tree234.c

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/*
 * 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;
    LOG123(("deleting item %d from tree %p\n", index, t));
    while (1) {
	while (n) {
	    int ki;
	    node234 *sub;

	    LOG123(("  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.
	     */
	    LOG123(("  moving to subtree %d\n", ki));
	    sub = n->kids[ki];
	    if (!sub->elems[1]) {
		LOG123(("  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);
		    LOG123(("  case 3a left\n"));
		    LOG123(("  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];
		    LOG123(("  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);
		    LOG123(("  case 3a right\n"));
		} else {
		    /*
		     * Case 3b. ki has only one element, and has no
		     * neighbor with more than one. So pick a
		     * neighbor 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;
		    LOG123(("  case 3b ki=%d\n", ki));

		    if (!n->elems[0]) {
			/*
			 * The root is empty and needs to be
			 * removed.
			 */
			LOG123(("  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]) {
	    LOG123(("  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;
	    LOG123(("  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;
	    LOG123(("  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;
	    LOG123(("  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;

	    LOG123(("  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) {
                LOG123(("  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