febd9a0f |
1 | /* |
d2371c81 |
2 | * tree234.c: reasonably generic counted 2-3-4 tree routines. |
3 | * |
4 | * This file is copyright 1999-2001 Simon Tatham. |
5 | * |
6 | * Permission is hereby granted, free of charge, to any person |
7 | * obtaining a copy of this software and associated documentation |
8 | * files (the "Software"), to deal in the Software without |
9 | * restriction, including without limitation the rights to use, |
10 | * copy, modify, merge, publish, distribute, sublicense, and/or |
11 | * sell copies of the Software, and to permit persons to whom the |
12 | * Software is furnished to do so, subject to the following |
13 | * conditions: |
14 | * |
15 | * The above copyright notice and this permission notice shall be |
16 | * included in all copies or substantial portions of the Software. |
17 | * |
18 | * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, |
19 | * EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES |
20 | * OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND |
21 | * NONINFRINGEMENT. IN NO EVENT SHALL SIMON TATHAM BE LIABLE FOR |
22 | * ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF |
23 | * CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN |
24 | * CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE |
25 | * SOFTWARE. |
febd9a0f |
26 | */ |
27 | |
28 | #include <stdio.h> |
29 | #include <stdlib.h> |
d2371c81 |
30 | #include <assert.h> |
dcbde236 |
31 | |
febd9a0f |
32 | #include "tree234.h" |
33 | |
d2371c81 |
34 | #define smalloc malloc |
35 | #define sfree free |
36 | |
dcbde236 |
37 | #define mknew(typ) ( (typ *) smalloc (sizeof (typ)) ) |
febd9a0f |
38 | |
39 | #ifdef TEST |
40 | #define LOG(x) (printf x) |
41 | #else |
42 | #define LOG(x) |
43 | #endif |
44 | |
d2371c81 |
45 | typedef struct node234_Tag node234; |
46 | |
febd9a0f |
47 | struct tree234_Tag { |
48 | node234 *root; |
49 | cmpfn234 cmp; |
50 | }; |
51 | |
52 | struct node234_Tag { |
53 | node234 *parent; |
54 | node234 *kids[4]; |
d2371c81 |
55 | int counts[4]; |
febd9a0f |
56 | void *elems[3]; |
57 | }; |
58 | |
59 | /* |
60 | * Create a 2-3-4 tree. |
61 | */ |
62 | tree234 *newtree234(cmpfn234 cmp) { |
63 | tree234 *ret = mknew(tree234); |
64 | LOG(("created tree %p\n", ret)); |
65 | ret->root = NULL; |
66 | ret->cmp = cmp; |
67 | return ret; |
68 | } |
69 | |
70 | /* |
71 | * Free a 2-3-4 tree (not including freeing the elements). |
72 | */ |
73 | static void freenode234(node234 *n) { |
74 | if (!n) |
75 | return; |
76 | freenode234(n->kids[0]); |
77 | freenode234(n->kids[1]); |
78 | freenode234(n->kids[2]); |
79 | freenode234(n->kids[3]); |
80 | sfree(n); |
81 | } |
82 | void freetree234(tree234 *t) { |
83 | freenode234(t->root); |
84 | sfree(t); |
85 | } |
86 | |
87 | /* |
d2371c81 |
88 | * Internal function to count a node. |
89 | */ |
90 | static int countnode234(node234 *n) { |
91 | int count = 0; |
92 | int i; |
93 | for (i = 0; i < 4; i++) |
94 | count += n->counts[i]; |
95 | for (i = 0; i < 3; i++) |
96 | if (n->elems[i]) |
97 | count++; |
98 | return count; |
99 | } |
100 | |
101 | /* |
102 | * Count the elements in a tree. |
103 | */ |
104 | int count234(tree234 *t) { |
105 | if (t->root) |
106 | return countnode234(t->root); |
107 | else |
108 | return 0; |
109 | } |
110 | |
111 | /* |
febd9a0f |
112 | * Add an element e to a 2-3-4 tree t. Returns e on success, or if |
113 | * an existing element compares equal, returns that. |
114 | */ |
d2371c81 |
115 | static void *add234_internal(tree234 *t, void *e, int index) { |
febd9a0f |
116 | node234 *n, **np, *left, *right; |
117 | void *orig_e = e; |
d2371c81 |
118 | int c, lcount, rcount; |
febd9a0f |
119 | |
120 | LOG(("adding node %p to tree %p\n", e, t)); |
121 | if (t->root == NULL) { |
122 | t->root = mknew(node234); |
123 | t->root->elems[1] = t->root->elems[2] = NULL; |
124 | t->root->kids[0] = t->root->kids[1] = NULL; |
125 | t->root->kids[2] = t->root->kids[3] = NULL; |
d2371c81 |
126 | t->root->counts[0] = t->root->counts[1] = 0; |
127 | t->root->counts[2] = t->root->counts[3] = 0; |
febd9a0f |
128 | t->root->parent = NULL; |
129 | t->root->elems[0] = e; |
130 | LOG((" created root %p\n", t->root)); |
131 | return orig_e; |
132 | } |
133 | |
134 | np = &t->root; |
135 | while (*np) { |
d2371c81 |
136 | int childnum; |
febd9a0f |
137 | n = *np; |
d2371c81 |
138 | LOG((" node %p: %p/%d [%p] %p/%d [%p] %p/%d [%p] %p/%d\n", |
139 | n, |
140 | n->kids[0], n->counts[0], n->elems[0], |
141 | n->kids[1], n->counts[1], n->elems[1], |
142 | n->kids[2], n->counts[2], n->elems[2], |
143 | n->kids[3], n->counts[3])); |
144 | if (index >= 0) { |
145 | if (!n->kids[0]) { |
146 | /* |
147 | * Leaf node. We want to insert at kid position |
148 | * equal to the index: |
149 | * |
150 | * 0 A 1 B 2 C 3 |
151 | */ |
152 | childnum = index; |
153 | } else { |
154 | /* |
155 | * Internal node. We always descend through it (add |
156 | * always starts at the bottom, never in the |
157 | * middle). |
158 | */ |
159 | do { /* this is a do ... while (0) to allow `break' */ |
160 | if (index <= n->counts[0]) { |
161 | childnum = 0; |
162 | break; |
163 | } |
164 | index -= n->counts[0] + 1; |
165 | if (index <= n->counts[1]) { |
166 | childnum = 1; |
167 | break; |
168 | } |
169 | index -= n->counts[1] + 1; |
170 | if (index <= n->counts[2]) { |
171 | childnum = 2; |
172 | break; |
173 | } |
174 | index -= n->counts[2] + 1; |
175 | if (index <= n->counts[3]) { |
176 | childnum = 3; |
177 | break; |
178 | } |
179 | return NULL; /* error: index out of range */ |
180 | } while (0); |
181 | } |
182 | } else { |
183 | if ((c = t->cmp(e, n->elems[0])) < 0) |
184 | childnum = 0; |
185 | else if (c == 0) |
186 | return n->elems[0]; /* already exists */ |
187 | else if (n->elems[1] == NULL || (c = t->cmp(e, n->elems[1])) < 0) |
188 | childnum = 1; |
189 | else if (c == 0) |
190 | return n->elems[1]; /* already exists */ |
191 | else if (n->elems[2] == NULL || (c = t->cmp(e, n->elems[2])) < 0) |
192 | childnum = 2; |
193 | else if (c == 0) |
194 | return n->elems[2]; /* already exists */ |
195 | else |
196 | childnum = 3; |
197 | } |
198 | np = &n->kids[childnum]; |
199 | LOG((" moving to child %d (%p)\n", childnum, *np)); |
