DES加密解密 Feistel算法网络结构详讲-Toy模板网

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简单知识导入：

DES算法是属于对称密码算法中的分组加算法。

分组加密和流密码加密是相对应的。

流密码是逐字节进行加密，即一个字节一个字节进行加密

分组加密算法也叫块加密，将明文分成固定字节块，对每个字节块分别进行加密，最后拼接在一起得到密文

密钥长64位,56位参与运算,其余8位为校验位。(8,16,24,32,40,48,56,64) 可以看出是8的倍数

当n个64位明文数据块都经过DES加密处理后,所得到的n个64位密文数据块串在一起就是密文。

具体过程：

$采用 F e i s t e l 算法 :$
$L_i = R_{i-1}$
$R_i = L_{i-1} \oplus f(R_{i-1}, K_i)$
DES加密解密 Feistel算法网络结构详讲,块密码,密码学,安全,算法,DES,Feistel

DES加密解密 Feistel算法网络结构详讲,块密码,密码学,安全,算法,DES,Feistel

$首先对64bits的明文进行初始IP置换，得到打乱顺序的64bits的数据\\接着对它对半分，得到32bits的L_{0}和R_{0}\\自此开始进入Feistel算法，知道R就很容易知道L,所以要解决就得从R入手，\\ 已知R_i = L_{i-1} \oplus f(R_{i-1}, K_i)，\\f()即使图中的F轮函数，进行完【4.P置换】后即可得到f(R_{i-1}, K_i)\\ 接着将其与L_{i-1}进行异或即可得到R_{i},自此第一轮运算结束，\\经过16轮相同的运算得到L_{16}和R_{16}，\\拼接在一起后进行最后的逆初始置换即可得到最终的密文$

下面我们对照图中一个一个来讲解

IP置换（ 64 − > 64 64->64 64−>64）

$按照一定的规则，将原来的64比特的二进制数进行重新排序，\\这里的规则即指初始置换表（最后的IP逆置换和此处置换过程一模一样，仅表不同），\\ 对照下面的表，将二进制明文的第58位放到第一位，\\第50位放到第二位，以此类推，最终得到重新排序的数据。自此，IP置换完成。$
DES加密解密 Feistel算法网络结构详讲,块密码,密码学,安全,算法,DES,Feistel

轮函数–E扩展置换（ 32 − > 48 32->48 32−>48）

$我觉得这里对照下面的图看是能看懂的，先将32bits的R分成8组，每组4bits,\\每组最前和最后各加1bits,不是随便加，按图中所给的表进行扩展$
DES加密解密 Feistel算法网络结构详讲,块密码,密码学,安全,算法,DES,Feistel

$不好理解的话举个例子：①中是32bits的数据，按照置换表进行置换后得到③④\\（此过程和最开始的IP置换的原理和过程是一样的）\\扩展后得到48bits$
DES加密解密 Feistel算法网络结构详讲,块密码,密码学,安全,算法,DES,Feistel

轮函数–与子密钥异或（ 48 − > 48 48->48 48−>48）

$上一步得到的拓展流与子密钥异或,子密钥均为48bits\\(为什么是48bits，文章最后来讲)$

轮函数–S盒压缩处理（ 48 − > 32 48->32 48−>32）

$首先先将48bits输入等分为8块，每块6bits,\\压缩后得到4位的输出，最终得到32bits输出，那如何进行压缩呢？（请移步下图的举例）$
DES加密解密 Feistel算法网络结构详讲,块密码,密码学,安全,算法,DES,Feistel

$举个例子：\\例如前6位111111，取头尾两个结合得到二进制11，转为10进制3，将其作为行数\\中间四位数据组合得到二进制1111，转为10进制15，将其作为列数，\\到S盒中相应的行列中去找，得到13，13转成的二进制数1101即为最终结果\\自此，6bits->4bits，48bits->32bits$
注意：行列都是从0开始，故等下代码阶段在S盒中找数,行列不需减一
DES加密解密 Feistel算法网络结构详讲,块密码,密码学,安全,算法,DES,Feistel

轮函数–P盒置换

$S盒所得结果再经过P盒置换（与初始IP置换过程相同）。\\至此，一次轮函数操作完毕$
DES加密解密 Feistel算法网络结构详讲,块密码,密码学,安全,算法,DES,Feistel

异或

$一次轮函数加密后的数据与L_{i-1}进行异或即可得到R_{i}（R_i = L_{i-1} \oplus f(R_{i-1}, K_i)）\\自此，一轮迭代完成$

IP逆置换

$进行16轮相同的运算，得到L_{16},R_{16}后将其进行拼接得到64bits的数据，\\将该64bits的数据进行IP逆置换后即可得到最终密文$

注意：以上所有置换表都是已知的，固定的，给定的，不可改的（但如果有人拿此出题进行修改就不好说了）

接下来讲讲密钥生成

密钥生成

$64bits密钥（种子密钥），首先通过PC-1表的置换得到56bits数据\\（里面无8的倍数，故该表的主要作用是去除校验码）\\平分分别得到28bits的C_{0},D_{0}\\根据移位次数表将C,D进行左移\\ 第一次：将C_{0},D_{0}左移1位得到C_{1},D_{1}\\ \hspace{4em}将C_{1},D_{1}进行拼接再通过PC-2表置换得到第一轮子密钥K_{1}\\ 第二次：将C_{1},D_{1}左移1位得到C_{2},D_{2}\\ \hspace{4em}将C_{2},D_{2}进行拼接再通过PC-2表置换得到第一轮子密钥K_{2} \\ 第三次：将C_{2},D_{2}左移2位得到C_{3},D_{3}\\ \hspace{4em}将C_{3},D_{3}进行拼接再通过PC-2表置换得到第一轮子密钥K_{3}\\ 以此类推，最终生成16轮子密钥$
注：经过PC-2，得到48位子密钥（56 --> 48）（9,18,22,25,35,38,43,54 not in PC-2）
DES加密解密 Feistel算法网络结构详讲,块密码,密码学,安全,算法,DES,Feistel

代码部分

# 本脚本名: __table.py
 
# 明文M 的初始置换 IP
IP = [58, 50, 42, 34, 26, 18, 10, 2,
      60, 52, 44, 36, 28, 20, 12, 4, 
      62, 54, 46, 38, 30, 22, 14, 6, 
      64, 56, 48, 40, 32, 24, 16, 8,
      57, 49, 41, 33, 25, 17,  9, 1, 
      59, 51, 43, 35, 27, 19, 11, 3, 
      61, 53, 45, 37, 29, 21, 13, 5, 
      63, 55, 47, 39, 31, 23, 15, 7] # 将明文中的第58位换到第一位
 
