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Let's recall how discrete logarithms are solved in strong elliptic curve groups. The basic idea is to iteratively walk through many combinations of the form $x_i = a_iP + b_iQ$ until we find a distinguished one, i.e., one that shares some common property (like the lowest $k$ bits of $x_i$ set to 0). We accumulate enough distinguished points until we find a ...

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That is definitely not a good hash function, but as you already realized, it is not a hash function actually. Your construction can be seen as PRF or an HMAC, and for both it is not secure either. A general problem is, that you allow the attacker way too little. Giving him only $h$, that's similar to a ciphertext only attack, which is just not enough in ...

1

Since attacker does not know $m$, he can't directly apply discrete logarithm methods. On the other hand, small message space allows to run discrete log algorithm on each possible $m$. There are subexponential algorithms for dlog, but I am not sure if they are directly applicable here. But the general BSGS algorithm will find $e$ in $sqrt(e)$ operations, so ...

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Why is diffie-hellman defined on a cyclic group[0]? Doesn't it work for any commutative operation which the inverse is hard to find? No, you need associativity as well; once you have that, your idea would work fine, once we find a semigroup (that's what we call sets with an operator that is associative) with the appropriate properties. That's the ...

2

Diffie-Hellman operates in a cyclic group by definition: the elements $g, g^a, g^b, g^{ab}$ are in the cyclic group generated by $g$. Technically, a monoid is sufficient, but since cryptography mostly operates in finite structures, you get a group anyway. In your example, you operate in the cyclic group $c\mathbf{Z}$, and as you were told in the comments, ...

6

In the basic fixed window method of performing point multiplication, we compute the value $nP$ (where $n$ is the integer we're multiplying by, and $P$ is the basis point) by finding the base $b$ representation $n = d_k b^k + d_{k-1} b^{k-1} + ... + d_1 b^1 + d_0 b^0$ (where $0 \le d_i < b$), and then computing first $1P, 2P, ..., (b-1)P$ and then \$nP = ...

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