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The sender and receiver share two different secret keys: s1 and s2 both y bit values. 'x' is a randomly generated y bit value. The reader sends to the tag a challenge y = x ⊕ s1, and the receiver responds with z = x ⊕ s2 after recovering x = y ⊕ s1.

s1 and s2 never change, x is randomly generated each complete transaction. Can you determine s1 and s2 through a single transaction? After multiple?

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  • $\begingroup$ It would be better if you call it one-time-tag. $\endgroup$ – kelalaka Mar 7 at 7:15
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If your goal is to ensure only the legitimate sender with knowledge of $s_1$ and $s_2$ can answer the challenge, then this protocol fails to achieve your goal: eavesdropping a single challenge/response pair $(y, z)$ is enough to forge any number of subsequent responses pairs with the same $s_1$ and $s_2$.

Specifically, $z = y \oplus s_1 \oplus s_2$, so to forge the correct response $z' = x' \oplus s_2$ to a subsequent challenge $y' = x' \oplus s_1$ it suffices to recover $s_1 \oplus s_2 = z \oplus y$ and then to compute $$z' = y' \oplus s_1 \oplus s_2 = x' \oplus s_1 \oplus s_1 \oplus s_2 = x' \oplus s_2$$ as expected.

If each $x$ is independently uniformly distributed, then you can't recover $s_1$ or $s_2$ from it (effectively, you have encrypted $s_1$ with the one-time pad $x$ in this scenario!), but seldom is the pad the object you're interested in protecting in the end!

So what are you really trying to accomplish here?

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  • $\begingroup$ I guess it come down to if there is cipher text reuse (But not key reuse) then is the one time pad secure - to which I believe it is as long the key is truely random $\endgroup$ – Dartuso Mar 7 at 4:15
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Squeamish Ossifrage gave a clear explanation on how to forge responses, it is perhape worth elaborating why you can not recover s1 or s2. We can recover s1^s2 as explained (s1^x)^(s2^x)=s1^s2 , so we will ask if we can discover s1, also discovering s2.

Given any transaction, every possible value for s1 corresponds to some possible value for x. So if x is uniform ly random every possible value of s1 is equally likely.

A second transaction will only tell us the difference between the x values, it will not give us any more information on s1 nor s2.

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