# 逆初始置换 IP^-1
IP_INV = [40, 8, 48, 16, 56, 24, 64, 32,
          39, 7, 47, 15, 55, 23, 63, 31,
          38, 6, 46, 14, 54, 22, 62, 30,
          37, 5, 45, 13, 53, 21, 61, 29,
          36, 4, 44, 12, 52, 20, 60, 28,
          35, 3, 43, 11, 51, 19, 59, 27,
          34, 2, 42, 10, 50, 18, 58, 26,
          33, 1, 41,  9, 49, 17, 57, 25]
 
# 右子密钥扩展表 32->48位 
E = [32, 1,   2,  3, 4,   5,
     4,  5,   6,  7, 8,   9,
     8,  9,  10, 11, 12, 13,
     12, 13, 14, 15, 16, 17, 
     16, 17, 18, 19, 20, 21, 
     20, 21, 22, 23, 24, 25, 
     24, 25, 26, 27, 28, 29, 
     28, 29, 30, 31, 32,  1]
 
# 轮结构中的F函数中的置换P
P = [16,  7, 20, 21,
     29, 12, 28, 17,
     1,  15, 23, 26, 
     5,  18, 31, 10, 
     2,   8, 24, 14, 
     32, 27,  3,  9, 
     19, 13, 30,  6,
     22, 11,  4, 25]
 
# 压缩置换: 64位母密钥->56位母密钥
PC_1 = [57, 49, 41, 33, 25, 17,  9,
        1,  58, 50, 42, 34, 26, 18, 
        10,  2, 59, 51, 43, 35, 27, 
        19, 11,  3, 60, 52, 44, 36, 
        63, 55, 47, 39, 31, 23, 15,
        7,  62, 54, 46, 38, 30, 22,
        14,  6, 61, 53, 45, 37, 29,
        21, 13,  5, 28, 20, 12,  4]
 
# 压缩置换: 56位左右密钥->48位子密钥
PC_2 = [14, 17, 11, 24,  1,  5,
        3,  28, 15,  6, 21, 10,
        23, 19, 12,  4, 26,  8, 
        16,  7, 27, 20, 13,  2, 
        41, 52, 31, 37, 47, 55,
        30, 40, 51, 45, 33, 48,
        44, 49, 39, 56, 34, 53,
        46, 42, 50, 36, 29, 32]
 
# 左右子密钥的循环移位表
ROTATE = [1, 1, 2, 2, 2, 2, 2, 2,
          1, 2, 2, 2, 2, 2, 2, 1]
 
# 8 个 S 盒
SBOX = [
   [14, 4, 13, 1, 2, 15, 11, 8, 3, 10, 6, 12, 5, 9, 0, 7,
    0, 15, 7, 4, 14, 2, 13, 1, 10, 6, 12, 11, 9, 5, 3, 8,
    4, 1, 14, 8, 13, 6, 2, 11, 15, 12, 9, 7, 3, 10, 5, 0,
    15, 12, 8, 2, 4, 9, 1, 7, 5, 11, 3, 14, 10, 0, 6, 13],
 
   [15, 1, 8, 14, 6, 11, 3, 4, 9, 7, 2, 13, 12, 0, 5, 10,
    3, 13, 4, 7, 15, 2, 8, 14, 12, 0, 1, 10, 6, 9, 11, 5,
    0, 14, 7, 11, 10, 4, 13, 1, 5, 8, 12, 6, 9, 3, 2, 15,
    13, 8, 10, 1, 3, 15, 4, 2, 11, 6, 7, 12, 0, 5, 14, 9],
 
   [10, 0, 9, 14, 6, 3, 15, 5, 1, 13, 12, 7, 11, 4, 2, 8,
    13, 7, 0, 9, 3, 4, 6, 10, 2, 8, 5, 14, 12, 11, 15, 1,
    13, 6, 4, 9, 8, 15, 3, 0, 11, 1, 2, 12, 5, 10, 14, 7,
    1, 10, 13, 0, 6, 9, 8, 7, 4, 15, 14, 3, 11, 5, 2, 12],
 
   [7, 13, 14, 3, 0, 6, 9, 10, 1, 2, 8, 5, 11, 12, 4, 15,
    13, 8, 11, 5, 6, 15, 0, 3, 4, 7, 2, 12, 1, 10, 14, 9,
    10, 6, 9, 0, 12, 11, 7, 13, 15, 1, 3, 14, 5, 2, 8, 4,
    3, 15, 0, 6, 10, 1, 13, 8, 9, 4, 5, 11, 12, 7, 2, 14],
 
   [2, 12, 4, 1, 7, 10, 11, 6, 8, 5, 3, 15, 13, 0, 14, 9,
    14, 11, 2, 12, 4, 7, 13, 1, 5, 0, 15, 10, 3, 9, 8, 6,
    4, 2, 1, 11, 10, 13, 7, 8, 15, 9, 12, 5, 6, 3, 0, 14,
    11, 8, 12, 7, 1, 14, 2, 13, 6, 15, 0, 9, 10, 4, 5, 3],
 
   [12, 1, 10, 15, 9, 2, 6, 8, 0, 13, 3, 4, 14, 7, 5, 11,
    10, 15, 4, 2, 7, 12, 9, 5, 6, 1, 13, 14, 0, 11, 3, 8,
    9, 14, 15, 5, 2, 8, 12, 3, 7, 0, 4, 10, 1, 13, 11, 6,
    4, 3, 2, 12, 9, 5, 15, 10, 11, 14, 1, 7, 6, 0, 8, 13],
 
   [4, 11, 2, 14, 15, 0, 8, 13, 3, 12, 9, 7, 5, 10, 6, 1,
    13, 0, 11, 7, 4, 9, 1, 10, 14, 3, 5, 12, 2, 15, 8, 6,
    1, 4, 11, 13, 12, 3, 7, 14, 10, 15, 6, 8, 0, 5, 9, 2,
    6, 11, 13, 8, 1, 4, 10, 7, 9, 5, 0, 15, 14, 2, 3, 12],
 
   [13, 2, 8, 4, 6, 15, 11, 1, 10, 9, 3, 14, 5, 0, 12, 7,
    1, 15, 13, 8, 10, 3, 7, 4, 12, 5, 6, 11, 0, 14, 9, 2,
    7, 11, 4, 1, 9, 12, 14, 2, 0, 6, 10, 13, 15, 3, 5, 8,
    2, 1, 14, 7, 4, 10, 8, 13, 15, 12, 9, 0, 3, 5, 6, 11]]

