What is the difference between information-theoretic and perfect types of security? I'm having a hard time pinning down an exact definition of the difference between  information-theoretic and perfect types of security.  A rigorous definition seems elusive...

A. Wikipedia puts the difference down to the perfect type being a (poorly) defined special case of the information-theoretic type. "...a cryptosystem to leak some information" - yet I thought that these two particular security definitions only applied to breaking the encryption, rather than peripheral things like leaking the time the message was sent, 
 it's length, etc.

B. An e-voting manufacturer uses entropy to quantify the difference as:-


$H(K | C) = H(K)$ – Perfect secrecy
$H(K | C) &lt; H(K)$ – Information-theoretic security


Yet #2 can be rewritten as $H(K | C) = H(K) - \delta$, and that leads to #2 equalling #1 if $\delta \to 0$.

C. And finally a link from What&#39;s the difference between perfect security and unconditional security? is suggesting that "perfect security is the same as information-theoretic security".
Information-theoretic security means that any algorithm (even unbounded) has a negligible probability of breaking the security property (in the security parameter). This is the same as unconditional security: it does not rely on any computational assumption, and is not limited to probabilistic polytime adversaries.
A perfectly secure protocol is such that any (possibly unbounded) adversary has probability $0$ of breaking the security property. It is a special case of information-theoretic security: every perfectly secure protocol is information-theoretically secure, but the converse is not true.
To take a simple example, the distinction often appears when some secret value is masked with a random value, and we ask how hard it is to distinguish the masked value from a uniformly random value. Consider the following protocol: $x$ is an integer, say, between $0$ and $n - 1$. The game is as follows: we first sample a random bit $b$. If $b = 0$, we send a random value $r \gets X$ to the (unbounded) adversary, sampled from some set $R$, while if $b=1$, we sample a random value $r \gets R$ and send $x + r$ to the adversary. Fix a security parameter $k$. We say that the protocol has perfect security if the adversary has probability exactly $1/2$ of guessing the value of $b$ given the input, and that the protocol has information-theoretic security if the adversary has probability $1/2 + \mu(k)$ of guessing the value of $b$, where $\mu$ is a negligible function.
Suppose we identify $[0, n-1]$ with $\mathbb{Z}_n$ and define $R$ to be $\mathbb{Z}_{n}$ as well. The computation of $x + r$ is done over $\mathbb{Z}_n$. In this case, the protocol is clearly perfectly secure, as sampling $r$ from $\mathbb{Z}_{n}$ and returning $x+r$ gives exactly the uniform distribution over $\mathbb{Z}_{n}$, for any $x$.
On the other hand, suppose we set $R = [0, 2^{k} \cdot n]$ and compute $x + r$ over the integers. Then, it is easy to show that any (possibly unbounded) adversary has probability at most $1/2^{k}$ of distinguishing a sample from $R$ from a sample from $x + R$ (the statistical distance between these sets is $1/2^{k}$). Since this is a negligible function in $k$, this variant satisfies information theoretic security, but not perfect security.