Loop Configuration

Figure 9.3: A loop configuration with $N=6$.

Table 9.3: Connection table for the star configuration in Fig. 9.3. The Left-half is the table given by Fig. 9.3 and the right-half is after sorting by particle indices.

before			after sorting
$i$	$\alpha_{i}$	$\beta_{i}$	$i$	$\alpha_{i}$	$\beta_{i}$
0	0	1	0	0	1
1	1	2	1	0	5	redundant
2	2	3	2	1	2
3	3	4	3	2	3
4	4	5	4	3	4
5	5	0	5	4	5

Figure 9.3 shows a loop configuration with $N=6$ particles. It is seen that the number of bonds $N_{b}$ is equal to $N$. This means that one of the bonds is redundant, that is, it can be derived from the other connector vectors. It is obvious that the connector vector for the 5th bond in Fig. 9.3, for example, can be obtained from the rest as

$$\bm{Q}_{5} := \bm{x}_{5} - \bm{x}_{0} = -\sum_{i=0}^{4}\left(\bm{x}_{i+1} - \bm{x}_{i}\right) = -\sum_{i=0}^{4}\bm{Q}_{i} .$$ (9.5)

Therefore, the number of independent connector vectors is $N-1$, as expected.

After sorting the connection table by the particle indices, we have the table shown in the right-half in Table 9.3. We drop the redundant bond $i=1$ and form the independent bond list bonds[] as

$${\tt bonds}[N-1] = \{0, 2, 3, 4, 5\}.$$ (9.6)

Figure 9.4: A loop configuration with extra connections from particle 0.

Table 9.4: Connection table for the star configuration in Fig. 9.4. The Left-half is the table given by Fig. 9.4 and the right-half is after sorting by particle indices.

before			after sorting
$i$	$\alpha_{i}$	$\beta_{i}$	$i$	$\alpha_{i}$	$\beta_{i}$
0	0	1	0	0	1
1	1	2	1	0	2	redundant
2	2	3	2	0	3	redundant
3	3	4	3	0	4	redundant
4	4	5	4	0	5	redundant
5	5	0	5	1	2
6	0	2	6	2	3
7	0	3	7	3	4
8	0	4	8	4	5

Figure 9.4 shows a loop configuration with extra connections with particle 0 to all the others. In this case, the number of bonds $N_{b}$ exceeds $N-1$ by 4 (because we add three more connections into the looped configuration in Fig. 9.3 which already has one redundant bond).

After sorting the connection table by the particle indices, we have the table shown in the right-half in Table 9.4. We drop the redundant bonds $i=1, 2, 3, 4$ and form the independent bond list bonds[] as

$${\tt bonds}[N-1] = \{0, 5, 6, 7, 8\}.$$ (9.7)

Figure 9.5: A loop configuration with extra connections skipping one particle.

Table 9.5: Connection table for the star configuration in Fig. 9.5. The Left-half is the table given by Fig. 9.5 and the right-half is after sorting by particle indices.

before			after sorting
$i$	$\alpha_{i}$	$\beta_{i}$	$i$	$\alpha_{i}$	$\beta_{i}$
0	0	1	0	0	1
1	1	2	1	0	2	redundant
2	2	3	2	0	4	redundant
3	3	4	3	0	5	redundant
4	4	5	4	1	2
5	5	0	5	1	3	redundant
6	0	2	6	1	5	redundant
7	1	3	7	2	3
8	2	4	8	2	4	redundant
9	3	5	9	3	4
10	4	0	10	3	5	redundant
11	5	1	11	4	5

Figure 9.5 shows another loop configuration with 6 extra connections skipping one particle. In this case, the number of bonds $N_{b}$ exceeds $N-1$ by 7.

After sorting the connection table by the particle indices, we have the table shown in the right-half in Table 9.5. We drop the redundant bonds shown in Table 9.5, bond list bonds[] as

$${\tt bonds}[N-1] = \{0, 4, 7, 9, 11\}.$$ (9.8)

Figure 9.6: A loop configuration with extra connections skipping one particle with $N$ particles in a straight form.

For general $N$, the the independent bond list in this type of configuration becomes

$${\tt bonds}[N-1] = \{0, 4, 7, 9, \cdots, 2i+3, \cdots, 2(N-2)+3\},$$ (9.9)

where $2i+3$ is the $i$th element (counting from 0) and $2(N-2)+3$ is the last ($N-2$)-th element. The difference between the 0th and first elements is 4, and that between the first and second is 3. Beyond this point, the differences are 2. To see this, let us see the configuration modified in a straight form in Fig. 9.6.

For the particle 0, there are 4 bonds and three of them are redundant. Therefore, in the sorted form, the second independent bond is the forth from the first independent bond. For the next particle 1, one of the 4 bonds is connecting to the particle 0 which is already taken into account in the sorted form. Among the rest three, two are redundant, so that the third independent bond is the third from the second. For the next particle 2, two of the 4 bonds are connecting either particle 0 or 1. Only one of the rest two bonds is redundant and the fourth independent bond is the second from the third. For the particles later, the situation is the same for the particle 2, so that the difference between the independent bonds are 2.