# 本脚本名: __util.py
 
from __table import *
import time
 
 
def permutate(bitstr, table):
    """ 通用置换运算
        @table: 置换盒
    """
    res = ""
    for i in table:   
        res += bitstr[i-1] # 下标从0开始，故减一
    
    return res
 
 
def leftShift(key, num): 
    """ 子密钥LR: 左循环移位
        @key: 待移位的子密钥
        @num: 列表左移位位数
    """
    return key[num:] + key[0:num]
 
 
def XOR(str1, str2):
    """ 两比特串进行异或 """
    res = ""
    for i in range(0, len(str1)):
        xor = int(str1[i]) ^ int(str2[i])
        res += str(xor)
    return res
 
 
def Sbox(binstr):
    """ S盒置换: 48位->32位 """
    box = 0 # S盒的索引
    res = ""
    for i in range(0, len(binstr), 6):
        tmp = binstr[i:i+6]
        row = int(tmp[0] + tmp[5], 2) # 头尾数据结合成的2进制数据转化为10进制作为行数
        col = int(tmp[1:5], 2)        # 中间4位数据结合成的2进制数据转换为10进制作为列数
 
        num = bin(SBOX[box][16*row+col])[2:].zfill(4) # 0是包含在里面的，所以不需要row-1,都是从0开始的。然后将在s盒中查到的数值转化为4位的2进制数
        # zfill(4) 不足4位左侧填充0以补全4位
        
        # 这样将48位分成8 * 6 ，然后6->4，最后得到8 * 4 = 32比特的数据
        res += num
        box += 1
 
    return res 
 
 
def createSubKey(key64):
    """ 64位母密钥->16轮的48位子密钥列表
        @key64: 输入的64位加密密钥
        @return: 16轮子密钥列表
    """
    subkey = []
    key56  = permutate(key64, PC_1)       # 64位密钥压缩置换，64bits变58bits(将校验位去除)
    C28b, D28b = key56[0:28], key56[28:]  # 56位密钥左右平分
    
    for i in range(16):
        C28b = leftShift(C28b, ROTATE[i]) # 左循环移位
        D28b = leftShift(D28b, ROTATE[i]) # 左循环移位
        subkey.append(permutate(C28b + D28b, PC_2)) # 得到一轮子密钥，加到子密钥subkey列表中去
 
    return subkey
 
 
def turnFunction(R32b, subkey):
    """ 轮结构中的F函数
        @R32b: F函数输入参数 32位的右明文
        @subkey: F函数输入参数 48位子密钥
    """
    R48b = permutate(R32b, E) # E盒扩展: 右明文32位->48位
    # E扩展置换，将32位输入扩展为48位输出（要与48比特的密钥进行运算（密钥中58位参与运算，但如何到48比特，密钥生成过程中会讲，如今已知的是K1到K16都是48比特的子密钥））
    R48b = XOR(R48b, subkey)  # 两异或:  扩展流与子密钥异或 R_(i-1) ^ K_i
    R32b = Sbox(R48b)         # S盒压缩: 压缩48位->32位
    Fout = permutate(R32b, P) # P盒置换
 
    return Fout # 至此，一次轮函数操作完毕
 
 
def hexToBit(hexstr):
    """ 十六进制字符串->二进制比特流(每字符4位) """
    binstr = ""
    for i in hexstr:
        binstr += bin(int(i,16))[2:].zfill(4)
    return binstr
 
 
def bitToHex(bitstr):
    """ 二进制比特流->十六进制字符串 """
    hexstr = ""
    for i in range(0, len(bitstr), 4):
        hexstr += hex(int(bitstr[i:i+4], 2))[2:].upper()
    return hexstr
 
 
def takeTime(func):
    """ 装饰器: 计算函数运行耗时"""
    def decorated(*args):
        startime = time.time()
        func(*args)
        endtime  = time.time()
        taketime = round((endtime - startime)*10**6)
        print(f"\n->运算耗时: {taketime} 微秒")
        return func(*args)
    return decorated

# 本脚本名: des.py
 
from __util import *
 
 
class DES():
    """ DES 加解密 """
    
    def __init__(self, scrkey):
        self.scrkey = hexToBit(scrkey) # 初始化密钥 16进制字符->比特流 ,也叫种子密钥
 
    def feistel(self, in64bit, flag):
        """ Feistel 网络结构算法
            @in64bit: 输入64位比特流
            @flag: 正向算法/反向算法，正向为1，反向为0
            单字符由8比特表示
        """
        bit64  = permutate(in64bit, IP)      # 初置换IP重排明文
        L32b, R32b = bit64[0:32], bit64[32:] # 64位明文左右平分
        
        subkey = createSubKey(self.scrkey)   # 生成所有的子密钥  
        if flag == 0: subkey = subkey[::-1]  # 解密用的逆子密钥
 
        for i in range(16):
            Fout = turnFunction(R32b, subkey[i])
            L32b, R32b = R32b, XOR(L32b, Fout)
        out64bit = permutate(R32b + L32b, IP_INV) # 左右交换 逆初始置换
 
        return out64bit
 
    @takeTime
    def encrypt(self, messag):
        """ DES加密
            @messag: 16进制明文字符串
            @return: 16进制密文字符串
        """
        mesbit = hexToBit(messag)
        cipbit = self.feistel(mesbit, 1)
        cipher = bitToHex(cipbit)
 
        return cipher
 
    @takeTime
    def decrypt(self, cipher):
        """ DES 解密
            @cipher: 16进制密文字符串
        """
        cipher = hexToBit(cipher)
        mesbit = self.feistel(cipher, 0)
        messag = bitToHex(mesbit)
 
        return messag
 
 
if __name__ == '__main__':
 
    scrkey = "133457799BBCDFF1" 
    messag = "0123456789ABCDEF"
    cipher = "85E813540F0AB405"
 
    crypto = DES(scrkey)
    print("16进制密文:", crypto.encrypt(messag))
    print("16进制明文:", crypto.decrypt(cipher))

参考

https://www.bilibili.com/video/BV1KQ4y127AT/?spm_id_from=333.880.my_history.page.click&vd_source=0bd1680dd495123dd54d157e522a2e86

