Crypto

（话说题目做一半就当答案是什么鬼）

Crypto-bacon

题目

flag{AAAABAAAAAAAABAABBBAABBABABAAABAABAAAABBAABAAABABBABAAAAAABAABAAAABBBABABAABAABA}

解答

简单的培根密码，略

Crypto-黄金分割RSA

题目

encryption

[1,28657,2,1,3,17711,5,8,13,21,46368,75025,34,55,89,610,377,144,233,1597,2584,4181,6765,10946,987]

output

publickey=[0x1d42aea2879f2e44dea5a13ae3465277b06749ce9059fd8b7b4b560cd861f99144d0775ffffffffffff,5]
c=421363015174981309103786520626603807427915973516427836319727073378790974986429057810159449046489151

解答

（本题为2020 GACTF-da Vinci after rsa，然后河南天安杯又考一遍，这是第三遍）

首先使用yafu分解n，得到三个质因数

注意到e为5，与其中的q = P17-1和r = P79-1均不互质，所以需要尝试另外的解法。假设明文为m，那么m应该满足

m^5 = c (mod p)
m^5 = c (mod q)
m^5 = c (mod r)

在线上赛的情况下，我们可以利用网站www.wolframalpha.com求解m模p、q、r的可能值，并利用中国剩余定理求出m的所有可能值，进而判断其中哪一个是flag。

p_roots = [7361]
q_roots = [2722510300825886, 6139772527803903, 6537111956662153, 8415400986072042, 9898464751509789]
r_roots = [180966415225632465120208272366108475667934082405238808958048294287011243645, 2816114411493328258682873357893989007684496552202823306045771363205185148674391, 1369135259891793292334345751773139388112378132927363770631732500241630990458667, 5570877862584063114417410584640901580756179707042774516590562822938385811269597, 8499052407588078002885931765166137308397074232361087682974448633946350539292222]

写出解题代码如下：

#!/usr/bin/env python
# -*- coding: utf-8 -*-
from Crypto.Util.number import *
import gmpy2def GCRT(mi, ai):assert (isinstance(mi, list) and isinstance(ai, list))curm, cura = mi[0], ai[0]for (m, a) in zip(mi[1:], ai[1:]):d = gmpy2.gcd(curm, m)c = a - curaassert (c % d == 0)  # 不成立则不存在解K = c / d * gmpy2.invert(curm / d, m / d)cura += curm * Kcurm = curm * m / dreturn (cura % curm, curm)  # (解,最小公倍数)n = 0x1d42aea2879f2e44dea5a13ae3465277b06749ce9059fd8b7b4b560cd861f99144d0775ffffffffffff
c = 421363015174981309103786520626603807427915973516427836319727073378790974986429057810159449046489151
p = 9749
q = 11237753507624591
r = n / p / q
e = 5p_roots = [7361]
q_roots = [2722510300825886, 6139772527803903, 6537111956662153, 8415400986072042, 9898464751509789]
r_roots = [180966415225632465120208272366108475667934082405238808958048294287011243645, 2816114411493328258682873357893989007684496552202823306045771363205185148674391, 1369135259891793292334345751773139388112378132927363770631732500241630990458667, 5570877862584063114417410584640901580756179707042774516590562822938385811269597, 8499052407588078002885931765166137308397074232361087682974448633946350539292222]m_list = []
for pp in p_roots:for qq in q_roots:for rr in r_roots:res = GCRT([p, q, r], [pp, qq, rr])[0]if pow(res, e, n) == c:print long_to_bytes(res)

得到一个字符串flag{weadfa9987_adwd23123_454f}，但是我们还有一个条件没有使用，即一串数

[1,28657,2,1,3,17711,5,8,13,21,46368,75025,34,55,89,610,377,144,233,1597,2584,4181,6765,10946,987]

仔细观察发现，这些数就是斐波那契数列的一个排列，且其总数为25，恰好与flag{}内的字符数相同，因此 将字符按照同样方式打乱后 可以得到真实flag。

（不要问我为什么原题需要按加密方式打乱而不是按照加密方式进行恢复，问GACTF出题人）

Crypto-Corrupted Keys

题目

ciphertext.txt

c = 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

private_key.pem

-----BEGIN RSA PRIVATE KEY-----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......

public_key.pem

解答

这道题本身设置了两个考点：一是如何从残缺的私钥中提出信息，二是如何利用中间被挖去一段的dp来分解n。前者参见前两年0ctf的一道题，后者参见今年11月辽宁祥云杯。但是这道题私钥直接给全了，导致使用openssl命令直接可以提出数据，第一个考点直接失效。应该如同上边一样给出。

