applied cryptography - protocols, algorithms, and source code in c

The final output is the concatenation of a b c d and

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x455a14ed) GG (a, b, c, d, M13, 5, 0xa9e3e905) GG (d, a, b, c, M2, 9, 0xfcefa3f8) GG (c, d, a, b, M7, 14, 0x676f02d9) GG (b, c, d, a, M12, 20, 0x8d2a4c8a) Round 3: HH (a, b, c, d, M5, 4, 0xfffa3942) HH (d, a, b, c, M8, 11, 0x8771f681) HH (c, d, a, b, M11, 16, 0x6d9d6122) HH (b, c, d, a, M14, 23, 0xfde5380c) HH (a, b, c, d, M1, 4, 0xa4beea44) HH (d, a, b, c, M4, 11, 0x4bdecfa9) HH (c, d, a, b, M7, 16, 0xf6bb4b60) HH (b, c, d, a, M10, 23, 0xbebfbc70) HH (a, b, c, d, M13, 4, 0x289b7ec6) HH (d, a, b, c, M0, 11, 0xeaa127fa) HH (c, d, a, b, M3, 16, 0xd4ef3085) HH (b, c, d, a, M6, 23, 0x04881d05) HH (a, b, c, d, M9, 4, 0xd9d4d039) HH (d, a, b, c, M12, 11, 0xe6db99e5) HH (c, d, a, b, M15, 16, 0x1fa27cf8) HH (b, c, d, a, M2, 23, 0xc4ac5665) Round 4: II (a, b, c, d, M0, 6, 0xf4292244) II (d, a, b, c, M7, 10, 0x432aff97) II (c, d, a, b, M14, 15, 0xab9423a7) II (b, c, d, a, M5, 21, 0xfc93a039) II (a, b, c, d, M12, 6, 0x655b59c3) II (d, a, b, c, M3, 10, 0x8f0ccc92) II (c, d, a, b, M10, 15, 0xffeff47d) II (b, c, d, a, M1, 21, 0x85845dd1) II (a, b, c, d, M8, 6, 0x6fa87e4f) II (d, a, b, c, M15, 10, 0xfe2ce6e0) II (c, d, a, b, M6, 15, 0xa3014314) II (b, c, d, a, M13, 21, 0x4e0811a1) II (a, b, c, d, M4, 6, 0xf7537e82) II (d, a, b, c, M11, 10, 0xbd3af235) II (c, d, a, b, M2, 15, 0x2ad7d2bb) II (b, c, d, a, M9, 21, 0xeb86d391) Those constants, ti, were chosen as follows: In step i, ti is the integer part of 232*abs(sin(i)), where i is in radians. After all of this, a, b, c, and d are added to A, B, C, D, respectively, and the algorithm continues with the next block of data. The final output is the concatenation of A, B, C, and D. Security of MD5 Ron Rivest outlined the improvements of MD5 over MD4 [1322]: 1. A fourth round has been added. 2. Each step now has a unique additive constant. 3. The function G in round 2 was changed from ((X¥ Y ) ¦ (X¥ Z ) ¦ (Y¥ Z )) to ((X¥ Z ) ¦ (Y¥ ¬ Z )) to make G less symmetric. 4. Each step now adds in the result of the previous step. This promotes a faster...
View Full Document

This note was uploaded on 10/18/2010 for the course MATH CS 301 taught by Professor Aliulger during the Fall '10 term at Koç University.

Ask a homework question - tutors are online