RSA 加密及一些攻击方式

原理

随机生成两个素数，p , q

令n = p*q

由欧拉公式计算出φ(n) = (p-1)(q-1)

规定e，使得e满足1<e<φ(n)，且gcd(e,φ(n)) = 1，一般e=65537或0x10001

此时就有了公钥=(e,n)

计算私钥

计算d，使得d满足ed≡1mod φ(n)，即称d是e在模φ(n)下的逆元

得到私钥=(d,n)

有关欧拉公式

• $n=p^{e\_1}\_1p^{e\_2}\_2\\dots p^{e\_k}\_k$n=pe_1_1pe_2_2dotspe_k_k
• $\\varphi(n)=n(1-\\frac{1}{p\_1})(1-\\frac{1}{p\_2})\\dots (1-\\frac{1}{p\_k})$varphi(n)=n(1−frac1p_1)(1−frac1p_2)dots(1−frac1p_k)
• $\\varphi(n)=p^{e\_1-1}\_1p^{e\_2-1}\_2\\dots p^{e\_k-1}\_k(p\_1-1)(p\_2-1)\\dots (p\_k-1)$varphi(n)=pe_1−1_1pe_2−1_2dotspe_k−1_k(p_1−1)(p_2−1)dots(p_k−1)
• 当$gcd(n,x)=1$gcd(n,x)=1时，满足$x^{\\varphi(n)}\\equiv 1\\ mod\\ n$xvarphi(n)equiv1modn

RSA加密和解密

规定m表示明文，c表示密文

加密：c≡me mod n

解密：m≡cd mod n

解密方式

正常解密

• 此方法适用于已知$p,q,e,c$p,q,e,c的情况下

可以使用RSA Tool 2工具解密或者编写如下脚本

from Crypto.Util.number import *
p = ?
q = ?
e = ?
c = ?
n = p*q
d = inverse(e,(p-1) * (q-1))
m = pow(c,d,n)
print(long_to_bytes(m))

dp,dq泄露

• 此方法使用于已知$p,q,d\_{p},d\_{q},c$p,q,d_p,d_q,c的情况下
• 然而使用此方法前可以先尝试常用e值，在发现无法确定e值的情况下再使用

实例BUUCTF-Crypto-21.RSA1

理论推导

• dp，dq泄露
• 理论推导 红色字 为重要结论

$\\because m \\equiv c^d(mod\\ n)$becausemequivcd(modn)

$\\therefore m = c^d+kn = c^d+k\\cdot pq$thereforem=cd+kn=cd+kcdotpq

$m\_{1} \\equiv c^d(mod\\ p)\\ ①\\ ,\\ m\_{2} \\equiv c^d(mod\\ q)\\ ②$m_1equivcd(modp)①,m_2equivcd(modq)②

$①\\Rightarrow c^d=m\_{1}+tp$①Rightarrowcd=m_1+tp

$代入②\\Rightarrow m\_{2}\\equiv m\_{1}+tp(mod\\ q)$代入②Rightarrowm_2equivm_1+tp(modq)

$\\therefore m\_{2}=m\_{1}+t\\cdot p+r\\cdot q\\ \\Rightarrow m\_{2}-m\_{1}=t\\cdot p+r\\cdot q$thereforem_2=m_1+tcdotp+rcdotqRightarrowm_2−m_1=tcdotp+rcdotq

$\\therefore m\_{2}-m\_{1}=t\\cdot p(mod\\ q)$thereforem_2−m_1=tcdotp(modq)

$\\therefore (m\_{2}-m\_{1})\\cdot p^{-1}\\equiv t(mod\\ q)$therefore(m_2−m_1)cdotp−1equivt(modq)

$\\Rightarrow t=(m\_{2}-m\_{1})\\cdot p^{-1}(mod\\ q)$Rightarrowt=(m_2−m_1)cdotp−1(modq)

\\therefore c^d=\[(m\_{2}-m\_{1})\\cdot p^{-1}(mod\\ q)\]p+m\_{1}

$\\because m\\equiv c^d(mod\\ n)$becausemequivcd(modn)

