diff options
Diffstat (limited to 'tools/plink/tree234.c')
| -rw-r--r-- | tools/plink/tree234.c | 2965 |
1 files changed, 1486 insertions, 1479 deletions
diff --git a/tools/plink/tree234.c b/tools/plink/tree234.c index 4e2da9dd6..f1c0c2edb 100644 --- a/tools/plink/tree234.c +++ b/tools/plink/tree234.c @@ -1,1479 +1,1486 @@ -/*
- * tree234.c: reasonably generic counted 2-3-4 tree routines.
- *
- * This file is copyright 1999-2001 Simon Tatham.
- *
- * Permission is hereby granted, free of charge, to any person
- * obtaining a copy of this software and associated documentation
- * files (the "Software"), to deal in the Software without
- * restriction, including without limitation the rights to use,
- * copy, modify, merge, publish, distribute, sublicense, and/or
- * sell copies of the Software, and to permit persons to whom the
- * Software is furnished to do so, subject to the following
- * conditions:
- *
- * The above copyright notice and this permission notice shall be
- * included in all copies or substantial portions of the Software.
- *
- * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
- * EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES
- * OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND
- * NONINFRINGEMENT. IN NO EVENT SHALL SIMON TATHAM BE LIABLE FOR
- * ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF
- * CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN
- * CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
- * SOFTWARE.
- */
-
-#include <stdio.h>
-#include <stdlib.h>
-#include <assert.h>
-
-#include "puttymem.h"
-#include "tree234.h"
-
-#ifdef TEST
-#define LOG(x) (printf x)
-#else
-#define LOG(x)
-#endif
-
-typedef struct node234_Tag node234;
-
-struct tree234_Tag {
- node234 *root;
- cmpfn234 cmp;
-};
-
-struct node234_Tag {
- node234 *parent;
- node234 *kids[4];
- int counts[4];
- void *elems[3];
-};
-
-/*
- * Create a 2-3-4 tree.
- */
-tree234 *newtree234(cmpfn234 cmp)
-{
- tree234 *ret = snew(tree234);
- LOG(("created tree %p\n", ret));
- ret->root = NULL;
- ret->cmp = cmp;
- return ret;
-}
-
-/*
- * Free a 2-3-4 tree (not including freeing the elements).
- */
-static void freenode234(node234 * n)
-{
- if (!n)
- return;
- freenode234(n->kids[0]);
- freenode234(n->kids[1]);
- freenode234(n->kids[2]);
- freenode234(n->kids[3]);
- sfree(n);
-}
-
-void freetree234(tree234 * t)
-{
- freenode234(t->root);
- sfree(t);
-}
-
-/*
- * Internal function to count a node.
- */
-static int countnode234(node234 * n)
-{
- int count = 0;
- int i;
- if (!n)
- return 0;
- for (i = 0; i < 4; i++)
- count += n->counts[i];
- for (i = 0; i < 3; i++)
- if (n->elems[i])
- count++;
- return count;
-}
-
-/*
- * Count the elements in a tree.
- */
-int count234(tree234 * t)
-{
- if (t->root)
- return countnode234(t->root);
- else
- return 0;
-}
-
-/*
- * Add an element e to a 2-3-4 tree t. Returns e on success, or if
- * an existing element compares equal, returns that.
- */
-static void *add234_internal(tree234 * t, void *e, int index)
-{
- node234 *n, **np, *left, *right;
- void *orig_e = e;
- int c, lcount, rcount;
-
- LOG(("adding node %p to tree %p\n", e, t));
- if (t->root == NULL) {
- t->root = snew(node234);
- t->root->elems[1] = t->root->elems[2] = NULL;
- t->root->kids[0] = t->root->kids[1] = NULL;
- t->root->kids[2] = t->root->kids[3] = NULL;
- t->root->counts[0] = t->root->counts[1] = 0;
- t->root->counts[2] = t->root->counts[3] = 0;
- t->root->parent = NULL;
- t->root->elems[0] = e;
- LOG((" created root %p\n", t->root));
- return orig_e;
- }
-
- n = NULL; /* placate gcc; will always be set below since t->root != NULL */
- np = &t->root;
- while (*np) {
- int childnum;
- n = *np;
- LOG((" node %p: %p/%d [%p] %p/%d [%p] %p/%d [%p] %p/%d\n",
- n,
- n->kids[0], n->counts[0], n->elems[0],
- n->kids[1], n->counts[1], n->elems[1],
- n->kids[2], n->counts[2], n->elems[2],
- n->kids[3], n->counts[3]));
- if (index >= 0) {
- if (!n->kids[0]) {
- /*
- * Leaf node. We want to insert at kid position
- * equal to the index:
- *
- * 0 A 1 B 2 C 3
- */
- childnum = index;
- } else {
- /*
- * Internal node. We always descend through it (add
- * always starts at the bottom, never in the
- * middle).
- */
- do { /* this is a do ... while (0) to allow `break' */
- if (index <= n->counts[0]) {
- childnum = 0;
- break;
- }
- index -= n->counts[0] + 1;
- if (index <= n->counts[1]) {
- childnum = 1;
- break;
- }
- index -= n->counts[1] + 1;
- if (index <= n->counts[2]) {
- childnum = 2;
- break;
- }
- index -= n->counts[2] + 1;
- if (index <= n->counts[3]) {
- childnum = 3;
- break;
- }
- return NULL; /* error: index out of range */
- } while (0);
- }
- } else {
- if ((c = t->cmp(e, n->elems[0])) < 0)
- childnum = 0;
- else if (c == 0)
- return n->elems[0]; /* already exists */
- else if (n->elems[1] == NULL
- || (c = t->cmp(e, n->elems[1])) < 0) childnum = 1;
- else if (c == 0)
- return n->elems[1]; /* already exists */
- else if (n->elems[2] == NULL
- || (c = t->cmp(e, n->elems[2])) < 0) childnum = 2;
- else if (c == 0)
- return n->elems[2]; /* already exists */
- else
- childnum = 3;
- }
- np = &n->kids[childnum];
- LOG((" moving to child %d (%p)\n", childnum, *np));
- }
-
- /*
- * We need to insert the new element in n at position np.
- */
- left = NULL;
- lcount = 0;
- right = NULL;
- rcount = 0;
- while (n) {
- LOG((" at %p: %p/%d [%p] %p/%d [%p] %p/%d [%p] %p/%d\n",
- n,
- n->kids[0], n->counts[0], n->elems[0],
- n->kids[1], n->counts[1], n->elems[1],
- n->kids[2], n->counts[2], n->elems[2],
- n->kids[3], n->counts[3]));
- LOG((" need to insert %p/%d [%p] %p/%d at position %d\n",
- left, lcount, e, right, rcount, np - n->kids));
- if (n->elems[1] == NULL) {
- /*
- * Insert in a 2-node; simple.
- */
- if (np == &n->kids[0]) {
- LOG((" inserting on left of 2-node\n"));
- n->kids[2] = n->kids[1];
- n->counts[2] = n->counts[1];
- n->elems[1] = n->elems[0];
- n->kids[1] = right;
- n->counts[1] = rcount;
- n->elems[0] = e;
- n->kids[0] = left;
- n->counts[0] = lcount;
- } else { /* np == &n->kids[1] */
- LOG((" inserting on right of 2-node\n"));
- n->kids[2] = right;
- n->counts[2] = rcount;
- n->elems[1] = e;
- n->kids[1] = left;
- n->counts[1] = lcount;
- }
- if (n->kids[0])
- n->kids[0]->parent = n;
- if (n->kids[1])
- n->kids[1]->parent = n;
- if (n->kids[2])
- n->kids[2]->parent = n;
- LOG((" done\n"));
- break;
- } else if (n->elems[2] == NULL) {
- /*
- * Insert in a 3-node; simple.
- */
- if (np == &n->kids[0]) {
- LOG((" inserting on left of 3-node\n"));
- n->kids[3] = n->kids[2];
- n->counts[3] = n->counts[2];
- n->elems[2] = n->elems[1];
- n->kids[2] = n->kids[1];
- n->counts[2] = n->counts[1];
- n->elems[1] = n->elems[0];
- n->kids[1] = right;
- n->counts[1] = rcount;
- n->elems[0] = e;
- n->kids[0] = left;
- n->counts[0] = lcount;
- } else if (np == &n->kids[1]) {
- LOG((" inserting in middle of 3-node\n"));
- n->kids[3] = n->kids[2];
- n->counts[3] = n->counts[2];
- n->elems[2] = n->elems[1];
- n->kids[2] = right;
- n->counts[2] = rcount;
- n->elems[1] = e;
- n->kids[1] = left;
- n->counts[1] = lcount;
- } else { /* np == &n->kids[2] */
- LOG((" inserting on right of 3-node\n"));
- n->kids[3] = right;
- n->counts[3] = rcount;
- n->elems[2] = e;
- n->kids[2] = left;
- n->counts[2] = lcount;
- }
- if (n->kids[0])
- n->kids[0]->parent = n;
- if (n->kids[1])
- n->kids[1]->parent = n;
- if (n->kids[2])
- n->kids[2]->parent = n;
- if (n->kids[3])
- n->kids[3]->parent = n;
- LOG((" done\n"));
- break;
- } else {
- node234 *m = snew(node234);
- m->parent = n->parent;
- LOG((" splitting a 4-node; created new node %p\n", m));
- /*
- * Insert in a 4-node; split into a 2-node and a
- * 3-node, and move focus up a level.
- *
- * I don't think it matters which way round we put the
- * 2 and the 3. For simplicity, we'll put the 3 first
- * always.
