hashtable

I expected the following two tuples >>> x = tuple(set([1, "a", "b", "c", "z", "f"])) >>> y = tuple(set(["a", "b", "c", "z", "f", 1])) to compare unequal, but they don't: >>> x == y >>> True Why is that? Zero Piraeus At first glance, it appears that x should always equal y , because two sets constructed from the same elements are always equal: >>> x = set([1, "a", "b", "c", "z", "f"]) >>> y = set(["a", "b", "c", "z", "f", 1]) >>> x {1, 'z', 'a', 'b', 'c', 'f'} >>> y {1, 'z', 'a', 'b', 'c', 'f'} >>> x == y True However , it is not always the case that tuples (or other ordered collections)

What does "bucket entries" mean in the context of a hashtable? A bucket is simply a fast-access location (like an array index) that is the the result of the hash function. The idea with hashing is to turn a complex input value into a different value which can be used to rapidly extract or store data. Consider the following hash function for mapping people's names into street addresses. First take the initials from the first and last name and turn them both into numeric values (0 through 25, from 'A' through 'Z'). Multiply the first by 26 and add the second, and this gives you a value from 0 to

I am trying to understand the open addressing method. I refer to T. H. Cormen's book on this topic, which states that deletion is difficult in open addressing. I am completely stuck at this paragraph: Deletion from an open-address hash table is difficult. When we delete a key from slot i , we cannot simply mark that slot as empty by storing NIL in it. Doing so might make it impossible to retrieve any key k during whose insertion we had probed slot i and found it occupied. I don't understand this. Please explain it with some examples. Assume hash(x) = hash(y) = hash(z) = i . And assume x was

myTable = {} myTable["foo"] = 12 myTable["bar"] = "blah" print(#myTable) -- this prints 0 Do I actually have to iterate through the items in the table to get the number of keys? numItems = 0 for k,v in pairs(myTable) do numItems = numItems + 1 end print(numItems) -- this prints 2 I experimented with both the # operator and table.getn(). I thought table.getn() would do what you wanted but as it turns out it's returning the same value as #, namely 0. It appears that dictionaries insert nil placeholders as necessary. Looping over the keys and counting them seems like the only way to get the

Just for practice (and not as a homework assignment) I have been trying to solve this problem (CLRS, 3rd edition, exercise 11.2-6): Suppose we have stored n keys in a hash table of size m, with collisions resolved by chaining, and that we know the length of each chain, including the length L of the longest chain. Describe a procedure that selects a key uniformly at random from among the keys in the hash table and returns it in expected time O(L * (1 + m/n)). What I thought so far is that the probability of each key being returned is 1/n. If we try to get a random value x between 1 to n, and

If I notice that a hash table (or any other data structure built on a hash table) is filling up, at what point should you build a new table with more buckets. And given n items in the table so far, how do you figure out how many buckets to use in the new one? So let's say I have 100 buckets. Should I reorganize it when there are 50 items in it? 500? 5000? Or should I look for the most-full bucket and key on that? Then when I hit that point how big do I make the new hash table? Related to this, if you know beforehand roughly how many items will go in, is there a way to compute the number of

I want to use a Structure like HashTable. Is there similar structure in Wolfram Mathematica ? Pillsy Update: Mathematica version 10 introduced the Association data structure ( tutorial ). There are a number of possibilities. The easiest possibility, which works well if you don't need to add or delete keys from your table, or change their associated values, is to construct a list of rules with the key on the left-hand side and the value on the right-hand side, and use Dispatch on it. If you do need to change the entries in your table, you can use the DownValues of a symbol as a hash table. This