代码来源：
https://blog.csdn.net/Alpherkin/article/details/121198150?ops_request_misc=%257B%2522request%255Fid%2522%253A%2522168929564016800211547104%2522%252C%2522scm%2522%253A%252220140713.130102334.pc%255Fall.%2522%257D&request_id=168929564016800211547104&biz_id=0&utm_medium=distribute.pc_search_result.none-task-blog-2_allfirst_rank_ecpm_v1~rank_v31_ecpm-7-121198150-null-null.142^v88control_2,239^v2insert_chatgpt&utm_term=feistel%E5%8A%A0%E8%A7%A3%E5%AF%86%E7%AE%97%E6%B3%95&spm=1018.2226.3001.4187

持续更新中…

[NCTF-2019-Crypto]Reverse

(于2023-09-10记录)
题目描述：

task.py

import os
import pyDes


flag = "NCTF{******************************************}"
key = os.urandom(8)

d = pyDes.des(key)
cipher = d.encrypt(flag.encode())

with open('cipher', 'wb') as f:
    f.write(cipher)

# Leak: d.Kn[10] == [0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1]

cipher.txt

汮S賳!Ⅳ?Z_l,?漋& hDa莸&5匼?oZ€|U€骸%#}巴

记录佬的分析：

大致流程图：
DES加密解密 Feistel算法网络结构详讲,块密码,密码学,安全,算法,DES,Feistel

leak中泄露出的Kn[10]应该是11组子密钥K11，逆一下从C11 D11得到K11的PC_2过程(即图中的PERMUTED CHOICE 2过程)就可得到C11 D11
不过C11 D11是56位，K11是48位，所以得爆一下另外8位

再沿着此流程顺下去(Ci,Di经过16次LEFT SHIFTS后会复原)，就可以恢复出所有子密钥。

exp:文章来源地址https://www.toymoban.com/news/detail-560953.html

from base64 import b64decode
from itertools import product
from DES import * 			 # https://github.com/soreatu/Cryptography/blob/master/DES.py


guess_8bit = list(product(range(2), repeat=8))
not_in_PC2 = [9,18,22,25,35,38,43,54]

def re_PC2(sbkey):
    # 48-bit -> 56-bit
    res = [0]*56
    for i in range(len(sbkey)):
        res[PC_2_table[i]-1] = sbkey[i]
    return res # ok

def guess_CiDi10(sbkey, t):
    res = re_PC2(sbkey)
    for i in range(8):
        res[not_in_PC2[i]-1] = guess_8bit[t][i]
    return res # ok

def guess_allsbkey(roundkey, r, t):
    sbkey = [[]]*16
    sbkey[r] = roundkey
    CiDi = guess_CiDi10(roundkey, t)
    Ci, Di = CiDi[:28], CiDi[28:]
    for i in range(r+1,r+16):
        Ci, Di = LR(Ci, Di, i%16)
        sbkey[i%16] = PC_2(Ci+Di)
    return sbkey # ok

def long_des_enc(c, k):
    assert len(c) % 8 == 0
    res = b''
    for i in range(0,len(c),8):
        res += DES_enc(c[i:i+8], k)
    return res

def try_des(cipher, roundkey):
    for t in range(256):
        allkey = guess_allsbkey(roundkey, 10, t)
        plain = long_des_enc(cipher, allkey[::-1])
        if plain.startswith(b'NCTF'):
            print(plain)

if __name__ == "__main__":
    cipher = b64decode(b'm0pT2YYUIaL0pjdaX2wsxwedViYAaBkZA0Rh3bUmNYVclBlvWoB8VYC6oSUjfbDN')
    sbkey10 = [0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1]
    try_des(cipher, sbkey10)

# b'NCTF{1t_7urn3d_0u7_7h47_u_2_g00d_@_r3v3rs3_1snt}'

[NepCTF-2023-Crypto]simple_des

(于2023-09-10记录)

题目描述：

from operator import add
from typing import List
from functools import reduce
from gmpy2 import *
from Crypto.Util.number import *


_IP = [57, 49, 41, 33, 25, 17, 9,  1,
	59, 51, 43, 35, 27, 19, 11, 3,
	61, 53, 45, 37, 29, 21, 13, 5,
	63, 55, 47, 39, 31, 23, 15, 7,
	56, 48, 40, 32, 24, 16, 8,  0,
	58, 50, 42, 34, 26, 18, 10, 2,
	60, 52, 44, 36, 28, 20, 12, 4,
	62, 54, 46, 38, 30, 22, 14, 6
]

def IP(plain: List[int]) -> List[int]:
    return [plain[x] for x in _IP]

__pc1 = [56, 48, 40, 32, 24, 16,  8,
	  0, 57, 49, 41, 33, 25, 17,
	  9,  1, 58, 50, 42, 34, 26,
	 18, 10,  2, 59, 51, 43, 35,
	 62, 54, 46, 38, 30, 22, 14,
	  6, 61, 53, 45, 37, 29, 21,
	 13,  5, 60, 52, 44, 36, 28,
	 20, 12,  4, 27, 19, 11,  3
]

__pc2 = [
	13, 16, 10, 23,  0,  4,
	 2, 27, 14,  5, 20,  9,
	22, 18, 11,  3, 25,  7,
	15,  6, 26, 19, 12,  1,
	40, 51, 30, 36, 46, 54,
	29, 39, 50, 44, 32, 47,
	43, 48, 38, 55, 33, 52,
	45, 41, 49, 35, 28, 31
]
ROTATIONS = [1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1]

def PC_1(key: List[int]) -> List[int]:
    return [key[x] for x in __pc1]

def PC_2(key: List[int]) -> List[int]:
    return [key[x] for x in __pc2]

def get_sub_key(key: List[int]) -> List[List[int]]:
    key = PC_1(key)
    L, R = key[:28], key[28:]

    sub_keys = []

    for i in range(16):
        for j in range(ROTATIONS[i]):
            L.append(L.pop(0))
            R.append(R.pop(0))

        combined = L + R
        sub_key = PC_2(combined)
        sub_keys.append(sub_key)
    print('LL=',L[:19])
    print('Rr=',R)
    return sub_keys

__ep = [31,  0,  1,  2,  3,  4,
		 3,  4,  5,  6,  7,  8,
		 7,  8,  9, 10, 11, 12,
		11, 12, 13, 14, 15, 16,
		15, 16, 17, 18, 19, 20,
		19, 20, 21, 22, 23, 24,
		23, 24, 25, 26, 27, 28,
		27, 28, 29, 30, 31,  0
] # 咦，看出来此表不同之处吗

