LSFR题目3级线性反馈移位寄存器在c31时有4种线性反馈函数,设其初始状态为(a1,a2,a3)(1,0,1),求各线性反馈函数的输出序列及周期,输出他们的密钥流,并利用生成的密钥流对明文0x0123456789ABCDEF进行加密,输出加密后的结果,再对密文进行解密,输出解密后的结果。已知寄存器长度n 3状态(a1, a2, a3)初始状态(1, 0, 1)条件c3 1LFSR 的反馈函数是fc1a1⊕c2a2⊕c3a3。因为c31所以fc1a1⊕c2a2⊕a3因为c1∈{0,1} c2∈{0,1}所以一共有4种情况对应4 个反馈函数。编号(c1,c2,c3)反馈函数周期1(0,0,1)a3很短2(0,1,1)a2⊕a3较短3(1,0,1)a1⊕a3可能最大4(1,1,1)a1⊕a2⊕a3最大周期7m序列状态更新原则(a1,a2,a3)→(a2,a3,f)每一轮输出值输出值为a1def lfsr_step(state, c1, c2):a1, a2, a3 statef (c1 a1) ^ (c2 a2) ^ a3return (a2, a3, f)def generate_sequence(init_state, c1, c2):state init_stateseen []output []while state not in seen:seen.append(state)output.append(state[0])state lfsr_step(state, c1, c2)return output, len(seen)def bits_to_bytes(bits):res []for i in range(0, len(bits), 8):byte 0for j in range(8):if i j len(bits):byte (byte 1) | bits[i j]res.append(byte)return resdef encrypt(hex_str, keystream_bits):data bytes.fromhex(hex_str[2:])ks bits_to_bytes(keystream_bits)return bytes([d ^ ks[i] for i, d in enumerate(data)]).hex()def decrypt(cipher_hex, keystream_bits):data bytes.fromhex(cipher_hex)ks bits_to_bytes(keystream_bits)return bytes([d ^ ks[i] for i, d in enumerate(data)]).hex()# # 主程序# init_state (1, 0, 1)plaintext 0x0123456789ABCDEFconfigs [(0, 0),(0, 1),(1, 0),(1, 1)]for c1, c2 in configs:print( * 40)print(f反馈函数: f {c1}*a1 ⊕ {c2}*a2 ⊕ a3)seq, period generate_sequence(init_state, c1, c2)print(输出序列:, seq)print(周期:, period)# 生成64bit密钥流keystream (seq * (64 // len(seq) 1))[:64]print(密钥流(bit):, keystream)# 加密cipher encrypt(plaintext, keystream)print(密文:, cipher)# 解密plain decrypt(cipher, keystream)print(解密:, plain)回到顶部NFLSR题目设n4,f(a1,a2,a3,a4)a1* a4^(a2 * a3),初始状态为(a1,a2,a3,a4)(1,1,0,1),求此非线性反馈移位寄存器的输出序列及周期,输出他的密钥流,并利用生成的密钥流对明文0x0123456789ABCDEF进行加密,输出加密后的结果,再对密文进行解密,输出解密后的结果。流密码原理 1生成密钥流keystream2明文 XOR 密钥流 → 密文 3密文 XOR 密钥流 → 明文。加密cipher plaintext XOR keystream 解密decrypt(cipher) plaintext寄存器长度n4状态(a1,a2,a3,a4)反馈函数f(a1,a2,a3,a4)a1⋅a4⊕(a2⋅a3)状态更新原则(a1,a2,a3,a4)→(a2,a3,a4,f)输出序列一般取输出a1(最左边)def nlfsr_step(state):a1, a2, a3, a4 state# 非线性反馈函数f (a1 a4) ^ (a2 a3)# 移位return (a2, a3, a4, f)def generate_sequence(init_state):state init_stateseen []output_seq []while state not in seen:seen.append(state)output_seq.append(state[0]) # 输出 a1state nlfsr_step(state)period len(seen)return output_seq, perioddef bits_to_bytes(bits):res []for i in range(0, len(bits), 8):byte 0for j in range(8):if i j len(bits):byte (byte 1) | bits[i j]res.append(byte)return resdef encrypt(plaintext_hex, keystream_bits):plaintext bytes.fromhex(plaintext_hex[2:])keystream_bytes bits_to_bytes(keystream_bits)cipher bytes([p ^ keystream_bytes[i] for i, p in enumerate(plaintext)])return cipher.hex()def decrypt(cipher_hex, keystream_bits):cipher bytes.fromhex(cipher_hex)keystream_bytes