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Maximum number of regions n lines can dissect a plane

Posted on 2008-06-19
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I want to find the maximum number of regions that n lines can dissect a plane into.
Let F(n) represent this number.

Through trial and error, I have found that:

F(0)=1
F(1)=2
F(2)=4
F(3)=7

From which I have found a pattern:

F(n) = n + F(n-1)  =>  F(n) = n(n+1)/2 + 1 .

This equation makes some sense, but (assuming that it's accurate for all n), how would I PROVE it to be true?
As all I've done so far, is throw some extrapolation on top of some trial-and-error. Hardly rigorous.

Is such a proof something which I'd be capable of 'getting'? Or, does it depend on some very high-level mathematics?

Thanks
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Question by:InteractiveMind
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by:Infinity08
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Your formula is correct :

        http://mathworld.wolfram.com/PlaneDivisionbyLines.html
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Infinity08 earned 1400 total points
ID: 21821480
Here's a proof by induction :

        http://www.jimloy.com/geometry/plane.htm

(no complicated maths ;) )
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by:Brian Utterback
Brian Utterback earned 600 total points
ID: 21821653
You can always add a line that intersects all of the previous lines and does not pass through any previous point of intersection. Any line that is not parallel to any of the existing lines will pass through all of them. Thus, when adding the N+1 line, it will intersect with all of the previous N lines at exactly N points. Each of these N points is the endpoint of a line segment that bisects an existing region. If we consider the two regions bisected by the line after it no longer intersects any other lines, then as we follow the new line, each of the N intersection points adds one new region, plus the region left over when there is no more intersection points. Thus, for an existing plane with N lines, forming R(N) regions, adding one more line produces N+1 lines with R(N+!) regions, where R(N+1) = R(N) + N + 1. If we change that to a form in N, we get R(N) = N + F(N-1), which is what you had.

Is that sufficient? The conversion of the recurrence relation into a closed form just follows normal methods.
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by:InteractiveMind
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Thanks
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