__p = [15,  6, 19, 20, 28, 11, 27, 16,
		0, 14, 22, 25,  4, 17, 30,  9,
		1,  7, 23, 13, 31, 26,  2,  8,
		18, 12, 29,  5, 21, 10,  3, 24
]

def EP(data: List[int]) -> List[int]:
    return [data[x] for x in __ep]

def P(data: List[int]) -> List[int]:
    return [data[x] for x in __p]

__s_box = [

	[
		[14,  4, 13,  1,  2, 15, 11,  8,  3, 10,  6, 12,  5,  9,  0,  7],
		[ 0, 15,  7,  4, 14,  2, 13,  1, 10,  6, 12, 11,  9,  5,  3,  8],
		[ 4,  1, 14,  8, 13,  6,  2, 11, 15, 12,  9,  7,  3, 10,  5,  0],
		[15, 12,  8,  2,  4,  9,  1,  7,  5, 11,  3, 14, 10,  0,  6, 13]
	],


	[
		[15,  1,  8, 14,  6, 11,  3,  4,  9,  7,  2, 13, 12,  0,  5, 10],
		[ 3, 13,  4,  7, 15,  2,  8, 14, 12,  0,  1, 10,  6,  9, 11,  5],
		[ 0, 14,  7, 11, 10,  4, 13,  1,  5,  8, 12,  6,  9,  3,  2, 15],
		[13,  8, 10,  1,  3, 15,  4,  2, 11,  6,  7, 12,  0,  5, 14,  9]
	],


	[
		[10,  0,  9, 14,  6,  3, 15,  5,  1, 13, 12,  7, 11,  4,  2,  8],
		[13,  7,  0,  9,  3,  4,  6, 10,  2,  8,  5, 14, 12, 11, 15,  1],
		[13,  6,  4,  9,  8, 15,  3,  0, 11,  1,  2, 12,  5, 10, 14,  7],
		[ 1, 10, 13,  0,  6,  9,  8,  7,  4, 15, 14,  3, 11,  5,  2, 12]
	],


	[
		[ 7, 13, 14,  3,  0,  6,  9, 10,  1,  2,  8,  5, 11, 12,  4, 15],
		[13,  8, 11,  5,  6, 15,  0,  3,  4,  7,  2, 12,  1, 10, 14,  9],
		[10,  6,  9,  0, 12, 11,  7, 13, 15,  1,  3, 14,  5,  2,  8,  4],
		[ 3, 15,  0,  6, 10,  1, 13,  8,  9,  4,  5, 11, 12,  7,  2, 14]
	],


	[
		[ 2, 12,  4,  1,  7, 10, 11,  6,  8,  5,  3, 15, 13,  0, 14,  9],
		[14, 11,  2, 12,  4,  7, 13,  1,  5,  0, 15, 10,  3,  9,  8,  6],
		[ 4,  2,  1, 11, 10, 13,  7,  8, 15,  9, 12,  5,  6,  3,  0, 14],
		[11,  8, 12,  7,  1, 14,  2, 13,  6, 15,  0,  9, 10,  4,  5,  3]
	],


	[
		[12,  1, 10, 15,  9,  2,  6,  8,  0, 13,  3,  4, 14,  7,  5, 11],
		[10, 15,  4,  2,  7, 12,  9,  5,  6,  1, 13, 14,  0, 11,  3,  8],
		[ 9, 14, 15,  5,  2,  8, 12,  3,  7,  0,  4, 10,  1, 13, 11,  6],
		[ 4,  3,  2, 12,  9,  5, 15, 10, 11, 14,  1,  7,  6,  0,  8, 13]
	],


	[
		[ 4, 11,  2, 14, 15,  0,  8, 13,  3, 12,  9,  7,  5, 10,  6,  1],
		[13,  0, 11,  7,  4,  9,  1, 10, 14,  3,  5, 12,  2, 15,  8,  6],
		[ 1,  4, 11, 13, 12,  3,  7, 14, 10, 15,  6,  8,  0,  5,  9,  2],
		[ 6, 11, 13,  8,  1,  4, 10,  7,  9,  5,  0, 15, 14,  2,  3, 12]
	],


	[
		[13,  2,  8,  4,  6, 15, 11,  1, 10,  9,  3, 14,  5,  0, 12,  7],
		[ 1, 15, 13,  8, 10,  3,  7,  4, 12,  5,  6, 11,  0, 14,  9,  2],
		[ 7, 11,  4,  1,  9, 12, 14,  2,  0,  6, 10, 13, 15,  3,  5,  8],
		[ 2,  1, 14,  7,  4, 10,  8, 13, 15, 12,  9,  0,  3,  5,  6, 11]
	]
]

def S_box(data: List[int]) -> List[int]:
    output = []
    for i in range(0, 48, 6):
        row = data[i] * 2 + data[i + 5]
        col = reduce(add, [data[i + j] * (2 ** (4 - j)) for j in range(1, 5)])
        output += [int(x) for x in format(__s_box[i // 6][row][col], '04b')]
    return output

def encrypt(plain: List[int], sub_keys: List[List[int]]) -> List[int]:
    plain = IP(plain)
    L, R = plain[:32], plain[32:]

    for i in range(16):
        prev_L = L # L0
        L = R      # L1 = R0
        expanded_R = EP(R)
        xor_result = [a ^ b for a, b in zip(expanded_R, sub_keys[i])]
        substituted = S_box(xor_result)
        permuted = P(substituted)

        R = [a ^ b for a, b in zip(permuted, prev_L)] # R1

    cipher = R + L
    cipher = [cipher[x] for x in [39,  7, 47, 15, 55, 23, 63, 31,
                                38,  6, 46, 14, 54, 22, 62, 30,
                                37,  5, 45, 13, 53, 21, 61, 29,
                                36,  4, 44, 12, 52, 20, 60, 28,
                                35,  3, 43, 11, 51, 19, 59, 27,
                                34,  2, 42, 10, 50, 18, 58, 26,
                                33,  1, 41,  9, 49, 17, 57, 25,
                                32,  0, 40,  8, 48, 16, 56, 24]]

    return cipher


def bitxor(plain1: List[int], plain2: List[List[int]]) -> List[int]:
    return [int(i) for i in bin(int(''.join(str(i) for i in plain1),2)^int(''.join(str(i) for i in plain2),2))[2:].zfill(64)]

#key的字母表为abcdefghijklmnopqrstuvwsyz
from secret import flag, key

t=[]
z=[[0]*64]
# z = [[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
key = reduce(add, [list(map(int, bin(key_byte)[2:].zfill(8))) for key_byte in key]) # 转为2进制列表，列表中每个1，0为int型
for i in range(0,3):
    pt = reduce(add, [list(map(int, bin(flag_byte)[2:].zfill(8))) for flag_byte in flag[ 8*i:8*(i+1) ]])
    ct = encrypt(pt, get_sub_key(bitxor(z[i],key)))
    z.append(pt)
    t += ct
print(t)


