*Dr. Carlo Pescio is a freelance consultant in Savona, Italy. He specializes in object technologies and is a member of the IEEE Computer Society and the ACM. You can contact him at pescio@ mbox.vol.it.*

Hashing is one of the oldest, simplest, and most powerful of all computer-science techniques. That doesn't keep it from being subtle, though. The choice of a particular hash function, or the key that you use for a hash, can have a phenomenal impact on performance. In his article "Hashing Rehashed," Andrew Binstock (DDJ, April 1996) pointed out the impacts of different kinds of collision resolution. This month, we'll see that you can sometimes completely eliminate collisions.

-Tim Kientzle

Hashing is a powerful and simple technique for accessing information: Each key is mathematically converted into a number, which is then used as an index into a table. In typical applications, it's common for two different keys to map to the same index, requiring some strategy for collision resolution. However, many applications use a static set of records; for instance, the table of keywords for a compiler, the contents of a dictionary, or the topics in a help file. In these cases, it's nice to have a perfect hashing function, one that maps different keys into different numbers. With a perfect function, you could retrieve the record (or detect that no record exists) in exactly one probe, without worrying about collisions. A special case of perfect hashing is minimal perfect hashing, where no empty locations in the table exist. Minimal perfect hashing satisfies the ideal of fast performance and low memory usage. However, finding perfect hashing functions is difficult. In this article, I'll present a simple algorithm to find perfect (and optionally minimal) hashing functions for a static set of data.

A family of perfect hashing functions was introduced by R. Chichelli in "** Minimal Perfect Hashing Made Simple**" (

Chichelli's hashing functions are very simple. Given word w, the hash value is Value(First_Letter(w)) + Value(Last_Letter(w)) + Length(w). The mapping function Value() is normally implemented as a table that maps characters into numbers. Figure 1 is an example of table and hash values of a minimal perfect-hashing function for the days of the week. Note that you only need to keep a small table for Value(), since only first and last letters need to be mapped. Calculating the hash value requires only two table accesses and two additions.

The original algorithm uses an exhaustive approach with backtracking. Every possible mapping is tried for Value(), and whenever an assignment results in a collision, the search path is pruned and another solution is tried. The algorithm is easy to code but can take exponential time (in the number of words) to calculate.

Chichelli suggested two ways to speed up this approach. One way is to sort the keys: Compute the frequency of each first and last letter, then sort the words according to the sum of the frequency for the first and last letters. This guarantees that the most common letters are assigned values early, and increases the probability that collisions will be found when the depth of the search tree is still small. This reduces the cost of backtracking.

This isn't always sufficient, however. Consider the set of keys:

{aa,ab,ac,ad,ae,af,ag,bd,cd,ce,de,fg},

which sorts to

{aa,ad,ac,ae,ab,af,ag,cd,de,bd,ce,fg}

Now, the value of the key cd is determined as soon as ac is given a value, but a test for collision on cd is not done until five more branches in the search tree are taken. This makes a collision on cd much more costly than it ought to be.

Chichelli suggested a second heuristic to address this problem. If a key in this sorted list has letters that have already occurred, the key is placed immediately after the one that completes its mapping. In the aforementioned example, cd will be moved immediately after ac. While doing so, you must maintain the original sorted order whenever possible. Listing One is a snippet of my C++ implementation.

There are five steps to Chichelli's algorithm:

1.Calculate the frequency of first and last letters.

2.Order the keys in decreasing order of the sum of the frequencies.

3.Reorder the keys from the beginning of the list, so that if a key's first and last letters have already appeared, that key is placed next to the completing key in the list.

4.Search cycle: For each key, if both values for first and last letters are already assigned, try to place the key in the table (step 5); otherwise, if only one value needs to be assigned, try all possible assignments. If both values must be assigned, use two nested loops to try all the possible assignments for the two letters.

5.Given the key, the length, and the Value mapping, compute the address in the table. Backtrack to step 4 if there is a collision; otherwise place the key in the table. If it is the last letter, a perfect hashing has been found; otherwise, recursively go into step 4 for the next key.

The table size is not fixed. If the table size is equal to the number of keys, you are looking for minimal perfect hashing; otherwise, some empty buckets remain at the end of the process.

Chichelli did not suggest any way to restrict the range of values to try for each letter. This range plays an important role in the algorithm. A small range may result in no solutions, and an overly large range may increase the width of the search tree unmanageably.

