9.10. Open Addressing, Deletion¶
9.10.1. Deletion in an open addressing hash table¶
When deleting records from a hash table, there are two important considerations.
Deleting a record must not hinder later searches. In other words, the search process must still pass through the newly emptied slot to reach records whose probe sequence passed through this slot. Thus, the delete process cannot simply mark the slot as empty, because this will isolate records further down the probe sequence.
We do not want to make positions in the hash table unusable because of deletion. The freed slot should be available to a future insertion.
Both of these problems can be resolved by placing a special mark in place of the deleted record, called a tombstone. The tombstone indicates that a record once occupied the slot but does so no longer. If a tombstone is encountered when searching along a probe sequence, the search procedure continues with the search. When a tombstone is encountered during insertion, that slot can be used to store the new record. However, to avoid inserting duplicate keys, it will still be necessary for the search procedure to follow the probe sequence until a truly empty position has been found, simply to verify that a duplicate is not in the table. However, the new record would actually be inserted into the slot of the first tombstone encountered.
Here is a practice exercise.
The use of tombstones allows searches to work correctly and allows reuse of deleted slots. However, after a series of intermixed insertion and deletion operations, some slots will contain tombstones. This will tend to lengthen the average distance from a record’s home position to the record itself, beyond where it could be if the tombstones did not exist. A typical database application will first load a collection of records into the hash table and then progress to a phase of intermixed insertions and deletions. After the table is loaded with the initial collection of records, the first few deletions will lengthen the average probe sequence distance for records (it will add tombstones). Over time, the average distance will reach an equilibrium point because insertions will tend to decrease the average distance by filling in tombstone slots. For example, after initially loading records into the database, the average path distance might be 1.2 (i.e., an average of 0.2 accesses per search beyond the home position will be required). After a series of insertions and deletions, this average distance might increase to 1.6 due to tombstones. This seems like a small increase, but it is three times longer on average beyond the home position than before deletions.
Two possible solutions to this problem are
Do a local reorganization upon deletion to try to shorten the average path length. For example, after deleting a key, continue to follow the probe sequence of that key and swap records further down the probe sequence into the slot of the recently deleted record (being careful not to remove any key from its probe sequence). This will not work for all collision resolution policies.
Periodically rehash the table by reinserting all records into a new hash table. Not only will this remove the tombstones, but it also provides an opportunity to place the most frequently accessed records into their home positions.
Note that since we are using a dynamic array when implementing hash tables, this can be viewed as a version of the second solution above (because all tombstones will be removed when the internal array is resized).
9.10.2. Simple implementation of deletion¶
Here is a simple implementation of deletion in a HashMap using tombstones.
public V remove(K key) {
int i = hashAndProbe(key);
KVPair<K,V> elem = internalTable[i];
if (elem == null)
return null;
V removed = elem.value;
elem.key = null;
elem.value = null;
mapSize--;
numDeleted++;
if (mapSize < MinLoadFactor * internalTable.length)
resizeTable((int) (internalTable.length / CapacityMultiplier));
return removed;
}
def remove(self, key):
i = self._hashAndProbe(key)
elem = self._internalTable[i]
if elem is None:
return None
removed = elem.value
elem.key = None
elem.value = None
self._mapSize -= 1
self._numDeleted += 1
if self._mapSize < self._minLoadFactor * len(self._internalTable):
self._resizeTable(len(self._internalTable) / self._capacityMultiplier);
return removed
Since we are using an internal array of KVPair, there are actually two possible empty entries, and we use this to encode the tombstones:
If the table cell is empty (
null
), then it is unoccupied.If the cell contains a KVPair, where the key is
null
, then it is a tombstone.
So, when we remove an entry, we do not remove the KVPair, but instead
set the key (and the value) to null
. This will make the cell a tombstone.
The current code has one problem:
Adding new entries will never make use of the tombstones, but will only insert into
completely empty cells.
It is possible to fix this by implementing a sligthly different version of hashAndProbe
,
which will only be used by the put
method.
This is left as an exercise to the reader.
9.10.3. Two load factors¶
When we have tombstones in our table, there are two possible ways of thinking about the load factor – depending on if we want to include the deleted cells or not. And both variants are useful!
When adding elements, we need to know if there are too few completely empty slots left, giving the load factor \(N + D / M\) (where \(N\) is the number of occupied cells and \(D\) the number of tombstones).
When deleting elements, we need to know if there are too few occupied slots, giving the load factor \(N / M\).
9.10.4. Hashing Deletion Summary Questions¶
Now here are some practice questions.
Congratulations! You have reached the end of the hashing tutorial. In summary, a properly tuned hashing system will return records with an average cost of less than two record accesses. This makes it the most effective way known to store a database of records to support exact-match queries. Unfortunately, hashing is not effective when implementing range queries, or answering questions like “Which record in the collection has the smallest key value?”