# Tag Info

## Hot answers tagged hash-tables

100

Consider the set of keys $K=\{0,1,...,100\}$ and a hash table where the number of buckets is $m=12$. Since $3$ is a factor of $12$, the keys that are multiples of $3$ will be hashed to buckets that are multiples of $3$: Keys $\{0,12,24,36,...\}$ will be hashed to bucket $0$. Keys $\{3,15,27,39,...\}$ will be hashed to bucket $3$. Keys $\{6,18,30,42,...\}$ ...

24

The hash function doesn't return some string such as mkwer. It directly returns the position of the item in the array. If, for example, your hash table has ten entries, the hash function will return an integer in the range 0–9.

15

Refresh your knowledge of binary! The $p$ lowest-order bits of $k$ are the last $p$ bits when $k$ is written out in binary (i.e., the $p$ rightmost bits). For example, if $p=3$ and $k=17$ then $k$ is $10001$ in binary and the three lowest-order bits are $001$. The point is that, in general, computing $k \bmod m$ is a relatively ...

15

Pathological data is supposed to be data that makes things go wrong in some way for your intended computation. It can be called pathological when it is rare enough in actual uses, so that things work OK most of the time. This can sometimes be made mathematically more precise (for example with probabilities), but the use of the word pathological in often ...

15

The entries of a hash table are stored in an array. However, you have misunderstood the application of the modulo operator to the hash values. If the hash table is stored in an array of size $n$, then the hash function is computed modulo $n$, regardless of how many items are currently stored in the table. So, in your example, if you were storing ...

14

The most obvious answer is that trees can be traversed in their natural order very efficiently. If you need to visit every element of a dictionary in alphabetical order, a tree can support this directly, where a hash table cannot. Another answer is that trees can be made immutable - where insertion and deletion only involve recreating a small number of ...

11

Pathological data is data that will make the algorithm perform bad. For hash tables, pathological data is data that causes collisions. That of course depends on the hash function being used. For example, if your hash function adds the characters together: hash("abcd") = 'a' + 'b' + 'c' + 'd'. Then pathological data looks like: {"abcd", "dcba", "cbda", ...}....

10

Binary search trees (BSTs) of various sorts and their variations are widely used data structures today, so they are hardly a "historical note". For example, both the .NET Framework and the Java Standard Library provide a tree-based implementation of a dictionary. A red-black tree no less in the latter case. One of the reasons for this is that tree-based ...

9

Simply compute the index of the permutation into the sorted list of all permutations and use that as your hash key. This can be achieved with a relatively simple algorithm: https://stackoverflow.com/questions/5131497/find-the-index-of-a-given-permutation-in-the-sorted-list-of-the-permutations-of Once you have that index, you can make a table with exactly 9! ...

9

When you're asking about "exact" memory usage, do consider that all of those pointers may not be necessary. To see why, consider that the number of binary trees with $n$ nodes is $C_{2n}$, where: $$C_i = \frac{1}{i+1} { 2i \choose i }$$ are the Catalan numbers. Using Stirling's approximation, we find: $$\log C_{2n} = 2n - O(\log n)$$ So to represent a ...

8

Hash-table usually do waste space. Many algorithms do, since time-space trade-offs are common, but they usually hide it better :). Like other algorithms, hash-tables do it to get better time performance. The first point is that you try to avoid collisions in your hash-table, because that keeps the access time cost constant (but collisions are usually ...

8

An hash set is an hash table. Using an hash set to handle collisions in an hash table is equivalent to using a bigger hash table, with an hashing function which is a combination of the hashing functions of both level. In other words, you'd probably be better with a bigger initial table (for instance there is no risk of resonance between the two hash ...

8

Because we generally use the RAM model of computation with uniform cost model when computing the running time of operations on a hash table, and the RAM model with uniform cost states that the time to do a single operation on an entire machine word is $O(1)$. Also, we generally assume that the hash value fits within a single machine word. Thus, the running ...

7

The reason you've never heard of hash tables being used like this is that hash tables are either "too much" or "not enough" in this situation. If the range of elements being sorted is small, then you can use counting sort, or something similar. But for counting sort, you would almost certainly want to use a simple array rather than a hash table. If you ...

7

SHA1 or SHA256, whichever you use, is for any practical purpose a random function. What you are observing is that random allocation is not as good as deterministic allocation. If you knew all the values in advance then you could indeed arrange that each cell would get exactly the same number of hits. Unfortunately, when you throw $n$ balls into $k$ bins, the ...

