Recovering a linear recurrence with the extended Euclidean algorithm

Правка en38, от adamant, 2022-04-08 17:06:22

Hi everyone!

The task of finding the minimum linear recurrence for the given starting sequence is typically solved with the Berlekamp-Massey algorithm. In this article I would like to highlight another possible approach, with the use of the extended Euclidean algorithm.

Tl'dr.

The procedure below is essentially a formalization of the extended Euclidean algorithm done on $$$F(x)$$$ and $$$x^{m+1}$$$.

If you need to find the minimum linear recurrence for a given sequence $$$F_0, F_1, \dots, F_m$$$, do the following:

Let $$$F(x) = F_m + F_{m-1} x + \dots + F_0 x^m$$$ be the generating function of the reversed $$$F$$$.

Compute the sequence of remainders $$$r_{-2}, r_{-1}, r_0, \dots, r_k$$$ such that $$$r_{-2} = F(x)$$$, $$$r_{-1}=x^{m+1}$$$ and

$$$r_{k} = r_{k-2} \mod r_{k-1}.$$$

Let $$$a_k(x)$$$ be a polynomial such that $$$r_k = r_{k-2} - a_k r_{k-1}$$$.

Compute the auxiliary sequence $$$q_{-2}, q_{-1}, q_0, \dots, q_k$$$ such that $$$q_{-2} = 1$$$, $$$q_{-1} = 0$$$ and

$$$q_{k} = q_{k-2} + a_k q_{k-1}.$$$

Pick $$$k$$$ to be the first index such that $$$\deg r_k < \deg q_k$$$. Let $$$q_k(x) = a_0 x^d - \sum\limits_{k=1}^d a_k x^{d-k}$$$, then it also holds that

$$$F_n = \sum\limits_{k=1}^d \frac{a_k}{a_0}F_{n-k}$$$

for any $$$n \geq d$$$ and $$$d$$$ is the minimum possible. Thus, $$$q_k(x)$$$ divided by $$$a_0$$$ is the characteristic polynomial of the minimum linear for $$$F$$$.

More generally, one can say for such $$$k$$$ that

$$$ F(x) \equiv \frac{(-1)^{k}r_k(x)}{q_k(x)} \pmod{x^{m+1}}. $$$


Linear recurrence interpolation

In the previous article on linear recurrences we derived that the generating function of the linear recurrence always looks like

$$$ G(x) = \frac{P(x)}{Q(x)}, $$$

where $$$P(x)$$$ and $$$Q(x)$$$ are polynomials and $$$\deg P < \deg Q$$$. In this representation, $$$Q(x) = 1 - \sum\limits_{k=1}^d a_k x^k$$$ corresponds to

$$$ F_n = \sum\limits_{k=1}^d a_k F_{n-k}. $$$

Typical competitive programming task of recovering linear recurrence is formulated as follows:


Find Linear Recurrence. You're given $$$F_0, \dots, F_{m}$$$. Find $$$a_1, \dots, a_d$$$ with minimum $$$d$$$ such that $$$F_n = \sum\limits_{k=1}^d a_k F_{n-k}$$$.
In formal power series terms it means that we're given $$$F(x) = F_0 + F_1 x + \dots + F_{m}x^{m}$$$ and we need to find $$$\frac{P(x)}{Q(x)}$$$ such that
$$$ F(x) \equiv \frac{P(x)}{Q(x)} \pmod{x^{m+1}} $$$

and $$$d$$$ is the minimum possible. Note that in this particular problem it is not required for $$$a_d$$$ to be $$$0$$$, hence $$$\deg Q \leq d$$$.

In this terms, as we will see later, what the problem asks us to find is essentially the Padé approximant of $$$F(x)$$$.

Padé approximants


Given a formal power series $$$f(x) = \sum\limits_{k=0}^\infty f_k x^k$$$, its Padé approximant of order $$$[p/q]$$$ is the rational function

$$$ R(x) = \frac{a_0 + a_1 x + \dots + a_p x^p}{b_0+b_1 x + \dots + b_q x^q}=\frac{P(x)}{Q(x)} $$$

such that $$$f(x) Q(x) \equiv P(x) \pmod{x^{p+q+1}}$$$. The Padé approximant is denoted $$$[p/q]_{f}(x) = R(x)$$$.

