Is the hash function defined in this exercise collision-resistant? I have this problem:


  "Let ${\rm Enc}_k(m)$ a safe block encryption system. We define a hash function $H(m)$ as follows:
  
  
  $m$ will be concatenated with $0$ until $|m|$ is a multiple of len(block);
  we break the sequence $m$ into $n$ blocks like $m = m_0 \mathop\| m_1 \mathop\| m_2 \mathop\| \dots \mathop\| m_{n-1}$;
  we apply this algorithm:  

c &larr; Encm0(m0)
for i = 1 to n-1 do
  d &larr; Encm0(mi)
  c &larr; c NOR d
end for
H(m) &larr; c
  


Where c NOR d denotes the bitwise operation NOT (c OR d), i.e. (for single bits) 1 NOR 0 = NOT (1 OR 0) = NOT 1 = 0.

Is this $H$ resistant to collisions?
The padding scheme isn't collision resistant.

For any message $m$ where $|m| \not\equiv 0 \pmod n$, there will always be a collision between $m$ and $m || 0$.