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author | Franklin Wei <frankhwei536@gmail.com> | 2016-11-20 15:16:41 -0500 |
---|---|---|
committer | Franklin Wei <me@fwei.tk> | 2016-12-18 18:13:22 +0100 |
commit | 1a6a8b52f7aa4e2da6f4c34a0c743c760b8cfd99 (patch) | |
tree | 8e7f2d6b0cbdb5d15c13457b2c3e1de69f598440 /apps/plugins/puzzles/penrose.c | |
parent | 3ee79724f6fb033d50e26ef37b33d3f8cedf0c5b (diff) | |
download | rockbox-1a6a8b52f7aa4e2da6f4c34a0c743c760b8cfd99.tar.gz rockbox-1a6a8b52f7aa4e2da6f4c34a0c743c760b8cfd99.zip |
Port of Simon Tatham's Puzzle Collection
Original revision: 5123b1bf68777ffa86e651f178046b26a87cf2d9
MIT Licensed. Some games still crash and others are unplayable due to
issues with controls. Still need a "real" polygon filling algorithm.
Currently builds one plugin per puzzle (about 40 in total, around 100K
each on ARM), but can easily be made to build a single monolithic
overlay (800K or so on ARM).
The following games are at least partially broken for various reasons,
and have been disabled on this commit:
Cube: failed assertion with "Icosahedron" setting
Keen: input issues
Mines: weird stuff happens on target
Palisade: input issues
Solo: input issues, occasional crash on target
Towers: input issues
Undead: input issues
Unequal: input and drawing issues (concave polys)
Untangle: input issues
Features left to do:
- In-game help system
- Figure out the weird bugs
Change-Id: I7c69b6860ab115f973c8d76799502e9bb3d52368
Diffstat (limited to 'apps/plugins/puzzles/penrose.c')
-rw-r--r-- | apps/plugins/puzzles/penrose.c | 629 |
1 files changed, 629 insertions, 0 deletions
diff --git a/apps/plugins/puzzles/penrose.c b/apps/plugins/puzzles/penrose.c new file mode 100644 index 0000000000..839b50a853 --- /dev/null +++ b/apps/plugins/puzzles/penrose.c | |||
@@ -0,0 +1,629 @@ | |||
1 | /* penrose.c | ||
2 | * | ||
3 | * Penrose tile generator. | ||
4 | * | ||
5 | * Uses half-tile technique outlined on: | ||
6 | * | ||
7 | * http://tartarus.org/simon/20110412-penrose/penrose.xhtml | ||
8 | */ | ||
9 | |||
10 | #include "rbassert.h" | ||
11 | #include <string.h> | ||
12 | #include <math.h> | ||
13 | #include <stdio.h> | ||
14 | |||
15 | #include "puzzles.h" /* for malloc routines, and PI */ | ||
16 | #include "penrose.h" | ||
17 | |||
18 | /* ------------------------------------------------------- | ||
19 | * 36-degree basis vector arithmetic routines. | ||
20 | */ | ||
21 | |||
22 | /* Imagine drawing a | ||
23 | * ten-point 'clock face' like this: | ||
24 | * | ||
25 | * -E | ||
26 | * -D | A | ||
27 | * \ | / | ||
28 | * -C. \ | / ,B | ||
29 | * `-._\|/_,-' | ||
30 | * ,-' /|\ `-. | ||
31 | * -B' / | \ `C | ||
32 | * / | \ | ||
33 | * -A | D | ||
34 | * E | ||
35 | * | ||
36 | * In case the ASCII art isn't clear, those are supposed to be ten | ||
37 | * vectors of length 1, all sticking out from the origin at equal | ||
38 | * angular spacing (hence 36 degrees). Our basis vectors are A,B,C,D (I | ||
39 | * choose them to be symmetric about the x-axis so that the final | ||
40 | * translation into 2d coordinates will also be symmetric, which I | ||
41 | * think will avoid minor rounding uglinesses), so our vector | ||
42 | * representation sets | ||