- */
- if (np == &n->kids[0]) {
- m->kids[0] = left;
- m->counts[0] = lcount;
- m->elems[0] = e;
- m->kids[1] = right;
- m->counts[1] = rcount;
- m->elems[1] = n->elems[0];
- m->kids[2] = n->kids[1];
- m->counts[2] = n->counts[1];
- e = n->elems[1];
- n->kids[0] = n->kids[2];
- n->counts[0] = n->counts[2];
- n->elems[0] = n->elems[2];
- n->kids[1] = n->kids[3];
- n->counts[1] = n->counts[3];
- } else if (np == &n->kids[1]) {
- m->kids[0] = n->kids[0];
- m->counts[0] = n->counts[0];
- m->elems[0] = n->elems[0];
- m->kids[1] = left;
- m->counts[1] = lcount;
- m->elems[1] = e;
- m->kids[2] = right;
- m->counts[2] = rcount;
- e = n->elems[1];
- n->kids[0] = n->kids[2];
- n->counts[0] = n->counts[2];
- n->elems[0] = n->elems[2];
- n->kids[1] = n->kids[3];
- n->counts[1] = n->counts[3];
- } else if (np == &n->kids[2]) {
- m->kids[0] = n->kids[0];
- m->counts[0] = n->counts[0];
- m->elems[0] = n->elems[0];
- m->kids[1] = n->kids[1];
- m->counts[1] = n->counts[1];
- m->elems[1] = n->elems[1];
- m->kids[2] = left;
- m->counts[2] = lcount;
- /* e = e; */
- n->kids[0] = right;
- n->counts[0] = rcount;
- n->elems[0] = n->elems[2];
- n->kids[1] = n->kids[3];
- n->counts[1] = n->counts[3];
- } else { /* np == &n->kids[3] */
- m->kids[0] = n->kids[0];
- m->counts[0] = n->counts[0];
- m->elems[0] = n->elems[0];
- m->kids[1] = n->kids[1];
- m->counts[1] = n->counts[1];
- m->elems[1] = n->elems[1];
- m->kids[2] = n->kids[2];
- m->counts[2] = n->counts[2];
- n->kids[0] = left;
- n->counts[0] = lcount;
- n->elems[0] = e;
- n->kids[1] = right;
- n->counts[1] = rcount;
- e = n->elems[2];
- }
- m->kids[3] = n->kids[3] = n->kids[2] = NULL;
- m->counts[3] = n->counts[3] = n->counts[2] = 0;
- m->elems[2] = n->elems[2] = n->elems[1] = NULL;
- if (m->kids[0])
- m->kids[0]->parent = m;
- if (m->kids[1])
- m->kids[1]->parent = m;
- if (m->kids[2])
- m->kids[2]->parent = m;
- if (n->kids[0])
- n->kids[0]->parent = n;
- if (n->kids[1])
- n->kids[1]->parent = n;
- LOG((" left (%p): %p/%d [%p] %p/%d [%p] %p/%d\n", m,
- m->kids[0], m->counts[0], m->elems[0],
- m->kids[1], m->counts[1], m->elems[1],
- m->kids[2], m->counts[2]));
- LOG((" right (%p): %p/%d [%p] %p/%d\n", n,
- n->kids[0], n->counts[0], n->elems[0],
- n->kids[1], n->counts[1]));
- left = m;
- lcount = countnode234(left);
- right = n;
- rcount = countnode234(right);
- }
- if (n->parent)
- np = (n->parent->kids[0] == n ? &n->parent->kids[0] :
- n->parent->kids[1] == n ? &n->parent->kids[1] :
- n->parent->kids[2] == n ? &n->parent->kids[2] :
- &n->parent->kids[3]);
- n = n->parent;
- }
-
- /*
- * If we've come out of here by `break', n will still be
- * non-NULL and all we need to do is go back up the tree
- * updating counts. If we've come here because n is NULL, we
- * need to create a new root for the tree because the old one
- * has just split into two. */
- if (n) {
- while (n->parent) {
- int count = countnode234(n);
- int childnum;
- childnum = (n->parent->kids[0] == n ? 0 :
- n->parent->kids[1] == n ? 1 :
- n->parent->kids[2] == n ? 2 : 3);
- n->parent->counts[childnum] = count;
- n = n->parent;
- }
- } else {
- LOG((" root is overloaded, split into two\n"));
- t->root = snew(node234);
- t->root->kids[0] = left;
- t->root->counts[0] = lcount;
- t->root->elems[0] = e;
- t->root->kids[1] = right;
- t->root->counts[1] = rcount;
- t->root->elems[1] = NULL;
- t->root->kids[2] = NULL;
- t->root->counts[2] = 0;
- t->root->elems[2] = NULL;
- t->root->kids[3] = NULL;
- t->root->counts[3] = 0;
- t->root->parent = NULL;
- if (t->root->kids[0])
- t->root->kids[0]->parent = t->root;
- if (t->root->kids[1])
- t->root->kids[1]->parent = t->root;
- LOG((" new root is %p/%d [%p] %p/%d\n",
- t->root->kids[0], t->root->counts[0],
- t->root->elems[0], t->root->kids[1], t->root->counts[1]));
- }
-
- return orig_e;
-}
-
-void *add234(tree234 * t, void *e)
-{
- if (!t->cmp) /* tree is unsorted */
- return NULL;
-
- return add234_internal(t, e, -1);
-}
-void *addpos234(tree234 * t, void *e, int index)
-{
- if (index < 0 || /* index out of range */
- t->cmp) /* tree is sorted */
- return NULL; /* return failure */
-
- return add234_internal(t, e, index); /* this checks the upper bound */
-}
-
-/*
- * Look up the element at a given numeric index in a 2-3-4 tree.
- * Returns NULL if the index is out of range.
- */
-void *index234(tree234 * t, int index)
-{
- node234 *n;
-
- if (!t->root)
- return NULL; /* tree is empty */
-
- if (index < 0 || index >= countnode234(t->root))
- return NULL; /* out of range */
-
- n = t->root;
-
- while (n) {
- if (index < n->counts[0])
- n = n->kids[0];
- else if (index -= n->counts[0] + 1, index < 0)
- return n->elems[0];
- else if (index < n->counts[1])
- n = n->kids[1];
- else if (index -= n->counts[1] + 1, index < 0)
- return n->elems[1];
- else if (index < n->counts[2])
- n = n->kids[2];
- else if (index -= n->counts[2] + 1, index < 0)
- return n->elems[2];
- else
- n = n->kids[3];
- }
-
- /* We shouldn't ever get here. I wonder how we did. */
- return NULL;
-}
-
-/*
- * Find an element e in a sorted 2-3-4 tree t. Returns NULL if not
- * found. e is always passed as the first argument to cmp, so cmp
- * can be an asymmetric function if desired. cmp can also be passed
- * as NULL, in which case the compare function from the tree proper
- * will be used.
- */
-void *findrelpos234(tree234 * t, void *e, cmpfn234 cmp,
- int relation, int *index)
-{
- node234 *n;
- void *ret;
- int c;
- int idx, ecount, kcount, cmpret;
-
- if (t->root == NULL)
- return NULL;
-
- if (cmp == NULL)
- cmp = t->cmp;
-
- n = t->root;
- /*
- * Attempt to find the element itself.
- */
- idx = 0;
- ecount = -1;
- /*
- * Prepare a fake `cmp' result if e is NULL.
- */
- cmpret = 0;
- if (e == NULL) {
- assert(relation == REL234_LT || relation == REL234_GT);
- if (relation == REL234_LT)
- cmpret = +1; /* e is a max: always greater */
- else if (relation == REL234_GT)
- cmpret = -1; /* e is a min: always smaller */
- }
- while (1) {
- for (kcount = 0; kcount < 4; kcount++) {
- if (kcount >= 3 || n->elems[kcount] == NULL ||
- (c = cmpret ? cmpret : cmp(e, n->elems[kcount])) < 0) {
- break;
- }
- if (n->kids[kcount])
- idx += n->counts[kcount];
- if (c == 0) {
- ecount = kcount;
- break;
- }
- idx++;
- }
- if (ecount >= 0)
- break;
- if (n->kids[kcount])
- n = n->kids[kcount];
- else
- break;
- }
-
- if (ecount >= 0) {
- /*
- * We have found the element we're looking for. It's
- * n->elems[ecount], at tree index idx. If our search
- * relation is EQ, LE or GE we can now go home.
- */
- if (relation != REL234_LT && relation != REL234_GT) {
- if (index)
- *index = idx;
- return n->elems[ecount];
- }
-
- /*
- * Otherwise, we'll do an indexed lookup for the previous
- * or next element. (It would be perfectly possible to
- * implement these search types in a non-counted tree by
- * going back up from where we are, but far more fiddly.)
- */
- if (relation == REL234_LT)
- idx--;
- else
- idx++;
- } else {
- /*
- * We've found our way to the bottom of the tree and we
- * know where we would insert this node if we wanted to:
- * we'd put it in in place of the (empty) subtree
- * n->kids[kcount], and it would have index idx
- *
- * But the actual element isn't there. So if our search
- * relation is EQ, we're doomed.
- */
- if (relation == REL234_EQ)
- return NULL;
-
- /*
- * Otherwise, we must do an index lookup for index idx-1
- * (if we're going left - LE or LT) or index idx (if we're
- * going right - GE or GT).
- */
- if (relation == REL234_LT || relation == REL234_LE) {
- idx--;
- }
- }
-
- /*
- * We know the index of the element we want; just call index234
- * to do the rest. This will return NULL if the index is out of
- * bounds, which is exactly what we want.
- */
- ret = index234(t, idx);
- if (ret && index)
- *index = idx;
- return ret;
-}
-void *find234(tree234 * t, void *e, cmpfn234 cmp)
-{
- return findrelpos234(t, e, cmp, REL234_EQ, NULL);
-}
-void *findrel234(tree234 * t, void *e, cmpfn234 cmp, int relation)
-{
- return findrelpos234(t, e, cmp, relation, NULL);
-}
-void *findpos234(tree234 * t, void *e, cmpfn234 cmp, int *index)
-{
- return findrelpos234(t, e, cmp, REL234_EQ, index);
-}
-
-/*
- * Delete an element e in a 2-3-4 tree. Does not free the element,
- * merely removes all links to it from the tree nodes.
- */
-static void *delpos234_internal(tree234 * t, int index)
-{
- node234 *n;
- void *retval;
- int ei = -1;
-
- retval = 0;
-
- n = t->root;
- LOG(("deleting item %d from tree %p\n", index, t));
- while (1) {
- while (n) {
- int ki;
- node234 *sub;
-
- LOG(
- (" node %p: %p/%d [%p] %p/%d [%p] %p/%d [%p] %p/%d index=%d\n",
- n, n->kids[0], n->counts[0], n->elems[0], n->kids[1],
- n->counts[1], n->elems[1], n->kids[2], n->counts[2],
- n->elems[2], n->kids[3], n->counts[3], index));
- if (index < n->counts[0]) {
- ki = 0;
- } else if (index -= n->counts[0] + 1, index < 0) {
- ei = 0;
- break;
- } else if (index < n->counts[1]) {
- ki = 1;
- } else if (index -= n->counts[1] + 1, index < 0) {
- ei = 1;
- break;
- } else if (index < n->counts[2]) {
- ki = 2;
- } else if (index -= n->counts[2] + 1, index < 0) {
- ei = 2;
- break;
- } else {
- ki = 3;
- }
- /*
- * Recurse down to subtree ki. If it has only one element,
- * we have to do some transformation to start with.
- */
- LOG((" moving to subtree %d\n", ki));
- sub = n->kids[ki];
- if (!sub->elems[1]) {
- LOG((" subtree has only one element!\n", ki));
- if (ki > 0 && n->kids[ki - 1]->elems[1]) {
- /*
- * Case 3a, left-handed variant. Child ki has
- * only one element, but child ki-1 has two or
- * more. So we need to move a subtree from ki-1
- * to ki.
- *
- * . C . . B .
- * / \ -> / \
- * [more] a A b B c d D e [more] a A b c C d D e
- */
- node234 *sib = n->kids[ki - 1];
- int lastelem = (sib->elems[2] ? 2 :
- sib->elems[1] ? 1 : 0);
- sub->kids[2] = sub->kids[1];
- sub->counts[2] = sub->counts[1];
- sub->elems[1] = sub->elems[0];
- sub->kids[1] = sub->kids[0];
- sub->counts[1] = sub->counts[0];
- sub->elems[0] = n->elems[ki - 1];
- sub->kids[0] = sib->kids[lastelem + 1];
- sub->counts[0] = sib->counts[lastelem + 1];
- if (sub->kids[0])
- sub->kids[0]->parent = sub;
- n->elems[ki - 1] = sib->elems[lastelem];
- sib->kids[lastelem + 1] = NULL;
- sib->counts[lastelem + 1] = 0;
- sib->elems[lastelem] = NULL;
- n->counts[ki] = countnode234(sub);
- LOG((" case 3a left\n"));
- LOG(
- (" index and left subtree count before adjustment: %d, %d\n",
- index, n->counts[ki - 1]));
- index += n->counts[ki - 1];
- n->counts[ki - 1] = countnode234(sib);
- index -= n->counts[ki - 1];
- LOG(
- (" index and left subtree count after adjustment: %d, %d\n",
- index, n->counts[ki - 1]));
- } else if (ki < 3 && n->kids[ki + 1]
- && n->kids[ki + 1]->elems[1]) {
- /*
- * Case 3a, right-handed variant. ki has only
- * one element but ki+1 has two or more. Move a
- * subtree from ki+1 to ki.
