/* Brute-force based perfect hash generator for small sets of integers. Just * fill the table below with the integer values, try to pad a little bit to * avoid too complicated divides, experiment with a few operations in the * hash function and reuse the output as-is to make your table. You may also * want to experiment with the random generator to use either one or two * distinct values for mul and key. */ #include #include /* warning no more than 32 distinct values! */ //#define CODES 21 //#define CODES 20 //#define CODES 19 //const int codes[CODES] = { 200,400,401,403,404,405,407,408,410,413,421,422,425,429,500,501,502,503,504}; #define CODES 32 const int codes[CODES] = { 200,400,401,403,404,405,407,408,410,413,421,422,425,429,500,501,502,503,504, /* padding entries below, which will fall back to the default code */ -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1}; unsigned mul, xor; unsigned bmul = 0, bxor = 0; static unsigned rnd32seed = 0x11111111U; static unsigned rnd32() { rnd32seed ^= rnd32seed << 13; rnd32seed ^= rnd32seed >> 17; rnd32seed ^= rnd32seed << 5; return rnd32seed; } /* the hash function to use in the target code. Try various combinations of * multiplies and xor, always folded with a modulo, and try to spot the * simplest operations if possible. Sometimes it may be worth adding a few * dummy codes to get a better modulo code. In this case, just add dummy * values at the end, but always distinct from the original ones. If the * number of codes is even, it might be needed to rotate left the result * before the modulo to compensate for lost LSBs. */ unsigned hash(unsigned i) { //return ((i * mul) - (i ^ xor)) % CODES; // more solutions //return ((i * mul) + (i ^ xor)) % CODES; // alternate //return ((i ^ xor) * mul) % CODES; // less solutions but still OK for sequences up to 19 long //return ((i * mul) ^ xor) % CODES; // less solutions but still OK for sequences up to 19 long i = i * mul; i >>= 5; //i = i ^ xor; //i = (i << 30) | (i >> 2); // rotate 2 right //i = (i << 2) | (i >> 30); // rotate 2 left //i |= i >> 20; //i += i >> 30; //i |= i >> 16; return i % CODES; //return ((i * mul) ^ xor) % CODES; // less solutions but still OK for sequences up to 19 long } int main(int argc, char **argv) { unsigned h, i, flag, best, tests; if (argc > 2) { mul = atol(argv[1]); xor = atol(argv[2]); for (i = 0; i < CODES && codes[i] >= 0; i++) printf("hash(%4u) = %4u // [%4u] = %4u\n", codes[i], hash(codes[i]), hash(codes[i]), codes[i]); return 0; } tests = 0; best = 0; while (/*best < CODES &&*/ ++tests) { mul = rnd32(); xor = mul; // works for some sequences up to 21 long //xor = rnd32(); // more solutions flag = 0; for (i = 0; i < CODES && codes[i] >= 0; i++) { h = hash(codes[i]); if (flag & (1 << h)) break; flag |= 1 << h; } if (i > best || (i == best && mul <= bmul && xor <= bxor)) { /* find the best code and try to find the smallest * parameters among the best ones (need to disable * best= 0; i++) printf("hash(%4u) = %2u // [%2u] = %4u\n", codes[i], hash(codes[i]), hash(codes[i]), codes[i]); }