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+/*
+ * 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"
+
+#define smalloc malloc
+#define sfree free
+
+#define snew(typ) ( (typ *) smalloc (sizeof (typ)) )
+
+#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;
+}
+
+/*
+ * Propagate a node overflow up a tree until it stops. Returns 0 or
+ * 1, depending on whether the root had to be split or not.
+ */
+static int add234_insert(node234 *left, void *e, node234 *right,
+ node234 **root, node234 *n, int ki) {
+ int lcount, rcount;
+ /*
+ * We need to insert the new left/element/right set in n at
+ * child position ki.
+ */
+ lcount = countnode234(left);
+ rcount = countnode234(right);
+ while (n) {
+ LOG((" at %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %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 \"%s\" %p/%d at position %d\n",
+ left, lcount, e, right, rcount, ki));
+ if (n->elems[1] == NULL) {
+ /*
+ * Insert in a 2-node; simple.
+ */
+ if (ki == 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 { /* ki == 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 (ki == 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 (ki == 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 { /* ki == 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 (ki == 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 (ki == 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 (ki == 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 { /* ki == 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 \"%s\" %p/%d \"%s\" %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 \"%s\" %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)
+ ki = (n->parent->kids[0] == n ? 0 :
+ n->parent->kids[1] == n ? 1 :
+ n->parent->kids[2] == n ? 2 : 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;
+ }
+ return 0; /* root unchanged */
+ } else {
+ LOG((" root is overloaded, split into two\n"));
+ (*root) = snew(node234);
+ (*root)->kids[0] = left; (*root)->counts[0] = lcount;
+ (*root)->elems[0] = e;
+ (*root)->kids[1] = right; (*root)->counts[1] = rcount;
+ (*root)->elems[1] = NULL;
+ (*root)->kids[2] = NULL; (*root)->counts[2] = 0;
+ (*root)->elems[2] = NULL;
+ (*root)->kids[3] = NULL; (*root)->counts[3] = 0;
+ (*root)->parent = NULL;
+ if ((*root)->kids[0]) (*root)->kids[0]->parent = (*root);
+ if ((*root)->kids[1]) (*root)->kids[1]->parent = (*root);
+ LOG((" new root is %p/%d \"%s\" %p/%d\n",
+ (*root)->kids[0], (*root)->counts[0],
+ (*root)->elems[0],
+ (*root)->kids[1], (*root)->counts[1]));
+ return 1; /* root moved */
+ }
+}
+
+/*
+ * 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;
+ int ki;
+ void *orig_e = e;
+ int c;
+
+ LOG(("adding element \"%s\" 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 = t->root;
+ while (n) {
+ LOG((" node %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %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
+ */
+ ki = index;
+ } else {
+ /*
+ * Internal node. We always descend through it (add
+ * always starts at the bottom, never in the
+ * middle).
+ */
+ if (index <= n->counts[0]) {
+ ki = 0;
+ } else if (index -= n->counts[0] + 1, index <= n->counts[1]) {
+ ki = 1;
+ } else if (index -= n->counts[1] + 1, index <= n->counts[2]) {
+ ki = 2;
+ } else if (index -= n->counts[2] + 1, index <= n->counts[3]) {
+ ki = 3;
+ } else
+ return NULL; /* error: index out of range */
+ }
+ } else {
+ if ((c = t->cmp(e, n->elems[0])) < 0)
+ ki = 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)
+ ki = 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)
+ ki = 2;
+ else if (c == 0)
+ return n->elems[2]; /* already exists */
+ else
+ ki = 3;
+ }
+ LOG((" moving to child %d (%p)\n", ki, n->kids[ki]));
+ if (!n->kids[ki])
+ break;
+ n = n->kids[ki];
+ }
+
+ add234_insert(NULL, e, NULL, &t->root, n, ki);
+
+ 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);
+}
+
+/*
+ * Tree transformation used in delete and split: move a subtree
+ * right, from child ki of a node to the next child. Update k and
+ * index so that they still point to the same place in the
+ * transformed tree. Assumes the destination child is not full, and
+ * that the source child does have a subtree to spare. Can cope if
+ * the destination child is undersized.
+ *
+ * . C . . B .
+ * / \ -> / \
+ * [more] a A b B c d D e [more] a A b c C d D e
+ *
+ * . C . . B .
+ * / \ -> / \
+ * [more] a A b B c d [more] a A b c C d
+ */
+static void trans234_subtree_right(node234 *n, int ki, int *k, int *index) {
+ node234 *src, *dest;
+ int i, srclen, adjust;
+
+ src = n->kids[ki];
+ dest = n->kids[ki+1];
+
+ LOG((" trans234_subtree_right(%p, %d):\n", n, ki));
+ LOG((" parent %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %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((" src %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %p/%d\n",
+ src,
+ src->kids[0], src->counts[0], src->elems[0],
+ src->kids[1], src->counts[1], src->elems[1],
+ src->kids[2], src->counts[2], src->elems[2],
+ src->kids[3], src->counts[3]));
+ LOG((" dest %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %p/%d\n",
+ dest,
+ dest->kids[0], dest->counts[0], dest->elems[0],
+ dest->kids[1], dest->counts[1], dest->elems[1],
+ dest->kids[2], dest->counts[2], dest->elems[2],
+ dest->kids[3], dest->counts[3]));
+ /*
+ * Move over the rest of the destination node to make space.
