OTP Implementation wp (Chinese ver.)

picoCTF - 2020 - otp write-up

前言

picoCTF是美国CMU大学举办的面向高中生的CTF竞赛,应该算是初学者水平的题。前几天做了里面的一道题感觉出的挺好的,也不算很难,有点小坑但不多。

题目来源和下载

来源:https://play.picoctf.org/events/3/challenges/challenge/92
好像每个人的下载后的文件都会不一样
我的文件(otp 和 flag.txt):otp.zip

工具

IDA Pro

分析

查壳

无壳

key 正确的条件

以下代码说明key必须要满足这两个条件:

  • length ([rbp+var_E8]) 等于 100
  • s1 等于 s2 ("mngjlepdcbcmjmmjipmmegfkjbicaemoemkkpjgnhgomlknmoepmfbcoffikhplmadmganmlojndmfahbhaancamdhfdkiancdjf")

In Disassembly,

.text:0000557E7FE4E99D                 cmp     [rbp+var_E8], 64h ; 'd'
.text:0000557E7FE4E9A4                 jnz     short loc_557E7FE4E9C6 ; if ( i == 100
.text:0000557E7FE4E9A4                                         ; line 35
.text:0000557E7FE4E9A6                 mov     eax, [rbp+var_E8] ;       && !strncmp(
.text:0000557E7FE4E9A6                                         ;             s1,
.text:0000557E7FE4E9A6                                         ;             "mngjlepdcbcmjmmjipmmegfkjbicaemoemkkpjgnhgomlknmoepmfbcoffikhplmadmganmlojndmfahbhaancamdhfdkiancdjf",
.text:0000557E7FE4E9A6                                         ;             0x64uLL) )
.text:0000557E7FE4E9A6                                         ; line 36-39
.text:0000557E7FE4E9AC                 movsxd  rdx, eax        ; n = i = [rbp+var_E8]
.text:0000557E7FE4E9AF                 lea     rax, [rbp+s1]
.text:0000557E7FE4E9B3                 lea     rsi, s2         ; "mngjlepdcbcmjmmjipmmegfkjbicaemoemkkpjg"...
.text:0000557E7FE4E9BA                 mov     rdi, rax        ; s1
.text:0000557E7FE4E9BD                 call    _strncmp        ; test if s1 equal s2
.text:0000557E7FE4E9C2                 test    eax, eax
.text:0000557E7FE4E9C4                 jz      short loc_557E7FE4E9D9 ; if i = 100 and s1 = s2,then key is correct

In Pseudocode,

if ( i == 100
  && !strncmp(
        s1,
        "mngjlepdcbcmjmmjipmmegfkjbicaemoemkkpjgnhgomlknmoepmfbcoffikhplmadmganmlojndmfahbhaancamdhfdkiancdjf",
        0x64uLL) )

分析算法 (输入: dest; 输出: s1)

