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 | |
dcbde236 |
9 | #include "puttymem.h" |
10 | |
febd9a0f |
11 | #include "tree234.h" |
12 | |
dcbde236 |
13 | #define mknew(typ) ( (typ *) smalloc (sizeof (typ)) ) |
14 | /* #define sfree free */ |
febd9a0f |
15 | |
16 | #ifdef TEST |
17 | #define LOG(x) (printf x) |
18 | #else |
19 | #define LOG(x) |
20 | #endif |
21 | |
22 | struct tree234_Tag { |
23 | node234 *root; |
24 | cmpfn234 cmp; |
25 | }; |
26 | |
27 | struct node234_Tag { |
28 | node234 *parent; |
29 | node234 *kids[4]; |
30 | void *elems[3]; |
31 | }; |
32 | |
33 | /* |
34 | * Create a 2-3-4 tree. |
35 | */ |
36 | tree234 *newtree234(cmpfn234 cmp) { |
37 | tree234 *ret = mknew(tree234); |
38 | LOG(("created tree %p\n", ret)); |
39 | ret->root = NULL; |
40 | ret->cmp = cmp; |
41 | return ret; |
42 | } |
43 | |
44 | /* |
45 | * Free a 2-3-4 tree (not including freeing the elements). |
46 | */ |
47 | static void freenode234(node234 *n) { |
48 | if (!n) |
49 | return; |
50 | freenode234(n->kids[0]); |
51 | freenode234(n->kids[1]); |
52 | freenode234(n->kids[2]); |
53 | freenode234(n->kids[3]); |
54 | sfree(n); |
55 | } |
56 | void freetree234(tree234 *t) { |
57 | freenode234(t->root); |
58 | sfree(t); |
59 | } |
60 | |
61 | /* |
62 | * Add an element e to a 2-3-4 tree t. Returns e on success, or if |
63 | * an existing element compares equal, returns that. |
64 | */ |
65 | void *add234(tree234 *t, void *e) { |
66 | node234 *n, **np, *left, *right; |
67 | void *orig_e = e; |
68 | int c; |
69 | |
70 | LOG(("adding node %p to tree %p\n", e, t)); |
71 | if (t->root == NULL) { |
72 | t->root = mknew(node234); |
73 | t->root->elems[1] = t->root->elems[2] = NULL; |
74 | t->root->kids[0] = t->root->kids[1] = NULL; |
75 | t->root->kids[2] = t->root->kids[3] = NULL; |
76 | t->root->parent = NULL; |
77 | t->root->elems[0] = e; |
78 | LOG((" created root %p\n", t->root)); |
79 | return orig_e; |
80 | } |
81 | |
82 | np = &t->root; |
83 | while (*np) { |
84 | n = *np; |
85 | LOG((" node %p: %p [%p] %p [%p] %p [%p] %p\n", |
86 | n, n->kids[0], n->elems[0], n->kids[1], n->elems[1], |
87 | n->kids[2], n->elems[2], n->kids[3])); |
88 | if ((c = t->cmp(e, n->elems[0])) < 0) |
89 | np = &n->kids[0]; |
90 | else if (c == 0) |
91 | return n->elems[0]; /* already exists */ |
92 | else if (n->elems[1] == NULL || (c = t->cmp(e, n->elems[1])) < 0) |
93 | np = &n->kids[1]; |
94 | else if (c == 0) |
95 | return n->elems[1]; /* already exists */ |
96 | else if (n->elems[2] == NULL || (c = t->cmp(e, n->elems[2])) < 0) |
97 | np = &n->kids[2]; |
98 | else if (c == 0) |
99 | return n->elems[2]; /* already exists */ |
100 | else |
101 | np = &n->kids[3]; |
102 | LOG((" moving to child %d (%p)\n", np - n->kids, *np)); |
103 | } |
104 | |
105 | /* |
106 | * We need to insert the new element in n at position np. |
107 | */ |
108 | left = NULL; |
109 | right = NULL; |
110 | while (n) { |
111 | LOG((" at %p: %p [%p] %p [%p] %p [%p] %p\n", |
112 | n, n->kids[0], n->elems[0], n->kids[1], n->elems[1], |
113 | n->kids[2], n->elems[2], n->kids[3])); |
114 | LOG((" need to insert %p [%p] %p at position %d\n", |
115 | left, e, right, np - n->kids)); |
116 | if (n->elems[1] == NULL) { |
117 | /* |
118 | * Insert in a 2-node; simple. |
119 | */ |
120 | if (np == &n->kids[0]) { |
121 | LOG((" inserting on left of 2-node\n")); |
122 | n->kids[2] = n->kids[1]; |
123 | n->elems[1] = n->elems[0]; |
124 | n->kids[1] = right; |
125 | n->elems[0] = e; |
126 | n->kids[0] = left; |
127 | } else { /* np == &n->kids[1] */ |
128 | LOG((" inserting on right of 2-node\n")); |
129 | n->kids[2] = right; |
130 | n->elems[1] = e; |
131 | n->kids[1] = left; |
132 | } |
133 | if (n->kids[0]) n->kids[0]->parent = n; |
134 | if (n->kids[1]) n->kids[1]->parent = n; |