+ */
+ dest->kids[3] = dest->kids[2]; dest->counts[3] = dest->counts[2];
+ dest->elems[2] = dest->elems[1];
+ dest->kids[2] = dest->kids[1]; dest->counts[2] = dest->counts[1];
+ dest->elems[1] = dest->elems[0];
+ dest->kids[1] = dest->kids[0]; dest->counts[1] = dest->counts[0];
+
+ /* which element to move over */
+ i = (src->elems[2] ? 2 : src->elems[1] ? 1 : 0);
+
+ dest->elems[0] = n->elems[ki];
+ n->elems[ki] = src->elems[i];
+ src->elems[i] = NULL;
+
+ dest->kids[0] = src->kids[i+1]; dest->counts[0] = src->counts[i+1];
+ src->kids[i+1] = NULL; src->counts[i+1] = 0;
+
+ if (dest->kids[0]) dest->kids[0]->parent = dest;
+
+ adjust = dest->counts[0] + 1;
+
+ n->counts[ki] -= adjust;
+ n->counts[ki+1] += adjust;
+
+ srclen = n->counts[ki];
+
+ if (k) {
+ LOG((" before: k,index = %d,%d\n", (*k), (*index)));
+ if ((*k) == ki && (*index) > srclen) {
+ (*index) -= srclen + 1;
+ (*k)++;
+ } else if ((*k) == ki+1) {
+ (*index) += adjust;
+ }
+ LOG((" after: k,index = %d,%d\n", (*k), (*index)));
+ }
+
+ LOG((" parent %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %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((" src %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %p/%d\n",
+ src,
+ src->kids[0], src->counts[0], src->elems[0],
+ src->kids[1], src->counts[1], src->elems[1],
+ src->kids[2], src->counts[2], src->elems[2],
+ src->kids[3], src->counts[3]));
+ LOG((" dest %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %p/%d\n",
+ dest,
+ dest->kids[0], dest->counts[0], dest->elems[0],
+ dest->kids[1], dest->counts[1], dest->elems[1],
+ dest->kids[2], dest->counts[2], dest->elems[2],
+ dest->kids[3], dest->counts[3]));
+}
+
+/*
+ * Tree transformation used in delete and split: move a subtree
+ * left, from child ki of a node to the previous child. Update k
+ * and index so that they still point to the same place in the
+ * transformed tree. Assumes the destination child is not full, and
+ * that the source child does have a subtree to spare. Can cope if
+ * the destination child is undersized.
+ *
+ * . B . . C .
+ * / \ -> / \
+ * a A b c C d D e [more] a A b B c d D e [more]
+ *
+ * . A . . B .
+ * / \ -> / \
+ * a b B c C d [more] a A b c C d [more]
+ */
+static void trans234_subtree_left(node234 *n, int ki, int *k, int *index) {
+ node234 *src, *dest;
+ int i, adjust;
+
+ src = n->kids[ki];
+ dest = n->kids[ki-1];
+
+ LOG((" trans234_subtree_left(%p, %d):\n", n, ki));
+ LOG((" parent %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %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((" dest %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %p/%d\n",
+ dest,
+ dest->kids[0], dest->counts[0], dest->elems[0],
+ dest->kids[1], dest->counts[1], dest->elems[1],
+ dest->kids[2], dest->counts[2], dest->elems[2],
+ dest->kids[3], dest->counts[3]));
+ LOG((" src %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %p/%d\n",
+ src,
+ src->kids[0], src->counts[0], src->elems[0],
+ src->kids[1], src->counts[1], src->elems[1],
+ src->kids[2], src->counts[2], src->elems[2],
+ src->kids[3], src->counts[3]));
+
+ /* where in dest to put it */
+ i = (dest->elems[1] ? 2 : dest->elems[0] ? 1 : 0);
+ dest->elems[i] = n->elems[ki-1];
+ n->elems[ki-1] = src->elems[0];
+
+ dest->kids[i+1] = src->kids[0]; dest->counts[i+1] = src->counts[0];
+
+ if (dest->kids[i+1]) dest->kids[i+1]->parent = dest;
+
+ /*
+ * Move over the rest of the source node.
+ */
+ src->kids[0] = src->kids[1]; src->counts[0] = src->counts[1];
+ src->elems[0] = src->elems[1];
+ src->kids[1] = src->kids[2]; src->counts[1] = src->counts[2];
+ src->elems[1] = src->elems[2];
+ src->kids[2] = src->kids[3]; src->counts[2] = src->counts[3];
+ src->elems[2] = NULL;
+ src->kids[3] = NULL; src->counts[3] = 0;
+
+ adjust = dest->counts[i+1] + 1;
+
+ n->counts[ki] -= adjust;
+ n->counts[ki-1] += adjust;
+
+ if (k) {
+ LOG((" before: k,index = %d,%d\n", (*k), (*index)));
+ if ((*k) == ki) {
+ (*index) -= adjust;
+ if ((*index) < 0) {
+ (*index) += n->counts[ki-1] + 1;
+ (*k)--;
+ }
+ }
+ LOG((" after: k,index = %d,%d\n", (*k), (*index)));
+ }
+
+ LOG((" parent %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %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((" dest %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %p/%d\n",
+ dest,
+ dest->kids[0], dest->counts[0], dest->elems[0],
+ dest->kids[1], dest->counts[1], dest->elems[1],
+ dest->kids[2], dest->counts[2], dest->elems[2],
+ dest->kids[3], dest->counts[3]));
+ LOG((" src %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %p/%d\n",
+ src,
+ src->kids[0], src->counts[0], src->elems[0],
+ src->kids[1], src->counts[1], src->elems[1],
+ src->kids[2], src->counts[2], src->elems[2],
+ src->kids[3], src->counts[3]));
+}
+
+/*
+ * Tree transformation used in delete and split: merge child nodes
+ * ki and ki+1 of a node. Update k and index so that they still
+ * point to the same place in the transformed tree. Assumes both
+ * children _are_ sufficiently small.