\\therefore {\\color{Red} m\\equiv \[\[(m\_{2}-m\_{1})\\cdot p^{-1}(mod\\ q)\]p+m\_{1}\]mod\\ n}

$d\\equiv d\_{p}mod(p-1)\\ , \\ d\\equiv d\_{q}mod(q-1)$dequivd_pmod(p−1),dequivd_qmod(q−1)

$m\_{1} \\equiv c^dmod\\ p\\ ,\\ m\_{2} \\equiv c^dmod\\ q$m_1equivcdmodp,m_2equivcdmodq

\\Rightarrow m\_{1} \\equiv c^\\left .d\_{p}\\ +k\_{1}\\cdot (p-1)\\right.\\ mod\\ p\\ ,\\ m\_{2} \\equiv c^\\left .d\_{q}\\ +k\_{2}\\cdot (q-1)\\right.\\ mod\\ p

\\because c^\\left.p-1\\right.\\equiv 1\\ mod\\ p

\\Rightarrow {\\color{Red} m\_{1}\\equiv c^\\left .d\_{p}\\right . mod\\ p\\ ,\\ m\_{2}\\equiv c^\\left .d\_{q}\\right . mod \\ q}

使用脚本解密

from Crypto.Util.number import *

p = ?
q = ?
dp = ?
dq = ?
c = ?

mp = pow(c,dp,p)
mq = pow(c,dq,q)

pi = inverse(p,q)

m = ((((mq-mp)pi)%q)p+mp)%(p*q)

print(long_to_bytes(m))

dp泄露

• 此方法使用于已知$e,n,d\_{p},c$e,n,d_p,c的情况下

实例BUUCTF-Crypto-27.RSA2

理论推导

• dp泄露
• 理论推导 红色字 为重要结论

$d\_{p} \\equiv d mod (p-1)$d_pequivdmod(p−1)

$\\Rightarrow d\\cdot e = k\_{1}(p-1)+d\_{p}\\cdot e$Rightarrowdcdote=k_1(p−1)+d_pcdote

$\\because d\\cdot e = 1mod(p-1)(q-1) \\Rightarrow d\\cdot e=k\_{2}(p-1)(q-1)+1$becausedcdote=1mod(p−1)(q−1)Rightarrowdcdote=k_2(p−1)(q−1)+1

$\\therefore {\\color{Red} (p-1)(k\_{2}(q-1)-k\_{1})+1 = d\_{p}e}$thereforecolorRed(p−1)(k_2(q−1)−k_1)+1=d_pe

$\\because d\_{p}<d$becaused_p<d

$\\therefore (k\_{2}(q-1)-k\_{1})=x\\in (1,e)$therefore(k_2(q−1)−k_1)=xin(1,e)

from Crypto.Util.number import *
e = 65537
n = ?
dp = ?
c = ?
a = dp*e-1
for x in range(2,e):
    if a%x == 0:
        p = a//x+1
        if n%p == 0:
            q = n//p
            break
d = inverse(e,(p-1)*(q-1))
m = pow(c,d,n)
print(long_to_bytes(m))

共模攻击

• 此方法使用于已知$n,e\_{1},e\_{2},c\_{1},c\_{2}$n,e_1,e_2,c_1,c_2的情况下
• 且$gcd(e\_{1},e\_{2})=1$gcd(e_1,e_2)=1

实例BUUCTF-Crypto-28.RSA3

理论推导

• 共模攻击
• 理论推导 红色字 为重要结论
• 存在两种密钥对同一明文进行加密
• 下述为针对此题特殊情况下的特殊推导，需满足$gcd(e\_{1},e\_{2})=1$gcd(e_1,e_2)=1

c\_{1}=m^\\left. e\_{1}\\right. mod\\ n\\ \\&\\ m=c\_{1}^\\left.d\_{1}\\right.mod\\ n

c\_{2}=m^\\left. e\_{2}\\right. mod\\ n\\ \\&\\ m=c\_{2}^\\left.d\_{2}\\right.mod\\ n

