# Subdividing the sides of a square to approximate a circle

I have read that you can approximate a sphere from an icosahedron by subdividing the faces and normalizing the points created by the subdivisions. Please see the accepted answer of this question: Algorithm for creating spheres.

Wouldn't it follow that you could create a circle by subdividing the sides of a square and normalizing the points? This doesn't seem to work.

Could someone explain to me why?

-
It does work. If it doesn't work for you, you're doing something wrong, but we can't tell what that is unless you tell us what you're doing and how it "doesn't seem to work". "Correct answers are all alike; every incorrect answer is incorrect in its own way." —not Leo Tolstoy –  Rahul Jan 1 '13 at 1:52
Sorry for not being clearer. I drew a square on a piece of graph paper. The vertices are (0, 1), (1, 0), (0, -1), (-1, 0). The midpoint of the segment (0, 1), (1, 0) is (0.5, 0.5). This point is normalized but does not lie on the circle that the four vertices of the square lie on. –  bwroga Jan 1 '13 at 2:12
The norm of $(0.5,0.5)$ is $\sqrt{0.5^2+0.5^2}=1/\sqrt2\ne1$, so the point is not normalized. You normalize it by dividing it by its norm, which gives $(1/\sqrt2,1/\sqrt2)$, whose norm is $1$ and which does lie on the unit circle. –  Rahul Jan 1 '13 at 2:37
I thought that a coordinate was normalized when its components added up to one. Is that incorrect? –  bwroga Jan 1 '13 at 2:52
Yes, that's incorrect. A coordinate is normalized when the squares of its components add up to one. –  mjqxxxx Jan 1 '13 at 3:20

Short answer: Your error is that the norm of a vector $(x,y)$ is not $x+y$, or $\lvert x\rvert+\lvert y\rvert$, but rather $\sqrt{x^2+y^2}$. A vector is normalized if its norm is $1$. If you start with $(\frac12,\frac12)$, its norm is $\sqrt{(\frac12)^2+(\frac12)^2} = \frac1{\sqrt2}$, so you have to divide it by $\frac1{\sqrt2}$ to get the normalized vector $(\frac1{\sqrt2},\frac1{\sqrt2})$.
In a geometry context, when someone says "norm" without further qualification, they invariably mean the Euclidean norm, or the $2$-norm, which for an $n$-dimensional vector $(x_1, x_2, \ldots, x_n)$ is $\sqrt{x_1^2+x_2^2+\cdots+x_n^2}$. Why? Because this is how lengths and distances work. Get a sheet of graph paper and mark the points $(0,0)$ and $(3,4)$. Measure the distance between them with a ruler: it's not $3+4=7$, it's $\sqrt{3^2+4^2}=5$. Or, connect the points $(0,0)$, $(3,0)$, and $(3,4)$: a wild right triangle appears! Cast Pythagoras to defeat the beast.
In general, though, a norm is just — roughly speaking — something that tells you how far away from zero a vector is, and there are lots of possible ways of measuring that. For example, you could just add up the absolute values of the components, $\lvert x_1\rvert+\lvert x_2\rvert+\cdots+\lvert x_n\rvert$, like you thought; this is known as the $1$-norm. Or you could take just the largest component, $\max(\lvert x_1\rvert,\lvert x_2\rvert,\ldots,\lvert x_n\rvert)$, which gives the maximum norm, or the $\infty$-norm. The thing is, though, that because different norms measure distance differently, the "unit spheres" made of points at distance $1$ have different shapes. In particular, the unit sphere for the $1$-norm is a square oriented at $45^\circ$ degrees, which is exactly what you started with, so normalizing the points on it doesn't change anything.