分析从 0000557E7FE4E89E 到 0000557E7FE4E99B 的汇编

In Disassembly,

.text:0000557E7FE4E89E ; ---------------------------------------------------------------------------
.text:0000557E7FE4E89E
.text:0000557E7FE4E89E loc_557E7FE4E89E:                       ; CODE XREF: main+14A↓j
.text:0000557E7FE4E89E                 cmp     [rbp+var_E8], 0 ; if ( i )
.text:0000557E7FE4E89E                                         ; line 21
.text:0000557E7FE4E8A5                 jnz     short loc_557E7FE4E8E4
.text:0000557E7FE4E8A7                 mov     eax, [rbp+var_E8] ; when i equal 0
.text:0000557E7FE4E8A7                                         ; line 30
.text:0000557E7FE4E8AD                 cdqe
.text:0000557E7FE4E8AF                 movzx   eax, [rbp+rax+dest]
.text:0000557E7FE4E8B7                 movsx   eax, al
.text:0000557E7FE4E8BA                 mov     edi, eax
.text:0000557E7FE4E8BC                 call    jumble
.text:0000557E7FE4E8C1                 mov     edx, eax
.text:0000557E7FE4E8C3                 mov     eax, edx
.text:0000557E7FE4E8C5                 sar     al, 7
.text:0000557E7FE4E8C8                 shr     al, 4
.text:0000557E7FE4E8CB                 add     edx, eax
.text:0000557E7FE4E8CD                 and     edx, 0Fh
.text:0000557E7FE4E8D0                 sub     edx, eax
.text:0000557E7FE4E8D2                 mov     eax, edx
.text:0000557E7FE4E8D4                 mov     edx, eax
.text:0000557E7FE4E8D6                 mov     eax, [rbp+var_E8]
.text:0000557E7FE4E8DC                 cdqe
.text:0000557E7FE4E8DE                 mov     [rbp+rax+s1], dl
.text:0000557E7FE4E8E2                 jmp     short loc_557E7FE4E935
.text:0000557E7FE4E8E4 ; ---------------------------------------------------------------------------
.text:0000557E7FE4E8E4
.text:0000557E7FE4E8E4 loc_557E7FE4E8E4:                       ; CODE XREF: main+97↑j
.text:0000557E7FE4E8E4                 mov     eax, [rbp+var_E8] ; when i not equal 0
.text:0000557E7FE4E8E4                                         ; line 23-26
.text:0000557E7FE4E8EA                 cdqe
.text:0000557E7FE4E8EC                 movzx   eax, [rbp+rax+dest]
.text:0000557E7FE4E8F4                 movsx   eax, al
.text:0000557E7FE4E8F7                 mov     edi, eax        ; edi = dest[i]
.text:0000557E7FE4E8F9                 call    jumble          ; line 23
.text:0000557E7FE4E8FE                 movsx   edx, al
.text:0000557E7FE4E901                 mov     eax, [rbp+var_E8]
.text:0000557E7FE4E907                 sub     eax, 1
.text:0000557E7FE4E90A                 cdqe
.text:0000557E7FE4E90C                 movzx   eax, [rbp+rax+s1]
.text:0000557E7FE4E911                 movsx   eax, al         ; eax = s1[i-1]
.text:0000557E7FE4E914                 add     edx, eax        ; edx = v5
.text:0000557E7FE4E914                                         ; line 24
.text:0000557E7FE4E916                 mov     eax, edx
.text:0000557E7FE4E918                 sar     eax, 1Fh
.text:0000557E7FE4E91B                 shr     eax, 1Ch        ; line 25
.text:0000557E7FE4E91E                 add     edx, eax
.text:0000557E7FE4E920                 and     edx, 0Fh
.text:0000557E7FE4E923                 sub     edx, eax        ; line 26
.text:0000557E7FE4E925                 mov     eax, edx
.text:0000557E7FE4E927                 mov     edx, eax
.text:0000557E7FE4E929                 mov     eax, [rbp+var_E8]
.text:0000557E7FE4E92F                 cdqe
.text:0000557E7FE4E931                 mov     [rbp+rax+s1], dl ; [rbp+rax+s1] = s1[i]
.text:0000557E7FE4E935
.text:0000557E7FE4E935 loc_557E7FE4E935:                       ; CODE XREF: main+D4↑j
.text:0000557E7FE4E935                 add     [rbp+var_E8], 1 ; ++i
.text:0000557E7FE4E93C
.text:0000557E7FE4E93C loc_557E7FE4E93C:                       ; CODE XREF: main+8B↑j
.text:0000557E7FE4E93C                 mov     eax, [rbp+var_E8] ; for ( i = 0; (unsigned int)valid_char(dest[i]); ++i )
.text:0000557E7FE4E93C                                         ; line 19-32
.text:0000557E7FE4E942                 cdqe
.text:0000557E7FE4E944                 movzx   eax, [rbp+rax+dest]
.text:0000557E7FE4E94C                 movsx   eax, al
.text:0000557E7FE4E94F                 mov     edi, eax        ; edi = dest[i]
.text:0000557E7FE4E951                 call    valid_char      ; valid_char(dest[i])
.text:0000557E7FE4E956                 test    eax, eax
.text:0000557E7FE4E958                 jnz     loc_557E7FE4E89E
.text:0000557E7FE4E95E                 mov     [rbp+var_E4], 0 ; j
.text:0000557E7FE4E968                 jmp     short loc_557E7FE4E98F
.text:0000557E7FE4E96A ; ---------------------------------------------------------------------------
.text:0000557E7FE4E96A
.text:0000557E7FE4E96A loc_557E7FE4E96A:                       ; CODE XREF: main+18D↓j
.text:0000557E7FE4E96A                 mov     eax, [rbp+var_E4]
.text:0000557E7FE4E970                 cdqe
.text:0000557E7FE4E972                 movzx   eax, [rbp+rax+s1]
.text:0000557E7FE4E977                 add     eax, 61h ; 'a'
.text:0000557E7FE4E97A                 mov     edx, eax
.text:0000557E7FE4E97C                 mov     eax, [rbp+var_E4]
.text:0000557E7FE4E982                 cdqe
.text:0000557E7FE4E984                 mov     [rbp+rax+s1], dl
.text:0000557E7FE4E988                 add     [rbp+var_E4], 1
.text:0000557E7FE4E98F
.text:0000557E7FE4E98F loc_557E7FE4E98F:                       ; CODE XREF: main+15A↑j
.text:0000557E7FE4E98F                 mov     eax, [rbp+var_E4] ; for ( j = 0; j < i; ++j )
.text:0000557E7FE4E98F                                         ; line 33-34
.text:0000557E7FE4E995                 cmp     eax, [rbp+var_E8]
.text:0000557E7FE4E99B                 jl      short loc_557E7FE4E96A

