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fgrieu can you please explain how to get this part of the question. R1(K1) = R2(K2) XOR R3 (K3). this is important to answer.


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Proposition aimed at 1/2/3/4 of the last section of the question: chained hybrid encryption. We'll use A fast symmetric authenticated encryption scheme, such as AES-GCM or ChaCha20-Poly1305 with 256-bit secret key, encryption with key $K$ noted $C=E_K(M)$ and decryption $M=D_K(C)$. An asymmetric encryption scheme capable of encrypting 256-bit messages, with ...


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The notion of imperfect (ramp) secret sharing allows one to share multiple secrets in one go with a lower overhead in shares than threshold secret sharing. I believe these might answer your question: Ramp Cheater Identification here Ramp Cheating Detection here, here and here


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Alternatively, is there a way to generate three random number sequences R1, R2, R3 from three keys K1, K2, K3 such that R1 XOR R2 XOR R3 = 0? That part's easy; we can just define: $$R1 = \text{SHAKE}(K1) \oplus \text{SHAKE}(K2)$$ $$R2 = \text{SHAKE}(K3) \oplus \text{SHAKE}(K1)$$ $$R3 = \text{SHAKE}(K2) \oplus \text{SHAKE}(K3)$$ (where $\text{SHAKE}$ can be, ...


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Assume we know $t'<t$ shares. Assume we are given some random values picked uniformly from the same field as the one used in SSS. Question: can we distinguish the random values from the shares with a non-negligible probability? It depends. Now, there are two possible ways to interpret your question (and while the answer is the same for both, the logic ...


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No we can't. We know that SSS is perfect that means knowledge of (t-1) or fewer shares provides no information about s -therefore about other shares-


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Each value of the degree $t-1$ full SSS polynomial $P$ evaluated at any $x$ in the finite field $F$ it is defined over is uniformly distributed in $F$. See this answer for example. Now assume $t’<t$ shares are revealed, say they are $S’=\{(x_1,P(x_1)),\ldots,(x_{t’},P(x_{t’}))\}.$ If the original set of shares was $S$ the nonempty set $T=S\setminus S’$ ...


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