Reversing this spiral function [duplicate] - algorithm

Question

I'm using Alberto Santini's solution to this question to get a spiral grid reference based on an items index
Algorithm for iterating over an outward spiral on a discrete 2D grid from the origin
It's not the accepted solution, but it's the best for my needs as it avoids using a loop.
It's working well, but what I want now is to do the inverse. Based on a known x and y coordinate return the index of a location.
This is as a precursor to returning the items surrounding a given location.

Answer 1

Pascal code:
if y * y >= x * x then begin
p := 4 * y * y - y - x;
if y < x then
p := p - 2 * (y - x)
end
else begin
p := 4 * x * x - y - x;
if y < x then
p := p + 2 *(y - x)
end;
Description: Left-upper semi-diagonal (0-4-16-36-64) contains squared layer number (4 * layer^2). External if-statement defines layer and finds (pre-)result for position in corresponding row or column of left-upper semi-plane, and internal if-statement corrects result for mirror position.

Answer 2

I don't know if there is a concise mathematical equation to derive what you want, but I have a solution that computes what you want in O(1) time per query. No loops like you wanted.
My approach :
(i) For any given point (x,y), find the number of points which lie in the square of side length (2*a-1), where a = Max( |x|, |y| ). These are the interior points. i.e, the number of points lying in all spirals NOT including current spiral.
This is nothing but ( 2*a -1 )*( 2*a -1 )
Eg : Consider the following diagram :
y
|
|
16 15 14 13 12
17 4 3 2 11
-- 18 5 0 1 10 --- x
19 6 7 8 9
20 21 22 23 24
|
|
For the point ( 2,1 ), a = 2. The interior points, here are labelled as 0, 1, 2, 3, 4, 5, 6, 7, 8 - The square with edge length 3
(ii) Now compute the points lying on the current spiral. The spiral has 4 "corner" points -
(a) The starting point ( where the current spiral starts )
(b) The point ( a, a )
(c) The point ( -a, a )
(d) The point ( -a, -a )
So, I compute the number of elements lying between each such pair [ i.e, between (a) and (b), (b) and (c), (c) and (d) ], such that all of these fall before the required input point in the spiral sequence. This can be done by simple subtraction of point co-ordinates.
This value, plus the number of interior points will give you the required answer.
I am not sure whether I have explained this very clearly. Do let me know if you require any clarifications or further explanation.
Attached is the JAVA code I wrote to test my logic. I am sorry but it is not very elegant, but it works :P
import java.io.IOException;
import java.util.Scanner;
class Pnt
{
int x, y;
public Pnt( int _x, int _y )
{
x = _x;
y = _y;
}
}
public class Spiral
{
static int interior( Pnt p ) // returns points within interior square of side length MAX( x, y ) - 1
{
int a = Math.max( Math.abs( p.x ), Math.abs( p.y ));
return ( 2*a - 1 )*( 2*a - 1 );
}
static Pnt startPnt( Pnt p ) // first point in that spiral
{
int a = Math.max( Math.abs( p.x ), Math.abs( p.y ));
// last pnt in prev spiral = [ a-1, -( a-1 ) ]
// next pnt = [ a, -( a-1 ) ]
return new Pnt( a, -( a-1 ));
}
static int offSetRow1( Pnt pStart, Pnt p )
{
return ( p.y - pStart.y ) + 1;
}
static int solve( Pnt curr )
{
// check location of curr
// It may lie on 1st row, 2nd row, 3rd or 4th row
int a = Math.max( Math.abs( curr.x ), Math.abs( curr.y ));
int off=0;
int interiorCnt = interior( curr );
Pnt start = startPnt( curr );
if( ( curr.x == a ) && ( curr.y >= start.y ) ) // row 1
{
off = offSetRow1( start, curr );
return off+interiorCnt;
}
if( curr.y == a ) // row 2
{
Pnt start2 = new Pnt( a, a );
int off1 = offSetRow1( start, start2 );
// now add diff in x-coordinates
int off2 = start2.x - curr.x;
off = off1 + off2;
return off+interiorCnt;
}
if( curr.x == -a ) // row 3
{
Pnt start2 = new Pnt( a, a );
int off1 = offSetRow1( start, start2 );
// now add diff in x-coordinates
Pnt start3 = new Pnt( -a, a );
int off2 = start2.x - start3.x;
// now add diff in y co-ordinates
int off3 = start3.y - curr.y;
off = off1 + off2 + off3;
return off+interiorCnt;
}
else // row 4
{
Pnt start2 = new Pnt( a, a );
int off1 = offSetRow1( start, start2 );
// now add diff in x-coordinates
Pnt start3 = new Pnt( -a, a );
int off2 = start2.x - start3.x;
// now add diff in y co-ordinates
int off3 = start3.y - curr.y;
Pnt start4 = new Pnt( -a, -a );
// add diff in x co-ordinates
int off4 = curr.x - start4.x;
off = off1 + off2 + off3 + off4;
return interiorCnt + off;
}
}
public static void main( String[] args ) throws IOException
{
Scanner s = new Scanner( System.in );
while( true )
{
int x = s.nextInt();
int y = s.nextInt();
Pnt curr = new Pnt( x, y );
System.out.println( solve( curr ));
}
}
}

Answer 3

I want to throw in my function since it's a bit more concise than the last solution but more complex than the first.
rather than have the indexes adjacent to each-other, my code opts for loops/layers where the first index of the next loop is always on the same axis.
like so:
23 24 9 10 11 +y
22 8 1 2 12
21 7 0 3 13
20 6 5 4 14
19 18 17 16 15 -y
-x +x
it has set directions and uses the smaller vec2 value as the offset from these NSEW axes
func translate_vector2_to_spiral_index(vec2):
#layer is the ring level the position is on
var layer = max(abs(vec2.x),abs(vec2.y))
if layer == 0:
return 0
#the total interior positions before the edge
var base_index = 0
var i = 0
while i < layer:
base_index += 8 * i
i+=1
var current_layer_total = 8 * i
#non_axis spaces at each corner (not directly any nesw axis)
var non_axis_spaces = (current_layer_total - 4)/4
#direct axes spaces on this layer
var N = 1
var E = N + non_axis_spaces + 1
var S = E + non_axis_spaces + 1
var W = S + non_axis_spaces + 1
var spiral_index = base_index
if abs(vec2.x) > abs(vec2.y):
if vec2.x < 0:
spiral_index+=W
spiral_index += vec2.y
elif vec2.x > 0:
spiral_index+=E
spiral_index -= vec2.y
else:
if vec2.y < 0:
spiral_index+=S
elif vec2.y > 0:
spiral_index+=N
#abs(y) must be equivalent to layers if x is 0
else:
if vec2.y < 0:
spiral_index+=S
spiral_index -= vec2.x
elif vec2.y > 0:
spiral_index
var x = N
x += vec2.x
#if x goes into the negative on the iteration axis (N) it's a subtraction from the layer total
if vec2.x < 0:
x = current_layer_total + 1 + vec2.x
spiral_index += x
else:
if vec2.x < 0:
spiral_index+=W
elif vec2.x > 0:
spiral_index+=E
#abs(x) must be equivalent to layers if y is 0
return spiral_index
there's probably a way to shorten this but i thought to throw this out there.