febd9a0f |
200 | } |
201 | |
202 | /* |
203 | * We need to insert the new element in n at position np. |
204 | */ |
d2371c81 |
205 | left = NULL; lcount = 0; |
206 | right = NULL; rcount = 0; |
febd9a0f |
207 | while (n) { |
d2371c81 |
208 | LOG((" at %p: %p/%d [%p] %p/%d [%p] %p/%d [%p] %p/%d\n", |
209 | n, |
210 | n->kids[0], n->counts[0], n->elems[0], |
211 | n->kids[1], n->counts[1], n->elems[1], |
212 | n->kids[2], n->counts[2], n->elems[2], |
213 | n->kids[3], n->counts[3])); |
214 | LOG((" need to insert %p/%d [%p] %p/%d at position %d\n", |
215 | left, lcount, e, right, rcount, np - n->kids)); |
febd9a0f |
216 | if (n->elems[1] == NULL) { |
217 | /* |
218 | * Insert in a 2-node; simple. |
219 | */ |
220 | if (np == &n->kids[0]) { |
221 | LOG((" inserting on left of 2-node\n")); |
d2371c81 |
222 | n->kids[2] = n->kids[1]; n->counts[2] = n->counts[1]; |
febd9a0f |
223 | n->elems[1] = n->elems[0]; |
d2371c81 |
224 | n->kids[1] = right; n->counts[1] = rcount; |
febd9a0f |
225 | n->elems[0] = e; |
d2371c81 |
226 | n->kids[0] = left; n->counts[0] = lcount; |
febd9a0f |
227 | } else { /* np == &n->kids[1] */ |
228 | LOG((" inserting on right of 2-node\n")); |
d2371c81 |
229 | n->kids[2] = right; n->counts[2] = rcount; |
febd9a0f |
230 | n->elems[1] = e; |
d2371c81 |
231 | n->kids[1] = left; n->counts[1] = lcount; |
febd9a0f |
232 | } |
233 | if (n->kids[0]) n->kids[0]->parent = n; |
234 | if (n->kids[1]) n->kids[1]->parent = n; |
235 | if (n->kids[2]) n->kids[2]->parent = n; |
236 | LOG((" done\n")); |
237 | break; |
238 | } else if (n->elems[2] == NULL) { |
239 | /* |
240 | * Insert in a 3-node; simple. |
241 | */ |
242 | if (np == &n->kids[0]) { |
243 | LOG((" inserting on left of 3-node\n")); |
d2371c81 |
244 | n->kids[3] = n->kids[2]; n->counts[3] = n->counts[2]; |
febd9a0f |
245 | n->elems[2] = n->elems[1]; |
d2371c81 |
246 | n->kids[2] = n->kids[1]; n->counts[2] = n->counts[1]; |
febd9a0f |
247 | n->elems[1] = n->elems[0]; |
d2371c81 |
248 | n->kids[1] = right; n->counts[1] = rcount; |
febd9a0f |
249 | n->elems[0] = e; |
d2371c81 |
250 | n->kids[0] = left; n->counts[0] = lcount; |
febd9a0f |
251 | } else if (np == &n->kids[1]) { |
252 | LOG((" inserting in middle of 3-node\n")); |
d2371c81 |
253 | n->kids[3] = n->kids[2]; n->counts[3] = n->counts[2]; |
febd9a0f |
254 | n->elems[2] = n->elems[1]; |
d2371c81 |
255 | n->kids[2] = right; n->counts[2] = rcount; |
febd9a0f |
256 | n->elems[1] = e; |
d2371c81 |
257 | n->kids[1] = left; n->counts[1] = lcount; |
febd9a0f |
258 | } else { /* np == &n->kids[2] */ |
259 | LOG((" inserting on right of 3-node\n")); |
d2371c81 |
260 | n->kids[3] = right; n->counts[3] = rcount; |
febd9a0f |
261 | n->elems[2] = e; |
d2371c81 |
262 | n->kids[2] = left; n->counts[2] = lcount; |
febd9a0f |
263 | } |
264 | if (n->kids[0]) n->kids[0]->parent = n; |
265 | if (n->kids[1]) n->kids[1]->parent = n; |
266 | if (n->kids[2]) n->kids[2]->parent = n; |
267 | if (n->kids[3]) n->kids[3]->parent = n; |
268 | LOG((" done\n")); |
269 | break; |
270 | } else { |
271 | node234 *m = mknew(node234); |
272 | m->parent = n->parent; |
273 | LOG((" splitting a 4-node; created new node %p\n", m)); |
274 | /* |
275 | * Insert in a 4-node; split into a 2-node and a |
276 | * 3-node, and move focus up a level. |
277 | * |
278 | * I don't think it matters which way round we put the |
279 | * 2 and the 3. For simplicity, we'll put the 3 first |
280 | * always. |
281 | */ |
282 | if (np == &n->kids[0]) { |
d2371c81 |
283 | m->kids[0] = left; m->counts[0] = lcount; |
febd9a0f |
284 | m->elems[0] = e; |
d2371c81 |
285 | m->kids[1] = right; m->counts[1] = rcount; |
febd9a0f |
286 | m->elems[1] = n->elems[0]; |
d2371c81 |
287 | m->kids[2] = n->kids[1]; m->counts[2] = n->counts[1]; |
febd9a0f |
288 | e = n->elems[1]; |
d2371c81 |
289 | n->kids[0] = n->kids[2]; n->counts[0] = n->counts[2]; |
febd9a0f |
290 | n->elems[0] = n->elems[2]; |
d2371c81 |
291 | n->kids[1] = n->kids[3]; n->counts[1] = n->counts[3]; |
febd9a0f |
292 | } else if (np == &n->kids[1]) { |
d2371c81 |
293 | m->kids[0] = n->kids[0]; m->counts[0] = n->counts[0]; |
febd9a0f |
294 | m->elems[0] = n->elems[0]; |
d2371c81 |
295 | m->kids[1] = left; m->counts[1] = lcount; |
febd9a0f |
296 | m->elems[1] = e; |
d2371c81 |
297 | m->kids[2] = right; m->counts[2] = rcount; |
febd9a0f |
298 | e = n->elems[1]; |
d2371c81 |
299 | n->kids[0] = n->kids[2]; n->counts[0] = n->counts[2]; |
febd9a0f |
300 | n->elems[0] = n->elems[2]; |
d2371c81 |
301 | n->kids[1] = n->kids[3]; n->counts[1] = n->counts[3]; |
febd9a0f |
302 | } else if (np == &n->kids[2]) { |
d2371c81 |
303 | m->kids[0] = n->kids[0]; m->counts[0] = n->counts[0]; |
febd9a0f |
304 | m->elems[0] = n->elems[0]; |
d2371c81 |
305 | m->kids[1] = n->kids[1]; m->counts[1] = n->counts[1]; |
febd9a0f |
306 | m->elems[1] = n->elems[1]; |
d2371c81 |
307 | m->kids[2] = left; m->counts[2] = lcount; |
febd9a0f |
308 | /* e = e; */ |
d2371c81 |
309 | n->kids[0] = right; n->counts[0] = rcount; |
febd9a0f |
310 | n->elems[0] = n->elems[2]; |
d2371c81 |
311 | n->kids[1] = n->kids[3]; n->counts[1] = n->counts[3]; |
febd9a0f |
312 | } else { /* np == &n->kids[3] */ |
d2371c81 |
313 | m->kids[0] = n->kids[0]; m->counts[0] = n->counts[0]; |
febd9a0f |
314 | m->elems[0] = n->elems[0]; |
d2371c81 |
315 | m->kids[1] = n->kids[1]; m->counts[1] = n->counts[1]; |
febd9a0f |
316 | m->elems[1] = n->elems[1]; |
d2371c81 |
317 | m->kids[2] = n->kids[2]; m->counts[2] = n->counts[2]; |
318 | n->kids[0] = left; n->counts[0] = lcount; |
febd9a0f |
319 | n->elems[0] = e; |
d2371c81 |
320 | n->kids[1] = right; n->counts[1] = rcount; |
febd9a0f |
321 | e = n->elems[2]; |
322 | } |
323 | m->kids[3] = n->kids[3] = n->kids[2] = NULL; |
d2371c81 |
324 | m->counts[3] = n->counts[3] = n->counts[2] = 0; |
febd9a0f |
325 | m->elems[2] = n->elems[2] = n->elems[1] = NULL; |
326 | if (m->kids[0]) m->kids[0]->parent = m; |
327 | if (m->kids[1]) m->kids[1]->parent = m; |
328 | if (m->kids[2]) m->kids[2]->parent = m; |
329 | if (n->kids[0]) n->kids[0]->parent = n; |
330 | if (n->kids[1]) n->kids[1]->parent = n; |
d2371c81 |
331 | LOG((" left (%p): %p/%d [%p] %p/%d [%p] %p/%d\n", m, |
332 | m->kids[0], m->counts[0], m->elems[0], |
