Can anyone explain how the MITM works on 3TDES (three distinct keys)?

I am typically interested in finding out why 3TDES has effective
key-length of 112-bit

In Peace,
Saqib Ali
http://www.full-disc-encryption.com

"Saqib Ali" <docbook.xml [at] gmail.com> wrote in news:1157590926.262613.165900
[at] e3g2000cwe.googlegroups.com:

> Can anyone explain how the MITM works on 3TDES (three distinct keys)?
>
> I am typically interested in finding out why 3TDES has effective
> key-length of 112-bit
>
> In Peace,
> Saqib Ali
> http://www.full-disc-encryption.com
>
>

http://en.wikipedia.org/wiki/Triple_DES

Regards,

> http://en.wikipedia.org/wiki/Triple_DES

This has NO explanation of how the MITM works or why the effective key
lenght is reduced to 112 bits.

In Peace,
Saqib Ali
http://www.full-disc-encryption.com

I think you should post your question to sci.crypt...

Kind regards
Ludovic

Saqib Ali wrote:
> Can anyone explain how the MITM works on 3TDES (three distinct keys)?
>
> I am typically interested in finding out why 3TDES has effective
> key-length of 112-bit
>

I wrote this ages ago for TechTarget, but it answers your question.

<http://searchsecurity.techtarget.com/ateQuestionNResponse/0,289625,sid14_cid591441_tax292741,00.html>

Jon

Jon wrote:
<http://searchsecurity.techtarget.com/ateQuestionNResponse/0,289625,sid14_cid591441_tax292741,00.html>

Thank you Sir. This is what I was looking for. :)

Another good explanation was given by Mark Wooding on sci.crypt:

----------------
So, triple DES involves three keys, K1, K2, K3. Write
single-DESencryption with a key K and plaintext block x as E(K, x), and
decryption as D(K, x). Triple DES encryption is E(K3, D(K2, E(K1,
x))).

Suppose you're given a plaintext block x and corresponding ciphertext
y. For each possible K3, compute D(K3, y), and store the result in a
table. This takes about 2^56 work, and uses 2^56 blocks of memory.

Now, for each pair K1, K2, compute D(K2, E(K1, x)). If this matches
one of the values in the table, find the corresponding K3, and test the
whole key against some other plaintext/ciphertext pairs. Continue
until you're done. This step takes no extra memory and requires 2^112
time.
--------------------

In Peace,
Saqib Ali
http://www.full-disc-encryption.com

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