Succinct Dynamic Data Structures (2001) [pdf]
imsc.res.in
imsc.res.in
Also I just started researching succinct dynamic graph data structures a few days ago and ended up on a very interesting paper if anyone is interested: https://arxiv.org/abs/1911.03195
Will also add this, lecture introducing succinct data structures: https://youtu.be/3Y2weLDiUWw
Ok last edit to this comment but I’m sure someone else will find this useful. I also found a good book (haven’t read it yet, but it doesn't seem to have much code examples if any) that covers Succinct Graph data structures and also Succinct Dyanmic Data structures: https://www.google.com/books/edition/Compact_Data_Structures...
We develop succinct data structures to represent (i) a sequence of values to support partial sum and select queries and update (changing values) and (ii) a dynamic array consisting of a sequence of elements which supports insertion, deletion and access of an element at any given index.
For the partial sums problem on n non-negative integers of k bits each, we support update operations in O(b) time and sum in O(logb n) time, for any parameter b, lgn/ lg lg n ≤ b ≤ nƐ for any fixed positive Ɛ < 1. The space used is kn+o(kn) bits and the time bounds are optimal. When b = lgn/ lg lg n or k = 1 (i.e., when we are dealing with a bit-vector), we can also support the select operation in the same time as the sum operation, but the update time becomes amortised.
For the dynamic array problem, we give two structures both using o(n) bits of extra space where n is the number of elements in the array: one supports lookup in constant worst case time and updates in O(nƐ) worst case time, and the other supports all operations in O(lg n/lg lg n) amortized time. The time bound of both these structures are optimal.
Picking on the O(1) time complexity for rank queries, if the superstructure is constant sized then surely each block would have super-constant size and thus couldn't be walked in O(1) time, but conversely if the superstructure has super-constant size it itself can't be walked in O(1) time. In all cases the runtime of such a rank operation is greater than O(1).
Would somebody be willing to ELI5 what I'm missing?
Also, at the cost of sounding like the CS equivalent of flat earthers, big-O notation is kind of broken, and specifically, it doesn't make a lot of sense to say that something is O(1). Even hashmap queries, if you think about it, can't really be O(1) for arbirarily large values of N.
The general interpretation of O(1) in theoretical CS literature is "we know it's bugged, but we promise we won't bugabuse".
- The vector represents a subset of N values, hence requires Z=N bits.
- The blocks themselves require at least Z bits to store, so to be succinct the superblocks must be o(Z) in space.
- Consequently, there are at most o(Z) superblocks.
- It follows that there are at most o(Z) blocks (since otherwise you would have a super-constant number to walk per block).
- Thus, each block takes up super-constant space (and the popcount component must increase in size as well).
- Your rank lookups require at least O(popcount-size) time, which is super-constant.
I think you're right though that the problem I was seeing just falls back to the general "we know it's bugged, but we promise we won't bugabuse" idea -- no function returning distinct values for N different inputs on a fixed-width register machine can execute in less than log(N) time, and I probably should have just stopped there and called it a day.
You are absolutely right that it can't be done on a fixed-width register machine, though.
Generally speaking, most papers assume arithmetic is O(1) time and integers fit in O(1) space because keeping track of the actual time and space of integer arithmetic gets unwieldy very quickly. But this model is super broken: if arbitrary arithmetic can be performed in O(1) then P = PSPACE. The "gentlemen's agreement" assumption is that operations that make numbers grow too big are forbidden.
I don't know much about this stuff (I'm way out of my depth as a matter of fact), but I'm glad I could be useful :)