If H is collision resistant then the following are collision resistant or not?
- H'(x) = H(x) || H(x)
- H'(x) = 1 || H(x)
My thoughts: I am confused in question 1. In case of different hash function(H'(x) = H1(x) || H2(x)) we can safely conclude that H'(x) is safe through proof by contradiction if any one of them is safe. But how to approach when same hash function is used.
In question 2, I am thinking that it is collision resistant with 1 appended in front of the generated hash value.
Need inputs on my analogy for question 2 and some hints for approaching question 1.