+ *
+ * . B . .
+ * / \ -> |
+ * a A b c C d a A b B c C d
+ *
+ * This routine can also cope with either child being undersized:
+ *
+ * . A . .
+ * / \ -> |
+ * a b B c a A b B c
+ *
+ * . A . .
+ * / \ -> |
+ * a b B c C d a A b B c C d
+ */
+static void trans234_subtree_merge(node234 *n, int ki, int *k, int *index) {
+ node234 *left, *right;
+ int i, leftlen, rightlen, lsize, rsize;
+
+ left = n->kids[ki]; leftlen = n->counts[ki];
+ right = n->kids[ki+1]; rightlen = n->counts[ki+1];
+
+ LOG((" trans234_subtree_merge(%p, %d):\n", n, ki));
+ LOG((" parent %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %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((" left %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %p/%d\n",
+ left,
+ left->kids[0], left->counts[0], left->elems[0],
+ left->kids[1], left->counts[1], left->elems[1],
+ left->kids[2], left->counts[2], left->elems[2],
+ left->kids[3], left->counts[3]));
+ LOG((" right %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %p/%d\n",
+ right,
+ right->kids[0], right->counts[0], right->elems[0],
+ right->kids[1], right->counts[1], right->elems[1],
+ right->kids[2], right->counts[2], right->elems[2],
+ right->kids[3], right->counts[3]));
+
+ assert(!left->elems[2] && !right->elems[2]); /* neither is large! */
+ lsize = (left->elems[1] ? 2 : left->elems[0] ? 1 : 0);
+ rsize = (right->elems[1] ? 2 : right->elems[0] ? 1 : 0);
+
+ left->elems[lsize] = n->elems[ki];
+
+ for (i = 0; i < rsize+1; i++) {
+ left->kids[lsize+1+i] = right->kids[i];
+ left->counts[lsize+1+i] = right->counts[i];
+ if (left->kids[lsize+1+i])
+ left->kids[lsize+1+i]->parent = left;
+ if (i < rsize)
+ left->elems[lsize+1+i] = right->elems[i];
+ }
+
+ n->counts[ki] += rightlen + 1;
+
+ sfree(right);
+
+ /*
+ * Move the rest of n up by one.
+ */
+ for (i = ki+1; i < 3; i++) {
+ n->kids[i] = n->kids[i+1];
+ n->counts[i] = n->counts[i+1];
+ }
+ for (i = ki; i < 2; i++) {
+ n->elems[i] = n->elems[i+1];
+ }
+ n->kids[3] = NULL;
+ n->counts[3] = 0;
+ n->elems[2] = NULL;
+
+ if (k) {
+ LOG((" before: k,index = %d,%d\n", (*k), (*index)));
+ if ((*k) == ki+1) {
+ (*k)--;
+ (*index) += leftlen + 1;
+ } else if ((*k) > ki+1) {
+ (*k)--;
+ }
+ LOG((" after: k,index = %d,%d\n", (*k), (*index)));
+ }
+
+ LOG((" parent %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %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((" merged %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %p/%d\n",
+ left,
+ left->kids[0], left->counts[0], left->elems[0],
+ left->kids[1], left->counts[1], left->elems[1],
+ left->kids[2], left->counts[2], left->elems[2],
+ left->kids[3], left->counts[3]));
+
+}
+
+/*
+ * 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 ki, i;
+
+ retval = NULL;
+
+ n = t->root; /* by assumption this is non-NULL */
+ LOG(("deleting item %d from tree %p\n", index, t));
+ while (1) {
+ node234 *sub;
+
+ LOG((" node %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %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 <= n->counts[1]) {
+ ki = 1;
+ } else if (index -= n->counts[1]+1, index <= n->counts[2]) {
+ ki = 2;
+ } else if (index -= n->counts[2]+1, index <= n->counts[3]) {
+ ki = 3;
+ } else {
+ assert(0); /* can't happen */
+ }
+
+ if (!n->kids[0])
+ break; /* n is a leaf node; we're here! */
+
+ /*
+ * Check to see if we've found our target element. If so,
+ * we must choose a new target (we'll use the old target's
+ * successor, which will be in a leaf), move it into the
+ * place of the old one, continue down to the leaf and
+ * delete the old copy of the new target.
+ */
+ if (index == n->counts[ki]) {
+ node234 *m;
+ LOG((" found element in internal node, index %d\n", ki));
+ assert(n->elems[ki]); /* must be a kid _before_ an element */
+ ki++; index = 0;
+ for (m = n->kids[ki]; m->kids[0]; m = m->kids[0])
+ continue;
+ LOG((" replacing with element \"%s\" from leaf node %p\n",
+ m->elems[0], m));
+ retval = n->elems[ki-1];
+ n->elems[ki-1] = m->elems[0];
+ }
+
+ /*
+ * 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"));
+ if (ki > 0 && n->kids[ki-1]->elems[1]) {
+ /*
+ * 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.