I've developed a general rule that works fine for most sets, resulting in a moderately small range that does not exclude solutions. Since you add the length of the word and you want to obtain small index values (including zero), most of the letters will be assigned to negative values. Also, the letter values will never be very large because the size of the table is an upper bound for the index.

I chose a range of [-MaxLen,TableSize-MinLen), where MaxLen is the maximum length of a key belonging to the set; MinLen, the minimum length of a key; and TableSize , the size of the hash table. The range is normally small (for the keys in Figure 1, assuming minimal hashing, it is equal to [-9,1)), but still effective.

This can be further improved by adapting the range dynamically. Consider what happens when one of the two letters has already been assigned a value. This occurs twice in the algorithm: when only one value must be assigned and in the inner of the two nested loops. In this case, you already know one of the values (say, v) and the length of the word, so you can recalculate the range as [-v - Len,TableSize -v - Len], where Len is the length of the key to which you are trying to assign an index. This optimization alone produced more than an order-of-magnitude improvement in the performance of the algorithm. I've frequently seen a speed up of 40 times or more.

Note that this dynamic range may occasionally extend the original, static one. For instance, in Figure 1, y is assigned a value of 3, outside the static range. Further optimization of the dynamic range can be obtained by considering which buckets are free. If you keep track of the first and last free buckets in the hash table, the dynamic range can be restricted to [FirstFree -v - Len, LastFree - v - Len]. Listing Two is my C++ implementation of steps 4 and 5. I also added some logic to stop when the first solution is found.

The result is a fairly quick algorithm. A minimal perfect hash for the 32 keywords of ANSI C (see Figure 2) can be found in less than one second on a Pentium/90, and the 46 noncolliding keywords of C++ need about four seconds. The complete C++ implementation of my algorithm is available electronically . It has been compiled and tested on several different architectures and operating systems, and it should work without changes on most systems.

Some words, such as the C++ keywords double and delete, result in a collision no matter which assignment is selected. There are several ways to overcome this problem: The first is to enhance the algorithm to consider letter positions other than the first and last. Another is to consider more than two letters. For instance, in "Finding Minimal Perfect Hash Functions," Haggard and Karplus (ACM SICCSE Bulletin, February 1986) tried using all letters of the word, with encouraging results.

Another idea is to partition the set of words into two or more sets. This is also effective when there are many words and the algorithm takes hours (or days) to calculate a perfect hash. The best idea would be to partition the keys into disjoint sets of first and last letters. This allows you to use a single hash table simply by adding an offset to each partition. Unfortunately, the distribution of letters usually does not allow such a partition, and you must often resort to multiple tables and, consequently, to multiple probes.

I'm currently investigating two promising ideas, one based on the concept of an articulation point in a graph, which should reduce backtracking, and another based on a look-ahead technique to prune dead paths more quickly. Both techniques should reduce the computational complexity of the algorithm in the average case, so that larger sets of keys can be conveniently mapped. If you are interested in the results of my research, drop me an e-mail message.

Letter Value f -8 m -4 s -9 t 7 w -6 y 3 Word Hash sunday 0 monday 5 tuesday 3 wednesday 6 thursday 4 friday 1 saturday 2

Letter Value Letter Value a -5 l 29 b -8 m -4 c -7 n -1 d -10 o 22 e 4 r 5 f 12 s 7 g -7 t 10 h 4 u 26 i 2 v 19 k 31 w 2 Word Hash Word Hash auto 21 int 15 break 28 long 26 case 1 register 18 char 2 return 10 const 8 short 22 continue 5 signed 3 default 7 sizeof 25 do 14 static 6 double 0 struct 23 else 12 switch 17 enum 4 typedef 29 extern 9 union 30 float 27 unsigned 24 for 20 void 13 goto 19 volatile 31 if 16 while 11

