Consider problem 3.1 from the required textbook for this course. that is, a 32-bit hash function defined as the concatenation of two 16-bit functions: xor and rxor. xor is described in section 3.2 as a simple vertical parity check, assuming that the hash is initialized to all zeros. rxor is described as a 1-bit left-shift rotation performed on the current hash value before each block (in sequence) is being xor-ed with it. that is, process each successive n-bit block of data: a. rotate the current hash value to the left by one bit. b. xor the block into the hash value. is it true that the following bit flips result in finding a message different than the original and that this new message "produces" the same hash? hint: write the original rxor hash and compare it to the final rxor hash you compute.