'''
i=0情况下的LL,Rr
LL= [0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
Rr= [0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0]
t=[0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0]
'''

题目实现了一个经典的DES加密算法，除了第一组，后面两组加密时都会密钥都会先和前一组的明文异或。题目给出了第一组加密时的最后一轮轮密钥（L部分少了9bits，可以爆破）

from base64 import b64decode
from itertools import product
from DES import * 			 # https://github.com/soreatu/Cryptography/blob/master/DES.py
from typing import List

guess_8bit = list(product(range(2), repeat=8))
not_in_PC2 = [9,18,22,25,35,38,43,54]

def re_PC2(sbkey):
    # 48-bit -> 56-bit
    res = [0]*56
    for i in range(len(sbkey)):
        res[PC_2_table[i]-1] = sbkey[i]
    return res # ok

def guess_CiDi10(sbkey, t):
    res = re_PC2(sbkey)
    for i in range(8):
        res[not_in_PC2[i]-1] = guess_8bit[t][i]
    return res # ok

def guess_allsbkey(roundkey, r, t):
    sbkey = [[]]*16
    sbkey[r] = roundkey
    CiDi = guess_CiDi10(roundkey, t)
    Ci, Di = CiDi[:28], CiDi[28:]
    for i in range(r+1,r+16):
        Ci, Di = LR(Ci, Di, i%16)
        sbkey[i%16] = PC_2(Ci+Di)
        if i%16 == 0:
            combined = Ci+Di
    return combined,sbkey # ok

def long_des_enc(c, k):
    assert len(c) % 8 == 0
    res = b''
    for i in range(0,len(c),8):
        res += DES_enc(c[i:i+8], k)
    return res

def try_des(cipher, roundkey):
    for t in range(256):
        combined,allkey = guess_allsbkey(roundkey, 15, t)
        plain = long_des_enc(cipher, allkey[::-1])
        if plain.startswith(b'Nep'):
            print(combined,plain)
            exit()

def PC_2(key: List[int]) -> List[int]:
    return [key[x] for x in __pc2]

__pc2 = [
    13, 16, 10, 23,  0,  4,
     2, 27, 14,  5, 20,  9,
    22, 18, 11,  3, 25,  7,
    15,  6, 26, 19, 12,  1,
    40, 51, 30, 36, 46, 54,
    29, 39, 50, 44, 32, 47,
    43, 48, 38, 55, 33, 52,
    45, 41, 49, 35, 28, 31
]

from Crypto.Util.number import *


tt=[0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0]
t = tt[:64]
t = ''.join(str(i) for i in t)
t= int(t,2)
t = long_to_bytes(t)

LL= [0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
Rr= [0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0]
from tqdm import *
for i in range(2**9-1,2**7,-1):
    tmp =  list(bin(i)[2:].rjust(9,'0'))
    L = LL + [ int(u) for u in tmp]
    R = Rr
    combined = L+R
    sub_key = PC_2(combined)
    try_des(t, sub_key)

# [0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0] b'NepCTF{N'

得到第一组明文：NepCTF{N,顺便输出了第一组加密过程中的第一轮轮密钥combined

然后拿着第一组的明文NepCTF{N，做一个密钥生成那里的 PC_1置换,再左右两边各自向左循环移位一下，再异或第一轮轮密钥combined，并不是key，不过也可以看成是一个不变的的特征，相当于key的一个小变换，处理得当可以直接用它而不需要用key(这里可以好好理解一下，表置换和移位并不会影响数据本身，故后面再来异或并不会改变结果)，就可以解密出第二组密文，以此类推得到第三组明文，全部拼接得到最终flag

from DES import *
from Crypto.Util.number import *

def long_des_enc(c, k):
    assert len(c) % 8 == 0
    res = b''
    for i in range(0,len(c),8):
        res += DES_enc(c[i:i+8], k)
    return res

def attack(p, c):
    combined = [0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0,
                1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0]

    p = bin(p)[2:].rjust(64, '0')
    p = [int(i) for i in p]

    # 对明文 PC_1 置换
    P = PC_1(p)

    # 对明文左右两部分进行第一轮的循环移位

    Ci, Di = LR(P[:28], P[28:], 0)
    P = Ci + Di
    CiDi = []
    for i in range(56):
        CiDi.append(P[i] ^ combined[i]) # 变换后的p与变换后的key异或，很妙（不需要直接用key）

    # 生成所有轮密钥
    Ci, Di = CiDi[:28], CiDi[28:]
    sbkey = [[]] * 16
    sbkey[0] = PC_2(Ci + Di)

    r = 1
    for i in range(r, r + 16):
        Ci, Di = LR(Ci, Di, i % 16)
        sbkey[i % 16] = PC_2(Ci + Di)

    # 解密密文
    c = ''.join(str(i) for i in c)
    c = int(c, 2)
    c = long_to_bytes(c)
    return long_des_enc(c, sbkey[::-1])


tt = [0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0,
      1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0,
      1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0,
      0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1,
      0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0,
      1, 0]
flag = b"NepCTF{N"
for i in range(2):
    t = tt[64 * i + 64:64 * i + 128]
    p = flag[i * 8:i * 8 + 8]
    flag += attack(bytes_to_long(p), t)
print(flag)

# b'NepCTF{Nep_d0wn_ddddd1s}'

第一次接触这种题，耐心看吧，看懂了真的很有成就感。~~但看不看得懂是一回事，做不做不出又是一回事，希望下次遇到这种题能别像如今这样无从下手了。~~

官方wp太复杂了，不想看

题目分析来源：

https://blog.soreatu.com/posts/intended-solution-to-crypto-problems-in-nctf-2019/#reverse909pt-2solvers

https://jayxv.github.io/2023/08/14/2023%20NepCTF/

[极客大挑战-2023] Simple3DES

(于2023-12-15记录)
题目描述：

from Crypto.Cipher import DES3
from Crypto.Util.number import *
import os
import random
import string
import hashlib

xor = lambda a,b: bytes([a[i % len(a)] ^ b[i % len(b)] for i in range(max(len(a), len(b)))])
pad = lambda msg,padlen: msg+chr((padlen-(len(msg)%padlen))).encode()*(padlen-(len(msg)%padlen))

