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Feb 28, 2021 at 5:08 comment added Manish Adhikari Forget it, ponzo already gave a very good hint and much simple one at that
Feb 28, 2021 at 4:57 comment added Manish Adhikari Sorry, I completely ignored the first question. My comment above was for second question, to check for solvability. Does any key (solution) exist for any arbitrary plain-text cipher-text pair can be checked by using key as variable vector instead.
Feb 28, 2021 at 4:50 comment added Manish Adhikari I think the better idea will be to use key $(a,b,c,d,e,f)$ as the variable vector $x$ with the last element = 1 as in $[a,b,c,d,e,f,1]^T$. Create matrix $A$ using plain texts and ciphertext is solution matrix $B$ in $Ax=B$. It will be easier to check for solvability for any plaintext matrix of the given form and any ciphertext that way.
Feb 27, 2021 at 23:03 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Jan 28, 2021 at 22:27 answer added poncho timeline score: 1
Jan 28, 2021 at 21:46 comment added kelalaka \begin{array}{ccccc} a & 1 & 0 & 0 & 0 & C_1\\ b & c & 0 & 0 & 0 & C_2 \\ 0 & 0 & d & e & 0 & C_3 \\ 0 & 0 & 0 & 1 & f & C_4 \end{array} See dummies.com/education/math/calculus/…
Jan 28, 2021 at 21:16 comment added Joshua Leung @kelalaka $$ \left[ \begin{array}{cccccc|c} a & 0 & 0 & 0 & 0 & 0 & P_1 \\ 0 & b & c & 0 & 0 & 0 & P_2 \\ 0 & 0 & 0 & d & e & 0 & P_3 \\ 0 & 0 & 0 & 0 & 0 & f & P_4 \end{array} \right] $$ something like this?
Jan 28, 2021 at 21:09 comment added kelalaka Sorry, but, basic linear algebra. Setup the equations and solve by Gaussian Elimination on $GF(2)$
Jan 28, 2021 at 21:06 history asked Joshua Leung CC BY-SA 4.0