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putty/tree234.c

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C
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/*
* 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 "defs.h"
#include "tree234.h"
#ifdef TEST
#define LOG(x) (printf x)
#define snew(type) ((type *)malloc(sizeof(type)))
#define snewn(n, type) ((type *)malloc((n) * sizeof(type)))
#define sresize(ptr, n, type) \
((type *)realloc(sizeof((type *)0 == (ptr)) ? (ptr) : (ptr), \
(n) * sizeof(type)))
#define sfree(ptr) free(ptr)
#else
#include "puttymem.h"
#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;
}
/*
* Internal function to return the number of elements in a node.
*/
static int elements234(node234 *n)
{
int i;
for (i = 0; i < 3; i++)
if (!n->elems[i])
break;
return i;
}
/*
* 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;
}
n = NULL; /* placate gcc; will always be set below since t->root != NULL */
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, (int)(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)
{
search234_state ss;
int reldir = (relation == REL234_LT || relation == REL234_LE ? -1 :
relation == REL234_GT || relation == REL234_GE ? +1 : 0);
bool equal_permitted = (relation != REL234_LT && relation != REL234_GT);
void *toret;
/* Only LT / GT relations are permitted with a null query element. */
assert(!(equal_permitted && !e));
if (cmp == NULL)
cmp = t->cmp;
search234_start(&ss, t);
while (ss.element) {
int cmpret;
if (e) {
cmpret = cmp(e, ss.element);
} else {
cmpret = -reldir; /* invent a fixed compare result */
}
if (cmpret == 0) {
/*
* We've found an element that compares exactly equal to
* the query element.
*/
if (equal_permitted) {
/* If our search relation permits equality, we've
* finished already. */
if (index)
*index = ss.index;
return ss.element;
} else {
/* Otherwise, pretend this element was slightly too
* big/small, according to the direction of search. */
cmpret = reldir;
}
}
search234_step(&ss, cmpret);
}
/*
* No element compares equal to the one we were after, but
* ss.index indicates the index that element would have if it were
* inserted.
*
* So if our search relation is EQ, we must simply return failure.
*/
if (relation == REL234_EQ)
return NULL;
/*
* Otherwise, we must do an index lookup for the previous index
* (if we're going left - LE or LT) or this index (if we're going
* right - GE or GT).
*/
if (relation == REL234_LT || relation == REL234_LE) {
ss.index--;
}
/*
* 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.
*/
toret = index234(t, ss.index);
if (toret && index)
*index = ss.index;
return toret;
}
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);
}
void search234_start(search234_state *state, tree234 *t)
{
state->_node = t->root;
state->_base = 0; /* index of first element in this node's subtree */
state->_last = -1; /* indicate that this node is not previously visted */
search234_step(state, 0);
}
void search234_step(search234_state *state, int direction)
{
node234 *node = state->_node;
int i;
if (!node) {
state->element = NULL;
state->index = 0;
return;
}
if (state->_last != -1) {
/*
* We're already pointing at some element of a node, so we
* should restrict to the elements left or right of it,
* depending on the requested search direction.
*/
assert(direction);
assert(node);
if (direction > 0)
state->_lo = state->_last + 1;
else
state->_hi = state->_last - 1;
if (state->_lo > state->_hi) {
/*
* We've run out of elements in this node, i.e. we've
* narrowed to nothing but a child pointer. Descend to
* that child, and update _base to the leftmost index of
* its subtree.
*/
for (i = 0; i < state->_lo; i++)
state->_base += 1 + node->counts[i];
state->_node = node = node->kids[state->_lo];
state->_last = -1;
}
}
if (state->_last == -1) {
/*
* We've just entered a new node - either because of the above
* code, or because we were called from search234_start - and
* anything in that node is a viable answer.
*/
state->_lo = 0;
state->_hi = node ? elements234(node)-1 : 0;
}
/*
* Now we've got something we can return.
*/
if (!node) {
state->element = NULL;
state->index = state->_base;
} else {
state->_last = (state->_lo + state->_hi) / 2;
state->element = node->elems[state->_last];
state->index = state->_base + state->_last;
for (i = 0; i <= state->_last; i++)
state->index += node->counts[i];
}
}
/*
* 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"));
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>
#include <string.h>
int n_errors = 0;
/*
* 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");
n_errors++;
}
/* 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);
}
}
void searchtest_recurse(search234_state ss, int lo, int hi,
char **expected, char *directionbuf,
char *directionptr)
{
*directionptr = '\0';
if (!ss.element) {
if (lo != hi) {
error("search234(%s) gave NULL for non-empty interval [%d,%d)",
directionbuf, lo, hi);
} else if (ss.index != lo) {
error("search234(%s) gave index %d should be %d",
directionbuf, ss.index, lo);
} else {
printf("%*ssearch234(%s) gave NULL,%d\n",
(int)(directionptr-directionbuf) * 2, "", directionbuf,
ss.index);
}
} else if (lo == hi) {
error("search234(%s) gave %s for empty interval [%d,%d)",
directionbuf, (char *)ss.element, lo, hi);
} else if (ss.element != expected[ss.index]) {
error("search234(%s) gave element %s should be %s",
directionbuf, (char *)ss.element, expected[ss.index]);
} else if (ss.index < lo || ss.index >= hi) {
error("search234(%s) gave index %d should be in [%d,%d)",
directionbuf, ss.index, lo, hi);
return;
} else {
search234_state next;
printf("%*ssearch234(%s) gave %s,%d\n",
(int)(directionptr-directionbuf) * 2, "", directionbuf,
(char *)ss.element, ss.index);
next = ss;
search234_step(&next, -1);
*directionptr = '-';
searchtest_recurse(next, lo, ss.index,
expected, directionbuf, directionptr+1);
next = ss;
search234_step(&next, +1);
*directionptr = '+';
searchtest_recurse(next, ss.index+1, hi,
expected, directionbuf, directionptr+1);
}
}
void searchtest(void)
{
char *expected[NSTR], *p;
char directionbuf[NSTR * 10];
int n;
search234_state ss;
printf("beginning searchtest:");
for (n = 0; (p = index234(tree, n)) != NULL; n++) {
expected[n] = p;
printf(" %d=%s", n, p);
}
printf(" count=%d\n", n);
search234_start(&ss, tree);
searchtest_recurse(ss, 0, n, expected, directionbuf, directionbuf);
}
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();
searchtest();
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();
searchtest();
}
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",
(const char *)array[j], j);
delpostest(j);
}
printf("%d errors found\n", n_errors);
return (n_errors != 0);
}
#endif
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