5.7: MD5 Algorithm in .NET

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Table 5.7: MD5 Algorithm
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M = (Yo, I.,. . ,YN-~), Y message to hash, after padding Each Y, is a 32-bit word and N is a multiple of 16
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MD5(M) // initialize ( A ,B , C,D ) = IV ( A , , D ) = (0x67452301,Oxef cdab89,Ox98badcfe,0x10325476) B f o r i = 0 t o N/16 - 1 // Copy block i into X xj = Y16i+j, f o r j = 0 t o 15 // Copy X to W Wj = X g ( j l , for j = 0 to 63 // initialize Q (Q-4, Q-3, Q-2, Q-1) = (A, D , C , B) // Rounds 0, 1, 2 and 3 RoundO(Q, W ) Roundl(Q, W ) Round2 (Q,W ) Round3(Q, W ) // Each addition is modulo 2 ( A ,B, ,D ) = (Q6o + Q-4, Q63 Q-I, Q6z + Q-2, Q6i + Q - 3 ) C next i return A,B , C , D end MD5
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6. It is claimed in [122] (without further explanation) that in MD5, the shift amounts in each round have been approximately optimized, to yield a faster avalanche effect . Also, the shifts employed in each round of MD5 are distinct, which is not the case in MD4.
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A single MD5 step is illustrated in Figure 5.5, where fi(A, C )= B,
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It may be instructive to compare Figure 5.5 to a step of the MD4 algorithm, which is illustrated in Figure 5.3. As in the MD4 attack of Section 5.3, here we denote rounds i through j of the MD5 function as MD5i..,j(A,B , C ,D , M ) , where ( A , C ,D ) are the B, initial values at step i and M is a 512-bit input block. Similar to MD4, we have MD5o...6 3 w , MI # h ( M ) ,
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F ( A ,B , C ) G(A,B,C) H(A,B,C) I ( A ,B , C )
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if 0 5 i 5 15 if 16 5 i 5 31 if325i547 if 48 5 i 5 63.
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HASH FUNCTIONS
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Table 5.8: MD5 Rounds RoundO(Q, W ) // steps 0 through 15
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i = 0 to 15 Qi = Q i - 1 + next i
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((Qi-4
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+ F(Qi-1,
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Q i - 2 , Qi-3)
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+ Wi + K i ) << s i ) <
end Round0
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Roundl(Q, W ) // steps 16 through 3 1 f o r i = 16 to 31
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next i
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= Qi-1
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+ ((Qi-4 + G(Qi-1, Q i - 2 , Q i - 3 ) + Wi+ K i ) << ~ <
end Round1
Round2(Q, W ) // steps 32 through 47 f o r i = 32 to 47
next i
= Qi-1
+ ( ( Q i - 4 + H(Qi-1, Qi-2,
Qi-3)
+ W + Ky) << s i ) i <
end Round2
RoundS(Q, W ) // steps 48 through 63 f o r i = 48 to 63
next i
= Qi-1+
((Qi-4
+ I(Qi-1,
Qi-2,
Qi-3)
&) << S i ) <
end Round3
where h is the MD5 hash function. In other words, the hash value is not just the output of this function, since there is a final transformation and the message is padded. Next, we consider the final transformation; for details on the padding, see Section 5.3. Define f(Iv, = ( Q ~ o , Q62, Q61) + IV, nil) Q63, where (Qfjg,Q 6 2 , Q 6 1 , &so) = MD50 ..63(IV,M ) and the addition is computed modulo 2'32,per 32-bit word. Then f is the MD5 compression function. If we hash a message M = (Mo,M I ) consisting of two 512-bit blocks, we have
h ( M ) = f ( f ( I Vb,f o ) , Ml).
In effect, f(IV, Mo) acts as the IV for the second iteration of the compression
5.4 MD5
Stepi 0 1 Shift si 7 12 S t e p i 16 17 Shift si 5 9 Step i 32 33 Shift si 4 11 SteD i 48 49 Shift s,I 6 10
2 3 4 5 6 7 8 9 10 11 12 13 14 15 17 22 7 12 17 22 7 12 17 22 7 12 17 22 18 14 35 16 50 15 19 20 35 23 51 21 20 5 36 4 52 6 21 9 37 11 53 10 22 14 38 16 54 15 23 20 39 23 55 21 24 5 40 4 56 6 25 9 41 11 57 10 26 14 42 16 58 15 27 20 43 23 58 21 28 5 44 4 60 6 29 9 45 11 61 10 30 14 46 16 62 15
31 20 47 23 63 21
Table 5.10: MD5 Input Word Order Stepil 0 a(z) 0 SteD i 16 ~ ( i )1 Step i 32 a(;) 5 Step i 48 a(i) 0
1 1 17 6 33 8 49 7
2 2 18 11 35 11 50 14
3 3 19 0 35 14 51 5
4 4 20 5 36 1 52 12
5 5 21 10 37 4 53 3
6 6 22 15 38 7 54 10
7 7 23 4 39 10 55 1
8 8 24 9
40 13 56 8
9 9 25 14
41 0 57 15
10 10 26 3 42 3 58 6
11 12 13 14 11 12 13 14 27 28 29 30 8 13 2 7
15 15 31 12 47 2 63 9
43 6 58 13
44 9 60 4
45 12 61 11
46 15 62 2
function f . Figure 5.1 illustrates the general case. At this point we have the necessary notation and background to begin discussing the MD5 attack. The MD4 attack in the previous section finds two 512-bit blocks that have the same MD4 hash value. The MD5 attack presented here, which is due to Wang [157],finds a pair of 1024-bit messages that have the same MD5 hash value. We denote the two 1024-bit messages as M = (M, ,,Mi) and 1 1 = (Mo,M I ) ,where each Ml and Mi is a 512-bit 1 block. As with MD4, each 512-bit block consists of 16 words, where each word is 32 bits. Initially, Wang gave a collision [155, 1561 without providing any explanation of the underlying technique. This lack of information led to an impressive attempt to reverse engineer the Chinese method [64]. Incredibly, this reverse engineering was successful enough that it yielded a more efficient attack [81] than Wang s, and it has provided the basis for subsequent improvements in the attack. The Chinese team did eventually release some details of their attack [157]. Minor flaws and incremental improvements have been reported by various