• md4算法实现原理深剖


    一、基本介绍

    MD系列算法是信息摘要三大算法中的一种,全称:Message Digest算法,按照规范版本分为MD2、MD4、MD5三种算法,目前最常用的是MD5版本算法。本文介绍MD4算法的实现原理。

    1990 年,罗纳德·李维斯特教授开发出较之 MD2 算法有着更高安全性的 MD4 算法。在这个算法中,我们仍需对信息进行数据补位。不同的是,这种补位使其信息的字节长度加上 448 个字节后能成为 512 的倍数(信息字节长度 mod 512 = 448)。此外,关于 MD4 算法的处理与 MD2 算法又有很大差别。但最终仍旧是会获得一个 128 位的散列值。MD4 算法对后续消息摘要算法起到了推动作用,许多比较有名的消息摘要算法都是在 MD4 算法的基础上发展而来的,如 MD5、SHA-1、RIPE-MD 和 HAVAL 算法等。

    二、实现原理

    有关 MD4 算法详情请参见 RFC 1320 http://www.ietf.org/rfc/rfc1320.txt,RFC 1320 是MD4算法的官方文档,其实现原理共分为5步:

    第1步:字节填充(Append Padding Bytes)

    数据先补上1个1比特,再补上k个0比特,使得补位后的数据比特数(n+1+k)满足(n+1+k) mod 512 = 448,k取最小正整数。

    第2步:追加长度信息(Append Length)

    数据比特位的数据长度追加到最后8字节中。

    第3步:初始化MD Buffer(Initialize MD Buffer)

    这一步最简单了,定义ABCD四个4字节数组,分别赋初值即可。

    uint32_t A = 0x67452301;	// [ 0x01, 0x23, 0x45, 0x67 ]
    uint32_t B = 0xEFCDAB89;	// [ 0x89, 0xAB, 0xCD, 0xEF ]
    uint32_t C = 0x98BADCFE;	// [ 0xFE, 0xDC, 0xBA, 0x98 ]
    uint32_t D = 0x10325476;	// [ 0x76, 0x54, 0x32, 0x10 ]    
    

    第4步:处理消息块(Process Message in 16-Byte Blocks)

    这个是MD4算法最核心的部分了,对第2步组装数据进行分块依次处理。

          Process each 16-word block. */
          For i = 0 to N/16-1 do
    
            /* Copy block i into X. */
            For j = 0 to 15 do
              Set X[j] to M[i*16+j].
            end /* of loop on j */
    
            /* Save A as AA, B as BB, C as CC, and D as DD. */
            AA = A
            BB = B
            CC = C
            DD = D
    
            /* Round 1. */
            /* Let [abcd k s] denote the operation
                 a = (a + F(b,c,d) + X[k]) <<< s. */
            /* Do the following 16 operations. */
            [ABCD  0  3]  [DABC  1  7]  [CDAB  2 11]  [BCDA  3 19]
            [ABCD  4  3]  [DABC  5  7]  [CDAB  6 11]  [BCDA  7 19]
            [ABCD  8  3]  [DABC  9  7]  [CDAB 10 11]  [BCDA 11 19]
            [ABCD 12  3]  [DABC 13  7]  [CDAB 14 11]  [BCDA 15 19]
    
            /* Round 2. */
            /* Let [abcd k s] denote the operation
                 a = (a + G(b,c,d) + X[k] + 5A827999) <<< s. */
            /* Do the following 16 operations. */
            [ABCD  0  3]  [DABC  4  5]  [CDAB  8  9]  [BCDA 12 13]
            [ABCD  1  3]  [DABC  5  5]  [CDAB  9  9]  [BCDA 13 13]
            [ABCD  2  3]  [DABC  6  5]  [CDAB 10  9]  [BCDA 14 13]
            [ABCD  3  3]  [DABC  7  5]  [CDAB 11  9]  [BCDA 15 13]
    
            /* Round 3. */
            /* Let [abcd k s] denote the operation
                 a = (a + H(b,c,d) + X[k] + 6ED9EBA1) <<< s. */
            /* Do the following 16 operations. */
            [ABCD  0  3]  [DABC  8  9]  [CDAB  4 11]  [BCDA 12 15]
            [ABCD  2  3]  [DABC 10  9]  [CDAB  6 11]  [BCDA 14 15]
            [ABCD  1  3]  [DABC  9  9]  [CDAB  5 11]  [BCDA 13 15]
            [ABCD  3  3]  [DABC 11  9]  [CDAB  7 11]  [BCDA 15 15]
    
            /* Then perform the following additions. (That is, increment each
               of the four registers by the value it had before this block
               was started.) */
            A = A + AA
            B = B + BB
            C = C + CC
            D = D + DD
    
          end /* of loop on i */
    

    第5步:输出(Output)

    这一步也非常简单,只需要将计算后的A、B、C、D进行拼接输出即可。

    三、示例讲解

     四、代码实现

    以下为C/C++代码实现:

    #include <string.h>
    #include <stdio.h>
    
    #define HASH_BLOCK_SIZE         64  /* 512 bits = 64 bytes */
    #define HASH_LEN_SIZE           8   /* 64 bits =  8 bytes */
    #define HASH_LEN_OFFSET         56  /* 64 bytes - 8 bytes */
    #define HASH_DIGEST_SIZE        16  /* 128 bits = 16 bytes */
    #define HASH_ROUND_NUM          64
    
    typedef unsigned char		uint8_t;
    typedef unsigned short int	uint16_t;
    typedef unsigned int		uint32_t;
    typedef unsigned long long	uint64_t;
    
    
    static uint32_t F(uint32_t X, uint32_t Y, uint32_t Z)
    {
    	return (X & Y) | ((~X) & Z);
    }
    static uint32_t G(uint32_t X, uint32_t Y, uint32_t Z)
    {
    	return (X & Y) | (X & Z) | (Y & Z);
    }
    static uint32_t H(uint32_t X, uint32_t Y, uint32_t Z)
    {
    	return X ^ Y ^ Z;
    }
    
    /* 循环向左移动offset个单位 */
    static uint32_t MoveLeft(uint32_t X, uint8_t offset)
    {
    	return (X << offset) | (X >> (32 - offset));
    }
    
    static uint32_t Round1(uint32_t A, uint32_t B, uint32_t C, uint32_t D, uint32_t M, uint32_t N)
    {
    	return MoveLeft(A + F(B, C, D) + M, N);
    }
    static uint32_t Round2(uint32_t A, uint32_t B, uint32_t C, uint32_t D, uint32_t M, uint32_t N)
    {
    	return MoveLeft(A + G(B, C, D) + M + 0x5A827999, N);
    }
    static uint32_t Round3(uint32_t A, uint32_t B, uint32_t C, uint32_t D, uint32_t M, uint32_t N)
    {
    	return MoveLeft(A + H(B, C, D) + M + 0x6ED9EBA1, N);
    }
    
    #define ASSERT_RETURN_INT(x, d) if(!(x)) { return d; }
    
    int md4(unsigned char *out, const unsigned char* in, const int inlen)
    {
    	ASSERT_RETURN_INT(out && in && (inlen > 0), 1);
    	int i = 0, j = 0, k = 0;
    	unsigned char L = 0, t = 0;
    
    	// step 1: 字节填充(Append Padding Bytes)
    	// 数据先补上1个1比特,再补上k个0比特,使得补位后的数据比特数(n+1+k)满足(n+1+k) mod 512 = 448,k取最小正整数
    	int iX = inlen / HASH_BLOCK_SIZE;
    	int iY = inlen % HASH_BLOCK_SIZE;
    	iX = (iY < HASH_LEN_OFFSET) ? iX : (iX + 1);
    
    	int iLen = (iX + 1) * HASH_BLOCK_SIZE;
    	unsigned char* M = malloc(iLen);
    	memcpy(M, in, inlen);
    	// 先补上1个1比特+7个0比特
    	M[inlen] = 0x80;
    	// 再补上(k-7)个0比特
    	for (i = inlen + 1; i < (iX * HASH_BLOCK_SIZE + HASH_LEN_OFFSET); i++)
    	{
    		M[i] = 0;
    	}
    
    	// step 2: 追加长度信息(Append Length)
    	uint64_t *pLen = (uint64_t*)(M + (iX * HASH_BLOCK_SIZE + HASH_LEN_OFFSET));
    	*pLen = inlen << 3;
    
    	// Step 3. 初始化MD Buffer(Initialize MD Buffer)
    	uint32_t A = 0x67452301;	// [ 0x01, 0x23, 0x45, 0x67 ]
    	uint32_t B = 0xEFCDAB89;	// [ 0x89, 0xAB, 0xCD, 0xEF ]
    	uint32_t C = 0x98BADCFE;	// [ 0xFE, 0xDC, 0xBA, 0x98 ]
    	uint32_t D = 0x10325476;	// [ 0x76, 0x54, 0x32, 0x10 ]
    
    	uint32_t X[HASH_BLOCK_SIZE / 4] = { 0 };
    
    	// step 4: 处理消息块(Process Message in 64-Byte Blocks)
    	for (i = 0; i < iLen / HASH_BLOCK_SIZE; i++)
    	{
    		/* Copy block i into X. */
    		for (j = 0; j < HASH_BLOCK_SIZE; j = j + 4)
    		{
    			uint32_t* temp = &M[i * HASH_BLOCK_SIZE + j];
    			X[j/4] = *temp;
    		}
    
    		/* Save A as AA, B as BB, C as CC, and D as DD. */
    		uint32_t AA = A;
    		uint32_t BB = B;
    		uint32_t CC = C;
    		uint32_t DD = D;
    
    		/* Round 1. */
    		/* Let [abcd k s] denote the operation
    		a = (a + F(b,c,d) + X[k]) <<< s. */
    