135 | if (n->kids[2]) n->kids[2]->parent = n; |
136 | LOG((" done\n")); |
137 | break; |
138 | } else if (n->elems[2] == NULL) { |
139 | /* |
140 | * Insert in a 3-node; simple. |
141 | */ |
142 | if (np == &n->kids[0]) { |
143 | LOG((" inserting on left of 3-node\n")); |
144 | n->kids[3] = n->kids[2]; |
145 | n->elems[2] = n->elems[1]; |
146 | n->kids[2] = n->kids[1]; |
147 | n->elems[1] = n->elems[0]; |
148 | n->kids[1] = right; |
149 | n->elems[0] = e; |
150 | n->kids[0] = left; |
151 | } else if (np == &n->kids[1]) { |
152 | LOG((" inserting in middle of 3-node\n")); |
153 | n->kids[3] = n->kids[2]; |
154 | n->elems[2] = n->elems[1]; |
155 | n->kids[2] = right; |
156 | n->elems[1] = e; |
157 | n->kids[1] = left; |
158 | } else { /* np == &n->kids[2] */ |
159 | LOG((" inserting on right of 3-node\n")); |
160 | n->kids[3] = right; |
161 | n->elems[2] = e; |
162 | n->kids[2] = left; |
163 | } |
164 | if (n->kids[0]) n->kids[0]->parent = n; |
165 | if (n->kids[1]) n->kids[1]->parent = n; |
166 | if (n->kids[2]) n->kids[2]->parent = n; |
167 | if (n->kids[3]) n->kids[3]->parent = n; |
168 | LOG((" done\n")); |
169 | break; |
170 | } else { |
171 | node234 *m = mknew(node234); |
172 | m->parent = n->parent; |
173 | LOG((" splitting a 4-node; created new node %p\n", m)); |
174 | /* |
175 | * Insert in a 4-node; split into a 2-node and a |
176 | * 3-node, and move focus up a level. |
177 | * |
178 | * I don't think it matters which way round we put the |
179 | * 2 and the 3. For simplicity, we'll put the 3 first |
180 | * always. |
181 | */ |
182 | if (np == &n->kids[0]) { |
183 | m->kids[0] = left; |
184 | m->elems[0] = e; |
185 | m->kids[1] = right; |
186 | m->elems[1] = n->elems[0]; |
187 | m->kids[2] = n->kids[1]; |
188 | e = n->elems[1]; |
189 | n->kids[0] = n->kids[2]; |
190 | n->elems[0] = n->elems[2]; |
191 | n->kids[1] = n->kids[3]; |
192 | } else if (np == &n->kids[1]) { |
193 | m->kids[0] = n->kids[0]; |
194 | m->elems[0] = n->elems[0]; |
195 | m->kids[1] = left; |
196 | m->elems[1] = e; |
197 | m->kids[2] = right; |
198 | e = n->elems[1]; |
199 | n->kids[0] = n->kids[2]; |
200 | n->elems[0] = n->elems[2]; |
201 | n->kids[1] = n->kids[3]; |
202 | } else if (np == &n->kids[2]) { |
203 | m->kids[0] = n->kids[0]; |
204 | m->elems[0] = n->elems[0]; |
205 | m->kids[1] = n->kids[1]; |
206 | m->elems[1] = n->elems[1]; |
207 | m->kids[2] = left; |
208 | /* e = e; */ |
209 | n->kids[0] = right; |
210 | n->elems[0] = n->elems[2]; |
211 | n->kids[1] = n->kids[3]; |
212 | } else { /* np == &n->kids[3] */ |
213 | m->kids[0] = n->kids[0]; |
214 | m->elems[0] = n->elems[0]; |
215 | m->kids[1] = n->kids[1]; |
216 | m->elems[1] = n->elems[1]; |
217 | m->kids[2] = n->kids[2]; |
218 | n->kids[0] = left; |
219 | n->elems[0] = e; |
220 | n->kids[1] = right; |
221 | e = n->elems[2]; |
222 | } |
223 | m->kids[3] = n->kids[3] = n->kids[2] = NULL; |
224 | m->elems[2] = n->elems[2] = n->elems[1] = NULL; |
225 | if (m->kids[0]) m->kids[0]->parent = m; |
226 | if (m->kids[1]) m->kids[1]->parent = m; |
227 | if (m->kids[2]) m->kids[2]->parent = m; |
228 | if (n->kids[0]) n->kids[0]->parent = n; |
229 | if (n->kids[1]) n->kids[1]->parent = n; |
230 | LOG((" left (%p): %p [%p] %p [%p] %p\n", m, |
231 | m->kids[0], m->elems[0], |
232 | m->kids[1], m->elems[1], |
233 | m->kids[2])); |
234 | LOG((" right (%p): %p [%p] %p\n", n, |
235 | n->kids[0], n->elems[0], |
236 | n->kids[1])); |
237 | left = m; |
238 | right = n; |
239 | } |
240 | if (n->parent) |
241 | np = (n->parent->kids[0] == n ? &n->parent->kids[0] : |
242 | n->parent->kids[1] == n ? &n->parent->kids[1] : |
243 | n->parent->kids[2] == n ? &n->parent->kids[2] : |
244 | &n->parent->kids[3]); |
245 | n = n->parent; |
246 | } |
247 | |
248 | /* |
249 | * If we've come out of here by `break', n will still be |
