febd9a0f |
1 | /* |
2 | * tree234.c: reasonably generic 2-3-4 tree routines. Currently |
3 | * supports insert, delete, find and iterate operations. |
4 | */ |
5 | |
6 | #include <stdio.h> |
7 | #include <stdlib.h> |
8 | |
9 | #include "tree234.h" |
10 | |
11 | #define mknew(typ) ( (typ *) malloc (sizeof (typ)) ) |
12 | #define sfree free |
13 | |
14 | #ifdef TEST |
15 | #define LOG(x) (printf x) |
16 | #else |
17 | #define LOG(x) |
18 | #endif |
19 | |
20 | struct tree234_Tag { |
21 | node234 *root; |
22 | cmpfn234 cmp; |
23 | }; |
24 | |
25 | struct node234_Tag { |
26 | node234 *parent; |
27 | node234 *kids[4]; |
28 | void *elems[3]; |
29 | }; |
30 | |
31 | /* |
32 | * Create a 2-3-4 tree. |
33 | */ |
34 | tree234 *newtree234(cmpfn234 cmp) { |
35 | tree234 *ret = mknew(tree234); |
36 | LOG(("created tree %p\n", ret)); |
37 | ret->root = NULL; |
38 | ret->cmp = cmp; |
39 | return ret; |
40 | } |
41 | |
42 | /* |
43 | * Free a 2-3-4 tree (not including freeing the elements). |
44 | */ |
45 | static void freenode234(node234 *n) { |
46 | if (!n) |
47 | return; |
48 | freenode234(n->kids[0]); |
49 | freenode234(n->kids[1]); |
50 | freenode234(n->kids[2]); |
51 | freenode234(n->kids[3]); |
52 | sfree(n); |
53 | } |
54 | void freetree234(tree234 *t) { |
55 | freenode234(t->root); |
56 | sfree(t); |
57 | } |
58 | |
59 | /* |
60 | * Add an element e to a 2-3-4 tree t. Returns e on success, or if |
61 | * an existing element compares equal, returns that. |
62 | */ |
63 | void *add234(tree234 *t, void *e) { |
64 | node234 *n, **np, *left, *right; |
65 | void *orig_e = e; |
66 | int c; |
67 | |
68 | LOG(("adding node %p to tree %p\n", e, t)); |
69 | if (t->root == NULL) { |
70 | t->root = mknew(node234); |
71 | t->root->elems[1] = t->root->elems[2] = NULL; |
72 | t->root->kids[0] = t->root->kids[1] = NULL; |
73 | t->root->kids[2] = t->root->kids[3] = NULL; |
74 | t->root->parent = NULL; |
75 | t->root->elems[0] = e; |
76 | LOG((" created root %p\n", t->root)); |
77 | return orig_e; |
78 | } |
79 | |
80 | np = &t->root; |
81 | while (*np) { |
82 | n = *np; |
83 | LOG((" node %p: %p [%p] %p [%p] %p [%p] %p\n", |
84 | n, n->kids[0], n->elems[0], n->kids[1], n->elems[1], |
85 | n->kids[2], n->elems[2], n->kids[3])); |
86 | if ((c = t->cmp(e, n->elems[0])) < 0) |
87 | np = &n->kids[0]; |
88 | else if (c == 0) |
89 | return n->elems[0]; /* already exists */ |
90 | else if (n->elems[1] == NULL || (c = t->cmp(e, n->elems[1])) < 0) |
91 | np = &n->kids[1]; |
92 | else if (c == 0) |
93 | return n->elems[1]; /* already exists */ |
94 | else if (n->elems[2] == NULL || (c = t->cmp(e, n->elems[2])) < 0) |
95 | np = &n->kids[2]; |
96 | else if (c == 0) |
97 | return n->elems[2]; /* already exists */ |
98 | else |
99 | np = &n->kids[3]; |
100 | LOG((" moving to child %d (%p)\n", np - n->kids, *np)); |
101 | } |
102 | |
103 | /* |
104 | * We need to insert the new element in n at position np. |
105 | */ |
106 | left = NULL; |
107 | right = NULL; |
108 | while (n) { |
109 | LOG((" at %p: %p [%p] %p [%p] %p [%p] %p\n", |
110 | n, n->kids[0], n->elems[0], n->kids[1], n->elems[1], |
111 | n->kids[2], n->elems[2], n->kids[3])); |
112 | LOG((" need to insert %p [%p] %p at position %d\n", |
113 | left, e, right, np - n->kids)); |
114 | if (n->elems[1] == NULL) { |
115 | /* |
116 | * Insert in a 2-node; simple. |
117 | */ |
118 | if (np == &n->kids[0]) { |
119 | LOG((" inserting on left of 2-node\n")); |
120 | n->kids[2] = n->kids[1]; |
121 | n->elems[1] = n->elems[0]; |
122 | n->kids[1] = right; |
123 | n->elems[0] = e; |
124 | n->kids[0] = left; |
125 | } else { /* np == &n->kids[1] */ |
126 | LOG((" inserting on right of 2-node\n")); |
127 | n->kids[2] = right; |
128 | n->elems[1] = e; |
129 | n->kids[1] = left; |
130 | } |
131 | if (n->kids[0]) n->kids[0]->parent = n; |
