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Concrete mathematics : a foundation for computer science

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(A pizza with Swiss<br />

cheese?)<br />

A region is convex<br />

if it includes all<br />

line segments between<br />

any two of its<br />

points. (That’s not<br />

what my dictionary<br />

says, but it’s what<br />

mathematicians<br />

believe.)<br />

1.2 LINES IN THE PLANE 5<br />

defined by n lines in the plane? This problem was first solved in 1826, by the<br />

Swiss mathematician Jacob Steiner [278].<br />

Again we start by looking at small cases, remembering to begin with the<br />

smallest of all. The plane with no lines has one region; with one line it has<br />

two regions; and with two lines it has four regions:<br />

(Each line extends infinitely in both directions.)<br />

Sure, we think, L, = 2”; of course! Adding a new line simply doubles<br />

the number of regions. Un<strong>for</strong>tunately this is wrong. We could achieve the<br />

doubling if the nth line would split each old region in two; certainly it can<br />

split an old region in at most two pieces, since each old region is convex. (A<br />

straight line can split a convex region into at most two new regions, which<br />

will also be convex.) But when we add the third line-the thick one in the<br />

diagram below- we soon find that it can split at most three of the old regions,<br />

no matter how we’ve placed the first two lines:<br />

Thus L3 = 4 + 3 = 7 is the best we can do.<br />

And after some thought we realize the appropriate generalization. The<br />

nth line (<strong>for</strong> n > 0) increases the number of regions by k if and only if it<br />

splits k of the old regions, and it splits k old regions if and only if it hits the<br />

previous lines in k- 1 different places. Two lines can intersect in at most one<br />

point. There<strong>for</strong>e the new line can intersect the n- 1 old lines in at most n- 1<br />

different points, and we must have k 6 n. We have established the upper<br />

bound<br />

L 6 L-1 +n,<br />

<strong>for</strong> n > 0.<br />

Furthermore it’s easy to show by induction that we can achieve equality in<br />

this <strong>for</strong>mula. We simply place the nth line in such a way that it’s not parallel<br />

to any of the others (hence it intersects them all), and such that it doesn’t go

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