For normalization purposes, we will demand $$$b_q = 1$$$, that is $$$\deg Q(x) = q$$$, but it is acceptable to have $$$\deg P(x) \leq p$$$.


Explanation: The definition of Padé approximants requires some normalization, as you could obtain different representations of the same rational function by multiplying its numerator and denominator with the same polynomial. Most commonly used normalization is $$$b_0 = 1$$$. However, to guarantee $$$q = \deg Q$$$ and to easier deal with possible trailing zeros in the recurrence, we will use the normalization $$$b_q = 1$$$.

These definitions are different in terms of whether the Padé approximant for the given normalization exists. However, when $$$\deg P < \deg Q$$$, they're equivalent in a sense that when $$$Q(x) = 1-b_1 x - \dots - b_q x^q$$$ defines the linear recurrence

$$$ F_n = \sum\limits_{k=1}^d b_k F_{n-k}, $$$

the reversed polynomial $$$x^q Q(x^{-1}) = b_q - b_{q-1}x - \dots - x^q$$$ defines the recurrence

$$$ F_n = \sum\limits_{k=1}^d b_k F_{n+k}. $$$
Formal proof

Thus to find the first kind recurrence for $$$F_0,F_1,\dots,F_m$$$, we could find the second kind recurrence for $$$F_m, F_{m-1}, \dots, F_0$$$ instead.


Online Judge — Rational Approximation. Given $$$f(x)$$$, compute $$$p(x)$$$ and $$$q(x)$$$ of degrees at most $$$m-1$$$ and $$$n-1$$$ such that

$$$ f(x)q(x) - p(x) \equiv 0 \pmod{x^{m+n}}. $$$

Extended Euclidean algorithm

Let's look again on the condition

$$$ F(x) Q(x) \equiv P(x) \pmod{x^{m+1}}. $$$

It translates into the following Bézout's identity:

$$$ F(x) Q(x) = P(x) + x^{m+1} K(x), $$$

where $$$K(x)$$$ is a formal power series. When $$$P(x)$$$ is divisible by $$$\gcd(F(x), x^{m+1})$$$ the solution to this equation can be found with the extended Euclidean algorithm. Turns out, the extended Euclidean algorithm can also be used to enumerate all $$$[p/q]_F$$$ with $$$p+q=m$$$.

Formalizing the algorithm

Let's formalize the extended Euclidean algorithm of $$$A(x)$$$ and $$$B(x)$$$. Starting with $$$r_{-2} = A$$$ and $$$r_{-1} = B$$$, we compute the sequence

$$$ r_{k} = r_{k-2} \mod r_{k-1} = r_{k-2} - a_{k} r_{k-1}. $$$

If you're familiar with continued fractions, you could recognize, that this sequence corresponds to the representation

$$$ \frac{A(x)}{B(x)} = a_0(x) + \frac{1}{a_1(x) + \frac{1}{a_2(x) + \frac{1}{\dots}}} = [a_0(x); a_1(x), a_2(x), \dots]. $$$

Same as with rational numbers, for such sequence it is possible to define the sequence of convergents

$$$ \frac{p_k(x)}{q_k(x)} = [a_0(x); a_1(x), \dots, a_k(x)] = \frac{a_{k}(x) p_{k-1}(x) + p_{k-2}(x)}{a_{k}(x) q_{k-1}(x) + q_{k-2}(x)}, $$$

starting with

$$$ \begin{matrix} \frac{p_{-2}(x)}{q_{-2}(x)} = \frac{0}{1}, & \frac{p_{-1}(x)}{q_{-1}(x)} = \frac{1}{0}, & \frac{p_{0}(x)}{q_{0}(x)} = \frac{a_0(x)}{1},& \dots \end{matrix} $$$

Now, one can prove the following identity:

$$$ (-1)^{k}r_k(x) = q_k(x) A(x) - p_k(x) B(x). $$$
Explanation

In other words, we get three sequences $$$r_k$$$, $$$p_k$$$ and $$$q_k$$$ that define a family of solutions to the Bézout's identity.

Using $$$A(x) = x^{m+1}$$$ and $$$B(x) = F(x)$$$, we conclude that

$$$ F(x) q_k(x) \equiv (-1)^k r_k(x) \pmod{x^{m+1}}. $$$

Enumerating approximants

Let's estimate the degrees of $$$r_k$$$ and $$$q_k$$$ to estimate how they relate with Padé approximants.

