CH. Xl] MISCELLANEOUS PROBLEMS 227 



to make AD the top, this rotation will be equivalent to 13 

 simple interchanges. For, if we keep the sixteenth square 

 always blank, then such a rotation would change any order 

 such as 



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

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

 which is equivalent to 13 simple interchanges. Hence it will 

 change the arrangement from one where a solution is impossible 

 to one where it is possible, and vice versa. 



Again, even if the initial order is one which makes a 

 solution impossible, yet if the first cell and not the last is left 

 blank it will be possible to arrange the fifteen counters in their 

 natural order. For, if we represent the blank cell by b, this 

 will be equivalent to changing the order 



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

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



this is a cyclical interchange of 16 things and therefore is 

 equivalent to 15 simple interchanges. Hence it will change 

 the arrangement from one where a solution is impossible to 

 one where it is possible, and vice versa. 



So, too, if it were permissible to turn the 6 and the 9 upside 

 down, thus changing them to 9 and 6 respectively, this would be 

 equivalent to one simple interchange, and therefore would change 

 an arrangement where a solution is impossible to one where 

 it is possible. 



It is evident that the above principles are applicable equally 

 to a rectangular box containing mn cells or spaces and mn — 1 

 counters which are numbered. Of course m may be equal to n. 

 If such a box is turned through a right angle, and m and n are 

 both even, it will be equivalent to mn — 3 simple interchanges — 

 and thus will change an impossible position to a possible one, 

 and vice versa — but unless both m and n are even the rotation 

 is equivalent to only an even number of interchanges. Similarly, 

 if either m or n is even, and it is impossible to solve the problem 

 when the last cell is left blank, then it will be possible to solve 

 it by leaving the first cell blank. 



15—2 



