_A Hash Function for Hash Table Lookup_ by Bob Jenkins Example 1: initialize(state) for (each block) { combine(state, block); mix(state); } return postprocess(state); Example 2: Typical hash function for (hash=0, i=0; i>27))^key[i]; return (hash % prime); Listing One typedef unsigned long int ub4; /* unsigned 4-byte quantities */ typedef unsigned char ub1; /* unsigned 1-byte quantities */ #define hashsize(n) ((ub4)1<<(n)) #define hashmask(n) (hashsize(n)-1) /* mix -- mix 3 32-bit values reversibly. For every delta with one or two bits set, and the deltas of all three high bits or all three low bits, whether the original value of a,b,c is almost all zero or is uniformly distributed, * If mix() is run forward or backward, at least 32 bits in a,b,c have at least 1/4 probability of changing. * If mix() is run forward, every bit of c will change between 1/3 and 2/3 of the time. (Well, 22/100 and 78/100 for some 2-bit deltas.) mix() takes 36 machine instructions, but only 18 cycles on a superscalar machine (like a Pentium or a Sparc). No faster mixer seems to work, that's the result of my brute-force search. There were about 2^68 hashes to choose from. I only tested about a billion of those. */ #define mix(a,b,c) \ { \ a -= b; a -= c; a ^= (c>>13); \ b -= c; b -= a; b ^= (a<<8); \ c -= a; c -= b; c ^= (b>>13); \ a -= b; a -= c; a ^= (c>>12); \ b -= c; b -= a; b ^= (a<<16); \ c -= a; c -= b; c ^= (b>>5); \ a -= b; a -= c; a ^= (c>>3); \ b -= c; b -= a; b ^= (a<<10); \ c -= a; c -= b; c ^= (b>>15); \ } /* hash() -- hash a variable-length key into a 32-bit value k : the key (the unaligned variable-length array of bytes) len : the length of the key, counting by bytes initval : can be any 4-byte value Returns a 32-bit value. Every bit of the key affects every bit of the return value. Every 1-bit and 2-bit delta achieves avalanche. About 6*len+35 instructions. The best hash table sizes are powers of 2. There is no need to do mod a prime (mod is sooo slow!). If you need less than 32 bits, use a bitmask. For example, if you need only 10 bits, do h = (h & hashmask(10)); In which case, the hash table should have hashsize(10) elements. If you are hashing n strings (ub1 **)k, do it like this: for (i=0, h=0; i= 12) { a += (k[0] +((ub4)k[1]<<8) +((ub4)k[2]<<16) +((ub4)k[3]<<24)); b += (k[4] +((ub4)k[5]<<8) +((ub4)k[6]<<16) +((ub4)k[7]<<24)); c += (k[8] +((ub4)k[9]<<8) +((ub4)k[10]<<16)+((ub4)k[11]<<24)); mix(a,b,c); k += 12; len -= 12; } /*------------------------------------- handle the last 11 bytes */ c += length; switch(len) /* all the case statements fall through */ { case 11: c+=((ub4)k[10]<<24); case 10: c+=((ub4)k[9]<<16); case 9 : c+=((ub4)k[8]<<8); /* the first byte of c is reserved for the length */ case 8 : b+=((ub4)k[7]<<24); case 7 : b+=((ub4)k[6]<<16); case 6 : b+=((ub4)k[5]<<8); case 5 : b+=k[4]; case 4 : a+=((ub4)k[3]<<24); case 3 : a+=((ub4)k[2]<<16); case 2 : a+=((ub4)k[1]<<8); case 1 : a+=k[0]; /* case 0: nothing left to add */ } mix(a,b,c); /*-------------------------------------------- report the result */ return c; } Listing Two /* Additive Hash */ int additive(char *key, int len, int prime) { int hash, i; for (hash=len, i=0; i>27)^key[i]; return (hash % prime); } /* Pearson's Hash */ char pearson(char *key, int len, char tab[256]) { char hash; int i; for (hash=len, i=0; i>24)^key[i]]; return (hash & mask); } /* Universal Hash */ int universal(char *key, int len, int mask, int tab[MAXBITS]) { int hash, i; for (hash=len, i=0; i<(length<<3); i+=8) { register char k = key[i>>3]; if (k&0x01) hash ^= tab[i+0]; if (k&0x02) hash ^= tab[i+1]; if (k&0x04) hash ^= tab[i+2]; if (k&0x08) hash ^= tab[i+3]; if (k&0x10) hash ^= tab[i+4]; if (k&0x20) hash ^= tab[i+5]; if (k&0x40) hash ^= tab[i+6]; if (k&0x80) hash ^= tab[i+7]; } return (hash & mask); } /* Zobrist Hash */ int zobrist( char *key, int len, int mask, int tab[MAXBYTES][256]) { int hash, i; for (hash=len, i=0; i