333 | m->kids[1], m->counts[1], m->elems[1], |
334 | m->kids[2], m->counts[2])); |
335 | LOG((" right (%p): %p/%d [%p] %p/%d\n", n, |
336 | n->kids[0], n->counts[0], n->elems[0], |
337 | n->kids[1], n->counts[1])); |
338 | left = m; lcount = countnode234(left); |
339 | right = n; rcount = countnode234(right); |
febd9a0f |
340 | } |
341 | if (n->parent) |
342 | np = (n->parent->kids[0] == n ? &n->parent->kids[0] : |
343 | n->parent->kids[1] == n ? &n->parent->kids[1] : |
344 | n->parent->kids[2] == n ? &n->parent->kids[2] : |
345 | &n->parent->kids[3]); |
346 | n = n->parent; |
347 | } |
348 | |
349 | /* |
350 | * If we've come out of here by `break', n will still be |
d2371c81 |
351 | * non-NULL and all we need to do is go back up the tree |
352 | * updating counts. If we've come here because n is NULL, we |
353 | * need to create a new root for the tree because the old one |
354 | * has just split into two. */ |
355 | if (n) { |
356 | while (n->parent) { |
357 | int count = countnode234(n); |
358 | int childnum; |
359 | childnum = (n->parent->kids[0] == n ? 0 : |
360 | n->parent->kids[1] == n ? 1 : |
361 | n->parent->kids[2] == n ? 2 : 3); |
362 | n->parent->counts[childnum] = count; |
363 | n = n->parent; |
364 | } |
365 | } else { |
febd9a0f |
366 | LOG((" root is overloaded, split into two\n")); |
367 | t->root = mknew(node234); |
d2371c81 |
368 | t->root->kids[0] = left; t->root->counts[0] = lcount; |
febd9a0f |
369 | t->root->elems[0] = e; |
d2371c81 |
370 | t->root->kids[1] = right; t->root->counts[1] = rcount; |
febd9a0f |
371 | t->root->elems[1] = NULL; |
d2371c81 |
372 | t->root->kids[2] = NULL; t->root->counts[2] = 0; |
febd9a0f |
373 | t->root->elems[2] = NULL; |
d2371c81 |
374 | t->root->kids[3] = NULL; t->root->counts[3] = 0; |
febd9a0f |
375 | t->root->parent = NULL; |
376 | if (t->root->kids[0]) t->root->kids[0]->parent = t->root; |
377 | if (t->root->kids[1]) t->root->kids[1]->parent = t->root; |
d2371c81 |
378 | LOG((" new root is %p/%d [%p] %p/%d\n", |
379 | t->root->kids[0], t->root->counts[0], |
380 | t->root->elems[0], |
381 | t->root->kids[1], t->root->counts[1])); |
febd9a0f |
382 | } |
383 | |
384 | return orig_e; |
385 | } |
386 | |
d2371c81 |
387 | void *add234(tree234 *t, void *e) { |
388 | if (!t->cmp) /* tree is unsorted */ |
389 | return NULL; |
390 | |
391 | return add234_internal(t, e, -1); |
392 | } |
393 | void *addpos234(tree234 *t, void *e, int index) { |
394 | if (index < 0 || /* index out of range */ |
395 | t->cmp) /* tree is sorted */ |
396 | return NULL; /* return failure */ |
397 | |
398 | return add234_internal(t, e, index); /* this checks the upper bound */ |
399 | } |
400 | |
febd9a0f |
401 | /* |
d2371c81 |
402 | * Look up the element at a given numeric index in a 2-3-4 tree. |
403 | * Returns NULL if the index is out of range. |
febd9a0f |
404 | */ |
d2371c81 |
405 | void *index234(tree234 *t, int index) { |
febd9a0f |
406 | node234 *n; |
febd9a0f |
407 | |
d2371c81 |
408 | if (!t->root) |
409 | return NULL; /* tree is empty */ |
febd9a0f |
410 | |
d2371c81 |
411 | if (index < 0 || index >= countnode234(t->root)) |
412 | return NULL; /* out of range */ |
febd9a0f |
413 | |
414 | n = t->root; |
d2371c81 |
415 | |
febd9a0f |
416 | while (n) { |
d2371c81 |
417 | if (index < n->counts[0]) |
febd9a0f |
418 | n = n->kids[0]; |
d2371c81 |
419 | else if (index -= n->counts[0] + 1, index < 0) |
febd9a0f |
420 | return n->elems[0]; |
d2371c81 |
421 | else if (index < n->counts[1]) |
febd9a0f |
422 | n = n->kids[1]; |
d2371c81 |
423 | else if (index -= n->counts[1] + 1, index < 0) |
febd9a0f |
424 | return n->elems[1]; |
d2371c81 |
425 | else if (index < n->counts[2]) |
febd9a0f |
426 | n = n->kids[2]; |
d2371c81 |
427 | else if (index -= n->counts[2] + 1, index < 0) |
febd9a0f |
428 | return n->elems[2]; |
429 | else |
430 | n = n->kids[3]; |
431 | } |
432 | |
d2371c81 |
433 | /* We shouldn't ever get here. I wonder how we did. */ |
434 | return NULL; |
435 | } |
436 | |
437 | /* |
438 | * Find an element e in a sorted 2-3-4 tree t. Returns NULL if not |
439 | * found. e is always passed as the first argument to cmp, so cmp |
440 | * can be an asymmetric function if desired. cmp can also be passed |
441 | * as NULL, in which case the compare function from the tree proper |
442 | * will be used. |
443 | */ |
444 | void *findrelpos234(tree234 *t, void *e, cmpfn234 cmp, |
445 | int relation, int *index) { |
446 | node234 *n; |
447 | void *ret; |
448 | int c; |
449 | int idx, ecount, kcount, cmpret; |
450 | |
451 | if (t->root == NULL) |
452 | return NULL; |
453 | |
454 | if (cmp == NULL) |
455 | cmp = t->cmp; |
456 | |
457 | n = t->root; |
febd9a0f |
458 | /* |
d2371c81 |
459 | * Attempt to find the element itself. |
febd9a0f |
460 | */ |
d2371c81 |
461 | idx = 0; |
462 | ecount = -1; |
463 | /* |
464 | * Prepare a fake `cmp' result if e is NULL. |
465 | */ |
466 | cmpret = 0; |
467 | if (e == NULL) { |
468 | assert(relation == REL234_LT || relation == REL234_GT); |
469 | if (relation == REL234_LT) |
470 | cmpret = +1; /* e is a max: always greater */ |
471 | else if (relation == REL234_GT) |
472 | cmpret = -1; /* e is a min: always smaller */ |
473 | } |
474 | while (1) { |
475 | for (kcount = 0; kcount < 4; kcount++) { |
476 | if (kcount >= 3 || n->elems[kcount] == NULL || |
477 | (c = cmpret ? cmpret : cmp(e, n->elems[kcount])) < 0) { |
478 | break; |
479 | } |
480 | if (n->kids[kcount]) idx += n->counts[kcount]; |
481 | if (c == 0) { |
482 | ecount = kcount; |
483 | break; |
484 | } |
485 | idx++; |
486 | } |
487 | if (ecount >= 0) |
488 | break; |
489 | if (n->kids[kcount]) |
490 | n = n->kids[kcount]; |
491 | else |
492 | break; |
493 | } |
494 | |
495 | if (ecount >= 0) { |
496 | /* |
497 | * We have found the element we're looking for. It's |
498 | * n->elems[ecount], at tree index idx. If our search |
499 | * relation is EQ, LE or GE we can now go home. |
500 | */ |
501 | if (relation != REL234_LT && relation != REL234_GT) { |
502 | if (index) *index = idx; |
503 | return n->elems[ecount]; |
504 | } |
505 | |
506 | /* |
507 | * Otherwise, we'll do an indexed lookup for the previous |
508 | * or next element. (It would be perfectly possible to |
