def random_polynomial(n): a = [random.randint(0, 100) for i in range(n)] x = random.randint(0, 100) y = a[0] for i in range(1, n): y = y * x + a[i] return y
时间: 2024-04-28 09:23:07 浏览: 15
这是一个随机生成多项式的函数,函数的输入是一个整数n,代表多项式的阶数,输出是在x取一个随机整数之后,根据多项式计算得到的y值。在函数中,a是一个长度为n的随机整数数组,x是一个随机整数,y的初始值为a[0],然后通过循环计算得到y的值。具体来说,对于每个i,都将y乘以x并加上a[i],最终得到的y就是多项式在x处的值。
相关问题
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:
```python
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:
```python
# 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。
基于python代码,实现500个节点进行随机多项式,计算运行时间和所占内存,然后将节点分区之后,再次计算得出时间和内存的对比结果,并绘图
以下是一个实现上述需求的 Python 代码示例:
```python
import random
import time
import psutil
import matplotlib.pyplot as plt
def random_polynomial(n):
a = [random.randint(0, 100) for i in range(n)]
x = random.randint(0, 100)
y = a[0]
for i in range(1, n):
y = y * x + a[i]
return y
def test_performance(n):
start_time = time.time()
start_memory = psutil.Process().memory_info().rss / 1024 / 1024
y = random_polynomial(n)
end_time = time.time()
end_memory = psutil.Process().memory_info().rss / 1024 / 1024
return (end_time - start_time, end_memory - start_memory)
def partition(lst, n):
return [lst[i:i + n] for i in range(0, len(lst), n)]
def test_performance_partitioned(n, k):
lst = [random.randint(0, 100) for i in range(n)]
partitions = partition(lst, n // k)
start_time = time.time()
start_memory = psutil.Process().memory_info().rss / 1024 / 1024
for p in partitions:
y = random_polynomial(len(p))
end_time = time.time()
end_memory = psutil.Process().memory_info().rss / 1024 / 1024
return (end_time - start_time, end_memory - start_memory)
n = 500
k = 10
time1, memory1 = test_performance(n)
time2, memory2 = test_performance_partitioned(n, k)
print(f"Performance without partitioning: time = {time1:.6f}s, memory = {memory1:.6f}MB")
print(f"Performance with partitioning: time = {time2:.6f}s, memory = {memory2:.6f}MB")
plt.bar(["Without partitioning", "With partitioning"], [time1, time2])
plt.title("Performance comparison")
plt.xlabel("Method")
plt.ylabel("Time (s)")
plt.show()
plt.bar(["Without partitioning", "With partitioning"], [memory1, memory2])
plt.title("Memory usage comparison")
plt.xlabel("Method")
plt.ylabel("Memory (MB)")
plt.show()
```
这段代码实现了以下功能:
1. 定义了一个 `random_polynomial` 函数,用于生成随机多项式并计算其值。
2. 定义了一个 `test_performance` 函数,用于测试生成随机多项式的时间和内存占用。
3. 定义了一个 `partition` 函数,用于将一个列表分成若干个子列表。
4. 定义了一个 `test_performance_partitioned` 函数,用于测试将节点分区后生成随机多项式的时间和内存占用。
5. 对于给定的节点数 `n` 和分区数 `k`,分别计算未分区和已分区的随机多项式生成时间和内存占用,并输出结果。
6. 使用 Matplotlib 库绘制了两个柱状图,分别展示未分区和已分区的时间和内存占用对比结果。
需要说明的是,由于随机多项式的生成是一个随机过程,因此每次运行程序得到的时间和内存占用结果可能会有所不同。
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