维纳攻击 一、攻击条件 设$$N=pq,q<p<2q,$$若：
$$d < \frac {1}{3}N^{\frac {1}{4}}$$
给出$$N,e(ed \equiv 1 \pmod {\lambda (N)}，$$则攻击者可以恢复d
针对维纳攻击的对策：
选择较大公钥 ：$$将e替换为e’=e+k \lambda(N),当e’足够大即e’>N^{\frac {3}{2}}$$中国剩余定理： $$选择d使得\quad dp \equiv d \pmod {p-1},dq \equiv d \pmod {q-1},但d本身不小，那么可以：$$
$$计算M_p \equiv C^dp \pmod p,Mq \equiv c^dq \pmod q$$
$$利用中国剩余定理：M \equiv M_p \pmod p,M\equiv M_q \pmod q，其中0 \leq M <N$$
$$由于d模 \lambda (N)可能很大，故维纳攻击并不适用$$
二、攻击原理 How to attack $$记G=gcd(p-1,q-1),\lambda(N)=lcm(p-1,q-1),即G \cdot \lambda(N)=\phi(N)$$
$$由ed \equiv 1 \pmod {\lambda (N)},设ed=K \cdot \lambda (N)+1 $$
$$则ed=\frac {K}{G}(p-1)(q-1)+1$$
$$记k=\frac {K}{gcd(K,G)},g=\frac {G}{gcd(K,G)},代入的：$$
$$ed=\frac {k}{g} (p-1)(q-1)+1$$
两边同时除以dpg得：
$$\frac {e}{pq}=\frac {k}{dg}(1-\frac {p+q-1- \frac {g}{k}}{pq})$$
$$故 \frac {e}{pq}略小于 \frac {k}{dg}$$
Proof $$已知ed-k \lambda (N)=1,那么：$$
$$\lvert \frac {e}{\lambda (N)}-\frac {k}{d} \rvert =\frac {1}{d \lambda (N)}$$
$$两边同时乘以 \frac {1}{G}得：$$
$$\lvert \frac {e}{\phi (N)}-\frac {k}{Gd} \rvert =\frac {1}{d \phi (N)}$$
$$也就是说 \frac {k}{Gd} 和 \frac {e}{\phi(N)}是很接近的，\phi (N）也可以用N来近似$$
$$又$$
$$\lvert p+q-1 \rvert <3 \sqrt {N} \ \lvert N-\phi(N) \rvert<3 \sqrt {N}$$
$$那么用N代替\phi(N)得到：$$
$$\lvert \frac {e}{N}-\frac {k}{Gd} \rvert=\lvert \frac{edG-kN}{NGd} \rvert =\lvert \frac{edG-k\phi(n)-kN+k\phi (N)}{NGd} \rvert \ \qquad \qquad \qquad \qquad \qquad \qquad=\lvert \frac {1-k(N-\phi(N))}{NGd} \rvert \ \qquad \qquad \qquad \qquad \qquad \qquad< \lvert \frac {-k(N-\phi(N))}{NGd} \rvert\ \qquad \qquad \qquad \qquad \qquad \qquad<\frac {3k \sqrt N}{NGd} \ \qquad \qquad \qquad \qquad \qquad \qquad \leq \frac {3k}{d \sqrt N}$$
$$由e<\lambda(N),k \lambda(N)d:$$
$$k\frac{1}{2d^2}$$
$$那么：$$
$$\lvert \frac{e}{N}-\frac{k}{Gd} \rvert<\frac {3k}{d\sqrt {N}}<\frac {1}{2d^2}$$
$$如果 \lvert x-\frac{a}{b} \rvert <\frac{1}{2b^2},那么x收敛于\frac {a}{b},即\frac {e}{N}收敛于\frac {k}{Gd},因此算法可以找到\frac {k}{Gd}$$
三、工具 https://github.com/orisano/owiener
1 2 3 4 5 6 7 8 9 10 11 12 import owienere = 30749686305802061816334591167284030734478031427751495527922388099381921172620569310945418007467306454160014597828390709770861577479329793948103408489494025272834473555854835044153374978554414416305012267643957838998648651100705446875979573675767605387333733876537528353237076626094553367977134079292593746416875606876735717905892280664538346000950343671655257046364067221469807138232820446015769882472160551840052921930357988334306659120253114790638496480092361951536576427295789429197483597859657977832368912534761100269065509351345050758943674651053419982561094432258103614830448382949765459939698951824447818497599 n = 