+ */
+ trans234_subtree_right(n, ki-1, &ki, &index);
+ } else if (ki < 3 && n->kids[ki+1] &&
+ n->kids[ki+1]->elems[1]) {
+ /*
+ * Child ki has only one element, but ki+1 has
+ * two or more. Move a subtree from ki+1 to ki.
+ */
+ trans234_subtree_left(n, ki+1, &ki, &index);
+ } else {
+ /*
+ * ki is small with only small neighbours. Pick a
+ * neighbour and merge with it.
+ */
+ trans234_subtree_merge(n, ki>0 ? ki-1 : ki, &ki, &index);
+ sub = n->kids[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 = NULL;
+ }
+ }
+ }
+
+ if (n)
+ n->counts[ki]--;
+ n = sub;
+ }
+
+ /*
+ * Now n is a leaf node, and ki marks the element number we
+ * want to delete. We've already arranged for the leaf to be
+ * bigger than minimum size, so let's just go to it.
+ */
+ assert(!n->kids[0]);
+ if (!retval)
+ retval = n->elems[ki];
+
+ for (i = ki; i < 2 && n->elems[i+1]; i++)
+ n->elems[i] = n->elems[i+1];
+ n->elems[i] = NULL;
+
+ /*
+ * It's just possible that we have reduced the leaf to zero
+ * size. This can only happen if it was the root - so destroy
+ * it and make the tree empty.
+ */
+ if (!n->elems[0]) {
+ LOG((" removed last element in tree, destroying empty root\n"));
+ assert(n == t->root);
+ sfree(n);
+ t->root = NULL;
+ }
+
+ return retval; /* finished! */
+}
+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. */
+}
+
+/*
+ * Join two subtrees together with a separator element between
+ * them, given their relative height.
+ *
+ * (Height<0 means the left tree is shorter, >0 means the right
+ * tree is shorter, =0 means (duh) they're equal.)
+ *
+ * It is assumed that any checks needed on the ordering criterion
+ * have _already_ been done.
+ *
+ * The value returned in `height' is 0 or 1 depending on whether the
+ * resulting tree is the same height as the original larger one, or
+ * one higher.
+ */
+static node234 *join234_internal(node234 *left, void *sep,
+ node234 *right, int *height) {
+ node234 *root, *node;
+ int relht = *height;
+ int ki;
+
+ LOG((" join: joining %p \"%s\" %p, relative height is %d\n",
+ left, sep, right, relht));
+ if (relht == 0) {
+ /*
+ * The trees are the same height. Create a new one-element
+ * root containing the separator and pointers to the two
+ * nodes.
+ */
+ node234 *newroot;
+ newroot = snew(node234);
+ newroot->kids[0] = left; newroot->counts[0] = countnode234(left);
+ newroot->elems[0] = sep;
+ newroot->kids[1] = right; newroot->counts[1] = countnode234(right);
+ newroot->elems[1] = NULL;
+ newroot->kids[2] = NULL; newroot->counts[2] = 0;
+ newroot->elems[2] = NULL;
+ newroot->kids[3] = NULL; newroot->counts[3] = 0;
+ newroot->parent = NULL;
+ if (left) left->parent = newroot;
+ if (right) right->parent = newroot;
+ *height = 1;
+ LOG((" join: same height, brand new root\n"));
+ return newroot;
+ }
+
+ /*
+ * This now works like the addition algorithm on the larger
+ * tree. We're replacing a single kid pointer with two kid
+ * pointers separated by an element; if that causes the node to
+ * overload, we split it in two, move a separator element up to
+ * the next node, and repeat.
+ */
+ if (relht < 0) {
+ /*
+ * Left tree is shorter. Search down the right tree to find
+ * the pointer we're inserting at.
+ */
+ node = root = right;
+ while (++relht < 0) {
+ node = node->kids[0];
+ }
+ ki = 0;
+ right = node->kids[ki];
+ } else {
+ /*
+ * Right tree is shorter; search down the left to find the
+ * pointer we're inserting at.
+ */
+ node = root = left;
+ while (--relht > 0) {
+ if (node->elems[2])
+ node = node->kids[3];
+ else if (node->elems[1])
+ node = node->kids[2];
+ else
+ node = node->kids[1];
+ }
+ if (node->elems[2])
+ ki = 3;
+ else if (node->elems[1])
+ ki = 2;
+ else
+ ki = 1;
+ left = node->kids[ki];
+ }
+
+ /*
+ * Now proceed as for addition.
