MD5 Algorithm Descrip on with an example We begin by supposing that we have a b‐bit message as input, and that we wish to find its message digest. Here b is an arbitrary nonnegative integer; b may be zero, it need not be a multiple of eight, and it may be arbitrarily large. We imagine the bits of the message written down as follows: m_0 m_1 ... m_{b‐1} The following five steps are performed to compute the message digest of the message. Step 1. Append Padding Bits Ex. : The message should be "Hello World!", which is of length 12 bytes, they are…
The message is "padded" (extended) so that its length (in bits) is congruent to 448, modulo 512. That is, the message is extended so that it is just 64 bits shy of being a mul ple of 512 bits long. Padding is always performed, even if the length of the message is already congruent to 448, modulo 512. Padding is performed as follows: a single "1" bit is appended to the message, and then "0" bits are appended so that the length in bits of the padded message becomes congruent to 448, modulo 512. In all, at least one bit and at most 512 bits are appended. Ex. In this case, the message fits a single 512 bit block, so we will add 448 – 12*8 = 352 padding bits (it will start with “1” then 351 “0”s) Step 2. Append Length A 64‐bit representation of b (the length of the message before the padding bits were added) is appended to the result of the previous step. In the unlikely event that b is greater than 2^64, then only the low‐order 64 bits of b are used. (These bits are appended as eight bytes and appended low‐order byte) Ex. : The length of message in bits is 96 bits == 00000000 00000000 00000000 00000000 00000000 00000000 00000000 01100000… Taking low order byte first will result in
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At this point the resulting message (after padding with bits and with b) has a length that is an exact mul ple of 512 bits. Equivalently, this message has a length that is an exact mul ple of 16 (32‐bit) words. Let M[0 ... N‐1] denote the words of the resul ng message, where N is a mul ple of 16. Ex. Now the only block available holds…
Step 3. Ini alize MD Buffer A four‐word buffer (A,B,C,D) is used to compute the message digest. Here each of A, B, C, D is a 32‐bit register. These registers are initialized to the following values in hexadecimal, low‐order bytes first): 32‐bit registers values (Hexa)
After conversion to in low‐order bytes first (Hexa)
After Conversion to in low‐order bytes first (Decimal)
word A: 01 23 45 67 word B: 89 ab cd ef word C: fe dc ba 98 word D: 76 54 32 10
0x67452301 0xefcdab89 0x98badcfe 0x10325476
a: 1732584193 b: 4023233417 c: 2562383102 d: 271733878
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Step 4. Process Message in 16‐Word Blocks We first define four auxiliary func ons that each take as input three 32‐bit words and produce as output one 32‐bit word. F(X,Y,Z) = XY v not(X) Z G(X,Y,Z) = XZ v Y not(Z) H(X,Y,Z) = X xor Y xor Z I(X,Y,Z) = Y xor (X v not(Z)) In each bit position F acts as a conditional: if X then Y else Z. The function F could have been defined using + instead of v since XY and not(X)Z will never have 1's in the same bit position.) It is interesting to note that if the bits of X, Y, and Z are independent and unbiased, the each bit of F(X,Y,Z) will be independent and unbiased. The functions G, H, and I are similar to the function F, in that they act in "bitwise parallel" to produce their output from the bits of X, Y, and Z, in such a manner that if the corresponding bits of X, Y, and Z are independent and unbiased, then each bit of G(X,Y,Z), H(X,Y,Z), and I(X,Y,Z) will be independent and unbiased. Note that the function H is the bit‐wise "xor" or "parity" function of its inputs. This step uses a 64‐element table T[1 ... 64] constructed from the sine func on. Let T[i] denote the i‐th element of the table, which is equal to the integer part of 4294967296 mes abs(sin(i)), where i is in
radians (switch your calculator to Radians mode instead of Degrees mode). The elements of the table are given in the appendix. Do the following: /* Process each 16-word block. */ For i = 0 to N/16-1 do /* Copy block i into X. */ Ex. Now “X” the 16 words each of 32 bits will hold, as the following loop does
For j = 0 to 15 do Set X[j] to M[i*16+j]. end /* of loop on j */
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/* 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 i] denote the operation a = b + ((a + F(b,c,d) + X[k] + T[i]) <<< s). */ /* Do the following 16 operations. */ [ABCD 0 7 1]
Ex. * Perform in Binary: F(b,c,d) = F(B,C,D) = BC v not(B) D = (4023233417^2562383102)V(NOT(4023233417)^ 271733878) = 2562383102 * (a + F(b,c,d) + X[k] + T[i]) = 1732584193+2562383102+1819043144+3614090360 =9728100799(10 01000011110101110000100110111111) The above is 34 bits, so we do modular additions, to get the results of only 32 bits 9728100799 MOD (2Power32) = 1138166207 (01000011110101110000100110111111) * ((a + F(b,c,d) + X[k] + T[i]) <<< s) = 1138166207 <<< 7 = 3951353761 OR 01000011110101110000100110111111 <<< 7 = 11101011100001001101111110100001 * Again Modular Addition with b: a = b + ((a + F(b,c,d) + X[k] + T[i]) <<< s) = (4023233417+3951353761)MOD(2Power32) = 3679619882 * Now a is 3679619882. b, c, and d are not changed. [DABC
