-- Test that the string hash map works properly with keys containing zero
-- bytes.
-- Keys with no central '1' are mostly duplicates. The unique keys
-- in this group are '', '\0', ...., '\0 x 34', to a total of 35. All other
-- keys are unique.
select count(*) = 18 * 18 * 17 + 35 
from (
    select key
    from (
        with 18 as n
        select repeat('\0', number % n)
            || repeat('1', intDiv(number, n) % n)
            || repeat('\0', intDiv(number, n * n) % n) key
        from numbers(18 * 18 * 18))
    group by key);