+ */
+ *height = add234_insert(left, sep, right, &root, node, ki);
+
+ return root;
+}
+static int height234(tree234 *t) {
+ int level = 0;
+ node234 *n = t->root;
+ while (n) {
+ level++;
+ n = n->kids[0];
+ }
+ return level;
+}
+tree234 *join234(tree234 *t1, tree234 *t2) {
+ int size2 = countnode234(t2->root);
+ if (size2 > 0) {
+ void *element;
+ int relht;
+
+ if (t1->cmp) {
+ element = index234(t2, 0);
+ element = findrelpos234(t1, element, NULL, REL234_GE, NULL);
+ if (element)
+ return NULL;
+ }
+
+ element = delpos234(t2, 0);
+ relht = height234(t1) - height234(t2);
+ t1->root = join234_internal(t1->root, element, t2->root, &relht);
+ t2->root = NULL;
+ }
+ return t1;
+}
+tree234 *join234r(tree234 *t1, tree234 *t2) {
+ int size1 = countnode234(t1->root);
+ if (size1 > 0) {
+ void *element;
+ int relht;
+
+ if (t2->cmp) {
+ element = index234(t1, size1-1);
+ element = findrelpos234(t2, element, NULL, REL234_LE, NULL);
+ if (element)
+ return NULL;
+ }
+
+ element = delpos234(t1, size1-1);
+ relht = height234(t1) - height234(t2);
+ t2->root = join234_internal(t1->root, element, t2->root, &relht);
+ t1->root = NULL;
+ }
+ return t2;
+}
+
+/*
+ * Split out the first <index> elements in a tree and return a
+ * pointer to the root node. Leave the root node of the remainder
+ * in t.
+ */
+static node234 *split234_internal(tree234 *t, int index) {
+ node234 *halves[2], *n, *sib, *sub;
+ node234 *lparent, *rparent;
+ int ki, pki, i, half, lcount, rcount;
+
+ n = t->root;
+ LOG(("splitting tree %p at point %d\n", t, index));
+
+ /*
+ * Easy special cases. After this we have also dealt completely
+ * with the empty-tree case and we can assume the root exists.
+ */
+ if (index == 0) /* return nothing */
+ return NULL;
+ if (index == countnode234(t->root)) { /* return the whole tree */
+ node234 *ret = t->root;
+ t->root = NULL;
+ return ret;
+ }
+
+ /*
+ * Search down the tree to find the split point.
+ */
+ lparent = rparent = NULL;
+ while (n) {
+ LOG((" node %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %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));
+ lcount = index;
+ rcount = countnode234(n) - lcount;
+ if (index <= n->counts[0]) {
+ ki = 0;
+ } else if (index -= n->counts[0]+1, index <= n->counts[1]) {
+ ki = 1;
+ } else if (index -= n->counts[1]+1, index <= n->counts[2]) {
+ ki = 2;
+ } else {
+ index -= n->counts[2]+1;
+ ki = 3;
+ }
+
+ LOG((" splitting at subtree %d\n", ki));
+ sub = n->kids[ki];
+
+ LOG((" splitting at child index %d\n", ki));
+
+ /*
+ * Split the node, put halves[0] on the right of the left
+ * one and halves[1] on the left of the right one, put the
+ * new node pointers in halves[0] and halves[1], and go up
+ * a level.
+ */
+ sib = snew(node234);
+ for (i = 0; i < 3; i++) {
+ if (i+ki < 3 && n->elems[i+ki]) {
+ sib->elems[i] = n->elems[i+ki];
+ sib->kids[i+1] = n->kids[i+ki+1];
+ if (sib->kids[i+1]) sib->kids[i+1]->parent = sib;
+ sib->counts[i+1] = n->counts[i+ki+1];
+ n->elems[i+ki] = NULL;
+ n->kids[i+ki+1] = NULL;
+ n->counts[i+ki+1] = 0;
+ } else {
+ sib->elems[i] = NULL;
+ sib->kids[i+1] = NULL;
+ sib->counts[i+1] = 0;
+ }
+ }
+ if (lparent) {
+ lparent->kids[pki] = n;
+ lparent->counts[pki] = lcount;
+ n->parent = lparent;
+ rparent->kids[0] = sib;
+ rparent->counts[0] = rcount;
+ sib->parent = rparent;
+ } else {
+ halves[0] = n;
+ n->parent = NULL;
+ halves[1] = sib;
+ sib->parent = NULL;
+ }
+ lparent = n;
+ rparent = sib;
+ pki = ki;
+ LOG((" left node %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %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((" right node %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %p/%d\n",
+ sib,
+ sib->kids[0], sib->counts[0], sib->elems[0],
+ sib->kids[1], sib->counts[1], sib->elems[1],
+ sib->kids[2], sib->counts[2], sib->elems[2],
+ sib->kids[3], sib->counts[3]));
+
+ n = sub;
+ }
+
+ /*
+ * We've come off the bottom here, so we've successfully split
+ * the tree into two equally high subtrees. The only problem is
+ * that some of the nodes down the fault line will be smaller
+ * than the minimum permitted size. (Since this is a 2-3-4
+ * tree, that means they'll be zero-element one-child nodes.)
+ */
+ LOG((" fell off bottom, lroot is %p, rroot is %p\n",
+ halves[0], halves[1]));
+ lparent->counts[pki] = rparent->counts[0] = 0;
+ lparent->kids[pki] = rparent->kids[0] = NULL;
+
+ /*
+ * So now we go back down the tree from each of the two roots,
+ * fixing up undersize nodes.
+ */
+ for (half = 0; half < 2; half++) {
+ /*
+ * Remove the root if it's undersize (it will contain only
+ * one child pointer, so just throw it away and replace it
+ * with its child). This might happen several times.
