tree234.c

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/*
 * $Id: tree234.c 84 2005-07-01 14:52:38Z bogdan_iancu $
 *
 * 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"
#include "../../mem/shm_mem.h"

//#define smalloc malloc
//#define sfree free
#define smalloc shm_malloc
#define sfree  shm_free

#define mknew(typ) ( (typ *) smalloc (sizeof (typ)) )

#ifdef TEST
#define LOG123(x) (printf x)
#else
#define LOG123(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);
    LOG123(("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)
{
    if(t == NULL)
      return;
    freenode234(t->root);
    sfree(t);
}

/*
 * Free a 2-3-4 tree (including freeing the elements with 'fn' function).
 */
static void free2node234(node234 *n, freefn fn )
{
    if (!n)
   return;
    free2node234(n->kids[0], fn);
    free2node234(n->kids[1], fn);
    free2node234(n->kids[2], fn);
    free2node234(n->kids[3], fn);
    fn(n->elems[0]);
    fn(n->elems[1]);
    fn(n->elems[2]);
    sfree(n);
}
void free2tree234(tree234 *t, freefn fn)
{
    if(t == NULL)
      return;
    free2node234(t->root, fn);
    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;

    LOG123(("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;
   LOG123(("  created root %p\n", t->root));
   return orig_e;
   }

   np = &t->root;
   n = *np;

   while (*np) {
   int childnum;
   n = *np;
   LOG123(("  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];
   LOG123(("  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) {
   LOG123(("  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]));
   LOG123(("  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]) {
      LOG123(("  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] */
      LOG123(("  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;
       LOG123(("  done\n"));
       break;
   } else if (n->elems[2] == NULL) {
       /*
        * Insert in a 3-node; simple.
        */
       if (np == &n->kids[0]) {
      LOG123(("  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]) {
      LOG123(("  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] */
      LOG123(("  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;
       LOG123(("  done\n"));
       break;
   } else {
       node234 *m = mknew(node234);
       m->parent = n->parent;
       LOG123(("  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;
       LOG123(("  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]));
       LOG123(("  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 {
   LOG123(("  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;
   LOG123(("  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;
    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
 *  - 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