$构造一对(s\_{1},s\_{2})满足e\_{1}s\_{1}+e\_{2}s\_{2}=1其中s\_{1},s\_{2}\\in Z,s\_{1}>0,s\_{2}<0$构造一对(s_1,s_2)满足e_1s_1+e_2s_2=1其中s_1,s_2inZ,s_1>0,s_2<0

c\_{1}=m^\\left. e\_{1}\\right. mod\\ n \\Rightarrow c\_{1}^\\left. s\_{1}\\right. =m^\\left. e\_{1}s\_{1}\\right. mod\\ n ①

c\_{2}=m^\\left. e\_{2}\\right. mod\\ n \\Rightarrow c\_{2}^\\left. s\_{2}\\right. =m^\\left. e\_{2}s\_{2}\\right. mod\\ n ②

①\\times ②\\Rightarrow c\_{1}^\\left. s\_{1}\\right. c\_{2}^\\left. s\_{2}\\right. mod\\ n = m^\\left. e\_{1}s\_{1}+e\_{2}s\_{2}\\right. mod\\ n

\\Rightarrow c\_{1}^\\left. s\_{1}\\right. c\_{2}^\\left. s\_{2}\\right. mod\\ n=m\\ mod\\ n

$\\because m=c^d mod\\ n \\therefore m<n$becausem=cdmodnthereforem<n

\\therefore {\\color{red}m=c\_{1}^\\left. s\_{1}\\right. c\_{2}^\\left. s\_{2}\\right. mod\\ n}

$需要注意的是推论到上面就结束了，但是实际计算中上述式子计算过于复杂，所以应当使用下述式子$需要注意的是推论到上面就结束了，但是实际计算中上述式子计算过于复杂，所以应当使用下述式子

{\\color{red}m=(c\_{1}^\\left. s\_{1}\\right. mod\\ n\\cdot c\_{2}^\\left. s\_{2}\\right. mod\\ n)mod\\ n}

$以及在计算s\_{1},s\_{2}过程中可以采用逆元的方式$以及在计算s_1,s_2过程中可以采用逆元的方式

$(e\_{1}-e{2})s\_{1}+e{2}(s\_{1}+s\_{2})=1$(e_1−e2)s_1+e2(s_1+s_2)=1

$(e\_{1}-e{2})s\_{1}\\equiv 1\\ mod\\ e\_{2}$(e_1−e2)s_1equiv1mode_2

s\_{1}=(e\_{1}-e{2})^\\left. -1\\right.

• 由此计算出一对$(s\_{1},s\_{2})$(s_1,s_2)，进而算出明文m
• 编写脚本

from Crypto.Util.number import *
n = ?
c1 = ?
c2 = ?
e1 = ?
e2 = ?
e1_e2 = e1-e2
s1 = inverse(e1_e2,e2)
s2 = (1-e1*s1)//e2
m = pow(c1,s1,n)*pow(c2,s2,n)%n
print(long_to_bytes(m))

低加密指数广播攻击

对于同一个明文$m$m，使用同一个低加密指数$e$e和一组不同的$n$n进行加密，得到一组$c\_i$c_i，则有

$c\_i = m^e \\mod n\_i$c_i=memodn_i

可以注意到对$(c\_i,n\_i)$(c_i,n_i)使用 CRT 合并那么就能得到$m^e$me

于是就能通过开$e$e次方根的方式得到$m$m

维纳攻击

• 此方法使用于$e$e过大或过小的情况下
• 本质：满足$d < \\frac{1}{3} N^{\\frac{1}{4} }$d<frac13Nfrac14则一定可以分解$N$N
• 大致解密过程是用RSAwienerHacker库来攻击得到d
• 板子

实例BUUCTF-Crypto-40.rsa2

from RSAwienerHacker import *
n = ?
e = ?
d = hack_RSA(e,n)

此攻击有两种方法可以实现，一种是用连分数实现，还有一种方法是用格基规约的方式实现的（据一大佬说，所有的连分数实现的攻击方法都能转化为格基规约的方式，我没有研究过这个问题）

