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result in a table, digest another message, compare it to the first one, and save the result in the table, and then digest another message, compare it to all previously saved values, and so on. Figure 5-4 illustrates these properties with a challenge: Find the message, or any message, and produce the given digest.
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Figure 5-4 Can you find the message we used to produce this digest (or any message that will produce it) If so, you will have found a collision in SHA-1
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By the way, you probably already know this, but for the sake of completeness, let s say it. A message is not necessarily a communication between two people. Any data you give to a digest algorithm is a message, even if it s not in human-readable form. Each byte of input is simply a byte of input, whether or not the byte is an ASCII character.
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When an algorithm violates one of the last two properties discussed in the preceding section, the result is a collision, the technical term to describe a situation in which two messages produce the same digest. A collision occurs when a second message produces the same digest as a previous message, or when two messages any two messages produce the same digest whatever that digest is. If two messages collide, they meet at the digest. Although the number of possible messages is virtually infinite, the number of possible digests is finite. With SHA-1, the number of possible digests is 2160. Clearly, there will be many messages that produce any one digest. To show that, let s use the time-honored mathematical tool known as the pigeonhole principle. Suppose you had a cabinet of pigeonholes (see Figure 5-5). Each pigeonhole corresponds to a digest. The zeroth pigeonhole is for the digest 00 00 . . . 00, the first is for 00 00 . . . 01, and so on, until you reach the last pigeonhole, the place for FF FF . . . FF. Now you start digesting messages. After you digest a message, place the message into the pigeonhole of the digest it produces. For example, the digest of the 1-byte message 00 is
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5b a9 3c 9d b0 cf f9 3f 52 b5 21 d7 42 0e 43 f6 ed a2 78 4f
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So you place message 00 into pigeonhole 5B A9 . . . 4F. The digest of message 01 is
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bf 8b 45 30 d8 d2 46 dd 74 ac 53 a1 34 71 bb a1 79 41 df f7
Message 01 goes into pigeonhole BF 8B . . . F7. Suppose you keep digesting messages, the next message being the preceding message plus 1. The sequence of messages is 00, 01, 02, . . ., FF, 01 00, 01 01, . . ., FF FF, 01 00 00, and so on. Suppose you did this for 2160 messages. The last message in the sequence would be
FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF
Figure 5-5
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The pigeonhole principle says that sooner or later some messages will collide in the same digest
Now suppose that each message produced a different digest. (For all we know, there were messages that produced the same digest, but for the sake of argument, let s say each message produced a different digest.) You had 2160 pigeonholes and 2160 messages, each message going into a different pigeonhole. This means that all pigeonholes are now occupied. Now consider Pao-Chi s message to Daniel (ordering four presses for Satomi). This 424-bit message is not a message you ve already examined. So far in this pigeonhole exercise, if you ve operated on a message, it s been 160 or fewer bits. To place Pao-Chi s message into a pigeonhole, you would place
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it into 46 73 . . . 5a. But that pigeonhole, like all the others, is already occupied. Which message it contains doesn t matter; you simply know it s occupied. You have a collision. Now consider that all possible messages includes messages of any size. Collisions exist, but no one can find a collision on demand (for some digest algorithms, no one has found any collision, even by accident). The worst possible scenario for a digest algorithm would be if someone could take any message and produce a similar message that produces the same digest. Figure 5-6 shows an example of that. One message mentions $1,000,000, and the second message mentions $1,500,000. If someone changes only the 5, the digests will not match. But what if someone could change the 5, change a few other things here and there, maybe add a phrase or two, and get the same digest
Figure 5-6 If a digest algorithm were predictable enough that an attacker could change a message slightly and produce the same digest, the algorithm would be broken
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