## November 29, 2008

### The Blogofractal.

Ron said...

Blogonauts? Blogodrome? Where are we, the PolitBlogo? Damn Commies!

Mark O said...

Finally, a Mayan blogger. Quick, what day is it?

dualdiagnosis said...

boobies!!

Eli Blake said...

The blogfractal is (like most true mathematical fractal objects) interdimensional. I calculate its Hausdorff dimension to be 1.47.

Using the formula m^d = n, where m is the factor increase in linear measurement, d is the dimension and n is the number of copies needed to create a larger replica. Hence the dimension of a square is 2, because for example you would need nine squares of equal size to replicate a square, if the linear dimension of the replicated square were increased three times (i.e. 3^2 = 9). Using this formula gives a dimension of 1.89* for the two-dimensional Cantor Set (more commonly known as a Sierpinski carpet.)

It isn't possible to create an exact replica of the Blogofractal, but after clipping the exact picture of the Blogofractal and the Sierpinski carpet and then obtaining an average luminosity for the whole picture by shrinking it down to one pixel size we get luminosity of 124 for the Sierpinski carpet and 179 for the Blogfractal. Subtracting from 240 we get a 'darkness' score of 116 and 61, respectively. This indicates that we have that there is about 1.90 times the degree of second dimensionality in the Sierpinski carpet as in the Blogofractal (a purely one dimensional object would be expected to show up as luminosity 240, pure white on a 2 dimensional drawing when reduced to size 1 pixel.) Dividing the degree of second dimensionality of the Sierpinski carpet (.89) by 1.90 we obtain 0.47.

Hence the dimensionality of the Blogfractal computes to 1.47.

*-- actually, log 8/log 3.

chickenlittle said...

Did you smoke a lot of blog back in the day Eli?

peter hoh said...

Rubik's Blog?

blake said...

I hate the Peano space
And the Koch curve
I fear the Cantor ternary set
Makes me wanna cry...

Michael_H said...

Shoot. I had 1.53, but that was just a quick off-the-top-of-my-head estimate.

Michael_H said...

Checking my calculations, I now see my error. I ascribed a higher algorithm to Lesbians than to Boobies. They should have had identical values.

Yep. 1.47 it is.

Ron said...

Checking my calculations, I now see my error. I ascribed a higher algorithm to Lesbians than to Boobies. They should have had identical values.

Yep. 1.47 it is.

This is the SF-Vegas Field Windage Error, as John Von Neumann named it. Totally understandable.