连分数是一种表示实数的方式，可以将一个实数表示为整数部分和分数部分的递归形式。连分数的形式为

$x = a\_0 + \\frac{1}{a\_1 + \\frac{1}{a\_2 + \\frac{1}{a\_3 + \\cdots}}}$x=a_0+frac1a_1+frac1a_2+frac1a_3+cdots

一般可以将这组$a\_i$a_i简写成\[a\_0;a\_1,a\_2,a\_3,\\cdots\]

在 RSA 的计算中，有$ed \\equiv 1 \\mod \\varphi(n)$edequiv1modvarphi(n)

于是能将其变形为$\\left|\\frac{e}{\\varphi(n)}-\\frac{k}{d}\\right|=\\frac{1}{d\\varphi(n)}$left∣fracevarphi(n)−frackdright∣=frac1dvarphi(n)

又由于$\\varphi(n)\\approx n$varphi(n)approxn

所以有$\\left|\\frac{e}{n}-\\frac{k}{d}\\right|=\\left|\\frac{ed-nk}{nd}\\right|=\\left|\\frac{1-k(n-\\varphi(n))}{nd}\\right|<\\left|\\frac{k(p+q-1)}{nd}\\right|$left∣fracen−frackdright∣=left∣fraced−nkndright∣=left∣frac1−k(n−varphi(n))ndright∣<left∣frack(p+q−1)ndright∣

令$p < q < 2p$p<q<2p，则有$p+q-1<3p<3\\sqrt{n}$p+q−1<3p<3sqrtn

同时有$k\\varphi(n) < ed$kvarphi(n)<ed，于是$k < d < \\frac{1}{3}n^{\\frac{1}{4}}$k<d<frac13nfrac14

带入后得到\\left|\\frac{e}{n}-\\frac{k}{d}\\right|<\\frac{\\frac{1}{3}n^{\\frac{1}{4}}\\cdot3\\sqrt{n}}{nd}=\\frac{1}{n^\\frac{1}{4}d}<\\frac{1}{2d^2}

Legendre’s theorem[1] 指出对于任意一个实数 $x$x，若有最简分数$\\frac{a}{b}$fracab其中$gcd(a,b)=1$gcd(a,b)=1，且满足

$\\left| x – \\frac{a}{b} \\right| < \\frac{1}{b^2}$left∣x–fracabright∣<frac1b2

那么该最简分数$\\frac{a}{b}$fracab一定是$x$x的某项渐进分数

于是对于$\\left|\\frac{e}{n}-\\frac{k}{d}\\right|<\\frac{1}{2d^2}$left∣fracen−frackdright∣<frac12d2只需要计算$\\frac{e}{n}$fracen的每一项的渐进分数，并且判断是否满足判别式，从而得到私钥$d$d

参考文献

[1] Dujella A. Continued fractions and RSA with small secret exponent[J]. arXiv preprint cs/0402052, 2004.

易得 $ed+k\\varphi(n)=1$ed+kvarphi(n)=1，所以 $ed-kn=k(p+q-1)+1$ed−kn=k(p+q−1)+1，显然有 $\\left|k\\right| < \\varphi(n) < n$left∣kright∣<varphi(n)<n

同时 $\\left|k\\right|=\\frac{ed-1}{\\varphi(n)}\\approx \\frac{ed}{n}$left∣kright∣=fraced−1varphi(n)approxfracedn，可以知道 k 和 d 的规模相近

构造这样的格，用 K 来配平格基的行列式

(d,k) \\left(\\begin{matrix} e & K \\\\ n & 0 \\\\ \\end{matrix}\\right) =(ed+kn,dK)

检查是否符合_Hermite_定理，由经验可得$dK\\le ed+kn$dKleed+kn

$\\left|\\left|\\vec{v}\\right|\\right|<\\frac{1}{2}{\\left|\\left|\\mathcal{L}\\right|\\right|}^{\\frac{1}{2}}$left∣left∣vecvright∣right∣<frac12left∣left∣mathcalLright∣right∣frac12

得到 $\\sqrt{(ed+kn)^2+dK^2}\\approx ed+kn<\\frac{1}{2}\\sqrt{nK}$sqrt(ed+kn)2+dK2approxed+kn<frac12sqrtnK

由于$ed+kn$ed+kn和$2dp$2dp规模相近，所以令$n$n的规模约为$2^{2x}$22x，$d$d的规模约为$2^y$2y，$K$K的规模约为$2^{2z}$22z，于是就有

\\left\\{\\begin{matrix} x+y+1 < x+z\\\\ x+y+1 > y+2z \\end{matrix}\\right.

