<P> On top of that, it is possible for the secondary hash function to evaluate to zero . For example, if we choose k = 5 with the following function: </P> <P> h 2 (k) = 5 − (k mod 7) (\ displaystyle h_ (2) (k) = 5 - (k \ mod 7)) </P> <P> The resulting sequence will always remain at the initial hash value . One possible solution is to change the secondary hash function to: </P> <P> h 2 (k) = (k mod 7) + 1 (\ displaystyle h_ (2) (k) = (k \ mod 7) + 1) </P>

What is the main advantage that open addressing hashing techniques have over separate chaining