使用 function 的遞歸歐幾里得 GCD 算法

Question

我在問我必須顯示的遞歸部分（在 function 中，還有另一個參數verbose: bool ）。 我需要計算兩個數字的 gcd。 但是如何顯示商和 rest 和操作？

像這樣：

By setting q the quotient of the integer division of a by b, let us calculate for example
the gcd of 90 and of 35 (gcd (90,35), so a = 90 and b = 35):
  a = b × q + r
As for 90/35 we have q = 2 and r = 20, then 90 = 35 × 2 + 20
According to Euclid, pgcd (90.35) = pgcd (35.20). So we start over with the new values:
  35 = 20 × 1 + 15
According to Euclid, pgcd (35.20) = pgcd (20.15). So we start over with the new values:
  20 = 15 × 1 + 5
According to Euclid, pgcd (20.15) = pgcd (15.5). So we start over with the new values:
  15 = 5 × 3 + 0
15 is divisible by 5 so we stop and pgcd (15.5) = pgcd (20.15) = pgcd (35.20) =
pgcd (90.35) = 5

Write the code of the euclid function (a: int, b: int, verbose: bool = False) -> int which
calculates the gcd of a and b and, if the verbose parameter is True, displays the
different steps of the algorithm of the calculation (the values ​​of a, b, q and r).

我試過這個，但這是完整的，我對此一無所知。 我什至不知道什么是什么順序。

我試過的：

def gcd(a,b, verbose = False):
    r = a%b
    q:int

    if r == 0:
        q = a//b
        print(b)
        return b
    else:
        verbose = True
        while r != 0:
            result = b*q+r
            a= result
            b = result
            r = b
        return result

num1 = int(input('chose a number'))
num2 = int(input('chose a second number'))

print(gcd(num1,num2, verbose = False))

這是 output：

“分配前”是的，但根據問題，我不知道把q放在哪里，除了if r== 0 then q = a//b 。

And I don't know how to make the recursive part how I'm supposed to loop on a = b*q + r and say that if r is not 0 then a is b and b is r and to do it until r is 0並打印 gcd，如果verbose為True我必須描述每個計算並打印a 、 b 、 r和q 。

Answer 1

您可以通過以下方式使用遞歸方法執行此操作：

def gcd(a,b,verbose=False):
    if b>a:
        return gcd(b,a,verbose)
    r=a%b
    if r==0:
        if verbose==True:
            q=a//b
            print("a=",a,"b=",b,"q=",q,"r=",r)
        return b
    else:
        if verbose==True:
            q=a//b
            print("a=",a,"b=",b,"q=",q,"r=",r)
        return gcd(b,r,verbose)

Answer 2

您的分配並沒有說您必須使用遞歸，事實上，如果您不使用並使用迭代（即循環）來代替它會更簡單。

這就是我的意思：

def gcd(a: int, b: int, verbose: bool = False) -> int:
    while b != 0:
        if verbose:
            print(f'{a=}, {b=}')
        q = a // b
        r = a % b
        a = b
        b = r
    return a

#num1 = int(input('chose a number'))
#num2 = int(input('chose a second number'))
num1, num2 = 90, 35  # Hardcode for testing.

Output：

a=90, b=35
a=35, b=20
a=20, b=15
a=15, b=5
gcd(num1, num2, verbose=True)=5