所以得到能攻击成功的约束条件约为 $K\\approx \\sqrt{n}$Kapproxsqrtn，d\\approx \\sqrt\[4\]{n}，然后规约，对得到的向量进行筛选，最终得到$d$d

# Sage
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

参考 CTF wiki

两个低解密指数

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)

三个低解密指数

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)

通用情况

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): # min u max v
    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
    # print(debug, end = ' ')
    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)]
    
    # split exp
    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
    # find e and N
    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))))
    
    # construct h
    mp = H(n)
    for val, en in l:
        cols[mp.index(en)] = val
    # print(cols)
    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)
    # print(alphaA, alphaB)
    D = []
    for i in range(len(S)):
        # print((alphaA-S[i].a), (alphaB - S[i].b))
        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 = ?   # n是d的比特位数
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)

Boneh Durfee攻击

• 这是维纳攻击的一种延伸，拓展了可以攻击的d的范围
• $d<n^{0.292}$d<n0.292

GitHub仓库，在此基础上修改

import time
from Crypto.Util.number import *

"""
Setting debug to true will display more informations
about the lattice, the bounds, the vectors...
"""
debug = True

"""
Setting strict to true will stop the algorithm (and
return (-1, -1)) if we don't have a correct 
upperbound on the determinant. Note that this 
doesn't necesseraly mean that no solutions 
will be found since the theoretical upperbound is
usualy far away from actual results. That is why
you should probably use `strict = False`
"""
strict = False

"""
This is experimental, but has provided remarkable results
so far. It tries to reduce the lattice as much as it can
while keeping its efficiency. I see no reason not to use
this option, but if things don't work, you should try
disabling it
"""
helpful_only = True
dimension_min = 7 # stop removing if lattice reaches that dimension

############################################
# Functions
##########################################

# display stats on helpful vectors
def helpful_vectors(BB, modulus):
    nothelpful = 0
    for ii in range(BB.dimensions()[0]):
        if BB[ii,ii] >= modulus:
            nothelpful += 1

    print (nothelpful, "/", BB.dimensions()[0], " vectors are not helpful")

# display matrix picture with 0 and X
def matrix_overview(BB, bound):
    for ii in range(BB.dimensions()[0]):
        a = ('%02d ' % ii)
        for jj in range(BB.dimensions()[1]):
            a += '0' if BB[ii,jj] == 0 else 'X'
            if BB.dimensions()[0] < 60:
                a += ' '
        if BB[ii, ii] >= bound:
            a += '~'
        print (a)

# tries to remove unhelpful vectors
# we start at current = n-1 (last vector)
def remove_unhelpful(BB, monomials, bound, current):
    # end of our recursive function
    if current == -1 or BB.dimensions()[0] <= dimension_min:
        return BB

    # we start by checking from the end
    for ii in range(current, -1, -1):
        # if it is unhelpful:
        if BB[ii, ii] >= bound:
            affected_vectors = 0
            affected_vector_index = 0
            # let's check if it affects other vectors
            for jj in range(ii + 1, BB.dimensions()[0]):
                # if another vector is affected:
                # we increase the count
                if BB[jj, ii] != 0:
                    affected_vectors += 1
                    affected_vector_index = jj

            # level:0
            # if no other vectors end up affected
            # we remove it
            if affected_vectors == 0:
                print ("* removing unhelpful vector", ii)
                BB = BB.delete_columns([ii])
                BB = BB.delete_rows([ii])
                monomials.pop(ii)
                BB = remove_unhelpful(BB, monomials, bound, ii-1)
                return BB

