Hi everyone!
This is yet another blog that I had drafted for quite some time, but was reluctant to publish. I decided to dig it up and complete to a more or less comprehensive state for the $300 contest.
Essentially, the blog tells how to combine CDQ technique for relaxed polynomial product computation ("online FFT") with linearization technique from Newton method (similar approach is used in the first example of the ODE blog post by jqdai0815). I will try to briefly cover the general idea of "online FFT" too and provide some examples, in case you're not well familiar with it. That being said...
Consider the following setting:
There is a differentiable function $$$F(x)$$$ such that $$$F(0)=0$$$ and a polynomial $$$f(x)$$$.
You want to compute first $$$n$$$ coefficients of a formal power series $$$g(x)$$$ such that $$$g(x) = F(f(x))$$$. However, the series $$$f(x)$$$ is not known in advance. Instead, the $$$k$$$-th coefficient of $$$f(x)$$$ is given to us after we compute the $$$k$$$-th coefficient of $$$g(x)$$$.
Looks familiar? No? Ok, let's make a brief detour first.
CDQ convolution
General idea of CDQ technique is described in the following simple scheme: To compute something on the $$$[l,r)$$$ interval,
- Compute it on $$$[l, m)$$$ for $$$m=\frac{l+r}{2}$$$,
- Compute the influence of $$$[l, m)$$$ onto $$$[m,r)$$$,
- Compute everything else in $$$[m, r)$$$ recursively,
- Merge the results.
This approach is very versatile, and In convolution context, it's commonly known as "online FFT". It has the following typical formulation:
Standard formulation
We want to compute a sequence $$$c_0, c_1, \dots, c_n$$$, such that
where $$$a_0, a_1, \dots, a_n$$$ and $$$b_0, b_1, \dots, b_n$$$ are not known in advance, but $$$a_k$$$ is revealed to you after you compute $$$c_k$$$.
In a more polynomial-like manner, we may formulate it as
where the $$$k$$$-th coefficient of the polynomials $$$A(x)$$$ and $$$B(x)$$$ is revealed to us as we compute the $$$k$$$-th coefficient of $$$C(x)$$$.
Examples
Polynomial exponent. Assume you want to compute $$$Q(x)=e^{P(x)}$$$ for a given $$$P(x)$$$.
Jetpack. You want to get from $$$(0, 0)$$$ to $$$(n, 0)$$$. On each step, you increase you $$$x$$$-coordinate by $$$1$$$, and your $$$y$$$ coordinate changes to $$$y+1$$$ if you use the jetpack or to $$$\max(0, y-1)$$$ if you don't. At the same time, you can only have $$$y > 0$$$ for at most $$$2k$$$ steps. How many ways are there to get to $$$(n, 0)$$$ under the given constraints?
Gennady Korotkevich Contest 5 — Bin. Find the number of full binary trees with $$$n$$$ leaves such that for every vertex with two children, the number of leaves in its left sub-tree doesn’t exceed the number of leaves in its right sub-tree by more than $$$k$$$.
Solution
Okay, how to solve this? Let's recall the convolution formula
Assume that we want to compute $$$c_k$$$ for $$$k$$$ in $$$[l, r)$$$ and we already know values of $$$c_k$$$, $$$a_i$$$ and $$$b_j$$$ for $$$i, j, k \in [l, m)$$$. Consider the contribution to $$$c_k$$$ for $$$k \in [m, r)$$$ of $$$(i, j)$$$ pairs such that both $$$i$$$ and $$$j$$$ are below $$$m$$$ and at least one of them is above or equal to $$$l$$$. For each such $$$a_i$$$, values of interest for $$$j$$$ range from $$$0$$$ to $$$\min(m, r-l)$$$.
Correspondingly, for each $$$b_j$$$, values of interest for $$$i$$$ range from $$$0$$$ to $$$\min(l, r-l)$$$. In both cases, the contribution may be computed with an appropriate convolution of polynomials of size at most $$$r-l$$$. Note that in the second case we used $$$\min(l, r-l)$$$ instead of $$$\min(m, r-l)$$$ to avoid double-counting pairs in which both $$$i$$$ and $$$j$$$ are $$$l$$$ or above.
We make two recursive calls and use extra $$$O(n \log n)$$$ time to consolidate them, making it for overall $$$O(n \log^2 n)$$$ time.
Generalization
Now back to the beginning. The examples above typically expect that the right-hand side is formulated in a way that is kind of similar to convolutions. But there are a lot of functions, for which it's not really possible to do so. Try the following ones in a similar setting (i.e. the coefficients of $$$f(x)$$$ are given after the coefficients of $$$g(x)$$$ are computed):
Those are the functions that are quite typical in formal power series (for their interpretation, see here). You may probably rephrase them in a convolution-like manner, so that CDQ is applicable, but you would need to do something ad hoc for each individual function. It doesn't seem very convenient, so it only makes sense to try and find some generic enough approach. What is a generic formulation here?
And you may note that what all these functions $$$F(\dots)$$$ have in common is that they're differentiable. Let's use it in a similar way to what we do in Newton's method and rewrite the right hand side in the following way:
This formula generally works for any $$$f_0$$$. In particular, let $$$f_0$$$ be first $$$n$$$ coefficients of $$$f$$$, then $$$(f-f_0)^2$$$ is divisible by $$$x^{2n}$$$. In other words,
In this formula, we still do not know $$$f(x)$$$ completely. But what's important is that it is no longer an argument of some generic function $$$F(\dots)$$$, so the right-hand side is now a linear function of $$$f(x)$$$! This is exactly the formulation for which we learned to apply CDQ convolution above, that is $$$g(x) = A(x) f(x) + B(x)$$$, where
are specific constant polynomials in this context. This sub-problem is solvable in $$$O(n \log^2 n)$$$ with the standard CDQ technique, and since it allows us to double the number of known coefficients at each step, the overall running time is also $$$O(n \log^2 n)$$$, assuming that we're able to compute $$$F(\dots)$$$ and $$$F'(\dots)$$$ in $$$O(n \log^2 n)$$$ as well.