flag = os.environ.get("FLAG", "SYC{Al3XEI_FAKE_FLAG}").encode()
sec = os.urandom(8)

banner = '|'*70

DEBUG = False 
def proof_of_work():
    if DEBUG:
        return True
    proof = ''.join([random.choice(string.ascii_letters+string.digits) for _ in range(20)])
    digest = hashlib.sha256(proof.encode()).hexdigest()
    print("sha256(XXXX+%s) == %s" % (proof[4:], digest))
    x = input("Give me XXXX: ")
    if len(x)!=4 or hashlib.sha256((x+proof[4:]).encode()).hexdigest() != digest:
        return False
    print("Right!")
    return True



def enc(msg,key):
    try:
        key = long_to_bytes(key)
        msg = xor(long_to_bytes(msg),sec)
        des = DES3.new(key,DES3.MODE_ECB)
        ct = xor(des.encrypt(pad(msg,8)),sec)
        return bytes_to_long(ct)
    except Exception as e:
        print(e)
        return Exception

def service():
    cnt = 0
    if not proof_of_work():
        exit()
    print(banner)
    print('Simple DES Encryption Service')
    print(banner)
    while cnt<2:
        print('1. Encrypt\n2. Get encrypted flag.')
        choice = int(input('> '))
        if choice == 1:
            print('Input msg:')
            msg = int(input('> ').strip())
            print('Input key:')
            key = int(input('> ').strip())
            print(enc(msg,key))
        elif choice == 2:
            print('Input key:')
            key = int(input('> ').strip())
            print(enc(bytes_to_long(flag),key))
        else:
            exit()
        cnt+=1
    print(banner)
    print('Bye!')
    exit()

try:
    service()
except Exception:
    print("Something goes wrong...\n")
    print(banner+'\n')
    exit()

题目分析：
3DES加密流程如下
DES加密解密 Feistel算法网络结构详讲,块密码,密码学,安全,算法,DES,Feistel

3DES也是三重DES，是为了增加DES的强度，将DES重复3次所得到的一种加密算法。
由于DES密钥长度实质上是56bit，因此三重DES的密钥长度就是3*56=168 bit

题目中
输入1，我们自己确定明文和密钥并得到明文加密后的密文
输入2，我们自己确定密钥并得到flag加密后的密文
两层循环，那就是1，2都输一遍喽

加密代码：

def enc(msg,key):
    try:
        key = long_to_bytes(key)
        msg = xor(long_to_bytes(msg),sec)
        des = DES3.new(key,DES3.MODE_ECB)
        ct = xor(des.encrypt(pad(msg,8)),sec)
        return bytes_to_long(ct)
    except Exception as e:
        print(e)
        return Exception

可以看出过程是：
明文和sec异或
异或后的数据进行填充，然后进行3DES加密
加密后的数据再与sec异或

其中sec我们不知道，不过注意sec为8bit
而后面明文会进行长度为8的填充

pad:

pad = lambda msg,padlen: msg+chr((padlen-(len(msg)%padlen))).encode()*(padlen-(len(msg)%padlen))

我们都知道DES为块加密，64位为一组进行加密，正好8字节
所以这里我们密文输入8的整数倍，那么可以知道最后8个为0
这样的话我们key知道，明文填充后最后8字节知道，那么可以通过加密函数得到明文填充后密文，又ct = sec ^ 明文填充后密文，其中ct知道，这样sec也就出来了

sec求出之后，flag也就差不多出了（后面很简单，应该知道怎么求，我就不说了）

如果密文直接输入空，那么3DES加密的数据就是padding(sec,8)，后面流程其实也差不多

看到一个非预期解，这里记录一下：弱密钥

就是说，在DES中，存在一些密钥，被处理后会使得16轮轮密钥完全相等，这种便称为弱密钥
这种的话，加密一次看起来很正常，不过再加密一次便能重新得到明文，反之亦然

四个弱密钥：

\x01\x01\x01\x01\x01\x01\x01\x01
\xFE\xFE\xFE\xFE\xFE\xFE\xFE\xFE
\xE0\xE0\xE0\xE0\xF1\xF1\xF1\xF1
\x1F\x1F\x1F\x1F\x0E\x0E\x0E\x0E

由于DES密钥长度实质上是56bit，因此三重DES的密钥长度就是3*56=168 bit
故我们可以构造如下密钥：

key = b'\x01\x01\x01\x01\x01\x01\x01\x01' + b'\xFE\xFE\xFE\xFE\xFE\xFE\xFE\xFE' + b'\x01\x01\x01\x01\x01\x01\x01\x01'

第一次输入2，输入上面的key，得到flag的加密值
第二次输入1，输入上面的key，再输入flag的加密值，得到flag
加密两次得到原值

想了解更多des的秘密，看这里
想了解des中的弱密钥，看这里

[moectf-2023] feistel

题目描述：

from Crypto.Util.number import *

def f(m, key):
    m = m ^ (m >> 4)
    m = m ^ (m << 5)
    m = m ^ (m >> 8)
    m ^= key
    m = (m * 1145 + 14) % 2**64
    m = (m * 1919 + 810) % 2**64
    m = (m * key) % 2**64
    return m

def dec(c,key):
    key = bytes_to_long(key)
    left = bytes_to_long(c[:8])
    right = bytes_to_long(c[8:])
    m_right = f(right,key) ^ left
    m_left = f(m_right,key) ^ right
    return long_to_bytes(m_left) + long_to_bytes(m_right)


def ecb_dec(c, key):
    clen = len(c)
    m = b""
    for i in range(clen // 16):
        m += dec(c[i * 16 : i * 16 + 16], key)
    return m

c = b'\x0b\xa7\xc6J\xf6\x80T\xc6\xfbq\xaa\xd8\xcc\x95\xad[\x1e\'W5\xce\x92Y\xd3\xa0\x1fL\xe8\xe1"^\xad'
key = b"wulidego"
flag = ecb_dec(c,key)
print(flag)