43 | * | ||
44 | * A = (1,0,0,0) | ||
45 | * B = (0,1,0,0) | ||
46 | * C = (0,0,1,0) | ||
47 | * D = (0,0,0,1) | ||
48 | * | ||
49 | * The fifth vector E looks at first glance as if it needs to be | ||
50 | * another basis vector, but in fact it doesn't, because it can be | ||
51 | * represented in terms of the other four. Imagine starting from the | ||
52 | * origin and following the path -A, +B, -C, +D: you'll find you've | ||
53 | * traced four sides of a pentagram, and ended up one E-vector away | ||
54 | * from the origin. So we have | ||
55 | * | ||
56 | * E = (-1,1,-1,1) | ||
57 | * | ||
58 | * This tells us that we can rotate any vector in this system by 36 | ||
59 | * degrees: if we start with a*A + b*B + c*C + d*D, we want to end up | ||
60 | * with a*B + b*C + c*D + d*E, and we substitute our identity for E to | ||
61 | * turn that into a*B + b*C + c*D + d*(-A+B-C+D). In other words, | ||
62 | * | ||
63 | * rotate_one_notch_clockwise(a,b,c,d) = (-d, d+a, -d+b, d+c) | ||
64 | * | ||
65 | * and you can verify for yourself that applying that operation | ||
66 | * repeatedly starting with (1,0,0,0) cycles round ten vectors and | ||
67 | * comes back to where it started. | ||
68 | * | ||
69 | * The other operation that may be required is to construct vectors | ||
70 | * with lengths that are multiples of phi. That can be done by | ||
71 | * observing that the vector C-B is parallel to E and has length 1/phi, | ||
72 | * and the vector D-A is parallel to E and has length phi. So this | ||
73 | * tells us that given any vector, we can construct one which points in | ||
74 | * the same direction and is 1/phi or phi times its length, like this: | ||
75 | * | ||
76 | * divide_by_phi(vector) = rotate(vector, 2) - rotate(vector, 3) | ||
77 | * multiply_by_phi(vector) = rotate(vector, 1) - rotate(vector, 4) | ||
78 | * | ||
79 | * where rotate(vector, n) means applying the above | ||
80 | * rotate_one_notch_clockwise primitive n times. Expanding out the | ||
81 | * applications of rotate gives the following direct representation in | ||
82 | * terms of the vector coordinates: | ||
83 | * | ||
84 | * divide_by_phi(a,b,c,d) = (b-d, c+d-b, a+b-c, c-a) | ||
85 | * multiply_by_phi(a,b,c,d) = (a+b-d, c+d, a+b, c+d-a) | ||
86 | * | ||
87 | * and you can verify for yourself that those two operations are | ||
88 | * inverses of each other (as you'd hope!). | ||
89 | * | ||
90 | * Having done all of this, testing for equality between two vectors is | ||
91 | * a trivial matter of comparing the four integer coordinates. (Which | ||
92 | * it _wouldn't_ have been if we'd kept E as a fifth basis vector, | ||
93 | * because then (-1,1,-1,1,0) and (0,0,0,0,1) would have had to be | ||
94 | * considered identical. So leaving E out is vital.) | ||
95 | */ | ||
96 | |||
97 | struct vector { int a, b, c, d; }; | ||
98 | |||
99 | static vector v_origin(void) | ||
100 | { | ||
101 | vector v; | ||
102 | v.a = v.b = v.c = v.d = 0; | ||
103 | return v; | ||
104 | } | ||
105 | |||
106 | /* We start with a unit vector of B: this means we can easily | ||
107 | * draw an isoceles triangle centred on the X axis. */ | ||
108 | #ifdef TEST_VECTORS | ||
109 | |||
110 | static vector v_unit(void) | ||