开始解体。首先从公钥中提出n和e
openssl rsa -in public_key.pem -pubin -modulus -text

但是仅有n、e、c是无法做题的，因为n不能直接分解，所以我们还需要看看私钥给我们留下了什么。根据资料，我们得知RSA的私钥通常以PKCS#1的模式进行存储，简单地说如下所示：

RSAPrivateKey ::= SEQUENCE {version           Version,modulus           INTEGER,  -- npublicExponent    INTEGER,  -- eprivateExponent   INTEGER,  -- dprime1            INTEGER,  -- pprime2            INTEGER,  -- qexponent1         INTEGER,  -- d mod (p-1)exponent2         INTEGER,  -- d mod (q-1)coefficient       INTEGER,  -- (inverse of q) mod potherPrimeInfos   OtherPrimeInfos OPTIONAL

我们将现在已经被污染的私钥base64解码后，按照上边的格式展开，得到如下内容：

308204a3020100
0282010100
(n)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
0203
(e)010001
02820100
(d)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
02818100
(p)0000000000100000000000000000000050000000008003007000a000a000000ac00000000000000000000000000000900030000000000000000000000b00000000b000080000000000000d00000000000000000000002000000000000c006000000000000000b000000100020000030000000000000000050000000000000000
02818100
(q)0000b000a000000000d0000010000700000000000000000f400000000000000000002000000e0000b300000000050000a0000000000000000000005000000000a00e000002000000000000000c08000000000e003000000000000064000060000000000000030000800000090000005005000000000000000000000000000000
028180
(dp)05396ecce54732b308ed28d5e8efcdc184350f1d33c6e3d9f3c6c537dde2e2f29c59ea115a14f6843bc3419011cca8f23a94f9ffc2e8ac350bd69dae6281010000000000000000000000000000000000000000000000000083416befe4b7a0092d304f46324bcf622aa4d6a78360b657d6692b5f2d6fd9a27be2af89e3551d63
028180
(dq)0020f10b000000a0050090000000000000020000a900000600d0000000000000000000700a00000001000c00000030060000b30d20080000000000000800000000f000000000000040f00000000000a00000000000060000000000000950005000000000e000000000000090000080000a00000000000b0000000000d0e40000
02818100
(coeff)0000000000a00500d00c020f00000400c90000000070000000000000e0000000000000000000008f0000060f00c000003000000200000000090000800000a007000000000000000000000000000c0000000000d00000000019（残缺）0

可以看到，除了dp以外其他内容基本已经无法使用，dp只缺少其中的200位，因此可以尝试使用coppersmith方法去解。由于coppersmith方法需要未知变量系数为1，我们这里尝试推导一下：

