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Cracking MD5 for $10M

I mentioned a few days ago that one of the "rump session" papers at the
crypto conference claimed that a machine could be built which would find
MD5 collisions for $10M in about 20 days.  I wanted to write a little
more detail about how this attack could work.  It is similar to a "meet
in the middle" (MITM) attack which Norm Hardy suggested here in July when
we were discussing double DES:

>There may be more than one way that MITM (meet in the middle) may be used
>to attack Double block cyphers. I assume the following attack. You know
>some block of plain-text P and corresponding cypher text C. You believe
>that C = E(k, E(j, P)) where E(k, p) is the encypherment of p with key k.
>D(k, E(k, p)) = p. You need to find keys k and j. Classic MITM is to
>produce a file A with records: <k, E(k, P)> for each k, and file B with
>records <j, D(j, C)> for each j. Sort both A and B on the second field.
>Pass over the sorted files looking for a record from file A whose second
>field is the same as a record in file B.

>To substantially shorten the ammount of tape used by a factor 2^n at the
>expense of evaluating C and D 2^n more often do the following:

>For m from 0 to 2^n-1 Do
>  Produce file A with records: <k, E(k, P)> for each k where
>    (the right n bits of E(k, P)) = m. (discarding other records)
>  Produce file B with records <j, D(j, C)> for each j where
>    (the right n bits of D(j, C)) = m
>  Sort files A and B on second field.
>  Pass over files looking for records from A that match records from b in the
>  second field.

The idea of saving only outputs where certain bits are constant is the
key to the "distinguished points" method which is used to save space with
only a modest cost in time.  The other key idea is that instead of
evaluating MD5(n) where n iterates on its own, you look for cycles in the
recurrence x = MD5(x).  Any cycle which is found which does not include
the x you start with will lead to a case where two values hash to the
same MD5 value.

For a trivial example, suppose the output of a formula like this consists
of the values 1,4,5,2,7,8,5,2,7,8,5,2,7,8,....  Here we have a four
element cycle which leads to two different predecessors for the value 5.

The brute-force way to solve this would be to save all outputs from the
formula, and with each new value to compare it with all earlier
values.  With MD5, which has a presumably random structure and 128 bits
of output, the birthday paradox suggests that you would have to create
and save about 2^64 output values before finding a match.  Creating
2^64 values might be possible today for the time and dollar values we
are talking about, but storing them appears to be out of the question,
as our earlier discussion of double DES (and other discussions of MITM
here) have made clear.

The distinguished points method reduces the space requirements by only
saving a fraction of the output values.  For example, in the list above,
we might only save multiples of 4.  This would lead to 4,8,8... and it
is easy to discover the match without nearly as much storage.  Note,
though, that 8 is not actually the value which has two predecessors, but
that once this match is discovered, you can go back to the previous
points (4 and 8 in this case) and run them forward more carefully,
looking for a match.

The other real advantage of the distinguished points method is that it
parallelizes very nicely.  Several machines can run x=MD5(x) with
different starting values, saving all of the distinguished outputs, and
we can look for matches between machines as well as in one machine.
Again, a match implies two different predecessors for the same value,
which is an MD5 collision.

With the size of MD5, suppose we generate 2^64 outputs but only save
those for which the low-order 32 bits are 0 as our distinguished points.
Only 1/2^32 of values will match, so we will end up with about 2^32
outputs, probably a manageable amount.  Chances are there will be a match
among that set.  We then go back to the previous distinguished points
before the match and work forward carefully to look for the exact pair of
values which lead to the same successor.  Distinguished points will be
about 2^32 apart so this step is easy and quick.  If you want to speed it
up still more you can do a recursive distinguished points pass for this
step using maybe d.p.'s with the low-order 16-bits of 0 and do it in two
steps that will both be very short.

The net result is that we have taken virtually no more time (the 2^64
creations of MD5 will dominate) and virtually no space (compared to 2^64
stored values) and we get the effect of a birthday attack.  This is
another cautionary data point about the risks of relying on space costs
for security rather than time costs.