2 
A ^ ( (A >> 2) + (A << 5) + C ) == B

如果 B 是 const 且 C 是变量，如何找到 A？（如果没有解决方案可以更改C）

A 是双字，B 是双字，C 是字节！= 0

Edit1：在 GalacticJello 的回答之后，我有另一个问题：有没有办法在没有循环的情况下做到这一点（简化表达式）？

为什么我需要这个：我正在尝试为

unsigned int X(const char* const in) { //strlen(in) is always < 127

  unsigned int result = 0x12345678; //just for an example
  for(int i = 0; in[i] != 0; ++i)
    result ^= (result >> 2) + (result << 5) + in[i];

  return result;
}

目前我有一个生成随机 C 然后搜索 A 的循环。（我使用生成随机值 [for A] 的循环搜索 A 并检查上述表达式是否为真）

Edit2：这是我当前用于搜索冲突的代码，我现在正在测试..

#include <stdio.h>
#include <conio.h>

using namespace std;


unsigned int originalHash(const char* const in) {

    unsigned int result = 0x12345678;
    for(int i = 0; in[i] != 0; ++i) {
        result = result ^ ((result >> 2) + (result << 5) + in[i]);
    }

    return result;

}


//A ^ ( (A >> 2) + (A << 5) + C ) == B
bool findSolutions(unsigned int inHash, char* _C, unsigned int* _A) { //Starts searching from *A and *C and writes there values on success.

    unsigned int C = *_C;
    if(C == 0) ++C;

    unsigned int A = *_A;

    for(C; C < 256; ++C) {

        for(A; A < 0xFFFFFFFF; ++A) {

            if((A ^ ( (A >> 2) + (A << 5) + C )) == inHash) {

                *_C = C;
                *_A = A;
                return true;

            }

        }

        A = 0;

    }

    return false;

}


bool findCollisions(unsigned int inHash, char* szOutStr) {

    const unsigned int REQ_HASH = 0x12345678;
    unsigned int prevHash = 0;
    int curChar = 0;

    do {

        printf("Loop Begin:\tI = %i | H = %08x | rH = %08x\n", curChar, inHash, REQ_HASH);

        if(!findSolutions(inHash, &szOutStr[curChar], &prevHash)) {

            printf("Unable to find solutions for %08x\n", inHash);

            if(curChar == 0) return false;
            --curChar;
            continue;

        }

        if(prevHash == REQ_HASH) {

            szOutStr[curChar] = 0;
            return true;

        }

        printf("Found solution:\tC = %02x (%c) | A = %08x\n", szOutStr[curChar], szOutStr[curChar], prevHash);

        char firstSolutionC = szOutStr[curChar];
        unsigned int firstSolutionA = prevHash;

        printf("Trying to find alternative solutions..\n");
        do {

            if(!findSolutions(inHash, &szOutStr[curChar], &prevHash)) {

                printf("Alternative solution not found!\n");
                break;

            }

            printf("Alternative solution found [%s valid]:\tC = %02x (%c) | A = %08x\n", prevHash == REQ_HASH ? "" : "not", szOutStr[curChar], szOutStr[curChar], prevHash);

            if(prevHash == REQ_HASH) {

                szOutStr[curChar] = 0;
                return true;

            }

            ++prevHash;

        } while(true);

        szOutStr[curChar] = firstSolutionC;
        prevHash = firstSolutionA;
        printf("Using first solution:\tC = %02x (%c) | A = %08x\n", szOutStr[curChar], szOutStr[curChar], prevHash);

        ++curChar;
        inHash = prevHash;

    } while(curChar < 127);

    return false;

}

int main(void) {

    char mask[] = "hQh";

    DWORD original = originalHash(mask);

    printf("%s == %08x\n", mask, original);

    char out[128];
    memset(out, 0, sizeof out);
    if(findCollisions(original, out))
        printf("%08x == %s\n", original, out);
    else
        printf("Unable to find collisions\n");

    getch();

    return 0;

}
4

1 回答 1

3

我只是想回答一下这个问题（对不起 C# 代码，但你应该能够理解要点）：

A ^ ( (A >> 2) + (A << 5) + C ) == B

static List<Tuple<uint, uint>> FindSolutions(uint B)
{
    var solutions = new List<Tuple<uint, uint>>();
    for (uint C = 0; C < uint.MaxValue; C++)
    {
        for (uint A = 0; A < uint.MaxValue; A++)
        {
            uint guess = A ^ ((A >> 2) + (A << 5) + C);
            if (guess == B)
                solutions.Add(new Tuple<uint,uint>(A, C));
        }
    }

    return solutions;
}

var solutions = FindSolutions(0x00000001);