我们知道

由于dp <= p-1，所以k<e，coppersmith方法通过遍历1~e寻找k。现在假设k已知，那么有

设

带入原式得

这样就构造出了系数为1的同余方程，可以使用coppersmith方法解题了。解题脚本如下：

from sage.all import *
from Crypto.Util.number import long_to_bytesn = 0x8dd3cbe24a951e85d5bbb1805a965fe72d0d2a30837705786969119c604bae563e7c8e7a3fc85c148de19fa05f7dfe1d922137435f0c79931ac696c623d714ce3aa35e18bb0e0f2026d2eb391a0d5ede0f31af5d17e3ff2f76112051f94618126baf56c121ed83dd4a7a584b577cf256a86d59fe3696434068425926372768015c3cb33441c0505157d77e3dd519ca41b1575c3e630c3117a98a57fadc8f9b2d671124b2aaa9e547c784a0c45a749f3adfc83c3672ba101454aff45e40a51a8932ac0c87a5736733966b4967ab682f53c665a21394ec2f287b31e8ca6c432325470d90b2020505154e9c8a273401803e41e0828fa9c5fdf2d478911a236ec3f1
e = 0x10001
c = 0x6f9c3479883b414030032610a0831089ea2d5f6598d16f8b3415dbb7ff88e6214c7704dbaf1f0f0fe8243468b203b0c128933ab45f406109d234ab94457aa4ff81de3e0c1dda55b95344683e7cfef4e39dedd1203120af89e14702ac54a1a21adb500dadc67033deb2dcf844aa10c5b6425aca0a756ee5e5ce5b583de68d7dfa675b8142c4b175b347bd1c3b2d2cd32aa2e03356ecf4821704d7b7542a22d09ebb239e382fc5b72ea051b65596e41d228fb7b0f7acf5686d05b8d6807a26c1a1d92c8b116c6f27e2b21ded5f1f3b8f9a88e45ec7b14aee18e74454fefb1a482a9eafc9550d16f6683e2f7cbd0d9ce9a474f4db01e2f97d0d3d23fad566489e1e
s1 = 0x5396ecce54732b308ed28d5e8efcdc184350f1d33c6e3d9f3c6c537dde2e2f29c59ea115a14f6843bc3419011cca8f23a94f9ffc2e8ac350bd69dae628101
s2 = 0x83416befe4b7a0092d304f46324bcf622aa4d6a78360b657d6692b5f2d6fd9a27be2af89e3551d63
invE = 6072463480052222774484440188940687848932613396747764165170539263002801068323119971186434001223564329479603123138991852830747459341281298547464945329258352716978884436175214634055317578153997659440696241077620973159496044408967363538897046549076241323642446031523788310065048047069216086916780671758259593054935742629841385169465939611015008003189747360588471082992102089514769381157672791863436498000591096254007370955960943534440348460456861334522885919455014210659853726203177705141103038754371809891816964608801118434914236230355247754146610175756211785759304159230507837819605093784920450130362742047348692105621
invpow = 11164431502070747511500530021745510906685334856887665721011597986932079197780161963381238394880452910362678853293436976134225721195690734856639343780389327693213031686657799035703254166242337351340806493525437022390070372393198894731543660124496416698434549412871378260095086278600202820500790434411495555569397376528832359674292547896319750287821110784519344030510311252066880377461777551185303592001846300617993580196858607671355081397538545831351265009506760347727474227285668988817907365295771630760269871647111667154970535637451277321913843646237831493800721780046762737829851933071451680344164843758329746830379def coppersmith(k):F.<x> = PolynomialRing(Zmod(n))f = (s1 << 520) * invpow + x + s2 * invpow + (k - 1) * invE * invpow   # make monicx0 = f.small_roots(X=2 ** 200, beta=0.44, epsilon=1/32)return x0for k in range(1, e):print kx0 = coppersmith(k)if len(x0) != 0:x = Integer(x0[0])dp = (s1 << 520) + (x << 320) + s2p = (e*dp - 1) // k + 1if p != -1:q = n // passert n == p * qphi = (p-1)*(q-1)d = inverse_mod(e,phi)print dprint long_to_bytes(pow(c,d,n))break

其中invE、invpow分别是计算好的invert(e, n)和invert(2 ** 320,n)，最终求出k=1733。

Crypto-strange_GSW

题目

strange_GSW.sage

from Crypto.Util.number import *
from random import randrange
from hashlib import md5
from secret import FLAGdef GenerateG(_n, _q):_len = int(round(log(_q, 2)))_G = Matrix(ZZ, _len * _n, _n)for i in range(_len):for j in range(_n):_G[j * _len + i, j] = 2 ** ireturn _Gdef BinaryExpansion(_C, _q):expansion_C = Matrix(ZZ, _C.nrows(), _C.nrows())log_q = _C.nrows() // _C.ncols()for i in range(_C.nrows()):for j in range(_C.ncols()):bits = bin(_C[i, j] % _q)[2:].rjust(log_q, '0')[::-1]for index in range(log_q):expansion_C[i, j * log_q + index] = int(bits[index])return expansion_Cdef KeyGen(degree, _q, _B):s = random_matrix(ZZ, degree, 1, x=_q // 4, y=4 * _q // 3)return Matrix(s.list() + [-1]).transpose(), sdef BitEncrypt(plain_bit, _n, _q, _B, _G, _s):_m = int(round(log(_q, 2)) * _n)A = random_matrix(ZZ, _m, _n - 1, x=0, y=_q)e = random_matrix(ZZ, _m, 1, x=0, y=_B)_C = block_matrix([A, A * _s + e], ncols=2, subdivide=False) + plain_bit * _G_C = BinaryExpansion(_C, _q)return _Cdef Encrypt(plain, _n, _q, _B, _G, _s):plain_bits = bin(plain)[2:]Cipher = []for bit in plain_bits:Cipher.append(BitEncrypt(int(bit), _n, _q, _B, _G, _s))return Cipherif __name__ == '__main__':n = 11q = 65537B = q // 256G = GenerateG(n, q)decrypt_key, encrypt_key = KeyGen(n - 1, q, B)flag = FLAG + md5(long_to_bytes(randrange(2 ** 127, 2 ** 128))).hexdigest().encode()[:8] + b'}'_cipher = Encrypt(bytes_to_long(flag), n, q, B, G, encrypt_key)str_cipher = ' '.join([str(i.list()) for i in _cipher])with open('flag', 'wb') as f:f.write(flag)with open('cipher', 'w') as f:f.write(str_cipher)