132 | if (n->kids[1]) n->kids[1]->parent = n; |
133 | if (n->kids[2]) n->kids[2]->parent = n; |
134 | LOG((" done\n")); |
135 | break; |
136 | } else if (n->elems[2] == NULL) { |
137 | /* |
138 | * Insert in a 3-node; simple. |
139 | */ |
140 | if (np == &n->kids[0]) { |
141 | LOG((" inserting on left of 3-node\n")); |
142 | n->kids[3] = n->kids[2]; |
143 | n->elems[2] = n->elems[1]; |
144 | n->kids[2] = n->kids[1]; |
145 | n->elems[1] = n->elems[0]; |
146 | n->kids[1] = right; |
147 | n->elems[0] = e; |
148 | n->kids[0] = left; |
149 | } else if (np == &n->kids[1]) { |
150 | LOG((" inserting in middle of 3-node\n")); |
151 | n->kids[3] = n->kids[2]; |
152 | n->elems[2] = n->elems[1]; |
153 | n->kids[2] = right; |
154 | n->elems[1] = e; |
155 | n->kids[1] = left; |
156 | } else { /* np == &n->kids[2] */ |
157 | LOG((" inserting on right of 3-node\n")); |
158 | n->kids[3] = right; |
159 | n->elems[2] = e; |
160 | n->kids[2] = left; |
161 | } |
162 | if (n->kids[0]) n->kids[0]->parent = n; |
163 | if (n->kids[1]) n->kids[1]->parent = n; |
164 | if (n->kids[2]) n->kids[2]->parent = n; |
165 | if (n->kids[3]) n->kids[3]->parent = n; |
166 | LOG((" done\n")); |
167 | break; |
168 | } else { |
169 | node234 *m = mknew(node234); |
170 | m->parent = n->parent; |
171 | LOG((" splitting a 4-node; created new node %p\n", m)); |
172 | /* |
173 | * Insert in a 4-node; split into a 2-node and a |
174 | * 3-node, and move focus up a level. |
175 | * |
176 | * I don't think it matters which way round we put the |
177 | * 2 and the 3. For simplicity, we'll put the 3 first |
178 | * always. |
179 | */ |
180 | if (np == &n->kids[0]) { |
181 | m->kids[0] = left; |
182 | m->elems[0] = e; |
183 | m->kids[1] = right; |
184 | m->elems[1] = n->elems[0]; |
185 | m->kids[2] = n->kids[1]; |
186 | e = n->elems[1]; |
187 | n->kids[0] = n->kids[2]; |
188 | n->elems[0] = n->elems[2]; |
189 | n->kids[1] = n->kids[3]; |
190 | } else if (np == &n->kids[1]) { |
191 | m->kids[0] = n->kids[0]; |
192 | m->elems[0] = n->elems[0]; |
193 | m->kids[1] = left; |
194 | m->elems[1] = e; |
195 | m->kids[2] = right; |
196 | e = n->elems[1]; |
197 | n->kids[0] = n->kids[2]; |
198 | n->elems[0] = n->elems[2]; |
199 | n->kids[1] = n->kids[3]; |
200 | } else if (np == &n->kids[2]) { |
201 | m->kids[0] = n->kids[0]; |
202 | m->elems[0] = n->elems[0]; |
203 | m->kids[1] = n->kids[1]; |
204 | m->elems[1] = n->elems[1]; |
205 | m->kids[2] = left; |
206 | /* e = e; */ |
207 | n->kids[0] = right; |
208 | n->elems[0] = n->elems[2]; |
209 | n->kids[1] = n->kids[3]; |
210 | } else { /* np == &n->kids[3] */ |
211 | m->kids[0] = n->kids[0]; |
212 | m->elems[0] = n->elems[0]; |
213 | m->kids[1] = n->kids[1]; |
214 | m->elems[1] = n->elems[1]; |
215 | m->kids[2] = n->kids[2]; |
216 | n->kids[0] = left; |
217 | n->elems[0] = e; |
218 | n->kids[1] = right; |
219 | e = n->elems[2]; |
220 | } |
221 | m->kids[3] = n->kids[3] = n->kids[2] = NULL; |
222 | m->elems[2] = n->elems[2] = n->elems[1] = NULL; |
223 | if (m->kids[0]) m->kids[0]->parent = m; |
224 | if (m->kids[1]) m->kids[1]->parent = m; |
225 | if (m->kids[2]) m->kids[2]->parent = m; |
226 | if (n->kids[0]) n->kids[0]->parent = n; |
227 | if (n->kids[1]) n->kids[1]->parent = n; |
228 | LOG((" left (%p): %p [%p] %p [%p] %p\n", m, |
229 | m->kids[0], m->elems[0], |
230 | m->kids[1], m->elems[1], |
231 | m->kids[2])); |
232 | LOG((" right (%p): %p [%p] %p\n", n, |
233 | n->kids[0], n->elems[0], |
234 | n->kids[1])); |
235 | left = m; |
236 | right = n; |
237 | } |
238 | if (n->parent) |
239 | np = (n->parent->kids[0] == n ? &n->parent->kids[0] : |
240 | n->parent->kids[1] == n ? &n->parent->kids[1] : |
241 | n->parent->kids[2] == n ? &n->parent->kids[2] : |
242 | &n->parent->kids[3]); |
243 | n = n->parent; |