It follows from the recurrence that $$$\deg q_{k} - \deg q_{k-1} = \deg a_k = \deg r_{k-2} - \deg r_{k-1}$$$, therefore

$$$ \deg q_k = \deg a_0 + \dots + \deg a_k $$$

and

$$$ \deg r_k = (m+1) - \deg a_0 - \dots - \deg a_{k} - \deg a_{k+1} = (m - \deg q_k) - (\deg a_{k+1}-1). $$$

Multiplying both the numerator and the denominator of $$$\frac{(-1)^k r_k}{q_k}$$$ by $$$1,x,x^2,\dots,x^{\deg a_{k+1}-1}$$$, we get the Padé approximants $$$[p/q]_F$$$ for all $$$q$$$ from $$$\deg q_k$$$ to $$$\deg q_k + \deg a_{k+1}-1 = \deg q_{k+1}-1$$$, while also maintaining the inequality $$$p \leq m - q$$$.

Joining it together for all $$$q_k$$$ we see that all $$$[(m-q)/q]_F$$$ for $$$q$$$ from $$$0$$$ to $$$m$$$ are covered.

Thus, if you find $$$k$$$ such that $$$\deg q_k \leq q$$$ and $$$\deg q_{k+1} > q$$$, assuming $$$p+q=m$$$, it will hold that

$$$ [p/q]_F = \frac{(-1)^k x^{q-\deg q_k} r_k}{x^{q - \deg q_k} q_k}. $$$

Note that the Padé approximant is not necessarily unique (for example, $$$0 = \frac{0}{1+x} = \frac{0}{1+2x}$$$). However, if

$$$ \begin{cases} f(x) Q_0(x) \equiv P_0(x) \pmod{x^{m+1}},\newline f(x) Q_1(x) \equiv P_1(x) \pmod{x^{m+1}}, \end{cases} $$$

then it is also true that

$$$ P_0 Q_1 \equiv f(x) Q_0(x) Q_1(x) \equiv P_1(x) Q_0(x) \pmod{x^{m+1}}. $$$

Assuming that both approximants satisfy the definition of $$$[(m-q)/q]_f$$$, we can conclude that

$$$ \deg P_0 = \deg P_1 \leq p $$$

and

$$$ \deg Q_0 = \deg Q_1 = q, $$$

otherwise the identity

$$$ P_0(x) Q_1(x) - P_1(x) Q_0(x) \equiv 0 \pmod{x^{m+1}} $$$

won't hold. That being said, $$$[(m-q)/q]_f$$$ found by the Euclidean algorithm provides the minimum possible $$$\deg P$$$ for a fixed $$$\deg Q = q$$$.

Finding the linear recurrence

In this notion, the minimum recurrence is defined by the first $$$k$$$ such that $$$\deg r_k < \deg q_k$$$ and has a characteristic polynomial $$$q_k$$$.

I have implemented the $$$O(n^2)$$$ algorithm as a part of my polynomial algorithms library:

// Returns the characteristic polynomial
// of the minimum linear recurrence for
// the first d+1 elements of the sequence
poly min_rec(int d = deg()) const {
    // R1 = reversed (Q(x) mod x^{d+1}), R2 = x^{d+1}
    auto R1 = mod_xk(d + 1).reverse(d + 1), R2 = xk(d + 1);
    auto Q1 = poly(T(1)), Q2 = poly(T(0));
    while(!R2.is_zero()) {
        auto [a, nR] = R1.divmod(R2); // R1 = a*R2 + nR, deg nR < deg R2
        tie(R1, R2) = make_tuple(R2, nR);
        tie(Q1, Q2) = make_tuple(Q2, Q1 + a * Q2);
        if(R2.deg() < Q2.deg()) {
            return Q2 / Q2.lead(); // guarantee that the highest coefficient is 1
        }
    }
    assert(0);
}

You can see this Library Judge submission for further details.

Also here is a library-free version, if you prefer it.

Half-GCD algorithm

The notion above provides the basis to construct the $$$O(n \log^2 n)$$$ divide and conquer algorithm of computing $$$\gcd(P, Q)$$$ in polynomials and finding the minimum linear recurrence. I have a truly marvelous demonstration of this proposition that this margin is, unfortunately, too narrow to contain. I hope, I will be able to formalize the process for good and write an article about it sometime...

