countnode234() should politely return 0 when passed NULL. Was

[u/mdw/putty] / tree234.c

/*
 * 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 mknew(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 = mknew(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 = mknew(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;
    }

    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 = mknew(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 = mknew(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 c;
            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>

#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;

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 = (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[] = {
    "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