244 | } |
245 | |
246 | /* |
247 | * If we've come out of here by `break', n will still be |
248 | * non-NULL and we've finished. If we've come here because n is |
249 | * NULL, we need to create a new root for the tree because the |
250 | * old one has just split into two. |
251 | */ |
252 | if (!n) { |
253 | LOG((" root is overloaded, split into two\n")); |
254 | t->root = mknew(node234); |
255 | t->root->kids[0] = left; |
256 | t->root->elems[0] = e; |
257 | t->root->kids[1] = right; |
258 | t->root->elems[1] = NULL; |
259 | t->root->kids[2] = NULL; |
260 | t->root->elems[2] = NULL; |
261 | t->root->kids[3] = NULL; |
262 | t->root->parent = NULL; |
263 | if (t->root->kids[0]) t->root->kids[0]->parent = t->root; |
264 | if (t->root->kids[1]) t->root->kids[1]->parent = t->root; |
265 | LOG((" new root is %p [%p] %p\n", |
266 | t->root->kids[0], t->root->elems[0], t->root->kids[1])); |
267 | } |
268 | |
269 | return orig_e; |
270 | } |
271 | |
272 | /* |
273 | * Find an element e in a 2-3-4 tree t. Returns NULL if not found. |
274 | * e is always passed as the first argument to cmp, so cmp can be |
275 | * an asymmetric function if desired. cmp can also be passed as |
276 | * NULL, in which case the compare function from the tree proper |
277 | * will be used. |
278 | */ |
279 | void *find234(tree234 *t, void *e, cmpfn234 cmp) { |
280 | node234 *n; |
281 | int c; |
282 | |
283 | if (t->root == NULL) |
284 | return NULL; |
285 | |
286 | if (cmp == NULL) |
287 | cmp = t->cmp; |
288 | |
289 | n = t->root; |
290 | while (n) { |
81d77872 |
291 | if ( (c = cmp(e, n->elems[0])) < 0) |
febd9a0f |
292 | n = n->kids[0]; |
293 | else if (c == 0) |
294 | return n->elems[0]; |
81d77872 |
295 | else if (n->elems[1] == NULL || (c = cmp(e, n->elems[1])) < 0) |
febd9a0f |
296 | n = n->kids[1]; |
297 | else if (c == 0) |
298 | return n->elems[1]; |
81d77872 |
299 | else if (n->elems[2] == NULL || (c = cmp(e, n->elems[2])) < 0) |
febd9a0f |
300 | n = n->kids[2]; |
301 | else if (c == 0) |
302 | return n->elems[2]; |
303 | else |
304 | n = n->kids[3]; |
305 | } |
306 | |
307 | /* |
308 | * We've found our way to the bottom of the tree and we know |
309 | * where we would insert this node if we wanted to. But it |
310 | * isn't there. |
311 | */ |
312 | return NULL; |
313 | } |
314 | |
315 | /* |
316 | * Delete an element e in a 2-3-4 tree. Does not free the element, |
317 | * merely removes all links to it from the tree nodes. |
318 | */ |
81d77872 |
319 | void del234(tree234 *t, void *e) { |
febd9a0f |
320 | node234 *n; |
321 | int ei = -1; |
322 | |
323 | n = t->root; |
324 | LOG(("deleting %p from tree %p\n", e, t)); |
325 | while (1) { |
326 | while (n) { |
327 | int c; |
328 | int ki; |
329 | node234 *sub; |
330 | |
331 | LOG((" node %p: %p [%p] %p [%p] %p [%p] %p\n", |
332 | n, n->kids[0], n->elems[0], n->kids[1], n->elems[1], |
333 | n->kids[2], n->elems[2], n->kids[3])); |
334 | if ((c = t->cmp(e, n->elems[0])) < 0) { |
335 | ki = 0; |
336 | } else if (c == 0) { |
337 | ei = 0; break; |
338 | } else if (n->elems[1] == NULL || (c = t->cmp(e, n->elems[1])) < 0) { |
339 | ki = 1; |
340 | } else if (c == 0) { |
341 | ei = 1; break; |
342 | } else if (n->elems[2] == NULL || (c = t->cmp(e, n->elems[2])) < 0) { |
343 | ki = 2; |
344 | } else if (c == 0) { |
345 | ei = 2; break; |
346 | } else { |
347 | ki = 3; |
348 | } |
349 | /* |
350 | * Recurse down to subtree ki. If it has only one element, |
351 | * we have to do some transformation to start with. |
352 | */ |
353 | LOG((" moving to subtree %d\n", ki)); |
354 | sub = n->kids[ki]; |
355 | if (!sub->elems[1]) { |
356 | LOG((" subtree has only one element!\n", ki)); |
357 | if (ki > 0 && n->kids[ki-1]->elems[1]) { |
358 | /* |
359 | * Case 3a, left-handed variant. Child ki has |