void WordSet :: Reorder() { // reorder items in the set, so that if an item first and last letter // occurred before, the item is moved upward in the list. This heuristic // maximizes the likelihood that collisions are found early in the // assignment of values. const unsigned NO_OCCURRENCE = MAXINT ; for( unsigned i = 1; i < card; i++ ) { const Word& w = *(set[ i ]) ; char first = w.First() ; char last = w.Last() ; unsigned firstOccurredIndex = NO_OCCURRENCE ; unsigned lastOccurredIndex = NO_OCCURRENCE ; for( unsigned j = 0; j < i; j++ ) { const Word& prev = *(set[ j ]) ; if( lastOccurredIndex == NO_OCCURRENCE && ( prev.Last() == last || prev.First() == last ) ) lastOccurredIndex = j ; if( firstOccurredIndex == NO_OCCURRENCE && ( prev.First() == first || prev.Last() == first ) ) firstOccurredIndex = j ; if( firstOccurredIndex != NO_OCCURRENCE && lastOccurredIndex != NO_OCCURRENCE ) { // both letters occurred before: move the item upward, by pushing down // the ones between the maximum occurence index and the item index // itself. This preserve the original ordering as much as possible unsigned firstToMoveDown = max( firstOccurredIndex, lastOccurredIndex ) + 1 ; unsigned lastToMoveDown = i - 1 ; Word* itemToMoveUp = set[ i ] ; for( unsigned k = lastToMoveDown; k >= firstToMoveDown; k-- ) set[ k + 1 ] = set[ k ] ; set[ firstToMoveDown ] = itemToMoveUp ; break ; // item 'i' handled } } } }

bool HashFinder :: Search( unsigned index ) { const Word& w = words[ index ] ; if( value[ w.First() ] != NO_VALUE && value[ w.Last() ] != NO_VALUE ) { // try to insert into the table int h = Hash( w ) ; if( h < 0 || h >= maxTableSize || ! buckets.IsFree( h ) ) return( FALSE ) ; else { buckets.MarkUsed( h ) ; index++ ; int card = words.Card() ; if( index == card ) { cout << "------------\n" ; cout << "perfect hashing found\n" ; for( int i = 0; i < CHAR_MAX; i++ ) if( value[ i ] != NO_VALUE ) cout << (char)i << " = " << value[ i ] << " " ; cout << endl ; for( i = 0; i < card; i++ ) cout << words[ i ] << " = " << Hash( words[ i ] ) << endl ; // calculate and print loading factor int used = 0 ; for( i = 0; i < maxTableSize; i++ ) if( ! buckets.IsFree( i ) ) used++ ; cout << "loading factor: " << (float)used/maxTableSize << endl; alues for the one without values // must find valid vtime found = found || res ; int valToAssign ; int valAssigned ; if( value[ w.First() ] == NO_VALUE ) { valToAssign = w.First() ; valAssigned = w.Last() ; } else { valToAssign = w.Last() ; valAssigned = w.First() ; } // see above for heuristics used here int minVal2 = buckets.FirstFree() - value[ valAssigned ] - w.Len(); int maxVal2 = buckets.LastFree() -value[valAssigned] - w.Len() + 1; assert( value[ valToAssign ] == NO_VALUE ) ; bool found = FALSE ; for( int i = minVal2; i < maxVal2 && ! stopNow; i++ ) { value[ valToAssign ] = i ; bool res = Search( index ) ; // will take first branch this // decide if it is time to stop now stopNow = stopFirst || ( used == maxTableSize && stopMinimal ); buckets.MarkFree( h ) ; // restore for backtracking return( TRUE ) ; } else { bool res = Search( index ) ; buckets.MarkFree( h ) ; // restore for backtracking return( res ) ; } } } else if( value[ w.First() ] == NO_VALUE && value[ w.Last() ] == NO_VALUE ) { // must find values for both first and last; int firstValToAssign = w.First() ; int lastValToAssign = w.Last() ; bool found = FALSE ; for( int i = minVal; i < maxVal && ! stopNow; i++ ) { value[ firstValToAssign ] = i ; // here we can change the heuristics: we already assigned one of // the values, so we have modified the boundaries. // Now we must have FirstFree <= to-assign + 'i' + len <= LastFree int minVal2 = buckets.FirstFree() - i - w.Len() ; int maxVal2 = buckets.LastFree() - i - w.Len() + 1 ; for( int j = minVal2; j < maxVal2 && ! stopNow; j++ ) { value[ lastValToAssign ] = j ; bool res = Search( index ) ; found = found || res ; } } value[ firstValToAssign ] = NO_VALUE ; // restore for backtracking value[ lastValToAssign ] = NO_VALUE ; return( found ) ; } else { } value[ valToAssign ] = NO_VALUE ; // restore for backtracking return( found ) ; } }