doesn't need stdc.h
[unix-history] / usr / src / lib / libc / stdlib / radixsort.c
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1/*-
2 * Copyright (c) 1990 The Regents of the University of California.
3 * All rights reserved.
4 *
5 * %sccs.include.redist.c%
6 */
7
8#if defined(LIBC_SCCS) && !defined(lint)
33f8130f 9static char sccsid[] = "@(#)radixsort.c 5.6 (Berkeley) %G%";
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10#endif /* LIBC_SCCS and not lint */
11
12#include <sys/types.h>
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13#include <limits.h>
14#include <stdlib.h>
15#include <stddef.h>
16
34849bd9 17/*
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18 * __rspartition is the cutoff point for a further partitioning instead
19 * of a shellsort. If it changes check __rsshell_increments. Both of
33f8130f 20 * these are exported, as the best values are data dependent.
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21 */
22#define NPARTITION 40
23int __rspartition = NPARTITION;
24int __rsshell_increments[] = { 4, 1, 0, 0, 0, 0, 0, 0 };
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25
26/*
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27 * Stackp points to context structures, where each structure schedules a
28 * partitioning. Radixsort exits when the stack is empty.
34849bd9 29 *
a5baf581 30 * If the buckets are placed on the stack randomly, the worst case is when
e061e641 31 * all the buckets but one contain (npartitions + 1) elements and the bucket
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32 * pushed on the stack last contains the rest of the elements. In this case,
33 * stack growth is bounded by:
dfad239c 34 *
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35 * limit = (nelements / (npartitions + 1)) - 1;
36 *
37 * This is a very large number, 52,377,648 for the maximum 32-bit signed int.
dfad239c 38 *
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39 * By forcing the largest bucket to be pushed on the stack first, the worst
40 * case is when all but two buckets each contain (npartitions + 1) elements,
41 * with the remaining elements split equally between the first and last
42 * buckets pushed on the stack. In this case, stack growth is bounded when:
dfad239c 43 *
a5baf581 44 * for (partition_cnt = 0; nelements > npartitions; ++partition_cnt)
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45 * nelements =
46 * (nelements - (npartitions + 1) * (nbuckets - 2)) / 2;
47 * The bound is:
48 *
49 * limit = partition_cnt * (nbuckets - 1);
a5baf581 50 *
e061e641 51 * This is a much smaller number, 4590 for the maximum 32-bit signed int.
34849bd9 52 */
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53#define NBUCKETS (UCHAR_MAX + 1)
54
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55typedef struct _stack {
56 u_char **bot;
57 int indx, nmemb;
58} CONTEXT;
59
34849bd9 60#define STACKPUSH { \
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61 stackp->bot = p; \
62 stackp->nmemb = nmemb; \
63 stackp->indx = indx; \
64 ++stackp; \
65}
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66#define STACKPOP { \
67 if (stackp == stack) \
68 break; \
69 --stackp; \
70 bot = stackp->bot; \
71 nmemb = stackp->nmemb; \
72 indx = stackp->indx; \
73}
74
75/*
76 * A variant of MSD radix sorting; see Knuth Vol. 3, page 177, and 5.2.5,
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77 * Ex. 10 and 12. Also, "Three Partition Refinement Algorithms, Paige
78 * and Tarjan, SIAM J. Comput. Vol. 16, No. 6, December 1987.
34849bd9 79 *
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80 * This uses a simple sort as soon as a bucket crosses a cutoff point,
81 * rather than sorting the entire list after partitioning is finished.
82 * This should be an advantage.
34849bd9 83 *
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84 * This is pure MSD instead of LSD of some number of MSD, switching to
85 * the simple sort as soon as possible. Takes linear time relative to
86 * the number of bytes in the strings.
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87 */
88radixsort(l1, nmemb, tab, endbyte)
89 u_char **l1, *tab, endbyte;
90 register int nmemb;
91{
92 register int i, indx, t1, t2;
93 register u_char **l2, **p, **bot, *tr;
dfad239c 94 CONTEXT *stack, *stackp;
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95 int c[NBUCKETS + 1], max;
96 u_char ltab[NBUCKETS];
33f8130f 97 static void shellsort();
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98
99 if (nmemb <= 1)
100 return(0);
101
a5baf581 102 /*
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103 * T1 is the constant part of the equation, the number of elements
104 * represented on the stack between the top and bottom entries.
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105 * It doesn't get rounded as the divide by 2 rounds down (correct
106 * for a value being subtracted). T2, the nelem value, has to be
107 * rounded up before each divide because we want an upper bound;
108 * this could overflow if nmemb is the maximum int.
109 */
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110 t1 = ((__rspartition + 1) * (NBUCKETS - 2)) >> 1;
111 for (i = 0, t2 = nmemb; t2 > __rspartition; i += NBUCKETS - 1)
5fa271fa 112 t2 = ((t2 + 1) >> 1) - t1;
dfad239c 113 if (i) {
a5baf581 114 if (!(stack = stackp = (CONTEXT *)malloc(i * sizeof(CONTEXT))))
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115 return(-1);
116 } else
117 stack = stackp = NULL;
118
34849bd9 119 /*
dfad239c 120 * There are two arrays, one provided by the user (l1), and the
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121 * temporary one (l2). The data is sorted to the temporary stack,
122 * and then copied back. The speedup of using index to determine
123 * which stack the data is on and simply swapping stacks back and
124 * forth, thus avoiding the copy every iteration, turns out to not
125 * be any faster than the current implementation.
