### jtnydv25's blog

By jtnydv25, history, 4 years ago,

In the static RMQ problem, one has to answer queries about the minimum in a range of a fixed array. A classic variation allows updates of the form $A_i \leftarrow x$. A good solution is segtree which solves it in $O(\log{n})$ query and update times. My question is : How fast can we update if we must answer queries in $O(1)$?

I can think of a solution which works in $O(\sqrt{n})$ per update :

We can answer queries in $O(1)$ using a sparse table. Since each element is a part of $O(n)$ ranges in the sparse table, we can clearly update a sparse table of an array of size $n$ in $O(n)$. Now, divide into $\sqrt{n}$ blocks. For each block, maintain the prefix and suffix minimums and also a sparse table. Clearly, a block can be updated in $O(\sqrt{n})$. On the upper level, maintain another sparse table which can answer minimum over a range of blocks. Clearly this can also be done in $O(\sqrt{n})$ and the query time is still $O(1)$. The memory used and precomputation done are both $O(n \log{n})$.

Can we do better, say $\text{polylog}(n)$, perhaps using more memory and/or precomputation? Is there a known lower bound?

• +90

 » 4 years ago, # |   +2 Updating the block sparse table, or the upper level sparse table, runs in $O(\sqrt{n} \log(n))$, not in $O(\sqrt{n})$.You approach reminds me a little of the Sqrt Tree by gepardo. That one does updates in true $O(\sqrt{n})$ and still answers queries in $O(1)$. And as a bonus it supports more complex operations than computing the maximum, like computing the sum in a range, for which the sparse table cannot compute the answer in $O(1)$. You can learn about it, see here or here
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 The second link contains also an idea, how to improve it even further with less precomputing/memory, $O(n \log^* n)$, so almost $O(n)$, if you can overlap ranges like you can do with min/max queries. I'm not 100% sure, but I think updates are still $O(\sqrt{n})$ though.
•  » » 4 years ago, # ^ | ← Rev. 2 →   +15 Each element is a part of $O(n)$ ranges in the sparse table of an array with $n$ entries. Proof : For the range $[j, j + 2^k - 1]$ to contain $i$, $i - 2^{k} + 1 \leq j \leq i$. So there are $\leq 2^k$ ranges of length $2^k$ that require a change. Summing up gives $O(n)$. Another way to see this is to subtract the left and right parts to get $n \log{n} - i \log{i} - (n-i) \log{(n-i)} = O(n)$So, the upper level sparse table takes $O(\sqrt{n})$ per update.Anyway, I also posted this on stackexchange, and as the answer says, we can recursively get $O(n^{\epsilon})$ for any constant $\epsilon > 0$, which is really cool.The possibility of a polylogarithmic update time is still a question though!
•  » » » 4 years ago, # ^ |   0 Oh, right. You are correct.Have you looked at the idea in the second article? It might have polylog updates. Basically splits the array into $log(n)$ segments, and do the same thing for each block recursively. And it keeps a sparse table for all every block in every layer. But I haven't read the article for a long time, so I don't understand all the details right now.
 » 4 years ago, # |   +16 Well... I don't think that's what you want, but you can get $O(k n^{\frac{2}{k}})$ for any fixed $k$.Let's build $k$-layer structure, $i$-th layer is divided into blocks of length $n^{\frac{i}{k}}$, each block is a union of $n^{\frac{1}{k}}$ blocks on previous layer. Each block stores minima on all $O((n^{\frac{1}{k}})^{2})$ segments of blocks of previous layer.I think you get the idea. Update is $O(k n^{\frac{2}{k}})$, get is $O(k)$ which is $O(1)$ for fixed $k$. With $k = \log n$ you get the segment tree.
 » 13 months ago, # |   +20 Is there a known lower bound?Maybe this is obvious but there is a lower bound of logN. If you had a better bound, say F(N), then you can use it to sort an array in O(NF(N)) (query min, update the element containing min, query again (O(N) times) until the array is sorted) and there is a well known bound of O(NlogN) on sorting.