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Usually, in public-key cryptography, the "key size" is implicitly referred to the size of the public key. In the case of NTRU, both public and private keys are conveyed by the same thing: polynomials defined over a specific polynomial ring. These polynomials can be represented as vectors in $\mathbb Z_q$ of size $N$. Therefore, raw public and private key ...

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RSA will never become "impractical" until it gets broken. (in polynomial-time) As the best current algorithm to break RSA is subexponential (general number field sieve / GNFS) but according to moore's law, computing power grows exponentially simply increasing key size will always be an option. But beyond 256-bit security ec crypto just is significantly ...

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The size of the key depends on the security level you want, it is not possible to say "you need exactly $n$ key bits if you have a block size of $m$ bits". Let's assume that for an ideal cipher, a block size of 32 bits means $2^{32}$ possible input values which can be permuted in $(2^{32})!$ ways. That means you would have at most $\log_2(2^{32}!)$ key ...

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