ref: 3b9cafa09f783ccadda14d11fc8b73dc496368c0
dir: /penrose.c/
/* penrose.c
*
* Penrose tile generator.
*
* Uses half-tile technique outlined on:
*
* http://tartarus.org/simon/20110412-penrose/penrose.xhtml
*/
#include <assert.h>
#include <string.h>
#ifdef NO_TGMATH_H
# include <math.h>
#else
# include <tgmath.h>
#endif
#include <stdio.h>
#include "puzzles.h" /* for malloc routines, and PI */
#include "penrose.h"
/* -------------------------------------------------------
* 36-degree basis vector arithmetic routines.
*/
/* Imagine drawing a
* ten-point 'clock face' like this:
*
* -E
* -D | A
* \ | /
* -C. \ | / ,B
* `-._\|/_,-'
* ,-' /|\ `-.
* -B' / | \ `C
* / | \
* -A | D
* E
*
* In case the ASCII art isn't clear, those are supposed to be ten
* vectors of length 1, all sticking out from the origin at equal
* angular spacing (hence 36 degrees). Our basis vectors are A,B,C,D (I
* choose them to be symmetric about the x-axis so that the final
* translation into 2d coordinates will also be symmetric, which I
* think will avoid minor rounding uglinesses), so our vector
* representation sets
*
* A = (1,0,0,0)
* B = (0,1,0,0)
* C = (0,0,1,0)
* D = (0,0,0,1)
*
* The fifth vector E looks at first glance as if it needs to be
* another basis vector, but in fact it doesn't, because it can be
* represented in terms of the other four. Imagine starting from the
* origin and following the path -A, +B, -C, +D: you'll find you've
* traced four sides of a pentagram, and ended up one E-vector away
* from the origin. So we have
*
* E = (-1,1,-1,1)
*
* This tells us that we can rotate any vector in this system by 36
* degrees: if we start with a*A + b*B + c*C + d*D, we want to end up
* with a*B + b*C + c*D + d*E, and we substitute our identity for E to
* turn that into a*B + b*C + c*D + d*(-A+B-C+D). In other words,
*
* rotate_one_notch_clockwise(a,b,c,d) = (-d, d+a, -d+b, d+c)
*
* and you can verify for yourself that applying that operation
* repeatedly starting with (1,0,0,0) cycles round ten vectors and
* comes back to where it started.
*
* The other operation that may be required is to construct vectors
* with lengths that are multiples of phi. That can be done by
* observing that the vector C-B is parallel to E and has length 1/phi,
* and the vector D-A is parallel to E and has length phi. So this
* tells us that given any vector, we can construct one which points in
* the same direction and is 1/phi or phi times its length, like this:
*
* divide_by_phi(vector) = rotate(vector, 2) - rotate(vector, 3)
* multiply_by_phi(vector) = rotate(vector, 1) - rotate(vector, 4)
*
* where rotate(vector, n) means applying the above
* rotate_one_notch_clockwise primitive n times. Expanding out the
* applications of rotate gives the following direct representation in
* terms of the vector coordinates:
*
* divide_by_phi(a,b,c,d) = (b-d, c+d-b, a+b-c, c-a)
* multiply_by_phi(a,b,c,d) = (a+b-d, c+d, a+b, c+d-a)
*
* and you can verify for yourself that those two operations are
* inverses of each other (as you'd hope!).
*
* Having done all of this, testing for equality between two vectors is
* a trivial matter of comparing the four integer coordinates. (Which
* it _wouldn't_ have been if we'd kept E as a fifth basis vector,
* because then (-1,1,-1,1,0) and (0,0,0,0,1) would have had to be
* considered identical. So leaving E out is vital.)
