#include <stdio.h> #include <ctype.h> #ifndef unix #define SHIFT 5 #define TABSIZE (int)(400000/(1<<SHIFT)) int *tab; /*honeywell loader deficiency*/ #else #define Tolower(c) (isupper(c)?tolower(c):c) /* ugh!!! */ #define SHIFT 4 #define TABSIZE 25000 /*(int)(400000/(1<<shift))--pdp11 compiler deficiency*/ short tab[TABSIZE]; #endif long p[] = { 399871, 399887, 399899, 399911, 399913, 399937, 399941, 399953, 399979, 399983, 399989, }; #define NP (sizeof(p)/sizeof(p[0])) #define NW 30 /* * Hash table for spelling checker has n bits. * Each word w is hashed by k different (modular) hash functions, hi. * The bits hi(w), i=1..k, are set for words in the dictionary. * Assuming independence, the probability that no word of a d-word * dictionary sets a particular bit is given by the Poisson formula * P = exp(-y)*y**0/0!, where y=d*k/n. * The probability that a random string is recognized as a word is then * (1-P)**k. For given n and d this is minimum when y=log(2), P=1/2, * whence one finds, for example, that a 25000-word dictionary in a * 400000-bit table works best with k=11. */ long pow2[NP][NW]; prime(argc, argv) register char **argv; { int i, j; long h; register long *lp; #ifndef unix if ((tab = (int *)calloc(sizeof(*tab), TABSIZE)) == NULL) return(0); #endif if (argc > 1) { FILE *f; if ((f = fopen(argv[1], "ri")) == NULL) return(0); if (fread((char *)tab, sizeof(*tab), TABSIZE, f) != TABSIZE) return(0); fclose(f); } for (i=0; i<NP; i++) { h = *(lp = pow2[i]) = 1<<14; for (j=1; j<NW; j++) h = *++lp = (h<<7) % p[i]; } return(1); } #define get(h) (tab[h>>SHIFT]&(1<<((int)h&((1<<SHIFT)-1)))) #define set(h) tab[h>>SHIFT] |= 1<<((int)h&((1<<SHIFT)-1))