bits_to_bytes(keystream_bits)plain bytes([c ^ keystream_bytes[i] for i, c in enumerate(cipher)])return plain.hex()# # 主流程# init_state (1, 1, 0, 1)# 生成序列seq, period generate_sequence(init_state)print(输出序列:, seq)print(周期:, period)# 生成足够长的密钥流明文8字节64bitkeystream_bits (seq * (64 // len(seq) 1))[:64]print(密钥流(bit):\n, keystream_bits)# 明文plaintext 0x0123456789ABCDEF# 加密cipher encrypt(plaintext, keystream_bits)print(密文:, cipher)# 解密decrypted decrypt(cipher, keystream_bits)print(解密结果:, decrypted)回到顶部ECC题目已知椭圆曲线E11(1,6)编程求其上所有的点。这个问题本质是在有限域 F11 上求椭圆曲线E:y2≡x3x6(mod11)的所有点。思路枚举 x∈[0,10]计算右边rhsx3x6(mod11)枚举 y∈[0,10]找满足以下 条件的解y2≡rhs(mod11)加上无穷远点 O#include stdio.h#define p 11#define a 1#define b 6typedef struct {int x;int y;} Point;int isOnCurve(Point point) {int left (point.y * point.y) % p;int right (point.x * point.x * point.x a * point.x b) % p;return left right;}int main() {int count 0;for (int x 0; x p; x) {for (int y 0; y p; y) {Point point {x, y};if (isOnCurve(point)) {printf((%d, %d)\n, point.x, point.y);count;}}}printf(Point at infinity: O\n); //注意必须添加无穷远点。printf(Total points (including O): %d\n, count 1);return 0;}p 11a 1b 6points []for x in range(p):rhs (x**3 a*x b) % pfor y in range(p):if (y*y) % p rhs:points.append((x, y))# 加上无穷远点points.append(O)print(椭圆曲线上的点)for pt in points:print(pt)print(\n点的总数, len(points))回到顶部背包题目编程实现背包密码的加密和解密假设选取的私钥为超递增背包向量 A[1, 3, 5, 11, 21, 44, 87, 175, 349, 701]选取模数k 1590和乘数t 43计算新的背包向量B t * A (mod k)作为公钥。利用上述密码算法对明文“SAUNA AND HEALTH”进行加密输出密文并对密文进行解密。实现一个公钥加密系统私钥超递增序列 A公钥B t·A mod k加密用 B 做“子集和”解密用 A 做“贪心恢复”加解密原理超递增序列私钥。A [1, 3, 5, 11, 21, 44, 87, 175, 349, 701] 。 满足每个元素 前面所有元素之和保证子集和可唯一反推贪心。↓公钥生成 Bi​(t⋅Ai​)mod k k 1590t 43i是下标↓加密。把字符 → 二进制10 bit对应 A 长度 C∑bi​⋅Bi​↓解密。1. 求逆元t^-1 modk2. 还原C′C⋅t^−1 modk3. 用 A 贪心恢复 bit# 私钥A [1, 3, 5, 11, 21, 44, 87, 175, 349, 701]# 参数k 1590t 43# 求逆元扩展欧几里得def modinv(a, m):def egcd(a, b):if b 0:return a, 1, 0g, x1, y1 egcd(b, a % b)return g, y1, x1 - (a // b) * y1g, x, _ egcd(a, m)if g ! 1:raise Exception(无逆元)return x % m# 生成公钥B [(t * ai) % k for ai in A]print(公钥 B:, B)# 字符转10bitdef char_to_bits(c):val ord(c)return [(val i) 1 for i in reversed(range(10))]# bits转字符def bits_to_char(bits):val 0for b in bits:val (val 1) | breturn chr(val)# 加密def encrypt(plaintext):ciphertext []for ch in plaintext:bits char_to_bits(ch)c sum(b * bi for b, bi in zip(bits, B))ciphertext.append(c)return ciphertext# 解密def decrypt(ciphertext):inv_t modinv(t, k)plaintext for c in ciphertext:c_prime (c * inv_t) % k# 贪心恢复bits [0] * len(A)for i in reversed(range(len(A))):if c_prime A[i]:bits[i] 1c_prime - A[i]plaintext bits_to_char(bits)return plaintext# # 主程序# plaintext SAUNA AND HEALTHprint(明文:, plaintext)cipher encrypt(plaintext)print(密文:, cipher)decrypted decrypt(cipher)print(解密:, decrypted)