509 | * implement these search types in a non-counted tree by |
510 | * going back up from where we are, but far more fiddly.) |
511 | */ |
512 | if (relation == REL234_LT) |
513 | idx--; |
514 | else |
515 | idx++; |
516 | } else { |
517 | /* |
518 | * We've found our way to the bottom of the tree and we |
519 | * know where we would insert this node if we wanted to: |
520 | * we'd put it in in place of the (empty) subtree |
521 | * n->kids[kcount], and it would have index idx |
522 | * |
523 | * But the actual element isn't there. So if our search |
524 | * relation is EQ, we're doomed. |
525 | */ |
526 | if (relation == REL234_EQ) |
527 | return NULL; |
528 | |
529 | /* |
530 | * Otherwise, we must do an index lookup for index idx-1 |
531 | * (if we're going left - LE or LT) or index idx (if we're |
532 | * going right - GE or GT). |
533 | */ |
534 | if (relation == REL234_LT || relation == REL234_LE) { |
535 | idx--; |
536 | } |
537 | } |
538 | |
539 | /* |
540 | * We know the index of the element we want; just call index234 |
541 | * to do the rest. This will return NULL if the index is out of |
542 | * bounds, which is exactly what we want. |
543 | */ |
544 | ret = index234(t, idx); |
545 | if (ret && index) *index = idx; |
546 | return ret; |
547 | } |
548 | void *find234(tree234 *t, void *e, cmpfn234 cmp) { |
549 | return findrelpos234(t, e, cmp, REL234_EQ, NULL); |
550 | } |
551 | void *findrel234(tree234 *t, void *e, cmpfn234 cmp, int relation) { |
552 | return findrelpos234(t, e, cmp, relation, NULL); |
553 | } |
554 | void *findpos234(tree234 *t, void *e, cmpfn234 cmp, int *index) { |
555 | return findrelpos234(t, e, cmp, REL234_EQ, index); |
febd9a0f |
556 | } |
557 | |
558 | /* |
559 | * Delete an element e in a 2-3-4 tree. Does not free the element, |
560 | * merely removes all links to it from the tree nodes. |
561 | */ |
d2371c81 |
562 | static void *delpos234_internal(tree234 *t, int index) { |
febd9a0f |
563 | node234 *n; |
d2371c81 |
564 | void *retval; |
febd9a0f |
565 | int ei = -1; |
566 | |
d2371c81 |
567 | retval = 0; |
568 | |
febd9a0f |
569 | n = t->root; |
d2371c81 |
570 | LOG(("deleting item %d from tree %p\n", index, t)); |
febd9a0f |
571 | while (1) { |
572 | while (n) { |
573 | int c; |
574 | int ki; |
575 | node234 *sub; |
576 | |
d2371c81 |
577 | LOG((" node %p: %p/%d [%p] %p/%d [%p] %p/%d [%p] %p/%d index=%d\n", |
578 | n, |
579 | n->kids[0], n->counts[0], n->elems[0], |
580 | n->kids[1], n->counts[1], n->elems[1], |
581 | n->kids[2], n->counts[2], n->elems[2], |
582 | n->kids[3], n->counts[3], |
583 | index)); |
584 | if (index < n->counts[0]) { |
febd9a0f |
585 | ki = 0; |
d2371c81 |
586 | } else if (index -= n->counts[0]+1, index < 0) { |
febd9a0f |
587 | ei = 0; break; |
d2371c81 |
588 | } else if (index < n->counts[1]) { |
febd9a0f |
589 | ki = 1; |
d2371c81 |
590 | } else if (index -= n->counts[1]+1, index < 0) { |
febd9a0f |
591 | ei = 1; break; |
d2371c81 |
592 | } else if (index < n->counts[2]) { |
febd9a0f |
593 | ki = 2; |
d2371c81 |
594 | } else if (index -= n->counts[2]+1, index < 0) { |
febd9a0f |
595 | ei = 2; break; |
596 | } else { |
597 | ki = 3; |
598 | } |
599 | /* |
600 | * Recurse down to subtree ki. If it has only one element, |
601 | * we have to do some transformation to start with. |
602 | */ |
603 | LOG((" moving to subtree %d\n", ki)); |
604 | sub = n->kids[ki]; |
605 | if (!sub->elems[1]) { |
606 | LOG((" subtree has only one element!\n", ki)); |
607 | if (ki > 0 && n->kids[ki-1]->elems[1]) { |
608 | /* |
609 | * Case 3a, left-handed variant. Child ki has |
610 | * only one element, but child ki-1 has two or |
611 | * more. So we need to move a subtree from ki-1 |
612 | * to ki. |
613 | * |
614 | * . C . . B . |
615 | * / \ -> / \ |
616 | * [more] a A b B c d D e [more] a A b c C d D e |
617 | */ |
618 | node234 *sib = n->kids[ki-1]; |
619 | int lastelem = (sib->elems[2] ? 2 : |
620 | sib->elems[1] ? 1 : 0); |
621 | sub->kids[2] = sub->kids[1]; |
d2371c81 |
622 | sub->counts[2] = sub->counts[1]; |
febd9a0f |
623 | sub->elems[1] = sub->elems[0]; |
624 | sub->kids[1] = sub->kids[0]; |
d2371c81 |
625 | sub->counts[1] = sub->counts[0]; |
febd9a0f |
626 | sub->elems[0] = n->elems[ki-1]; |
627 | sub->kids[0] = sib->kids[lastelem+1]; |
d2371c81 |
628 | sub->counts[0] = sib->counts[lastelem+1]; |
100122a9 |
629 | if (sub->kids[0]) sub->kids[0]->parent = sub; |
febd9a0f |
630 | n->elems[ki-1] = sib->elems[lastelem]; |
631 | sib->kids[lastelem+1] = NULL; |
d2371c81 |
632 | sib->counts[lastelem+1] = 0; |
febd9a0f |
633 | sib->elems[lastelem] = NULL; |
d2371c81 |
634 | n->counts[ki] = countnode234(sub); |
febd9a0f |
635 | LOG((" case 3a left\n")); |
d2371c81 |
636 | LOG((" index and left subtree count before adjustment: %d, %d\n", |
637 | index, n->counts[ki-1])); |
638 | index += n->counts[ki-1]; |
639 | n->counts[ki-1] = countnode234(sib); |
640 | index -= n->counts[ki-1]; |
641 | LOG((" index and left subtree count after adjustment: %d, %d\n", |
642 | index, n->counts[ki-1])); |
febd9a0f |
643 | } else if (ki < 3 && n->kids[ki+1] && |
644 | n->kids[ki+1]->elems[1]) { |
645 | /* |
646 | * Case 3a, right-handed variant. ki has only |
647 | * one element but ki+1 has two or more. Move a |
648 | * subtree from ki+1 to ki. |
649 | * |
650 | * . B . . C . |
651 | * / \ -> / \ |
652 | * a A b c C d D e [more] a A b B c d D e [more] |
653 | */ |
654 | node234 *sib = n->kids[ki+1]; |
655 | int j; |
656 | sub->elems[1] = n->elems[ki]; |
657 | sub->kids[2] = sib->kids[0]; |
d2371c81 |
658 | sub->counts[2] = sib->counts[0]; |
100122a9 |
659 | if (sub->kids[2]) sub->kids[2]->parent = sub; |
febd9a0f |
660 | n->elems[ki] = sib->elems[0]; |
661 | sib->kids[0] = sib->kids[1]; |
d2371c81 |
662 | sib->counts[0] = sib->counts[1]; |
febd9a0f |
663 | for (j = 0; j < 2 && sib->elems[j+1]; j++) { |
664 | sib->kids[j+1] = sib->kids[j+2]; |
d2371c81 |
665 | sib->counts[j+1] = sib->counts[j+2]; |
febd9a0f |
666 | sib->elems[j] = sib->elems[j+1]; |
667 | } |
668 | sib->kids[j+1] = NULL; |
d2371c81 |
669 | sib->counts[j+1] = 0; |
febd9a0f |
670 | sib->elems[j] = NULL; |
d2371c81 |
671 | n->counts[ki] = countnode234(sub); |