111 | { | ||
112 | vector v; | ||
113 | |||
114 | v.b = 1; | ||
115 | v.a = v.c = v.d = 0; | ||
116 | return v; | ||
117 | } | ||
118 | |||
119 | #endif | ||
120 | |||
121 | #define COS54 0.5877852 | ||
122 | #define SIN54 0.8090169 | ||
123 | #define COS18 0.9510565 | ||
124 | #define SIN18 0.3090169 | ||
125 | |||
126 | /* These two are a bit rough-and-ready for now. Note that B/C are | ||
127 | * 18 degrees from the x-axis, and A/D are 54 degrees. */ | ||
128 | double v_x(vector *vs, int i) | ||
129 | { | ||
130 | return (vs[i].a + vs[i].d) * COS54 + | ||
131 | (vs[i].b + vs[i].c) * COS18; | ||
132 | } | ||
133 | |||
134 | double v_y(vector *vs, int i) | ||
135 | { | ||
136 | return (vs[i].a - vs[i].d) * SIN54 + | ||
137 | (vs[i].b - vs[i].c) * SIN18; | ||
138 | |||
139 | } | ||
140 | |||
141 | static vector v_trans(vector v, vector trans) | ||
142 | { | ||
143 | v.a += trans.a; | ||
144 | v.b += trans.b; | ||
145 | v.c += trans.c; | ||
146 | v.d += trans.d; | ||
147 | return v; | ||
148 | } | ||
149 | |||
150 | static vector v_rotate_36(vector v) | ||
151 | { | ||
152 | vector vv; | ||
153 | vv.a = -v.d; | ||
154 | vv.b = v.d + v.a; | ||
155 | vv.c = -v.d + v.b; | ||
156 | vv.d = v.d + v.c; | ||
157 | return vv; | ||
158 | } | ||
159 | |||
160 | static vector v_rotate(vector v, int ang) | ||
161 | { | ||
162 | int i; | ||
163 | |||
164 | assert((ang % 36) == 0); | ||
165 | while (ang < 0) ang += 360; | ||
166 | ang = 360-ang; | ||
167 | for (i = 0; i < (ang/36); i++) | ||
168 | v = v_rotate_36(v); | ||
169 | return v; | ||
170 | } | ||
171 | |||
172 | #ifdef TEST_VECTORS | ||
173 | |||
174 | static vector v_scale(vector v, int sc) | ||
175 | { | ||
176 | v.a *= sc; | ||
177 | v.b *= sc; | ||
178 | v.c *= sc; | ||
179 | v.d *= sc; | ||
180 | return v; | ||
181 | } | ||
182 | |||
183 | #endif | ||
184 | |||
185 | static vector v_growphi(vector v) | ||
186 | { | ||
187 | vector vv; | ||
188 | vv.a = v.a + v.b - v.d; | ||
189 | vv.b = v.c + v.d; | ||
190 | vv.c = v.a + v.b; | ||
191 | vv.d = v.c + v.d - v.a; | ||
192 | return vv; | ||
193 | } | ||
194 | |||
195 | static vector v_shrinkphi(vector v) | ||
196 | { | ||
197 | vector vv; | ||
198 | vv.a = v.b - v.d; | ||
199 | vv.b = v.c + v.d - v.b; | ||
200 | vv.c = v.a + v.b - v.c; | ||
201 | vv.d = v.c - v.a; | ||
202 | return vv; | ||
203 | } | ||
204 | |||
205 | #ifdef TEST_VECTORS | ||
206 | |||
207 | static const char *v_debug(vector v) | ||
208 | { | ||
209 | static char buf[255]; | ||
210 | sprintf(buf, | ||
211 | "(%d,%d,%d,%d)[%2.2f,%2.2f]", | ||
212 | v.a, v.b, v.c, v.d, v_x(&v,0), v_y(&v,0)); | ||
213 | return buf; | ||
214 | } | ||
215 | |||
216 | #endif | ||
217 | |||
218 | /* ------------------------------------------------------- | ||
219 | * Tiling routines. | ||
220 | */ | ||
221 | |||
222 | static vector xform_coord(vector vo, int shrink, vector vtrans, int ang) | ||
223 | { | ||
224 | if (shrink < 0) | ||
225 | vo = v_shrinkphi(vo); | ||
226 | else if (shrink > 0) | ||
227 | vo = v_growphi(vo); | ||
228 | |||
229 | vo = v_rotate(vo, ang); | ||
230 | vo = v_trans(vo, vtrans); | ||
231 | |||
232 | return vo; | ||
233 | } | ||
234 | |||
235 | |||
236 | #define XFORM(n,o,s,a) vs[(n)] = xform_coord(v_edge, (s), vs[(o)], (a)) | ||
