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John Smith and Sandra Dee share the same hash value of 02, causing a hash collision. In computer science, a hash collision or hash clash [1] is when two distinct pieces of data in a hash table share the same hash value. The hash value in this case is derived from a hash function which takes a data input and returns a fixed length of bits. [2]
A hash of n bits can be broken in 2 n/2 time steps (evaluations of the hash function). Mathematically stated, a collision attack finds two different messages m1 and m2, such that hash(m1) = hash(m2). In a classical collision attack, the attacker has no control over the content of either message, but they are arbitrarily chosen by the algorithm.
In words, when given an x, it is not possible to find another x' such that the hashing function would create a collision. A hash function has strong collision resistance when, given a hashing function H, no arbitrary x and x' can be found where H(x)=H(x'). In words, no two x's can be found where the hashing function would create a collision.
Collisions originally reported in 2004, [13] followed up by cryptanalysis paper in 2005. [19] RadioGatún: Up to 2 608 [20] 2 704: 2008-12-04 For a word size w between 1-64 bits, the hash provides a security claim of 2 9.5w. The attack can find a collision in 2 11w time. [21] RIPEMD-160 2 80: 48 of 80 rounds (2 51 time) 2006 Paper. [22] SHA-0: ...
A hash table uses a hash function to compute an index, also called a hash code, into an array of buckets or slots, from which the desired value can be found. During lookup, the key is hashed and the resulting hash indicates where the corresponding value is stored. A map implemented by a hash table is called a hash map.
The probability of a collision depends mainly on the hash length (see birthday attack). Thus, the concern arises that data corruption can occur if a hash collision occurs, and additional means of verification are not used to verify whether there is a difference in data, or not. Both in-line and post-process architectures may offer bit-for-bit ...
As an example, consider the scenario in which a teacher with a class of 30 students (n = 30) asks for everybody's birthday (for simplicity, ignore leap years) to determine whether any two students have the same birthday (corresponding to a hash collision as described further). Intuitively, this chance may seem small.
This means that all data keys land in the same bin, making hashing useless. Furthermore, a deterministic hash function does not allow for rehashing: sometimes the input data turns out to be bad for the hash function (e.g. there are too many collisions), so one would like to change the hash function.