672 | n->counts[ki+1] = countnode234(sib); |
febd9a0f |
673 | LOG((" case 3a right\n")); |
674 | } else { |
675 | /* |
676 | * Case 3b. ki has only one element, and has no |
677 | * neighbour with more than one. So pick a |
678 | * neighbour and merge it with ki, taking an |
679 | * element down from n to go in the middle. |
680 | * |
681 | * . B . . |
682 | * / \ -> | |
683 | * a A b c C d a A b B c C d |
684 | * |
685 | * (Since at all points we have avoided |
686 | * descending to a node with only one element, |
687 | * we can be sure that n is not reduced to |
688 | * nothingness by this move, _unless_ it was |
689 | * the very first node, ie the root of the |
690 | * tree. In that case we remove the now-empty |
691 | * root and replace it with its single large |
692 | * child as shown.) |
693 | */ |
694 | node234 *sib; |
695 | int j; |
696 | |
d2371c81 |
697 | if (ki > 0) { |
febd9a0f |
698 | ki--; |
d2371c81 |
699 | index += n->counts[ki] + 1; |
700 | } |
febd9a0f |
701 | sib = n->kids[ki]; |
702 | sub = n->kids[ki+1]; |
703 | |
704 | sub->kids[3] = sub->kids[1]; |
d2371c81 |
705 | sub->counts[3] = sub->counts[1]; |
febd9a0f |
706 | sub->elems[2] = sub->elems[0]; |
707 | sub->kids[2] = sub->kids[0]; |
d2371c81 |
708 | sub->counts[2] = sub->counts[0]; |
febd9a0f |
709 | sub->elems[1] = n->elems[ki]; |
710 | sub->kids[1] = sib->kids[1]; |
d2371c81 |
711 | sub->counts[1] = sib->counts[1]; |
100122a9 |
712 | if (sub->kids[1]) sub->kids[1]->parent = sub; |
febd9a0f |
713 | sub->elems[0] = sib->elems[0]; |
714 | sub->kids[0] = sib->kids[0]; |
d2371c81 |
715 | sub->counts[0] = sib->counts[0]; |
100122a9 |
716 | if (sub->kids[0]) sub->kids[0]->parent = sub; |
febd9a0f |
717 | |
d2371c81 |
718 | n->counts[ki+1] = countnode234(sub); |
719 | |
febd9a0f |
720 | sfree(sib); |
721 | |
722 | /* |
723 | * That's built the big node in sub. Now we |
724 | * need to remove the reference to sib in n. |
725 | */ |
726 | for (j = ki; j < 3 && n->kids[j+1]; j++) { |
727 | n->kids[j] = n->kids[j+1]; |
d2371c81 |
728 | n->counts[j] = n->counts[j+1]; |
febd9a0f |
729 | n->elems[j] = j<2 ? n->elems[j+1] : NULL; |
730 | } |
731 | n->kids[j] = NULL; |
d2371c81 |
732 | n->counts[j] = 0; |
febd9a0f |
733 | if (j < 3) n->elems[j] = NULL; |
2d56b16f |
734 | LOG((" case 3b ki=%d\n", ki)); |
febd9a0f |
735 | |
736 | if (!n->elems[0]) { |
737 | /* |
738 | * The root is empty and needs to be |
739 | * removed. |
740 | */ |
741 | LOG((" shifting root!\n")); |
742 | t->root = sub; |
743 | sub->parent = NULL; |
744 | sfree(n); |
745 | } |
746 | } |
747 | } |
748 | n = sub; |
749 | } |
d2371c81 |
750 | if (!retval) |
751 | retval = n->elems[ei]; |
752 | |
febd9a0f |
753 | if (ei==-1) |
d2371c81 |
754 | return NULL; /* although this shouldn't happen */ |
febd9a0f |
755 | |
756 | /* |
757 | * Treat special case: this is the one remaining item in |
758 | * the tree. n is the tree root (no parent), has one |
759 | * element (no elems[1]), and has no kids (no kids[0]). |
760 | */ |
761 | if (!n->parent && !n->elems[1] && !n->kids[0]) { |
762 | LOG((" removed last element in tree\n")); |
763 | sfree(n); |
764 | t->root = NULL; |
d2371c81 |
765 | return retval; |
febd9a0f |
766 | } |
767 | |
768 | /* |
769 | * Now we have the element we want, as n->elems[ei], and we |
770 | * have also arranged for that element not to be the only |
771 | * one in its node. So... |
772 | */ |
773 | |
774 | if (!n->kids[0] && n->elems[1]) { |
775 | /* |
776 | * Case 1. n is a leaf node with more than one element, |
777 | * so it's _really easy_. Just delete the thing and |
778 | * we're done. |
779 | */ |
780 | int i; |
781 | LOG((" case 1\n")); |
a4a19e73 |
782 | for (i = ei; i < 2 && n->elems[i+1]; i++) |
febd9a0f |
783 | n->elems[i] = n->elems[i+1]; |
784 | n->elems[i] = NULL; |
d2371c81 |
785 | /* |
786 | * Having done that to the leaf node, we now go back up |
787 | * the tree fixing the counts. |
788 | */ |
789 | while (n->parent) { |
790 | int childnum; |
791 | childnum = (n->parent->kids[0] == n ? 0 : |
792 | n->parent->kids[1] == n ? 1 : |
793 | n->parent->kids[2] == n ? 2 : 3); |
794 | n->parent->counts[childnum]--; |
795 | n = n->parent; |
796 | } |
797 | return retval; /* finished! */ |
febd9a0f |
798 | } else if (n->kids[ei]->elems[1]) { |
799 | /* |
800 | * Case 2a. n is an internal node, and the root of the |
801 | * subtree to the left of e has more than one element. |
802 | * So find the predecessor p to e (ie the largest node |
803 | * in that subtree), place it where e currently is, and |
804 | * then start the deletion process over again on the |
805 | * subtree with p as target. |
806 | */ |
807 | node234 *m = n->kids[ei]; |
808 | void *target; |
809 | LOG((" case 2a\n")); |
810 | while (m->kids[0]) { |
811 | m = (m->kids[3] ? m->kids[3] : |
812 | m->kids[2] ? m->kids[2] : |
813 | m->kids[1] ? m->kids[1] : m->kids[0]); |
814 | } |
815 | target = (m->elems[2] ? m->elems[2] : |
816 | m->elems[1] ? m->elems[1] : m->elems[0]); |
817 | n->elems[ei] = target; |
d2371c81 |
818 | index = n->counts[ei]-1; |
febd9a0f |
819 | n = n->kids[ei]; |
febd9a0f |
820 | } else if (n->kids[ei+1]->elems[1]) { |
821 | /* |
822 | * Case 2b, symmetric to 2a but s/left/right/ and |
823 | * s/predecessor/successor/. (And s/largest/smallest/). |
824 | */ |
825 | node234 *m = n->kids[ei+1]; |
826 | void *target; |
827 | LOG((" case 2b\n")); |
828 | while (m->kids[0]) { |
829 | m = m->kids[0]; |
830 | } |
831 | target = m->elems[0]; |
832 | n->elems[ei] = target; |
833 | n = n->kids[ei+1]; |
d2371c81 |
834 | index = 0; |
febd9a0f |
835 | } else { |
836 | /* |
837 | * Case 2c. n is an internal node, and the subtrees to |
838 | * the left and right of e both have only one element. |
839 | * So combine the two subnodes into a single big node |
840 | * with their own elements on the left and right and e |
841 | * in the middle, then restart the deletion process on |
842 | * that subtree, with e still as target. |
843 | */ |
844 | node234 *a = n->kids[ei], *b = n->kids[ei+1]; |
845 | int j; |
846 | |
847 | LOG((" case 2c\n")); |
848 | a->elems[1] = n->elems[ei]; |
849 | a->kids[2] = b->kids[0]; |
d2371c81 |
850 | a->counts[2] = b->counts[0]; |
100122a9 |
