enter image description here

A scanner encrypts a 32 bit message 'x' using a secret 32 bit key 's1' and sends the resulting cipher y= x XOR s1 to an RFID tag. The Tag decrypts the message 'x' ,and then encrypts the message 'x' using its secret 32 bit key 's2' and sends the resulting cipher z= x XOR s2 back to the Scanner.

Assuming that I know what the hexadecimal values for the ciphers y and z are during every transmission. How can I find out what the keys s1 and s2 are using this limited information, but having access to the 'y' and 'z' values from multiple transmissions?

I understand that, because there are two different keys, but that the keys remain the same during multiple transmissions, that there is maybe some way to figure out the value of the keys.

I know that if I XOR y and z I get: 'y XOR z = s1 XOR s2'. I am not sure how to use this information or if it is relevant.

Sorry for any misused terms, I am new to cryptanalysis.

  • $\begingroup$ @Squeamish Ossifrage quite similar, however the big difference is whether the keys themselves can be discerned, as opposed to forging a correct response.As for determining whether the message 'x' is randomly distributed, we can assume that x is a 32 bit binary message, where each bit of the message is either a 0 or a 1. Does this information change anything? We know the messages Y and Z from eavesdropping, and Y XOR Z = S1 XOR S2. so if bit 1 of Y is a 1 and bit 1 of Z is a 1 then one of bit 1 of s1 XOR S2 is a 1, while the other is a 0. $\endgroup$
    – SeesSound
    Mar 9 '19 at 4:37
  • $\begingroup$ Why do you care whether s1 or s2 can be recovered? Why are those values in particular important to your application, except insofar as they serve (or, in this case, fail) to prevent forgery of the RFID tags? $\endgroup$ Mar 9 '19 at 4:41
  • $\begingroup$ (The original question also asked for recovery of s1 and s2; except for the wording of the prose this still appears to me to be an exact duplicate of the previous question.) $\endgroup$ Mar 9 '19 at 4:42
  • $\begingroup$ it’s worth mentioning that RFID security is not implemented in that fashion $\endgroup$
    – b degnan
    Mar 9 '19 at 12:35

Browse other questions tagged or ask your own question.