rc4 key recovery (2 messages, same partial key) - stream

Question

I have two messages that are encrypted with the same partial key. For example:
C1 = RC4(M1, "(VARIABLE_DATA)XXXXYYYY")
C2 = RC4(M2, "(VARIABLE_DATA)XXXXYYYY")
Is it possible with RC4, if C1 and C2 are known to atleast recover the partial key of "XXXXYYYY" since that never changes?

Answer 1

I think there's some confusion about your question. The way stream ciphers work is by generating a keystream that gets (usually) exclusive-or'ed with the message. You are correct that if you use the same key and IV, and thus the same keystream, that this leaks information about the messages.
Here, K is the key stream generated by RC4:
C1 = K ^ M1
C2 = K ^ M2
And by rearranging:
C1 ^ C2 = (K ^ M1) ^ (K ^ M2)
the keystream cancels out here, and you're left with
C1 ^ C2 = M1 ^ M2
Since the attacker knows the two ciphertext values, he can compute the difference of the two messages. If the attacker knows one of the inputs (perhaps a fixed header), he can compute the second message.
M2 = (C1 ^ C2) ^ M1
There's also some statistical tests using cribs, if the messages are natural language.
To answer your question, RC4 should generate an entirely different keystream under related keys, so this attack won't work. There are other attacks against the key scheduling algorithm though, and plenty of reasons to prefer an alternative to RC4.
If you're asking about recovering the initial key from the keystream, there are a few

Answer 2

In general keys of uncompromised encryption techniques are only recoverable using brute force, which in turn requires some method of verifying that the decryption was successful.

Answer 3

Trying to solve this exact question i've stumbled across several threads with almost the same question and absolutely the same answer... which didn't work for me. BUT! The answers were absolutely correct and #mfanto gives very accurate description of what needs to be done (though the brackets make no sense as u can see in my code)!
Here is my C code which worked for me:
#include <stdio.h>
#include <string.h>
int main(int argc, char *argv[])
{
if(argc != 4)
{
printf("usage: ./rc4_pa cipher1 cipher2 message1\n");
return 1;
}
char *c1 = argv[1];
char *c2 = argv[2];
char *m1 = argv[3];
int len_c1 = strlen(c1);
int len_m1 = strlen(m1);
char m2[len_m1 + 1];
m2[len_m1] = '\0';
for(int i = 0; i < len_m1; i++)
{
m2[i] = c1[i] ^ m1[i] ^ c2[i];
}
printf("decrypted: %s\n", m2);
}
Why my code didn't work out of the box? I've got my cipher text from a web server and usually some chars of a ciphertext are not really printable. The only way to pass them further is to encode once more. In my case that was base64.
save the code to rc4_pa.c and make rc4_pa than use it like this
$ ./rc4_pa $(echo L1Gd8F5g | base64 -d) $(echo MFuD8FVg | base64 -d) hello
hope someone else might find it helpful.