连分数和 Pell 方程 - 数值问题

Question

数学背景

Continued fractions are a way to represent numbers (rational or not), with a basic recursion formula 计算一下。给定一个数字 r，我们定义 r[0]=r 并且有：

for n in range(0..N):
    a[n] = floor(r[n])
    if r[n] == [an]: break
    r[n+1] = 1 / (r[n]-a[n])

其中 a 是最终表示。我们还可以通过

定义一系列convergents

h[-2,-1] = [0, 1]
k[-2, -1] = [1, 0]
h[n] = a[n]*h[n-1]+h[n-2]
k[n] = a[n]*k[n-1]+k[n-2]

其中 h[n]/k[n] 收敛于 r.

Pell's equation is a problem of the form x^2-D*y^2=1 where all numbers are integers and D is not a perfect square in our case. A solution for a given D that minimizes x is given by continued fractions。基本上，对于上面的等式，保证这个（基本）解是 x=h[n] 和 y=k[n] 找到的最低 n 解决了 [=57 的连分数扩展中的方程=]sqrt(D).

问题

我无法让这个简单的算法适用于 D=61。我首先注意到它没有解决 100 个系数的 Pell 方程，所以我将它与 Wolfram Alpha 的 convergents and continued fraction representation 进行了比较，并注意到第 20 个元素失败了 - 与我得到的 4 相比，表示形式是 3 ，产生不同的收敛性 - h[20]=335159612 在 Wolfram 上与 425680601 相比对我而言。

我在两个系统上测试了下面的代码，两种语言（虽然公平地说，Python 是我猜的幕后 C），并得到相同的结果 - 循环 20 上的差异。我'你会注意到收敛仍然是准确和收敛的！ 为什么我得到的结果与 Wolfram Alpha 不同，是否可以修复？

为了测试，这里有一个 Python 程序来求解 D=61 的 Pell 方程，打印前 20 个收敛点和连分数表示 cf（以及一些额外的不需要的绒毛）：

from math import floor, sqrt  # Can use mpmath here as well.


def continued_fraction(D, count=100, thresh=1E-12, verbose=False):
    cf = []
    h = (0, 1)
    k = (1, 0)
    r = start = sqrt(D)
    initial_count = count
    x = (1+thresh+start)*start
    y = start
    while abs(x/y - start) > thresh and count:
        i = int(floor(r))
        cf.append(i)
        f = r - i
        x, y = i*h[-1] + h[-2], i*k[-1] + k[-2]
        if verbose is True or verbose == initial_count-count:
            print(f'{x}\u00B2-{D}x{y}\u00B2 = {x**2-D*y**2}')
        if x**2 - D*y**2 == 1:
            print(f'{x}\u00B2-{D}x{y}\u00B2 = {x**2-D*y**2}')
            print(cf)
            return
        count -= 1
        r = 1/f
        h = (h[1], x)
        k = (k[1], y)

    print(cf)
    raise OverflowError(f"Converged on {x} {y} with count {count} and diff {abs(start-x/y)}!")

continued_fraction(61, count=20, verbose=True, thresh=-1)  # We don't want to stop on account of thresh in this example

一个c程序做同样的事情：

#include<stdio.h>
#include<math.h>
#include<stdlib.h>

int main() {
    long D = 61;
    double start = sqrt(D);
    long h[] = {0, 1};
    long k[] = {1, 0};
    int count = 20;
    float thresh = 1E-12;
    double r = start;
    long x = (1+thresh+start)*start;
    long y = start;
    while(abs(x/(double)y-start) > -1 && count) {
        long i = floor(r);
        double f = r - i;
        x = i * h[1] + h[0];
        y = i * k[1] + k[0];
        printf("%ld\u00B2-%ldx%ld\u00B2 = %lf\n", x, D, y, x*x-D*y*y);
        r = 1/f;
        --count;
        h[0] = h[1];
        h[1] = x;
        k[0] = k[1];
        k[1] = y;
    }
    return 0;
}

Answer 1

mpmath,python的多精度库可以使用。请注意所有重要的数字都是 mp 格式。

在下面的代码中，x, y and i 是标准的多精度整数。 r 和 f 是多精度实数。请注意，初始计数设置为高于 20。

from mpmath import mp, mpf

mp.dps = 50  # precision in number of decimal digits

def continued_fraction(D, count=22, thresh=mpf(1E-12), verbose=False):
    cf = []
    h = (0, 1)
    k = (1, 0)
    r = start = mp.sqrt(D)
    initial_count = count
    x = 0 # some dummy starting values, they will be overwritten early in the while loop
    y = 1
    while abs(x/y - start) > thresh and count > 0:
        i = int(mp.floor(r))
        cf.append(i)
        x, y = i*h[-1] + h[-2], i*k[-1] + k[-2]
        if verbose or initial_count == count:
            print(f'{x}\u00B2-{D}x{y}\u00B2 = {x**2-D*y**2}')
        if x**2 - D*y**2 == 1:
            print(f'{x}\u00B2-{D}x{y}\u00B2 = {x**2-D*y**2}')
            print(cf)
            return
        count -= 1
        f = r - i
        r = 1/f
        h = (h[1], x)
        k = (k[1], y)

    print(cf)
    raise OverflowError(f"Converged on {x} {y} with count {count} and diff {abs(start-x/y)}!")

continued_fraction(61, count=22, verbose=True, thresh=mpf(1e-100))

输出类似于 wolfram 的：

...
335159612²-61x42912791² = 3
1431159437²-61x183241189² = -12
1766319049²-61x226153980² = 1
[7, 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14, 1, 4, 3, 1, 2, 2, 1, 3, 4, 1]

连分数和 Pell 方程 - 数值问题

Continued fractions and Pell's equation - numerical issues

math

precision

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