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Diffstat (limited to 'intl/tsearch.c')
| -rw-r--r-- | intl/tsearch.c | 684 |
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diff --git a/intl/tsearch.c b/intl/tsearch.c new file mode 100644 index 0000000..d549dd4 --- /dev/null +++ b/intl/tsearch.c @@ -0,0 +1,684 @@ +/* Copyright (C) 1995, 1996, 1997, 2000, 2006 Free Software Foundation, Inc. + Contributed by Bernd Schmidt <crux@Pool.Informatik.RWTH-Aachen.DE>, 1997. + + NOTE: The canonical source of this file is maintained with the GNU C + Library. Bugs can be reported to bug-glibc@gnu.org. + + This program is free software; you can redistribute it and/or modify it + under the terms of the GNU Library General Public License as published + by the Free Software Foundation; either version 2, or (at your option) + any later version. + + This program is distributed in the hope that it will be useful, + but WITHOUT ANY WARRANTY; without even the implied warranty of + MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU + Library General Public License for more details. + + You should have received a copy of the GNU Library General Public + License along with this program; if not, write to the Free Software + Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, + USA. */ + +/* Tree search for red/black trees. + The algorithm for adding nodes is taken from one of the many "Algorithms" + books by Robert Sedgewick, although the implementation differs. + The algorithm for deleting nodes can probably be found in a book named + "Introduction to Algorithms" by Cormen/Leiserson/Rivest. At least that's + the book that my professor took most algorithms from during the "Data + Structures" course... + + Totally public domain. */ + +/* Red/black trees are binary trees in which the edges are colored either red + or black. They have the following properties: + 1. The number of black edges on every path from the root to a leaf is + constant. + 2. No two red edges are adjacent. + Therefore there is an upper bound on the length of every path, it's + O(log n) where n is the number of nodes in the tree. No path can be longer + than 1+2*P where P is the length of the shortest path in the tree. + Useful for the implementation: + 3. If one of the children of a node is NULL, then the other one is red + (if it exists). + + In the implementation, not the edges are colored, but the nodes. The color + interpreted as the color of the edge leading to this node. The color is + meaningless for the root node, but we color the root node black for + convenience. All added nodes are red initially. + + Adding to a red/black tree is rather easy. The right place is searched + with a usual binary tree search. Additionally, whenever a node N is + reached that has two red successors, the successors are colored black and + the node itself colored red. This moves red edges up the tree where they + pose less of a problem once we get to really insert the new node. Changing + N's color to red may violate rule 2, however, so rotations may become + necessary to restore the invariants. Adding a new red leaf may violate + the same rule, so afterwards an additional check is run and the tree + possibly rotated. + + Deleting is hairy. There are mainly two nodes involved: the node to be + deleted (n1), and another node that is to be unchained from the tree (n2). + If n1 has a successor (the node with a smallest key that is larger than + n1), then the successor becomes n2 and its contents are copied into n1, + otherwise n1 becomes n2. + Unchaining a node may violate rule 1: if n2 is black, one subtree is + missing one black edge afterwards. The algorithm must try to move this + error upwards towards the root, so that the subtree that does not have + enough black edges becomes the whole tree. Once that happens, the error + has disappeared. It may not be necessary to go all the way up, since it + is possible that rotations and recoloring can fix the error before that. + + Although the deletion algorithm must walk upwards through the tree, we + do not store parent pointers in the nodes. Instead, delete allocates a + small array of parent pointers and fills it while descending the tree. + Since we know that the length of a path is O(log n), where n is the number + of nodes, this is likely to use less memory. */ + +/* Tree rotations look like this: + A C + / \ / \ + B C A G + / \ / \ --> / \ + D E F G B F + / \ + D E + + In this case, A has been rotated left. This preserves the ordering of the + binary tree. */ + +#include <config.h> + +/* Specification. */ +#ifdef IN_LIBINTL +# include "tsearch.h" +#else +# include <search.h> +#endif + +#include <stdlib.h> + +typedef int (*__compar_fn_t) (const void *, const void *); +typedef void (*__action_fn_t) (const void *, VISIT, int); + +#ifndef weak_alias +# define __tsearch tsearch +# define __tfind tfind +# define __tdelete tdelete +# define __twalk twalk +#endif + +#ifndef internal_function +/* Inside GNU libc we mark some function in a special way. In other + environments simply ignore the marking. */ +# define internal_function +#endif + +typedef struct node_t +{ + /* Callers expect this to be the first element in the structure - do not + move! */ + const void *key; + struct node_t *left; + struct node_t *right; + unsigned int red:1; +} *node; +typedef const struct node_t *const_node; + +#undef DEBUGGING + +#ifdef DEBUGGING + +/* Routines to check tree invariants. */ + +#include <assert.h> + +#define CHECK_TREE(a) check_tree(a) + +static void +check_tree_recurse (node p, int d_sofar, int d_total) +{ + if (p == NULL) + { + assert (d_sofar == d_total); + return; + } + + check_tree_recurse (p->left, d_sofar + (p->left && !p->left->red), d_total); + check_tree_recurse (p->right, d_sofar + (p->right && !p->right->red), d_total); + if (p->left) + assert (!(p->left->red && p->red)); + if (p->right) + assert (!(p->right->red && p->red)); +} + +static void +check_tree (node root) +{ + int cnt = 0; + node p; + if (root == NULL) + return; + root->red = 0; + for(p = root->left; p; p = p->left) + cnt += !p->red; + check_tree_recurse (root, 0, cnt); +} + + +#else + +#define CHECK_TREE(a) + +#endif + +/* Possibly "split" a node with two red successors, and/or fix up two red + edges in a row. ROOTP is a pointer to the lowest node we visited, PARENTP + and GPARENTP pointers to its parent/grandparent. P_R and GP_R contain the + comparison values that determined which way was taken in the tree to reach + ROOTP. MODE is 1 if we need not do the split, but must check for two red + edges between GPARENTP and ROOTP. */ +static void +maybe_split_for_insert (node *rootp, node *parentp, node *gparentp, + int p_r, int gp_r, int mode) +{ + node root = *rootp; + node *rp, *lp; + rp = &(*rootp)->right; + lp = &(*rootp)->left; + + /* See if we have to split this node (both successors red). */ + if (mode == 1 + || ((*rp) != NULL && (*lp) != NULL && (*rp)->red && (*lp)->red)) + { + /* This node becomes red, its successors black. */ + root->red = 1; + if (*rp) + (*rp)->red = 0; + if (*lp) + (*lp)->red = 0; + + /* If the parent of this node is also red, we have to do + rotations. */ + if (parentp != NULL && (*parentp)->red) + { + node gp = *gparentp; + node p = *parentp; + /* There are two main cases: + 1. The edge types (left or right) of the two red edges differ. + 2. Both red edges are of the same type. + There exist two symmetries of each case, so there is a total of + 4 cases. */ + if ((p_r > 0) != (gp_r > 0)) + { + /* Put the child at the top of the tree, with its parent + and grandparent as successors. */ + p->red = 1; + gp->red = 1; + root->red = 0; + if (p_r < 0) + { + /* Child is left of parent. */ + p->left = *rp; + *rp = p; + gp->right = *lp; + *lp = gp; + } + else + { + /* Child is right of parent. */ + p->right = *lp; + *lp = p; + gp->left = *rp; + *rp = gp; + } + *gparentp = root; + } + else + { + *gparentp = *parentp; + /* Parent becomes the top of the tree, grandparent and + child are its successors. */ + p->red = 0; + gp->red = 1; + if (p_r < 0) + { + /* Left edges. */ + gp->left = p->right; + p->right = gp; + } + else + { + /* Right edges. */ + gp->right = p->left; + p->left = gp; + } + } + } + } +} + +/* Find or insert datum into search tree. + KEY is the key to be located, ROOTP is the address of tree root, + COMPAR the ordering function. */ +void * +__tsearch (const void *key, void **vrootp, __compar_fn_t compar) +{ + node q; + node *parentp = NULL, *gparentp = NULL; + node *rootp = (node *) vrootp; + node *nextp; + int r = 0, p_r = 0, gp_r = 0; /* No they might not, Mr Compiler. */ + + if (rootp == NULL) + return NULL; + + /* This saves some additional tests below. */ + if (*rootp != NULL) + (*rootp)->red = 0; + + CHECK_TREE (*rootp); + + nextp = rootp; + while (*nextp != NULL) + { + node root = *rootp; + r = (*compar) (key, root->key); + if (r == 0) + return root; + + maybe_split_for_insert (rootp, parentp, gparentp, p_r, gp_r, 0); + /* If that did any rotations, parentp and gparentp are now