- *
- * . B . . C .
- * / \ -> / \
- * a A b c C d D e [more] a A b B c d D e [more]
- */
- node234 *sib = n->kids[ki + 1];
- int j;
- sub->elems[1] = n->elems[ki];
- sub->kids[2] = sib->kids[0];
- sub->counts[2] = sib->counts[0];
- if (sub->kids[2])
- sub->kids[2]->parent = sub;
- n->elems[ki] = sib->elems[0];
- sib->kids[0] = sib->kids[1];
- sib->counts[0] = sib->counts[1];
- for (j = 0; j < 2 && sib->elems[j + 1]; j++) {
- sib->kids[j + 1] = sib->kids[j + 2];
- sib->counts[j + 1] = sib->counts[j + 2];
- sib->elems[j] = sib->elems[j + 1];
- }
- sib->kids[j + 1] = NULL;
- sib->counts[j + 1] = 0;
- sib->elems[j] = NULL;
- n->counts[ki] = countnode234(sub);
- n->counts[ki + 1] = countnode234(sib);
- LOG((" case 3a right\n"));
- } else {
- /*
- * Case 3b. ki has only one element, and has no
- * neighbour with more than one. So pick a
- * neighbour and merge it with ki, taking an
- * element down from n to go in the middle.
- *
- * . B . .
- * / \ -> |
- * a A b c C d a A b B c C d
- *
- * (Since at all points we have avoided
- * descending to a node with only one element,
- * we can be sure that n is not reduced to
- * nothingness by this move, _unless_ it was
- * the very first node, ie the root of the
- * tree. In that case we remove the now-empty
- * root and replace it with its single large
- * child as shown.)
- */
- node234 *sib;
- int j;
-
- if (ki > 0) {
- ki--;
- index += n->counts[ki] + 1;
- }
- sib = n->kids[ki];
- sub = n->kids[ki + 1];
-
- sub->kids[3] = sub->kids[1];
- sub->counts[3] = sub->counts[1];
- sub->elems[2] = sub->elems[0];
- sub->kids[2] = sub->kids[0];
- sub->counts[2] = sub->counts[0];
- sub->elems[1] = n->elems[ki];
- sub->kids[1] = sib->kids[1];
- sub->counts[1] = sib->counts[1];
- if (sub->kids[1])
- sub->kids[1]->parent = sub;
- sub->elems[0] = sib->elems[0];
- sub->kids[0] = sib->kids[0];
- sub->counts[0] = sib->counts[0];
- if (sub->kids[0])
- sub->kids[0]->parent = sub;
-
- n->counts[ki + 1] = countnode234(sub);
-
- sfree(sib);
-
- /*
- * That's built the big node in sub. Now we
- * need to remove the reference to sib in n.
- */
- for (j = ki; j < 3 && n->kids[j + 1]; j++) {
- n->kids[j] = n->kids[j + 1];
- n->counts[j] = n->counts[j + 1];
- n->elems[j] = j < 2 ? n->elems[j + 1] : NULL;
- }
- n->kids[j] = NULL;
- n->counts[j] = 0;
- if (j < 3)
- n->elems[j] = NULL;
- LOG((" case 3b ki=%d\n", ki));
-
- if (!n->elems[0]) {
- /*
- * The root is empty and needs to be
- * removed.
- */
- LOG((" shifting root!\n"));
- t->root = sub;
- sub->parent = NULL;
- sfree(n);
- }
- }
- }
- n = sub;
- }
- if (!retval)
- retval = n->elems[ei];
-
- if (ei == -1)
- return NULL; /* although this shouldn't happen */
-
- /*
- * Treat special case: this is the one remaining item in
- * the tree. n is the tree root (no parent), has one
- * element (no elems[1]), and has no kids (no kids[0]).
- */
- if (!n->parent && !n->elems[1] && !n->kids[0]) {
- LOG((" removed last element in tree\n"));
- sfree(n);
- t->root = NULL;
- return retval;
- }
-
- /*
- * Now we have the element we want, as n->elems[ei], and we
- * have also arranged for that element not to be the only
- * one in its node. So...
- */
-
- if (!n->kids[0] && n->elems[1]) {
- /*
- * Case 1. n is a leaf node with more than one element,
- * so it's _really easy_. Just delete the thing and
- * we're done.
- */
- int i;
- LOG((" case 1\n"));
- for (i = ei; i < 2 && n->elems[i + 1]; i++)
- n->elems[i] = n->elems[i + 1];
- n->elems[i] = NULL;
- /*
- * Having done that to the leaf node, we now go back up
- * the tree fixing the counts.
- */
- while (n->parent) {
- int childnum;
- childnum = (n->parent->kids[0] == n ? 0 :
- n->parent->kids[1] == n ? 1 :
- n->parent->kids[2] == n ? 2 : 3);
- n->parent->counts[childnum]--;
- n = n->parent;
- }
- return retval; /* finished! */
- } else if (n->kids[ei]->elems[1]) {
- /*
- * Case 2a. n is an internal node, and the root of the
- * subtree to the left of e has more than one element.
- * So find the predecessor p to e (ie the largest node
- * in that subtree), place it where e currently is, and
- * then start the deletion process over again on the
- * subtree with p as target.
- */
- node234 *m = n->kids[ei];
- void *target;
- LOG((" case 2a\n"));
- while (m->kids[0]) {
- m = (m->kids[3] ? m->kids[3] :
- m->kids[2] ? m->kids[2] :
- m->kids[1] ? m->kids[1] : m->kids[0]);
- }
- target = (m->elems[2] ? m->elems[2] :
- m->elems[1] ? m->elems[1] : m->elems[0]);
- n->elems[ei] = target;
- index = n->counts[ei] - 1;
- n = n->kids[ei];
- } else if (n->kids[ei + 1]->elems[1]) {
- /*
- * Case 2b, symmetric to 2a but s/left/right/ and
- * s/predecessor/successor/. (And s/largest/smallest/).
- */
- node234 *m = n->kids[ei + 1];
- void *target;
- LOG((" case 2b\n"));
- while (m->kids[0]) {
- m = m->kids[0];
- }
- target = m->elems[0];
- n->elems[ei] = target;
- n = n->kids[ei + 1];
- index = 0;
- } else {
- /*
- * Case 2c. n is an internal node, and the subtrees to
- * the left and right of e both have only one element.
- * So combine the two subnodes into a single big node
- * with their own elements on the left and right and e
- * in the middle, then restart the deletion process on
- * that subtree, with e still as target.
- */
- node234 *a = n->kids[ei], *b = n->kids[ei + 1];
- int j;
-
- LOG((" case 2c\n"));
- a->elems[1] = n->elems[ei];
- a->kids[2] = b->kids[0];
- a->counts[2] = b->counts[0];
- if (a->kids[2])
- a->kids[2]->parent = a;
- a->elems[2] = b->elems[0];
- a->kids[3] = b->kids[1];
- a->counts[3] = b->counts[1];
- if (a->kids[3])
- a->kids[3]->parent = a;
- sfree(b);
- n->counts[ei] = countnode234(a);
- /*
- * That's built the big node in a, and destroyed b. Now
- * remove the reference to b (and e) in n.
- */
- for (j = ei; j < 2 && n->elems[j + 1]; j++) {
- n->elems[j] = n->elems[j + 1];
- n->kids[j + 1] = n->kids[j + 2];
- n->counts[j + 1] = n->counts[j + 2];
- }
- n->elems[j] = NULL;
- n->kids[j + 1] = NULL;
- n->counts[j + 1] = 0;
- /*
- * It's possible, in this case, that we've just removed
- * the only element in the root of the tree. If so,
- * shift the root.
- */
- if (n->elems[0] == NULL) {
- LOG((" shifting root!\n"));
- t->root = a;
- a->parent = NULL;
- sfree(n);
- }
- /*
- * Now go round the deletion process again, with n
- * pointing at the new big node and e still the same.
- */
- n = a;
- index = a->counts[0] + a->counts[1] + 1;
- }
- }
-}
-void *delpos234(tree234 * t, int index)
-{
- if (index < 0 || index >= countnode234(t->root))
- return NULL;
- return delpos234_internal(t, index);
-}
-void *del234(tree234 * t, void *e)
-{
- int index;
- if (!findrelpos234(t, e, NULL, REL234_EQ, &index))
- return NULL; /* it wasn't in there anyway */
- return delpos234_internal(t, index); /* it's there; delete it. */
-}
-
-#ifdef TEST
-
-/*
- * Test code for the 2-3-4 tree. This code maintains an alternative
- * representation of the data in the tree, in an array (using the
- * obvious and slow insert and delete functions). After each tree
- * operation, the verify() function is called, which ensures all
- * the tree properties are preserved:
- * - node->child->parent always equals node
- * - tree->root->parent always equals NULL
- * - number of kids == 0 or number of elements + 1;
- * - tree has the same depth everywhere
- * - every node has at least one element
- * - subtree element counts are accurate
- * - any NULL kid pointer is accompanied by a zero count
- * - in a sorted tree: ordering property between elements of a
- * node and elements of its children is preserved
- * and also ensures the list represented by the tree is the same
- * list it should be. (This last check also doubly verifies the
- * ordering properties, because the `same list it should be' is by
- * definition correctly ordered. It also ensures all nodes are
- * distinct, because the enum functions would get caught in a loop
- * if not.)
- */
-
-#include <stdarg.h>
-
-/*
- * Error reporting function.
- */
-void error(char *fmt, ...)
-{
- va_list ap;
- printf("ERROR: ");
- va_start(ap, fmt);
- vfprintf(stdout, fmt, ap);
- va_end(ap);
- printf("\n");
-}
-
-/* The array representation of the data. */
-void **array;
-int arraylen, arraysize;
-cmpfn234 cmp;
-
-/* The tree representation of the same data. */
-tree234 *tree;
-
-typedef struct {
- int treedepth;
- int elemcount;
-} chkctx;
-
-int chknode(chkctx * ctx, int level, node234 * node,
- void *lowbound, void *highbound)
-{
- int nkids, nelems;
- int i;
- int count;
-
- /* Count the non-NULL kids. */
- for (nkids = 0; nkids < 4 && node->kids[nkids]; nkids++);
- /* Ensure no kids beyond the first NULL are non-NULL. */
- for (i = nkids; i < 4; i++)
- if (node->kids[i]) {
- error("node %p: nkids=%d but kids[%d] non-NULL",
- node, nkids, i);
- } else if (node->counts[i]) {
- error("node %p: kids[%d] NULL but count[%d]=%d nonzero",
- node, i, i, node->counts[i]);
- }
-
- /* Count the non-NULL elements. */
- for (nelems = 0; nelems < 3 && node->elems[nelems]; nelems++);
- /* Ensure no elements beyond the first NULL are non-NULL. */
- for (i = nelems; i < 3; i++)
- if (node->elems[i]) {
- error("node %p: nelems=%d but elems[%d] non-NULL",
- node, nelems, i);
- }
-
- if (nkids == 0) {
- /*
- * If nkids==0, this is a leaf node; verify that the tree
- * depth is the same everywhere.
