What happens when we hash already hashed values, concatenated together?

cn flag

I read on the page 16 of On the Security of Hash Function Combiners that

the classical combiner for collision-resistance simply concatenates the outputs of both hash functions $Comb_{\mathbin\|}(M) = H_0(M) \mathbin\| H_1(M)$ in order to ensure collision resistance as long as either of H0, H1 obeys the property.

Consider H, a secure internal hash function with 256-bit inputs and 128-bit outputs

My question is, if we hash the concatenation again with the same hash function used before,

Like so

H(H(M)∥H(M))) (pardon my lack of knowledge of what latex is)

what happens to the collision resistance? does it become

$$2^{(n/2)} \cdot 2^n$$

or does it become

$$2^{(n/2)} + 2^n$$

or am I going the wrong direction to find if it even improves or makes it worse? The hash function used all three times is the same hash function, which I am assuming to be ideal/robust and collision resistant.

kelalaka avatar
in flag
Err, what is the page? Your quote is not perfect. And we have $\LaTeX$/MathJax enabled on our site. It seems that you want to learn about hash design, my advice read the [Blake book]( And, I remember such a question...
kelalaka avatar
in flag
Note that; the question is not clear. `If we hash the concatenation again` you mean with $H_1,H_2$, or another independent hash function $H_3$? Why do we need such desing while we can use $SHA-3, Shake, Blake$ etc?
kelalaka avatar
in flag
[Multicollisions in Iterated Hash Functions. Application to Cascaded Constructions Antoine Joux]( and the [outer hash doesn't change the internal collisions](
kelalaka avatar
in flag
[The Not So Short Introduction to LATEX 2ε](
ph flag
Seconding kelalaka's question: the quoted text talks about 2 different hashes but your question is about a single hash? Are you really asking about the collision resistance of applying that construct to a single hash? Why?

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