250 | * non-NULL and we've finished. If we've come here because n is |
251 | * NULL, we need to create a new root for the tree because the |
252 | * old one has just split into two. |
253 | */ |
254 | if (!n) { |
255 | LOG((" root is overloaded, split into two\n")); |
256 | t->root = mknew(node234); |
257 | t->root->kids[0] = left; |
258 | t->root->elems[0] = e; |
259 | t->root->kids[1] = right; |
260 | t->root->elems[1] = NULL; |
261 | t->root->kids[2] = NULL; |
262 | t->root->elems[2] = NULL; |
263 | t->root->kids[3] = NULL; |
264 | t->root->parent = NULL; |
265 | if (t->root->kids[0]) t->root->kids[0]->parent = t->root; |
266 | if (t->root->kids[1]) t->root->kids[1]->parent = t->root; |
267 | LOG((" new root is %p [%p] %p\n", |
268 | t->root->kids[0], t->root->elems[0], t->root->kids[1])); |
269 | } |
270 | |
271 | return orig_e; |
272 | } |
273 | |
274 | /* |
275 | * Find an element e in a 2-3-4 tree t. Returns NULL if not found. |
276 | * e is always passed as the first argument to cmp, so cmp can be |
277 | * an asymmetric function if desired. cmp can also be passed as |
278 | * NULL, in which case the compare function from the tree proper |
279 | * will be used. |
280 | */ |
281 | void *find234(tree234 *t, void *e, cmpfn234 cmp) { |
282 | node234 *n; |
283 | int c; |
284 | |
285 | if (t->root == NULL) |
286 | return NULL; |
287 | |
288 | if (cmp == NULL) |
289 | cmp = t->cmp; |
290 | |
291 | n = t->root; |
292 | while (n) { |
81d77872 |
293 | if ( (c = cmp(e, n->elems[0])) < 0) |
febd9a0f |
294 | n = n->kids[0]; |
295 | else if (c == 0) |
296 | return n->elems[0]; |
81d77872 |
297 | else if (n->elems[1] == NULL || (c = cmp(e, n->elems[1])) < 0) |
febd9a0f |
298 | n = n->kids[1]; |
299 | else if (c == 0) |
300 | return n->elems[1]; |
81d77872 |
301 | else if (n->elems[2] == NULL || (c = cmp(e, n->elems[2])) < 0) |
febd9a0f |
302 | n = n->kids[2]; |
303 | else if (c == 0) |
304 | return n->elems[2]; |
305 | else |
306 | n = n->kids[3]; |
307 | } |
308 | |
309 | /* |
310 | * We've found our way to the bottom of the tree and we know |
311 | * where we would insert this node if we wanted to. But it |
312 | * isn't there. |
313 | */ |
314 | return NULL; |
315 | } |
316 | |
317 | /* |
318 | * Delete an element e in a 2-3-4 tree. Does not free the element, |
319 | * merely removes all links to it from the tree nodes. |
320 | */ |
81d77872 |
321 | void del234(tree234 *t, void *e) { |
febd9a0f |
322 | node234 *n; |
323 | int ei = -1; |
324 | |
325 | n = t->root; |
326 | LOG(("deleting %p from tree %p\n", e, t)); |
327 | while (1) { |
328 | while (n) { |
329 | int c; |
330 | int ki; |
331 | node234 *sub; |
332 | |
333 | LOG((" node %p: %p [%p] %p [%p] %p [%p] %p\n", |
334 | n, n->kids[0], n->elems[0], n->kids[1], n->elems[1], |
335 | n->kids[2], n->elems[2], n->kids[3])); |
336 | if ((c = t->cmp(e, n->elems[0])) < 0) { |
337 | ki = 0; |
338 | } else if (c == 0) { |
339 | ei = 0; break; |
340 | } else if (n->elems[1] == NULL || (c = t->cmp(e, n->elems[1])) < 0) { |
341 | ki = 1; |
342 | } else if (c == 0) { |
343 | ei = 1; break; |
344 | } else if (n->elems[2] == NULL || (c = t->cmp(e, n->elems[2])) < 0) { |
345 | ki = 2; |
346 | } else if (c == 0) { |
347 | ei = 2; break; |
348 | } else { |
349 | ki = 3; |
350 | } |
351 | /* |
352 | * Recurse down to subtree ki. If it has only one element, |
353 | * we have to do some transformation to start with. |
354 | */ |
355 | LOG((" moving to subtree %d\n", ki)); |
356 | sub = n->kids[ki]; |
357 | if (!sub->elems[1]) { |
358 | LOG((" subtree has only one element!\n", ki)); |
359 | if (ki > 0 && n->kids[ki-1]->elems[1]) { |
360 | /* |
361 | * Case 3a, left-handed variant. Child ki has |
362 | * only one element, but child ki-1 has two or |
