问题:Python中的模块化乘法逆函数

一些标准的Python模块是否包含用于计算数字(即诸如)的模数乘法逆的函数?Google似乎对此没有任何好的暗示。y = invmod(x, p)x*y == 1 (mod p)

当然,可以提出扩展的欧几里得算法的自酿10线性算法,但是为什么要重新发明轮子呢?

例如,Java的BigIntegerhas modInverse方法。Python没有类似的东西吗?

Does some standard Python module contain a function to compute modular multiplicative inverse of a number, i.e. a number y = invmod(x, p) such that x*y == 1 (mod p)? Google doesn’t seem to give any good hints on this.

Of course, one can come up with home-brewed 10-liner of extended Euclidean algorithm, but why reinvent the wheel.

For example, Java’s BigInteger has modInverse method. Doesn’t Python have something similar?


回答 0

也许有人会觉得这很有用(来自Wikibooks):

def egcd(a, b):
    if a == 0:
        return (b, 0, 1)
    else:
        g, y, x = egcd(b % a, a)
        return (g, x - (b // a) * y, y)

def modinv(a, m):
    g, x, y = egcd(a, m)
    if g != 1:
        raise Exception('modular inverse does not exist')
    else:
        return x % m

Maybe someone will find this useful (from wikibooks):

def egcd(a, b):
    if a == 0:
        return (b, 0, 1)
    else:
        g, y, x = egcd(b % a, a)
        return (g, x - (b // a) * y, y)

def modinv(a, m):
    g, x, y = egcd(a, m)
    if g != 1:
        raise Exception('modular inverse does not exist')
    else:
        return x % m

回答 1

如果您的模数是素数(称为p),则可以简单地计算:

y = x**(p-2) mod p  # Pseudocode

或者在Python中:

y = pow(x, p-2, p)

这是已经在Python中实现了一些数论功能的人:http : //www.math.umbc.edu/~campbell/Computers/Python/numbthy.html

这是在提示符下完成的示例:

m = 1000000007
x = 1234567
y = pow(x,m-2,m)
y
989145189L
x*y
1221166008548163L
x*y % m
1L

If your modulus is prime (you call it p) then you may simply compute:

y = x**(p-2) mod p  # Pseudocode

Or in Python proper:

y = pow(x, p-2, p)

Here is someone who has implemented some number theory capabilities in Python: http://www.math.umbc.edu/~campbell/Computers/Python/numbthy.html

Here is an example done at the prompt:

m = 1000000007
x = 1234567
y = pow(x,m-2,m)
y
989145189L
x*y
1221166008548163L
x*y % m
1L

回答 2

您可能还需要查看gmpy模块。它是Python和GMP多重精度库之间的接口。gmpy提供了一个invert函数,可以完全满足您的需求:

>>> import gmpy
>>> gmpy.invert(1234567, 1000000007)
mpz(989145189)

更新的答案

如@hyh所示,gmpy.invert()如果不存在逆,则返回0。符合GMP mpz_invert()功能的行为。gmpy.divm(a, b, m)提供的一般解决方案a=bx (mod m)

>>> gmpy.divm(1, 1234567, 1000000007)
mpz(989145189)
>>> gmpy.divm(1, 0, 5)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ZeroDivisionError: not invertible
>>> gmpy.divm(1, 4, 8)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ZeroDivisionError: not invertible
>>> gmpy.divm(1, 4, 9)
mpz(7)

divm()gcd(b,m) == 1不存在乘法逆时,将返回一个解决方案,并引发异常。

免责声明:我是gmpy库的当前维护者。

更新的答案2

现在,当反函数不存在时,gmpy2会正确引发一个异常:

>>> import gmpy2

>>> gmpy2.invert(0,5)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ZeroDivisionError: invert() no inverse exists

You might also want to look at the gmpy module. It is an interface between Python and the GMP multiple-precision library. gmpy provides an invert function that does exactly what you need:

>>> import gmpy
>>> gmpy.invert(1234567, 1000000007)
mpz(989145189)

Updated answer

As noted by @hyh , the gmpy.invert() returns 0 if the inverse does not exist. That matches the behavior of GMP’s mpz_invert() function. gmpy.divm(a, b, m) provides a general solution to a=bx (mod m).

