Collision resistance in a hash function I'm confused about whether the following hash function is collision resistant. 
$x \sqsubseteq y $ denotes that x is equal to y or x precedes y.

$ H:\{0, 1\}^* \rightarrow \{0, 1\}^n$ 

If $x \sqsubseteq y $ then $h(x) \sqsubseteq h(y) $

Does this mean H is collision resistant?
Does this mean H is collision resistant?


No, it cannot be collision resistant.

Consider $H(x)$ over all possible values $x \in \{ 0, 1 \}^k$ for $k \ggg n$.  There are $2^k$ possible inputs, and only $2^n$ possible outputs; furthermore, the preimages for any specific output must be contiguous.

Hence, if we consider $x$ stepping from the minimal possible value $0^k$ through consecutive values to the maximal value $1^k$, then $H(x)$ must be split up into (at most) $2^{n}$ ranges, where any two inputs in the same range will generate the same output; for those $2^k$ inputs, there are at most $2^n-1$ places where adjacent inputs produce different outputs.

Hence, if we select a random $r \in \{ 0, 1 \}^{k-1}$, and consider $H( r \mathbin\Vert 0 ), H( r \mathbin\Vert 1)$, then with high probability (at least $1 - 2^{n - k - 1}$), those two hashes will be the same, and that's a collision...