- */
- if (ctx->treedepth < 0)
- ctx->treedepth = level; /* we didn't know the depth yet */
- else if (ctx->treedepth != level)
- error("node %p: leaf at depth %d, previously seen depth %d",
- node, level, ctx->treedepth);
- } else {
- /*
- * If nkids != 0, then it should be nelems+1, unless nelems
- * is 0 in which case nkids should also be 0 (and so we
- * shouldn't be in this condition at all).
- */
- int shouldkids = (nelems ? nelems + 1 : 0);
- if (nkids != shouldkids) {
- error("node %p: %d elems should mean %d kids but has %d",
- node, nelems, shouldkids, nkids);
- }
- }
-
- /*
- * nelems should be at least 1.
- */
- if (nelems == 0) {
- error("node %p: no elems", node, nkids);
- }
-
- /*
- * Add nelems to the running element count of the whole tree.
- */
- ctx->elemcount += nelems;
-
- /*
- * Check ordering property: all elements should be strictly >
- * lowbound, strictly < highbound, and strictly < each other in
- * sequence. (lowbound and highbound are NULL at edges of tree
- * - both NULL at root node - and NULL is considered to be <
- * everything and > everything. IYSWIM.)
- */
- if (cmp) {
- for (i = -1; i < nelems; i++) {
- void *lower = (i == -1 ? lowbound : node->elems[i]);
- void *higher =
- (i + 1 == nelems ? highbound : node->elems[i + 1]);
- if (lower && higher && cmp(lower, higher) >= 0) {
- error("node %p: kid comparison [%d=%s,%d=%s] failed",
- node, i, lower, i + 1, higher);
- }
- }
- }
-
- /*
- * Check parent pointers: all non-NULL kids should have a
- * parent pointer coming back to this node.
- */
- for (i = 0; i < nkids; i++)
- if (node->kids[i]->parent != node) {
- error("node %p kid %d: parent ptr is %p not %p",
- node, i, node->kids[i]->parent, node);
- }
-
-
- /*
- * Now (finally!) recurse into subtrees.
- */
- count = nelems;
-
- for (i = 0; i < nkids; i++) {
- void *lower = (i == 0 ? lowbound : node->elems[i - 1]);
- void *higher = (i >= nelems ? highbound : node->elems[i]);
- int subcount =
- chknode(ctx, level + 1, node->kids[i], lower, higher);
- if (node->counts[i] != subcount) {
- error("node %p kid %d: count says %d, subtree really has %d",
- node, i, node->counts[i], subcount);
- }
- count += subcount;
- }
-
- return count;
-}
-
-void verify(void)
-{
- chkctx ctx;
- int i;
- void *p;
-
- ctx.treedepth = -1; /* depth unknown yet */
- ctx.elemcount = 0; /* no elements seen yet */
- /*
- * Verify validity of tree properties.
- */
- if (tree->root) {
- if (tree->root->parent != NULL)
- error("root->parent is %p should be null", tree->root->parent);
- chknode(&ctx, 0, tree->root, NULL, NULL);
- }
- printf("tree depth: %d\n", ctx.treedepth);
- /*
- * Enumerate the tree and ensure it matches up to the array.
- */
- for (i = 0; NULL != (p = index234(tree, i)); i++) {
- if (i >= arraylen)
- error("tree contains more than %d elements", arraylen);
- if (array[i] != p)
- error("enum at position %d: array says %s, tree says %s",
- i, array[i], p);
- }
- if (ctx.elemcount != i) {
- error("tree really contains %d elements, enum gave %d",
- ctx.elemcount, i);
- }
- if (i < arraylen) {
- error("enum gave only %d elements, array has %d", i, arraylen);
- }
- i = count234(tree);
- if (ctx.elemcount != i) {
- error("tree really contains %d elements, count234 gave %d",
- ctx.elemcount, i);
- }
-}
-
-void internal_addtest(void *elem, int index, void *realret)
-{
- int i, j;
- void *retval;
-
- if (arraysize < arraylen + 1) {
- arraysize = arraylen + 1 + 256;
- array = sresize(array, arraysize, void *);
- }
-
- i = index;
- /* now i points to the first element >= elem */
- retval = elem; /* expect elem returned (success) */
- for (j = arraylen; j > i; j--)
- array[j] = array[j - 1];
- array[i] = elem; /* add elem to array */
- arraylen++;
-
- if (realret != retval) {
- error("add: retval was %p expected %p", realret, retval);
- }
-
- verify();
-}
-
-void addtest(void *elem)
-{
- int i;
- void *realret;
-
- realret = add234(tree, elem);
-
- i = 0;
- while (i < arraylen && cmp(elem, array[i]) > 0)
- i++;
- if (i < arraylen && !cmp(elem, array[i])) {
- void *retval = array[i]; /* expect that returned not elem */
- if (realret != retval) {
- error("add: retval was %p expected %p", realret, retval);
- }
- } else
- internal_addtest(elem, i, realret);
-}
-
-void addpostest(void *elem, int i)
-{
- void *realret;
-
- realret = addpos234(tree, elem, i);
-
- internal_addtest(elem, i, realret);
-}
-
-void delpostest(int i)
-{
- int index = i;
- void *elem = array[i], *ret;
-
- /* i points to the right element */
- while (i < arraylen - 1) {
- array[i] = array[i + 1];
- i++;
- }
- arraylen--; /* delete elem from array */
-
- if (tree->cmp)
- ret = del234(tree, elem);
- else
- ret = delpos234(tree, index);
-
- if (ret != elem) {
- error("del returned %p, expected %p", ret, elem);
- }
-
- verify();
-}
-
-void deltest(void *elem)
-{
- int i;
-
- i = 0;
- while (i < arraylen && cmp(elem, array[i]) > 0)
- i++;
- if (i >= arraylen || cmp(elem, array[i]) != 0)
- return; /* don't do it! */
- delpostest(i);
-}
-
-/* A sample data set and test utility. Designed for pseudo-randomness,
- * and yet repeatability. */
-
-/*
- * This random number generator uses the `portable implementation'
- * given in ANSI C99 draft N869. It assumes `unsigned' is 32 bits;
- * change it if not.
- */
-int randomnumber(unsigned *seed)
-{
- *seed *= 1103515245;
- *seed += 12345;
- return ((*seed) / 65536) % 32768;
-}
-
-int mycmp(void *av, void *bv)
-{
- char const *a = (char const *) av;
- char const *b = (char const *) bv;
- return strcmp(a, b);
-}
-
-#define lenof(x) ( sizeof((x)) / sizeof(*(x)) )
-
-char *strings[] = {
- "a", "ab", "absque", "coram", "de",
- "palam", "clam", "cum", "ex", "e",
- "sine", "tenus", "pro", "prae",
- "banana", "carrot", "cabbage", "broccoli", "onion", "zebra",
- "penguin", "blancmange", "pangolin", "whale", "hedgehog",
- "giraffe", "peanut", "bungee", "foo", "bar", "baz", "quux",
- "murfl", "spoo", "breen", "flarn", "octothorpe",
- "snail", "tiger", "elephant", "octopus", "warthog", "armadillo",
- "aardvark", "wyvern", "dragon", "elf", "dwarf", "orc", "goblin",
- "pixie", "basilisk", "warg", "ape", "lizard", "newt", "shopkeeper",
- "wand", "ring", "amulet"
-};
-
-#define NSTR lenof(strings)
-
-int findtest(void)
-{
- const static int rels[] = {
- REL234_EQ, REL234_GE, REL234_LE, REL234_LT, REL234_GT
- };
- const static char *const relnames[] = {
- "EQ", "GE", "LE", "LT", "GT"
- };
- int i, j, rel, index;
- char *p, *ret, *realret, *realret2;
- int lo, hi, mid, c;
-
- for (i = 0; i < NSTR; i++) {
- p = strings[i];
- for (j = 0; j < sizeof(rels) / sizeof(*rels); j++) {
- rel = rels[j];
-
- lo = 0;
- hi = arraylen - 1;
- while (lo <= hi) {
- mid = (lo + hi) / 2;
- c = strcmp(p, array[mid]);
- if (c < 0)
- hi = mid - 1;
- else if (c > 0)
- lo = mid + 1;
- else
- break;
- }
-
- if (c == 0) {
- if (rel == REL234_LT)
- ret = (mid > 0 ? array[--mid] : NULL);
- else if (rel == REL234_GT)
- ret = (mid < arraylen - 1 ? array[++mid] : NULL);
- else
- ret = array[mid];
- } else {
- assert(lo == hi + 1);
- if (rel == REL234_LT || rel == REL234_LE) {
- mid = hi;
- ret = (hi >= 0 ? array[hi] : NULL);
- } else if (rel == REL234_GT || rel == REL234_GE) {
- mid = lo;
- ret = (lo < arraylen ? array[lo] : NULL);
- } else
- ret = NULL;
- }
-
- realret = findrelpos234(tree, p, NULL, rel, &index);
- if (realret != ret) {
- error("find(\"%s\",%s) gave %s should be %s",
- p, relnames[j], realret, ret);
- }
- if (realret && index != mid) {
- error("find(\"%s\",%s) gave %d should be %d",
- p, relnames[j], index, mid);
- }
- if (realret && rel == REL234_EQ) {
- realret2 = index234(tree, index);
- if (realret2 != realret) {
- error("find(\"%s\",%s) gave %s(%d) but %d -> %s",
- p, relnames[j], realret, index, index, realret2);
- }
- }
-#if 0
- printf("find(\"%s\",%s) gave %s(%d)\n", p, relnames[j],
- realret, index);
-#endif
- }
- }
-
- realret = findrelpos234(tree, NULL, NULL, REL234_GT, &index);
- if (arraylen && (realret != array[0] || index != 0)) {
- error("find(NULL,GT) gave %s(%d) should be %s(0)",
- realret, index, array[0]);
- } else if (!arraylen && (realret != NULL)) {
- error("find(NULL,GT) gave %s(%d) should be NULL", realret, index);
- }
-
- realret = findrelpos234(tree, NULL, NULL, REL234_LT, &index);
- if (arraylen
- && (realret != array[arraylen - 1] || index != arraylen - 1)) {
- error("find(NULL,LT) gave %s(%d) should be %s(0)", realret, index,
- array[arraylen - 1]);
- } else if (!arraylen && (realret != NULL)) {
- error("find(NULL,LT) gave %s(%d) should be NULL", realret, index);
- }
-}
-
-int main(void)
-{
- int in[NSTR];
- int i, j, k;
- unsigned seed = 0;
-
- for (i = 0; i < NSTR; i++)
- in[i] = 0;
- array = NULL;
- arraylen = arraysize = 0;
- tree = newtree234(mycmp);
- cmp = mycmp;
-
- verify();
- for (i = 0; i < 10000; i++) {
- j = randomnumber(&seed);
- j %= NSTR;
- printf("trial: %d\n", i);
- if (in[j]) {
- printf("deleting %s (%d)\n", strings[j], j);
- deltest(strings[j]);
- in[j] = 0;
- } else {
- printf("adding %s (%d)\n", strings[j], j);
- addtest(strings[j]);
- in[j] = 1;
- }
- findtest();
- }
-
- while (arraylen > 0) {
- j = randomnumber(&seed);
- j %= arraylen;
- deltest(array[j]);
- }
-
- freetree234(tree);
-
- /*
- * Now try an unsorted tree. We don't really need to test
- * delpos234 because we know del234 is based on it, so it's
- * already been tested in the above sorted-tree code; but for
- * completeness we'll use it to tear down our unsorted tree
- * once we've built it.