360 | * only one element, but child ki-1 has two or |
361 | * more. So we need to move a subtree from ki-1 |
362 | * to ki. |
363 | * |
364 | * . C . . B . |
365 | * / \ -> / \ |
366 | * [more] a A b B c d D e [more] a A b c C d D e |
367 | */ |
368 | node234 *sib = n->kids[ki-1]; |
369 | int lastelem = (sib->elems[2] ? 2 : |
370 | sib->elems[1] ? 1 : 0); |
371 | sub->kids[2] = sub->kids[1]; |
372 | sub->elems[1] = sub->elems[0]; |
373 | sub->kids[1] = sub->kids[0]; |
374 | sub->elems[0] = n->elems[ki-1]; |
375 | sub->kids[0] = sib->kids[lastelem+1]; |
100122a9 |
376 | if (sub->kids[0]) sub->kids[0]->parent = sub; |
febd9a0f |
377 | n->elems[ki-1] = sib->elems[lastelem]; |
378 | sib->kids[lastelem+1] = NULL; |
379 | sib->elems[lastelem] = NULL; |
380 | LOG((" case 3a left\n")); |
381 | } else if (ki < 3 && n->kids[ki+1] && |
382 | n->kids[ki+1]->elems[1]) { |
383 | /* |
384 | * Case 3a, right-handed variant. ki has only |
385 | * one element but ki+1 has two or more. Move a |
386 | * subtree from ki+1 to ki. |
387 | * |
388 | * . B . . C . |
389 | * / \ -> / \ |
390 | * a A b c C d D e [more] a A b B c d D e [more] |
391 | */ |
392 | node234 *sib = n->kids[ki+1]; |
393 | int j; |
394 | sub->elems[1] = n->elems[ki]; |
395 | sub->kids[2] = sib->kids[0]; |
100122a9 |
396 | if (sub->kids[2]) sub->kids[2]->parent = sub; |
febd9a0f |
397 | n->elems[ki] = sib->elems[0]; |
398 | sib->kids[0] = sib->kids[1]; |
399 | for (j = 0; j < 2 && sib->elems[j+1]; j++) { |
400 | sib->kids[j+1] = sib->kids[j+2]; |
401 | sib->elems[j] = sib->elems[j+1]; |
402 | } |
403 | sib->kids[j+1] = NULL; |
404 | sib->elems[j] = NULL; |
405 | LOG((" case 3a right\n")); |
406 | } else { |
407 | /* |
408 | * Case 3b. ki has only one element, and has no |
409 | * neighbour with more than one. So pick a |
410 | * neighbour and merge it with ki, taking an |
411 | * element down from n to go in the middle. |
412 | * |
413 | * . B . . |
414 | * / \ -> | |
415 | * a A b c C d a A b B c C d |
416 | * |
417 | * (Since at all points we have avoided |
418 | * descending to a node with only one element, |
419 | * we can be sure that n is not reduced to |
420 | * nothingness by this move, _unless_ it was |
421 | * the very first node, ie the root of the |
422 | * tree. In that case we remove the now-empty |
423 | * root and replace it with its single large |
424 | * child as shown.) |
425 | */ |
426 | node234 *sib; |
427 | int j; |
428 | |
429 | if (ki > 0) |
430 | ki--; |
431 | sib = n->kids[ki]; |
432 | sub = n->kids[ki+1]; |
433 | |
434 | sub->kids[3] = sub->kids[1]; |
435 | sub->elems[2] = sub->elems[0]; |
436 | sub->kids[2] = sub->kids[0]; |
437 | sub->elems[1] = n->elems[ki]; |
438 | sub->kids[1] = sib->kids[1]; |
100122a9 |
439 | if (sub->kids[1]) sub->kids[1]->parent = sub; |
febd9a0f |
440 | sub->elems[0] = sib->elems[0]; |
441 | sub->kids[0] = sib->kids[0]; |
100122a9 |
442 | if (sub->kids[0]) sub->kids[0]->parent = sub; |
febd9a0f |
443 | |
444 | sfree(sib); |
445 | |
446 | /* |
447 | * That's built the big node in sub. Now we |
448 | * need to remove the reference to sib in n. |
449 | */ |
450 | for (j = ki; j < 3 && n->kids[j+1]; j++) { |
451 | n->kids[j] = n->kids[j+1]; |
452 | n->elems[j] = j<2 ? n->elems[j+1] : NULL; |
453 | } |
454 | n->kids[j] = NULL; |
455 | if (j < 3) n->elems[j] = NULL; |
2d56b16f |
456 | LOG((" case 3b ki=%d\n", ki)); |
febd9a0f |
457 | |
458 | if (!n->elems[0]) { |
459 | /* |
460 | * The root is empty and needs to be |
461 | * removed. |
462 | */ |
463 | LOG((" shifting root!\n")); |
464 | t->root = sub; |
465 | sub->parent = NULL; |
466 | sfree(n); |
467 | } |
468 | } |
469 | } |
470 | n = sub; |
471 | } |
472 | if (ei==-1) |
473 | return; /* nothing to do; `already removed' */ |
474 | |
475 | /* |
476 | * Treat special case: this is the one remaining item in |
