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hw7-sol - AMS 345/CSE 355(Fall 2010 Joe Mitchell...

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AMS 345/CSE 355 (Fall, 2010) Joe Mitchell COMPUTATIONAL GEOMETRY Homework Set # 7 – Solution Notes (1). [20 points] For each of the following sign vectors, draw an arrangement of 5 distinct lines, { 1 ,...,ℓ 5 } , that has a face (either a 2-face (cell), a 1-face (edge), or a 0-face (vertex)) with the corresponding sign vector (where “+” means lying strictly above a line, “-” means strictly below, and “0” means on the line). Clearly label each line and highlight the face. Give the dimension of each such face. Also, say for each case whether the arrangement is simple or not. (You should make your example simple if it is possible to do so.) NOTE: Many different arrangements are possible to illustrate each case; chances are, your figure looks different from mine! (a). (+,+,+,-,-); This is a 2-face, since it lies interior to all 5 half-planes. The arrangement in the example is simple; see below, left. The 2-face is highlighted in light blue. (b). (0,+,0,-,0); This is a 0-face (vertex), since it lies on 2 or more lines. The arrangement cannot be simple, since this 0-face is common to 3 lines and in a simple arrangement no three lines can pass through a common point. See below, middle. The vertex is highlighted in light blue. (c). (0,-,+,-,+); This is a 1-face (edge), since it lies on exactly one line (one “0”). The arrangement in the example is simple. See below, right. The edge is highlighted in light blue. 5 1 4 2 3 1 4 3 5 2 1 4 3 2 5 (d). (-,-,-,+,0); This is a 1-face (edge), since it lies on exactly one line (one “0”). The arrangement in the example is simple. See below, left. The edge is highlighted in light blue. (e). (0,0,0,-,0); This is a 0-face (vertex), since it lies on 2 or more lines. The arrangement cannot be simple, since this 0-face is common to 4 lines and in a simple arrangement no three lines can pass through a common point. See below, right. The vertex is highlighted in light blue. 1 3 2 5 4 1 3 2 4 5 (Optional question for extra thought: Is it possible to find one set of 5 lines whose arrangement contains faces with all five sign vectors?) NO, it is not possible. There are several ways to see this. One is to note that the existence of a face with sign vector (0,0,0,-,0) implies that lines 1 , 2 , 3 , 5 must all pass through a common point (that vertex, v , with that sign vector); then, there can be no face with sign vector (0,+,0,-,0), since such a face (0-face) lies at the intersection of 1 , 3 , and 5 (because of the three 0’s), meaning it lies at point v = 1 2 3 5 , contradicting the “+” in the sign vector associated with 2 .
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