Does a hash collision involve two inputs with same hash?

The wide-scale use of cryptographic hash functions has also taken root in the security community where hash functions are also used to identify artifacts. It is common to refer to malware, binary artifacts, and library and reference objects by the 32 hex digits comprising its MD5 sum. The range of a hash function is a finite (bounded) set of integers R, but the domain is an infinite set of strings. Therefore hash collisions are possible, and among a set of n objects, there is some probability that any two of them will have a common hash value.

As we have seen in previous videos, it happens sometimes that two keys yield the same hash value for a given table size. This is called a “hash collision” or just “collision.”

Hash is typically used as a many-to-one function, with the number of potential inputs (size of input domain) much larger that the number of potential output values ("range"), making collisions inevitable ("pigeonhole principle"). For the cryptographic hash functions (CHFs), the output is a ...

Much like symmetric-key ciphers are vulnerable to brute force attacks, every cryptographic hash function is inherently vulnerable to collisions using a birthday attack. Due to the birthday problem, these attacks are much faster than a brute force would be.

Due to the pigeonhole principle ... but how hard they are to find. ... Example: Input1: "hello world" → Hash: a123 Input2: "totally different" → Hash: a123 (Collision!) Definition: Finding any two different inputs that ...

When two or more keys have the same hash value, a collision happens. To handle this collision, we use Collision Resolution Techniques. ... The idea behind Separate Chaining is to make each cell of the hash table point to a linked list of records ...

In 2017, Google researchers generated two different PDF files with identical SHA-1 hashes, finally proving what cryptographers had warned about for years: hash functions don’t create truly unique fingerprints ( Stevens et al., 2017 ). This “SHAttered” attack required 9 quintillion SHA-1 computations, which is the equivalent to 6,500 years of single-CPU computation.

In a recent very long thread (with 63 posts) on the NetBackup Mailing List, someone took issue with a quote of mine from my January article, “The Skinny on Deduplication.” In that article, I pointed out that the odds of having two different blocks of data have the same hash (known as a hash collision) are 1:2^160, which is an astronomical number. They said that what’s important is the probability of a hash collision in a given environment, and those odds increase with the size of the environment. They told me to read up on the Birthday Paradox to see what they’re talking about.

(Which is why its called a collision'). That proves the hash method used cannot be relied on any more to identify different data and keep data and hash isolated, and its time to move to a more advanced hash instead.