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If H is collision resistant then the following are collision resistant or not?

  1. H'(x) = H(x) || H(x)
  2. 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.

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  • $\begingroup$ Since this is HW, here the hint due to our HW policy. Consider that $H'$ is not collision-resistant. What can you deduce? Same hint for 2. , too. $\endgroup$ – kelalaka Feb 28 at 17:53
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    $\begingroup$ @kelalaka Here is what I am thinking regarding question 1. Let H'(x) is not collision resistant. Therefore, for different inputs x and y, H'(x) = H'(y). Which can be further written as H(x) || H(x) = H(y) || H(y). This implies as H(x) = H(y), which is a contradiction. Therefore, it is collision resistant. $\endgroup$ – Nathan Anatoli Feb 28 at 18:04
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    $\begingroup$ In question 2, we will get, 1 || H(x) = 1 || H(y), again the second part is H(x) = H(y), which is a contradiction. Therefore, it is collision resistant. $\endgroup$ – Nathan Anatoli Feb 28 at 18:16
  • $\begingroup$ That's all. This question is going to be closed. You can keep it or delete it. Have fun. $\endgroup$ – kelalaka Feb 28 at 18:17
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    $\begingroup$ Thank you very much for the motivation $\endgroup$ – Nathan Anatoli Feb 28 at 18:17