On a two dimensional grid is there a formula I can use to spiral coordinates in an outward pattern?
Here’s a recipe for finding the coordinates of your position after $n$ steps along the spiral.
It’s simpler to number the positions on the spiral starting at $0$: position $0$ is $\langle 0,0\rangle$, the origin, position $1$ is $\langle 1,0\rangle$, position $2$ is $\langle 1,-1\rangle$, and so on. Using $R,D,L$, and $U$ to indicate steps Right, Down, Left, and Up, respectively, we see the following pattern:
$$RD\,|LLUU\,\|\,RRRDDD\,|LLLLUUUU\,\|\,RRRRRDDDDD\,|LLLLLLUUUUUU\;\|\dots\;,$$
or with exponents to denote repetition, $R^1D^1|L^2U^2\|R^3D^3|L^4U^4\|R^5D^5|L^6U^6\|\dots\;$. I’ll call each $RDLU$ group a block; the first block is the initial $RDLLUU$, and I’ve displayed the first three full blocks above.
Clearly the first $m$ blocks comprise a total of $2\sum_{k=1}^mk=m(m+1)$ steps. It’s also not hard to see that the $k$-th block is $R^{2k+1}D^{2k+1}L^{2k+2}U^{2k+2}$, so that the net effect of the block is to move you one step up and to the left. Since the starting position after $0$ blocks is $\langle 0,0\rangle$, the starting position after $k$ full blocks is $\langle -k,k\rangle$.
Suppose that you’ve taken $n$ steps. There is a unique even integer $2k$ such that $$2k(2k+1)<n\le(2k+2)(2k+3)\;;$$ at this point you’ve gone through $k$ blocks plus an additional $n-2k(2k+1)$ steps. After some straightforward but slightly tedious algebra we find that you’re at
$$\begin{cases} \langle n-4k^2-3k,k\rangle,&\text{if }2k(2k+1)<n\le(2k+1)^2\\ \langle k+1,4k^2+5k+1-n\rangle,&\text{if }(2k+1)^2<n\le 2(k+1)(2k+1)\\ \langle 4k^2+7k+3-n,-k-1\rangle,&\text{if }2(k+1)(2k+1)<n\le4(k+1)^2\\ \langle -k-1,n-4k^2-9k-5\rangle,&\text{if }4(k+1)^2<n\le2(k+1)(2k+3)\;. \end{cases}$$
To find $k$ easily, let $m=\lfloor\sqrt n\rfloor$. If $m$ is odd, $k=\frac12(m-1)$. If $m$ is even, and $n\ge m(m+1)$, then $k=\frac{m}2$; otherwise, $k=\frac{m}2-1$.
Here is some code that finds the $n$-th point in the spiral. Unfortunately it spirals the other way but perhaps it helps anyway.
function spiral(n)
k=ceil((sqrt(n)-1)/2)
t=2*k+1
m=t^2
t=t-1
if n>=m-t then return k-(m-n),-k else m=m-t end
if n>=m-t then return -k,-k+(m-n) else m=m-t end
if n>=m-t then return -k+(m-n),k else return k,k-(m-n-t) end
end
See http://upload.wikimedia.org/wikipedia/commons/1/1d/Ulam_spiral_howto_all_numbers.svg.
If you are looking for a no-if solution and a formula, I was able to find this one:
$A = ||x| - |y|| + |x| + |y|;$
$R = A^2 + sgn(x + y + 0.1)*(A + x - y) + 1;$
$x, y \in \mathbb{Z}$
$sgn$ — sign function
<?php
$n = 4;
$from = -intval($n / 2) - 1;
$to = -$from + ($n % 2) - 2;
for ($x = $to; $x > $from; $x--) {
for ($y = $to; $y > $from; $y--) {
$result = pow((abs(abs($x) - abs($y)) + abs($x) + abs($y)), 2) + abs($x + $y + 0.1) / ($x + $y + 0.1) * (abs(abs($x) - abs($y)) + abs($x) + abs($y) + $x - $y) + 1;
echo $result . "\t";
}
echo "\n";
}
which prints
7 8 9 10
6 1 2 11
5 4 3 12
16 15 14 13
https://repl.it/repls/DarkslategraySteepBluebottle