            # level:1
            # if just one was affected we check
            # if it is affecting someone else
            elif affected_vectors == 1:
                affected_deeper = True
                for kk in range(affected_vector_index + 1, BB.dimensions()[0]):
                    # if it is affecting even one vector
                    # we give up on this one
                    if BB[kk, affected_vector_index] != 0:
                        affected_deeper = False
                # remove both it if no other vector was affected and
                # this helpful vector is not helpful enough
                # compared to our unhelpful one
                if affected_deeper and abs(bound - BB[affected_vector_index, affected_vector_index]) < abs(bound - BB[ii, ii]):
                    print ("* removing unhelpful vectors", ii, "and", affected_vector_index)
                    BB = BB.delete_columns([affected_vector_index, ii])
                    BB = BB.delete_rows([affected_vector_index, ii])
                    monomials.pop(affected_vector_index)
                    monomials.pop(ii)
                    BB = remove_unhelpful(BB, monomials, bound, ii-1)
                    return BB
    # nothing happened
    return BB

""" 
Returns:
* 0,0   if it fails
* -1,-1 if `strict=true`, and determinant doesn't bound
* x0,y0 the solutions of `pol`
"""
def boneh_durfee(pol, modulus, mm, tt, XX, YY):
    """
    Boneh and Durfee revisited by Herrmann and May
    
    finds a solution if:
    * d < N^delta
    * |x| < e^delta
    * |y| < e^0.5
    whenever delta < 1 - sqrt(2)/2 ~ 0.292
    """

    # substitution (Herrman and May)
    PR.<u, x, y> = PolynomialRing(ZZ)
    Q = PR.quotient(x*y + 1 - u) # u = xy + 1
    polZ = Q(pol).lift()

    UU = XX*YY + 1

    # x-shifts
    gg = []
    for kk in range(mm + 1):
        for ii in range(mm - kk + 1):
            xshift = x^ii * modulus^(mm - kk) * polZ(u, x, y)^kk
            gg.append(xshift)
    gg.sort()

    # x-shifts list of monomials
    monomials = []
    for polynomial in gg:
        for monomial in polynomial.monomials():
            if monomial not in monomials:
                monomials.append(monomial)
    monomials.sort()
    
    # y-shifts (selected by Herrman and May)
    for jj in range(1, tt + 1):
        for kk in range(floor(mm/tt) * jj, mm + 1):
            yshift = y^jj * polZ(u, x, y)^kk * modulus^(mm - kk)
            yshift = Q(yshift).lift()
            gg.append(yshift) # substitution
    
    # y-shifts list of monomials
    for jj in range(1, tt + 1):
        for kk in range(floor(mm/tt) * jj, mm + 1):
            monomials.append(u^kk * y^jj)

    # construct lattice B
    nn = len(monomials)
    BB = Matrix(ZZ, nn)
    for ii in range(nn):
        BB[ii, 0] = gg[ii](0, 0, 0)
        for jj in range(1, ii + 1):
            if monomials[jj] in gg[ii].monomials():
                BB[ii, jj] = gg[ii].monomial_coefficient(monomials[jj]) * monomials[jj](UU,XX,YY)

    # Prototype to reduce the lattice
    if helpful_only:
        # automatically remove
        BB = remove_unhelpful(BB, monomials, modulus^mm, nn-1)
        # reset dimension
        nn = BB.dimensions()[0]
        if nn == 0:
            print ("failure")
            return 0,0

    # check if vectors are helpful
    if debug:
        helpful_vectors(BB, modulus^mm)
    
    # check if determinant is correctly bounded
    det = BB.det()
    bound = modulus^(mm*nn)
    if det >= bound:
        print ("We do not have det < bound. Solutions might not be found.")
        print ("Try with highers m and t.")
        if debug:
            diff = (log(det) - log(bound)) / log(2)
            print ("size det(L) - size e^(m*n) = ", floor(diff))
        if strict:
            return -1, -1
    else:
        print ("det(L) < e^(m*n) (good! If a solution exists < N^delta, it will be found)")

    # display the lattice basis
    if debug:
        matrix_overview(BB, modulus^mm)

    # LLL
    if debug:
        print ("optimizing basis of the lattice via LLL, this can take a long time")

    BB = BB.LLL()

    if debug:
        print ("LLL is done!")