851 | if (a->kids[2]) a->kids[2]->parent = a; |
febd9a0f |
852 | a->elems[2] = b->elems[0]; |
853 | a->kids[3] = b->kids[1]; |
d2371c81 |
854 | a->counts[3] = b->counts[1]; |
100122a9 |
855 | if (a->kids[3]) a->kids[3]->parent = a; |
febd9a0f |
856 | sfree(b); |
d2371c81 |
857 | n->counts[ei] = countnode234(a); |
febd9a0f |
858 | /* |
859 | * That's built the big node in a, and destroyed b. Now |
860 | * remove the reference to b (and e) in n. |
861 | */ |
862 | for (j = ei; j < 2 && n->elems[j+1]; j++) { |
863 | n->elems[j] = n->elems[j+1]; |
864 | n->kids[j+1] = n->kids[j+2]; |
d2371c81 |
865 | n->counts[j+1] = n->counts[j+2]; |
febd9a0f |
866 | } |
867 | n->elems[j] = NULL; |
868 | n->kids[j+1] = NULL; |
d2371c81 |
869 | n->counts[j+1] = 0; |
e9e9556d |
870 | /* |
871 | * It's possible, in this case, that we've just removed |
872 | * the only element in the root of the tree. If so, |
873 | * shift the root. |
874 | */ |
875 | if (n->elems[0] == NULL) { |
876 | LOG((" shifting root!\n")); |
877 | t->root = a; |
878 | a->parent = NULL; |
879 | sfree(n); |
880 | } |
febd9a0f |
881 | /* |
882 | * Now go round the deletion process again, with n |
883 | * pointing at the new big node and e still the same. |
884 | */ |
885 | n = a; |
d2371c81 |
886 | index = a->counts[0] + a->counts[1] + 1; |
febd9a0f |
887 | } |
888 | } |
889 | } |
d2371c81 |
890 | void *delpos234(tree234 *t, int index) { |
891 | if (index < 0 || index >= countnode234(t->root)) |
febd9a0f |
892 | return NULL; |
d2371c81 |
893 | return delpos234_internal(t, index); |
febd9a0f |
894 | } |
d2371c81 |
895 | void *del234(tree234 *t, void *e) { |
896 | int index; |
897 | if (!findrelpos234(t, e, NULL, REL234_EQ, &index)) |
898 | return NULL; /* it wasn't in there anyway */ |
899 | return delpos234_internal(t, index); /* it's there; delete it. */ |
febd9a0f |
900 | } |
901 | |
902 | #ifdef TEST |
903 | |
2d56b16f |
904 | /* |
905 | * Test code for the 2-3-4 tree. This code maintains an alternative |
906 | * representation of the data in the tree, in an array (using the |
907 | * obvious and slow insert and delete functions). After each tree |
7aa7c43a |
908 | * operation, the verify() function is called, which ensures all |
d2371c81 |
909 | * the tree properties are preserved: |
910 | * - node->child->parent always equals node |
911 | * - tree->root->parent always equals NULL |
912 | * - number of kids == 0 or number of elements + 1; |
913 | * - tree has the same depth everywhere |
914 | * - every node has at least one element |
915 | * - subtree element counts are accurate |
916 | * - any NULL kid pointer is accompanied by a zero count |
917 | * - in a sorted tree: ordering property between elements of a |
918 | * node and elements of its children is preserved |
919 | * and also ensures the list represented by the tree is the same |
920 | * list it should be. (This last check also doubly verifies the |
921 | * ordering properties, because the `same list it should be' is by |
922 | * definition correctly ordered. It also ensures all nodes are |
923 | * distinct, because the enum functions would get caught in a loop |
924 | * if not.) |
2d56b16f |
925 | */ |
926 | |
927 | #include <stdarg.h> |
928 | |
d2371c81 |
929 | #define srealloc realloc |
930 | |
2d56b16f |
931 | /* |
932 | * Error reporting function. |
933 | */ |
934 | void error(char *fmt, ...) { |
935 | va_list ap; |
936 | printf("ERROR: "); |
937 | va_start(ap, fmt); |
938 | vfprintf(stdout, fmt, ap); |
939 | va_end(ap); |
940 | printf("\n"); |
941 | } |
942 | |
943 | /* The array representation of the data. */ |
944 | void **array; |
945 | int arraylen, arraysize; |
946 | cmpfn234 cmp; |
947 | |
948 | /* The tree representation of the same data. */ |
949 | tree234 *tree; |
950 | |
951 | typedef struct { |
952 | int treedepth; |
953 | int elemcount; |
954 | } chkctx; |
955 | |
d2371c81 |
956 | int chknode(chkctx *ctx, int level, node234 *node, |
2d56b16f |
957 | void *lowbound, void *highbound) { |
958 | int nkids, nelems; |
959 | int i; |
d2371c81 |
960 | int count; |
2d56b16f |
961 | |
962 | /* Count the non-NULL kids. */ |
963 | for (nkids = 0; nkids < 4 && node->kids[nkids]; nkids++); |
964 | /* Ensure no kids beyond the first NULL are non-NULL. */ |
965 | for (i = nkids; i < 4; i++) |
966 | if (node->kids[i]) { |
967 | error("node %p: nkids=%d but kids[%d] non-NULL", |
968 | node, nkids, i); |
d2371c81 |
969 | } else if (node->counts[i]) { |
970 | error("node %p: kids[%d] NULL but count[%d]=%d nonzero", |
971 | node, i, i, node->counts[i]); |
972 | } |
2d56b16f |
973 | |
974 | /* Count the non-NULL elements. */ |
975 | for (nelems = 0; nelems < 3 && node->elems[nelems]; nelems++); |
976 | /* Ensure no elements beyond the first NULL are non-NULL. */ |
977 | for (i = nelems; i < 3; i++) |
978 | if (node->elems[i]) { |
979 | error("node %p: nelems=%d but elems[%d] non-NULL", |
980 | node, nelems, i); |
981 | } |
982 | |
983 | if (nkids == 0) { |
984 | /* |
985 | * If nkids==0, this is a leaf node; verify that the tree |
986 | * depth is the same everywhere. |
987 | */ |
988 | if (ctx->treedepth < 0) |
989 | ctx->treedepth = level; /* we didn't know the depth yet */ |
990 | else if (ctx->treedepth != level) |
991 | error("node %p: leaf at depth %d, previously seen depth %d", |
992 | node, level, ctx->treedepth); |
993 | } else { |
994 | /* |
995 | * If nkids != 0, then it should be nelems+1, unless nelems |
996 | * is 0 in which case nkids should also be 0 (and so we |
997 | * shouldn't be in this condition at all). |
998 | */ |
999 | int shouldkids = (nelems ? nelems+1 : 0); |
1000 | if (nkids != shouldkids) { |
1001 | error("node %p: %d elems should mean %d kids but has %d", |
1002 | node, nelems, shouldkids, nkids); |
1003 | } |
1004 | } |
1005 | |
1006 | /* |
1007 | * nelems should be at least 1. |
1008 | */ |
1009 | if (nelems == 0) { |
1010 | error("node %p: no elems", node, nkids); |
1011 | } |
1012 | |
1013 | /* |
d2371c81 |
1014 | * Add nelems to the running element count of the whole tree. |
2d56b16f |
1015 | */ |
1016 | ctx->elemcount += nelems; |
1017 | |
1018 | /* |
1019 | * Check ordering property: all elements should be strictly > |