如果 B 是您的常数（在这种情况下为 0x00000001），那么 A 和 C 的前几个解决方案是：

A = 0x8b439581, B= 0x00000001, C = 0x00000000
(0x8b439581 ^ ((22d0e560) + (6872B020) + 0)) == 0x00000001
0x8b439581 ^ (0x8b439580) == 0x00000001
0x00000001 == 0x00000001

其他：

A = 0x9ba5e354, B= 0x00000001, C = 0x00000000
A = 0x00000000, B= 0x00000001, C = 0x00000000
A = 0x6a7ef9db, B= 0x00000001, C = 0x00000004

... ETC。

编辑：

要查找冲突，您可以简单地通过键空间执行蛮力。

再次对 C# 代码表示抱歉：

static uint originalHash(string input)
{
   unt result = 0x12345678;
   for (int i = 0; i < input.Length; i++)            
      result ^= (result >> 2) + (result << 5) + input[i];

   return result;
}


var charset = new string(Enumerable.Range(1, 255).Select(i => (char)i).ToArray());

var hits = new List<string>();
var hashToFind = originalHash("hQh");

for (int wordNum = 1; wordNum < int.MaxValue; wordNum++)
{
    var word = Utils.NumberToString(wordNum, charset);
    var guess = originalHash(word);
    if (guess == hashToFind)
    {
        Console.WriteLine("Found: " + word);
        hits.Add(word);
    }
}

运行上面的代码在一两分钟后给了我以下碰撞：

cê cë1 côÌ cõí cö c÷¦ cøG cùh dÌ1 dÍ dÒí dÓÌ dÖh d×G dئ dÙ e¬ e1 e²Ì e³í e¶G e·he¸ e¹¦ f1 ff¦ f fh fG fí fÌ gm gn1 gq gr¦ gsG gth gwÌ gxí hO1 hP hQh hRG hS¦ hT hUí hVÌ i/ i01 i1G i2h i3 i4¦ i5Ì i6í

那些翻译得不好，这是字节值：

{ 0x63, 0xEA, 0x10, }
{ 0x63, 0xEB, 0x31, }
{ 0x63, 0xF4, 0xCC, }
{ 0x63, 0xF5, 0xED, }
{ 0x63, 0xF6, 0x85, }
{ 0x63, 0xF7, 0xA6, }
{ 0x63, 0xF8, 0x47, }
{ 0x63, 0xF9, 0x68, }
{ 0x64, 0xCC, 0x31, }
{ 0x64, 0xCD, 0x10, }
{ 0x64, 0xD2, 0xED, }
{ 0x64, 0xD3, 0xCC, }
{ 0x64, 0xD6, 0x68, }
{ 0x64, 0xD7, 0x47, }
{ 0x64, 0xD8, 0xA6, }
{ 0x64, 0xD9, 0x85, }
{ 0x65, 0xAC, 0x10, }
{ 0x65, 0xAD, 0x31, }
{ 0x65, 0xB2, 0xCC, }
{ 0x65, 0xB3, 0xED, }
{ 0x65, 0xB6, 0x47, }
{ 0x65, 0xB7, 0x68, }
{ 0x65, 0xB8, 0x85, }
{ 0x65, 0xB9, 0xA6, }
{ 0x66, 0x8D, 0x31, }
{ 0x66, 0x8E, 0x10, }
{ 0x66, 0x91, 0xA6, }
{ 0x66, 0x92, 0x85, }
{ 0x66, 0x93, 0x68, }
{ 0x66, 0x94, 0x47, }
{ 0x66, 0x97, 0xED, }
{ 0x66, 0x98, 0xCC, }
{ 0x67, 0x6D, 0x10, }
{ 0x67, 0x6E, 0x31, }
{ 0x67, 0x71, 0x85, }
{ 0x67, 0x72, 0xA6, }
{ 0x67, 0x73, 0x47, }
{ 0x67, 0x74, 0x68, }
{ 0x67, 0x77, 0xCC, }
{ 0x67, 0x78, 0xED, }
{ 0x68, 0x4F, 0x31, }
{ 0x68, 0x50, 0x10, }
{ 0x68, 0x51, 0x68, }
{ 0x68, 0x52, 0x47, }
{ 0x68, 0x53, 0xA6, }
{ 0x68, 0x54, 0x85, }
{ 0x68, 0x55, 0xED, }
{ 0x68, 0x56, 0xCC, }
{ 0x69, 0x2F, 0x10, }
{ 0x69, 0x30, 0x31, }
{ 0x69, 0x31, 0x47, }
{ 0x69, 0x32, 0x68, }
{ 0x69, 0x33, 0x85, }
{ 0x69, 0x34, 0xA6, }
{ 0x69, 0x35, 0xCC, }
{ 0x69, 0x36, 0xED, }
于 2013-02-19T21:25:26.807 回答