363 | * more. So we need to move a subtree from ki-1 |
364 | * to ki. |
365 | * |
366 | * . C . . B . |
367 | * / \ -> / \ |
368 | * [more] a A b B c d D e [more] a A b c C d D e |
369 | */ |
370 | node234 *sib = n->kids[ki-1]; |
371 | int lastelem = (sib->elems[2] ? 2 : |
372 | sib->elems[1] ? 1 : 0); |
373 | sub->kids[2] = sub->kids[1]; |
374 | sub->elems[1] = sub->elems[0]; |
375 | sub->kids[1] = sub->kids[0]; |
376 | sub->elems[0] = n->elems[ki-1]; |
377 | sub->kids[0] = sib->kids[lastelem+1]; |
100122a9 |
378 | if (sub->kids[0]) sub->kids[0]->parent = sub; |
febd9a0f |
379 | n->elems[ki-1] = sib->elems[lastelem]; |
380 | sib->kids[lastelem+1] = NULL; |
381 | sib->elems[lastelem] = NULL; |
382 | LOG((" case 3a left\n")); |
383 | } else if (ki < 3 && n->kids[ki+1] && |
384 | n->kids[ki+1]->elems[1]) { |
385 | /* |
386 | * Case 3a, right-handed variant. ki has only |
387 | * one element but ki+1 has two or more. Move a |
388 | * subtree from ki+1 to ki. |
389 | * |
390 | * . B . . C . |
391 | * / \ -> / \ |
392 | * a A b c C d D e [more] a A b B c d D e [more] |
393 | */ |
394 | node234 *sib = n->kids[ki+1]; |
395 | int j; |
396 | sub->elems[1] = n->elems[ki]; |
397 | sub->kids[2] = sib->kids[0]; |
100122a9 |
398 | if (sub->kids[2]) sub->kids[2]->parent = sub; |
febd9a0f |
399 | n->elems[ki] = sib->elems[0]; |
400 | sib->kids[0] = sib->kids[1]; |
401 | for (j = 0; j < 2 && sib->elems[j+1]; j++) { |
402 | sib->kids[j+1] = sib->kids[j+2]; |
403 | sib->elems[j] = sib->elems[j+1]; |
404 | } |
405 | sib->kids[j+1] = NULL; |
406 | sib->elems[j] = NULL; |
407 | LOG((" case 3a right\n")); |
408 | } else { |
409 | /* |
410 | * Case 3b. ki has only one element, and has no |
411 | * neighbour with more than one. So pick a |
412 | * neighbour and merge it with ki, taking an |
413 | * element down from n to go in the middle. |
414 | * |
415 | * . B . . |
416 | * / \ -> | |
417 | * a A b c C d a A b B c C d |
418 | * |
419 | * (Since at all points we have avoided |
420 | * descending to a node with only one element, |
421 | * we can be sure that n is not reduced to |
422 | * nothingness by this move, _unless_ it was |
423 | * the very first node, ie the root of the |
424 | * tree. In that case we remove the now-empty |
425 | * root and replace it with its single large |
426 | * child as shown.) |
427 | */ |
428 | node234 *sib; |
429 | int j; |
430 | |
431 | if (ki > 0) |
432 | ki--; |
433 | sib = n->kids[ki]; |
434 | sub = n->kids[ki+1]; |
435 | |
436 | sub->kids[3] = sub->kids[1]; |
437 | sub->elems[2] = sub->elems[0]; |
438 | sub->kids[2] = sub->kids[0]; |
439 | sub->elems[1] = n->elems[ki]; |
440 | sub->kids[1] = sib->kids[1]; |
100122a9 |
441 | if (sub->kids[1]) sub->kids[1]->parent = sub; |
febd9a0f |
442 | sub->elems[0] = sib->elems[0]; |
443 | sub->kids[0] = sib->kids[0]; |
100122a9 |
444 | if (sub->kids[0]) sub->kids[0]->parent = sub; |
febd9a0f |
445 | |
446 | sfree(sib); |
447 | |
448 | /* |
449 | * That's built the big node in sub. Now we |
450 | * need to remove the reference to sib in n. |
451 | */ |
452 | for (j = ki; j < 3 && n->kids[j+1]; j++) { |
453 | n->kids[j] = n->kids[j+1]; |
454 | n->elems[j] = j<2 ? n->elems[j+1] : NULL; |
455 | } |
456 | n->kids[j] = NULL; |
457 | if (j < 3) n->elems[j] = NULL; |
2d56b16f |
458 | LOG((" case 3b ki=%d\n", ki)); |
febd9a0f |
459 | |
460 | if (!n->elems[0]) { |
461 | /* |
462 | * The root is empty and needs to be |
463 | * removed. |
464 | */ |
465 | LOG((" shifting root!\n")); |
466 | t->root = sub; |
467 | sub->parent = NULL; |
468 | sfree(n); |
469 | } |
470 | } |
471 | } |
472 | n = sub; |
473 | } |
474 | if (ei==-1) |
475 | return; /* nothing to do; `already removed' */ |
476 | |
477 | /* |
478 | * Treat special case: this is the one remaining item in |
