# How are the cipher, the key and the initial message (that is not encrypted) are releted?

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?

• 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? Jan 4, 2022 at 14:41
• @kelalaka what do you mean what is the aim of this? Jan 4, 2022 at 15:22
• Just use RSA-KEM to encapsulate random key per user and encrypt with AES-GCM or see Libsodium... Jan 4, 2022 at 15:33
• @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 Jan 4, 2022 at 16:57
• 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? Jan 4, 2022 at 17:44

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.

• 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. Jan 6, 2022 at 12:28
• 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. Jan 6, 2022 at 12:49