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38. When implementing a dictionary ('I want to look up customer data by their customer IDs'), the typical data structures used are hash tables and binary search trees. I know for instance that the C++ STL library implements dictionaries (they call them maps) using (balanced) binary search trees, and the .NET framework uses hash tables under the ...
Many hash table implementations found in programming languages (such as Java's HashMap or Python's dict) dynamically increase the size of the hash table once the number of items reaches a certain threshold. In these cases, resizing the hash table means allocating a new, larger memory area, re-hashing all the items and moving them from the old ...
So it doesn't matter if a hash function has uniform distribution. For example, hash values of 0, 5, and 10 are uniformly distributed, but when inserted into a hash table of potential size 5, they all collide in bucket 0. It would be better to say the hash % table size should be uniformly distributed, not the hash itself. $\endgroup$ –
With 10,000 records, I'd expect easily 100 records with the same hash code. There is no way on earth you can get the expected number of comparisons below 20 with this hash code. Take a decent hash code, calculated from all characters in the name. Then you want something like 15,000 to 20,000 slots. 8 comparisons on average is something I would ...
For each slot of the first-level hash table (which contains more than 1 keys), we will need to decide a hash-function. Since we want perfect-hashing, we want this function to give no collision. Theorem 11.9 (quoted below) guides us that it is in-fact possible to find such a function in a universal-class, with probability atleast 1/2 (meaning ...
Given an open-address hash table with $\alpha$ < 1, the expected number of probes in a successful search is at most $\frac{1}{\alpha}\ln\frac{1}{1-\alpha}$ I read this in a book and the proof starts by saying. Searching for k follows the same probe sequence as inserting it.
Benubird. 183 1 5. 3. 1) Hashtables are not ordered. 2) Hashtables don't have O(1) O (1) worst-case operations. – Raphael. Jun 18, 2015 at 10:38. @Raphael 1) They are not ordered, but if you require the the only data stored in it is a series of numbers, and you know the min and max number, then you know all of the entries present (although ...
Because a nearly full hash table will give too many collisions and a big (empty) hash table is a waste of space, many implementations allow the hash table to grow (and shrink !) as needed. The growing of a table can involve a full copy of all items (and possibly a reshuffle), because the storage needs to be continuous for performance reasons.
$\begingroup$ Note that the ordering in the search trees is according to System.identityHashCode(Object x), which returns the value that Object.hashCode() would return; if the class doesn't overwrite hashCode(), then o.hashCode() == System.identityHashCode(o), and the secondary structures degenerate to lists.
PS. I have never seen an implementation that doesn’t store all items in the table. In general it’s better to generate locations based on the hash code, and not each location based on the previous location, so items with different hash codes can hit the same first slot but then follow different paths.