1020 | * lowbound, strictly < highbound, and strictly < each other in |
1021 | * sequence. (lowbound and highbound are NULL at edges of tree |
1022 | * - both NULL at root node - and NULL is considered to be < |
1023 | * everything and > everything. IYSWIM.) |
1024 | */ |
d2371c81 |
1025 | if (cmp) { |
1026 | for (i = -1; i < nelems; i++) { |
1027 | void *lower = (i == -1 ? lowbound : node->elems[i]); |
1028 | void *higher = (i+1 == nelems ? highbound : node->elems[i+1]); |
1029 | if (lower && higher && cmp(lower, higher) >= 0) { |
1030 | error("node %p: kid comparison [%d=%s,%d=%s] failed", |
1031 | node, i, lower, i+1, higher); |
1032 | } |
1033 | } |
2d56b16f |
1034 | } |
1035 | |
1036 | /* |
1037 | * Check parent pointers: all non-NULL kids should have a |
1038 | * parent pointer coming back to this node. |
1039 | */ |
1040 | for (i = 0; i < nkids; i++) |
1041 | if (node->kids[i]->parent != node) { |
1042 | error("node %p kid %d: parent ptr is %p not %p", |
1043 | node, i, node->kids[i]->parent, node); |
1044 | } |
1045 | |
1046 | |
1047 | /* |
1048 | * Now (finally!) recurse into subtrees. |
1049 | */ |
d2371c81 |
1050 | count = nelems; |
1051 | |
2d56b16f |
1052 | for (i = 0; i < nkids; i++) { |
1053 | void *lower = (i == 0 ? lowbound : node->elems[i-1]); |
1054 | void *higher = (i >= nelems ? highbound : node->elems[i]); |
d2371c81 |
1055 | int subcount = chknode(ctx, level+1, node->kids[i], lower, higher); |
1056 | if (node->counts[i] != subcount) { |
1057 | error("node %p kid %d: count says %d, subtree really has %d", |
1058 | node, i, node->counts[i], subcount); |
1059 | } |
1060 | count += subcount; |
2d56b16f |
1061 | } |
d2371c81 |
1062 | |
1063 | return count; |
2d56b16f |
1064 | } |
1065 | |
1066 | void verify(void) { |
1067 | chkctx ctx; |
2d56b16f |
1068 | int i; |
1069 | void *p; |
1070 | |
1071 | ctx.treedepth = -1; /* depth unknown yet */ |
1072 | ctx.elemcount = 0; /* no elements seen yet */ |
1073 | /* |
1074 | * Verify validity of tree properties. |
1075 | */ |
d2371c81 |
1076 | if (tree->root) { |
1077 | if (tree->root->parent != NULL) |
1078 | error("root->parent is %p should be null", tree->root->parent); |
2d56b16f |
1079 | chknode(&ctx, 0, tree->root, NULL, NULL); |
d2371c81 |
1080 | } |
2d56b16f |
1081 | printf("tree depth: %d\n", ctx.treedepth); |
1082 | /* |
1083 | * Enumerate the tree and ensure it matches up to the array. |
1084 | */ |
d2371c81 |
1085 | for (i = 0; NULL != (p = index234(tree, i)); i++) { |
2d56b16f |
1086 | if (i >= arraylen) |
1087 | error("tree contains more than %d elements", arraylen); |
1088 | if (array[i] != p) |
1089 | error("enum at position %d: array says %s, tree says %s", |
1090 | i, array[i], p); |
1091 | } |
d2371c81 |
1092 | if (ctx.elemcount != i) { |
2d56b16f |
1093 | error("tree really contains %d elements, enum gave %d", |
d2371c81 |
1094 | ctx.elemcount, i); |
2d56b16f |
1095 | } |
1096 | if (i < arraylen) { |
1097 | error("enum gave only %d elements, array has %d", i, arraylen); |
1098 | } |
d2371c81 |
1099 | i = count234(tree); |
1100 | if (ctx.elemcount != i) { |
1101 | error("tree really contains %d elements, count234 gave %d", |
1102 | ctx.elemcount, i); |
1103 | } |
2d56b16f |
1104 | } |
1105 | |
d2371c81 |
1106 | void internal_addtest(void *elem, int index, void *realret) { |
2d56b16f |
1107 | int i, j; |
d2371c81 |
1108 | void *retval; |
2d56b16f |
1109 | |
1110 | if (arraysize < arraylen+1) { |
1111 | arraysize = arraylen+1+256; |
dcbde236 |
1112 | array = (array == NULL ? smalloc(arraysize*sizeof(*array)) : |
1113 | srealloc(array, arraysize*sizeof(*array))); |
2d56b16f |
1114 | } |
1115 | |
d2371c81 |
1116 | i = index; |
2d56b16f |
1117 | /* now i points to the first element >= elem */ |
d2371c81 |
1118 | retval = elem; /* expect elem returned (success) */ |
1119 | for (j = arraylen; j > i; j--) |
1120 | array[j] = array[j-1]; |
1121 | array[i] = elem; /* add elem to array */ |
1122 | arraylen++; |
2d56b16f |
1123 | |
2d56b16f |
1124 | if (realret != retval) { |
1125 | error("add: retval was %p expected %p", realret, retval); |
1126 | } |
1127 | |
1128 | verify(); |
1129 | } |
1130 | |
d2371c81 |
1131 | void addtest(void *elem) { |
2d56b16f |
1132 | int i; |
d2371c81 |
1133 | void *realret; |
1134 | |
1135 | realret = add234(tree, elem); |
2d56b16f |
1136 | |
1137 | i = 0; |
1138 | while (i < arraylen && cmp(elem, array[i]) > 0) |
1139 | i++; |
d2371c81 |
1140 | if (i < arraylen && !cmp(elem, array[i])) { |
1141 | void *retval = array[i]; /* expect that returned not elem */ |
1142 | if (realret != retval) { |
1143 | error("add: retval was %p expected %p", realret, retval); |
1144 | } |
1145 | } else |
1146 | internal_addtest(elem, i, realret); |
1147 | } |
1148 | |
1149 | void addpostest(void *elem, int i) { |
1150 | void *realret; |
1151 | |
1152 | realret = addpos234(tree, elem, i); |
1153 | |
1154 | internal_addtest(elem, i, realret); |
1155 | } |
1156 | |
1157 | void delpostest(int i) { |
1158 | int index = i; |
1159 | void *elem = array[i], *ret; |
1160 | |
1161 | /* i points to the right element */ |
1162 | while (i < arraylen-1) { |
1163 | array[i] = array[i+1]; |
1164 | i++; |
2d56b16f |
1165 | } |
d2371c81 |
1166 | arraylen--; /* delete elem from array */ |
1167 | |
1168 | if (tree->cmp) |
1169 | ret = del234(tree, elem); |
1170 | else |
1171 | ret = delpos234(tree, index); |
2d56b16f |
1172 | |
d2371c81 |
1173 | if (ret != elem) { |
1174 | error("del returned %p, expected %p", ret, elem); |
1175 | } |
2d56b16f |
1176 | |
1177 | verify(); |
febd9a0f |
1178 | } |
2d56b16f |
1179 | |
d2371c81 |
1180 | void deltest(void *elem) { |
1181 | int i; |
1182 | |
1183 | i = 0; |
1184 | while (i < arraylen && cmp(elem, array[i]) > 0) |
1185 | i++; |
1186 | if (i >= arraylen || cmp(elem, array[i]) != 0) |
1187 | return; /* don't do it! */ |
1188 | delpostest(i); |
1189 | } |
1190 | |
2d56b16f |
1191 | /* A sample data set and test utility. Designed for pseudo-randomness, |
1192 | * and yet repeatability. */ |
1193 | |
1194 | /* |
1195 | * This random number generator uses the `portable implementation' |
1196 | * given in ANSI C99 draft N869. It assumes `unsigned' is 32 bits; |
1197 | * change it if not. |
1198 | */ |