109966163992903243770643456296093759130737510333736483352345488643432614201030629970207047930115652268531222079508230987041869779760776072105738457123387124961036111210544028669181361694095594938869077306417325203381820822917059651429857093388618818437282624857927551285811542685269229705594166370426152128895901914709902037365652575730201897361139518816164746228733410283595236405985958414491372301878718635708605256444921222945267625853091126691358833453283744166617463257821375566155675868452032401961727814314481343467702299949407935602389342183536222842556906657001984320973035314726867840698884052182976760066141 d = owiener.attack(e, n) if d is None : print ("Failed" ) else : print ("Hacked d={}" .format (d))
1 2 3 4 5 6 7 8 9 10 11 12 13 14 from Crypto.Util.number import * n = ? e = ? c = ? Ge = Matrix(ZZ, 2, 2, [e, isqrt(n), n, 0]) L = Ge.LLL() for row in L: if row[1]%isqrt(n) == 0: d = row[1]//isqrt(n) m = pow(c, d, n) print(long_to_bytes(m)) break
扩展维纳攻击 guo $$对于两个e的情况，考虑：$$
$$e_1d_1g-k_1(p-1)(q-1)=g \ e_2d_2g-k_2(p-1)(q-1)=g$$
$$进而推出：$$
$$k_2d_1e_1-k_1d_2e_2=k_2-k_1$$
$$两边同时除以k_2d_1e_2:$$
$$\frac {e_1}{e_2}-\frac{k_1d_2}{k_2d_1}=\frac {k_2-k_1}{k_2d_1e_2}$$
$$设d_i
$$\frac {k_1d_2}{k_2d_1}是\frac {e_1}{e_2}的连分数近似，当k_2和d_1最多为N^\alpha而且g很小，有：$$
$$\alpha <\frac {1}{3}-\epsilon(\epsilon >0)$$
$$然而得到了\frac {k_1d_2}{k_2d_1}还是无法分解$$
一、攻击原理 $$记d_ige_i-k_iN=g+k_is为维纳等式,记W_i,k_id_je_j-k_jd_ie_i=k_i-k_j为郭等式G_{i,j}$$
待补充
二、两个小解密指数 构造 $$选取关系W_1,G_{1,2}.W_1W_2，有：$$
$$d_1 g e_1- k_1 N = g+k_1s$$
$$k_1d_2e_2-k_2d_1e_1=k_1-k_2$$
$$d_1d_2g^2e_1e_2-d_1gk_2e_1N-d_2gk_1e_2N+k_1k_2N^2=(g+k_1s)(g+k_2s)$$
$$将第一个式子乘上k_2,那么便可构造格:$$
$$\begin {pmatrix} k_1k_2&d_1gk_2&d_2gk_1&d_1d_2g^2 \end {pmatrix} \begin {pmatrix} 1&-N&0&N^2 \ &e_1&-e_1&-e_1N \ &&e_2&-e_2N \ &&&e_1e_2 \end {pmatrix} = $$
$$\begin {pmatrix} k_1k_2 &k_2(g+k_1s)&g(k_1-k_2)&(g+k_1s)(g+k)2s) \end {pmatrix}$$
$$等式右边向量的各方向上大小为N^{2\alpha_2},N^{\frac {1}{2+2\alpha_2}},N^{\alpha_2},N^{1+2\alpha_2}，为了使大小相等，我们可以考虑构造矩阵：$$
$$D=\begin {pmatrix} N&&&\ &N^{\frac{1}{2}}&&\ &&N^{1+\alpha_2}&\ &&&1 \end {pmatrix}$$
$$最终构造的矩阵为：$$
$$L_2=\begin {pmatrix} 1&-N&0&N^2 \ &e_1&-e_1&-e_1N \ &&e_2&-e_2N \ &&&e_1e_2 \end {pmatrix}*D$$
$$检查一下向量\mathbf b有$$
$$| \mathbf bL_2|<2N^{1+2\alpha_2}$$
$$那么便可以使用格基规约算法LLL得到\mathbf b，进而求解\frac {\mathbf b[1]}{\mathbf b[0]}即\frac{d_1g}{k_1},那么：$$
$$\phi (N)=\frac {edg}{k}-\frac{g}{k}$$
$$假设这些格中最短向量长度为\Delta^{\frac{1}{4}-\epsilon} (\Delta =det(L_2)=N^{\frac{13}{2+\alpha_2}})$$
$$通常情况下，随机格中最短向量的长度大约为闵可夫斯基界：2\Delta^{\frac{1}{n}}$$
$$所以当\mathbf b是最短向量时：$$
$$N^{1+2\alpha_2}<\frac{1}{c_2}(N^{\frac{13}{2+\alpha_2}})^{\frac{1}{4}}$$
$$其中c_2是一个小常数（忽略），对于N的指数：$$
$$1+2\alpha_2<\frac {13}{4(2+\alpha_2)} \Rightarrow \alpha_2< \frac{5}{14}$$
$$因此，当\alpha_2<\frac{5}{14}-\epsilon’时目标向量比闵可夫斯基界短$$
模版 1 2 3 4 5 6 7 8 9 10 11 12 13 from Crypto.Util.number import * e1 = ? e2 = ? N = ? a = 5/14 D = diagonal_matrix(ZZ, [N, int(N^(1/2)), int(N^(1+a)), 1]) M = matrix(ZZ, [[1, -N, 0, N^2], [0, e1, -e1, -e1*N], [0, 0, e2, -e2*N], [0, 0, 0, e1*e2]])*D L = M.LLL() t = vector(ZZ, L[0]) x = t * M^(-1) phi = int(x[1]/x[0]*e1) d = inverse(e,phi)
三、三个小解密指数 构造 之后补充