- */
- tree = newtree234(NULL);
- cmp = NULL;
- verify();
- for (i = 0; i < 1000; i++) {
- printf("trial: %d\n", i);
- j = randomnumber(&seed);
- j %= NSTR;
- k = randomnumber(&seed);
- k %= count234(tree) + 1;
- printf("adding string %s at index %d\n", strings[j], k);
- addpostest(strings[j], k);
- }
- while (count234(tree) > 0) {
- printf("cleanup: tree size %d\n", count234(tree));
- j = randomnumber(&seed);
- j %= count234(tree);
- printf("deleting string %s from index %d\n", array[j], j);
- delpostest(j);
- }
-
- return 0;
-}
-
-#endif
+/* + * tree234.c: reasonably generic counted 2-3-4 tree routines. + * + * This file is copyright 1999-2001 Simon Tatham. + * + * Permission is hereby granted, free of charge, to any person + * obtaining a copy of this software and associated documentation + * files (the "Software"), to deal in the Software without + * restriction, including without limitation the rights to use, + * copy, modify, merge, publish, distribute, sublicense, and/or + * sell copies of the Software, and to permit persons to whom the + * Software is furnished to do so, subject to the following + * conditions: + * + * The above copyright notice and this permission notice shall be + * included in all copies or substantial portions of the Software. + * + * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, + * EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES + * OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND + * NONINFRINGEMENT. IN NO EVENT SHALL SIMON TATHAM BE LIABLE FOR + * ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF + * CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN + * CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE + * SOFTWARE. + */ + +#include <stdio.h> +#include <stdlib.h> +#include <assert.h> + +#include "tree234.h" + +#ifdef TEST +#define LOG(x) (printf x) +#define snew(type) ((type *)malloc(sizeof(type))) +#define snewn(n, type) ((type *)malloc((n) * sizeof(type))) +#define sresize(ptr, n, type) \ + ((type *)realloc(sizeof((type *)0 == (ptr)) ? (ptr) : (ptr), \ + (n) * sizeof(type))) +#define sfree(ptr) free(ptr) +#else +#include "puttymem.h" +#define LOG(x) +#endif + +typedef struct node234_Tag node234; + +struct tree234_Tag { + node234 *root; + cmpfn234 cmp; +}; + +struct node234_Tag { + node234 *parent; + node234 *kids[4]; + int counts[4]; + void *elems[3]; +}; + +/* + * Create a 2-3-4 tree. + */ +tree234 *newtree234(cmpfn234 cmp) +{ + tree234 *ret = snew(tree234); + LOG(("created tree %p\n", ret)); + ret->root = NULL; + ret->cmp = cmp; + return ret; +} + +/* + * Free a 2-3-4 tree (not including freeing the elements). + */ +static void freenode234(node234 * n) +{ + if (!n) + return; + freenode234(n->kids[0]); + freenode234(n->kids[1]); + freenode234(n->kids[2]); + freenode234(n->kids[3]); + sfree(n); +} + +void freetree234(tree234 * t) +{ + freenode234(t->root); + sfree(t); +} + +/* + * Internal function to count a node. + */ +static int countnode234(node234 * n) +{ + int count = 0; + int i; + if (!n) + return 0; + for (i = 0; i < 4; i++) + count += n->counts[i]; + for (i = 0; i < 3; i++) + if (n->elems[i]) + count++; + return count; +} + +/* + * Count the elements in a tree. + */ +int count234(tree234 * t) +{ + if (t->root) + return countnode234(t->root); + else + return 0; +} + +/* + * Add an element e to a 2-3-4 tree t. Returns e on success, or if + * an existing element compares equal, returns that. + */ +static void *add234_internal(tree234 * t, void *e, int index) +{ + node234 *n, **np, *left, *right; + void *orig_e = e; + int c, lcount, rcount; + + LOG(("adding node %p to tree %p\n", e, t)); + if (t->root == NULL) { + t->root = snew(node234); + t->root->elems[1] = t->root->elems[2] = NULL; + t->root->kids[0] = t->root->kids[1] = NULL; + t->root->kids[2] = t->root->kids[3] = NULL; + t->root->counts[0] = t->root->counts[1] = 0; + t->root->counts[2] = t->root->counts[3] = 0; + t->root->parent = NULL; + t->root->elems[0] = e; + LOG((" created root %p\n", t->root)); + return orig_e; + } + + n = NULL; /* placate gcc; will always be set below since t->root != NULL */ + np = &t->root; + while (*np) { + int childnum; + n = *np; + LOG((" node %p: %p/%d [%p] %p/%d [%p] %p/%d [%p] %p/%d\n", + n, + n->kids[0], n->counts[0], n->elems[0], + n->kids[1], n->counts[1], n->elems[1], + n->kids[2], n->counts[2], n->elems[2], + n->kids[3], n->counts[3])); + if (index >= 0) { + if (!n->kids[0]) { + /* + * Leaf node. We want to insert at kid position + * equal to the index: + * + * 0 A 1 B 2 C 3 + */ + childnum = index; + } else { + /* + * Internal node. We always descend through it (add + * always starts at the bottom, never in the + * middle). + */ + do { /* this is a do ... while (0) to allow `break' */ + if (index <= n->counts[0]) { + childnum = 0; + break; + } + index -= n->counts[0] + 1; + if (index <= n->counts[1]) { + childnum = 1; + break; + } + index -= n->counts[1] + 1; + if (index <= n->counts[2]) { + childnum = 2; + break; + } + index -= n->counts[2] + 1; + if (index <= n->counts[3]) { + childnum = 3; + break; + } + return NULL; /* error: index out of range */ + } while (0); + } + } else { + if ((c = t->cmp(e, n->elems[0])) < 0) + childnum = 0; + else if (c == 0) + return n->elems[0]; /* already exists */ + else if (n->elems[1] == NULL + || (c = t->cmp(e, n->elems[1])) < 0) childnum = 1; + else if (c == 0) + return n->elems[1]; /* already exists */ + else if (n->elems[2] == NULL + || (c = t->cmp(e, n->elems[2])) < 0) childnum = 2; + else if (c == 0) + return n->elems[2]; /* already exists */ + else + childnum = 3; + } + np = &n->kids[childnum]; + LOG((" moving to child %d (%p)\n", childnum, *np)); + } + + /* + * We need to insert the new element in n at position np. + */ + left = NULL; + lcount = 0; + right = NULL; + rcount = 0; + while (n) { + LOG((" at %p: %p/%d [%p] %p/%d [%p] %p/%d [%p] %p/%d\n", + n, + n->kids[0], n->counts[0], n->elems[0], + n->kids[1], n->counts[1], n->elems[1], + n->kids[2], n->counts[2], n->elems[2], + n->kids[3], n->counts[3])); + LOG((" need to insert %p/%d [%p] %p/%d at position %d\n", + left, lcount, e, right, rcount, (int)(np - n->kids))); + if (n->elems[1] == NULL) { + /* + * Insert in a 2-node; simple. + */ + if (np == &n->kids[0]) { + LOG((" inserting on left of 2-node\n")); + n->kids[2] = n->kids[1]; + n->counts[2] = n->counts[1]; + n->elems[1] = n->elems[0]; + n->kids[1] = right; + n->counts[1] = rcount; + n->elems[0] = e; + n->kids[0] = left; + n->counts[0] = lcount; + } else { /* np == &n->kids[1] */ + LOG((" inserting on right of 2-node\n")); + n->kids[2] = right; + n->counts[2] = rcount; + n->elems[1] = e; + n->kids[1] = left; + n->counts[1] = lcount; + } + if (n->kids[0]) + n->kids[0]->parent = n; + if (n->kids[1]) + n->kids[1]->parent = n; + if (n->kids[2]) + n->kids[2]->parent = n; + LOG((" done\n")); + break; + } else if (n->elems[2] == NULL) { + /* + * Insert in a 3-node; simple. + */ + if (np == &n->kids[0]) { + LOG((" inserting on left of 3-node\n")); + n->kids[3] = n->kids[2]; + n->counts[3] = n->counts[2]; + n->elems[2] = n->elems[1]; + n->kids[2] = n->kids[1]; + n->counts[2] = n->counts[1]; + n->elems[1] = n->elems[0]; + n->kids[1] = right; + n->counts[1] = rcount; + n->elems[0] = e; + n->kids[0] = left; + n->counts[0] = lcount; + } else if (np == &n->kids[1]) { + LOG((" inserting in middle of 3-node\n")); + n->kids[3] = n->kids[2]; + n->counts[3] = n->counts[2]; + n->elems[2] = n->elems[1]; + n->kids[2] = right; + n->counts[2] = rcount; + n->elems[1] = e; + n->kids[1] = left; + n->counts[1] = lcount; + } else { /* np == &n->kids[2] */ + LOG((" inserting on right of 3-node\n")); + n->kids[3] = right; + n->counts[3] = rcount; + n->elems[2] = e; + n->kids[2] = left; + n->counts[2] = lcount; + } + if (n->kids[0]) + n->kids[0]->parent = n; + if (n->kids[1]) + n->kids[1]->parent = n; + if (n->kids[2]) + n->kids[2]->parent = n; + if (n->kids[3]) + n->kids[3]->parent = n; + LOG((" done\n")); + break; + } else { + node234 *m = snew(node234); + m->parent = n->parent; + LOG((" splitting a 4-node; created new node %p\n", m)); + /* + * Insert in a 4-node; split into a 2-node and a + * 3-node, and move focus up a level. + * + * I don't think it matters which way round we put the + * 2 and the 3. For simplicity, we'll put the 3 first + * always. + */ + if (np == &n->kids[0]) { + m->kids[0] = left; + m->counts[0] = lcount; + m->elems[0] = e; + m->kids[1] = right; + m->counts[1] = rcount; + m->elems[1] = n->elems[0]; + m->kids[2] = n->kids[1]; + m->counts[2] = n->counts[1]; + e = n->elems[1]; + n->kids[0] = n->kids[2]; + n->counts[0] = n->counts[2]; + n->elems[0] = n->elems[2]; + n->kids[1] = n->kids[3]; + n->counts[1] = n->counts[3]; + } else if (np == &n->kids[1]) { + m->kids[0] = n->kids[0]; + m->counts[0] = n->counts[0]; + m->elems[0] = n->elems[0]; + m->kids[1] = left; + m->counts[1] = lcount; + m->elems[1] = e; + m->kids[2] = right; + m->counts[2] = rcount; + e = n->elems[1]; + n->kids[0] = n->kids[2]; + n->counts[0] = n->counts[2]; + n->elems[0] = n->elems[2]; + n->kids[1] = n->kids[3]; + n->counts[1] = n->counts[3]; + } else if (np == &n->kids[2]) { + m->kids[0] = n->kids[0]; + m->counts[0] = n->counts[0]; + m->elems[0] = n->elems[0]; + m->kids[1] = n->kids[1]; + m->counts[1] = n->counts[1]; + m->elems[1] = n->elems[1]; + m->kids[2] = left; + m->counts[2] = lcount; + /* e = e; */ + n->kids[0] = right; + n->counts[0] = rcount; + n->elems[0] = n->elems[2]; + n->kids[1] = n->kids[3]; + n->counts[1] = n->counts[3]; + } else { /* np == &n->kids[3] */ + m->kids[0] = n->kids[0]; + m->counts[0] = n->counts[0]; + m->elems[0] = n->elems[0]; + m->kids[1] = n->kids[1]; + m->counts[1] = n->counts[1]; + m->elems[1] = n->elems[1]; + m->kids[2] = n->kids[2]; + m->counts[2] = n->counts[2]; + n->kids[0] = left; + n->counts[0] = lcount; + n->elems[0] = e; + n->kids[1] = right; + n->counts[1] = rcount; + e = n->elems[2]; + } + m->kids[3] = n->kids[3] = n->kids[2] = NULL; + m->counts[3] = n->counts[3] = n->counts[2] = 0; + m->elems[2] = n->elems[2] = n->elems[1] = NULL; + if (m->kids[0]) + m->kids[0]->parent = m; + if (m->kids[1]) + m->kids[1]->parent = m; + if (m->kids[2]) + m->kids[2]->parent = m; + if (n->kids[0]) + n->kids[0]->parent = n; + if (n->kids[1]) + n->kids[1]->parent = n; + LOG((" left (%p): %p/%d [%p] %p/%d [%p] %p/%d\n", m, + m->kids[0], m->counts[0], m->elems[0], + m->kids[1], m->counts[1], m->elems[1], + m->kids[2], m->counts[2])); + LOG((" right (%p): %p/%d [%p] %p/%d\n", n, + n->kids[0], n->counts[0], n->elems[0], + n->kids[1], n->counts[1])); + left = m; + lcount = countnode234(left); + right = n; + rcount = countnode234(right); + } + if (n->parent) + np = (n->parent->kids[0] == n ? &n->parent->kids[0] : + n->parent->kids[1] == n ? &n->parent->kids[1] : + n->parent->kids[2] == n ? &n->parent->kids[2] : + &n->parent->kids[3]); + n = n->parent; + } + + /* + * If we've come out of here by `break', n will still be + * non-NULL and all we need to do is go back up the tree + * updating counts. If we've come here because n is NULL, we + * need to create a new root for the tree because the old one + * has just split into two. */ + if (n) { + while (n->parent) { + int count = countnode234(n); + int childnum; + childnum = (n->parent->kids[0] == n ? 0 : + n->parent->kids[1] == n ? 1 : + n->parent->kids[2] == n ? 2 : 3); + n->parent->counts[childnum] = count; + n = n->parent; + } + } else { + LOG((" root is overloaded, split into two\n")); + t->root = snew(node234); + t->root->kids[0] = left; + t->root->counts[0] = lcount; + t->root->elems[0] = e; + t->root->kids[1] = right; + t->root->counts[1] = rcount; + t->root->elems[1] = NULL; + t->root->kids[2] = NULL; + t->root->counts[2] = 0; + t->root->elems[2] = NULL; + t->root->kids[3] = NULL; + t->root->counts[3] = 0; + t->root->parent = NULL; + if (t->root->kids[0]) + t->root->kids[0]->parent = t->root; + if (t->root->kids[1]) + t->root->kids[1]->parent = t->root; + LOG((" new root is %p/%d [%p] %p/%d\n", + t->root->kids[0], t->root->counts[0], + t->root->elems[0], t->root->kids[1], t->root->counts[1])); + } + + return orig_e; +} + +void *add234(tree234 * t, void *e) +{ + if (!t->cmp) /* tree is unsorted */ + return NULL; + + return add234_internal(t, e, -1); +} +void *addpos234(tree234 * t, void *e, int index) +{ + if (index < 0 || /* index out of range */ + t->cmp) /* tree is sorted */ + return NULL; /* return failure */ + + return add234_internal(t, e, index); /* this checks the upper bound */ +} + +/* + * Look up the element at a given numeric index in a 2-3-4 tree. + * Returns NULL if the index is out of range. + */ +void *index234(tree234 * t, int index) +{ + node234 *n; + + if (!t->root) + return NULL; /* tree is empty */ + + if (index < 0 || index >= countnode234(t->root)) + return NULL; /* out of range */ + + n = t->root; + + while (n) { + if (index < n->counts[0]) + n = n->kids[0]; + else if (index -= n->counts[0] + 1, index < 0) + return n->elems[0]; + else if (index < n->counts[1]) + n = n->kids[1]; + else if (index -= n->counts[1] + 1, index < 0) + return n->elems[1]; + else if (index < n->counts[2]) + n = n->kids[2]; + else if (index -= n->counts[2] + 1, index < 0) + return n->elems[2]; + else + n = n->kids[3]; + } + + /* We shouldn't ever get here. I wonder how we did. */ + return NULL; +} + +/* + * Find an element e in a sorted 2-3-4 tree t. Returns NULL if not + * found. e is always passed as the first argument to cmp, so cmp + * can be an asymmetric function if desired. cmp can also be passed + * as NULL, in which case the compare function from the tree proper + * will be used. + */ +void *findrelpos234(tree234 * t, void *e, cmpfn234 cmp, + int relation, int *index) +{ + node234 *n; + void *ret; + int c; + int idx, ecount, kcount, cmpret; + + if (t->root == NULL) + return NULL; + + if (cmp == NULL) + cmp = t->cmp; + + n = t->root; + /* + * Attempt to find the element itself. + */ + idx = 0; + ecount = -1; + /* + * Prepare a fake `cmp' result if e is NULL. + */ + cmpret = 0; + if (e == NULL) { + assert(relation == REL234_LT || relation == REL234_GT); + if (relation == REL234_LT) + cmpret = +1; /* e is a max: always greater */ + else if (relation == REL234_GT) + cmpret = -1; /* e is a min: always smaller */ + } + while (1) { + for (kcount = 0; kcount < 4; kcount++) { + if (kcount >= 3 || n->elems[kcount] == NULL || + (c = cmpret ? cmpret : cmp(e, n->elems[kcount])) < 0) { + break; + } + if (n->kids[kcount]) + idx += n->counts[kcount]; + if (c == 0) { + ecount = kcount; + break; + } + idx++; + } + if (ecount >= 0) + break; + if (n->kids[kcount]) + n = n->kids[kcount]; + else + break; + } + + if (ecount >= 0) { + /* + * We have found the element we're looking for. It's + * n->elems[ecount], at tree index idx. If our search + * relation is EQ, LE or GE we can now go home. + */ + if (relation != REL234_LT && relation != REL234_GT) { + if (index) + *index = idx; + return n->elems[ecount]; + } + + /* + * Otherwise, we'll do an indexed lookup for the previous + * or next element. (It would be perfectly possible to + * implement these search types in a non-counted tree by + * going back up from where we are, but far more fiddly.) + */ + if (relation == REL234_LT) + idx--; + else + idx++; + } else { + /* + * We've found our way to the bottom of the tree and we + * know where we would insert this node if we wanted to: + * we'd put it in in place of the (empty) subtree + * n->kids[kcount], and it would have index idx + * + * But the actual element isn't there. So if our search + * relation is EQ, we're doomed. + */ + if (relation == REL234_EQ) + return NULL; + + /* + * Otherwise, we must do an index lookup for index idx-1 + * (if we're going left - LE or LT) or index idx (if we're + * going right - GE or GT). + */ + if (relation == REL234_LT || relation == REL234_LE) { + idx--; + } + } + + /* + * We know the index of the element we want; just call index234 + * to do the rest. This will return NULL if the index is out of + * bounds, which is exactly what we want. + */ + ret = index234(t, idx); + if (ret && index) + *index = idx; + return ret; +} +void *find234(tree234 * t, void *e, cmpfn234 cmp) +{ + return findrelpos234(t, e, cmp, REL234_EQ, NULL); +} +void *findrel234(tree234 * t, void *e, cmpfn234 cmp, int relation) +{ + return findrelpos234(t, e, cmp, relation, NULL); +} +void *findpos234(tree234 * t, void *e, cmpfn234 cmp, int *index) +{ + return findrelpos234(t, e, cmp, REL234_EQ, index); +} + +/* + * Delete an element e in a 2-3-4 tree. Does not free the element, + * merely removes all links to it from the tree nodes. + */ +static void *delpos234_internal(tree234 * t, int index) +{ + node234 *n; + void *retval; + int ei = -1; + + retval = 0; + + n = t->root; + LOG(("deleting item %d from tree %p\n", index, t)); + while (1) { + while (n) { + int ki; + node234 *sub; + + LOG( + (" node %p: %p/%d [%p] %p/%d [%p] %p/%d [%p] %p/%d index=%d\n", + n, n->kids[0], n->counts[0], n->elems[0], n->kids[1], + n->counts[1], n->elems[1], n->kids[2], n->counts[2], + n->elems[2], n->kids[3], n->counts[3], index)); + if (index < n->counts[0]) { + ki = 0; + } else if (index -= n->counts[0] + 1, index < 0) { + ei = 0; + break; + } else if (index < n->counts[1]) { + ki = 1; + } else if (index -= n->counts[1] + 1, index < 0) { + ei = 1; + break; + } else if (index < n->counts[2]) { + ki = 2; + } else if (index -= n->counts[2] + 1, index < 0) { + ei = 2; + break; + } else { + ki = 3; + } + /* + * Recurse down to subtree ki. If it has only one element, + * we have to do some transformation to start with. + */ + LOG((" moving to subtree %d\n", ki)); + sub = n->kids[ki]; + if (!sub->elems[1]) { + LOG((" subtree has only one element!\n", ki)); + if (ki > 0 && n->kids[ki - 1]->elems[1]) { + /* + * Case 3a, left-handed variant. Child ki has + * only one element, but child ki-1 has two or + * more. So we need to move a subtree from ki-1 + * to ki. + * + * . C . . B . + * / \ -> / \ + * [more] a A b B c d D e [more] a A b c C d D e + */ + node234 *sib = n->kids[ki - 1]; + int lastelem = (sib->elems[2] ? 2 : + sib->elems[1] ? 