Answers to Chapter 4. Exercise 4A. b vertex U V W X Y Z valency There are more than 2 nodes of odd degree so the. vertex A B C D E F

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1 nswers to hapter xercise 1 a vertex valency 2 2 There are nodes with odd valency so the graph is neither ulerian nor Semi-ulerian. vertex J K valency 2 There are precisely 2 nodes of odd degree ( and ) so the graph is semi-ulerian. possile route starting at and finishing at is: K J K. c vertex L M N P Q R valency ll vertices have even valency, so the graph is ulerian. possile route starting and finishing at L is: L M N P M R P Q R L. 2 a i vertex valency vertex U V W X Y Z valency There are more than 2 nodes of odd degree so the graph is not traversale. 5 n each case there are either zero, or an even numer of, vertices with odd valency. 6 f a graph is traversale we will approach each vertex on one edge and must leave on a different one. in edge out edge a This means that the edges must e in pairs at each vertex, an out edge for each in edge, and since the graph is traversale there will e no edges left over. So the valency of each vertex will e even if we return to the start. or routes that start and finish at different vertices there will e an unpaired out edge from the start vertex which will e alanced y an unpaired in edge at the finish vertex. So these two vertices will have odd valency, ut all others will e even. 7 a ii vertex valency 2 2 i possile route is:. ii possile route is:. a i vertex R S T U V W valency Precisely 2 vertices of odd valency (T and U) so semi-ulerian. ii vertex J K L M N valency 2 2 Precisely two nodes of odd degree (J and L) so semi- ulerian. i possile route starting of T and finishing at U is: T R S U W V T U. ii possile route starts at J and finishes at L: J K L M J M N N L. a vertex valency There are more than 2 vertices of odd degree so the graph is not traversale. c vertex valency 5 There are more than two odd nodes, so the graph is not traversale. 8th ridge We will start of and finish at so these still need to have odd valency. We can only have two odd valencies so and must have even valencies (see tale). We need to change the valency of and of. So we uild a ridge from to. vertex valency with 7 ridges odd odd odd odd valency wanted odd even odd even 1

2 NSWRS d 9th ridge We will start at and finish at so these vertices need to e the two vertices with odd valency. We need and to have even valency (see tale). We need to change the valency of node and of node. So we uild a ridge from to. vertex valency with 8 ridges odd even odd even valency wanted even odd odd even e 10th ridge ll vertices now need to have even valency. This means we need to change the valencies of nodes and. So the 10th ridge needs to e uilt from to. vertex valency with 9 ridges even odd odd even valency wanted even even even even xercise 1 a ll valencies are even, so the network is traversale and can return to its start. possile route is:. length of route weight of network 285 The valencies of and are odd, the rest are even. We must repeat the shortest path etween and, which is the direct path. We add this extra arc to the diagram. possile route is:. length of route weight of network arc c The degrees of and are odd, the rest are even. We must repeat the shortest path from to. y inspection this is, length 260. We add these extra arcs to the diagram. possile route is:. length of route weight of network d The order of and are odd, the rest are even. We must repeat the shortest path from to. y inspection this is, length 18. We add these arcs to the diagram. possile route is:. length of route weight of network a Odd valencies at,, and. onsidering all possile pairings and their weights least weight Shortest route We need to repeat arcs to, is. and. weight of network dding and to the diagram gives. possile route is:. 2

3 NSWRS Odd valencies at,, and onsidering all possile pairings and their weights least weight We need to repeat arcs and. weight of network dding and to the diagram gives. Shortest route from to is. possile route is:. c Odd degrees at,, and. onsidering all possile pairings and their weight Shortest route least weight We need to repeat arcs and. weight of network dding and to the diagram gives. dding these to the diagram gives possile route is:. length We will still have two odd valencies. We need to select the pair that gives the least path. rom part a our six choices are (250), (200), (50), (80), (00) and (180). The shortest is (180) so we choose to repeat this. t is the other two vertices ( and ) that will e our start and finish. or example, start at, finish at length of route a ach arc must e traversed twice, whereas in the standard prolem each arc need only e visited once. This has the same effect as douling up all the edges The length of the route 2 weight of network km Odd nodes,,,. onsidering all possile pairings least weight We need to repeat arcs and. dding these to the network Shortest routes is is possile route is. a Odd degrees are,, and. onsidering all possile pairings and their weight least weight We need to repeat arcs and. possile route is. c f is omitted and ecome even and the only odd valencies are at and. We must repeat the shortest path etween and,. We no longer need to travel along so we can sutract 5 from the weight of the network. The new length (89 5) 1 97 km

4 NSWRS Mixed xercise 1 a vertex J degree least weight Shortest routes: to is ; to is Repeat and dd these to the network to get onsidering all possile pairings and their weight 7 15 via , and should e repeated. Shortest routes: is ; is ; is ; is. dding these arcs to the network gives possile route is J J. c length Odd vertices Q, R, T, V onsidering all possile pairings and their weight QR TV QT RV least weight QV RT Shortest route from Q to T is QST. Repeat arcs QS, ST and RV. dd these to the network P J possile route is. c Length a vertex valency Odd valencies at,, and. onsidering all possile complete pairings and their weight least weight Shortest routes: is ; is, is ; is, is. Repeat, and,. dding these arcs to the network gives Q V U R possile route is P Q S T Q S T R Q V R V T U V P. length of route a vertex degree Odd valencies at,, and. T S possile route is:. c length km d f is included and now have even valency. Only and have odd valency. So the shortest path from to needs to e repeated. Length of new route 51. path from to km This is (slightly) shorter than the previous route so choose to grit since it saves 0.1 km.

5 NSWRS 5 a The route inspection algorithm (method as shown in main text page 69) Odd valencies,,,. onsidering all possile complete pairings and their weight least weight Shortest routes: is ; is, is Repeat, and dding these arcs to the network gives Odd vertices,,, onsidering all complete pairings least weight The shortest route is. Repeat and. dding these arcs to the network gives J possile route is:. length m c This would make the start and the finish. We would have to repeat the shortest path etween and only. New route m So this would decrease the total distance y 18 m. 6 a The route inspection algorithm description in main text on page 69. possile route is: J J K. c length of route d i We will still have to repeat the shortest path etween a pair of the odd nodes. We will choose the pair that requires the shortest path. The shortest path of the six is (10). We will use and as the start and finish nodes. ii 259 e ach edge, having two ends, contriutes two to the sum of valencies for the network. Therefore the sum 2 numer of edges The sum is even so any odd valencies must occur in pairs. K 5