479 | * the tree. n is the tree root (no parent), has one |
480 | * element (no elems[1]), and has no kids (no kids[0]). |
481 | */ |
482 | if (!n->parent && !n->elems[1] && !n->kids[0]) { |
483 | LOG((" removed last element in tree\n")); |
484 | sfree(n); |
485 | t->root = NULL; |
486 | return; |
487 | } |
488 | |
489 | /* |
490 | * Now we have the element we want, as n->elems[ei], and we |
491 | * have also arranged for that element not to be the only |
492 | * one in its node. So... |
493 | */ |
494 | |
495 | if (!n->kids[0] && n->elems[1]) { |
496 | /* |
497 | * Case 1. n is a leaf node with more than one element, |
498 | * so it's _really easy_. Just delete the thing and |
499 | * we're done. |
500 | */ |
501 | int i; |
502 | LOG((" case 1\n")); |
a4a19e73 |
503 | for (i = ei; i < 2 && n->elems[i+1]; i++) |
febd9a0f |
504 | n->elems[i] = n->elems[i+1]; |
505 | n->elems[i] = NULL; |
506 | return; /* finished! */ |
507 | } else if (n->kids[ei]->elems[1]) { |
508 | /* |
509 | * Case 2a. n is an internal node, and the root of the |
510 | * subtree to the left of e has more than one element. |
511 | * So find the predecessor p to e (ie the largest node |
512 | * in that subtree), place it where e currently is, and |
513 | * then start the deletion process over again on the |
514 | * subtree with p as target. |
515 | */ |
516 | node234 *m = n->kids[ei]; |
517 | void *target; |
518 | LOG((" case 2a\n")); |
519 | while (m->kids[0]) { |
520 | m = (m->kids[3] ? m->kids[3] : |
521 | m->kids[2] ? m->kids[2] : |
522 | m->kids[1] ? m->kids[1] : m->kids[0]); |
523 | } |
524 | target = (m->elems[2] ? m->elems[2] : |
525 | m->elems[1] ? m->elems[1] : m->elems[0]); |
526 | n->elems[ei] = target; |
527 | n = n->kids[ei]; |
528 | e = target; |
529 | } else if (n->kids[ei+1]->elems[1]) { |
530 | /* |
531 | * Case 2b, symmetric to 2a but s/left/right/ and |
532 | * s/predecessor/successor/. (And s/largest/smallest/). |
533 | */ |
534 | node234 *m = n->kids[ei+1]; |
535 | void *target; |
536 | LOG((" case 2b\n")); |
537 | while (m->kids[0]) { |
538 | m = m->kids[0]; |
539 | } |
540 | target = m->elems[0]; |
541 | n->elems[ei] = target; |
542 | n = n->kids[ei+1]; |
543 | e = target; |
544 | } else { |
545 | /* |
546 | * Case 2c. n is an internal node, and the subtrees to |
547 | * the left and right of e both have only one element. |
548 | * So combine the two subnodes into a single big node |
549 | * with their own elements on the left and right and e |
550 | * in the middle, then restart the deletion process on |
551 | * that subtree, with e still as target. |
552 | */ |
553 | node234 *a = n->kids[ei], *b = n->kids[ei+1]; |
554 | int j; |
555 | |
556 | LOG((" case 2c\n")); |
557 | a->elems[1] = n->elems[ei]; |
558 | a->kids[2] = b->kids[0]; |
100122a9 |
559 | if (a->kids[2]) a->kids[2]->parent = a; |
febd9a0f |
560 | a->elems[2] = b->elems[0]; |
561 | a->kids[3] = b->kids[1]; |
100122a9 |
562 | if (a->kids[3]) a->kids[3]->parent = a; |
febd9a0f |
563 | sfree(b); |
564 | /* |
565 | * That's built the big node in a, and destroyed b. Now |
566 | * remove the reference to b (and e) in n. |
567 | */ |
568 | for (j = ei; j < 2 && n->elems[j+1]; j++) { |
569 | n->elems[j] = n->elems[j+1]; |
570 | n->kids[j+1] = n->kids[j+2]; |
571 | } |
572 | n->elems[j] = NULL; |
573 | n->kids[j+1] = NULL; |
e9e9556d |
574 | /* |
575 | * It's possible, in this case, that we've just removed |
576 | * the only element in the root of the tree. If so, |
577 | * shift the root. |
578 | */ |
579 | if (n->elems[0] == NULL) { |
580 | LOG((" shifting root!\n")); |
581 | t->root = a; |
582 | a->parent = NULL; |
583 | sfree(n); |
584 | } |
febd9a0f |
585 | /* |
586 | * Now go round the deletion process again, with n |
587 | * pointing at the new big node and e still the same. |