+ */
+ while (halves[half] && !halves[half]->elems[0]) {
+ LOG((" root %p is undersize, throwing away\n", halves[half]));
+ halves[half] = halves[half]->kids[0];
+ sfree(halves[half]->parent);
+ halves[half]->parent = NULL;
+ LOG((" new root is %p\n", halves[half]));
+ }
+
+ n = halves[half];
+ while (n) {
+ void (*toward)(node234 *n, int ki, int *k, int *index);
+ int ni, merge;
+
+ /*
+ * Now we have a potentially undersize node on the
+ * right (if half==0) or left (if half==1). Sort it
+ * out, by merging with a neighbour or by transferring
+ * subtrees over. At this time we must also ensure that
+ * nodes are bigger than minimum, in case we need an
+ * element to merge two nodes below.
+ */
+ LOG((" node %p: %p/%d \"%s\" %p/%d \"%s\" %p/%d \"%s\" %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 (half == 1) {
+ ki = 0; /* the kid we're interested in */
+ ni = 1; /* the neighbour */
+ merge = 0; /* for merge: leftmost of the two */
+ toward = trans234_subtree_left;
+ } else {
+ ki = (n->kids[3] ? 3 : n->kids[2] ? 2 : 1);
+ ni = ki-1;
+ merge = ni;
+ toward = trans234_subtree_right;
+ }
+
+ sub = n->kids[ki];
+ if (sub && !sub->elems[1]) {
+ /*
+ * This node is undersized or minimum-size. If we
+ * can merge it with its neighbour, we do so;
+ * otherwise we must be able to transfer subtrees
+ * over to it until it is greater than minimum
+ * size.
+ */
+ int undersized = (!sub->elems[0]);
+ LOG((" child %d is %ssize\n", ki,
+ undersized ? "under" : "minimum-"));
+ LOG((" neighbour is %s\n",
+ n->kids[ni]->elems[2] ? "large" :
+ n->kids[ni]->elems[1] ? "medium" : "small"));
+ if (!n->kids[ni]->elems[1] ||
+ (undersized && !n->kids[ni]->elems[2])) {
+ /*
+ * Neighbour is small, or possibly neighbour is
+ * medium and we are undersize.
+ */
+ trans234_subtree_merge(n, merge, NULL, NULL);
+ sub = n->kids[merge];
+ if (!n->elems[0]) {
+ /*
+ * n is empty, and hence must have been the
+ * root and needs to be removed.
+ */
+ assert(!n->parent);
+ LOG((" shifting root!\n"));
+ halves[half] = sub;
+ halves[half]->parent = NULL;
+ sfree(n);
+ }
+ } else {
+ /* Neighbour is big enough to move trees over. */
+ toward(n, ni, NULL, NULL);
+ if (undersized)
+ toward(n, ni, NULL, NULL);
+ }
+ }
+ n = sub;
+ }
+ }
+
+ t->root = halves[1];
+ return halves[0];
+}
+tree234 *splitpos234(tree234 *t, int index, int before) {
+ tree234 *ret;
+ node234 *n;
+ int count;
+
+ count = countnode234(t->root);
+ if (index < 0 || index > count)
+ return NULL; /* error */
+ ret = newtree234(t->cmp);
+ n = split234_internal(t, index);
+ if (before) {
+ /* We want to return the ones before the index. */
+ ret->root = n;
+ } else {
+ /*
+ * We want to keep the ones before the index and return the
+ * ones after.
+ */
+ ret->root = t->root;
+ t->root = n;
+ }
+ return ret;
+}
+tree234 *split234(tree234 *t, void *e, cmpfn234 cmp, int rel) {
+ int before;
+ int index;
+
+ assert(rel != REL234_EQ);
+
+ if (rel == REL234_GT || rel == REL234_GE) {
+ before = 1;
+ rel = (rel == REL234_GT ? REL234_LE : REL234_LT);
+ } else {
+ before = 0;
+ }
+ if (!findrelpos234(t, e, cmp, rel, &index))
+ index = 0;
+
+ return splitpos234(t, index+1, before);
+}
+
+static node234 *copynode234(node234 *n, copyfn234 copyfn, void *copyfnstate) {
+ int i;
+ node234 *n2 = snew(node234);
+
+ for (i = 0; i < 3; i++) {
+ if (n->elems[i] && copyfn)
+ n2->elems[i] = copyfn(copyfnstate, n->elems[i]);
+ else
+ n2->elems[i] = n->elems[i];
+ }
+
+ for (i = 0; i < 4; i++) {
+ if (n->kids[i]) {
+ n2->kids[i] = copynode234(n->kids[i], copyfn, copyfnstate);
+ n2->kids[i]->parent = n2;
+ } else {
+ n2->kids[i] = NULL;
+ }
+ n2->counts[i] = n->counts[i];
+ }
+
+ return n2;
+}
+tree234 *copytree234(tree234 *t, copyfn234 copyfn, void *copyfnstate) {
+ tree234 *t2;
+
+ t2 = newtree234(t->cmp);
+ if (t->root) {
+ t2->root = copynode234(t->root, copyfn, copyfnstate);
+ t2->root->parent = NULL;
+ } else
+ t2->root = NULL;
+
+ return t2;
+}
+
+#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>
+
+#define srealloc realloc
+
+/*
+ * 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;
+
+/*
+ * Routines to provide a diagnostic printout of a tree. Currently
+ * relies on every element in the tree being a one-character string
+ * :-)
+ */
+typedef struct {
+ char **levels;
+} dispctx;
+
+int dispnode(node234 *n, int level, dispctx *ctx) {
+ if (level == 0) {
+ int xpos = strlen(ctx->levels[0]);
+ int len;
+
+ if (n->elems[2])
+ len = sprintf(ctx->levels[0]+xpos, " %s%s%s",
+ n->elems[0], n->elems[1], n->elems[2]);
+ else if (n->elems[1])
+ len = sprintf(ctx->levels[0]+xpos, " %s%s",
+ n->elems[0], n->elems[1]);
+ else
