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The correct answer is(e square solution of problem

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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 14-dimensional spaces we have a cross-section of B which is a hollow 3-dimensional octahedron, and so is 2-dimensional. In other words B is a stack of 2-dimensional objects. The only additional parameter in B is the thickness of the stack. So B is 2 + 1 = 3-dimensional. 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 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 connectof even valency....
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The correct answer is(e square Solution of problem 1.4 We...

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