garbage. + That doesn't matter, because the values they contain are never + used again in that case. */ + + nextp = r < 0 ? &root->left : &root->right; + if (*nextp == NULL) + break; + + gparentp = parentp; + parentp = rootp; + rootp = nextp; + + gp_r = p_r; + p_r = r; + } + + q = (struct node_t *) malloc (sizeof (struct node_t)); + if (q != NULL) + { + *nextp = q; /* link new node to old */ + q->key = key; /* initialize new node */ + q->red = 1; + q->left = q->right = NULL; + + if (nextp != rootp) + /* There may be two red edges in a row now, which we must avoid by + rotating the tree. */ + maybe_split_for_insert (nextp, rootp, parentp, r, p_r, 1); + } + + return q; +} +#ifdef weak_alias +weak_alias (__tsearch, tsearch) +#endif + + +/* Find datum in search tree. + KEY is the key to be located, ROOTP is the address of tree root, + COMPAR the ordering function. */ +void * +__tfind (key, vrootp, compar) + const void *key; + void *const *vrootp; + __compar_fn_t compar; +{ + node *rootp = (node *) vrootp; + + if (rootp == NULL) + return NULL; + + CHECK_TREE (*rootp); + + while (*rootp != NULL) + { + node root = *rootp; + int r; + + r = (*compar) (key, root->key); + if (r == 0) + return root; + + rootp = r < 0 ? &root->left : &root->right; + } + return NULL; +} +#ifdef weak_alias +weak_alias (__tfind, tfind) +#endif + + +/* Delete node with given key. + KEY is the key to be deleted, ROOTP is the address of the root of tree, + COMPAR the comparison function. */ +void * +__tdelete (const void *key, void **vrootp, __compar_fn_t compar) +{ + node p, q, r, retval; + int cmp; + node *rootp = (node *) vrootp; + node root, unchained; + /* Stack of nodes so we remember the parents without recursion. It's + _very_ unlikely that there are paths longer than 40 nodes. The tree + would need to have around 250.000 nodes. */ + int stacksize = 100; + int sp = 0; + node *nodestack[100]; + + if (rootp == NULL) + return NULL; + p = *rootp; + if (p == NULL) + return NULL; + + CHECK_TREE (p); + + while ((cmp = (*compar) (key, (*rootp)->key)) != 0) + { + if (sp == stacksize) + abort (); + + nodestack[sp++] = rootp; + p = *rootp; + rootp = ((cmp < 0) + ? &(*rootp)->left + : &(*rootp)->right); + if (*rootp == NULL) + return NULL; + } + + /* This is bogus if the node to be deleted is the root... this routine + really should return an integer with 0 for success, -1 for failure + and errno = ESRCH or something. */ + retval = p; + + /* We don't unchain the node we want to delete. Instead, we overwrite + it with its successor and unchain the successor. If there is no + successor, we really unchain the node to be deleted. */ + + root = *rootp; + + r = root->right; + q = root->left; + + if (q == NULL || r == NULL) + unchained = root; + else + { + node *parent = rootp, *up = &root->right; + for (;;) + { + if (sp == stacksize) + abort (); + nodestack[sp++] = parent; + parent = up; + if ((*up)->left == NULL) + break; + up = &(*up)->left; + } + unchained = *up; + } + + /* We know that either the left or right successor of UNCHAINED is NULL. + R becomes the other one, it is chained into the parent of UNCHAINED. */ + r = unchained->left; + if (r == NULL) + r = unchained->right; + if (sp == 0) + *rootp = r; + else + { + q = *nodestack[sp-1]; + if (unchained == q->right) + q->right = r; + else + q->left = r; + } + + if (unchained != root) + root->key = unchained->key; + if (!unchained->red) + { + /* Now we lost a black edge, which means that the number of black + edges on every path is no longer constant. We must balance the + tree. */ + /* NODESTACK now contains all parents of R. R is likely to be NULL + in the first iteration. */ + /* NULL nodes are considered black throughout - this is necessary for + correctness. */ + while (sp > 0 && (r == NULL || !r->red)) + { + node *pp = nodestack[sp - 1]; + p = *pp; + /* Two symmetric cases. */ + if (r == p->left) + { + /* Q is R's brother, P is R's parent. The subtree with root + R has one black edge less than the subtree with root Q. */ + q = p->right; + if (q->red) + { + /* If Q is red, we know that P is black. We rotate P left + so that Q becomes the top node in the tree, with P below + it. P is colored red, Q is colored black. + This action does not change the black edge count for any + leaf in the tree, but we will be able to recognize one + of the following situations, which all require that Q + is black. */ + q->red = 0; + p->red = 1; + /* Left rotate p. */ + p->right = q->left; + q->left = p; + *pp = q; + /* Make sure pp is right if the case