hypothesis of the induction, the order of $A^{'}_{i1}$ is $n-1$, then $$\begin{split}
\operatorname{ham}(A^{'}_{i1})&=\sum\limits_{\{k^{'}_2 \cdots,
k^{'}_{n-1}\}}A^{'}_{i1}(k^{'}_2,i-1)\cdots
A^{'}_{i1}(k^{'}_{n-1},k^{'}_{n-2})A^{'}_{i1}(i-1,k^{'}_{n-1})\\
&=\sum\limits_{\{k_2 \cdots,
k_{n-1}\}}A^{'}_{i1}(k_2-1,i-1)A^{'}_{i1}(k_{n-1}-1,k_{n-2}-1)A^{'}_{i1}(i-1,k_{n-1}-1)\\
&=\sum\limits_{\{k_2 \cdots, k_{n-1}\}}A(k_2,i)\cdots
A(k_{n-1},k_{n-2})A(1,k_{n-1}),
\end{split}$$ where $\{k^{'}_2 \cdots, k^{'}_{n-1}\}$ and $\{k_2 \cdots,
k_{n-1}\}$ go over all the permutations of $\{1,\cdots,
n-1\}/\{i-1\}$ and $\{2,\cdots, n\}/\{i\}$ respectively. This completes the proof of Lemma 4.    $\Box$\
\
**Hamiltonian Recovery** Let $A=(A(i,j))_{n\times n}$ be an $n\times n$ positive matrix. The following procedure is applied to selecting elements from $A$ (The first two steps are given explicitly). We call this procedure Selecting Hamiltonian Cycle(SHC for simplicity).\
\
**Step 1.** Let $A^{1}=A$. Choose a natural number $1<j_1\leq
n$, denote $\pi(1)=j_1$ and select $A^1(\pi(1),1)$.\
**Step 2.** Let $A^{2}=(A^{1}_{j_11})^{'}$. Choose a natural number $1<j_2\leq n-1$, denote $\pi(2)=j_2$ and select $A^2(\pi(2),1)$.\
Similarly $A^{k}$, $\pi(k)$ and $A^k(\pi(k),1)$, $1\leq k\leq n-1$ can be obtained in Step k iteratively. Since $A^{n}$ has only one entry, let $A^{n}=(A^{n-1}_{j_{n-1}1})^{'}$ , $\pi(n)=j_n=1$ and select $A^n(\pi(n),1)$.\
By Lemma 4, the set of selected elements $A^k(\pi(k),1)$, $k=1,\cdots,n$, from the above procedure forms the edge weight of a Hamiltonian cycle in $G$ if $A=A_G$. If $\pi(1),\pi(2),\cdots,\pi(n)$ is given, we provide a simple algorithm to determine which Hamiltonian cycle in $G$ is selected. This process is called Hamiltonian Recovery.

The 