For example, it assumes that hash function calculation is O(1) for keys of varying lengths, but it's basically never that.

Hash tables are not O(1) lookup in the size of the key, the are O(1) lookup in the number of elements.

They then compare this element-count time bound to a kind of trie whose time bound is expressed in terms of size of key.

But ~all hash functions take O(k) time where k is the size of the key, because they look at all the bits of the key.

Example: "Exact-match lookups are normal O(1) hash map lookups. Predecessor / successor searches use binary chop on the length of the key. Where a qp-trie search is O(k), where k is the length of the key, a p-fast trie search is O(log k)."

The O(1) in the first sentence is about number of elements, and cannot be compared to the other time bounds, which are about length of key. The exact match lookups are O(k), and thus at best as fast as the qp-trie, unless you use a hash function that is guaranteed to look at less than a constant factor of all of the bits of the key. They don't say this, or give an example of one that they want you to use. In order to be faster than the qp trie, they would have to, at minimum, use a hash function that is log(k)[1]

Additionally, if you look at the idea for search:

" To search, start by splitting the query string at its end into prefix + final chunk of bits. Look up the prefix in the hash map and check the chunk’s bit in the bitmap. If it’s set, you can return the corresponding leaf object because it’s either an exact match or the nearest predecessor."

Unless i'm missing something (maybe i am, it's hard to tell without even pseudocode), this is also O(k) already.

Maybe this is a good idea, maybe it's a bad one, but it's really hard to tell what it's supposed to be fast at or not, and to better analyze the timebounds, without pseudocode of some sort.

[1] This actually is worse than i gloss over, because hash tables are O(1) lookup in amortized time and assume ~even distribution by the hash function in the table to achieve this. But if we are limited to hash functions that take log(k) time, it's not obvious that this distribution is achievable or guaranteed. So it's not even clear it would still be O(1) in the number of elements either. Maybe it is, but at a glance this would seem to imply you can achieve guaranteed perfect hashing without having to look at all the bits of the key. If you did this I would think i could always construct keys that only differ in the bits you don't look at and ruin your distribution, making it no longer O(1) in the number of elements for lookups.

Again, i haven't thought tremendously hard about this, so feel free to tell me i'm wrong :)