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

Reviewed-on: ccyj/mimajingsai#14

src/demo.py

@@ -2,8 +2,7 @@ from tpre import *
 
 # 1
 pk_a, sk_a = GenerateKeyPair(1, ())
-m = b'hello world'
-m = int.from_bytes(m)
+m = b"hello world"
 
 # 2
 capsule_ct = Encrypt(pk_a, m)
@@ -11,11 +10,11 @@ capsule_ct = Encrypt(pk_a, m)
 # 3
 pk_b, sk_b = GenerateKeyPair(1, ())
 
-N = 20
-T = 10
+N = 5
+T = 2
 
 # 5
-rekeys =  GenerateReKey(sk_a, pk_b, N, T)
+rekeys = GenerateReKey(sk_a, pk_b, N, T)
 
 # 7
 cfrag_cts = []
@@ -23,9 +22,9 @@ cfrag_cts = []
 for rekey in rekeys:
     cfrag_ct = ReEncrypt(rekey, capsule_ct)
     cfrag_cts.append(cfrag_ct)
-    
+
 # 9
 cfrags = mergecfrag(cfrag_cts)
 m = DecryptFrags(sk_b, pk_b, pk_a, cfrags)
 
-
+print(m)

src/tpre.py

@@ -271,10 +271,18 @@ def hash6(triple_G: Tuple[point, point, point]) -> 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
     for i in range(T):
         res += f_modulus[i] * pow(x, i)
+    res = res % sm2p256v1.P
     return res
 
 
@@ -324,6 +332,7 @@ def Encapsulate(pk_A: point) -> Tuple[int, capsule]:
     E = multiply(g, r)
     V = multiply(g, u)
     s = u + r * hash2((E, V))
+    s = s % sm2p256v1.P
     pk_A_ru = multiply(pk_A, r + u)
     K = KDF(pk_A_ru)
     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
         Sx.append(sxi)
     bis = []  #  b ==> λ
-    j = 1
-    i = 1
     bi = 1
     for i in range(len(cFrags)):
         for j in range(len(cFrags)):