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According to the books that i have read, it says that S.H.A(Secure Hash Algorithm) is collision resistant.But if the input space is a 1024 bit number and the output space is a 512 bit message digest then shouldn't it be colliding for (2^1024)/(2^512) times? As the range is lesser than the domain being mapped there should have been collisions. please explain where i am going wrong.
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The first element is the hash function. The MD5 function is now considered very insecure: it is easy to reverse with current processing power. The SHA1, SHA256, and SHA512 functions are no longer considered secure, either, and PBKDF2 is considered acceptable.
Since executing a brute-force attack of this size is considered computationally infeasible, SHA-256 can be considered collision-resistant, for now at least.
Cryptographic hash functions are usually designed to be collision resistant. However, many hash functions that were once thought to be collision resistant were later broken. MD5 and SHA-1 in particular both have published techniques more efficient than brute force for finding collisions.
In particular, cryptographic hash functions exhibit these three properties: They are “collision-free.” This means that no two input hashes should map to the same output hash. They can be hidden. It should be difficult to guess the input value for a hash function from its output.
The chance for a collision does not depend on the input size. The chance to a 512-bit hash collision is 1.4×10^77, see Probability table
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Maybe your book has also mentioned the definition of collision resistance? It does not mean that no collisions are created (which is clearly not the case), but that given a hash you are not able to create a message easily that produces this hash.
a hash function H is collision resistant if it is hard to find two inputs that hash to the same output; that is, two inputs a and b such that H(a) = H(b), and a ≠ b
From Wikipedia
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