Since deciding whether a graph contains a Hamiltonian cycle is $NP$-complete, the counting problem which counts the number of such cycles of a graph is $NP$-hard.
Is it also $PP$-hard in the sense that $PP\subseteq P^{\#HAM-CYCLE}$?
Since deciding whether a graph contains a Hamiltonian cycle is $NP$-complete, the counting problem which counts the number of such cycles of a graph is $NP$-hard.
Is it also $PP$-hard in the sense that $PP\subseteq P^{\#HAM-CYCLE}$?
In this paper, Liskiewicz et al. state their Lemma 4 as follows:
Lemma 4: The problem of counting Hamiltonian paths in planar graphs of max-deg $\Delta=3$ is $\#P$-complete under $\leq^p_{r-shift}$-reductions.
And, the definition of $\leq^p_{r-shift}$-reductions is as follows:
DEFINITION: Polynomial-time Right-bit-Shift Reduction
Let $f,g:\Sigma^*\rightarrow\mathbb{N}$, $f$ is poly-time right-bit-shift reducible to $g$, denoted $f\leq^p_{r-shift}g$, if there exists a poly-time computable function $R_3:\Sigma^*\rightarrow\mathbb{N}-\{0\}$ and a polynomial-time computable function $R_1:\Sigma^*\rightarrow\Sigma^*$, such that $f(x)=g(R_1(x))\mathrm{div}\ 2^{R_3(x)}$, for all $x$.
So, yes, $PP\subseteq P^{\#HAM-CYCLE}$.