smax's blog

By smax, 7 weeks ago, In English
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7 weeks ago, # |
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Thank you for the blog! The text is very easy to read in my opinion. I would like to add something.

You provided one algorithm of finding the $$$k$$$-th term of a linear recurrence, which works in $$$O(M(n)\log{k})$$$, where $$$M(n)$$$ is the time complexity of polynomial multiplication. It is actually not the fastest algorithm (anymore), check out this paper: what you described seems to be Fiduccia's algorithm, and it takes $$$3M(n)\log{k} + O(n\log{k})$$$ time to evaluate (according to this paper). Its authors propose another algorithm, which runs in $$$2M(n)\log(k+1) + M(n)$$$. It works as follows:

Let $$$Q(x)$$$ be the characteristic polynomial of our recurrence, and $$$F(x) = \sum_{i=0}^{\infty}a_ix^i$$$ be the generating formal power series of our sequence. Then it can be seen that all nonzero terms of $$$F(x)Q(x)$$$ are of at most $$$(n-1)$$$-st power. This means that $$$F(x) = P(x)/Q(x)$$$ for some polynomial $$$P(x)$$$. Moreover, we know what $$$P(x)$$$ is: it is basically the first $$$n$$$ terms of $$$F(x)Q(x)$$$, that is, can be found in one multiplication of $$$a_0 + \ldots + a_{n-1}x^{n-1}$$$ and $$$Q(x)$$$, and then trimming to the proper degree.

So what remains is to find $$$[x^k]\dfrac{P(x)}{Q(x)}$$$ (recall that $$$[x^k]G(x)$$$ is the $$$k$$$-th coefficient of a formal power series or a polynomial $$$G(x)$$$). We do this recursively:

$$$[x^k]\dfrac{P(x)}{Q(x)} = [x^k]\dfrac{P(x)Q(-x)}{Q(x)Q(-x)} = [x^k]\dfrac{S(x)}{T(x^2)} = [x^k]\dfrac{S_0(x^2) + xS_1(x^2)}{T(x^2)}.$$$

Here $S(x)$ is $$$P(x)Q(-x)$$$, $$$S_0(x)$$$ and $$$S_1(x)$$$ are polynomials composed of only even/odd-indexed terms of $$$S(x)$$$, and we notice that $$$Q(x)Q(-x)$$$ is an even function, thus its polynomial representation contains only even-indexed terms, which represent a polynomial $$$T(x)$$$.

Now, if $$$k$$$ is even, then

$$$[x^k]\dfrac{S_0(x^2) + xS_1(x^2)}{T(x^2)} = [x^k]\dfrac{S_0(x^2)}{T(x^2)} = [x^{k/2}]\dfrac{S_0(x)}{T(x)},$$$


$$$[x^k]\dfrac{S_0(x^2) + xS_1(x^2)}{T(x^2)} = [x^k]\dfrac{xS_1(x^2)}{T(x^2)} = [x^{(k-1)/2}]\dfrac{S_1(x)}{T(x)}.$$$

Thus we divided $k$ by $$$2$$$ with the cost of only two multiplications. Moreover, one can probably even use fewer FFTs to compute $$$Q(x)Q(-x)$$$ and maybe apply other tricks to speed this up.

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    7 weeks ago, # ^ |
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    Woah that's a neat trick, thanks for sharing!

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    15 hours ago, # ^ |
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    Interestingly, this approach (per Schönhage's article, inspired by Graeffe's method) also applies to reciprocals.

    $$$\frac{1}{Q(x)} = \frac{Q(-x)}{Q(-x)Q(x)} = \frac{Q(-x)}{T(x^2)}$$$

    If we want to calculate the result $$$\bmod x^{n+1}$$$, only first $$$\lfloor n/2\rfloor$$$ coefficients of $$$\frac{1}{T(x)}$$$ matter, which allows half-sized reduction for calculating $$$\frac{1}{Q(x)}$$$. Seemingly no significant constant-time improvements here, but might be simpler to comprehend than typical Newton iteration.

7 weeks ago, # |
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In the first code shown you write that if $$$n$$$ is $$$1$$$ (which means $$$f(x) = x - c_0$$$) then $$$a = [c[0]]$$$ without mentioning it. For some of us it might not be trivial that $$$x^k \equiv c_0^k \; mod(x - c_0)$$$. If someone else didn't get that, notice that $$$x^k - r^k = (x-r)(x^{k-1} + x^{k-2}r + x^{k-3}r^2 + \dots + xr^{k-2} + r^{k-1})$$$. It can be extended to arbitrary polynomials to get $$$f(x) \equiv f(k) \; mod (x-k)$$$, which is known as polynomial remainder theorem.

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    7 weeks ago, # ^ |
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    Good catch, I did forget to explain that part of the code. You can also think of it as $$$x^k \bmod (x - c_0) = (x \bmod (x - c_0))^k \bmod (x - c_0) = c_0^k \bmod (x - c_0)$$$. $$$x \bmod (x - c_0) = c_0$$$ by doing one round of long division.