477 | * the tree. n is the tree root (no parent), has one |
478 | * element (no elems[1]), and has no kids (no kids[0]). |
479 | */ |
480 | if (!n->parent && !n->elems[1] && !n->kids[0]) { |
481 | LOG((" removed last element in tree\n")); |
482 | sfree(n); |
483 | t->root = NULL; |
484 | return; |
485 | } |
486 | |
487 | /* |
488 | * Now we have the element we want, as n->elems[ei], and we |
489 | * have also arranged for that element not to be the only |
490 | * one in its node. So... |
491 | */ |
492 | |
493 | if (!n->kids[0] && n->elems[1]) { |
494 | /* |
495 | * Case 1. n is a leaf node with more than one element, |
496 | * so it's _really easy_. Just delete the thing and |
497 | * we're done. |
498 | */ |
499 | int i; |
500 | LOG((" case 1\n")); |
a4a19e73 |
501 | for (i = ei; i < 2 && n->elems[i+1]; i++) |
febd9a0f |
502 | n->elems[i] = n->elems[i+1]; |
503 | n->elems[i] = NULL; |
504 | return; /* finished! */ |
505 | } else if (n->kids[ei]->elems[1]) { |
506 | /* |
507 | * Case 2a. n is an internal node, and the root of the |
508 | * subtree to the left of e has more than one element. |
509 | * So find the predecessor p to e (ie the largest node |
510 | * in that subtree), place it where e currently is, and |
511 | * then start the deletion process over again on the |
512 | * subtree with p as target. |
513 | */ |
514 | node234 *m = n->kids[ei]; |
515 | void *target; |
516 | LOG((" case 2a\n")); |
517 | while (m->kids[0]) { |
518 | m = (m->kids[3] ? m->kids[3] : |
519 | m->kids[2] ? m->kids[2] : |
520 | m->kids[1] ? m->kids[1] : m->kids[0]); |
521 | } |
522 | target = (m->elems[2] ? m->elems[2] : |
523 | m->elems[1] ? m->elems[1] : m->elems[0]); |
524 | n->elems[ei] = target; |
525 | n = n->kids[ei]; |
526 | e = target; |
527 | } else if (n->kids[ei+1]->elems[1]) { |
528 | /* |
529 | * Case 2b, symmetric to 2a but s/left/right/ and |
530 | * s/predecessor/successor/. (And s/largest/smallest/). |
531 | */ |
532 | node234 *m = n->kids[ei+1]; |
533 | void *target; |
534 | LOG((" case 2b\n")); |
535 | while (m->kids[0]) { |
536 | m = m->kids[0]; |
537 | } |
538 | target = m->elems[0]; |
539 | n->elems[ei] = target; |
540 | n = n->kids[ei+1]; |
541 | e = target; |
542 | } else { |
543 | /* |
544 | * Case 2c. n is an internal node, and the subtrees to |
545 | * the left and right of e both have only one element. |
546 | * So combine the two subnodes into a single big node |
547 | * with their own elements on the left and right and e |
548 | * in the middle, then restart the deletion process on |
549 | * that subtree, with e still as target. |
550 | */ |
551 | node234 *a = n->kids[ei], *b = n->kids[ei+1]; |
552 | int j; |
553 | |
554 | LOG((" case 2c\n")); |
555 | a->elems[1] = n->elems[ei]; |
556 | a->kids[2] = b->kids[0]; |
100122a9 |
557 | if (a->kids[2]) a->kids[2]->parent = a; |
febd9a0f |
558 | a->elems[2] = b->elems[0]; |
559 | a->kids[3] = b->kids[1]; |
100122a9 |
560 | if (a->kids[3]) a->kids[3]->parent = a; |
febd9a0f |
561 | sfree(b); |
562 | /* |
563 | * That's built the big node in a, and destroyed b. Now |
564 | * remove the reference to b (and e) in n. |
565 | */ |
566 | for (j = ei; j < 2 && n->elems[j+1]; j++) { |
567 | n->elems[j] = n->elems[j+1]; |
568 | n->kids[j+1] = n->kids[j+2]; |
569 | } |
570 | n->elems[j] = NULL; |
571 | n->kids[j+1] = NULL; |
e9e9556d |
572 | /* |
573 | * It's possible, in this case, that we've just removed |
574 | * the only element in the root of the tree. If so, |
575 | * shift the root. |
576 | */ |
577 | if (n->elems[0] == NULL) { |
578 | LOG((" shifting root!\n")); |
579 | t->root = a; |
580 | a->parent = NULL; |
581 | sfree(n); |
582 | } |
febd9a0f |
583 | /* |
584 | * Now go round the deletion process again, with n |
