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authormarha <marha@users.sourceforge.net>2010-11-19 12:14:18 +0000
committermarha <marha@users.sourceforge.net>2010-11-19 12:14:18 +0000
commit6be86147f292b3178413bc644853ad80b620042e (patch)
treed972cbd73289e9b6e9574c5fd65c6830f67861c6 /tools/plink/tree234.c
parent111cb82886d25b0b7faa526ce411cc8ef02235a6 (diff)
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Reintegrate tools from trunk
Diffstat (limited to 'tools/plink/tree234.c')
-rw-r--r--tools/plink/tree234.c1479
1 files changed, 1479 insertions, 0 deletions
diff --git a/tools/plink/tree234.c b/tools/plink/tree234.c
new file mode 100644
index 000000000..4e2da9dd6
--- /dev/null
+++ b/tools/plink/tree234.c
@@ -0,0 +1,1479 @@
+/*
+ * 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