180 



UNICURSAL PROBLEMS 



[CH. IX 



in the route given above the double-two can be put between 

 0-2 and 2-3 or between 1-2 and 2-4). Hence the total 

 number of unicursal arrangements of the 15 dominoes is 2"a. 

 If we arrange the dominoes in a straight line, then as we 

 may begin with any of the 15 dominoes, the total number of 

 arrangements is 15. 2 8 . a. 



We have next to find the number of unicursal routes of 

 the pentagon delineated above in figure A. At the node 

 there are four paths which may be coupled in three pairs. If 

 1 and 2 are coupled, as also 3 and 4, we get figure B. 

 If 1 and 3 are coupled, as also 2 and 4, we get figure G. 

 If 1 and 4 are coupled, as also 2 and 3, we get figure D. 



Figure B. Figure G. Figure D. 



Let us denote the number of ways of describing figure B by b, 

 of describing figure G by c, and so on. The effect of suppressing 

 the node in the pentagon A is to give us three quadrilaterals, 

 B, G, D. And, in the above notation, we have a = b + c + d. 



Take any one of these quadrilaterals, for instance D. We 

 can suppress the node 1 in it by coupling the four paths which 

 meet there in pairs. If we couple 1 2 with the upper of the 

 paths 1 4, as also 1 3 with the lower of the paths 1 4, we get 



Figure E. Figure F. 



the figure E. If we couple 1 2 with the lower of the paths 1 4, 

 as also 1 3 with the upper of the paths 1 4, we again get the 

 figure E. If we couple 1 2 and 1 3, as also the two paths 1 4, 

 we get the figure F. Then as above, d = 2e +f. Similarly 

 b = 2e +/, and c = 2e +/. Hence o = b + c + d = 6e + 3/. 