1199 | int randomnumber(unsigned *seed) { |
1200 | *seed *= 1103515245; |
1201 | *seed += 12345; |
1202 | return ((*seed) / 65536) % 32768; |
febd9a0f |
1203 | } |
1204 | |
2d56b16f |
1205 | int mycmp(void *av, void *bv) { |
1206 | char const *a = (char const *)av; |
1207 | char const *b = (char const *)bv; |
febd9a0f |
1208 | return strcmp(a, b); |
1209 | } |
1210 | |
2d56b16f |
1211 | #define lenof(x) ( sizeof((x)) / sizeof(*(x)) ) |
1212 | |
1213 | char *strings[] = { |
1214 | "a", "ab", "absque", "coram", "de", |
1215 | "palam", "clam", "cum", "ex", "e", |
1216 | "sine", "tenus", "pro", "prae", |
1217 | "banana", "carrot", "cabbage", "broccoli", "onion", "zebra", |
1218 | "penguin", "blancmange", "pangolin", "whale", "hedgehog", |
1219 | "giraffe", "peanut", "bungee", "foo", "bar", "baz", "quux", |
1220 | "murfl", "spoo", "breen", "flarn", "octothorpe", |
1221 | "snail", "tiger", "elephant", "octopus", "warthog", "armadillo", |
1222 | "aardvark", "wyvern", "dragon", "elf", "dwarf", "orc", "goblin", |
1223 | "pixie", "basilisk", "warg", "ape", "lizard", "newt", "shopkeeper", |
1224 | "wand", "ring", "amulet" |
1225 | }; |
1226 | |
1227 | #define NSTR lenof(strings) |
1228 | |
d2371c81 |
1229 | int findtest(void) { |
1230 | const static int rels[] = { |
1231 | REL234_EQ, REL234_GE, REL234_LE, REL234_LT, REL234_GT |
1232 | }; |
1233 | const static char *const relnames[] = { |
1234 | "EQ", "GE", "LE", "LT", "GT" |
1235 | }; |
1236 | int i, j, rel, index; |
1237 | char *p, *ret, *realret, *realret2; |
1238 | int lo, hi, mid, c; |
1239 | |
1240 | for (i = 0; i < NSTR; i++) { |
1241 | p = strings[i]; |
1242 | for (j = 0; j < sizeof(rels)/sizeof(*rels); j++) { |
1243 | rel = rels[j]; |
1244 | |
1245 | lo = 0; hi = arraylen-1; |
1246 | while (lo <= hi) { |
1247 | mid = (lo + hi) / 2; |
1248 | c = strcmp(p, array[mid]); |
1249 | if (c < 0) |
1250 | hi = mid-1; |
1251 | else if (c > 0) |
1252 | lo = mid+1; |
1253 | else |
1254 | break; |
1255 | } |
1256 | |
1257 | if (c == 0) { |
1258 | if (rel == REL234_LT) |
1259 | ret = (mid > 0 ? array[--mid] : NULL); |
1260 | else if (rel == REL234_GT) |
1261 | ret = (mid < arraylen-1 ? array[++mid] : NULL); |
1262 | else |
1263 | ret = array[mid]; |
1264 | } else { |
1265 | assert(lo == hi+1); |
1266 | if (rel == REL234_LT || rel == REL234_LE) { |
1267 | mid = hi; |
1268 | ret = (hi >= 0 ? array[hi] : NULL); |
1269 | } else if (rel == REL234_GT || rel == REL234_GE) { |
1270 | mid = lo; |
1271 | ret = (lo < arraylen ? array[lo] : NULL); |
1272 | } else |
1273 | ret = NULL; |
1274 | } |
1275 | |
1276 | realret = findrelpos234(tree, p, NULL, rel, &index); |
1277 | if (realret != ret) { |
1278 | error("find(\"%s\",%s) gave %s should be %s", |
1279 | p, relnames[j], realret, ret); |
1280 | } |
1281 | if (realret && index != mid) { |
1282 | error("find(\"%s\",%s) gave %d should be %d", |
1283 | p, relnames[j], index, mid); |
1284 | } |
1285 | if (realret && rel == REL234_EQ) { |
1286 | realret2 = index234(tree, index); |
1287 | if (realret2 != realret) { |
1288 | error("find(\"%s\",%s) gave %s(%d) but %d -> %s", |
1289 | p, relnames[j], realret, index, index, realret2); |
1290 | } |
1291 | } |
1292 | #if 0 |
1293 | printf("find(\"%s\",%s) gave %s(%d)\n", p, relnames[j], |
1294 | realret, index); |
1295 | #endif |
1296 | } |
1297 | } |
1298 | |
1299 | realret = findrelpos234(tree, NULL, NULL, REL234_GT, &index); |
1300 | if (arraylen && (realret != array[0] || index != 0)) { |
1301 | error("find(NULL,GT) gave %s(%d) should be %s(0)", |
1302 | realret, index, array[0]); |
1303 | } else if (!arraylen && (realret != NULL)) { |
1304 | error("find(NULL,GT) gave %s(%d) should be NULL", |
1305 | realret, index); |
1306 | } |
1307 | |
1308 | realret = findrelpos234(tree, NULL, NULL, REL234_LT, &index); |
1309 | if (arraylen && (realret != array[arraylen-1] || index != arraylen-1)) { |
1310 | error("find(NULL,LT) gave %s(%d) should be %s(0)", |
1311 | realret, index, array[arraylen-1]); |
1312 | } else if (!arraylen && (realret != NULL)) { |
1313 | error("find(NULL,LT) gave %s(%d) should be NULL", |
1314 | realret, index); |
1315 | } |
1316 | } |
1317 | |
febd9a0f |
1318 | int main(void) { |
2d56b16f |
1319 | int in[NSTR]; |
d2371c81 |
1320 | int i, j, k; |
2d56b16f |
1321 | unsigned seed = 0; |
1322 | |
1323 | for (i = 0; i < NSTR; i++) in[i] = 0; |
1324 | array = NULL; |
1325 | arraylen = arraysize = 0; |
1326 | tree = newtree234(mycmp); |
1327 | cmp = mycmp; |
1328 | |
1329 | verify(); |
1330 | for (i = 0; i < 10000; i++) { |
1331 | j = randomnumber(&seed); |
1332 | j %= NSTR; |
1333 | printf("trial: %d\n", i); |
1334 | if (in[j]) { |
1335 | printf("deleting %s (%d)\n", strings[j], j); |
1336 | deltest(strings[j]); |
1337 | in[j] = 0; |
1338 | } else { |
1339 | printf("adding %s (%d)\n", strings[j], j); |
1340 | addtest(strings[j]); |
1341 | in[j] = 1; |
1342 | } |
d2371c81 |
1343 | findtest(); |
2d56b16f |
1344 | } |
1345 | |
1346 | while (arraylen > 0) { |
1347 | j = randomnumber(&seed); |
1348 | j %= arraylen; |
1349 | deltest(array[j]); |
1350 | } |
1351 | |
d2371c81 |
1352 | freetree234(tree); |
1353 | |
1354 | /* |
1355 | * Now try an unsorted tree. We don't really need to test |
1356 | * delpos234 because we know del234 is based on it, so it's |
1357 | * already been tested in the above sorted-tree code; but for |
1358 | * completeness we'll use it to tear down our unsorted tree |
1359 | * once we've built it. |
1360 | */ |
1361 | tree = newtree234(NULL); |
1362 | cmp = NULL; |
1363 | verify(); |
1364 | for (i = 0; i < 1000; i++) { |
1365 | printf("trial: %d\n", i); |
1366 | j = randomnumber(&seed); |
1367 | j %= NSTR; |
1368 | k = randomnumber(&seed); |
1369 | k %= count234(tree)+1; |
1370 | printf("adding string %s at index %d\n", strings[j], k); |
1371 | addpostest(strings[j], k); |
1372 | } |
1373 | while (count234(tree) > 0) { |
1374 | printf("cleanup: tree size %d\n", count234(tree)); |
1375 | j = randomnumber(&seed); |
1376 | j %= count234(tree); |
1377 | printf("deleting string %s from index %d\n", array[j], j); |
1378 | delpostest(j); |
1379 | } |
1380 | |
2d56b16f |
1381 | return 0; |
febd9a0f |
1382 | } |
2d56b16f |
1383 | |
febd9a0f |
1384 | #endif |