Wednesday, January 13, 2010

xkcd fixed point problem

Self-Description: The contents of any one panel are dependent on the contents of every panel including itself. The graph of panel dependencies is complete and bidirectional, and each node has a loop. The mouseover text has two hundred and forty-two characters.
I'm a big fan of xkcd. Something that bothers me about these graphs though. If the lines (including the lines used to form the letters) were Euclidean lines (in that they have no width and hence no area), then there should be no ink at all. If you think about the pie chart in the first panel, and say it is x% full, where x%=x/C (where W is the area of the wedge and C is the area of the pie), then if I is the area of the whole image then since C < I, you get W/I < W/C, hence contradiction. So what you need to balance this equation, is that the lines must have area, in fact we can calculate how much. Let L be the area of the lines, then W/C = (W+L)/I which we can solve for W to get

IW = CW + CL
W = CL / (I-C)

The bar chart complicated this a bit, but we assume that away for now.
Post a Comment

Amazon Contextual Product Ads