There will (on average) be one 16-byte message that maps to each of the possible MD5s. And there will be one 32-byte message that maps to each of the MD5s. And one 48-byte message that maps to each of the MD5s. And so on.
I don't follow this logic. There are a possible of 2^128 possible 128bit hashes. 2^128 == 8^16, the number of possible 16-byte messages. So that indeeds maps, on average 1 message to each hash.

But there are 8^32 possible 32 byte messages. Mapping to a possible 2^128 128bit hashes. Which means that, on average, 8^16 messages map to each hash. All with the same length.

So, for messages of length 0 .. 2^64, there will be (on average) 2^59 messages that will map to each of the MD5s.
I don't think you are right. Just looking at messages of 2^64 bytes, you have 8^(2^64) different messages, or 2^(3*2^64). Mapping to a space of 2^128. Which means that on average, 2^(3*2^64 - 128) messages map to the same hash. Or 2^55340232221128654720 different messages. Which is a really big number.

In reply to Re: (OT)Speculation: 128-bit digest + 64-bit length (192-bits) is more reliable and unique than a 256-digest alone. by JavaFan
in thread (OT)Speculation: 128-bit digest + 64-bit length (192-bits) is more reliable and unique than a 256-digest alone. by BrowserUk

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