126 */
127 if (!(l2 = (u_char **)malloc(sizeof(u_char *) * nmemb)))
128 return(-1);
129
34849bd9 130 /*
dfad239c 131 * Tr references a table of sort weights; multiple entries may
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132 * map to the same weight; EOS char must have the lowest weight.
133 */
134 if (tab)
135 tr = tab;
136 else {
137 tr = ltab;
138 for (t1 = 0, t2 = endbyte; t1 < t2; ++t1)
139 tr[t1] = t1 + 1;
140 tr[t2] = 0;
e061e641 141 for (t1 = endbyte + 1; t1 < NBUCKETS; ++t1)
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142 tr[t1] = t1;
143 }
144
dfad239c 145 /* First sort is entire stack */
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146 bot = l1;
147 indx = 0;
148
149 for (;;) {
dfad239c 150 /* Clear bucket count array */
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151 bzero((char *)c, sizeof(c));
152
153 /*
dfad239c 154 * Compute number of items that sort to the same bucket
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155 * for this index.
156 */
33f8130f 157 for (p = bot, i = nmemb; --i >= 0;)
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158 ++c[tr[(*p++)[indx]]];
159
160 /*
dfad239c 161 * Sum the number of characters into c, dividing the temp
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162 * stack into the right number of buckets for this bucket,
163 * this index. C contains the cumulative total of keys
164 * before and included in this bucket, and will later be
e061e641 165 * used as an index to the bucket. c[NBUCKETS] contains
34849bd9 166 * the total number of elements, for determining how many
dfad239c 167 * elements the last bucket contains. At the same time
e061e641 168 * find the largest bucket so it gets pushed first.
34849bd9 169 */
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170 for (i = max = t1 = 0, t2 = __rspartition; i <= NBUCKETS; ++i) {
171 if (c[i] > t2) {
172 t2 = c[i];
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173 max = i;
174 }
e061e641 175 t1 = c[i] += t1;
dfad239c 176 }
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177
178 /*
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179 * Partition the elements into buckets; c decrements through
180 * the bucket, and ends up pointing to the first element of
181 * the bucket.
34849bd9 182 */
33f8130f 183 for (i = nmemb; --i >= 0;) {
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184 --p;
185 l2[--c[tr[(*p)[indx]]]] = *p;
186 }
187
dfad239c 188 /* Copy the partitioned elements back to user stack */
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189 bcopy(l2, bot, nmemb * sizeof(u_char *));
190
191 ++indx;
192 /*
dfad239c 193 * Sort buckets as necessary; don't sort c[0], it's the
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194 * EOS character bucket, and nothing can follow EOS.
195 */
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196 for (i = max; i; --i) {
197 if ((nmemb = c[i + 1] - (t1 = c[i])) < 2)
198 continue;
199 p = bot + t1;
200 if (nmemb > __rspartition)
201 STACKPUSH
202 else
33f8130f 203 shellsort(p, indx, nmemb, tr);
dfad239c 204 }
e061e641 205 for (i = max + 1; i < NBUCKETS; ++i) {
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206 if ((nmemb = c[i + 1] - (t1 = c[i])) < 2)
207 continue;
208 p = bot + t1;
209 if (nmemb > __rspartition)
210 STACKPUSH
211 else
33f8130f 212 shellsort(p, indx, nmemb, tr);
34849bd9 213 }
dfad239c 214 /* Break out when stack is empty */
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215 STACKPOP
216 }
217
218 free((char *)l2);
219 free((char *)stack);
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220 return(0);
221}
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222
223/*
224 * Shellsort (diminishing increment sort) from Data Structures and
225 * Algorithms, Aho, Hopcraft and Ullman, 1983 Edition, page 290;
226 * see also Knuth Vol. 3, page 84. The increments are selected from
227 * formula (8), page 95. Roughly O(N^3/2).
228 */
229static void
230shellsort(p, indx, nmemb, tr)
231 register u_char **p, *tr;
232 register int indx, nmemb;
233{
234 register u_char ch, *s1, *s2;
235 register int incr, *incrp, t1, t2;
236
237 for (incrp = __rsshell_increments; incr = *incrp++;)
238 for (t1 = incr; t1 < nmemb; ++t1)
239 for (t2 = t1 - incr; t2 >= 0;) {
240 s1 = p[t2] + indx;
241 s2 = p[t2 + incr] + indx;
242 while ((ch = tr[*s1++]) == tr[*s2] && ch)
243 ++s2;
244 if (ch > tr[*s2]) {
245 s1 = p[t2];
246 p[t2] = p[t2 + incr];
247 p[t2 + incr] = s1;
248 t2 -= incr;
249 } else
250 break;
251 }
252}