113 lines
4.2 KiB
Python
113 lines
4.2 KiB
Python
import hashlib
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import secrets
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from mnemonic import Mnemonic # 仅用于标准的助记词转换
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class SentinelKeyEngine:
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# 使用第 13 个梅森素数 (2^521 - 1)，远大于 128-bit 熵，确保有限域安全
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PRIME = 2**521 - 1
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def __init__(self):
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self.mnemo = Mnemonic("english")
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def generate_vault_keys(self):
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"""
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1. 生成原始 12 助记词 (Master Key)
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"""
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words = self.mnemo.generate(strength=128)
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entropy = self.mnemo.to_entropy(words)
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return words, entropy
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def split_to_shares(self, entropy):
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"""
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2. SSS (3,2) 门限分片逻辑
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公式: f(x) = S + a*x (直线方程，S为秘密，a为随机斜率)
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我们将秘密 S 分成 3 份，任选 2 份即可恢复。
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注意：必须在有限域 GF(PRIME) 下进行运算以保证完善保密性。
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"""
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# 将熵转换为大整数
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secret_int = int.from_bytes(entropy, 'big')
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# 生成一个随机系数 a (安全性需与秘密强度一致)
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# a 必须在 [0, PRIME-1] 范围内
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a = secrets.randbelow(self.PRIME)
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# 定义 3 个点: x=1, x=2, x=3
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# Share = (x, f(x))
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def f(x): return (secret_int + a * x) % self.PRIME
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share1 = (1, f(1)) # 手机分片
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share2 = (2, f(2)) # 云端分片
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share3 = (3, f(3)) # 传承卡分片
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return [share1, share2, share3]
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def recover_from_shares(self, share_a, share_b):
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"""
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3. 恢复逻辑：拉格朗日插值还原
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已知 (x1, y1) 和 (x2, y2)，求 f(0) 即秘密 S
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公式: S = (x2*y1 - x1*y2) / (x2 - x1)
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在有限域下，除法变为乘以模逆: S = (x2*y1 - x1*y2) * (x2 - x1)^-1 mod P
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"""
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x1, y1 = share_a
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x2, y2 = share_b
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# 计算分子
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numerator = (x2 * y1 - x1 * y2) % self.PRIME
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# 计算分母的模逆 (x2 - x1)
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denominator = (x2 - x1) % self.PRIME
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inv_denominator = pow(denominator, -1, self.PRIME)
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# 还原常数项 S
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secret_int = (numerator * inv_denominator) % self.PRIME
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# 转回字节并生成助记词
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# 注意：secret_int 可能略小于 16 字节（高位为0），需要补齐
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# 但由于 entropy 原始就是 16 字节，这里直接转换即可
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try:
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recovered_entropy = secret_int.to_bytes(16, 'big')
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except OverflowError:
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# 理论上不应发生，除非计算出的 secret_int 大于 128 bit (即原始 entropy 大于 128 bit)
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# 这里为了健壮性，如果原始 entropy 是 16 字节，这里应该也是。
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# 如果 PRIME 很大，secret_int 还是原来的值。
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recovered_entropy = secret_int.to_bytes((secret_int.bit_length() + 7) // 8, 'big')
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return self.mnemo.to_mnemonic(recovered_entropy)
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if __name__ == "__main__":
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# --- Sentinel 协议业务流程模拟 ---
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engine = SentinelKeyEngine()
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# [生前]：初始化金库
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master_words, entropy = engine.generate_vault_keys()
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print(f"【1. 生成原始助记词】: {master_words}")
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shares = engine.split_to_shares(entropy)
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print(f"【2. SSS 分片完成】:")
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print(f" - 分片1 (手机安全区): {shares[0]}")
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print(f" - 分片2 (Sentinel云): {shares[1]}")
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print(f" - 分片3 (传承卡单词): {shares[2]}")
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print("-" * 50)
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# [死后/传承]：模拟用户失联，触发被动验证
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# 假设继承人拿着卡片 (Share 3)，向服务器请求分片 (Share 2)
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successor_share = shares[2]
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server_share = shares[1]
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# 执行恢复
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recovered_words = engine.recover_from_shares(shares[0], shares[1])
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print(f"【1. 手机+云 : {recovered_words}")
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recovered_words = engine.recover_from_shares(shares[0], shares[2])
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print(f"【2. 手机+传承卡 : {recovered_words}")
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recovered_words = engine.recover_from_shares(shares[1], shares[2])
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print(f"【3. 云+传承卡 : {recovered_words}")
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# 校验一致性
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assert recovered_words == master_words
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print("\n结果：恢复出的助记词与原始完全一致。")
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with open("words.txt", "w") as f:
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f.write("%s\n"%master_words) |