226 MISCELLANEOUS PROBLEMS [CH. XI 



3 and 4, 5 and 6, 7 and 8, 9 and 10, 11 and 12. Hence the one 

 can be deduced from the other by moving the counters about 

 in the box. 



If however in the second diagram the order of the last 

 three counters had been 13, 15, 14, then it would have required 

 seven interchanges of counters to bring them into the order 

 given in the first diagram. Hence in this case the problem 

 would be insoluble. 



The easiest way of finding the number of simple inter- 

 changes necessary in order to obtain one given arrangement 

 from another is to make the transformation by a series of cycles. 

 For example, suppose that we take the counbers in the box in 

 any definite order, such as taking the successive rows from left 

 to right, and suppose the original order and the final order to 

 be respectively 



1, 13, 2, 3, 5, 7, 12, 8, 15, 6, 9, 4, 11, 10, 14, 

 and 11, 2, 3, 4, 5, 6, 7, 1, 9, 10, 13, 12, 8, 14, 15. 



We can deduce the second order from the first by 12 simple 

 interchanges. The simplest way of seeing this is to arrange the 

 process in three separate cycles as follows : — 



1, 11, 8; 13, 2, 3, 4, 12, 7, 6, 10, 14, 15, 9; 5. 

 11, 8, 1; 2, 3, 4, 12, 7, 6, 10, 14, 15, 9, 13; 5. 



Thus, if in the first row of figures 11 is substituted for 1, then 

 8 for 11, then 1 for 8, we have made a cyclical interchange of 

 3 numbers, which is equivalent to 2 simple interchanges (namely, 

 interchanging 1 and 11, and then 1 and 8). Thus the whole 

 process is equivalent to one cyclical interchange of 3 numbers, 

 another of 11 numbers, and anothei of 1 number. Hence it is 

 equivalent to (2 + 10 + 0) simple interchanges. This is an even 

 number, and thus one of these orders can be deduced from the 

 other by moving the counters about in the box. 



It is obvious that, if the initial order is the same as the 

 required order except that the last three counters are in the 

 order 15, 14, 13, it would require one interchange to put them 

 in the order 13, 14, 15 ; hence the problem is insoluble. 



If however the box is turned through a right angle, so as 