In Pseudocode,

for ( i = 0; (unsigned int)valid_char(dest[i]); ++i )// test if dest[i] is 1-9 or a-f
{
  if ( i )
  {
    v4 = jumble(dest[i]);
    v5 = s1[i - 1] + v4;
    v6 = (unsigned int)((s1[i - 1] + v4) >> 31) >> 28;
    s1[i] = ((v6 + v5) & 0xF) - v6;
  }
  else
  {
    s1[0] = (char)jumble(dest[0]) % 16;
  }
}
for ( j = 0; j < i; ++j )
  s1[j] += 97;

通过分析得知输入为dest,输出为s1
接下来就可以在制作逆向算法时直接把部分以上算法直接复制进去用于穷举了。

求 Key [Java Code (Reversely:s2 => s1 => dest)] (关键部分)

写出Java代码:通过已知的s1(也就是s2)求出dest(正确的key)

求 dest[0]

s2[0] = "m" = 109
s1[0] = s2[0] - 97 = 12

输出所有可能的s1[0]
寻找当s1[0]等于12时,dest[0]的值

public static void main(String[] args)
{
        System.out.println("dest[0]|s1[0]"); 
        int[] a1s = {48,49,50,51,52,53,54,55,56,57,97,98,99,100,101,102};
        for(int i = 0; i < 16; i++)
        {
            System.out.println(a1s[i] + "|" +jumble(a1s[i]) % 16);                   
        }
}
    
public static int jumble(int a1)
{
        int v2 = a1;
        if ( a1 > 96 )                                // a-f
            v2 = a1 + 9;                                // j-o
        int v3 = 2 * (v2 % 16);
        if ( (char)v3 > 15 )
            ++v3;
        return v3;
}

Output:

dest[0]|s1[0]
48|0
49|2
50|4
51|6
52|8
53|10
54|12
55|14
56|1
57|3
97|5
98|7
99|9
100|11
101|13
102|15

结论:当 dest[0] 等于 54 (char “6”), s1[0] 等于 12.

Find dest[1]

s2[1] = "n" = 110

输出所有可能的s1[1]
寻找当s1[1]等于110时,dest[1]的值

public static void main(String[] args)
{
        s1[0] = 12;
        System.out.println("dest[1]|s1[1]"); 
        int[] a1s = {48,49,50,51,52,53,54,55,56,57,97,98,99,100,101,102};
        for(int i = 0; i < 16; i++)
        {
                getMS1IndexNotEqual0(a1s[i],1);
                System.out.println(a1s[i] + "|" +s1[1]);                   
        }
}

static int[] s1 = new int[100];

//find dest[not equal 0]
//mD = dest[i]
public static int getMS1IndexNotEqual0(int mD, int i)
{
        if(i == 0)
        {
                return -1;
        }
        int v4 = jumble(mD);
        int v5 = s1[i - 1] + v4;
        int v6 = (v5 >> 31) >>> 28;
        s1[i] = ((v6 + v5) & 0xF) - v6;
        s1[i] += 97;
        return s1[i];
}
    
public static int jumble(int a1)
{
        int v2 = a1;
        if ( a1 > 96 )                                // a-f
            v2 = a1 + 9;                                // j-o
        int v3 = 2 * (v2 % 16);
        if ( (char)v3 > 15 )
            ++v3;
        return v3;
}

Output:

dest[1]|s1[1]
48|109
49|111
50|97
51|99
52|101
53|103
54|105
55|107
56|110
57|112
97|98
98|100
99|102
100|104
101|106
102|108

结论:当 dest[1] 等于 56 (char “8”), s1[1] 等于 110.

