New answers tagged complexity
1
The immediately obvious way to do this is to use RSA in a large-public-exponent format.
That is, the server selects an RSA modulus, and a large public exponent (say, $e = 2^{2^{30}}+1$) [1]; if the RSA primes are safe primes, this practically eliminates the possibility that $e$ is not relative prime to either $p-1$ or $q-1$. The server would internally ...
1
For simplicity assume that the Merkle tree is a complete binary tree. Let the number of data blocks is $n$ which are linked to the leaf nodes. Therefore the total number of tree nodes are $|nodes| = n + n/2+ \cdots +1$. If we assume that $n =2^k$ for simplicity than $$|nodes| = 1 + 2 + 2^2 + \cdots + 2^k = \frac{2^k-1}{2-1} = 2^k-1 = n-1.$$ In total, we have ...
2
There are 4294967296 nonces, because it is 32-bit.
Most of the time, none of the nonces will produce a "Golden Block".
If none of the nonces work, the miner must make a new block (with a different Merkle root (256 bits) which means the block has different transactions in it) and try to mine that block instead.
Currently, about 99.999999994% of blocks don'...
5
Yes, this is feasible. Just generate sufficiently many nonce, rnd values, and you will eventually stumble upon a prime (or a number that can be factored into a large prime and a number of small prime factors). This is how most probabilistic prime generators operate.
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