below tries to use + it. */ + nodestack[sp++] = pp = &q->left; + q = p->right; + } + /* We know that Q can't be NULL here. We also know that Q is + black. */ + if ((q->left == NULL || !q->left->red) + && (q->right == NULL || !q->right->red)) + { + /* Q has two black successors. We can simply color Q red. + The whole subtree with root P is now missing one black + edge. Note that this action can temporarily make the + tree invalid (if P is red). But we will exit the loop + in that case and set P black, which both makes the tree + valid and also makes the black edge count come out + right. If P is black, we are at least one step closer + to the root and we'll try again the next iteration. */ + q->red = 1; + r = p; + } + else + { + /* Q is black, one of Q's successors is red. We can + repair the tree with one operation and will exit the + loop afterwards. */ + if (q->right == NULL || !q->right->red) + { + /* The left one is red. We perform the same action as + in maybe_split_for_insert where two red edges are + adjacent but point in different directions: + Q's left successor (let's call it Q2) becomes the + top of the subtree we are looking at, its parent (Q) + and grandparent (P) become its successors. The former + successors of Q2 are placed below P and Q. + P becomes black, and Q2 gets the color that P had. + This changes the black edge count only for node R and + its successors. */ + node q2 = q->left; + q2->red = p->red; + p->right = q2->left; + q->left = q2->right; + q2->right = q; + q2->left = p; + *pp = q2; + p->red = 0; + } + else + { + /* It's the right one. Rotate P left. P becomes black, + and Q gets the color that P had. Q's right successor + also becomes black. This changes the black edge + count only for node R and its successors. */ + q->red = p->red; + p->red = 0; + + q->right->red = 0; + + /* left rotate p */ + p->right = q->left; + q->left = p; + *pp = q; + } + + /* We're done. */ + sp = 1; + r = NULL; + } + } + else + { + /* Comments: see above. */ + q = p->left; + if (q->red) + { + q->red = 0; + p->red = 1; + p->left = q->right; + q->right = p; + *pp = q; + nodestack[sp++] = pp = &q->right; + q = p->left; + } + if ((q->right == NULL || !q->right->red) + && (q->left == NULL || !q->left->red)) + { + q->red = 1; + r = p; + } + else + { + if (q->left == NULL || !q->left->red) + { + node q2 = q->right; + q2->red = p->red; + p->left = q2->right; + q->right = q2->left; + q2->left = q; + q2->right = p; + *pp = q2; + p->red = 0; + } + else + { + q->red = p->red; + p->red = 0; + q->left->red = 0; + p->left = q->right; + q->right = p; + *pp = q; + } + sp = 1; + r = NULL; + } + } + --sp; + } + if (r != NULL) + r->red = 0; + } + + free (unchained); + return retval; +} +#ifdef weak_alias +weak_alias (__tdelete, tdelete) +#endif + + +/* Walk the nodes of a tree. + ROOT is the root of the tree to be walked, ACTION the function to be + called at each node. LEVEL is the level of ROOT in the whole tree. */ +static void +internal_function +trecurse (const void *vroot, __action_fn_t action, int level) +{ + const_node root = (const_node) vroot; + + if (root->left == NULL && root->right == NULL) + (*action) (root, leaf, level); + else + { + (*action) (root, preorder, level); + if (root->left != NULL) + trecurse (root->left, action, level + 1); + (*action) (root, postorder, level); + if (root->right != NULL) + trecurse (root->right, action, level + 1); + (*action) (root, endorder, level); + } +} + + +/* Walk the nodes of a tree. + ROOT is the root of the tree to be walked, ACTION the function to be + called at each node. */ +void +__twalk (const void *vroot, __action_fn_t action) +{ + const_node root = (const_node) vroot; + + CHECK_TREE (root); + + if (root != NULL && action != NULL) + trecurse (root, action, 0); +} +#ifdef weak_alias +weak_alias (__twalk, twalk) +#endif + + +#ifdef _LIBC + +/* The standardized functions miss an important functionality: the + tree cannot be removed easily. We provide a function to do this. */ +static void +internal_function +tdestroy_recurse (node root, __free_fn_t freefct) +{ + if (root->left != NULL) + tdestroy_recurse (root->left, freefct); + if (root->right != NULL) + tdestroy_recurse (root->right, freefct); + (*freefct) ((void *) root->key); + /* Free the node itself. */ + free (root); +} + +void +__tdestroy (void *vroot, __free_fn_t freefct) +{ + node root = (node) vroot; + + CHECK_TREE (root); + + if (root != NULL) + tdestroy_recurse (root, freefct); +} +weak_alias (__tdestroy, tdestroy) + +#endif /* _LIBC */ |