+ len = sprintf(ctx->levels[0]+xpos, " %s",
+ n->elems[0]);
+ return xpos + 1 + (len-1) / 2;
+ } else {
+ int xpos[4], nkids;
+ int nodelen, mypos, myleft, x, i;
+
+ xpos[0] = dispnode(n->kids[0], level-3, ctx);
+ xpos[1] = dispnode(n->kids[1], level-3, ctx);
+ nkids = 2;
+ if (n->kids[2]) {
+ xpos[2] = dispnode(n->kids[2], level-3, ctx);
+ nkids = 3;
+ }
+ if (n->kids[3]) {
+ xpos[3] = dispnode(n->kids[3], level-3, ctx);
+ nkids = 4;
+ }
+
+ if (nkids == 4)
+ mypos = (xpos[1] + xpos[2]) / 2;
+ else if (nkids == 3)
+ mypos = xpos[1];
+ else
+ mypos = (xpos[0] + xpos[1]) / 2;
+ nodelen = nkids * 2 - 1;
+ myleft = mypos - ((nodelen-1)/2);
+ assert(myleft >= xpos[0]);
+ assert(myleft + nodelen-1 <= xpos[nkids-1]);
+
+ x = strlen(ctx->levels[level]);
+ while (x <= xpos[0] && x < myleft)
+ ctx->levels[level][x++] = ' ';
+ while (x < myleft)
+ ctx->levels[level][x++] = '_';
+ if (nkids==4)
+ x += sprintf(ctx->levels[level]+x, ".%s.%s.%s.",
+ n->elems[0], n->elems[1], n->elems[2]);
+ else if (nkids==3)
+ x += sprintf(ctx->levels[level]+x, ".%s.%s.",
+ n->elems[0], n->elems[1]);
+ else
+ x += sprintf(ctx->levels[level]+x, ".%s.",
+ n->elems[0]);
+ while (x < xpos[nkids-1])
+ ctx->levels[level][x++] = '_';
+ ctx->levels[level][x] = '\0';
+
+ x = strlen(ctx->levels[level-1]);
+ for (i = 0; i < nkids; i++) {
+ int rpos, pos;
+ rpos = xpos[i];
+ if (i > 0 && i < nkids-1)
+ pos = myleft + 2*i;
+ else
+ pos = rpos;
+ if (rpos < pos)
+ rpos++;
+ while (x < pos && x < rpos)
+ ctx->levels[level-1][x++] = ' ';
+ if (x == pos)
+ ctx->levels[level-1][x++] = '|';
+ while (x < pos || x < rpos)
+ ctx->levels[level-1][x++] = '_';
+ if (x == pos)
+ ctx->levels[level-1][x++] = '|';
+ }
+ ctx->levels[level-1][x] = '\0';
+
+ x = strlen(ctx->levels[level-2]);
+ for (i = 0; i < nkids; i++) {
+ int rpos = xpos[i];
+
+ while (x < rpos)
+ ctx->levels[level-2][x++] = ' ';
+ ctx->levels[level-2][x++] = '|';
+ }
+ ctx->levels[level-2][x] = '\0';
+
+ return mypos;
+ }
+}
+
+void disptree(tree234 *t) {
+ dispctx ctx;
+ char *leveldata;
+ int width = count234(t);
+ int ht = height234(t) * 3 - 2;
+ int i;
+
+ if (!t->root) {
+ printf("[empty tree]\n");
+ }
+
+ leveldata = smalloc(ht * (width+2));
+ ctx.levels = smalloc(ht * sizeof(char *));
+ for (i = 0; i < ht; i++) {
+ ctx.levels[i] = leveldata + i * (width+2);
+ ctx.levels[i][0] = '\0';
+ }
+
+ (void) dispnode(t->root, ht-1, &ctx);
+
+ for (i = ht; i-- ;)
+ printf("%s\n", ctx.levels[i]);
+
+ sfree(ctx.levels);
+ sfree(leveldata);
+}
+
+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 verifytree(tree234 *tree, void **array, int arraylen) {
+ 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 verify(void) { verifytree(tree, array, arraylen); }
+
+void internal_addtest(void *elem, int index, void *realret) {
+ int i, j;
+ void *retval;
+
+ if (arraysize < arraylen+1) {
+ arraysize = arraylen+1+256;
+ array = (array == NULL ? smalloc(arraysize*sizeof(*array)) :
+ srealloc(array, arraysize*sizeof(*array)));
+ }
+
+ 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[] = {
+ "0", "2", "3", "I", "K", "d", "H", "J", "Q", "N", "n", "q", "j", "i",
+ "7", "G", "F", "D", "b", "x", "g", "B", "e", "v", "V", "T", "f", "E",
+ "S", "8", "A", "k", "X", "p", "C", "R", "a", "o", "r", "O", "Z", "u",
+ "6", "1", "w", "L", "P", "M", "c", "U", "h", "9", "t", "5", "W", "Y",
+ "m", "s", "l", "4",
+#if 0
+ "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"
+#endif
+};
+
+#define NSTR lenof(strings)
+
+void findtest(void) {
+ static const int rels[] = {
+ REL234_EQ, REL234_GE, REL234_LE, REL234_LT, REL234_GT
+ };
+ static const 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 < (int)NSTR; i++) {
+ p = strings[i];
+ for (j = 0; j < (int)(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);
+ }
+}
+
+void splittest(tree234 *tree, void **array, int arraylen) {
+ int i;
+ tree234 *tree3, *tree4;
+ for (i = 0; i <= arraylen; i++) {
+ tree3 = copytree234(tree, NULL, NULL);
+ tree4 = splitpos234(tree3, i, 0);
+ verifytree(tree3, array, i);
+ verifytree(tree4, array+i, arraylen-i);
+ join234(tree3, tree4);
+ freetree234(tree4); /* left empty by join */
+ verifytree(tree3, array, arraylen);
+ freetree234(tree3);
+ }
+}
+
+int main(void) {
+ int in[NSTR];
+ int i, j, k;
+ int tworoot, tmplen;
+ unsigned seed = 0;
+ tree234 *tree2, *tree3, *tree4;
+ int c;
+
+ setvbuf(stdout, NULL, _IOLBF, 0);
+
+ for (i = 0; i < (int)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;
+ }
+ disptree(tree);
+ 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 we have this tree in its full form, we'll take a copy
+ * of it to use in split and join testing.
