Question

对于一个大学项目，我想实现Bellard算法来计算Fortran中pi的n位数字。我在math.stackexchange上偶然发现了这个问题：https://math.stackexchange.com/questions/1776840/confusion-with-bellards-algorithm-for-pi

有了这个问题的答案，我设法实现了代码，但是我没有得到结果，也弄不清楚我在做什么错了：

PROGRAM pi

IMPLICIT NONE

INTEGER :: find_number_of_primes, number_primes, last_digit, digit, N, &
eps, i, j, multiplicity, test
REAL :: log2, base, v_max, m, s, v, b, A, pi_sum
INTEGER, ALLOCATABLE :: primes(:)   

digit = 3
eps = 2
base = 10
N = INT((digit+eps)*log2(base))
pi_sum = 0

number_primes = find_number_of_primes(2*N)
ALLOCATE(primes(number_primes))
CALL get_primes(number_primes, primes)

DO i=1,SIZE(primes)
    v_max = INT(log(REAL(2*N))/log(REAL(primes(i))))
    m = primes(i)**v_max
    s = 0
    v = 0
    b = 1
    A = 1
    DO j = 1,(N)
        b = MOD((j/(primes(i)**multiplicity(digit, j)) * b), m)
        A = MOD((2*j-1)/(primes(i)**multiplicity(primes(i), (2*j-1)))*A, m)
        v = v - multiplicity(primes(i),j)+multiplicity(primes(i),(2*j-1))
        IF (v > 0) THEN
            s = MOD(s*j*b*(A**(-1))*a**(v_max-v), m)
        END IF
    END DO
    s = MOD(s * base**(digit-1), m)
    pi_sum = pi_sum + MOD((s/m), REAL(1))
END DO

PRINT*, pi_sum  

END PROGRAM

自定义功能find_number_of_primes，get_primes，log2和multiplicity可以正常工作。第一个查找给定间隔中的质数，第二个返回数组中的质数，第三个计算

log_2(x) = log(x)/log(2)

，最后一个通过检查第二个数字除以第一个数字的频率，直到除数的其余部分不再为零，来计算多重性（我想这就是所谓的）。这是源代码：

对数：

real function log2(x)
implicit none
real, intent(in) :: x

log2 = log(x) / log(2.)
end function

查找间隔中的质数：

integer function find_number_of_primes(limit) result(number_primes)
implicit none
INTEGER, INTENT(IN) :: limit
INTEGER :: i, j
LOGICAL :: is_prime

number_primes = 0

DO i = 3,(limit-1)
    is_prime = .TRUE.
    DO j = 2, (i-1)
        IF (MODULO(i, j) == 0) THEN
            is_prime = .FALSE.
            EXIT
        END IF
    END DO
    IF (is_prime .EQV. .TRUE.) THEN
    number_primes = number_primes + 1
    END IF
END DO
end function find_number_of_primes

获取实际质数：

SUBROUTINE get_primes(number_primes, primes)
IMPLICIT NONE
INTEGER, INTENT(IN) :: number_primes
INTEGER :: i, found_primes, j
LOGICAL:: is_prime
INTEGER, INTENT(OUT) :: primes(number_primes)
i = 3
found_primes = 0
DO
is_prime = .TRUE.
DO j = 2, (i-1)
    IF (MODULO(i,j) == 0) THEN
        is_prime = .FALSE.
    END IF
END DO
IF (is_prime .EQV. .TRUE.) THEN
    found_primes = found_primes + 1
    primes(found_primes) = i
    IF (found_primes == number_primes) EXIT
END IF
i = i + 1
END DO


end SUBROUTINE get_primes

integer function multiplicity(a, b)
implicit none
INTEGER, INTENT(IN) :: a, b

multiplicity = 0
DO
    multiplicity = multiplicity + 1
    IF (MOD(b, (a**(multiplicity+1))) /= 0) THEN
        EXIT
    END IF
END DO
end function multiplicity

整个文件的废纸:： https://pastebin.com/0F4zQYaH

现在我链接的问题中，在内环中 b 的计算中， a ^ {v（n，k）} 分母，但在回答问题时，分母中的术语是 a ^ {v（a，k）} 。此外，问题的内循环从 1 到 N ，而答案的循环从 1 到 2N 。

最终结果是NaN，这是digit = 2和eps = 1的一些控制台输出：

************ i =           1 ***************************
 v_max =   2.00000000
 m =   9.00000000
 ------------- j =           1 -------------------------------
 b =   0.00000000
 A =   0.00000000
 v =   0.00000000
 ------------- j =           2 -------------------------------
 b =   0.00000000
 A =   0.00000000
 v =   0.00000000
 ------------- j =           3 -------------------------------
 b =   0.00000000
 A =   0.00000000
 v =   0.00000000
 ------------- j =           4 -------------------------------
 b =   0.00000000
 A =   0.00000000
 v =   0.00000000
 ------------- j =           5 -------------------------------
 b =   0.00000000
 A =   0.00000000
 v =   1.00000000
 v > 0 --> s =              NaN
 ------------- j =           6 -------------------------------
 b =   0.00000000
 A =   0.00000000
 v =   1.00000000
 v > 0 --> s =              NaN

一个完整的输出：https://pastebin.com/ucJNg6Vi

有人可以帮我弄清楚我在做什么错吗？

Answer 1

您的代码中有一个错误：

v = v - multiplicity(primes(i),j)+multiplicity(primes(i),(2*j-1))

应该是

v = v - multiplicity(primes(i),j)- multiplicity(primes(i),(2*j-1))

还

        s = MOD(s*j*b*(A**(-1))*a**(v_max-v), m)

应该是

        s = MOD(s + j*b*(A**(-1))*a**(v_max-v), m)

Bellard计算Pi的算法

1 个答案: