【密码学——基础理论与应用】李子臣编著 第十二章 SM3密码杂凑算法 课后习题

题目

逐题解析

12.1

512bit 256bit

12.2\12.3

12.4

W[16] = 9092E200 W[17] = 00000000 W[18] = 000C0606 W[19] = 719C70ED

12.5

W[20] = 00000000 W[21] = 8001801F W[22] = 939F7DA9 W[23] = 00000000

12.6

W′[16] = 9092E200 W′[17] = 8001801 W′[18] = 93937BAF W′[19] = 719C70ED

12.7

填充结果前64位： 0110000101100010011000110110010001100001011000100110001101100100
填充结果末64位： 0000000000000000000000000000000000000000000000000000001000000000

12.8

A<<<12=7380166F<<<12=0000 0001 0110 0110 1111 0111 0011 1000=0166F738

12.9

B=7380166F C=29657292 D=172442D7

代码

免责声明：我从来不敢保证结果是对的，我只敢保证代码是我认真审的。

#-------------- 基础函数 ----------------#
def str_to_bin(msg: str) -> str:return ''.join(f'{ord(c):08b}' for c in msg)def padding_sm3(msg: str) -> str:m_bin = str_to_bin(msg)l = len(m_bin)m_bin += '1'k = (448 - (l + 1)) % 512m_bin += '0' * kl_bin = f'{l:064b}'m_bin += l_binreturn m_bindef left_rotate(x, n):n = n % 32return ((x << n) | (x >> (32 - n))) & 0xFFFFFFFFdef split_blocks(m_bin: str, block_size=512):return [m_bin[i:i + block_size] for i in range(0, len(m_bin), block_size)]def binstr_to_int_list(block_512bit: str):return [int(block_512bit[i:i+32], 2) for i in range(0, 512, 32)]#-------------- 扩展函数 ----------------#
def P1(X):return X ^ left_rotate(X, 15) ^ left_rotate(X, 23)def P0(X):return X ^ left_rotate(X, 9) ^ left_rotate(X, 17)def message_expand(block_512bit: str):W = binstr_to_int_list(block_512bit)for j in range(16, 68):term = P1(W[j - 16] ^ W[j - 9] ^ left_rotate(W[j - 3], 15))term = term ^ left_rotate(W[j - 13], 7) ^ W[j - 6]W.append(term & 0xFFFFFFFF)W_ = [(W[j] ^ W[j + 4]) & 0xFFFFFFFF for j in range(64)]return W, W_def print_expansion(W, W_):print("=== W[0..67] ===")for i in range(68):print(f"W[{i:02}] = {W[i]:08X}")print("\n=== W′[0..63] ===")for i in range(64):print(f"W′[{i:02}] = {W_[i]:08X}")#-------------- 压缩函数 ----------------#
def FF(X, Y, Z, j):return (X ^ Y ^ Z) if j < 16 else ((X & Y) | (X & Z) | (Y & Z))def GG(X, Y, Z, j):return (X ^ Y ^ Z) if j < 16 else ((X & Y) | (~X & Z))def CF(V, B):A, B_, C, D, E, F_, G, H = VW, W_ = message_expand(B)print("\n>>> [压缩函数初始化] <<<")print(f"A={A:08X} B={B_:08X} C={C:08X} D={D:08X} E={E:08X} F={F_:08X} G={G:08X} H={H:08X}")for j in range(64):Tj = 0x79CC4519 if j < 16 else 0x7A879D8ASS1 = left_rotate((left_rotate(A, 12) + E + left_rotate(Tj, j)) & 0xFFFFFFFF, 7)SS2 = SS1 ^ left_rotate(A, 12)TT1 = (FF(A, B_, C, j) + D + SS2 + W_[j]) & 0xFFFFFFFFTT2 = (GG(E, F_, G, j) + H + SS1 + W[j]) & 0xFFFFFFFFD = CC = left_rotate(B_, 9)B_ = AA = TT1H = GG = left_rotate(F_, 19)F_ = EE = P0(TT2)print(f"j={j:02} → A={A:08X} B={B_:08X} C={C:08X} D={D:08X} E={E:08X} F={F_:08X} G={G:08X} H={H:08X}")print(">>> [压缩函数结束] <<<")return [(v ^ n) & 0xFFFFFFFF for v, n in zip(V, [A, B_, C, D, E, F_, G, H])]#-------------- 哈希函数主控 ----------------#
def sm3_hash(msg: str, visual=True):IV = [0x7380166F, 0x4914B2B9, 0x172442D7, 0xDA8A0600,0xA96F30BC, 0x163138AA, 0xE38DEE4D, 0xB0EB0E4E]padded = padding_sm3(msg)blocks = split_blocks(padded)print("原始长度：", len(str_to_bin(msg)), "bits")print("填充后总长度：", len(padded), "bits")print("填充后分组数量：", len(blocks))print("填充结果前64位：", padded[:64])print("填充结果末64位：", padded[-64:])V = IV[:]for i, block in enumerate(blocks):print(f"\n>>> 第 {i} 块 <<<")W, W_ = message_expand(block)if visual:print_expansion(W, W_)V = CF(V, block)digest = ''.join(f'{x:08x}' for x in V)print("\n最终 SM3 摘要为：")print(digest)return digest#-------------- 测试入口 ----------------#
if __name__ == "__main__":msg = "abc"  # 可以改为 input("请输入消息: ")sm3_hash(msg)