For my first log, I'll tell ya about something arbitrary that I find cool.
As a child, you must've known the age-old children's puzzle - Draw the following without lifting pen from paper and without repeating any line.
The age old solution is - you fold the paper and use the backside as a bridge to get from one point to another. It's a very general trick, and can be used to draw anything without lifting pen from paper.
Let's look deeper. Using graph theory, it's easy to show that the above figure has no path that uses each edge exactly once. (Just read euler paths on wiki.) The basic idea is that if there's such a path, every time it visits a vertex, it must leave the vertex using a different edge, and so the degree of every vertex must be even, except for the starting and ending vertex (if they are not same), in which case, they both must have odd degrees. In the above figure, the four ends of the square all have odd degrees, which makes it impossible for such a path to exist.
Actually, the above condition on degrees can be shown to be both necessary and sufficient for an Euler path to exist. So, consider the following figure:
(Same as figure 1, except each line is drawn twice.) Clearly, every vertex has even degree, as I just doubled the degree of every vertex. So, this figure has an Euler cycle.
This gives us a new solution to the age-old problem of drawing figure 1. You draw figure 2 instead, and then use the axiom - "What I can do twice, I can do once" ! Neat huh? My new solution too is a general trick; it can be used to draw anything connected. Ta-da!
The age old solution is - you fold the paper and use the backside as a bridge to get from one point to another. It's a very general trick, and can be used to draw anything without lifting pen from paper.Let's look deeper. Using graph theory, it's easy to show that the above figure has no path that uses each edge exactly once. (Just read euler paths on wiki.) The basic idea is that if there's such a path, every time it visits a vertex, it must leave the vertex using a different edge, and so the degree of every vertex must be even, except for the starting and ending vertex (if they are not same), in which case, they both must have odd degrees. In the above figure, the four ends of the square all have odd degrees, which makes it impossible for such a path to exist.
Actually, the above condition on degrees can be shown to be both necessary and sufficient for an Euler path to exist. So, consider the following figure:
(Same as figure 1, except each line is drawn twice.) Clearly, every vertex has even degree, as I just doubled the degree of every vertex. So, this figure has an Euler cycle.This gives us a new solution to the age-old problem of drawing figure 1. You draw figure 2 instead, and then use the axiom - "What I can do twice, I can do once" ! Neat huh? My new solution too is a general trick; it can be used to draw anything connected. Ta-da!
2 comments:
wtf??
when did u start blogging ?
i knew this puzzle & the paper solution but never thought of relating it to graph theory... awesome.. keep blogging.. :)
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