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.
 */

/* Modifications made for speed/specialization for the ReuseDistance
 * library by Michael Laurenzano in 2012. michaell@sdsc.edu
 */

#include <stdio.h>
#include <stdlib.h>
#include <assert.h>

#include "tree234.h"
#include "ReuseDistance.hpp"

#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

#define reusecmp(va, vb) (va->__seq - vb->__seq)

typedef struct node234_Tag node234;

struct tree234_Tag {
    node234 *root;
};

struct node234_Tag {
    node234 *parent;
    node234 *kids[4];
    int counts[4];
    ReuseEntry* elems[3];
};

/*
 * Create a 2-3-4 tree.
 */
tree234 *newtree234(){
    tree234 *ret = mknew(tree234);
    LOG(("created tree %p\n", ret));
    ret->root = NULL;
    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 inline ReuseEntry* add234_internal(tree234 *t, ReuseEntry* e, int index) {
    node234 *n, **np, *left, *right;
    ReuseEntry* 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 = reusecmp(e, n->elems[0])) < 0)
		childnum = 0;
	    else if (c == 0)
		return n->elems[0];	       /* already exists */
	    else if (n->elems[1] == NULL || (c = reusecmp(e, n->elems[1])) < 0)
		childnum = 1;
	    else if (c == 0)
		return n->elems[1];	       /* already exists */
	    else if (n->elems[2] == NULL || (c = reusecmp(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;
}

ReuseEntry* add234(tree234 *t, ReuseEntry* e) {
    return add234_internal(t, e, -1);
}

/*
 * Look up the element at a given numeric index in a 2-3-4 tree.
 * Returns NULL if the index is out of range.
 */
ReuseEntry* 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.
 */
ReuseEntry* findrelpos234(tree234 *t, ReuseEntry* e, int *index) {
    node234 *n;
    ReuseEntry* ret;
    int c;
    int idx, ecount, kcount, cmpret;

    if (t->root == NULL)
	return NULL;

    n = t->root;
    /*
     * Attempt to find the element itself.
     */
    idx = 0;
    ecount = -1;
    /*
     * Prepare a fake `cmp' result if e is NULL.
     */
    cmpret = 0;
    while (1) {
	for (kcount = 0; kcount < 4; kcount++) {
	    if (kcount >= 3 || n->elems[kcount] == NULL ||
		(c = cmpret ? cmpret : reusecmp(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) {
        if (index) *index = idx;
        return n->elems[ecount];
    } else {
        return NULL;
    }

    /*
     * 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;
}

/*
 * 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 inline ReuseEntry* delpos234_internal(tree234 *t, int index) {
    node234 *n;
    ReuseEntry* 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];
	    ReuseEntry* 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];
	    ReuseEntry* 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;
	}
    }
}

ReuseEntry* delpos234(tree234 *t, int index) {
    if (index < 0 || index >= countnode234(t->root))
	return NULL;
    return delpos234_internal(t, index);
}

ReuseEntry* del234(tree234 *t, ReuseEntry* e) {
    int index;
    if (!findrelpos234(t, e, &index))
	return NULL;		       /* it wasn't in there anyway */
    return delpos234_internal(t, index); /* it's there; delete it. */
}