### Kapt's blog

By Kapt, history, 10 days ago,

So, about three weeks ago a fast ($\mathcal{O}(n\log^2(n))$) algorithm for composition and compositional inverse of formal power series has been published (and discovered some time earlier) by noshi91, here is the original blog (in Japanese). Previous best algorithm was $\mathcal{O}((n\log n)^{3/2})$. I hope someone more competent than myself will make an in-deapth blog about this later, I'll link it here as soon as I see it. For now I'll give a "quick" retelling of the algorithm.

UPD: here is a blog that focuses on implementing polynomial composition. It also uses a somewhat different approach, that doesn't explicitly use Tellegen's principle.

UPD 2: Article by noshi91 and Elegia

# Bostan-Mori algorithm

We want to compute $[x^N]\frac{P(x)}{Q(x)}$. To do that we will use an observation that for any polynomial $Q(x)$ the polynomial $Q(x)Q(-x)$ is even, so it is equal so $V(x^2)$ for some $V$. Also let $U(x)=P(x)Q(-x), U(x)=U_e(x^2)+xU_o(x^2)$.

$[x^N]\dfrac{P(x)}{Q(x)}=[x^N]\dfrac{P(x)Q(-x)}{Q(x)Q(-x)}=[x^N]\dfrac{U_e(x^2)+xU_o(x^2)}{V(x^2)}=[x^N]\left( \dfrac{U_e}{V}(x^2)+x\dfrac{U_o}{V}(x^2) \right)$

So if $N$ is even $[x^N]\dfrac{P(x)}{Q(x)} = [x^{\frac{N}{2}}]\dfrac{U_e(x)}{V(x)}$ and if $N$ is odd $[x^N]\dfrac{P(x)}{Q(x)}=[x^{\frac{N-1}{2}}]\dfrac{U_o(x)}{V(x)}$. Note that $deg(V)=deg(Q)$ and $deg(U_i) \approx \dfrac{deg(P)+deg(Q)}{2}$, so if initially $deg(P) \leqslant n$ and $deg(Q) \leqslant n$, those inequalities still hold after we make $P=U_i, Q=V, N = \left[\frac{N}{2} \right]$. So each step can be made in $2$ multiplications of polynomials of degree $\leqslant n$ and there are $\log(N)$ steps untill we have $N=0$ and the task is trivial, so the algorithm works in $\mathcal{O}(n\log n \log N)$.

## Application to bivariate polynomials

It turns out that this algorithm works great with some bivariate polynomials as well.

Now we want to find $[x^k]\frac{P(x, y)}{Q(x, y)}$ where $deg_y(P), deg_y(Q) \leqslant 1$. Note that we don't need any coeffitients with $[x^{>k}]$ in $P$ and $Q$, so we can assume $deg_x(P), deg_x(Q) \leqslant k$. Now if we make one step of Bostan-Mori algorithm we will double limits on degrees of $P$ and $Q$ in respect to $y$ and halve $k$, so after we take $P$ and $Q$ modulo $x^{k+1}$ again, we'll halve their degrees in respect to $x$. So the product of degrees in respect to $x$ and $y$ stays the same and we'll be able to find $U$ and $V$ on each step in $\mathcal{O}(k\log k)$ and solve the problem in $\mathcal{O}(k\log^2(k))$ total time.

# Compositional inverse

From the "Coefficient of fixed $x^k$ in $f(x)^i$ " example of Lagrange inversion theorem from this blog by zscoder we know a connection between $[x^k]\dfrac{1}{1-yf(x)}$ and $\left(\dfrac{x}{f^{(-1)}(x)}\right)^k \mod x^k$, where $f^{(-1)}(f(x))=x$. Specifically,

$[y^i]\left([x^k]\dfrac{1}{1-yf(x)} \right) = \dfrac{i}{k}[x^{k-i}] \left(\left(\dfrac{x}{f^{(-1)}(x)}\right)^k \right)$

so we can find either one from the other in linear time. We can now find the first value with Bostan-Mori algorithm in $\mathcal{O}(k\log^2(k))$, and getting $f^{(-1)}$ from the second value is easy (take $ln$, multiply by $-\frac{1}{k}$, take $exp$, multiply by $x$), we can now find $f^{(-1)} \mod x^k$ in $\mathcal{O}(k\log^2(k))$ time.

# Composition

Finding $H(F(x)) \mod x^n$ is a linear transformation of vector of coefficients of $H$. If we transpose this transformation we'll get some linear transformation of $n$-dimansional vector $\overline{c}$ into a $deg(H)+1$-dimansional one. If $C(x)$ is the polynomial whose coeffitients correspond to coordinates of $\overline{c}$, then the transposed transformation is finding $[x^{n-1}]\dfrac{rev(C(x))}{1-yF(x)} \mod y^{deg(H)+1}$, where $rev(C(x))$ is $x^{n-1}C(\frac{1}{x})$ or just reversed $C$. This can be found in $\mathcal{O}(n\log^2(n))$ with Bostan-Mori algorithm, so by Tellegen's principle the original problem of polynomial composition can be solved in $\mathcal{O}(n\log^2(n))$ with the same constant factor as well.

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