| 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 |
projects / u / mdw / putty / blame