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Unformatted text preview: The correct answer is (e) . square Solution of problem 1.4: We have a stack of 14 dimensional spaces making up the 15 dimensional space. In each of these 14dimensional spaces we have a crosssection of B which is a hollow 3dimensional octahedron, and so is 2dimensional. In other words B is a stack of 2dimensional objects. The only additional parameter in B is the thickness of the stack. So B is 2 + 1 = 3dimensional. The correct answer is (a) . square 12 Solution of problem 1.5: D can not be distorted into T . Indeed, if there was such a distortion, then it will distort T with any point deleted into D with some point deleted. But if we delete the crossing point of T , we will be left with an object consisting of three connected pieces, while deleting any point from D leaves a single piece. Three connected pieces can not be distorted continuously into a single connected piece, so D can not be distorted into T . For the same reason D can not be distorted into Y . Similarly D can not be distorted into N . Indeed  removing any point from D leaves a single connected piece, while removing any point from N leaves two connected pieces. Also neither Y nor T can be distorted into N . Removing the cross ing point from either Y or T will leave three connected pieces, while removing any point from N leaves two connected pieces. Finally Y can clearly be distorted into T by taking the two top portions of Y and pushing them continuously down until we get a horizontal cross bar. Therefore the correct answer is (e) . square Solution of problem 1.6: The pairs of pants has three boundary circles. A torus and a sphere have no boundary so they can not be topologically equivalent to a pair of pants. Similarly a hemisphere and a torus with one hole have one boundary circles, so they are not equivalent to a pair of pants. A cylinder has two boundary circles so it can not be topologically equivalent to a pair of pants. So the only possibility is the sphere with three holes which indeed has three boundary circles. The correct answer is (f) . square Solution of problem 1.7: A rectangular sheet of fabric with two parallel boundaries identified with a flip of orientation produces a M¨ obius band, and so has a single boundary. Our sheet of fabric has an additional circular hole and so after the identification we will get a M¨obius band with a hole: 13 boundary of the hole original boundary This surface has two boundary edges: the original single boundary edge of the M¨obius band and the new boundary edge of the circular hole. Thus the correct choice is (c) . square Solution of problem 1.8: The graph has four 3valent 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 3valent vertices into vertices of even valency. Since we have four 3valent vertices we can connectof even valency....
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 Spring '08
 schneps
 Math, Planar graph, Euler

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