7

But then the lookup time is no longer constant Not worst-case constant -- which it never is for (basic) hashtables -- but it is still average-case constant, provided the usual assumptions on input distribution and hashing function. Why not use a hash set instead of a linked list? And how do you implement that one? You have created a circular definition. ...

7

The easiest way is to construct a static hash table $T$ containing all the collisions, in the following form: for each set of keys $S$ which are supposed to map to the same value, single out some $x \in S$, and put all other $y \in S$ in the table with an entry stating "$x$". Now take a good hash function $h$, and construct a new one as follows: On input $... 6 Hash function calculates array position from given string. If this is perfect hash it means that there are for sure no collisions, the most probably array is at least twice bigger than number of elements. For example I will give very poor hash for letters, just to ilustrate mechanism: 0)$x = 0;$1) for each character in string take ascii value, subtract 'a'... 6 The load factor denotes the average expected length of a chain, therefore it is interesting for an average case analysis, not the worst case analysis. That's why on average you expect needing constant time for search operations. In the worst case however, all your elements hash to the same location and are part of one long chain of size n. Then, it depends ... 6 An easy way to visualize this is to imagine a hash table of size$n$(implemented with chaining) that contains all of the elements of$U$(even though this is unrealistic in practice because$U$typically has massive size). Since$|U| >> n$, all of the elements of$U$do not fit into the hash table; therefore, there will be collisions. Consider, for ... 6 You are right now thinking of a data structure from which just three operations are expected, Insertion Lookup Deletion But if you extend these range of operations, to let's say finding number of elements greater than a certain value, one can see how BST's can be useful. A BST can still manage this operation in$\log n$time but a hash table can't. ... 6 This is a nice question. In the comparison model or, what is more general, the algebraic decision-tree model, the problem of element distinctness has a lower bound of$\Theta(n\log n)$time-complexity in the worst case as said in this Wikipedia article. So there is no algorithm to count distinct elements in linear time in the worst case, even without ... 5 The short answer is that the terms were coined by different people for different purposes, and it's because hash tables historically had two different contexts: in-memory data structures (I believe that compiler symbol tables was one of the first "killer app" uses) and external-memory data structures (file-based databases). According to Knuth, the first ... 5 I assume under "size of slot" you mean length of chain in hash table with separate chaining collision resolution. From Knuth, TAoCP, Vol. 3, 6.4, ex. 34 probability of k-length chain in hash table with$M$slots and$N$entries is$\binom{N}{k} \left ( M-1 \right )^{N-k} / M^N$Assuming "ideal" hash function, however practical real-world hash functions ... 5 Adding permutations isn't about preventing slow servers from becoming bottlenecks, rather it's about dispersing a convoy once one forms behind a slow server. Because of the way tract locations are hashed, sequential reads of any blob always walk the tract locator table in the same order. Suppose you have six tract servers and your tract locator table looks ... 5 A hash function is a pseudorandom function with a constant range. Ideally, one would like two central properties: The hash function should be easy (fast) to compute. The probability that two inputs$x,y$hash to the same value is roughly$2^{-n}$, where$n$is the output length. The second property isn't stated rigorously. There are several ways to state ... 5 The algorithm you give is exponential time, not linear. If you're given$nb$-bit entries, the size of your input is$nb$bits but the algorithm takes time$\Theta(2^b)$, which is exponential in the input length. In particular, your algorithm takes$2^k$steps to sort the roughly$2k$-bit input$\{0, 2^k\}$. 5 Yes, after storing many items in a distributed hash table spread over a hundred computers, if hypothetically we used the sort of hash function popular for in-RAM hash table, adding another computer would cause every item to be re-hashed and nearly every item to move from one computer to another, which could take days. So instead we use special hash ... 5 You absolutely can do this. You just have to be careful with how you set things up. There's a type of hash table called a dynamic perfect hash table that, with some modifications, is essentially what you're describing. It works by having a two-layer hash structure, where collisions in the top level are resolved by building a hash table for the second level ... 5 Let$U = [m]$, and let$h$be the identity function. If you insist that$|U| > m$, then you can take$U = [m+1]$, and consider the functions$h_i$, for$i \in [m]$, given by $$h_i(x) = \begin{cases} x & \text{if } x \neq m+1, \\ i & \text{if } x = m+1. \end{cases}$$ The same approach can be used for arbitrary$|U|$: fix the first$m\$ ...

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