Merge pull request 'main' (#14) from sangge/mimajingsai:main into main

Reviewed-on: ccyj/mimajingsai#14
This commit is contained in:
ccyj 2023-10-18 17:35:17 +08:00
commit d1856f053b
2 changed files with 16 additions and 10 deletions

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@ -2,8 +2,7 @@ from tpre import *
# 1 # 1
pk_a, sk_a = GenerateKeyPair(1, ()) pk_a, sk_a = GenerateKeyPair(1, ())
m = b'hello world' m = b"hello world"
m = int.from_bytes(m)
# 2 # 2
capsule_ct = Encrypt(pk_a, m) capsule_ct = Encrypt(pk_a, m)
@ -11,8 +10,8 @@ capsule_ct = Encrypt(pk_a, m)
# 3 # 3
pk_b, sk_b = GenerateKeyPair(1, ()) pk_b, sk_b = GenerateKeyPair(1, ())
N = 20 N = 5
T = 10 T = 2
# 5 # 5
rekeys = GenerateReKey(sk_a, pk_b, N, T) rekeys = GenerateReKey(sk_a, pk_b, N, T)
@ -28,4 +27,4 @@ for rekey in rekeys:
cfrags = mergecfrag(cfrag_cts) cfrags = mergecfrag(cfrag_cts)
m = DecryptFrags(sk_b, pk_b, pk_a, cfrags) m = DecryptFrags(sk_b, pk_b, pk_a, cfrags)
print(m)

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@ -271,10 +271,18 @@ def hash6(triple_G: Tuple[point, point, point]) -> int:
def f(x: int, f_modulus: list, T: int) -> int: def f(x: int, f_modulus: list, T: int) -> int:
""" """ '''
功能: 通过多项式插值来实现信息的分散和重构
例如: 随机生成一个多项式f(x)=4x+5,质数P=11,其中f(0)=5,将多项式的系数分别分配给两个人,例如第一个人得到(1, 9),第二个人得到(2, 2).如果两个人都收集到了这两个点,那么可以使用拉格朗日插值法恢复原始的多项式,进而得到秘密信息"5"
param:
x, f_modulus(多项式系数列表), T(门限)
return:
res
'''
res = 0 res = 0
for i in range(T): for i in range(T):
res += f_modulus[i] * pow(x, i) res += f_modulus[i] * pow(x, i)
res = res % sm2p256v1.P
return res return res
@ -324,6 +332,7 @@ def Encapsulate(pk_A: point) -> Tuple[int, capsule]:
E = multiply(g, r) E = multiply(g, r)
V = multiply(g, u) V = multiply(g, u)
s = u + r * hash2((E, V)) s = u + r * hash2((E, V))
s = s % sm2p256v1.P
pk_A_ru = multiply(pk_A, r + u) pk_A_ru = multiply(pk_A, r + u)
K = KDF(pk_A_ru) K = KDF(pk_A_ru)
capsule = (E, V, s) capsule = (E, V, s)
@ -408,8 +417,6 @@ def DecapsulateFrags(sk_B: int, pk_B: point, pk_A: point, cFrags: list) -> int:
sxi = hash5(id, D) # id 节点的编号 sxi = hash5(id, D) # id 节点的编号
Sx.append(sxi) Sx.append(sxi)
bis = [] # b ==> λ bis = [] # b ==> λ
j = 1
i = 1
bi = 1 bi = 1
for i in range(len(cFrags)): for i in range(len(cFrags)):
for j in range(len(cFrags)): for j in range(len(cFrags)):