Man in middle attack

2 Jul 2002

      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@subdimension.com> wrote:
...
From: "gfgs pedo" <jtrjtrjtr2001@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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