题目分析：
feistel加密结构，加解密过程一样，就是轮密钥相反
而这里每轮密钥都是wulidego，所以说这个加解密是完全一样的

exp：

from Crypto.Util.number import *

round = 2

def f(m, key):
    m = m ^ (m >> 4)
    m = m ^ (m << 5)
    m = m ^ (m >> 8)
    m ^= key
    m = (m * 1145 + 14) % 2**64
    m = (m * 1919 + 810) % 2**64
    m = (m * key) % 2**64
    return m

def enc(m, key, round):
    key = bytes_to_long(key)
    left = bytes_to_long(m[:8])
    right = bytes_to_long(m[8:])
    for i in range(round):
        left, right = right, f(right, key) ^ left
    left, right = right, left
    return long_to_bytes(left).rjust(8, b"\x00") + long_to_bytes(right).rjust(8, b"\x00")

def padding(m):
    mlen = len(m)
    pad = 16 - mlen % 16
    return m + pad * bytes([pad])

def ecb_enc(m, key):
    m = padding(m)
    mlen = len(m)
    c = b""
    for i in range(mlen // 16):
        c += enc(m[i * 16 : i * 16 + 16], key, round) # round = 2
    return c

def ecb_dec(m, key):
    m = padding(m)
    mlen = len(m)
    c = b""
    for i in range(mlen // 16):
        c += enc(m[i * 16 : i * 16 + 16], key, round) # round = 2
    return c

c = b'\x0b\xa7\xc6J\xf6\x80T\xc6\xfbq\xaa\xd8\xcc\x95\xad[\x1e\'W5\xce\x92Y\xd3\xa0\x1fL\xe8\xe1"^\xad'
print(ecb_enc(c,b'wulidego'))

# moectf{M@g1cA1_Encr1tion!!!}

[moectf-2023] feistel_promax

题目描述：

from Crypto.Util.number import *
from os import urandom

round = 2
flag = open("./secret", "rb").read().strip()

def f(m, key):
    m = m ^ (m >> 4)
    m = m ^ (m << 5)
    m = m ^ (m >> 8)
    m ^= key
    m = (m * 1145 + 14) % 2**64
    m = (m * 1919 + 810) % 2**64
    m = (m * key) % 2**64
    return m


def enc(m, key, round):
    key = bytes_to_long(key)
    left = bytes_to_long(m[:8])
    right = bytes_to_long(m[8:])
    for i in range(round):
        left, right = right, f(right, key) ^ left
    left, right = right, left
    return long_to_bytes(left).rjust(8, b"\x00") + long_to_bytes(right).rjust(8, b"\x00")


def padding(m):
    mlen = len(m)
    pad = 16 - mlen % 16
    return m + pad * bytes([pad])


def ecb_enc(m, key):
    mlen = len(m)
    c = b""
    for i in range(mlen // 16):
        c += enc(m[i * 16 : i * 16 + 16], key, round)
    return c


key = urandom(8)

print(ecb_enc(padding(flag), key))
# b'B\xf5\xd8gy\x0f\xaf\xc7\xdf\xabn9\xbb\xd0\xe3\x1e0\x9eR\xa9\x1c\xb7\xad\xe5H\x8cC\x07\xd5w9Ms\x03\x06\xec\xb4\x8d\x80\xcb}\xa9\x8a\xcc\xd1W\x82[\xd3\xdc\xb4\x83P\xda5\xac\x9e\xb0)\x98R\x1c\xb3h'

相比于上一题，这题key不知，仅根据此条件根本无从下手，不过观察仔细的话会发现最后一行，加密的主体是padding(flag)，而加密过程中又进行了一次padding，所以说进行了两次padding
故：

padding(padding(m)) = b'****' + b'\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10'

又有以下过程：
$\begin{gathered} \text{初始}:m[:8]+m[8:] \\ \text{第一轮}:m[8:]+f(m[8:],key)\otimes m[:8] \\ \text{第二轮}:f(m[8:],key)\otimes m[:8]+f(f(m[8:],key)\otimes m[:8],key)\otimes m[8:] \\ \text{再交换最后得到}:f(f(m[8:],key)\otimes m[:8],key)\otimes m[8:]+\boxed{f(m[8:],key)\otimes m[:8]} \end{gathered}$
不过最后还是没很大思路，参考此文：https://dexterjie.github.io/2023/09/25/%E8%B5%9B%E9%A2%98%E5%A4%8D%E7%8E%B0/2023MoeCTF/#feistel-promax
我们以最后一组来分析，最后一组加密明文：b'\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10'
故 $f(m[8:],key)\otimes m[:8]$ 中m[8:] = m[:8]
所以能有以下爆破

exp:

from Crypto.Util.number import *

round = 2

def f(m, key):
    m = m ^ (m >> 4)
    m = m ^ (m << 5)
    m = m ^ (m >> 8)
    m ^= key
    m = (m * 1145 + 14) % 2**64
    m = (m * 1919 + 810) % 2**64
    m = (m * key) % 2**64
    return m

def dec(m, key, round):
    key = bytes_to_long(key)
    left = bytes_to_long(m[:8])
    right = bytes_to_long(m[8:])
    for i in range(round):
        left, right = right, f(right, key) ^ left
    left, right = right, left
    return long_to_bytes(left).rjust(8, b"\x00") + long_to_bytes(right).rjust(8, b"\x00")

def ecb_dec(c, key):
    clen = len(c)
    m = b""
    for i in range(clen // 16):
        m += dec(c[i * 16 : i * 16 + 16], key, 2)
    return m

m = b'\x10\x10\x10\x10\x10\x10\x10\x10'
ct = b'B\xf5\xd8gy\x0f\xaf\xc7\xdf\xabn9\xbb\xd0\xe3\x1e0\x9eR\xa9\x1c\xb7\xad\xe5H\x8cC\x07\xd5w9Ms\x03\x06\xec\xb4\x8d\x80\xcb}\xa9\x8a\xcc\xd1W\x82[\xd3\xdc\xb4\x83P\xda5\xac\x9e\xb0)\x98R\x1c\xb3h'
c = bytes_to_long(ct[-8:]) ^ bytes_to_long(m)
bin_c = bin(c)[2:].rjust(64,'0') # = f(M[8:],key)，其中M = m * 2

k1 = [b'']
for i in range(1,5):
    k2 = []
    for KEY in k1:
        for j in range(2**16):
            key1 = long_to_bytes(j) + KEY
            cc = f(bytes_to_long(m),bytes_to_long(key1)) # = f(M[8:],key)
            bin_cc = bin(cc)[2:].rjust(64,'0')
            if bin_cc[-16*i:] == bin_c[-16*i:]:
                k2.append(key1)
                k1 = k2

# print(k1)
# k1= [b'4t*zFD\xac\xb4', b'\xb4t*zFD\xac\xb4', b'\\q\x0f\x00w\xeb\xc1"', b'\xdcq\x0f\x00w\xeb\xc1"']
for KEY in k1:
    flag = ecb_dec(ct,KEY)
    print(flag)

# moectf{F_func_1s_n1t_Ve5y_$EcU%e}