> From: Valdis.Kletnieks (at) vt (dot) edu [email concealed] [mailto:Valdis.Kletnieks (at) vt (dot) edu [email concealed]]
> Sent: Thursday, December 11, 2003 12:06 PM
>
> On Thu, 11 Dec 2003 07:37:02 GMT, Nick Cleaton said:
>
> > Even if the attacker knows or controls every other byte in
> > the packet and thus controls the checksum before the final 16
> > bits go in, the final checksum is as unpredictable as those 16
> > bits.

> However, it's a trivial matter to take the original text, the
> replacement text, and compute an original such that the checksum
> comes out "the same".

True, but irrelevant to the problem at hand, where the attacker has neither
the original checksum nor the original text. Michal's question was whether
an attacker who controls

- the checksum
- part, but not all, of the text

can set the checksum so as to have a better than 1/65536 possibility of
having it correct. Nick's response was no, if the attacker cannot control
as little as 16 bits of the text. As you pointed out yourself, any partial
checksum value can be mapped to any final checksum value by adding the final
two bytes.

--
Michael Wojcik
Principal Software Systems Developer, Micro Focus

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