*/
struct vector { int a, b, c, d; };
static vector v_origin(void)
{
vector v;
v.a = v.b = v.c = v.d = 0;
return v;
}
/* We start with a unit vector of B: this means we can easily
* draw an isoceles triangle centred on the X axis. */
#ifdef TEST_VECTORS
static vector v_unit(void)
{
vector v;
v.b = 1;
v.a = v.c = v.d = 0;
return v;
}
#endif
#define COS54 0.5877852
#define SIN54 0.8090169
#define COS18 0.9510565
#define SIN18 0.3090169
/* These two are a bit rough-and-ready for now. Note that B/C are
* 18 degrees from the x-axis, and A/D are 54 degrees. */
double v_x(vector *vs, int i)
{
return (vs[i].a + vs[i].d) * COS54 +
(vs[i].b + vs[i].c) * COS18;
}
double v_y(vector *vs, int i)
{
return (vs[i].a - vs[i].d) * SIN54 +
(vs[i].b - vs[i].c) * SIN18;
}
static vector v_trans(vector v, vector trans)
{
v.a += trans.a;
v.b += trans.b;
v.c += trans.c;
v.d += trans.d;
return v;
}
static vector v_rotate_36(vector v)
{
vector vv;
vv.a = -v.d;
vv.b = v.d + v.a;
vv.c = -v.d + v.b;
vv.d = v.d + v.c;
return vv;
}
static vector v_rotate(vector v, int ang)
{
int i;
assert((ang % 36) == 0);
while (ang < 0) ang += 360;
ang = 360-ang;
for (i = 0; i < (ang/36); i++)
v = v_rotate_36(v);
return v;
}
#ifdef TEST_VECTORS
static vector v_scale(vector v, int sc)
{
v.a *= sc;
v.b *= sc;
v.c *= sc;
v.d *= sc;
return v;
}
#endif
static vector v_growphi(vector v)
{
vector vv;
vv.a = v.a + v.b - v.d;
vv.b = v.c + v.d;
vv.c = v.a + v.b;
vv.d = v.c + v.d - v.a;
return vv;
}
static vector v_shrinkphi(vector v)
{
vector vv;
vv.a = v.b - v.d;
vv.b = v.c + v.d - v.b;
vv.c = v.a + v.b - v.c;
vv.d = v.c - v.a;
return vv;
}
#ifdef TEST_VECTORS
static const char *v_debug(vector v)
{
static char buf[255];
sprintf(buf,
"(%d,%d,%d,%d)[%2.2f,%2.2f]",
v.a, v.b, v.c, v.d, v_x(&v,0), v_y(&v,0));
return buf;
}
#endif
/* -------------------------------------------------------
* Tiling routines.
*/
static vector xform_coord(vector vo, int shrink, vector vtrans, int ang)
{
if (shrink < 0)
vo = v_shrinkphi(vo);
else if (shrink > 0)
vo = v_growphi(vo);
vo = v_rotate(vo, ang);
vo = v_trans(vo, vtrans);
return vo;
}
#define XFORM(n,o,s,a) vs[(n)] = xform_coord(v_edge, (s), vs[(o)], (a))
static int penrose_p2_small(penrose_state *state, int depth, int flip,
vector v_orig, vector v_edge);
static int penrose_p2_large(penrose_state *state, int depth, int flip,
vector v_orig, vector v_edge)
{
vector vv_orig, vv_edge;
#ifdef DEBUG_PENROSE
{
vector vs[3];
vs[0] = v_orig;
XFORM(1, 0, 0, 0);
XFORM(2, 0, 0, -36*flip);
state->new_tile(state, vs, 3, depth);
}
#endif
if (flip > 0) {
vector vs[4];
vs[0] = v_orig;
XFORM(1, 0, 0, -36);
XFORM(2, 0, 0, 0);
XFORM(3, 0, 0, 36);
state->new_tile(state, vs, 4, depth);
}
if (depth >= state->max_depth) return 0;
vv_orig = v_trans(v_orig, v_rotate(v_edge, -36*flip));
vv_edge = v_rotate(v_edge, 108*flip);
penrose_p2_small(state, depth+1, flip,
v_orig, v_shrinkphi(v_edge));
penrose_p2_large(state, depth+1, flip,
vv_orig, v_shrinkphi(vv_edge));
penrose_p2_large(state, depth+1, -flip,
vv_orig, v_shrinkphi(vv_edge));
return 0;
}
static int penrose_p2_small(penrose_state *state, int depth, int flip,
vector v_orig, vector v_edge)
{
vector vv_orig;
#ifdef DEBUG_PENROSE
{
vector vs[3];
vs[0] = v_orig;
XFORM(1, 0, 0, 0);
XFORM(2, 0, -1, -36*flip);
state->new_tile(state, vs, 3, depth);
}
#endif
if (flip > 0) {
vector vs[4];
vs[0] = v_orig;
XFORM(1, 0, 0, -72);
XFORM(2, 0, -1, -36);
XFORM(3, 0, 0, 0);
state->new_tile(state, vs, 4, depth);
}
if (depth >= state->max_depth) return 0;
vv_orig = v_trans(v_orig, v_edge);
penrose_p2_large(state, depth+1, -flip,
v_orig, v_shrinkphi(v_rotate(v_edge, -36*flip)));
penrose_p2_small(state, depth+1, flip,
vv_orig, v_shrinkphi(v_rotate(v_edge, -144*flip)));
return 0;
}
static int penrose_p3_small(penrose_state *state, int depth, int flip,
vector v_orig, vector v_edge);
static int penrose_p3_large(penrose_state *state, int depth, int flip,
vector v_orig, vector v_edge)
{
vector vv_orig;
#ifdef DEBUG_PENROSE
{
vector vs[3];
vs[0] = v_orig;
XFORM(1, 0, 1, 0);
XFORM(2, 0, 0, -36*flip);
state->new_tile(state, vs, 3, depth);
}