+ */
+ tree2 = copytree234(tree, NULL, NULL);
+ verifytree(tree2, array, arraylen);/* check the copy is accurate */
+ /*
+ * Split tests. Split the tree at every possible point and
+ * check the resulting subtrees.
+ */
+ tworoot = (!tree2->root->elems[1]);/* see if it has a 2-root */
+ splittest(tree2, array, arraylen);
+ /*
+ * Now do the split test again, but on a tree that has a 2-root
+ * (if the previous one didn't) or doesn't (if the previous one
+ * did).
+ */
+ tmplen = arraylen;
+ while ((!tree2->root->elems[1]) == tworoot) {
+ delpos234(tree2, --tmplen);
+ }
+ printf("now trying splits on second tree\n");
+ splittest(tree2, array, tmplen);
+ freetree234(tree2);
+
+ /*
+ * Back to the main testing of uncounted trees.
+ */
+ 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", (char *)array[j], j);
+ delpostest(j);
+ }
+ freetree234(tree);
+
+ /*
+ * Finally, do some testing on split/join on _sorted_ trees. At
+ * the same time, we'll be testing split on very small trees.
+ */
+ tree = newtree234(mycmp);
+ cmp = mycmp;
+ arraylen = 0;
+ for (i = 0; i < 17; i++) {
+ tree2 = copytree234(tree, NULL, NULL);
+ splittest(tree2, array, arraylen);
+ freetree234(tree2);
+ if (i < 16)
+ addtest(strings[i]);
+ }
+ freetree234(tree);
+
+ /*
+ * Test silly cases of join: join(emptytree, emptytree), and
+ * also ensure join correctly spots when sorted trees fail the
+ * ordering constraint.
+ */
+ tree = newtree234(mycmp);
+ tree2 = newtree234(mycmp);
+ tree3 = newtree234(mycmp);
+ tree4 = newtree234(mycmp);
+ assert(mycmp(strings[0], strings[1]) < 0); /* just in case :-) */
+ add234(tree2, strings[1]);
+ add234(tree4, strings[0]);
+ array[0] = strings[0];
+ array[1] = strings[1];
+ verifytree(tree, array, 0);
+ verifytree(tree2, array+1, 1);
+ verifytree(tree3, array, 0);
+ verifytree(tree4, array, 1);
+
+ /*
+ * So:
+ * - join(tree,tree3) should leave both tree and tree3 unchanged.
+ * - joinr(tree,tree2) should leave both tree and tree2 unchanged.
+ * - join(tree4,tree3) should leave both tree3 and tree4 unchanged.
+ * - join(tree, tree2) should move the element from tree2 to tree.
+ * - joinr(tree4, tree3) should move the element from tree4 to tree3.
+ * - join(tree,tree3) should return NULL and leave both unchanged.
+ * - join(tree3,tree) should work and create a bigger tree in tree3.
+ */
+ assert(tree == join234(tree, tree3));
+ verifytree(tree, array, 0);
+ verifytree(tree3, array, 0);
+ assert(tree2 == join234r(tree, tree2));
+ verifytree(tree, array, 0);
+ verifytree(tree2, array+1, 1);
+ assert(tree4 == join234(tree4, tree3));
+ verifytree(tree3, array, 0);
+ verifytree(tree4, array, 1);
+ assert(tree == join234(tree, tree2));
+ verifytree(tree, array+1, 1);
+ verifytree(tree2, array, 0);
+ assert(tree3 == join234r(tree4, tree3));
+ verifytree(tree3, array, 1);
+ verifytree(tree4, array, 0);
+ assert(NULL == join234(tree, tree3));
+ verifytree(tree, array+1, 1);
+ verifytree(tree3, array, 1);
+ assert(tree3 == join234(tree3, tree));
+ verifytree(tree3, array, 2);
+ verifytree(tree, array, 0);
+
+ return 0;
+}
+
+#endif
+
+#if 0 /* sorted list of strings might be useful */
+{
+ "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x",
+}
+#endif
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