When using the table of keys directly, any key could be a target. How do we mark an empty entry?
With a power-of-two hash table size, hashes of 0 and -1 always correspond to the first and last entry, so keys with those hashes will only ever be compared to the first and last 15 entries
The above scheme is independent of the hash table size
Our hash functions can be inverted with enough mathematics, but with only 2*32 = 4.3e9 possible keys brute force takes just a few seconds