585 | * pointing at the new big node and e still the same. |
586 | */ |
587 | n = a; |
588 | } |
589 | } |
590 | } |
591 | |
592 | /* |
593 | * Iterate over the elements of a tree234, in order. |
594 | */ |
595 | void *first234(tree234 *t, enum234 *e) { |
596 | node234 *n = t->root; |
597 | if (!n) |
598 | return NULL; |
599 | while (n->kids[0]) |
600 | n = n->kids[0]; |
601 | e->node = n; |
602 | e->posn = 0; |
603 | return n->elems[0]; |
604 | } |
605 | |
606 | void *next234(enum234 *e) { |
607 | node234 *n = e->node; |
608 | int pos = e->posn; |
609 | |
610 | if (n->kids[pos+1]) { |
611 | n = n->kids[pos+1]; |
612 | while (n->kids[0]) |
613 | n = n->kids[0]; |
614 | e->node = n; |
615 | e->posn = 0; |
616 | return n->elems[0]; |
617 | } |
618 | |
6aff1005 |
619 | if (pos < 2 && n->elems[pos+1]) { |
620 | e->posn = pos+1; |
621 | return n->elems[e->posn]; |
febd9a0f |
622 | } |
623 | |
624 | do { |
625 | node234 *nn = n->parent; |
626 | if (nn == NULL) |
627 | return NULL; /* end of tree */ |
628 | pos = (nn->kids[0] == n ? 0 : |
629 | nn->kids[1] == n ? 1 : |
630 | nn->kids[2] == n ? 2 : 3); |
631 | n = nn; |
632 | } while (pos == 3 || n->kids[pos+1] == NULL); |
633 | |
634 | e->node = n; |
635 | e->posn = pos; |
636 | return n->elems[pos]; |
637 | } |
638 | |
639 | #ifdef TEST |
640 | |
2d56b16f |
641 | /* |
642 | * Test code for the 2-3-4 tree. This code maintains an alternative |
643 | * representation of the data in the tree, in an array (using the |
644 | * obvious and slow insert and delete functions). After each tree |
7aa7c43a |
645 | * operation, the verify() function is called, which ensures all |
646 | * the tree properties are preserved (node->child->parent always |
647 | * equals node; number of kids == 0 or number of elements + 1; |
648 | * ordering property between elements of a node and elements of its |
649 | * children is preserved; tree has the same depth everywhere; every |
650 | * node has at least one element) and also ensures the list |
651 | * represented by the tree is the same list it should be. (This |
652 | * last check also verifies the ordering properties, because the |
653 | * `same list it should be' is by definition correctly ordered. It |
654 | * also ensures all nodes are distinct, because the enum functions |
655 | * would get caught in a loop if not.) |
2d56b16f |
656 | */ |
657 | |
658 | #include <stdarg.h> |
659 | |
660 | /* |
661 | * Error reporting function. |
662 | */ |
663 | void error(char *fmt, ...) { |
664 | va_list ap; |
665 | printf("ERROR: "); |
666 | va_start(ap, fmt); |
667 | vfprintf(stdout, fmt, ap); |
668 | va_end(ap); |
669 | printf("\n"); |
670 | } |
671 | |
672 | /* The array representation of the data. */ |
673 | void **array; |
674 | int arraylen, arraysize; |
675 | cmpfn234 cmp; |
676 | |
677 | /* The tree representation of the same data. */ |
678 | tree234 *tree; |
679 | |
680 | typedef struct { |
681 | int treedepth; |
682 | int elemcount; |
683 | } chkctx; |
684 | |
685 | void chknode(chkctx *ctx, int level, node234 *node, |
686 | void *lowbound, void *highbound) { |
687 | int nkids, nelems; |
688 | int i; |
689 | |
690 | /* Count the non-NULL kids. */ |
691 | for (nkids = 0; nkids < 4 && node->kids[nkids]; nkids++); |
692 | /* Ensure no kids beyond the first NULL are non-NULL. */ |
693 | for (i = nkids; i < 4; i++) |
694 | if (node->kids[i]) { |
695 | error("node %p: nkids=%d but kids[%d] non-NULL", |
696 | node, nkids, i); |
697 | } |
698 | |
699 | /* Count the non-NULL elements. */ |
700 | for (nelems = 0; nelems < 3 && node->elems[nelems]; nelems++); |
701 | /* Ensure no elements beyond the first NULL are non-NULL. */ |
702 | for (i = nelems; i < 3; i++) |
703 | if (node->elems[i]) { |
704 | error("node %p: nelems=%d but elems[%d] non-NULL", |
705 | node, nelems, i); |
