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Suppose that $m$ is a message that someone player $i$ wants to send to a network of other players $j\neq -i$. The player to prevent his message from cheating by others uses an encyrpstion scheme. Say $$g:M\times Y \to X$$ denotes a cipher where $Y$ is the key and $X$ a code that makes the message to look random. The standard assumptions to be made are that $|Y|\geq |M|$ and $g(\cdot,y)$ is a bijection namely every pair of $(m,y)$ is associated with only one $x$. My question is how are the key $y$, the code $x$, and the message $m$ are associated? for example if we could make some operations among $g$, $y$ and $m$, what would that be? could we claim that $x\oplus y \underbrace{=}_{?}m$? or somehting like this?

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  • $\begingroup$ What is the origin of this Question? You did not define $g(\cdot,y)$ other than saying it is a bijection. What is the aim of this? $\endgroup$
    – kelalaka
    Jan 4, 2022 at 14:41
  • $\begingroup$ @kelalaka what do you mean what is the aim of this? $\endgroup$
    – Nav89
    Jan 4, 2022 at 15:22
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    $\begingroup$ Just use RSA-KEM to encapsulate random key per user and encrypt with AES-GCM or see Libsodium... $\endgroup$
    – kelalaka
    Jan 4, 2022 at 15:33
  • $\begingroup$ @kelalaka I have no idea what is RSA-KEM and AES-GCM... cryptography is not my field, so explain to me what are these schemes. I only know group theory that I was taught in an introductory course as undergraduate $\endgroup$
    – Nav89
    Jan 4, 2022 at 16:57
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    $\begingroup$ Well, domain and codomain are really dependent on the KDF: Just a Hash, HKDF, Password based... My humble advice for you reading some into dictionary books? A heavily math based An Introduction to Mathematical Cryptography and/or Introduction to Modern Cryptography: Third Edition and/or A Graduate Course in Applied Cryptography ( free book) and some free lectures? $\endgroup$
    – kelalaka
    Jan 4, 2022 at 17:44

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Taking into account the book. I write here an example. Suppose, that we have a mechanism of communication $\mathcal{M}=(g,h)$ such that $\mathcal{M}$ is defined over $(Y,M,X)$, where $Y$ is the key, $M$ the message and $X$ the cipher spaces respectively. To simplify the problem even more I assume that $Y=M=L=\{0,1\}^l=G$ instead of an arbitrarily finite field $\mathbb{F}^n$ and write below

$$g(y,m)=x,\quad\text{is the encrypted message, which by definition equals $x$}$$

$$h(y,x)=m,\quad\text{is the decrypted message, which by definition equals $m$}$$

So, indeed $(y,x)$ is defined to be associated with only one $m$ and hence $g(y,\cdot)$ is bijective by definition. To anser the question how are they associated, when someone knows both $x$ and $y$, then indeed $x\oplus_{G} y=m$

In order to decrypt the message we have that

$$h(y,x)=h(y,g(y,m))=y\oplus_G x=m$$

where $\oplus_{G}$ is the operation of $+$ as it is defined in the finite field $G$. And hence we have show that the calculation that you ask for, it holds by definition.

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  • $\begingroup$ Anyone who has to add a comment or thinks that I am understanding something wrong you can say this to me. But I think that this is the simplest explanation under the Shannon mechanism for perfect security. $\endgroup$ Jan 6, 2022 at 12:28
  • $\begingroup$ Well, it seems ok to me...and after taking a look at the books mentions by @kelalaka I think that this is the case. So if the specialists here think that your answer is fine, I will accept it as the answer that solved my problem. $\endgroup$
    – Nav89
    Jan 6, 2022 at 12:49

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