1 : 0); + sub->kids[2] = sub->kids[1]; + sub->counts[2] = sub->counts[1]; + sub->elems[1] = sub->elems[0]; + sub->kids[1] = sub->kids[0]; + sub->counts[1] = sub->counts[0]; + sub->elems[0] = n->elems[ki - 1]; + sub->kids[0] = sib->kids[lastelem + 1]; + sub->counts[0] = sib->counts[lastelem + 1]; + if (sub->kids[0]) + sub->kids[0]->parent = sub; + n->elems[ki - 1] = sib->elems[lastelem]; + sib->kids[lastelem + 1] = NULL; + sib->counts[lastelem + 1] = 0; + sib->elems[lastelem] = NULL; + n->counts[ki] = countnode234(sub); + LOG((" case 3a left\n")); + LOG( + (" index and left subtree count before adjustment: %d, %d\n", + index, n->counts[ki - 1])); + index += n->counts[ki - 1]; + n->counts[ki - 1] = countnode234(sib); + index -= n->counts[ki - 1]; + LOG( + (" index and left subtree count after adjustment: %d, %d\n", + index, n->counts[ki - 1])); + } else if (ki < 3 && n->kids[ki + 1] + && n->kids[ki + 1]->elems[1]) { + /* + * Case 3a, right-handed variant. ki has only + * one element but ki+1 has two or more. Move a + * subtree from ki+1 to ki. + * + * . B . . C . + * / \ -> / \ + * a A b c C d D e [more] a A b B c d D e [more] + */ + node234 *sib = n->kids[ki + 1]; + int j; + sub->elems[1] = n->elems[ki]; + sub->kids[2] = sib->kids[0]; + sub->counts[2] = sib->counts[0]; + if (sub->kids[2]) + sub->kids[2]->parent = sub; + n->elems[ki] = sib->elems[0]; + sib->kids[0] = sib->kids[1]; + sib->counts[0] = sib->counts[1]; + for (j = 0; j < 2 && sib->elems[j + 1]; j++) { + sib->kids[j + 1] = sib->kids[j + 2]; + sib->counts[j + 1] = sib->counts[j + 2]; + sib->elems[j] = sib->elems[j + 1]; + } + sib->kids[j + 1] = NULL; + sib->counts[j + 1] = 0; + sib->elems[j] = NULL; + n->counts[ki] = countnode234(sub); + n->counts[ki + 1] = countnode234(sib); + LOG((" case 3a right\n")); + } else { + /* + * Case 3b. ki has only one element, and has no + * neighbour with more than one. So pick a + * neighbour and merge it with ki, taking an + * element down from n to go in the middle. + * + * . B . . + * / \ -> | + * a A b c C d a A b B c C d + * + * (Since at all points we have avoided + * descending to a node with only one element, + * we can be sure that n is not reduced to + * nothingness by this move, _unless_ it was + * the very first node, ie the root of the + * tree. In that case we remove the now-empty + * root and replace it with its single large + * child as shown.) + */ + node234 *sib; + int j; + + if (ki > 0) { + ki--; + index += n->counts[ki] + 1; + } + sib = n->kids[ki]; + sub = n->kids[ki + 1]; + + sub->kids[3] = sub->kids[1]; + sub->counts[3] = sub->counts[1]; + sub->elems[2] = sub->elems[0]; + sub->kids[2] = sub->kids[0]; + sub->counts[2] = sub->counts[0]; + sub->elems[1] = n->elems[ki]; + sub->kids[1] = sib->kids[1]; + sub->counts[1] = sib->counts[1]; + if (sub->kids[1]) + sub->kids[1]->parent = sub; + sub->elems[0] = sib->elems[0]; + sub->kids[0] = sib->kids[0]; + sub->counts[0] = sib->counts[0]; + if (sub->kids[0]) + sub->kids[0]->parent = sub; + + n->counts[ki + 1] = countnode234(sub); + + sfree(sib); + + /* + * That's built the big node in sub. Now we + * need to remove the reference to sib in n. + */ + for (j = ki; j < 3 && n->kids[j + 1]; j++) { + n->kids[j] = n->kids[j + 1]; + n->counts[j] = n->counts[j + 1]; + n->elems[j] = j < 2 ? n->elems[j + 1] : NULL; + } + n->kids[j] = NULL; + n->counts[j] = 0; + if (j < 3) + n->elems[j] = NULL; + LOG((" case 3b ki=%d\n", ki)); + + if (!n->elems[0]) { + /* + * The root is empty and needs to be + * removed. + */ + LOG((" shifting root!\n")); + t->root = sub; + sub->parent = NULL; + sfree(n); + } + } + } + n = sub; + } + if (!retval) + retval = n->elems[ei]; + + if (ei == -1) + return NULL; /* although this shouldn't happen */ + + /* + * Treat special case: this is the one remaining item in + * the tree. n is the tree root (no parent), has one + * element (no elems[1]), and has no kids (no kids[0]). + */ + if (!n->parent && !n->elems[1] && !n->kids[0]) { + LOG((" removed last element in tree\n")); + sfree(n); + t->root = NULL; + return retval; + } + + /* + * Now we have the element we want, as n->elems[ei], and we + * have also arranged for that element not to be the only + * one in its node. So... + */ + + if (!n->kids[0] && n->elems[1]) { + /* + * Case 1. n is a leaf node with more than one element, + * so it's _really easy_. Just delete the thing and + * we're done. + */ + int i; + LOG((" case 1\n")); + for (i = ei; i < 2 && n->elems[i + 1]; i++) + n->elems[i] = n->elems[i + 1]; + n->elems[i] = NULL; + /* + * Having done that to the leaf node, we now go back up + * the tree fixing the counts. + */ + while (n->parent) { + int childnum; + childnum = (n->parent->kids[0] == n ? 0 : + n->parent->kids[1] == n ? 1 : + n->parent->kids[2] == n ? 2 : 3); + n->parent->counts[childnum]--; + n = n->parent; + } + return retval; /* finished! */ + } else if (n->kids[ei]->elems[1]) { + /* + * Case 2a. n is an internal node, and the root of the + * subtree to the left of e has more than one element. + * So find the predecessor p to e (ie the largest node + * in that subtree), place it where e currently is, and + * then start the deletion process over again on the + * subtree with p as target. + */ + node234 *m = n->kids[ei]; + void *target; + LOG((" case 2a\n")); + while (m->kids[0]) { + m = (m->kids[3] ? m->kids[3] : + m->kids[2] ? m->kids[2] : + m->kids[1] ? m->kids[1] : m->kids[0]); + } + target = (m->elems[2] ? m->elems[2] : + m->elems[1] ? m->elems[1] : m->elems[0]); + n->elems[ei] = target; + index = n->counts[ei] - 1; + n = n->kids[ei]; + } else if (n->kids[ei + 1]->elems[1]) { + /* + * Case 2b, symmetric to 2a but s/left/right/ and + * s/predecessor/successor/. (And s/largest/smallest/). + */ + node234 *m = n->kids[ei + 1]; + void *target; + LOG((" case 2b\n")); + while (m->kids[0]) { + m = m->kids[0]; + } + target = m->elems[0]; + n->elems[ei] = target; + n = n->kids[ei + 1]; + index = 0; + } else { + /* + * Case 2c. n is an internal node, and the subtrees to + * the left and right of e both have only one element. + * So combine the two subnodes into a single big node + * with their own elements on the left and right and e + * in the middle, then restart the deletion process on + * that subtree, with e still as target. + */ + node234 *a = n->kids[ei], *b = n->kids[ei + 1]; + int j; + + LOG((" case 2c\n")); + a->elems[1] = n->elems[ei]; + a->kids[2] = b->kids[0]; + a->counts[2] = b->counts[0]; + if (a->kids[2]) + a->kids[2]->parent = a; + a->elems[2] = b->elems[0]; + a->kids[3] = b->kids[1]; + a->counts[3] = b->counts[1]; + if (a->kids[3]) + a->kids[3]->parent = a; + sfree(b); + n->counts[ei] = countnode234(a); + /* + * That's built the big node in a, and destroyed b. Now + * remove the reference to b (and e) in n. + */ + for (j = ei; j < 2 && n->elems[j + 1]; j++) { + n->elems[j] = n->elems[j + 1]; + n->kids[j + 1] = n->kids[j + 2]; + n->counts[j + 1] = n->counts[j + 2]; + } + n->elems[j] = NULL; + n->kids[j + 1] = NULL; + n->counts[j + 1] = 0; + /* + * It's possible, in this case, that we've just removed + * the only element in the root of the tree. If so, + * shift the root. + */ + if (n->elems[0] == NULL) { + LOG((" shifting root!\n")); + t->root = a; + a->parent = NULL; + sfree(n); + } + /* + * Now go round the deletion process again, with n + * pointing at the new big node and e still the same. + */ + n = a; + index = a->counts[0] + a->counts[1] + 1; + } + } +} +void *delpos234(tree234 * t, int index) +{ + if (index < 0 || index >= countnode234(t->root)) + return NULL; + return delpos234_internal(t, index); +} +void *del234(tree234 * t, void *e) +{ + int index; + if (!findrelpos234(t, e, NULL, REL234_EQ, &index)) + return NULL; /* it wasn't in there anyway */ + return delpos234_internal(t, index); /* it's there; delete it. */ +} + +#ifdef TEST + +/* + * Test code for the 2-3-4 tree. This code maintains an alternative + * representation of the data in the tree, in an array (using the + * obvious and slow insert and delete functions). After each tree + * operation, the verify() function is called, which ensures all + * the tree properties are preserved: + * - node->child->parent always equals node + * - tree->root->parent always equals NULL + * - number of kids == 0 or number of elements + 1; + * - tree has the same depth everywhere + * - every node has at least one element + * - subtree element counts are accurate + * - any NULL kid pointer is accompanied by a zero count + * - in a sorted tree: ordering property between elements of a + * node and elements of its children is preserved + * and also ensures the list represented by the tree is the same + * list it should be. (This last check also doubly verifies the + * ordering properties, because the `same list it should be' is by + * definition correctly ordered. It also ensures all nodes are + * distinct, because the enum functions would get caught in a loop + * if not.) + */ + +#include <stdarg.h> + +/* + * Error reporting function. + */ +void error(char *fmt, ...) +{ + va_list ap; + printf("ERROR: "); + va_start(ap, fmt); + vfprintf(stdout, fmt, ap); + va_end(ap); + printf("\n"); +} + +/* The array representation of the data. */ +void **array; +int arraylen, arraysize; +cmpfn234 cmp; + +/* The tree representation of the same data. */ +tree234 *tree; + +typedef struct { + int treedepth; + int elemcount; +} chkctx; + +int chknode(chkctx * ctx, int level, node234 * node, + void *lowbound, void *highbound) +{ + int nkids, nelems; + int i; + int count; + + /* Count the non-NULL kids. */ + for (nkids = 0; nkids < 4 && node->kids[nkids]; nkids++); + /* Ensure no kids beyond the first NULL are non-NULL. */ + for (i = nkids; i < 4; i++) + if (node->kids[i]) { + error("node %p: nkids=%d but kids[%d] non-NULL", + node, nkids, i); + } else if (node->counts[i]) { + error("node %p: kids[%d] NULL but count[%d]=%d nonzero", + node, i, i, node->counts[i]); + } + + /* Count the non-NULL elements. */ + for (nelems = 0; nelems < 3 && node->elems[nelems]; nelems++); + /* Ensure no elements beyond the first NULL are non-NULL. */ + for (i = nelems; i < 3; i++) + if (node->elems[i]) { + error("node %p: nelems=%d but elems[%d] non-NULL", + node, nelems, i); + } + + if (nkids == 0) { + /* + * If nkids==0, this is a leaf node; verify that the tree + * depth is the same everywhere. + */ + if (ctx->treedepth < 0) + ctx->treedepth = level; /* we didn't know the depth yet */ + else if (ctx->treedepth != level) + error("node %p: leaf at depth %d, previously seen depth %d", + node, level, ctx->treedepth); + } else { + /* + * If nkids != 0, then it should be nelems+1, unless nelems + * is 0 in which case nkids should also be 0 (and so we + * shouldn't be in this condition at all). + */ + int