>>> gmpy.divm(1, 1234567, 1000000007)
mpz(989145189)
>>> gmpy.divm(1, 0, 5)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ZeroDivisionError: not invertible
>>> gmpy.divm(1, 4, 8)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ZeroDivisionError: not invertible
>>> gmpy.divm(1, 4, 9)
mpz(7)

divm() will return a solution when gcd(b,m) == 1 and raises an exception when the multiplicative inverse does not exist.

Disclaimer: I’m the current maintainer of the gmpy library.

Updated answer 2

gmpy2 now properly raises an exception when the inverse does not exists:

>>> import gmpy2

>>> gmpy2.invert(0,5)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ZeroDivisionError: invert() no inverse exists

回答 3

从3.8开始,python的pow()函数可以采用一个模数和一个负整数。看这里。他们如何使用它的情况是

>>> pow(38, -1, 97)
23
>>> 23 * 38 % 97 == 1
True

As of 3.8 pythons pow() function can take a modulus and a negative integer. See here. Their case for how to use it is

>>> pow(38, -1, 97)
23
>>> 23 * 38 % 97 == 1
True

回答 4

这是CodeFights的一线工具;它是最短的解决方案之一:

MMI = lambda A, n,s=1,t=0,N=0: (n < 2 and t%N or MMI(n, A%n, t, s-A//n*t, N or n),-1)[n<1]

-1如果A在中没有乘法逆,它将返回n

用法:

MMI(23, 99) # returns 56
MMI(18, 24) # return -1

该解决方案使用扩展的欧几里得算法

Here is a one-liner for CodeFights; it is one of the shortest solutions:

MMI = lambda A, n,s=1,t=0,N=0: (n < 2 and t%N or MMI(n, A%n, t, s-A//n*t, N or n),-1)[n<1]

It will return -1 if A has no multiplicative inverse in n.

Usage:

MMI(23, 99) # returns 56
MMI(18, 24) # return -1

The solution uses the Extended Euclidean Algorithm.


回答 5

Sympy,为象征性的数学Python模块,具有内置式模块反函数,如果你不希望实现自己的(或者,如果您已经在使用Sympy):

from sympy import mod_inverse

mod_inverse(11, 35) # returns 16
mod_inverse(15, 35) # raises ValueError: 'inverse of 15 (mod 35) does not exist'

这似乎没有在Sympy网站上记录,但这里是文档字符串:Github上的Sympy mod_inverse文档字符串

Sympy, a python module for symbolic mathematics, has a built-in modular inverse function if you don’t want to implement your own (or if you’re using Sympy already):

from sympy import mod_inverse

mod_inverse(11, 35) # returns 16
mod_inverse(15, 35) # raises ValueError: 'inverse of 15 (mod 35) does not exist'

This doesn’t seem to be documented on the Sympy website, but here’s the docstring: Sympy mod_inverse docstring on Github


回答 6

这是我的代码,可能有点草率,但是无论如何它似乎对我有用。

# a is the number you want the inverse for
# b is the modulus

def mod_inverse(a, b):
    r = -1
    B = b
    A = a
    eq_set = []
    full_set = []
    mod_set = []

    #euclid's algorithm
    while r!=1 and r!=0:
        r = b%a
        q = b//a
        eq_set = [r, b, a, q*-1]
        b = a
        a = r
        full_set.append(eq_set)

    for i in range(0, 4):
        mod_set.append(full_set[-1][i])

    mod_set.insert(2, 1)
    counter = 0

    #extended euclid's algorithm
    for i in range(1, len(full_set)):
        if counter%2 == 0:
            mod_set[2] = full_set[-1*(i+1)][3]*mod_set[4]+mod_set[2]
            mod_set[3] = full_set[-1*(i+1)][1]

        elif counter%2 != 0:
            mod_set[4] = full_set[-1*(i+1)][3]*mod_set[2]+mod_set[4]
            mod_set[1] = full_set[-1*(i+1)][1]

        counter += 1

    if mod_set[3] == B:
        return mod_set[2]%B
    return mod_set[4]%B

Here is my code, it might be sloppy but it seems to work for me anyway.

# a is the number you want the inverse for
# b is the modulus

def mod_inverse(a, b):
    r = -1
    B = b
    A = a
    eq_set = []
    full_set = []
    mod_set = []

    #euclid's algorithm
    while r!=1 and r!=0:
        r = b%a
        q = b//a
        eq_set = [r, b, a, q*-1]
        b = a
        a = r
        full_set.append(eq_set)

    for i in range(0, 4):
        mod_set.append(full_set[-1][i])

    mod_set.insert(2, 1)
    counter = 0

    #extended euclid's algorithm
    for i in range(1, len(full_set)):
        if counter%2 == 0:
            mod_set[2] = full_set[-1*(i+1)][3]*mod_set[4]+mod_set[2]
            mod_set[3] = full_set[-1*(i+1)][1]

        elif counter%2 != 0:
            mod_set[4] = full_set[-1*(i+1)][3]*mod_set[2]+mod_set[4]
            mod_set[1] = full_set[-1*(i+1)][1]

        counter += 1

    if mod_set[3] == B:
        return mod_set[2]%B
    return mod_set[4]%B

回答 7

上面的代码无法在python3中运行,并且与GCD变体相比效率较低。但是,此代码非常透明。它触发了我创建一个更紧凑的版本:

def imod(a, n):
 c = 1
 while (c % a > 0):
     c += n
 return c // a

The code above will not run in python3 and is less efficient compared to the GCD variants. However, this code is very transparent. It triggered me to create a more compact version:

def imod(a, n):
 c = 1
 while (c % a > 0):
     c += n
 return c // a

回答 8

这是一个简洁的1-liner,无需使用任何外部库即可执行此操作。

# Given 0<a<b, returns the unique c such that 0<c<b and a*c == gcd(a,b) (mod b).
# In particular, if a,b are relatively prime, returns the inverse of a modulo b.
def invmod(a,b): return 0 if a==0 else 1 if b%a==0 else b - invmod(b%a,a)*b//a

请注意,这实际上只是egcd,经过精简后仅返回单个感兴趣的系数。

Here is a concise 1-liner that does it, without using any external libraries.

# Given 0<a<b, returns the unique c such that 0<c<b and a*c == gcd(a,b) (mod b).
# In particular, if a,b are relatively prime, returns the inverse of a modulo b.
def invmod(a,b): return 0 if a==0 else 1 if b%a==0 else b - invmod(b%a,a)*b//a

Note that this is really just egcd, streamlined to return only the single coefficient of interest.


回答 9

为了弄清楚模块化乘法逆,我建议使用扩展欧几里得算法,如下所示:

def multiplicative_inverse(a, b):
    origA = a
    X = 0
    prevX = 1
    Y = 1
    prevY = 0
    while b != 0:
        temp = b
        quotient = a/b
        b = a%b
        a = temp
        temp = X
        a = prevX - quotient * X
        prevX = temp
        temp = Y
        Y = prevY - quotient * Y
        prevY = temp

    return origA + prevY

To figure out the modular multiplicative inverse I recommend using the Extended Euclidean Algorithm like this:

def multiplicative_inverse(a, b):
    origA = a
    X = 0
    prevX = 1
    Y = 1
    prevY = 0
    while b != 0:
        temp = b
        quotient = a/b
        b = a%b
        a = temp
        temp = X
        a = prevX - quotient * X
        prevX = temp
        temp = Y
        Y = prevY - quotient * Y
        prevY = temp

