帮我用python写出N值为41的NTRUsign签名算法，函数自己编写

好的，以下是一个N值为41的NTRUSign签名算法的Python实现示例代码：

import random

# NTRUSign签名算法实现
class NTRUSign:
    # 参数定义
    class Params:
        def __init__(self):
            self.N = 41    # N值
            self.q = 2048  # 模数
            self.p = 47    # 置换多项式的系数
            self.d = 7     # 多项式f和g的度数
            self.nu = 3    # 噪声的上界
            self.beta = 1  # 公钥中的参数

        # 生成密钥对
        def generate_key_pair(self):
            # 生成随机多项式f和g
            f = self.generate_random_poly()
            g = self.generate_random_poly()

            # 计算私钥
            h = f.invert(g, self.q)
            priv_key = (f, g, h)

            # 计算公钥
            f_inv = f.invert()
            pub_key = (self.beta * f_inv * g) % self.q

            return priv_key, pub_key

        # 生成随机多项式
        def generate_random_poly(self):
            coeffs = [random.randint(-self.nu, self.nu) for i in range(self.d)]
            return Poly(coeffs, self.q)

    # 多项式类
    class Poly:
        def __init__(self, coeffs, q):
            self.coeffs = coeffs
            self.q = q

        # 多项式加法
        def __add__(self, other):
            coeffs = [(self.coeffs[i] + other.coeffs[i]) % self.q
                      for i in range(len(self.coeffs))]
            return Poly(coeffs, self.q)

        # 多项式减法
        def __sub__(self, other):
            coeffs = [(self.coeffs[i] - other.coeffs[i]) % self.q
                      for i in range(len(self.coeffs))]
            return Poly(coeffs, self.q)

        # 多项式乘法
        def __mul__(self, other):
            coeffs = [0] * (len(self.coeffs) + len(other.coeffs) - 1)
            for i in range(len(self.coeffs)):
                for j in range(len(other.coeffs)):
                    coeffs[i+j] += (self.coeffs[i] * other.coeffs[j])
            coeffs = [c % self.q for c in coeffs]
            return Poly(coeffs, self.q)

        # 多项式取反
        def __neg__(self):
            coeffs = [-c % self.q for c in self.coeffs]
            return Poly(coeffs, self.q)

        # 多项式求逆
        def invert(self, mod_poly=None):
            if not mod_poly:
                mod_poly = Poly([1, 0, 1], self.q)
            r = self
            t = Poly([0], self.q)
            new_t = Poly([1], self.q)
            while not r.is_zero():
                q, _ = divmod(mod_poly, r)
                t, new_t = new_t, t - q * new_t
                r, mod_poly = mod_poly % r, r
            if mod_poly.coeffs[0] != 1:
                raise ZeroDivisionError("Polynomial not invertible")
            return t * mod_poly.invert()

        # 多项式是否为0
        def is_zero(self):
            return all(c == 0 for c in self.coeffs)

        # 多项式转为字节串
        def to_bytes(self):
            return b"".join([c.to_bytes(2, byteorder="big") for c in self.coeffs])

        # 从字节串中还原多项式
        @classmethod
        def from_bytes(cls, data, q):
            coeffs = [int.from_bytes(data[i:i+2], byteorder="big", signed=True) % q
                      for i in range(0, len(data), 2)]
            return cls(coeffs, q)

    # 签名
    @staticmethod
    def sign(priv_key, message):
        f, g, h = priv_key
        r = NTRUSign.Params().generate_random_poly()
        e = NTRUSign.Params().generate_random_poly()
        s = (h * r + e - f * Poly(list(message), NTRUSign.Params().q)) % NTRUSign.Params().q
        return r.to_bytes() + s.to_bytes()

    # 验证签名
    @staticmethod
    def verify(pub_key, message, signature):
        f_inv = Poly([NTRUSign.Params().beta], NTRUSign.Params().q) * pub_key
        r = Poly.from_bytes(signature[:84], NTRUSign.Params().q)
        s = Poly.from_bytes(signature[84:], NTRUSign.Params().q)
        m_calc = (f_inv * s + r).coeffs[0:NTRUSign.Params().d]
        return list(message) == m_calc

使用示例：

# 生成密钥对
params = NTRUSign.Params()
priv_key, pub_key = params.generate_key_pair()

# 签名
message = b"Hello, world!"
signature = NTRUSign.sign(priv_key, message)

# 验证签名
is_valid = NTRUSign.verify(pub_key, message, signature)
print("Signature is valid:", is_valid)

需要注意的是，此代码仅作为示例，实际使用中需要根据具体情况进行修改和完善。