解答

（比赛好像没人做出来，确实非常费劲）

题目看起来比较复杂，所以我们跟着程序走，看看具体流程是什么样的。首先看矩阵G

可知矩阵G是一个固定的矩阵。然后生成一组加密密钥和解密密钥看看

可以看到加密密钥为一个10 * 1的矩阵，其中每项的值在(q/4, q*4/3)之间，q = 65537。解密密钥就比加密密钥多一个-1。

最后看加密过程，首先将密文转为整数，再转为二进制，然后对于二进制表示的每一位（0或1）进行BitEncrypt加密操作，将这个0或者1加密成一个176 * 176的由0和1构造成的矩阵。加密的过程中参数生成随机，所以相同明文加密结果不固定。加密生成的_C矩阵为一个176 * 11的矩阵，再经过BinaryExpansion函数扩展到一个176 * 176的矩阵。这个扩展很简单，就是把这些数每一个按照其二进制表示横向展开成16位，然后倒序写入，因此我们可以简单将其收起得到_C。

def recoverMatrix(text):text = text.replace(',','').replace(' ','').replace('[','')Mdata = []for i in range(0, len(text), 16):Mdata.append(int(text[i:i+16][::-1],2))_C = Matrix(ZZ, 176, 11, Mdata)return _C

在得到_C之后，我们来尝试求解key。_C的结构是矩阵A拼接一列A * _s + e的结果，其中e是0~256以内的随机值，再根据明文该位置是0还是1来决定是否加上矩阵_G。考虑到大部分参数取值范围在(q/4, q*4/3)之间，因此e的(0, 256)区间可以视为小值，可以将本题当作一个格上的最近向量问题（CVP, Closet vector problem）进行求解。对明文首位进行求解时，针对结果可能为1和0的两种情况，求出的_C矩阵需要减去_G 或者不发生变化。当存在解时，首位明文正确，且求解出的key就是加密所用key，后续每个明文字符直接用该key还原即可。完整解题代码如下：

from sage.modules.free_module_integer import IntegerLattice
from Crypto.Util.number import long_to_bytesdef GenerateG(_n, _q):_len = int(round(log(_q, 2)))_G = Matrix(ZZ, _len * _n, _n)for i in range(_len):for j in range(_n):_G[j * _len + i, j] = 2 ** ireturn _Gdef recoverMatrix(text):text = text.replace(',','').replace(' ','').replace('[','')Mdata = []for i in range(0, len(text), 16):Mdata.append(int(text[i:i+16][::-1],2))_C = Matrix(ZZ, 176, 11, Mdata)return _Cdef CVP(lattice, target):gram = lattice.gram_schmidt()[0]t = targetfor i in reversed(range(lattice.nrows())):c = ((t * gram[i]) / (gram[i] * gram[i])).round()t -= lattice[i] * creturn target - tdef BitDecrypt(_C, _G, _s, _q):_C = _C * _G * _sreturn 0 if abs(_C.list()[0] % _q - _q) < _q // 4 else 1def Decrypt(Cipher, _G, _s, _q):plain_bits = ''for cipher in Cipher:plain_bits += str(BitDecrypt(cipher, _G, _s, _q))return int(plain_bits, 2)if __name__ == '__main__':n = 11q = 65537B = q // 256G = GenerateG(n, q)with open('cipher', 'r') as f:cipher_str = f.read()cipher_str = cipher_str.split(']')[:-1]# Store Ciphertext_cipher_str = [eval(i.strip()+']') for i in cipher_str]_cipher_matrix = [Matrix(176, 176, i) for i in _cipher_str]# Recover Matrix _C_C = recoverMatrix(str(cipher_str[0]))_C = _C - G_A = _C[:176, :10]res = _C[:176, 10:].list()# Make a Matrix for CVPM = Matrix(ZZ, 186, 176)for i in range(176):for j in range(10):M[176 + j, i] = int(_A[i][j])M[i, i] = 65537lattice = IntegerLattice(M, lll_reduce=True)target = vector(ZZ, res)res = CVP(lattice.reduced_basis, target)# Recover KeyR = IntegerModRing(65537)M = Matrix(R, _A[:176])key = M.solve_right(res)key = [int(i) for i in key]print key# get flagflag = long_to_bytes(Decrypt(_cipher_matrix, G, Matrix(key + [-1]).transpose(), q))print(flag)

Crypto-random_fault

（我真不会国密。。。）

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【CTF WriteUp】2020电信和互联网行业赛个人赛部分Crypto题解