588 | */ |
589 | n = a; |
590 | } |
591 | } |
592 | } |
593 | |
594 | /* |
595 | * Iterate over the elements of a tree234, in order. |
596 | */ |
597 | void *first234(tree234 *t, enum234 *e) { |
598 | node234 *n = t->root; |
599 | if (!n) |
600 | return NULL; |
601 | while (n->kids[0]) |
602 | n = n->kids[0]; |
603 | e->node = n; |
604 | e->posn = 0; |
605 | return n->elems[0]; |
606 | } |
607 | |
608 | void *next234(enum234 *e) { |
609 | node234 *n = e->node; |
610 | int pos = e->posn; |
611 | |
612 | if (n->kids[pos+1]) { |
613 | n = n->kids[pos+1]; |
614 | while (n->kids[0]) |
615 | n = n->kids[0]; |
616 | e->node = n; |
617 | e->posn = 0; |
618 | return n->elems[0]; |
619 | } |
620 | |
6aff1005 |
621 | if (pos < 2 && n->elems[pos+1]) { |
622 | e->posn = pos+1; |
623 | return n->elems[e->posn]; |
febd9a0f |
624 | } |
625 | |
626 | do { |
627 | node234 *nn = n->parent; |
628 | if (nn == NULL) |
629 | return NULL; /* end of tree */ |
630 | pos = (nn->kids[0] == n ? 0 : |
631 | nn->kids[1] == n ? 1 : |
632 | nn->kids[2] == n ? 2 : 3); |
633 | n = nn; |
634 | } while (pos == 3 || n->kids[pos+1] == NULL); |
635 | |
636 | e->node = n; |
637 | e->posn = pos; |
638 | return n->elems[pos]; |
639 | } |
640 | |
641 | #ifdef TEST |
642 | |
2d56b16f |
643 | /* |
644 | * Test code for the 2-3-4 tree. This code maintains an alternative |
645 | * representation of the data in the tree, in an array (using the |
646 | * obvious and slow insert and delete functions). After each tree |
7aa7c43a |
647 | * operation, the verify() function is called, which ensures all |
648 | * the tree properties are preserved (node->child->parent always |
649 | * equals node; number of kids == 0 or number of elements + 1; |
650 | * ordering property between elements of a node and elements of its |
651 | * children is preserved; tree has the same depth everywhere; every |
652 | * node has at least one element) and also ensures the list |
653 | * represented by the tree is the same list it should be. (This |
654 | * last check also verifies the ordering properties, because the |
655 | * `same list it should be' is by definition correctly ordered. It |
656 | * also ensures all nodes are distinct, because the enum functions |
657 | * would get caught in a loop if not.) |
2d56b16f |
658 | */ |
659 | |
660 | #include <stdarg.h> |
661 | |
662 | /* |
663 | * Error reporting function. |
664 | */ |
665 | void error(char *fmt, ...) { |
666 | va_list ap; |
667 | printf("ERROR: "); |
668 | va_start(ap, fmt); |
669 | vfprintf(stdout, fmt, ap); |
670 | va_end(ap); |
671 | printf("\n"); |
672 | } |
673 | |
674 | /* The array representation of the data. */ |
675 | void **array; |
676 | int arraylen, arraysize; |
677 | cmpfn234 cmp; |
678 | |
679 | /* The tree representation of the same data. */ |
680 | tree234 *tree; |
681 | |
682 | typedef struct { |
683 | int treedepth; |
684 | int elemcount; |
685 | } chkctx; |
686 | |
687 | void chknode(chkctx *ctx, int level, node234 *node, |
688 | void *lowbound, void *highbound) { |
689 | int nkids, nelems; |
690 | int i; |
691 | |
692 | /* Count the non-NULL kids. */ |
693 | for (nkids = 0; nkids < 4 && node->kids[nkids]; nkids++); |
694 | /* Ensure no kids beyond the first NULL are non-NULL. */ |
695 | for (i = nkids; i < 4; i++) |
696 | if (node->kids[i]) { |
697 | error("node %p: nkids=%d but kids[%d] non-NULL", |
698 | node, nkids, i); |
699 | } |
700 | |
701 | /* Count the non-NULL elements. */ |
702 | for (nelems = 0; nelems < 3 && node->elems[nelems]; nelems++); |
703 | /* Ensure no elements beyond the first NULL are non-NULL. */ |
704 | for (i = nelems; i < 3; i++) |
705 | if (node->elems[i]) { |
706 | error("node %p: nelems=%d but elems[%d] non-NULL", |
707 | node, nelems, i); |
708 | } |
709 | |
710 | if (nkids == 0) { |
