Applied Finite Mathematics HW Solutions 82

Applied Finite Mathematics HW Solutions 82 - 80 EXERCISES...

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Unformatted text preview: 80 - EXERCISES 4.3 (d) The edges traced out are shown in the left graph below with arrows indicating direction. It is a path. Since it returns to the same node, it is also a circuit. (e) This is impossible; nodes E and F do not have an edge joining them. ' (f) Since the edge joining nodes G and E is repeated, this is not a path. (g) The edges traced ”out are shown in the right graph below with arrows indicating direction. It is a path. Since it traces every edge of the graph, it is an Euler path. a A s A a c 4‘" G F G 8. (a) Since this graph has four nodes with odd degree, there cannot be an Euler path or circuit. There are many Hamiltonian circuits, one of which is shown in the left figure below. (b) Since all nodes of the graph have even degree, there are Euler circuits, one of which is shown in the middle figure below. There is no Euler path. There are many Hamiltonian circuits. (c) Since this graph has exactly two nodes with odd degree, there are Euler paths but no Euler. circuit. One is shown in the right figure below. There is no Hamiltonian circuit. 9. We use one node for each of the islands and one node for each of the river banks (figure to the right). Since every node has even degree, there is an Euler circuit for the graph. Hence, it is possible to ' walk around the city starting and ending at the same point and crossing each bridge exactly once. ' ' 10. Suppose we represent rooms and the outside by nodes and doorways by edges (left figure below). To answer the question, we ask whether there is an Euler path from one node representing the outside to the other node representing the outside. Since all other nodes have even degree, there is indeed an Euler path from outside to outside. ' For the second floor plan, we again represent rooms and the outside by nodes and doorways . by edges (right figure above). To answer the question, we ask whether there is an Euler path ' from one node representing the outside to the other node representing the outside. Since there are two nodes with degree three, as well as two nodes with degree one, there is no Euler path from outside to outside. _ _l__2_l L _ _ L--- --_..-.I;.....A :. -..........I.._.... ..:.oL -....-_'..l.c 3-... L..4I..ATY..I..mliu ArltnnkAka DAA‘Idm EnakmmnrinM-im in ..a n. h. l—‘ull L. «spinnumkikflazl ...
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