Interesting question. I'll work on it now.

Interesting challenge. Might work on it later this day.

thanks ^.^

Solution:

if (abs(x)>abs(y))
{
if (x < 0) return 4*x*x-x+y; else return 4*x*x-3*x-y;
}
else
{
if (y < 0) return 4*y*y+y-x; else return 4*y*y+3*y+x;
}

0 lies at co-ordinate (0,0)
1 lies at co-ordinate (1,0)
2 lies at co-ordinate (1,1)
The numbers 1 to 8 inclusive lie on ring 1.
The numbers 9 to 24 inclusive lie on ring 2.

The numbers along the x-axis (that is: 1, 10, 27, 52, 85 et cetera) are defined by 4n^2 - 3n where n is the ring number.

The formula for working out which ring n a number x lies on is as follows: n = ipart( ( 3 + sqrt(9 + 16x) ) / 8 )
Mathematical ipart() is the same as programming floor()

That's the maths behind it anyway. sirXemic has got the implementation.

Wut.

Also, wut @ your maths. I didn't approach it that way =P
EDIT: I just realized your maths is useful for calculating the other way around.

Whoops, I mean ipart not abs.

genius :o

I wonder how to make this work in 3 dimensions.

I forgot that I need to reverse this function... any help?

"return 4*x*x-x+y; else return 4*x*x-3*x-y;"

Just wondering, does the x-3 need to be in parentheses or not? Otherwise, you could have -12*x*x*x-y.

To both of you - work it out yourselves. You're not gonna get better at this unless you do some work.

I wonder how you would make a 3D spiral similar to that 2D one.


Of course not. They would be there if they needed to.

Maybe I could run the function twice:

Ingenious! Juju, I applaud your ingenuity.