711 | /* |
712 | * If nkids==0, this is a leaf node; verify that the tree |
713 | * depth is the same everywhere. |
714 | */ |
715 | if (ctx->treedepth < 0) |
716 | ctx->treedepth = level; /* we didn't know the depth yet */ |
717 | else if (ctx->treedepth != level) |
718 | error("node %p: leaf at depth %d, previously seen depth %d", |
719 | node, level, ctx->treedepth); |
720 | } else { |
721 | /* |
722 | * If nkids != 0, then it should be nelems+1, unless nelems |
723 | * is 0 in which case nkids should also be 0 (and so we |
724 | * shouldn't be in this condition at all). |
725 | */ |
726 | int shouldkids = (nelems ? nelems+1 : 0); |
727 | if (nkids != shouldkids) { |
728 | error("node %p: %d elems should mean %d kids but has %d", |
729 | node, nelems, shouldkids, nkids); |
730 | } |
731 | } |
732 | |
733 | /* |
734 | * nelems should be at least 1. |
735 | */ |
736 | if (nelems == 0) { |
737 | error("node %p: no elems", node, nkids); |
738 | } |
739 | |
740 | /* |
741 | * Add nelems to the running element count of the whole tree |
742 | * (to ensure the enum234 routines see them all). |
743 | */ |
744 | ctx->elemcount += nelems; |
745 | |
746 | /* |
747 | * Check ordering property: all elements should be strictly > |
748 | * lowbound, strictly < highbound, and strictly < each other in |
749 | * sequence. (lowbound and highbound are NULL at edges of tree |
750 | * - both NULL at root node - and NULL is considered to be < |
751 | * everything and > everything. IYSWIM.) |
752 | */ |
753 | for (i = -1; i < nelems; i++) { |
754 | void *lower = (i == -1 ? lowbound : node->elems[i]); |
755 | void *higher = (i+1 == nelems ? highbound : node->elems[i+1]); |
756 | if (lower && higher && cmp(lower, higher) >= 0) { |
757 | error("node %p: kid comparison [%d=%s,%d=%s] failed", |
758 | node, i, lower, i+1, higher); |
759 | } |
760 | } |
761 | |
762 | /* |
763 | * Check parent pointers: all non-NULL kids should have a |
764 | * parent pointer coming back to this node. |
765 | */ |
766 | for (i = 0; i < nkids; i++) |
767 | if (node->kids[i]->parent != node) { |
768 | error("node %p kid %d: parent ptr is %p not %p", |
769 | node, i, node->kids[i]->parent, node); |
770 | } |
771 | |
772 | |
773 | /* |
774 | * Now (finally!) recurse into subtrees. |
775 | */ |
776 | for (i = 0; i < nkids; i++) { |
777 | void *lower = (i == 0 ? lowbound : node->elems[i-1]); |
778 | void *higher = (i >= nelems ? highbound : node->elems[i]); |
779 | chknode(ctx, level+1, node->kids[i], lower, higher); |
780 | } |
781 | } |
782 | |
783 | void verify(void) { |
784 | chkctx ctx; |
785 | enum234 e; |
786 | int i; |
787 | void *p; |
788 | |
789 | ctx.treedepth = -1; /* depth unknown yet */ |
790 | ctx.elemcount = 0; /* no elements seen yet */ |
791 | /* |
792 | * Verify validity of tree properties. |
793 | */ |
794 | if (tree->root) |
795 | chknode(&ctx, 0, tree->root, NULL, NULL); |
796 | printf("tree depth: %d\n", ctx.treedepth); |
797 | /* |
798 | * Enumerate the tree and ensure it matches up to the array. |
799 | */ |
800 | for (i = 0, p = first234(tree, &e); |
801 | p; |
802 | i++, p = next234(&e)) { |
803 | if (i >= arraylen) |
804 | error("tree contains more than %d elements", arraylen); |
805 | if (array[i] != p) |
806 | error("enum at position %d: array says %s, tree says %s", |
807 | i, array[i], p); |
808 | } |
809 | if (i != ctx.elemcount) { |
810 | error("tree really contains %d elements, enum gave %d", |
811 | i, ctx.elemcount); |
812 | } |
813 | if (i < arraylen) { |
814 | error("enum gave only %d elements, array has %d", i, arraylen); |
815 | } |
816 | } |
817 | |
818 | void addtest(void *elem) { |
819 | int i, j; |
820 | void *retval, *realret; |
821 | |
822 | if (arraysize < arraylen+1) { |
823 | arraysize = arraylen+1+256; |
