CH. IX] UNICURSAL PROBLEMS 173 



First. The number of odd nodes in a closed network is 

 even. 



Suppose the number of branches to be b. Therefore the 

 number of hooks is 2b. Let k n be the number of nodes of 

 the nth order. Since a node of the nth order is one at which 

 n branches meet, there are n hooks there. Also since the figure 

 is closed, n cannot be less than 2. 



.♦. 2k a +3k i + 4<k i +... + nk n + ... = 26. 



Hence 8k, + 5k s + ... is even. 



,\ k„+ k, + ... is even. 



Second. A figure which has no odd node can be described 

 unicursally in a re-entrant route. 



Since the route is to be re-entrant it will make no difference 

 where it commences. Suppose that we start from a node A. 

 Every time our route takes us through a node we use up one 

 hook in entering it and one in leaving it. There are no odd 

 nodes, therefore the number of hooks at every node is even: 

 hence, if we reach any node except A, we shall always find 

 a hook which will take us into a branch previously untraversed. 

 Hence the route will take us finally to the node A from which 

 we started. If there are more than two hooks at A, we can 

 continue the route over one of the branches from A previously 

 untraversed, but in the same way as before we shall finally 

 come back to A. 



It remains to show that we can arrange our route so as 

 to make it cover all the branches. Suppose each branch of 

 the network to be represented by a string with a hook at each 

 end, and that at each node all the hooks there are fastened 

 together. The number of hooks at each node is even, and if 

 they are unfastened they can be re-coupled together in pairs, 

 the arrangement of the pairs being immaterial. The whole 

 network will then form one or more closed curves, since now 

 each node consists merely of two ends hooked together. 



If this random coupling gives us one single curve then the 

 proposition is proved; for starting at any point we shall go 

 along every branch and come back to the initial point. But 



