/*- * Copyright (c) 1990 The Regents of the University of California. * All rights reserved. * * This code is derived from software contributed to Berkeley by * Dan Bernstein at New York University. * * %sccs.include.redist.c% */ #if defined(LIBC_SCCS) && !defined(lint) static char sccsid[] = "@(#)radixsort.c 5.10 (Berkeley) 09/26/91"; #endif /* LIBC_SCCS and not lint */ #include #include #include #include #include /* * __rspartition is the cutoff point for a further partitioning instead * of a shellsort. If it changes check __rsshell_increments. Both of * these are exported, as the best values are data dependent. */ #define NPARTITION 40 int __rspartition = NPARTITION; int __rsshell_increments[] = { 4, 1, 0, 0, 0, 0, 0, 0 }; /* * Stackp points to context structures, where each structure schedules a * partitioning. Radixsort exits when the stack is empty. * * If the buckets are placed on the stack randomly, the worst case is when * all the buckets but one contain (npartitions + 1) elements and the bucket * pushed on the stack last contains the rest of the elements. In this case, * stack growth is bounded by: * * limit = (nelements / (npartitions + 1)) - 1; * * This is a very large number, 52,377,648 for the maximum 32-bit signed int. * * By forcing the largest bucket to be pushed on the stack first, the worst * case is when all but two buckets each contain (npartitions + 1) elements, * with the remaining elements split equally between the first and last * buckets pushed on the stack. In this case, stack growth is bounded when: * * for (partition_cnt = 0; nelements > npartitions; ++partition_cnt) * nelements = * (nelements - (npartitions + 1) * (nbuckets - 2)) / 2; * The bound is: * * limit = partition_cnt * (nbuckets - 1); * * This is a much smaller number, 4590 for the maximum 32-bit signed int. */ #define NBUCKETS (UCHAR_MAX + 1) typedef struct _stack { const u_char **bot; int indx, nmemb; } CONTEXT; #define STACKPUSH { \ stackp->bot = p; \ stackp->nmemb = nmemb; \ stackp->indx = indx; \ ++stackp; \ } #define STACKPOP { \ if (stackp == stack) \ break; \ --stackp; \ bot = stackp->bot; \ nmemb = stackp->nmemb; \ indx = stackp->indx; \ } /* * A variant of MSD radix sorting; see Knuth Vol. 3, page 177, and 5.2.5, * Ex. 10 and 12. Also, "Three Partition Refinement Algorithms, Paige * and Tarjan, SIAM J. Comput. Vol. 16, No. 6, December 1987. * * This uses a simple sort as soon as a bucket crosses a cutoff point, * rather than sorting the entire list after partitioning is finished. * This should be an advantage. * * This is pure MSD instead of LSD of some number of MSD, switching to * the simple sort as soon as possible. Takes linear time relative to * the number of bytes in the strings. */ int radixsort(l1, nmemb, tab, endbyte) const u_char **l1; register int nmemb; const u_char *tab; u_int endbyte; { register int i, indx, t1, t2; register const u_char **l2; register const u_char **p; register const u_char **bot; register const u_char *tr; CONTEXT *stack, *stackp; int c[NBUCKETS + 1], max; u_char ltab[NBUCKETS]; static void shellsort(); if (nmemb <= 1) return(0); /* * T1 is the constant part of the equation, the number of elements * on the stack other than the two largest, worst-case buckets. */ t1 = (__rspartition + 1) * (NBUCKETS - 2); for (i = 0, t2 = nmemb; t2 > __rspartition; i += NBUCKETS - 1) t2 = (t2 - t1) >> 1; if (i) { if (!(stack = stackp = (CONTEXT *)malloc(i * sizeof(CONTEXT)))) return(-1); } else stack = stackp = NULL; /* * There are two arrays, one provided by the user (l1), and the * temporary one (l2). The data is sorted to the temporary stack, * and then copied back. The speedup of using index to determine * which stack the data is on and simply swapping stacks back and * forth, thus avoiding the copy every iteration, turns out to not * be any faster than the current implementation. */ if (!(l2 = (const u_char **)malloc(sizeof(u_char *) * nmemb))) return(-1); /* * Tr references a table of sort weights; multiple entries may * map to the same weight; EOS char must have the lowest weight. */ if (tab) tr = tab; else { for (t1 = 0, t2 = endbyte; t1 < t2; ++t1) ltab[t1] = t1 + 1; ltab[t2] = 0; for (t1 = endbyte + 1; t1 < NBUCKETS; ++t1) ltab[t1] = t1; tr = ltab; } /* First sort is entire stack */ bot = l1; indx = 0; for (;;) { /* Clear bucket count array */ bzero((char *)c, sizeof(c)); /* * Compute number of items that sort to the same bucket * for this index. */ for (p = bot, i = nmemb; --i >= 0;) ++c[tr[(*p++)[indx]]]; /* * Sum the number of characters into c, dividing the temp * stack into the right number of buckets for this bucket, * this index. C contains the cumulative total of keys * before and included in this bucket, and will later be * used as an index to the bucket. c[NBUCKETS] contains * the total number of elements, for determining how many * elements the last bucket contains. At the same time * find the largest bucket so it gets pushed first. */ for (i = max = t1 = 0, t2 = __rspartition; i <= NBUCKETS; ++i) { if (c[i] > t2) { t2 = c[i]; max = i; } t1 = c[i] += t1; } /* * Partition the elements into buckets; c decrements through * the bucket, and ends up pointing to the first element of * the bucket. */ for (i = nmemb; --i >= 0;) { --p; l2[--c[tr[(*p)[indx]]]] = *p; } /* Copy the partitioned elements back to user stack */ bcopy(l2, bot, nmemb * sizeof(u_char *)); ++indx; /* * Sort buckets as necessary; don't sort c[0], it's the * EOS character bucket, and nothing can follow EOS. */ for (i = max; i; --i) { if ((nmemb = c[i + 1] - (t1 = c[i])) < 2) continue; p = bot + t1; if (nmemb > __rspartition) STACKPUSH else shellsort(p, indx, nmemb, tr); } for (i = max + 1; i < NBUCKETS; ++i) { if ((nmemb = c[i + 1] - (t1 = c[i])) < 2) continue; p = bot + t1; if (nmemb > __rspartition) STACKPUSH else shellsort(p, indx, nmemb, tr); } /* Break out when stack is empty */ STACKPOP } free((char *)l2); free((char *)stack); return(0); } /* * Shellsort (diminishing increment sort) from Data Structures and * Algorithms, Aho, Hopcraft and Ullman, 1983 Edition, page 290; * see also Knuth Vol. 3, page 84. The increments are selected from * formula (8), page 95. Roughly O(N^3/2). */ static void shellsort(p, indx, nmemb, tr) register u_char **p, *tr; register int indx, nmemb; { register u_char ch, *s1, *s2; register int incr, *incrp, t1, t2; for (incrp = __rsshell_increments; incr = *incrp++;) for (t1 = incr; t1 < nmemb; ++t1) for (t2 = t1 - incr; t2 >= 0;) { s1 = p[t2] + indx; s2 = p[t2 + incr] + indx; while ((ch = tr[*s1++]) == tr[*s2] && ch) ++s2; if (ch > tr[*s2]) { s1 = p[t2]; p[t2] = p[t2 + incr]; p[t2 + incr] = s1; t2 -= incr; } else break; } }