shouldkids = (nelems ? nelems + 1 : 0); + if (nkids != shouldkids) { + error("node %p: %d elems should mean %d kids but has %d", + node, nelems, shouldkids, nkids); + } + } + + /* + * nelems should be at least 1. + */ + if (nelems == 0) { + error("node %p: no elems", node, nkids); + } + + /* + * Add nelems to the running element count of the whole tree. + */ + ctx->elemcount += nelems; + + /* + * Check ordering property: all elements should be strictly > + * lowbound, strictly < highbound, and strictly < each other in + * sequence. (lowbound and highbound are NULL at edges of tree + * - both NULL at root node - and NULL is considered to be < + * everything and > everything. IYSWIM.) + */ + if (cmp) { + for (i = -1; i < nelems; i++) { + void *lower = (i == -1 ? lowbound : node->elems[i]); + void *higher = + (i + 1 == nelems ? highbound : node->elems[i + 1]); + if (lower && higher && cmp(lower, higher) >= 0) { + error("node %p: kid comparison [%d=%s,%d=%s] failed", + node, i, lower, i + 1, higher); + } + } + } + + /* + * Check parent pointers: all non-NULL kids should have a + * parent pointer coming back to this node. + */ + for (i = 0; i < nkids; i++) + if (node->kids[i]->parent != node) { + error("node %p kid %d: parent ptr is %p not %p", + node, i, node->kids[i]->parent, node); + } + + + /* + * Now (finally!) recurse into subtrees. + */ + count = nelems; + + for (i = 0; i < nkids; i++) { + void *lower = (i == 0 ? lowbound : node->elems[i - 1]); + void *higher = (i >= nelems ? highbound : node->elems[i]); + int subcount = + chknode(ctx, level + 1, node->kids[i], lower, higher); + if (node->counts[i] != subcount) { + error("node %p kid %d: count says %d, subtree really has %d", + node, i, node->counts[i], subcount); + } + count += subcount; + } + + return count; +} + +void verify(void) +{ + chkctx ctx; + int i; + void *p; + + ctx.treedepth = -1; /* depth unknown yet */ + ctx.elemcount = 0; /* no elements seen yet */ + /* + * Verify validity of tree properties. + */ + if (tree->root) { + if (tree->root->parent != NULL) + error("root->parent is %p should be null", tree->root->parent); + chknode(&ctx, 0, tree->root, NULL, NULL); + } + printf("tree depth: %d\n", ctx.treedepth); + /* + * Enumerate the tree and ensure it matches up to the array. + */ + for (i = 0; NULL != (p = index234(tree, i)); i++) { + if (i >= arraylen) + error("tree contains more than %d elements", arraylen); + if (array[i] != p) + error("enum at position %d: array says %s, tree says %s", + i, array[i], p); + } + if (ctx.elemcount != i) { + error("tree really contains %d elements, enum gave %d", + ctx.elemcount, i); + } + if (i < arraylen) { + error("enum gave only %d elements, array has %d", i, arraylen); + } + i = count234(tree); + if (ctx.elemcount != i) { + error("tree really contains %d elements, count234 gave %d", + ctx.elemcount, i); + } +} + +void internal_addtest(void *elem, int index, void *realret) +{ + int i, j; + void *retval; + + if (arraysize < arraylen + 1) { + arraysize = arraylen + 1 + 256; + array = sresize(array, arraysize, void *); + } + + i = index; + /* now i points to the first element >= elem */ + retval = elem; /* expect elem returned (success) */ + for (j = arraylen; j > i; j--) + array[j] = array[j - 1]; + array[i] = elem; /* add elem to array */ + arraylen++; + + if (realret != retval) { + error("add: retval was %p expected %p", realret, retval); + } + + verify(); +} + +void addtest(void *elem) +{ + int i; + void *realret; + + realret = add234(tree, elem); + + i = 0; + while (i < arraylen && cmp(elem, array[i]) > 0) + i++; + if (i < arraylen && !cmp(elem, array[i])) { + void *retval = array[i]; /* expect that returned not elem */ + if (realret != retval) { + error("add: retval was %p expected %p", realret, retval); + } + } else + internal_addtest(elem, i, realret); +} + +void addpostest(void *elem, int i) +{ + void *realret; + + realret = addpos234(tree, elem, i); + + internal_addtest(elem, i, realret); +} + +void delpostest(int i) +{ + int index = i; + void *elem = array[i], *ret; + + /* i points to the right element */ + while (i < arraylen - 1) { + array[i] = array[i + 1]; + i++; + } + arraylen--; /* delete elem from array */ + + if (tree->cmp) + ret = del234(tree, elem); + else + ret = delpos234(tree, index); + + if (ret != elem) { + error("del returned %p, expected %p", ret, elem); + } + + verify(); +} + +void deltest(void *elem) +{ + int i; + + i = 0; + while (i < arraylen && cmp(elem, array[i]) > 0) + i++; + if (i >= arraylen || cmp(elem, array[i]) != 0) + return; /* don't do it! */ + delpostest(i); +} + +/* A sample data set and test utility. Designed for pseudo-randomness, + * and yet repeatability. */ + +/* + * This random number generator uses the `portable implementation' + * given in ANSI C99 draft N869. It assumes `unsigned' is 32 bits; + * change it if not. + */ +int randomnumber(unsigned *seed) +{ + *seed *= 1103515245; + *seed += 12345; + return ((*seed) / 65536) % 32768; +} + +int mycmp(void *av, void *bv) +{ + char const *a = (char const *) av; + char const *b = (char const *) bv; + return strcmp(a, b); +} + +#define lenof(x) ( sizeof((x)) / sizeof(*(x)) ) + +char *strings[] = { + "a", "ab", "absque", "coram", "de", + "palam", "clam", "cum", "ex", "e", + "sine", "tenus", "pro", "prae", + "banana", "carrot", "cabbage", "broccoli", "onion", "zebra", + "penguin", "blancmange", "pangolin", "whale", "hedgehog", + "giraffe", "peanut", "bungee", "foo", "bar", "baz", "quux", + "murfl", "spoo", "breen", "flarn", "octothorpe", + "snail", "tiger", "elephant", "octopus", "warthog", "armadillo", + "aardvark", "wyvern", "dragon", "elf", "dwarf", "orc", "goblin", + "pixie", "basilisk", "warg", "ape", "lizard", "newt", "shopkeeper", + "wand", "ring", "amulet" +}; + +#define NSTR lenof(strings) + +int findtest(void) +{ + const static int rels[] = { + REL234_EQ, REL234_GE, REL234_LE, REL234_LT, REL234_GT + }; + const static char *const relnames[] = { + "EQ", "GE", "LE", "LT", "GT" + }; + int i, j, rel, index; + char *p, *ret, *realret, *realret2; + int lo, hi, mid, c; + + for (i = 0; i < NSTR; i++) { + p = strings[i]; + for (j = 0; j < sizeof(rels) / sizeof(*rels); j++) { + rel = rels[j]; + + lo = 0; + hi = arraylen - 1; + while (lo <= hi) { + mid = (lo + hi) / 2; + c = strcmp(p, array[mid]); + if (c < 0) + hi = mid - 1; + else if (c > 0) + lo = mid + 1; + else + break; + } + + if (c == 0) { + if (rel == REL234_LT) + ret = (mid > 0 ? array[--mid] : NULL); + else if (rel == REL234_GT) + ret = (mid < arraylen - 1 ? array[++mid] : NULL); + else + ret = array[mid]; + } else { + assert(lo == hi + 1); + if (rel == REL234_LT || rel == REL234_LE) { + mid = hi; + ret = (hi >= 0 ? array[hi] : NULL); + } else if (rel == REL234_GT || rel == REL234_GE) { + mid = lo; + ret = (lo < arraylen ? array[lo] : NULL); + } else + ret = NULL; + } + + realret = findrelpos234(tree, p, NULL, rel, &index); + if (realret != ret) { + error("find(\"%s\",%s) gave %s should be %s", + p, relnames[j], realret, ret); + } + if (realret && index != mid) { + error("find(\"%s\",%s) gave %d should be %d", + p, relnames[j], index, mid); + } + if (realret && rel == REL234_EQ) { + realret2 = index234(tree, index); + if (realret2 != realret) { + error("find(\"%s\",%s) gave %s(%d) but %d -> %s", + p, relnames[j], realret, index, index, realret2); + } + } +#if 0 + printf("find(\"%s\",%s) gave %s(%d)\n", p, relnames[j], + realret, index); +#endif + } + } + + realret = findrelpos234(tree, NULL, NULL, REL234_GT, &index); + if (arraylen && (realret != array[0] || index != 0)) { + error("find(NULL,GT) gave %s(%d) should be %s(0)", + realret, index, array[0]); + } else if (!arraylen && (realret != NULL)) { + error("find(NULL,GT) gave %s(%d) should be NULL", realret, index); + } + + realret = findrelpos234(tree, NULL, NULL, REL234_LT, &index); + if (arraylen + && (realret != array[arraylen - 1] || index != arraylen - 1)) { + error("find(NULL,LT) gave %s(%d) should be %s(0)", realret, index, + array[arraylen - 1]); + } else if (!arraylen && (realret != NULL)) { + error("find(NULL,LT) gave %s(%d) should be NULL", realret, index); + } +} + +int main(void) +{ + int in[NSTR]; + int i, j, k; + unsigned seed = 0; + + for (i = 0; i < NSTR; i++) + in[i] = 0; + array = NULL; + arraylen = arraysize = 0; + tree = newtree234(mycmp); + cmp = mycmp; + + verify(); + for (i = 0; i < 10000; i++) { + j = randomnumber(&seed); + j %= NSTR; + printf("trial: %d\n", i); + if (in[j]) { + printf("deleting %s (%d)\n", strings[j], j); + deltest(strings[j]); + in[j] = 0; + } else { + printf("adding %s (%d)\n", strings[j], j); + addtest(strings[j]); + in[j] = 1; + } + findtest(); + } + + while (arraylen > 0) { + j = randomnumber(&seed); + j %= arraylen; + deltest(array[j]); + } + + freetree234(tree); + + /* + * Now try an unsorted tree. We don't really need to test + * delpos234 because we know del234 is based on it, so it's + * already been tested in the above sorted-tree code; but for + * completeness we'll use it to tear down our unsorted tree + * once we've built it. + */ + tree = newtree234(NULL); + cmp = NULL; + verify(); + for (i = 0; i < 1000; i++) { + printf("trial: %d\n", i); + j = randomnumber(&seed); + j %= NSTR; + k = randomnumber(&seed); + k %= count234(tree) + 1; + printf("adding string %s at index %d\n", strings[j], k); + addpostest(strings[j], k); + } + while (count234(tree) > 0) { + printf("cleanup: tree size %d\n", count234(tree)); + j = randomnumber(&seed); + j %= count234(tree); + printf("deleting string %s from index %d\n", + (const char *)array[j], j); + delpostest(j); + } + + return 0; +} + +#endif |