Answers 363and also Band C, you get Figure 3; if you link A and C, and also Band D,
you get Figure 4. In each case Band D are "odd nodes" (points from which
you can proceed in an odd number of ways, three), so in every route you
A
must start and finish at B or D, to go over every line once, and once only.
Therefore, Tompkins must live at B or D: we will say B, and place Johnson
at D. There are 44 routes by scheme 2, 44 by scheme 3, and 44 by scheme 4,
making 132 in all, not counting reverse routes as different. Taking Figure 2,
and calling the outside curved lines 0, if you start BOAB, BOAC, BAOB, or
BAC, there are 6 ways of continuing in each case. If you start BOAD, BAD,
BCOD, BCA, or BCD, there are always 4 ways of continuing. In the case of
Figure 3, BOCA, BOCB, BCA, or BCOB give 6 ways. BOCD, BAOD, BAC,
BAD, or BCD give 4 ways each. Similarly, in the case of Figure 4.