Thus the correct choice is c square solution of

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Thus the correct choice is (c) . square Solution of problem 1.8: The graph has four 3-valent vertices and four vertices of even valency. By Euler circuit theorem a connected graph admits an Euler circuit if and only if all vertices have even valency. So we must add some edges to convert the 3-valent vertices into vertices of even valency. Since we have four 3-valent vertices we can connect them in pairs by two new edges. Since one edge can not connect four vertices, two is the minimal number of edges we can add. One possible way to do this is The correct answer is (c) . square 14
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Solution of problem 1.9: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Solution of problem 1.10: (1) is True . For instance the following dis- connected graph has the required properties: (2) is True . Each of the connected pieces of G has Euler characteristic 2 by the Euler characteritic theorem. Denote the connected pieces of our graph by G 1 , G 2 , G 3 , and G 4 . We can build G in stages by starting with G 1 and adding to it the other pieces one at a time. It is convenient to introduce notation for the resulting graph at each stage: G 1 the graph at the first stage; G 12 the graph at the second stage, i.e. the graph consisting of the connected pieces G 1 and G 2 ; G 123 the graph at the third stage, i.e. the graph consisting of the con- nected pieces G 1 , G 2 , and G 3 ; 15
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G the final graph, i.e. the graph consisting of the connected pieces G 1 , G 2 , G 3 , and G 4 . The Euler characteristic of G 1 is 2. If we add to it the piece G 2 then since G 1 and G 2 are disjoint the resulting graph G 12 has V ( G 1 )+ V ( G 2 ) vertices and E ( G 1 )+ E ( G 2 ) edges. For the regions in which G 1 and G 2 split the plane we have two possibilities: either G 1 is entirely contained in one of the regions for G 2 , or G 2 is entirely contained in one of the regions for G 1 . In the first case the outside region of G 1 is the same as the region of G 2 in which G 1 is contained, in the second case the outside region of G 2 is the same as the region of G 1 in which G 2 is contained. Hence G 12 splits theplane in F ( G 1 ) + F ( G 2 ) - 1 regions. Therefore the Euler characteristic of of G 12 is one less than the sum of the Euler characteristics of G 1 and G 2 . In other words G 12 has Euler characteristic 2 + 2 - 1 = 3. Next observe that by the same reasoning when we add G 3 to G 12 , the resulting graph G 123 will have V ( G 12 )+ V ( G 3 ) vertices, E ( G 12 )+ E ( G 3 ) edges, and splits the plane into F ( G 12 ) + F ( G 3 ) - 1 regions. Therefore the Euler characteristic of G 123 is one less than the sum of the Euler characteristics of G 12 and G 3 . Thus G 123 has Euler characteristic 3 + 2 - 1 = 4.
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