dcbde236 |
824 | array = (array == NULL ? smalloc(arraysize*sizeof(*array)) : |
825 | srealloc(array, arraysize*sizeof(*array))); |
2d56b16f |
826 | } |
827 | |
828 | i = 0; |
829 | while (i < arraylen && cmp(elem, array[i]) > 0) |
830 | i++; |
831 | /* now i points to the first element >= elem */ |
832 | if (i < arraylen && !cmp(elem, array[i])) |
833 | retval = array[i]; /* expect that returned not elem */ |
834 | else { |
835 | retval = elem; /* expect elem returned (success) */ |
836 | for (j = arraylen; j > i; j--) |
837 | array[j] = array[j-1]; |
838 | array[i] = elem; /* add elem to array */ |
839 | arraylen++; |
840 | } |
841 | |
842 | realret = add234(tree, elem); |
843 | if (realret != retval) { |
844 | error("add: retval was %p expected %p", realret, retval); |
845 | } |
846 | |
847 | verify(); |
848 | } |
849 | |
850 | void deltest(void *elem) { |
851 | int i; |
852 | |
853 | i = 0; |
854 | while (i < arraylen && cmp(elem, array[i]) > 0) |
855 | i++; |
856 | /* now i points to the first element >= elem */ |
857 | if (i >= arraylen || cmp(elem, array[i]) != 0) |
858 | return; /* don't do it! */ |
859 | else { |
860 | while (i < arraylen-1) { |
861 | array[i] = array[i+1]; |
862 | i++; |
863 | } |
864 | arraylen--; /* delete elem from array */ |
865 | } |
866 | |
867 | del234(tree, elem); |
868 | |
869 | verify(); |
febd9a0f |
870 | } |
2d56b16f |
871 | |
872 | /* A sample data set and test utility. Designed for pseudo-randomness, |
873 | * and yet repeatability. */ |
874 | |
875 | /* |
876 | * This random number generator uses the `portable implementation' |
877 | * given in ANSI C99 draft N869. It assumes `unsigned' is 32 bits; |
878 | * change it if not. |
879 | */ |
880 | int randomnumber(unsigned *seed) { |
881 | *seed *= 1103515245; |
882 | *seed += 12345; |
883 | return ((*seed) / 65536) % 32768; |
febd9a0f |
884 | } |
885 | |
2d56b16f |
886 | int mycmp(void *av, void *bv) { |
887 | char const *a = (char const *)av; |
888 | char const *b = (char const *)bv; |
febd9a0f |
889 | return strcmp(a, b); |
890 | } |
891 | |
2d56b16f |
892 | #define lenof(x) ( sizeof((x)) / sizeof(*(x)) ) |
893 | |
894 | char *strings[] = { |
895 | "a", "ab", "absque", "coram", "de", |
896 | "palam", "clam", "cum", "ex", "e", |
897 | "sine", "tenus", "pro", "prae", |
898 | "banana", "carrot", "cabbage", "broccoli", "onion", "zebra", |
899 | "penguin", "blancmange", "pangolin", "whale", "hedgehog", |
900 | "giraffe", "peanut", "bungee", "foo", "bar", "baz", "quux", |
901 | "murfl", "spoo", "breen", "flarn", "octothorpe", |
902 | "snail", "tiger", "elephant", "octopus", "warthog", "armadillo", |
903 | "aardvark", "wyvern", "dragon", "elf", "dwarf", "orc", "goblin", |
904 | "pixie", "basilisk", "warg", "ape", "lizard", "newt", "shopkeeper", |
905 | "wand", "ring", "amulet" |
906 | }; |
907 | |
908 | #define NSTR lenof(strings) |
909 | |
febd9a0f |
910 | int main(void) { |
2d56b16f |
911 | int in[NSTR]; |
912 | int i, j; |
913 | unsigned seed = 0; |
914 | |
915 | for (i = 0; i < NSTR; i++) in[i] = 0; |
916 | array = NULL; |
917 | arraylen = arraysize = 0; |
918 | tree = newtree234(mycmp); |
919 | cmp = mycmp; |
920 | |
921 | verify(); |
922 | for (i = 0; i < 10000; i++) { |
923 | j = randomnumber(&seed); |
924 | j %= NSTR; |
925 | printf("trial: %d\n", i); |
926 | if (in[j]) { |
927 | printf("deleting %s (%d)\n", strings[j], j); |
928 | deltest(strings[j]); |
929 | in[j] = 0; |
930 | } else { |
931 | printf("adding %s (%d)\n", strings[j], j); |
932 | addtest(strings[j]); |
933 | in[j] = 1; |
934 | } |
935 | } |
936 | |
937 | while (arraylen > 0) { |
938 | j = randomnumber(&seed); |
939 | j %= arraylen; |
940 | deltest(array[j]); |
941 | } |
942 | |
943 | return 0; |
febd9a0f |
944 | } |
2d56b16f |
945 | |
febd9a0f |
946 | #endif |