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1 / / A poor hash function when tablesize is large 2 unsigned int hash( const string & key, int tablesize 3 { 4 unsigned int hashVal = 0;
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for( int i = 0; i < keylength( hashVal += keycharAt( i ) ; return hashVal % tablesize;
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A bad hash function if tableSize is large
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does not distribute the keys well For instance, suppose that tablesize is 10,000Also suppose that all keys are 8 or fewer characters long Because an ASCII c h a r is an integer between 0 and 127, the hash function can assume values only between 0 and 1016 (127 x 8) This restriction certainly does not permit an equitable distribution Any speed gained by the quickness of the hash function calculation is more than offset by the effort taken to resolve a larger than expected number of collisions However, a reasonable alternative is described in Exercise 2015 Finally, note that 0 is a possible result of the hash function, so hash tables are indexed starting at 0
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The table runsfrom 0 to tablesize-1
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203 Linear Probing
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Now that we have a hash function, we need to decide what to do when a collision occurs Specifically, if X hashes out to a position that is already occupied, where do we place it The simplest possible strategy is linear probing, or searching sequentially in the array until we find an empty cell The search wraps around from the last position to the first, if necessary Figure 204 shows the result of inserting the keys 89, 18, 49, 58, and 9 in a hash table when linear probing is used We assume a hash function that returns the key X mod the size of the table Figure 204 includes the result of the hash function The first collision occurs when 49 is inserted; the 49 is put in the next available spot-namely, spot 0, which is open Then 58 collides with 18, 89, and 49 before an empty spot is found three slots away in position 1 The collision for element 9 i s resolved similarly So long as the table is large enough, a free cell can always be found However, the time needed to find a free cell can get to be quite long For example, if there is only one free cell left in the table, we may have to search the entire table to find it On average we would expect to have to search half the table to find it, which is far from
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in linearprobing, are resolved by sequentially scanning an array (with wraparound) until an empty cell is found
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After insert 89 After insert 18 After insert 49 After insert 58 After insert 9
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The find algorithm follows the same probe sequence as the insert algorithm
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the constant time per access that we are hoping for But, if the table is kept relatively empty, insertions should not be so costly We discuss this approach shortly The f i n d algorithm merely follows the same path as the i n s e r t algorithm If it reaches an empty slot, the item we are searching for is not found; otherwise, it finds the match eventually For example to find 58, we start at slot 8 (as indicated by the hash function) We see an item, but it is the wrong one, so we try slot 9 Again, we have an item, but it is the wrong one, so we try slot 0 and then slot 1 until we find a match A f i n d for 19 would involve trying slots 9, 0, 1, and 2 before finding the empty cell in slot 3 Thus 19 is not found Standard deletion cannot be performed because, as with a binary search tree, an item in the hash table not only represents itself, but it also connects other items by serving as a placeholder during collision resolution Thus, if we removed 89 from the hash table, virtually all the remaining f i n d operations would fail Consequently, we implement lazy deletion, or marking items as deleted rather than physically removing them from the table This information is recorded in an extra data member Each item is either active or deleted
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