# Hashing Algorithm: Deleting an element in linear probing

While using Linear probing method to implement hashing, when we delete and element, the position of the deleted element is declared as a tombstone/ mark it as deleted. Why can't we just shift all the elements from the current position until the next empty element is encountered? Won't this save the time when too many tombstones are encountered later and a rehashing might be required? Am I missing something? Please tell me if I need to clarify my question.

Thanks a lot!

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Best explained by counter example:

``````HASH(A) = 10
HASH(B) = 10
HASH(C) = 12

myHash.insert(A); // A goes to slot 10

10   11   12
+----+----+----+
| A  |    |    |
+----+----+----+

myHash.insert(B); // B goes to slot 10, but that is taken, so it sees 11 is empty and inserts there

10   11   12
+----+----+----+
| A  | B  |    |
+----+----+----+

myHash.insert(C); // C hashes to slot 12 and gets put there

10   11   12
+----+----+----+
| A  | B  | C  |
+----+----+----+

myHash.remove(B); // We remove B, and shift the ones after it back

10   11   12
+----+----+----+
| A  | C  |    |
+----+----+----+

myHash.contains(C) == false; // C hashes to slot 12, but there is nothing there.  So therefore the hash does not contain C (but it does CONTRADICTION).
``````

Therefore, by proof of contradiction, simply sliding the succeeding elements upon removal will result in incorrectness.

EDIT: C should have said that it hashes to 12

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Yes, got it. Thanks! – P Ramesh Aug 23 '13 at 18:05
@BobFincheirmer You example doesn't explain what happens if we remove A and then check hash.contains(B). – duros Aug 23 '13 at 19:17
@duros It would work in that situation, but the point of the counter-example is to show it does not work in all situations – Bob Fincheimer Aug 26 '13 at 15:16