237 | |||
238 | static int penrose_p2_small(penrose_state *state, int depth, int flip, | ||
239 | vector v_orig, vector v_edge); | ||
240 | |||
241 | static int penrose_p2_large(penrose_state *state, int depth, int flip, | ||
242 | vector v_orig, vector v_edge) | ||
243 | { | ||
244 | vector vv_orig, vv_edge; | ||
245 | |||
246 | #ifdef DEBUG_PENROSE | ||
247 | { | ||
248 | vector vs[3]; | ||
249 | vs[0] = v_orig; | ||
250 | XFORM(1, 0, 0, 0); | ||
251 | XFORM(2, 0, 0, -36*flip); | ||
252 | |||
253 | state->new_tile(state, vs, 3, depth); | ||
254 | } | ||
255 | #endif | ||
256 | |||
257 | if (flip > 0) { | ||
258 | vector vs[4]; | ||
259 | |||
260 | vs[0] = v_orig; | ||
261 | XFORM(1, 0, 0, -36); | ||
262 | XFORM(2, 0, 0, 0); | ||
263 | XFORM(3, 0, 0, 36); | ||
264 | |||
265 | state->new_tile(state, vs, 4, depth); | ||
266 | } | ||
267 | if (depth >= state->max_depth) return 0; | ||
268 | |||
269 | vv_orig = v_trans(v_orig, v_rotate(v_edge, -36*flip)); | ||
270 | vv_edge = v_rotate(v_edge, 108*flip); | ||
271 | |||
272 | penrose_p2_small(state, depth+1, flip, | ||
273 | v_orig, v_shrinkphi(v_edge)); | ||
274 | |||
275 | penrose_p2_large(state, depth+1, flip, | ||
276 | vv_orig, v_shrinkphi(vv_edge)); | ||
277 | penrose_p2_large(state, depth+1, -flip, | ||
278 | vv_orig, v_shrinkphi(vv_edge)); | ||
279 | |||
280 | return 0; | ||
281 | } | ||
282 | |||
283 | static int penrose_p2_small(penrose_state *state, int depth, int flip, | ||
284 | vector v_orig, vector v_edge) | ||
285 | { | ||
286 | vector vv_orig; | ||
287 | |||
288 | #ifdef DEBUG_PENROSE | ||
289 | { | ||
290 | vector vs[3]; | ||
291 | vs[0] = v_orig; | ||
292 | XFORM(1, 0, 0, 0); | ||
293 | XFORM(2, 0, -1, -36*flip); | ||
294 | |||
295 | state->new_tile(state, vs, 3, depth); | ||
296 | } | ||
297 | #endif | ||
298 | |||
299 | if (flip > 0) { | ||
300 | vector vs[4]; | ||
301 | |||
302 | vs[0] = v_orig; | ||
303 | XFORM(1, 0, 0, -72); | ||
304 | XFORM(2, 0, -1, -36); | ||
305 | XFORM(3, 0, 0, 0); | ||
306 | |||
307 | state->new_tile(state, vs, 4, depth); | ||
308 | } | ||
309 | |||
310 | if (depth >= state->max_depth) return 0; | ||
311 | |||
312 | vv_orig = v_trans(v_orig, v_edge); | ||
313 | |||
314 | penrose_p2_large(state, depth+1, -flip, | ||
315 | v_orig, v_shrinkphi(v_rotate(v_edge, -36*flip))); | ||
316 | |||
317 | penrose_p2_small(state, depth+1, flip, | ||
318 | vv_orig, v_shrinkphi(v_rotate(v_edge, -144*flip))); | ||
319 | |||
320 | return 0; | ||
321 | } | ||
322 | |||
323 | static int penrose_p3_small(penrose_state *state, int depth, int flip, | ||
324 | vector v_orig, vector v_edge); | ||
325 | |||
326 | static int penrose_p3_large(penrose_state *state, int depth, int flip, | ||
327 | vector v_orig, vector v_edge) | ||
328 | { | ||
329 | vector vv_orig; | ||
330 | |||
331 | #ifdef DEBUG_PENROSE | ||
332 | { | ||
333 | vector vs[3]; | ||
334 | vs[0] = v_orig; | ||
335 | XFORM(1, 0, 1, 0); | ||
336 | XFORM(2, 0, 0, -36*flip); | ||
337 | |||
338 | state->new_tile(state, vs, 3, depth); | ||
339 | } | ||
340 | #endif | ||
341 | |||
342 | if (flip > 0) { | ||
343 | vector vs[4]; | ||
344 | |||
345 | vs[0] = v_orig; | ||
346 | XFORM(1, 0, 0, -36); | ||
347 | XFORM(2, 0, 1, 0); | ||
