### Table 8 and so it is clear that each GOG arising from this action contains a Hamilton cycle.

2007

"... In PAGE 15: ...GOG arising from this action contains a Hamilton cycle. Graphs corresponding to row 9 of Table 1: The relevant graphs are given in Table8 , and so it is clear that each GOG arising from this action contains a Hamilton cycle. Graphs corresponding to row 10 of Table 1: The relevant graphs are given in Table 9 and Figure 9, and so it is clear that each GOG arising from this action contains a Hamilton cycle.... ..."

### Table 1 Running time to nd a Hamilton cycle in the middle levels graph

"... In PAGE 10: ...The results are summarized in Table1 . We measured elapsed time using a timer with a resolution of 1 second.... In PAGE 10: ... Runs that start and complete in the same second show a time of 0 seconds. Table1 also shows earlier results obtained with the SS heuristic on a 2.4GHz Intel Pentium 4 system with 512MB of RAM.... ..."

### Table 1: Running time for the heuristic to nd a Hamilton path in Rn (and thus a Hamilton cycle in the middle two levels of Bn.)

1999

"... In PAGE 15: ...ach k 15. The case k = 15, which has about 9.6M vertices in the reduced graph and 155M vertices in the full graph, took almost 3 weeks on 400MHz P-II with 192MB RAM. The timing results are summarized in Table1 . For the middle two levels graph, the number of edges, E is O(V log V )whereV is the number of vertices.... ..."

Cited by 5

### Table 3. A Hamilton cycle and the corresponding mapping h of a 4-cube. h(x) x h(x) x h(x) x h(x) x

1993

"... In PAGE 24: ... 9. One can easily obtain a Hamilton cycle shown in Table3 . For each node x, the corresponding h(x) is also shown in Table 3.... In PAGE 24: ...onsider a 4-cube shown in Fig. 9. One can easily obtain a Hamilton cycle shown in Table 3. For each node x, the corresponding h(x) is also shown in Table3 . Suppose we have the multicast set K = f0011; 0100; 0111; 1100; 1010; 1111g.... ..."

Cited by 34

### Table 1. A Hamilton cycle and the corresponding mapping h of a 4 4 mesh. h(x) x h(x) x h(x) x h(x) x

1993

"... In PAGE 23: ...hown in Fig. 8. Obviously, C = (0; 1; 2; 3; 7; 6; 5; 9; 10; 11; 15; 14; 13; 12; 8; 4; 0) is a Hamilton cycle in G. For each node x (0 x 15), Table1 shows the corresponding h(x). Consider the multicast set K = f9; 0; 1; 6; 12g, where u0 = 9 is the source.... ..."

Cited by 34

### Table 9: Relevant graphs cor- responding to the action of

2007

"... In PAGE 15: ... Graphs corresponding to row 9 of Table 1: The relevant graphs are given in Table 8, and so it is clear that each GOG arising from this action contains a Hamilton cycle. Graphs corresponding to row 10 of Table 1: The relevant graphs are given in Table9 and Figure 9, and so it is clear that each GOG arising from this action contains a Hamilton cycle. Graphs corresponding to row 11 of Table 1: Lemma 4.... ..."

### Table 4: Execution times in seconds when running Au- rora Hamilton on the Sequent and on the emulator.

1994

"... In PAGE 6: ... The memory is set to respond in one cycle (133 ns), and the networkis set to have negligible delay. The results of running Aurora (a parallel Prolog system [14]) on the DDM and on the Sequent are summarised in Table4 . The error is around 5%, indicating that the emulator gives good results in this case as well.... ..."

Cited by 15

### Table 4: Execution times in seconds when running Au- rora Hamilton on the Sequent and on the emulator.

1994

"... In PAGE 6: ... The memory is set to respond in one cycle (133 ns), and the networkis set to have negligible delay. The results of running Aurora (a parallel Prolog system [14]) on the DDM and on the Sequent are summarised in Table4 . The error is around 5%, indicating that the emulator gives good results in this case as well.... ..."

Cited by 15