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This doesn't add new security as much as it just shifts it.
Encryption algorithms are carefully studied. Hmm... I didn't make that emphatic enough. Encryption algorithms are C A R E F U L L Y studied. There. That's better.
There are all sorts of tiny nuances to be had when designing an algorithm. A famous example were some of the S-Boxes in DES which ...
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Suppose you have two cryptosystems $A$ and $B$ with $n$-bit keys. Maybe they're both secure at what they aim to do; maybe they aren't. Say they both take about the same cost to implement.
You are proposing using an $(n + 1)$-bit key, where the extra bit selects between $A$ or $B$. You have now doubled the cost to implement your system.
Does this add any ...
7
Suppose that one of your cipher suites has some weaknesses. Then the adversary can play with your network to force you to select the weak cipher to exploit the weakness. This is highly done in TLS. Therefore your system can turn into a single case during an attack.
Also, the attacker can store all off your communication. If somehow they can break one of ...
2
You can do something like this. See Automated Analysis and Synthesis of Authenticated Encryption Schemes. They model (authenticated) encryption as a circuit, develop a certain basic set of primitives, and type inference rules to "automatically" prove authenticity + privacy.
That being said, this is all defined relative to a single block cipher I believe, so ...
1
..how does this formula $(aG+bG) = (a+b)G$ work in ECDSA?
Perfectly well. It follows from the definition of $kG$ as $\overbrace{G+\cdots+G}^{k\text{ times}}$, associativity and commutativity of point addition. Notice that operator $+$ in $(aG+bG)$ and $G+\cdots+G$ is elliptic curve point addition, while operator $+$ in $(a+b)$ is addition in $\Bbb Z$ (...
1
Let $p$ and $q$ be secret uniform random primes, and define $n = pq$. Consider $s = 2^{2^t} \bmod n$.
Without knowledge of $p$ or $q$, the best way we know to find $s$ is to start with $2$ and square it $t$ times successively. We can tune $t$ to force the other party to take arbitrarily many steps.
With knowledge of $p$ and $q$, we can use the shortcut $2^...
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