#endif
if (flip > 0) {
vector vs[4];
vs[0] = v_orig;
XFORM(1, 0, 0, -36);
XFORM(2, 0, 1, 0);
XFORM(3, 0, 0, 36);
state->new_tile(state, vs, 4, depth);
}
if (depth >= state->max_depth) return 0;
vv_orig = v_trans(v_orig, v_edge);
penrose_p3_large(state, depth+1, -flip,
vv_orig, v_shrinkphi(v_rotate(v_edge, 180)));
penrose_p3_small(state, depth+1, flip,
vv_orig, v_shrinkphi(v_rotate(v_edge, -108*flip)));
vv_orig = v_trans(v_orig, v_growphi(v_edge));
penrose_p3_large(state, depth+1, flip,
vv_orig, v_shrinkphi(v_rotate(v_edge, -144*flip)));
return 0;
}
static int penrose_p3_small(penrose_state *state, int depth, int flip,
vector v_orig, vector v_edge)
{
vector vv_orig;
#ifdef DEBUG_PENROSE
{
vector vs[3];
vs[0] = v_orig;
XFORM(1, 0, 0, 0);
XFORM(2, 0, 0, -36*flip);
state->new_tile(state, vs, 3, depth);
}
#endif
if (flip > 0) {
vector vs[4];
vs[0] = v_orig;
XFORM(1, 0, 0, -36);
XFORM(3, 0, 0, 0);
XFORM(2, 3, 0, -36);
state->new_tile(state, vs, 4, depth);
}
if (depth >= state->max_depth) return 0;
/* NB these two are identical to the first two of p3_large. */
vv_orig = v_trans(v_orig, v_edge);
penrose_p3_large(state, depth+1, -flip,
vv_orig, v_shrinkphi(v_rotate(v_edge, 180)));
penrose_p3_small(state, depth+1, flip,
vv_orig, v_shrinkphi(v_rotate(v_edge, -108*flip)));
return 0;
}
/* -------------------------------------------------------
* Utility routines.
*/
double penrose_side_length(double start_size, int depth)
{
return start_size / pow(PHI, depth);
}
/*
* It turns out that an acute isosceles triangle with sides in ratio 1:phi:phi
* has an incentre which is conveniently 2*phi^-2 of the way from the apex to
* the base. Why's that convenient? Because: if we situate the incentre of the
* triangle at the origin, then we can place the apex at phi^-2 * (B+C), and
* the other two vertices at apex-B and apex-C respectively. So that's an acute
* triangle with its long sides of unit length, covering a circle about the
* origin of radius 1-(2*phi^-2), which is conveniently enough phi^-3.
*
* (later mail: this is an overestimate by about 5%)
*/
int penrose(penrose_state *state, int which, int angle)
{
vector vo = v_origin();
vector vb = v_origin();
vo.b = vo.c = -state->start_size;
vo = v_shrinkphi(v_shrinkphi(vo));
vb.b = state->start_size;
vo = v_rotate(vo, angle);
vb = v_rotate(vb, angle);
if (which == PENROSE_P2)
return penrose_p2_large(state, 0, 1, vo, vb);
else
return penrose_p3_small(state, 0, 1, vo, vb);
}
/*
* We're asked for a MxN grid, which just means a tiling fitting into roughly
* an MxN space in some kind of reasonable unit - say, the side length of the
* two-arrow edges of the tiles. By some reasoning in a previous email, that
* means we want to pick some subarea of a circle of radius 3.11*sqrt(M^2+N^2).
* To cover that circle, we need to subdivide a triangle large enough that it
* contains a circle of that radius.
*
* Hence: start with those three vectors marking triangle vertices, scale them
* all up by phi repeatedly until the radius of the inscribed circle gets
* bigger than the target, and then recurse into that triangle with the same
* recursion depth as the number of times you scaled up. That will give you
* tiles of unit side length, covering a circle big enough that if you randomly
* choose an orientation and coordinates within the circle, you'll be able to
* get any valid piece of Penrose tiling of size MxN.
*/
#define INCIRCLE_RADIUS 0.22426 /* phi^-3 less 5%: see above */
void penrose_calculate_size(int which, int tilesize, int w, int h,
double *required_radius, int *start_size, int *depth)
{
double rradius, size;
int n = 0;
/*
* Fudge factor to scale P2 and P3 tilings differently. This
* doesn't seem to have much relevance to questions like the
* average number of tiles per unit area; it's just aesthetic.
*/
if (which == PENROSE_P2)
tilesize = tilesize * 3 / 2;
else
tilesize = tilesize * 5 / 4;
rradius = tilesize * 3.11 * sqrt((double)(w*w + h*h));
size = tilesize;
while ((size * INCIRCLE_RADIUS) < rradius) {
n++;
size = size * PHI;
}
*start_size = (int)size;
*depth = n;
*required_radius = rradius;
}