通过编程自动求 dest[1-100] (dest[1] - dest[100])

public static void auto()
{
        s1[0] = 12;
        String s2 = "mngjlepdcbcmjmmjipmmegfkjbicaemoemkkpjgnhgomlknmoepmfbcoffikhplmadmganmlojndmfahbhaancamdhfdkiancdjf";
        char[] a1s = {48,49,50,51,52,53,54,55,56,57,97,98,99,100,101,102};
        boolean hasResult = false;
        for(int i2 = 1; i2 < 100; i2++)
        {
                hasResult = false;
            for(int i = 0; i < 16; i++)
            {
                    int s1Add97 = getMS1IndexNotEqual0(a1s[i],i2);
                    if(s2.charAt(i2) == (char)s1Add97)
                    {
                            hasResult = true;
                            System.out.print(a1s[i]);
                            break;
                    }
                    
            }
            if(hasResult == false)
                    System.out.println("NO");
        }
}

static int[] s1 = new int[100];

//find dest[not equal 0]
//mD = dest[i]
public static int getMS1IndexNotEqual0(int mD, int i)
{
        if(i == 0)
        {
                return -1;
        }
        int v4 = jumble(mD);
        int v5 = s1[i - 1] + v4;
        int v6 = (v5 >> 31) >>> 28;
        s1[i] = ((v6 + v5) & 0xF) - v6;
        //s1[i] += 97;
        return (s1[i]+97);
}

Output:

8c91cd2ff85e90efbe041faf4b572413470a5eb5f47ff9f13dec686b091e46829c55eff9d23ccdb53c0ea76b277b74ea836

dest结论

第一字符是 "6",
所以完整的dest是 "68c91cd2ff85e90efbe041faf4b572413470a5eb5f47ff9f13dec686b091e46829c55eff9d23ccdb53c0ea76b277b74ea836",也就是key

测试 key

congrats.png

求最终 Flag

逆向得知key是 "68c91cd2ff85e90efbe041faf4b572413470a5eb5f47ff9f13dec686b091e46829c55eff9d23ccdb53c0ea76b277b74ea836"
flag.txt中的内容是 "18a07fbdbcd1af759895328ec4d82d2b411dc7876c34a0ab61eda8f2efa5bb0f198a3aa0ac47ff9a0cf3d913d3138678ce4b"
异或这两个值,得到 "7069636f4354467b63757374306d5f6a756d626c33735f3472336e745f345f67304f645f316433415f33336561643136667d"

将异或的结果 ("7069636f4354467b63757374306d5f6a756d626c33735f3472336e745f345f67304f645f316433415f33336561643136667d") 从十六进制转成ASCII, 得到 "picoCTF{cust0m_jumbl3s_4r3nt_4_g0Od_1d3A_33ead16f}", 也就是最终的Flag

java code

public static void main (String[] args) {
        String str1 = "68c91cd2ff85e90efbe041faf4b572413470a5eb5f47ff9f13dec686b091e46829c55eff9d23ccdb53c0ea76b277b74ea836";
        String str2 = "18a07fbdbcd1af759895328ec4d82d2b411dc7876c34a0ab61eda8f2efa5bb0f198a3aa0ac47ff9a0cf3d913d3138678ce4b";
        String xorRes = doXor(str1,str2);
        String flag = hexToString(xorRes);
}

public static String doXor(String str1, String str2)
{
        String xorRes = StringXor(str1, str2);
        System.out.println(xorRes);
        return xorRes;
}

public static String StringXor(String str1, String str2) {
        BigInteger big1 = new BigInteger(str1, 16);
        BigInteger big2 = new BigInteger(str2, 16);
        return big1.xor(big2).toString(16);
        
}

public static String hexToString(String hex)
{
    byte[] s = DatatypeConverter.parseHexBinary(hex);
    String ret = new String(s);
    System.out.println(ret);
    return ret;
}

Output:

7069636f4354467b63757374306d5f6a756d626c33735f3472336e745f345f67304f645f316433415f33336561643136667d
picoCTF{cust0m_jumbl3s_4r3nt_4_g0Od_1d3A_33ead16f}

标签:picoctf, picoctf_2020mini

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