348 | XFORM(3, 0, 0, 36); | ||
349 | |||
350 | state->new_tile(state, vs, 4, depth); | ||
351 | } | ||
352 | if (depth >= state->max_depth) return 0; | ||
353 | |||
354 | vv_orig = v_trans(v_orig, v_edge); | ||
355 | |||
356 | penrose_p3_large(state, depth+1, -flip, | ||
357 | vv_orig, v_shrinkphi(v_rotate(v_edge, 180))); | ||
358 | |||
359 | penrose_p3_small(state, depth+1, flip, | ||
360 | vv_orig, v_shrinkphi(v_rotate(v_edge, -108*flip))); | ||
361 | |||
362 | vv_orig = v_trans(v_orig, v_growphi(v_edge)); | ||
363 | |||
364 | penrose_p3_large(state, depth+1, flip, | ||
365 | vv_orig, v_shrinkphi(v_rotate(v_edge, -144*flip))); | ||
366 | |||
367 | |||
368 | return 0; | ||
369 | } | ||
370 | |||
371 | static int penrose_p3_small(penrose_state *state, int depth, int flip, | ||
372 | vector v_orig, vector v_edge) | ||
373 | { | ||
374 | vector vv_orig; | ||
375 | |||
376 | #ifdef DEBUG_PENROSE | ||
377 | { | ||
378 | vector vs[3]; | ||
379 | vs[0] = v_orig; | ||
380 | XFORM(1, 0, 0, 0); | ||
381 | XFORM(2, 0, 0, -36*flip); | ||
382 | |||
383 | state->new_tile(state, vs, 3, depth); | ||
384 | } | ||
385 | #endif | ||
386 | |||
387 | if (flip > 0) { | ||
388 | vector vs[4]; | ||
389 | |||
390 | vs[0] = v_orig; | ||
391 | XFORM(1, 0, 0, -36); | ||
392 | XFORM(3, 0, 0, 0); | ||
393 | XFORM(2, 3, 0, -36); | ||
394 | |||
395 | state->new_tile(state, vs, 4, depth); | ||
396 | } | ||
397 | if (depth >= state->max_depth) return 0; | ||
398 | |||
399 | /* NB these two are identical to the first two of p3_large. */ | ||
400 | vv_orig = v_trans(v_orig, v_edge); | ||
401 | |||
402 | penrose_p3_large(state, depth+1, -flip, | ||
403 | vv_orig, v_shrinkphi(v_rotate(v_edge, 180))); | ||
404 | |||
405 | penrose_p3_small(state, depth+1, flip, | ||
406 | vv_orig, v_shrinkphi(v_rotate(v_edge, -108*flip))); | ||
407 | |||
408 | return 0; | ||
409 | } | ||
410 | |||
411 | /* ------------------------------------------------------- | ||
412 | * Utility routines. | ||
413 | */ | ||
414 | |||
415 | double penrose_side_length(double start_size, int depth) | ||
416 | { | ||
417 | return start_size / pow(PHI, depth); | ||
418 | } | ||
419 | |||
420 | void penrose_count_tiles(int depth, int *nlarge, int *nsmall) | ||
421 | { | ||
422 | /* Steal sgt's fibonacci thingummy. */ | ||
423 | } | ||
424 | |||
425 | /* | ||
426 | * It turns out that an acute isosceles triangle with sides in ratio 1:phi:phi | ||
427 | * has an incentre which is conveniently 2*phi^-2 of the way from the apex to | ||
428 | * the base. Why's that convenient? Because: if we situate the incentre of the | ||
429 | * triangle at the origin, then we can place the apex at phi^-2 * (B+C), and | ||
430 | * the other two vertices at apex-B and apex-C respectively. So that's an acute | ||
431 | * triangle with its long sides of unit length, covering a circle about the | ||
432 | * origin of radius 1-(2*phi^-2), which is conveniently enough phi^-3. | ||
433 | * | ||
434 | * (later mail: this is an overestimate by about 5%) | ||
435 | */ | ||
436 | |||
437 | int penrose(penrose_state *state, int which, int angle) | ||
438 | { | ||
439 | vector vo = v_origin(); | ||
440 | vector vb = v_origin(); | ||
441 | |||
