Spiral positions in cartesian coordinates
Ludovicus
spiral to its integer cartesian coordinates? Example:
The origin of coordinates is the same zero of spiral.
________________________________________
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| 16 | 15 | 14 |
13 | 12 |
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| 17 | 4 | 3 |
2 | 11 |
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---------------------------------------------------------------------
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| 18 | 5 | 0 |
1 | 10 |
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---------------------------------------------------------------------
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| 19 | 6 | 7 |
8 | 9 |
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---------------------------------------------------------------------
Ludovicus
> Is there an algorithm to transform the numbers of an Ulam
> spiral to its integer cartesian coordinates? Example:
> The origin of coordinates is the same zero of spiral.
>
> | | | | | |
> | | | | | |
> | 16 | 15 | 14 | 13 | 12 |
> | | | | | |
> | | | | | |
> ---------------------------------------------------------------------
> | | | | | |
> | | | | | |
> | 17 | 4 | 3 | 2 | 11 |
> | | | | | |
> | | | | | |
> | | | | | |
> | | | | | |
> | 18 | 5 | 0 | 1 | 10 |
> | | | | | |
> | | | | | |
> | | | | | |
> | | | | | |
> | 19 | 6 | 7 | 8 | 9 |
> | | | | | |
> | | | | | |
> | | | | | |
> | | | | | |
> | 20 | 21 | 22 | 23 | 24 |
> | | | | | |
> | | | | | |
> ---------------------------------------------------------------------
JEMebius
Of course there is such an algorithm. If it is ever to be caught in a single formula, this
formula necessarily involves Floor and/or Ceiling functions.
Hint:
Departing from 0 in the upward left (NW) direction one has the even integral squares.
Departing from 1 in the SO direction one has the odd integral squares.
Departing from 0 in the SW direction one has 0 = 0*1, 6 = 2*3, 20 = 4*5, 42 = 6*7 etc.
Departing from 0 in the NE direction one has 0 = -1*0, 2 = 1*2, 12 = 3*4, 30 = 5*6 etc.
Good luck: Johan E. Mebius
Ludovicus
Thanks for that information. But the question was:
If I have the number on the spiral which are the
cartesian coordinates?.
Otherwise, having the coordinates which is the number?.
Departing from zero in East direcction and counterclockwise.
(Origin of coordinates in center of cell number zero.)
Examples:
n X Y
11 2 1
14 0 2
21 -1 -2
24 2 -2
Ludovicus
James Waldby
> On Jun 1, 6:27 pm, JEMebius wrote:
>> > On Jun 1, 4:10 pm, Ludovicus wrote:
>> >> Is there an algorithm to transform the numbers of an Ulam spiral to
>> >> its integer cartesian coordinates? Example: The origin of
>> >> coordinates is the same zero of spiral.
16__15__14__13__12
17___4___3___2__11
18___5___0___1__10
19___6___7___8___9
20__21__22__23__24 ]
>> Of course there is such an algorithm. If it is ever to be caught in a
>> single formula, this formula necessarily involves Floor and/or Ceiling
>> functions.
...
>> Departing from 0 in the upward left (NW) direction one has the even
>> integral squares. Departing from 1 in the SO direction one has the odd
>> integral squares. Departing from 0 in the SW direction one has 0 = 0*1,
>> 6 = 2*3, 20 = 4*5, 42 = 6*7 etc. Departing from 0 in the NE direction
>> one has 0 = -1*0, 2 = 1*2, 12 = 3*4, 30 = 5*6 etc.
> Thanks for that information. But the question was: If I have the
> number on the spiral which are the cartesian coordinates?.
> Otherwise, having the coordinates which is the number?. [...]
[Eg
n X Y
11 2 1
19 -2 -1
24 2 -2 ]
The rules that Johan mentioned allow calculating such coordinates,
and are clear enough (except for the 'SO' typo that should be 'SE').
From the rules: Given n, compute s = floor(sqrt(n)).
Case 0, s even: let b = s/2.
Case 0.a, n < s*(s+1): x,y = -b, b-n+s^2.
Case 0.b, n >= s*(s+1): x,y = -b+n-s*(s+1), -b.
Case 1, s odd: let b = (s+1)/2.
Case 1.a, n < s*(s+1): (etc)
Two example calcs:
If n=19, s=4, b=2, and 19 < 4*5. By case 0.a,
x = -b = -2, y = b-n+s^2 = 2-19+16 = -1; x,y = -2,-1.
If n=24, s=4, b=2, and 24 > 4*5. By case 0.b,
x = -b+n-s*(s+1) = -2+24-20 = 2, y = -b = -2; x,y = 2,-2.
For the other question, coordinates-to-number, you could
also have 4 cases, N-S-E-W, like (|x| > |y|: x>0; x<0)
and (|y| >= |x|: y>0; y<0).
--
jiw
Rob Johnson
Index to position
-----------------
n -> (x,y)
0 -> (0,0)
for n > 0,
sqrt(n)+1
m = floor( --------- )
2
k = n - 4m(m-1)
1 <= k <= 2m -> (x,y) = (m,k-m)
2m <= k <= 4m -> (x,y) = (3m-k,m)
4m <= k <= 6m -> (x,y) = (-m,5m-k)
6m <= k <= 8m -> (x,y) = (k-7m,-m)
Position to index
-----------------
(x,y) -> n
m = max(|x|,|y|)
x = m -> n = 4m(m-1) + m + y except if y = -m
y = m -> n = 4m(m-1) + 3m - x
x = -m -> n = 4m(m-1) + 5m - y
y = -m -> n = 4m(m-1) + 7m + x
Rob Johnson <r...@trash.whim.org>
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{
float x = abs(p.x);
float y = abs(p.y);
bool q = x > y;
x = q ? x : y;
y = q ? p.x + p.y : p.x - p.y;
y = abs(y) + 4. * x * x + 1.;
x *= 2.;
return q
? (p.x > 0. ? y - x - x : y)
: (p.y > 0. ? y - x : y + x);
}
vec2 inverse_ulam(float u)
{
float r = sqrt(u);
float m = mod(r, 1.);
float p = mod(r * .5, 1.) > .5 ? 1. : -1.;
float s = p * 1.5 - m * p * 2.;
float x = m < .5 ? r * .5 * p : r * s;
float y = m > .5 ? r * .5 * p : r * p - r * s;
return vec2(x, y);
}
I wrote such a thing, though I’ve not made use of it yet. This code is in GLSL, a simple c subset for GPU. You can copy and paste this on glslsandbox.com to test it out.
Rob’s solution looks more numerically stable and elegant than my hacks, although I do enjoy my application of the residual for the inverse.
What are you using this for, Rob?