    		/* Do the following 16 operations.
    		[ABCD  0  3][DABC  1  7][CDAB  2 11][BCDA  3 19]
    		[ABCD  4  3][DABC  5  7][CDAB  6 11][BCDA  7 19]
    		[ABCD  8  3][DABC  9  7][CDAB 10 11][BCDA 11 19]
    		[ABCD 12  3][DABC 13  7][CDAB 14 11][BCDA 15 19]
    		*/
    		A = Round1(A, B, C, D, X[0], 3);  D = Round1(D, A, B, C, X[1], 7);  C = Round1(C, D, A, B, X[2], 11);  B = Round1(B, C, D, A, X[3], 19);
    		A = Round1(A, B, C, D, X[4], 3);  D = Round1(D, A, B, C, X[5], 7);  C = Round1(C, D, A, B, X[6], 11);  B = Round1(B, C, D, A, X[7], 19);
    		A = Round1(A, B, C, D, X[8], 3);  D = Round1(D, A, B, C, X[9], 7);  C = Round1(C, D, A, B, X[10], 11); B = Round1(B, C, D, A, X[11], 19);
    		A = Round1(A, B, C, D, X[12], 3); D = Round1(D, A, B, C, X[13], 7); C = Round1(C, D, A, B, X[14], 11); B = Round1(B, C, D, A, X[15], 19);
    
    		/* Round 2. */
    		/* Let [abcd k s] denote the operation
    		a = (a + G(b,c,d) + X[k] + 5A827999) <<< s. */
    
    		/* Do the following 16 operations.
    		[ABCD  0  3][DABC  4  5][CDAB  8  9][BCDA 12 13]
    		[ABCD  1  3][DABC  5  5][CDAB  9  9][BCDA 13 13]
    		[ABCD  2  3][DABC  6  5][CDAB 10  9][BCDA 14 13]
    		[ABCD  3  3][DABC  7  5][CDAB 11  9][BCDA 15 13] 
    		*/
    		A = Round2(A, B, C, D, X[0], 3);  D = Round2(D, A, B, C, X[4], 5);  C = Round2(C, D, A, B, X[8], 9);  B = Round2(B, C, D, A, X[12], 13);
    		A = Round2(A, B, C, D, X[1], 3);  D = Round2(D, A, B, C, X[5], 5);  C = Round2(C, D, A, B, X[9], 9);  B = Round2(B, C, D, A, X[13], 13);
    		A = Round2(A, B, C, D, X[2], 3);  D = Round2(D, A, B, C, X[6], 5);  C = Round2(C, D, A, B, X[10], 9); B = Round2(B, C, D, A, X[14], 13);
    		A = Round2(A, B, C, D, X[3], 3);  D = Round2(D, A, B, C, X[7], 5);  C = Round2(C, D, A, B, X[11], 9); B = Round2(B, C, D, A, X[15], 13);
    
    		/* Round 3. */
    		/* Let [abcd k s] denote the operation
    		a = (a + H(b,c,d) + X[k] + 6ED9EBA1) <<< s. */
    		/* Do the following 16 operations. 
    		[ABCD  0  3][DABC  8  9][CDAB  4 11][BCDA 12 15]
    		[ABCD  2  3][DABC 10  9][CDAB  6 11][BCDA 14 15]
    		[ABCD  1  3][DABC  9  9][CDAB  5 11][BCDA 13 15]
    		[ABCD  3  3][DABC 11  9][CDAB  7 11][BCDA 15 15]
    		*/
    		A = Round3(A, B, C, D, X[0], 3);  D = Round3(D, A, B, C, X[8], 9);  C = Round3(C, D, A, B, X[4], 11);  B = Round3(B, C, D, A, X[12], 15);
    		A = Round3(A, B, C, D, X[2], 3);  D = Round3(D, A, B, C, X[10], 9); C = Round3(C, D, A, B, X[6], 11);  B = Round3(B, C, D, A, X[14], 15);
    		A = Round3(A, B, C, D, X[1], 3);  D = Round3(D, A, B, C, X[9], 9);  C = Round3(C, D, A, B, X[5], 11);  B = Round3(B, C, D, A, X[13], 15);
    		A = Round3(A, B, C, D, X[3], 3);  D = Round3(D, A, B, C, X[11], 9); C = Round3(C, D, A, B, X[7], 11);  B = Round3(B, C, D, A, X[15], 15);
    
    		/* Then perform the following additions. (That is, increment each
    		of the four registers by the value it had before this block
    		was started.) */
    		A = A + AA;
    		B = B + BB;
    		C = C + CC;
    		D = D + DD;
    	}
    
    	// step 5: 输出ABCD
    	memcpy(out + 0,  &A, 4);
    	memcpy(out + 4,  &B, 4);
    	memcpy(out + 8,  &C, 4);
    	memcpy(out + 12, &D, 4);
    	free(M);
    
    	return 0;
    }
    
    int main()
    {
    	unsigned char digest[16] = { 0 };
    
    	md4(digest, "Hello World!", strlen("Hello World!"));
    
    	return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Kingfans/p/16552308.html
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