706 | } |
707 | |
708 | if (nkids == 0) { |
709 | /* |
710 | * If nkids==0, this is a leaf node; verify that the tree |
711 | * depth is the same everywhere. |
712 | */ |
713 | if (ctx->treedepth < 0) |
714 | ctx->treedepth = level; /* we didn't know the depth yet */ |
715 | else if (ctx->treedepth != level) |
716 | error("node %p: leaf at depth %d, previously seen depth %d", |
717 | node, level, ctx->treedepth); |
718 | } else { |
719 | /* |
720 | * If nkids != 0, then it should be nelems+1, unless nelems |
721 | * is 0 in which case nkids should also be 0 (and so we |
722 | * shouldn't be in this condition at all). |
723 | */ |
724 | int shouldkids = (nelems ? nelems+1 : 0); |
725 | if (nkids != shouldkids) { |
726 | error("node %p: %d elems should mean %d kids but has %d", |
727 | node, nelems, shouldkids, nkids); |
728 | } |
729 | } |
730 | |
731 | /* |
732 | * nelems should be at least 1. |
733 | */ |
734 | if (nelems == 0) { |
735 | error("node %p: no elems", node, nkids); |
736 | } |
737 | |
738 | /* |
739 | * Add nelems to the running element count of the whole tree |
740 | * (to ensure the enum234 routines see them all). |
741 | */ |
742 | ctx->elemcount += nelems; |
743 | |
744 | /* |
745 | * Check ordering property: all elements should be strictly > |
746 | * lowbound, strictly < highbound, and strictly < each other in |
747 | * sequence. (lowbound and highbound are NULL at edges of tree |
748 | * - both NULL at root node - and NULL is considered to be < |
749 | * everything and > everything. IYSWIM.) |
750 | */ |
751 | for (i = -1; i < nelems; i++) { |
752 | void *lower = (i == -1 ? lowbound : node->elems[i]); |
753 | void *higher = (i+1 == nelems ? highbound : node->elems[i+1]); |
754 | if (lower && higher && cmp(lower, higher) >= 0) { |
755 | error("node %p: kid comparison [%d=%s,%d=%s] failed", |
756 | node, i, lower, i+1, higher); |
757 | } |
758 | } |
759 | |
760 | /* |
761 | * Check parent pointers: all non-NULL kids should have a |
762 | * parent pointer coming back to this node. |
763 | */ |
764 | for (i = 0; i < nkids; i++) |
765 | if (node->kids[i]->parent != node) { |
766 | error("node %p kid %d: parent ptr is %p not %p", |
767 | node, i, node->kids[i]->parent, node); |
768 | } |
769 | |
770 | |
771 | /* |
772 | * Now (finally!) recurse into subtrees. |
773 | */ |
774 | for (i = 0; i < nkids; i++) { |
775 | void *lower = (i == 0 ? lowbound : node->elems[i-1]); |
776 | void *higher = (i >= nelems ? highbound : node->elems[i]); |
777 | chknode(ctx, level+1, node->kids[i], lower, higher); |
778 | } |
779 | } |
780 | |
781 | void verify(void) { |
782 | chkctx ctx; |
783 | enum234 e; |
784 | int i; |
785 | void *p; |
786 | |
787 | ctx.treedepth = -1; /* depth unknown yet */ |
788 | ctx.elemcount = 0; /* no elements seen yet */ |
789 | /* |
790 | * Verify validity of tree properties. |
791 | */ |
792 | if (tree->root) |
793 | chknode(&ctx, 0, tree->root, NULL, NULL); |
794 | printf("tree depth: %d\n", ctx.treedepth); |
795 | /* |
796 | * Enumerate the tree and ensure it matches up to the array. |
797 | */ |
798 | for (i = 0, p = first234(tree, &e); |
799 | p; |
800 | i++, p = next234(&e)) { |
801 | if (i >= arraylen) |
802 | error("tree contains more than %d elements", arraylen); |
803 | if (array[i] != p) |
804 | error("enum at position %d: array says %s, tree says %s", |
805 | i, array[i], p); |
806 | } |
807 | if (i != ctx.elemcount) { |
808 | error("tree really contains %d elements, enum gave %d", |
809 | i, ctx.elemcount); |
810 | } |
811 | if (i < arraylen) { |
812 | error("enum gave only %d elements, array has %d", i, arraylen); |
813 | } |
814 | } |
815 | |
816 | void addtest(void *elem) { |
817 | int i, j; |
818 | void *retval, *realret; |
819 | |
820 | if (arraysize < arraylen+1) { |
