
Also, note that the XOR function is its own inverse. That is,
Moreover, if then you can left-multiply by as follows:
(associativity of XOR)
.
Hence, you must have
its an extremely long cryptography equation for encoding messages, i've been working on it for the last two hours, and everytime i come to a conclusion close to this:
(C⊞K)⊕(V⊞K) = 0
i know both C and V, now i need to know K, what can i do? i don't really know how to interact with the XOR and the addition mod togather
ok how about this:
i didnt really think enough about the 0, and i see where your going at, i should have mentioned earlier that this is the equation:
P = (C⊟K )⊕(V⊟K )
⊟ : Modular Subtraction
⊕ : XOR
the right hand side was quite long but i can simplify it to a 64 bit block P, now i need to tell K (which is the key)
i am not really certain the the distributive property works here, and that is precisely the problem, i have nothing to work on here ... how about this:
given these two equations, do you think there is a different approach to solve for X,Y:
X=(V⊟Y)⊕P
X=(D⊟Y)⊕P
again, all data here are 64 bit data, ⊕ is bitwise XOR, and ⊟ is subtraction mod 2^64
Interesting discussion on this thread, and I certainly don't claim to be an expert on any of this, but I see a problem that I'll explain by example.
Consider 6-bit data, where our operations are mod 2^6.
Let C = 101010
Let V = 011110
Now consider K1 = 001000 and K2 = 000010
So
A1 = C - K1 = 100010
B1 = V - K1 = 010110
P1 = A1 XOR B1 = 110100
A2 = C - K2 = 101000
B2 = V - K2 = 011100
P2 = A2 XOR B2 = 110100
So as you can see two different values of K yielded the same P. From this I am led to believe that we cannot determine K in general without guesswork.
Didn't see this until after I made my post. Very flattered but I think guru is too strong of a word in my case!