    # transform vector i & j -> polynomials 1 & 2
    if debug:
        print ("looking for independent vectors in the lattice")
    found_polynomials = False
    
    for pol1_idx in range(nn - 1):
        for pol2_idx in range(pol1_idx + 1, nn):
            # for i and j, create the two polynomials
            PR.<w,z> = PolynomialRing(ZZ)
            pol1 = pol2 = 0
            for jj in range(nn):
                pol1 += monomials[jj](w*z+1,w,z) * BB[pol1_idx, jj] / monomials[jj](UU,XX,YY)
                pol2 += monomials[jj](w*z+1,w,z) * BB[pol2_idx, jj] / monomials[jj](UU,XX,YY)

            # resultant
            PR.<q> = PolynomialRing(ZZ)
            rr = pol1.resultant(pol2)

            # are these good polynomials?
            if rr.is_zero() or rr.monomials() == [1]:
                continue
            else:
                print ("found them, using vectors", pol1_idx, "and", pol2_idx)
                found_polynomials = True
                break
        if found_polynomials:
            break

    if not found_polynomials:
        print ("no independant vectors could be found. This should very rarely happen...")
        return 0, 0
    
    rr = rr(q, q)

    # solutions
    soly = rr.roots()

    if len(soly) == 0:
        print ("Your prediction (delta) is too small")
        return 0, 0

    soly = soly[0][0]
    ss = pol1(q, soly)
    solx = ss.roots()[0][0]

    #
    return solx, soly

def example(N,e,delta,m=None):
    ############################################
    # How To Use This Script
    ##########################################

    #
    # Lattice (tweak those values)
    #

    # you should tweak this (after a first run), (e.g. increment it until a solution is found)
    m = 4 if m == None else m # size of the lattice (bigger the better/slower)

    # you need to be a lattice master to tweak these
    t = int((1-2*delta) * m)  # optimization from Herrmann and May
    X = 2*floor(N^delta)  # this _might_ be too much
    Y = floor(N^(1/2))    # correct if p, q are ~ same size

    #
    # Don't touch anything below
    #

    # Problem put in equation
    P.<x,y> = PolynomialRing(ZZ)
    A = int((N+1)/2)
    pol = 1 + x * (A + y)

    #
    # Find the solutions!
    #

    # Checking bounds
    if debug:
        print ("=== checking values ===")
        print ("* delta:", delta)
        print ("* delta < 0.292", delta < 0.292)
        print ("* size of e:", int(log(e)/log(2)))
        print ("* size of N:", int(log(N)/log(2)))
        print ("* m:", m, ", t:", t)

    # boneh_durfee
    if debug:
        print ("=== running algorithm ===")
        start_time = time.time()

    solx, soly = boneh_durfee(pol, e, m, t, X, Y)

    # found a solution?
    if solx > 0:
        print ("=== solution found ===")
        if False:
            print ("x:", solx)
            print ("y:", soly)

        d = int(pol(solx, soly) / e)
        print ("private key found:", d)
    else:
        print ("=== no solution was found ===")

    if debug:
        print("=== %s seconds ===" % (time.time() - start_time))
    return d

if __name__ == "__main__":
    n = ?
    e = ?
    c = ?
    # the hypothesis on the private exponent (the theoretical maximum is 0.292)
    delta = 0.28 # this means that d < N^delta
    d = example(n,e,delta)
    print(long_to_bytes(int(pow(c,d,n))))

e和φ不互质

1.二次剩余+CRT

• 此方法适用于$e$e非质，即$gcd(e,\\varphi )=2$gcd(e,varphi)=2
• 要求解任意模数二次剩余先要求解奇素数模数二次剩余–具体理论查看此链
• 据此推导过程编写脚本