442 | vo.b = vo.c = -state->start_size; | ||
443 | vo = v_shrinkphi(v_shrinkphi(vo)); | ||
444 | |||
445 | vb.b = state->start_size; | ||
446 | |||
447 | vo = v_rotate(vo, angle); | ||
448 | vb = v_rotate(vb, angle); | ||
449 | |||
450 | if (which == PENROSE_P2) | ||
451 | return penrose_p2_large(state, 0, 1, vo, vb); | ||
452 | else | ||
453 | return penrose_p3_small(state, 0, 1, vo, vb); | ||
454 | } | ||
455 | |||
456 | /* | ||
457 | * We're asked for a MxN grid, which just means a tiling fitting into roughly | ||
458 | * an MxN space in some kind of reasonable unit - say, the side length of the | ||
459 | * two-arrow edges of the tiles. By some reasoning in a previous email, that | ||
460 | * means we want to pick some subarea of a circle of radius 3.11*sqrt(M^2+N^2). | ||
461 | * To cover that circle, we need to subdivide a triangle large enough that it | ||
462 | * contains a circle of that radius. | ||
463 | * | ||
464 | * Hence: start with those three vectors marking triangle vertices, scale them | ||
465 | * all up by phi repeatedly until the radius of the inscribed circle gets | ||
466 | * bigger than the target, and then recurse into that triangle with the same | ||
467 | * recursion depth as the number of times you scaled up. That will give you | ||
468 | * tiles of unit side length, covering a circle big enough that if you randomly | ||
469 | * choose an orientation and coordinates within the circle, you'll be able to | ||
470 | * get any valid piece of Penrose tiling of size MxN. | ||
471 | */ | ||
472 | #define INCIRCLE_RADIUS 0.22426 /* phi^-3 less 5%: see above */ | ||
473 | |||
474 | void penrose_calculate_size(int which, int tilesize, int w, int h, | ||
475 | double *required_radius, int *start_size, int *depth) | ||
476 | { | ||
477 | double rradius, size; | ||
478 | int n = 0; | ||
479 | |||
480 | /* | ||
481 | * Fudge factor to scale P2 and P3 tilings differently. This | ||
482 | * doesn't seem to have much relevance to questions like the | ||
483 | * average number of tiles per unit area; it's just aesthetic. | ||
484 | */ | ||
485 | if (which == PENROSE_P2) | ||
486 | tilesize = tilesize * 3 / 2; | ||
487 | else | ||
488 | tilesize = tilesize * 5 / 4; | ||
489 | |||
490 | rradius = tilesize * 3.11 * sqrt((double)(w*w + h*h)); | ||
491 | size = tilesize; | ||
492 | |||
493 | while ((size * INCIRCLE_RADIUS) < rradius) { | ||
494 | n++; | ||
495 | size = size * PHI; | ||
496 | } | ||
497 | |||
498 | *start_size = (int)size; | ||
499 | *depth = n; | ||
500 | *required_radius = rradius; | ||
501 | } | ||
502 | |||
503 | /* ------------------------------------------------------- | ||
504 | * Test code. | ||
505 | */ | ||
506 | |||
507 | #ifdef TEST_PENROSE | ||
508 | |||
509 | #include <stdio.h> | ||
510 | #include <string.h> | ||
511 | |||
512 | int show_recursion = 0; | ||
513 | int ntiles, nfinal; | ||
514 | |||
515 | int test_cb(penrose_state *state, vector *vs, int n, int depth) | ||
516 | { | ||
517 | int i, xoff = 0, yoff = 0; | ||
518 | double l = penrose_side_length(state->start_size, depth); | ||
519 | double rball = l / 10.0; | ||
520 | const char *col; | ||
521 | |||
522 | ntiles++; | ||
523 | if (state->max_depth == depth) { | ||
524 | col = n == 4 ? "black" : "green"; | ||
