Unformatted text preview: EXERCISES 4.3 81 11. The root diagram is shown below. There are six Hamiltonian circuits. Three of them are
the same as other three, just in the reverse direction. When we calculate the length of the
three distinct ones we obtain the following table ' Hamiltonian Circuit Length (1,2,4,3,1) 320 + 350 + 400 + 300 = 1370
(1,2,3,4,1) _ 320 + 500 + 400 + 200 = 1420
(1,3,2,4,1) 300 + 500 + 350 + 200 = 1350‘ Thus, the shortest routes are (1,3,2,4,1) and its reverse (1,4,2,3,1). 12. Since every node has even degree, the only way for an Euler circuit not to exist is fer the
graph not to be connected. An example of such a graph is in part (c) of Example 14.4. 13. The root diagram ibr the graph is shown below. The only two Hamiltonian circuits are
(1: 2: 61 3, 7; 4, 5) 1) and (1, 514: 7331 6, 211) 0 e o e
E! G U 9 e t) (I
G o o E: o E: 0 00 0 0 a la (1 0 a
950 0 o e a case 9009900000 806 z.
o no as one o c o o' {I one oeoe 0 00 90 o o
a e or. as; o . o o of; 14. The root diagram for the graph is shown below. There are eight Hamiltonian circuits, four
of which are the other four in the opposite direction. They are tabulate below.
Hamiltonian Circuit Length (1,2,3,4,5,1) 4 +4 + 5 + 1 +2 = 16
(1,2,3,5,4,1) 4 + 4 + 3 + 1 + 2 z 14
(1,2,5,3,4,1) 4 + 2 + 3 + 5 + 2 = 16
(1,4,3,2,5,1) 2 + 5 +4 + 2 +2 =15 The shortest Hamiltonian circuits are (1, 2, 3, 5,4, 1) and its reverse (1,4,5,3,2,1), both giving14.
o
e o 9
9 9 e 9 6 (=3 9
.0 9 o 9 a a (3 new a to as on e
I) (3 0 (J O 090 630 E) (199 0‘) 0 (9
n I] I] I1 [I I] [I n 15. (a) The plow can accomplish the task if there is an Euler path or an Euler circuit for the
graph. Since the middle two nodes at the bottom of the graph have degree three and all
other nodes have degree two or four, there is an Euler path for the graph. The garage must
be at one of the two middle nodes on the bottom of the graph. . u :  .1.“ r 34.1.. n 4...».. Flu01.». ..unAndhm in new! ear in ﬁr” 24. elridlv nmhihﬂet‘l. ...
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 Fall '11
 Dr.Cornell
 Math

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