As a teaser, here's an example of the problem, that (probably) requires Half-GCD:


Library Checker — Inv of Polynomials. You're given $$$f(x)$$$ and $$$h(x)$$$. Compute $$$f^{-1}(x)$$$ modulo $$$h(x)$$$.


Let $$$\frac{f(x)}{h(x)} = [a_0; a_1, \dots, a_k]$$$ and $$$\frac{p_{k-1}}{q_{k-1}} = [a_0; a_1, \dots, a_{k-1}]$$$, then $$$f q_{k-1} - h p_{k-1} = (-1)^{k-2} r_k = (-1)^{k-2} \gcd(f, p)$$$.

Therefore, if $$$r_{k} = \gcd(f, h)$$$ is non-constant, the inverse doesn't exist. Otherwise, the inverse is $$$(-1)^{k-2} q_{k-1}(x)$$$.

In the problem, $$$n \leq 5 \cdot 10^4$$$, therefore you need to do something better than $$$O(n^2)$$$ Euclidean algorithm.

Теги berlekamp-massey, tutorial, linear recurrence, continued fraction

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en45 Английский adamant 2022-07-22 14:30:59 12
en44 Английский adamant 2022-07-22 14:30:30 31
en43 Английский adamant 2022-07-21 21:44:21 7000
en42 Английский adamant 2022-07-21 16:37:28 18
en41 Английский adamant 2022-04-08 17:18:50 140
en40 Английский adamant 2022-04-08 17:17:39 491
en39 Английский adamant 2022-04-08 17:07:37 37
en38 Английский adamant 2022-04-08 17:06:22 21
en37 Английский adamant 2022-04-08 17:04:53 773
en36 Английский adamant 2022-04-08 15:39:20 10
en35 Английский adamant 2022-04-08 15:38:39 5
en34 Английский adamant 2022-04-08 15:27:29 74
en33 Английский adamant 2022-04-08 15:24:39 22
en32 Английский adamant 2022-04-08 15:22:01 5090 explaining on pade approximants, b0=1 and bq=1
en31 Английский adamant 2022-04-07 22:03:19 253 rephrase tldr
en30 Английский adamant 2022-04-06 23:37:11 13
en29 Английский adamant 2022-04-06 22:19:33 334 problem example
en28 Английский adamant 2022-04-06 21:22:31 13
en27 Английский adamant 2022-04-06 21:21:31 105
en26 Английский adamant 2022-04-06 21:14:41 1082 Tldr
en25 Английский adamant 2022-04-06 18:50:52 34
en24 Английский adamant 2022-04-06 18:50:22 114
en23 Английский adamant 2022-04-06 17:23:32 1953
en22 Английский adamant 2022-04-06 17:05:46 1249
en21 Английский adamant 2022-04-06 16:43:26 89
en20 Английский adamant 2022-04-06 16:32:56 838 explanation
en19 Английский adamant 2022-04-06 14:57:04 216 it seems it is not necessarily unique?
en18 Английский adamant 2022-04-06 14:02:54 509 less continued fractions, they probably scare away people
en17 Английский adamant 2022-04-06 13:59:17 85
en16 Английский adamant 2022-04-06 13:57:34 47
en15 Английский adamant 2022-04-06 13:56:19 9
en14 Английский adamant 2022-04-06 13:53:57 1338 finally got the hang of it
en13 Английский adamant 2022-04-06 06:46:22 113 some edge case still need to be considered (published)
en12 Английский adamant 2022-04-06 06:00:12 0 (saved to drafts)
en11 Английский adamant 2022-04-06 03:49:58 15
en10 Английский adamant 2022-04-06 03:41:10 345
en9 Английский adamant 2022-04-06 03:23:06 77
en8 Английский adamant 2022-04-06 03:14:13 5
en7 Английский adamant 2022-04-06 02:13:21 86
en6 Английский adamant 2022-04-06 01:37:01 13
en5 Английский adamant 2022-04-06 01:28:10 57
en4 Английский adamant 2022-04-06 01:23:00 29
en3 Английский adamant 2022-04-06 00:10:26 9 I'm not sure
en2 Английский adamant 2022-04-05 23:29:13 22
en1 Английский adamant 2022-04-05 23:27:31 6370 Initial revision (published)