模版 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 e1 = ? e2 = ? e3 = ? n = ? a = 2 /5 D = diagonal_matrix(ZZ,[int (n^1.5 ), n, int (n^(a+1.5 )), int (n^0.5 ), int (n^(a+1.5 )), int (n^(a+1 )), int (n^(a+1 )), 1 ]) L = Matrix(ZZ,[[1 , -n, 0 , n^2 , 0 , 0 , 0 , -n^3 ], [0 , e1, -e1, -n*e1, -e1, 0 , n*e1, n^2 *e1], [0 , 0 , e2, -n*e2, 0 , n*e2, 0 , n^2 *e2], [0 , 0 , 0 , e1*e2, 0 , -e1*e2, -e1*e2, -n*e1*e2], [0 , 0 , 0 , 0 , e3, -n*e3, -n*e3, n^2 *e3], [0 , 0 , 0 , 0 , 0 , e1*e3, 0 , -n*e1*e3], [0 , 0 , 0 , 0 , 0 , 0 , e2*e3, -n*e2*e3], [0 , 0 , 0 , 0 , 0 , 0 , 0 , e1*e2*e3]]) * D Ge = L.LLL()[0 ]x = vector(ZZ, Ge) / L phi = int (x[1 ]/x[0 ]*e1)d = inverse(e,phi)
四、四个小解密指数 构造 之后补充
模版 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 e1 = ? e2 = ? e3 = ? n = ? a = 2 /5 D = diagonal_matrix(ZZ,[int (n^1.5 ), n, int (n^(a+1.5 )), int (n^0.5 ), int (n^(a+1.5 )), int (n^(a+1 )), int (n^(a+1 )), 1 ]) L = Matrix(ZZ,[[1 , -n, 0 , n^2 , 0 , 0 , 0 , -n^3 ], [0 , e1, -e1, -n*e1, -e1, 0 , n*e1, n^2 *e1], [0 , 0 , e2, -n*e2, 0 , n*e2, 0 , n^2 *e2], [0 , 0 , 0 , e1*e2, 0 , -e1*e2, -e1*e2, -n*e1*e2], [0 , 0 , 0 , 0 , e3, -n*e3, -n*e3, n^2 *e3], [0 , 0 , 0 , 0 , 0 , e1*e3, 0 , -n*e1*e3], [0 , 0 , 0 , 0 , 0 , 0 , e2*e3, -n*e2*e3], [0 , 0 , 0 , 0 , 0 , 0 , 0 , e1*e2*e3]]) * D Ge = L.LLL()[0 ]x = vector(ZZ, Ge) / L phi = int (x[1 ]/x[0 ]*e1)d = inverse(e,phi)
五、一般情况 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 from Crypto.Util.number import *isdigit = lambda x: ord ('0' ) <= ord (x) <= ord ('9' ) def my_permutations (g, n ): sub = [] res = [] def dfs (s, prev ): if len (s) == n: res.append(s[::]) for i in g: if i in s or i < prev: continue s.append(i) dfs(s, max (prev, i)) s.remove(i) dfs(sub, 0 ) return res class X3NNY (object ): def __init__ (self, exp1, exp2 ): self .exp1 = exp1 self .exp2 = exp2 def __mul__ (self, b ): return X3NNY(self .exp1 * b.exp1, self .exp2 * b.exp2) def __repr__ (self ): return '%s = %s' % (self .exp1.expand().collect_common_factors(), self .exp2) class X_Complex (object ): def __init__ (self, exp ): i = 0 s = '%s' % exp while i < len (s): if isdigit(s[i]): num = 0 while i < len (s) and isdigit(s[i]): num = num*10 + int (s[i]) i += 1 if i >= len (s): self .b = num elif s[i] == '*' : self .a = num i += 2 elif s[i] == '/' : i += 1 r = 0 while i < len (s) and isdigit(s[i]): r = r*10 + int (s[i]) i += 1 self .b = num/r else : i += 1 if not hasattr (self , 'a' ): self .a = 1 if not hasattr (self , 'b' ): self .b = 0 def WW (e, d, k, g, N, s ): return X3NNY(e*d*g-k*N, g+k*s) def GG (e1, e2, d1, d2, k1, k2 ): return X3NNY(e1*d1*k2- e2*d2*k1, k2 - k1) def W (i ): e = eval ("e%d" % i) d = eval ("d%d" % i) k = eval ("k%d" % i) return WW(e, d, k, g, N, s) def G (i, j ): e1 = eval ("e%d" % i) d1 = eval ("d%d" % i) k1 = eval ("k%d" % i) e2 = eval ("e%d" % j) d2 = eval ("d%d" % j) k2 = eval ("k%d" % j) return GG(e1, e2, d1, d2, k1, k2) def R (e, sn ): ret = X3NNY(1 , 1 ) n = max (e) nn = len (e) l = set (i for i in range (1 , n+1 )) debug = '' u, v = 0 , 0 for i in e: if i == 1 : ret *= W(1 ) debug += 'W(%d)' % i nn -= 1 l.remove(1 ) u += 1 elif i > min (l) and len (l) >= 2 *nn: ret *= G(min (l), i) nn -= 1 debug += 'G(%d, %d)' % (min (l), i) l.remove(min (l)) l.remove(i) v += 1 else : ret *= W(i) l.remove(i) debug += 'W(%d)' % i nn -= 1 u += 1 return ret, u/2 + (sn - v) * a def H (n ): if n == 0 : return [0 ] if n == 2 : return [(), (1 ,), (2 ,), (1 , 2 )] ret = [] for i in range (3 , n+1 ): ret.append((i,)) for j in range (1 , i): for k in my_permutations(range (1 , i), j): ret.append(tuple (k + [i])) return H(2 ) + ret def CC (exp, n ): cols = [0 for i in range (1 <<n)] texps = ('%s' % exp.exp1.expand()).strip().split(' - ' ) ops = [] exps = [] for i in range (len (texps)): if texps[i].find(' + ' ) != -1 : tmp = texps[i].split(' + ' ) ops.append(0 ) exps.append(tmp[0 ]) for i in range (1 , len (tmp)): ops.append(1 ) exps.append(tmp[i]) else : ops.append(0 ) exps.append(texps[i]) if exps[0 ][0 ] == '-' : for i in range (len (exps)): ops[i] = 1 -ops[i] exps[0 ] = exps[0 ][1 :] else : ops[0 ] = 1 l = [] for i in range (len (exps)): tmp = 1 if ops[i] else -1 en = [] j = 0 while j < len (exps[i]): if exps[i][j] == 'e' : num = 0 j += 1 while isdigit(exps[i][j]): num = num*10 + int (exps[i][j]) j += 1 tmp *= eval ('e%d' % num) en.append(num) elif exps[i][j] == 'N' : j += 1 num = 0 if exps[i][j] == '^' : j += 1 while isdigit(exps[i][j]): num = num*10 + int (exps[i][j]) j += 1 if num == 0 : num = 1 tmp *= eval ('N**%d' % num) else : j += 1 if tmp == 1 or tmp == -1 : l.append((0 , ())) else : l.append((tmp, tuple (sorted (en)))) mp = H(n) for val, en in l: cols[mp.index(en)] = val return cols def EWA (n, elist, NN, alpha ): mp = H(n) var('a' ) S = [X_Complex(n*a)] cols = [[1 if i == 0 else 0 for i in range (2 ^n)]] for i in mp[1 :]: eL, s = R(i, n) cols.append(CC(eL, n)) S.append(X_Complex(s)) alphaA,alphaB = 0 , 0 for i in S: alphaA = max (i.a, alphaA) alphaB = max (i.b, alphaB) D = [] for i in range (len (S)): D.append( int (NN^((alphaA-S[i].a)*alpha + (alphaB - S[i].b))) ) kw = {'N' : NN} for i in range (len (elist)): kw['e%d' % (i+1 )] = elist[i] B = Matrix(ZZ, Matrix(cols).T(**kw)) * diagonal_matrix(ZZ, D) L = B.LLL(0.5 ) v = Matrix(ZZ, L[0 ]) x = v * B**(-1 ) phi = int (x[0 ,1 ]/x[0 ,0 ]*elist[0 ]) return phi def attack (NN, elist, alpha ): phi = EWA(len (elist), elist, NN, alpha) print (phi) return phi n = ? NN = ? elist = [e1,e2,e3, ...,en] alpha = n / int (NN).bit_length() for i in range (1 , len (elist)+1 ): var("e%d" % i) var("d%d" % i) var("k%d" % i) g, N, s = var('g' ), var('N' ), var('s' ) for i in range (len (elist)): elist[i] = Integer(elist[i]) phi = attack(NN, elist, alpha) d = inverse(e, phi)
参考 https://ctf-wiki.org/crypto/asymmetric/rsa/d_attacks/rsa_extending_wiener/#_3
https://seandictionary.top/rsa/
感谢