821 | arraysize = arraylen+1+256; |
822 | array = (array == NULL ? malloc(arraysize*sizeof(*array)) : |
823 | realloc(array, arraysize*sizeof(*array))); |
824 | } |
825 | |
826 | i = 0; |
827 | while (i < arraylen && cmp(elem, array[i]) > 0) |
828 | i++; |
829 | /* now i points to the first element >= elem */ |
830 | if (i < arraylen && !cmp(elem, array[i])) |
831 | retval = array[i]; /* expect that returned not elem */ |
832 | else { |
833 | retval = elem; /* expect elem returned (success) */ |
834 | for (j = arraylen; j > i; j--) |
835 | array[j] = array[j-1]; |
836 | array[i] = elem; /* add elem to array */ |
837 | arraylen++; |
838 | } |
839 | |
840 | realret = add234(tree, elem); |
841 | if (realret != retval) { |
842 | error("add: retval was %p expected %p", realret, retval); |
843 | } |
844 | |
845 | verify(); |
846 | } |
847 | |
848 | void deltest(void *elem) { |
849 | int i; |
850 | |
851 | i = 0; |
852 | while (i < arraylen && cmp(elem, array[i]) > 0) |
853 | i++; |
854 | /* now i points to the first element >= elem */ |
855 | if (i >= arraylen || cmp(elem, array[i]) != 0) |
856 | return; /* don't do it! */ |
857 | else { |
858 | while (i < arraylen-1) { |
859 | array[i] = array[i+1]; |
860 | i++; |
861 | } |
862 | arraylen--; /* delete elem from array */ |
863 | } |
864 | |
865 | del234(tree, elem); |
866 | |
867 | verify(); |
febd9a0f |
868 | } |
2d56b16f |
869 | |
870 | /* A sample data set and test utility. Designed for pseudo-randomness, |
871 | * and yet repeatability. */ |
872 | |
873 | /* |
874 | * This random number generator uses the `portable implementation' |
875 | * given in ANSI C99 draft N869. It assumes `unsigned' is 32 bits; |
876 | * change it if not. |
877 | */ |
878 | int randomnumber(unsigned *seed) { |
879 | *seed *= 1103515245; |
880 | *seed += 12345; |
881 | return ((*seed) / 65536) % 32768; |
febd9a0f |
882 | } |
883 | |
2d56b16f |
884 | int mycmp(void *av, void *bv) { |
885 | char const *a = (char const *)av; |
886 | char const *b = (char const *)bv; |
febd9a0f |
887 | return strcmp(a, b); |
888 | } |
889 | |
2d56b16f |
890 | #define lenof(x) ( sizeof((x)) / sizeof(*(x)) ) |
891 | |
892 | char *strings[] = { |
893 | "a", "ab", "absque", "coram", "de", |
894 | "palam", "clam", "cum", "ex", "e", |
895 | "sine", "tenus", "pro", "prae", |
896 | "banana", "carrot", "cabbage", "broccoli", "onion", "zebra", |
897 | "penguin", "blancmange", "pangolin", "whale", "hedgehog", |
898 | "giraffe", "peanut", "bungee", "foo", "bar", "baz", "quux", |
899 | "murfl", "spoo", "breen", "flarn", "octothorpe", |
900 | "snail", "tiger", "elephant", "octopus", "warthog", "armadillo", |
901 | "aardvark", "wyvern", "dragon", "elf", "dwarf", "orc", "goblin", |
902 | "pixie", "basilisk", "warg", "ape", "lizard", "newt", "shopkeeper", |
903 | "wand", "ring", "amulet" |
904 | }; |
905 | |
906 | #define NSTR lenof(strings) |
907 | |
febd9a0f |
908 | int main(void) { |
2d56b16f |
909 | int in[NSTR]; |
910 | int i, j; |
911 | unsigned seed = 0; |
912 | |
913 | for (i = 0; i < NSTR; i++) in[i] = 0; |
914 | array = NULL; |
915 | arraylen = arraysize = 0; |
916 | tree = newtree234(mycmp); |
917 | cmp = mycmp; |
918 | |
919 | verify(); |
920 | for (i = 0; i < 10000; i++) { |
921 | j = randomnumber(&seed); |
922 | j %= NSTR; |
923 | printf("trial: %d\n", i); |
924 | if (in[j]) { |
925 | printf("deleting %s (%d)\n", strings[j], j); |
926 | deltest(strings[j]); |
927 | in[j] = 0; |
928 | } else { |
929 | printf("adding %s (%d)\n", strings[j], j); |
930 | addtest(strings[j]); |
931 | in[j] = 1; |
932 | } |
933 | } |
934 | |
935 | while (arraylen > 0) { |
936 | j = randomnumber(&seed); |
937 | j %= arraylen; |
938 | deltest(array[j]); |
939 | } |
940 | |
941 | return 0; |
febd9a0f |
942 | } |
2d56b16f |
943 | |
febd9a0f |
944 | #endif |