Please refer to the paper Migacz, S., et al. (2019). Parallel implementation of a sequential Markov chain in Monte Carlo simulations of physical systems with pairwise interactions. Journal of Chemical Theory and Computation, 15(5), 2797–2806.
Suppose I have $N=5$ particles: $A(1,4)$, $B(2,2)$, $C(3,1)$, $D(4,3)$, and $E(0,0)$.
Suppose $MovingSet=\{A, C, D\}$, and $FixedSet=\{B, E\}$.
$\begin{array} {r | c | c | c | c | c} & A & C & D & B & E \\ \hline A & \times & E_{AC} & E_{AD} & E_{AB} & E_{AE} \\ \hline C & - & \times & E_{CD} & E_{CB} & E_{CE} \\ \hline D & - & - & \times & E_{DB} & E_{DE} \\ \hline B & - & - & - & \times & E_{BE} \\ \hline E & - & - & - & - & \times \\ \hline \end{array}$
Here is the extracted equation $(4)$ rewritten in MathJax:
$$E_{\text{fixed}}^{\text{moving}} = \sum_{i=1}^{M} \sum_{j=M+1}^{N} E(X_i^{0} \lor X_j^{0})$$ $$= \sum_{i=1}^{M-1} \sum_{j=i+1}^{M} \{ E(X_i^{0}, X_j^{0}) \lor E(X_i^{1}, X_j^{0}) \} \tag{4}$$
The ranges mentioned in the formula $(4)$ don't cover all elements.
For example, in the case of $E_{(moving, fixed)}^{00}$ the formula (4) takes the following form:
$$E_{(moving, fixed)}^{00} = \sum_{i=1}^{M-1} \sum_{j=i+1}^{M} E(X_{i}^{0}, X_{j}^{0}) \tag{4.1}$$
So, $MovingSet$'s size = $M=3$, $FixedSet$'s size=$(N-M)=2$, and so, we have $(3 \times 2)=6$ combinations for $E^{00}$:
- $A$ vs. $B$
- $A$ vs. $E$
- $C$ vs. $B$
- $C$ vs. $E$
- $D$ vs. $B$
- $D$ vs. $E$
Now, the formula #4 (1-indexed) says that:
=> $i$ goes from $1$ to $M-1$ (I.e, from $1$ to $2$).
=> $j$ goes from $i+1$ to $M$ (I.e., from $2$ to $3$ when $i=1$. $j$ goes from $3$ to $3$ when $i=2$).
So, we can have the following three combinations only:
- $i=1$, $j=2$
- $i=1$, $j=3$
- $i=2$, $j=3$
So, the nested looping statement fails if we use the ranges for $i$ and $j$ indices mentioned in the formula $(4)$.
So, how can we include the missing values in our computation?