实例[0xGame 2024] Number-Theory-CRT

from Crypto.Util.number import *
import sympy
from sympy.ntheory.modular import crt
c = ?
e = ?
n = ?
p = ?
q = ?
phi = (p-1) * (q-1)
print(GCD(e,phi)) # 发现公约数2,分析得到二次剩余
 
def find_quadratic_residues(a, p):
    # 首先检查 a 是否是模 p 下的二次剩余
    if not sympy.is_quad_residue(a, p):
        return None  # 如果 a 不是二次剩余，返回 None
    
    # 使用 sympy 的 nthroot_mod 找到一个解
    x = sympy.nthroot_mod(a, 2, p, all_roots=False)
    
    # 计算另一个解
    second_solution = p - x
    
    return (x, second_solution)
 
x1 = find_quadratic_residues(c,p)       # 求解模p下的二次剩余
x2 = find_quadratic_residues(c,q)       # 求解模q下的二次剩余
 
for i in x1:
    for j in x2:
        remainders = [i,j]
        mods = [p,q]
        m_ = crt(mods, remainders)[0]   # CRT合并得到模n的二次剩余解
        c_ = m_%n
 
        e_ = e//2
        d = inverse(e_,phi)
        m = pow(c_,d,n)
        print(long_to_bytes(m))

2.用Sage求解有限域上开方

• 理论来说应该可以解决任意公约数的不互质问题，但是运行时间较长
• 自认为Method 1适用于gcd小、n大的情况，Method 2适用于gcd大、n小的情况

from Crypto.Util.number import *
from sympy.ntheory.modular import crt

p = ?
q = ?
n = ?
c = ?
e = ?

phi = (p-1)*(q-1)
gcd = GCD(e,phi)
d = inverse(e//gcd,phi)

# Method 1 求解同余方程
R.<x> = PolynomialRing(Zmod(p))
f = x^gcd - c
res1 = f.roots(multiplicities=False)

R.<x> = PolynomialRing(Zmod(q))
f = x^gcd - c
res2 = f.roots(multiplicities=False)

# Method 2 开根号
res1 = Zmod(p)(c).nth_root(gcd, all=True)
res2 = Zmod(q)(c).nth_root(gcd, all=True)

for i in res1:
    for j in res2:
        m = crt([p,q],[int(i),int(j)])
        if m is not None:
            try:
                print(long_to_bytes(int(pow(m[0],d,n))).decode())
            except Exception as e:
                continue

3.AMM算法

当$e$e和 $\\varphi(n)$varphi(n)不互质时，便无法通过逆元计算出$d$d，此时可以使用 AMM 算法计算得到明文

例如 $c=m^e \\mod n$c=memodn，如果 $gcd(e,\\varphi(n))=g$gcd(e,varphi(n))=g，那么就有 $c=m^{e’g^t} \\mod n$c=me’gtmodn，只要计算模$n$n下$c$c的 $g^t$gt 次方根即可得到能够正常解密的密文了

AMM算法就是计算有限域下的开方运算，然后用CRT合并即可

二次剩余

二次剩余是指对于一个模数 $n$n，如果存在整数 $a$a 使得 $a^2 \\equiv x \\mod n$a2equivxmodn，则称 $x$x 是模 $n$n 的二次剩余。否则称为二次非剩余。

Euler判别法：对于任意一个数$x$x和奇素数$p$p，若 $x^{\\frac{p-1}{2}} \\equiv 1 \\mod p$xfracp−12equiv1modp，则$x$x为二次剩余。若 $x^{\\frac{p-1}{2}} \\equiv -1 \\mod p$xfracp−12equiv−1modp，则$x$x为二次非剩余。

已知$x$x，求$a$a的时候可以这样看

能注意到对于奇素数$p$p一定能有这样的表示$p-1=2^t\\cdot$p−1=2tcdot

当$t=1$t=1时

$x^s\\equiv 1 \\mod p$xsequiv1modp

$\\Rightarrow x^{\\frac{s+1}{2}}\\equiv x^{\\frac{1}{2}} \\mod p$Rightarrowxfracs+12equivxfrac12modp

$\\because \\pm a\\equiv x^{\\frac{1}{2}} \\mod p$becausepmaequivxfrac12modp