    return origA + prevY

回答 10

我从该线程尝试了不同的解决方案,最后我使用了一个:

def egcd(a, b):
    lastremainder, remainder = abs(a), abs(b)
    x, lastx, y, lasty = 0, 1, 1, 0
    while remainder:
        lastremainder, (quotient, remainder) = remainder, divmod(lastremainder, remainder)
        x, lastx = lastx - quotient*x, x
        y, lasty = lasty - quotient*y, y
    return lastremainder, lastx * (-1 if a < 0 else 1), lasty * (-1 if b < 0 else 1)


def modinv(a, m):
    g, x, y = self.egcd(a, m)
    if g != 1:
        raise ValueError('modinv for {} does not exist'.format(a))
    return x % m

Python中的Modular_inverse

I try different solutions from this thread and in the end I use this one:

def egcd(a, b):
    lastremainder, remainder = abs(a), abs(b)
    x, lastx, y, lasty = 0, 1, 1, 0
    while remainder:
        lastremainder, (quotient, remainder) = remainder, divmod(lastremainder, remainder)
        x, lastx = lastx - quotient*x, x
        y, lasty = lasty - quotient*y, y
    return lastremainder, lastx * (-1 if a < 0 else 1), lasty * (-1 if b < 0 else 1)


def modinv(a, m):
    g, x, y = self.egcd(a, m)
    if g != 1:
        raise ValueError('modinv for {} does not exist'.format(a))
    return x % m

Modular_inverse in Python


回答 11

好吧,我在python中没有函数,但是我在C中具有可以轻松转换为python的函数,在下面的c函数中,扩展的欧几里得算法用于计算逆mod。

int imod(int a,int n){
int c,i=1;
while(1){
    c = n * i + 1;
    if(c%a==0){
        c = c/a;
        break;
    }
    i++;
}
return c;}

Python函数

def imod(a,n):
  i=1
  while True:
    c = n * i + 1;
    if(c%a==0):
      c = c/a
      break;
    i = i+1

  return c

从下面的链接C程序中引用了上面的C函数,以找到两个相对质数的模乘逆

Well, I don’t have a function in python but I have a function in C which you can easily convert to python, in the below c function extended euclidian algorithm is used to calculate inverse mod.

int imod(int a,int n){
int c,i=1;
while(1){
    c = n * i + 1;
    if(c%a==0){
        c = c/a;
        break;
    }
    i++;
}
return c;}

Python Function

def imod(a,n):
  i=1
  while True:
    c = n * i + 1;
    if(c%a==0):
      c = c/a
      break;
    i = i+1

  return c

Reference to the above C function is taken from the following link C program to find Modular Multiplicative Inverse of two Relatively Prime Numbers


回答 12

从cpython实现源代码

def invmod(a, n):
    b, c = 1, 0
    while n:
        q, r = divmod(a, n)
        a, b, c, n = n, c, b - q*c, r
    # at this point a is the gcd of the original inputs
    if a == 1:
        return b
    raise ValueError("Not invertible")

根据此代码上方的注释,它可以返回小的负值,因此您可以潜在地检查是否为负,并在返回b之前将n添加为负。

from the cpython implementation source code:

def invmod(a, n):
    b, c = 1, 0
    while n:
        q, r = divmod(a, n)
        a, b, c, n = n, c, b - q*c, r
    # at this point a is the gcd of the original inputs
    if a == 1:
        return b
    raise ValueError("Not invertible")

according to the comment above this code, it can return small negative values, so you could potentially check if negative and add n when negative before returning b.


回答 13

截至2017年1月23日,上面的许多链接已断开。我找到了这个实现:https : //courses.csail.mit.edu/6.857/2016/files/ffield.py

Many of the links above are broken as for 1/23/2017. I found this implementation: https://courses.csail.mit.edu/6.857/2016/files/ffield.py


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