525 | nfinal++; | ||
526 | } else { | ||
527 | if (!show_recursion) | ||
528 | return 0; | ||
529 | col = n == 4 ? "red" : "blue"; | ||
530 | } | ||
531 | if (n != 4) yoff = state->start_size; | ||
532 | |||
533 | printf("<polygon points=\""); | ||
534 | for (i = 0; i < n; i++) { | ||
535 | printf("%s%f,%f", (i == 0) ? "" : " ", | ||
536 | v_x(vs, i) + xoff, v_y(vs, i) + yoff); | ||
537 | } | ||
538 | printf("\" style=\"fill: %s; fill-opacity: 0.2; stroke: %s\" />\n", col, col); | ||
539 | printf("<ellipse cx=\"%f\" cy=\"%f\" rx=\"%f\" ry=\"%f\" fill=\"%s\" />", | ||
540 | v_x(vs, 0) + xoff, v_y(vs, 0) + yoff, rball, rball, col); | ||
541 | |||
542 | return 0; | ||
543 | } | ||
544 | |||
545 | void usage_exit(void) | ||
546 | { | ||
547 | fprintf(stderr, "Usage: penrose-test [--recursion] P2|P3 SIZE DEPTH\n"); | ||
548 | exit(1); | ||
549 | } | ||
550 | |||
551 | int main(int argc, char *argv[]) | ||
552 | { | ||
553 | penrose_state ps; | ||
554 | int which = 0; | ||
555 | |||
556 | while (--argc > 0) { | ||
557 | char *p = *++argv; | ||
558 | if (!strcmp(p, "-h") || !strcmp(p, "--help")) { | ||
559 | usage_exit(); | ||
560 | } else if (!strcmp(p, "--recursion")) { | ||
561 | show_recursion = 1; | ||
562 | } else if (*p == '-') { | ||
563 | fprintf(stderr, "Unrecognised option '%s'\n", p); | ||
564 | exit(1); | ||
565 | } else { | ||
566 | break; | ||
567 | } | ||
568 | } | ||
569 | |||
570 | if (argc < 3) usage_exit(); | ||
571 | |||
572 | if (strcmp(argv[0], "P2") == 0) which = PENROSE_P2; | ||
573 | else if (strcmp(argv[0], "P3") == 0) which = PENROSE_P3; | ||
574 | else usage_exit(); | ||
575 | |||
576 | ps.start_size = atoi(argv[1]); | ||
577 | ps.max_depth = atoi(argv[2]); | ||
578 | ps.new_tile = test_cb; | ||
579 | |||
580 | ntiles = nfinal = 0; | ||
581 | |||
582 | printf("\ | ||
583 | <?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"no\"?>\n\ | ||
584 | <!DOCTYPE svg PUBLIC \"-//W3C//DTD SVG 20010904//EN\"\n\ | ||
585 | \"http://www.w3.org/TR/2001/REC-SVG-20010904/DTD/svg10.dtd\">\n\ | ||
586 | \n\ | ||
587 | <svg xmlns=\"http://www.w3.org/2000/svg\"\n\ | ||
588 | xmlns:xlink=\"http://www.w3.org/1999/xlink\">\n\n"); | ||
589 | |||
590 | printf("<g>\n"); | ||
591 | penrose(&ps, which); | ||
592 | printf("</g>\n"); | ||
593 | |||
594 | printf("<!-- %d tiles and %d leaf tiles total -->\n", | ||
595 | ntiles, nfinal); | ||
596 | |||
597 | printf("</svg>"); | ||
598 | |||
599 | return 0; | ||
600 | } | ||
601 | |||
602 | #endif | ||
603 | |||
604 | #ifdef TEST_VECTORS | ||
605 | |||
606 | static void dbgv(const char *msg, vector v) | ||
607 | { | ||
608 | printf("%s: %s\n", msg, v_debug(v)); | ||
609 | } | ||
610 | |||
611 | int main(int argc, const char *argv[]) | ||
612 | { | ||
613 | vector v = v_unit(); | ||
614 | |||
615 | dbgv("unit vector", v); | ||
616 | v = v_rotate(v, 36); | ||
617 | dbgv("rotated 36", v); | ||
618 | v = v_scale(v, 2); | ||
619 | dbgv("scaled x2", v); | ||
620 | v = v_shrinkphi(v); | ||
621 | dbgv("shrunk phi", v); | ||
622 | v = v_rotate(v, -36); | ||
623 | dbgv("rotated -36", v); | ||
624 | |||
625 | return 0; | ||
626 | } | ||
627 | |||
628 | #endif | ||
629 | /* vim: set shiftwidth=4 tabstop=8: */ | ||