Recovering a linear recurrence with the extended Euclidean algorithm

Revision en18, by adamant, 2022-04-06 14:02:54

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.

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 $$$d = \deg Q(x)$$$ is the minimum possible. In other words, what we're asked to do is to find 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

$$$ R(x) = \frac{a_0 + a_1 x + \dots + a_p x^p}{1+b_1 x + \dots + b_q x^q}=\frac{P(x)}{Q(x)} $$$

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


By definition, for Padé approximants generally holds

$$$ [p/q]_f \equiv [(p+k)/(q+l)]_f \pmod {x^{p+q+1}}. $$$

From this follows that the generating function of the linear recurrence is also the Padé approximant $$$[\deg P / \deg Q]_F$$$. Moreover, since

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

it also holds that $$$\frac{P(x)}{Q(x)} = [(\deg P + \alpha)/(\deg Q + \beta)]_F$$$ for any $$$\alpha + \beta \leq m - \deg P - \deg Q$$$.

Therefore, the needed fraction $$$\frac{P(x)}{Q(x)}$$$ can be found among $$$[1/m]_F, [2/(m-1)]_F,\dots,[m/1]_F$$$.

Extended Euclidean algorithm

Let's look again on the condition

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

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 $$$p+q=m$$$.

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

$$$ r_{k+1} = r_{k-1} \mod r_{k} = r_{k-1} - a_k r_k. $$$

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

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

Note that $$$\deg r_k$$$ monotonously decreases until $$$r_k=0$$$ while $$$\deg q_k$$$ monotonously increases until $$$q_k = \frac{B(x)}{\gcd(B(x), x^{m+1})}$$$.

Moreover, from the way the transition formulas were defined, we can assert that

$$$ \deg q_{k+1} - \deg q_k = a_k = \deg r_{k-1} - \deg r_k. $$$

From this follows that 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

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

Note that it is not guaranteed that $$$q_k(0) = 1$$$. Moreover, it is possible that $$$q_k(0)=0$$$. This is due to the fact that with the Euclidean algorithm we can guarantee the normal form of the denominator

$$$ Q(x) = x^n - \sum\limits_{k=1}^d a_k x^{n-k} $$$

rather than

$$$ Q(x) = 1 - \sum\limits_{k=1}^d a_k x^k. $$$

To circumvent this, one should execute the algorithm for $$$x^m F(x^{-1}) = F_{m} + F_{m-1} x + \dots + F_0 x^m$$$ and reverse the resulting $$$Q(x)$$$ in the end. In this notion, the minimum recurrence is defined by the first $$$k$$$ such that $$$\deg r_k < \deg q_k$$$.

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.

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...

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

History

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