Man in middle attack

[gfgs pedo]

hi,

> The only thing that might, as far as I can see,
> succeed (with a high
> probability) would be for everyone to hash the
> *next* half - meaning that,
> together with half 2 of message N, there will be the
> hash of half one of
> message N + 1. However, I don't see how this would
> be possible for an
> interactive communication...

As far as i can extend the previous attack,i.e faking
1 packet for interlock protocol in the above 1 you
propose,extending the same attack it only takes
Mallory
one and a half faked packets to launch a succefull
attack on the above proposal.

let 
A=Alice
M=Mallory
B=Bob

let 
1:1  indicate 1 st packet ,1st half
1:2  indicate 1 st packet , 2nd half

2:1  indicate 2 nd packet, 1st half
2:2  indicate 2nd packet , 2nd half 

and so on


so we are now have  1:2 and 2:1 as one complete
message
and so on

No:     A               M                B



1                    A->1:1           M->1:1   

2    M->1:1          B->1:1           

3                    A->1:2           M->1:2 

4    M->1:2          B->1:2          

5                    A->2:1           M->2:1

6    M->2:1          B->2:1

7                    A->2:2           ******
 


The blank spaces corresponding to each row indicates
that it is a sender and the other 2 are receivers.

Once Mallory receives A->2:2 ,he has 2 full packets in
hand and has faked 1 and a half packets(Step 7)

**** indicates that it is now the earler packet Bob
receives of Alice after Mallory's manupilation.
I hope that table will give some clarity.

now he can send Bob the original message of Alice.

So I think the above suggested protocol will not work.
Mallory can still get away with his scheme


Regards Data.


--- Marcel Popescu <mdpopescu at subdimension.com> wrote:
> From: "gfgs pedo" <jtrjtrjtr2001 at yahoo.com>
> 
> > One solution suggested against the man in the
> middle
> > attack is using the interlock protocol
> 
> This is the one I vaguely recalled, thank you.
> 
> > All mallory would have to do is send the half of
> the
> > (n th) packet when he receives the half of (n+1)th
> > packet since the 1 st packet was faked by mallory.
> 
> Interesting attack... assuming that a one-block
> delay doesn't look
> suspicious.
> 
> What if every message except the very first one has
> a hash of the previously
> received message?
> 
> A -> (M ->) B: half 1 of message A1
> B -> (M ->) A: half 1 of message B1 | hash (half 1
> of message A1)
> A -> (M ->) B: half 2 of message A1 | hash (half 1
> of message B1)
> B -> (M ->) A: half 2 of message B1 | hash (half 2
> of message A1)
> A -> (M ->) B: half 1 of message A2 | hash (half 2
> of message B1)
> ... and so on
> 
> Nah... won't work; since M captures A1 and B1, he
> can compute the hashes for
> both the initial bogus message and the (delayed)
> genuine ones. Same if they
> try hasing all the previous messages.
> 
> What if they send the hash of the *other* half? (The
> program splitting the
> messages already has the full ones.)
> 
> A -> (M ->) B: half 1 of message A1 | hash (half 2
> of message A1)
> B -> (M ->) A: half 1 of message B1 | hash (half 2
> of message B1)
> A -> (M ->) B: half 2 of message A1 | hash (half 1
> of message A1)
> B -> (M ->) A: half 2 of message B1 | hash (half 1
> of message B1)
> ... and so on
> 
> Nope, no good... M fakes the first message in both
> direction, and then he
> always has a good one, so he can compute the hashes.
> 
> The only thing that might, as far as I can see,
> succeed (with a high
> probability) would be for everyone to hash the
> *next* half - meaning that,
> together with half 2 of message N, there will be the
> hash of half one of
> message N + 1. However, I don't see how this would
> be possible for an
> interactive communication...
> 
> Thanks,
> Mark
> 
> 


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