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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 "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;
    }

    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

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