1 12
2]
* Just as if you call [abcd k s i] by passing [DABC 1 12 2], meaning in the expression mentioned in the comment above, you will replace a by D, b by A, c by B d by C, k by 1, s by 12, and i by 2 * Now c is 2562383102 and the rest of registers is unchanged [ABCD 4 [ABCD 8 [ABCD 12
7 5] 7 9] 7 13]
[DABC 5 12 6] [DABC 9 12 10] [DABC 13 12 14]
[CDAB 2 17 3] [CDAB 6 17 7] [CDAB 10 17 11] [CDAB 14 17 15]
/* Round 2. */ /* Let [abcd k s i] denote the operation a = b + ((a + G(b,c,d) + X[k] + T[i]) /* Do the following 16 operations. */ [ABCD 1 5 17] [DABC 6 9 18] [CDAB 11 [ABCD 5 5 21] [DABC 10 9 22] [CDAB 15 [ABCD 9 5 25] [DABC 14 9 26] [CDAB 3 [ABCD 13 5 29] [DABC 2 9 30] [CDAB 7 /* Round 3. */ /* Let [abcd k s t] denote the operation a = b + ((a + H(b,c,d) + X[k] + T[i]) /* Do the following 16 operations. */ [ABCD 5 4 33] [DABC 8 11 34] [CDAB 11 [ABCD 1 4 37] [DABC 4 11 38] [CDAB 7 [ABCD 13 4 41] [DABC 0 11 42] [CDAB 3 [ABCD 9 4 45] [DABC 12 11 46] [CDAB 15 /* Round 4. */ /* Let [abcd k s t] denote the operation a = b + ((a + I(b,c,d) + X[k] + T[i]) /* Do the following 16 operations. */ [ABCD 0 6 49] [DABC 7 10 50] [CDAB 14 [ABCD 12 6 53] [DABC 3 10 54] [CDAB 10 [ABCD 8 6 57] [DABC 15 10 58] [CDAB 6 [ABCD 4 6 61] [DABC 11 10 62] [CDAB 2
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[BCDA 3 22 4] [BCDA 7 22 8] [BCDA 11 22 12] [BCDA 15 22 16]
<<< s). */ 14 14 14 14
19] 23] 27] 31]
[BCDA 0 [BCDA 4 [BCDA 8 [BCDA 12
20 20 20 20
20] 24] 28] 32]
<<< s). */ 16 16 16 16
35] 39] 43] 47]
[BCDA 14 23 36] [BCDA 10 23 40] [BCDA 6 23 44] [BCDA 2 23 48]
<<< s). */ 15 15 15 15
51] 55] 59] 63]
[BCDA 5 21 [BCDA 1 21 [BCDA 13 21 [BCDA 9 21
52] 56] 60] 64]
Round/Function Iteration
1 / FF
2 / GG
3 / HH
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44
Changed Register a d c b a d c b a d c b a d c b a d c b a d c b a d c b a d c b a d c b a d c b a d c b 5
Value 3679619882 2081685101 133919401 2426329734 1168620452 4019849227 370807664 420445366 4198619016 2676785399 1993335089 1109401451 4048899853 3950087495 1946913403 2833700088 260718522 1355334187 80112311 3466307461 396033290 1763835213 4239641319 3052862749 2616217882 921013414 2418186433 2064486443 1988550793 146954281 198383918 1503938765 1829143082 2652512423 44713168 3452888323 4152347168 2590106416 4138040678 3946422880 1824650282 3516852032 823396667 206704551
45 a 1129083802 46 d 1795018062 47 c 2728674280 48 b 2145583183 49 a 3393064847 50 d 3765494986 51 c 2359501063 52 b 1432248893 53 a 449791293 54 d 411940788 55 c 548752943 56 b 1539193972 4 / II 57 a 3430658958 58 d 3442781692 59 c 1141842014 60 b 924561567 a 538764524 61 d 2077978953 62 c 4184057696 63 b 1186497226 64 The last four rows hold the last values of the four registers… /* 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
Ex. At last Modular Addition with the original values A = A + AA = (1732584193 + 538764524 ) MOD(2Power32) = 2271348717 = 10000111 01100010 00000111 11101101 = 0x876207ED B = B + BB = (4023233417 + 1186497226) MOD(2Power32) = 914763347 = 00110110 10000110 00101110 01010011 = 0x36862E53 C = C + CC = (2562383102 + 4184057696) MOD(2Power32) = 2451473502 = 10010010 00011110 10000100 01011110 = 0x921E845E D = D + DD = (271733878 + 2077978953) MOD(2Power32) = 2349712831 = 10001100 00001101 11000101 10111111 = 0x8C0DC5BF Ex. This is for the only first 512 bit block, now continue to the next 512-bit blocks. But this is not in this example ☺. We are done manipulating our “Hello World!” message. end /* of loop on i */
Step 5. Output The message digest produced as output is A, B, C, D. That is, we begin with the low‐order byte of A, and end with the high‐order byte of D.
Ex. After we are done with the above loop, we take A, B, C, and D and reorder their bytes: (Each register is 32 bits (4 Bytes))
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A 4 3 2 1 B 8 7 6 5 C 12 11 10 9 D 16 15 14 13 * Now the MD5 digest is stored as the above bytes in ascending order (1, 2, 3, …) (Remember that a single byte is equivalent to 2 Hexadecimals) * The MD5 digest for the above example is 0xED076287532E86365E841E92BFC50D8C Summary The MD5 message‐digest algorithm is simple to implement, and provides a "fingerprint" or message digest of a message of arbitrary length. It is conjectured that the difficulty of coming up with two messages having the same message digest is on the order of 2^64 opera ons, and that the difficulty of coming up with any message having a given message digest is on the order of 2^128 opera ons. The MD5 algorithm has been carefully scru nized for weaknesses. It is, however, a relatively new algorithm and further security analysis is of course justified, as is the case with any new proposal of this sort. References MD5 Specifica ons: h p://www.ie .org/rfc/rfc1321.txt MD5 Implementa on in C#: h p://www.flowgroup.fr/download/md5.zip
Good Luck Dr. Hossam Mr. Bishoy Moussa
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