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I'm wondering why hash constructions aren't parallel like CTR mode for encryption?

For example: We have ideal block cipher E, where we encrypt message block with counter and xor output block with message block. Then we xor (sum) all blocks and encrypt it with length of message and xor with sum of all blocks to get hash. If last message block is shorter than block size it is padded with zeroes.

$H_{0} = E_{0}(m_{0}) \oplus m_{0}$

$H_{1} = E_{1}(m_{1}) \oplus m_{1}$

$...$

$H_{n} = E_{n}(m_{n}) \oplus m_{n}$

$H_{sum} = H_{0} \oplus H_{1} \oplus ...\oplus H_{n}$

$H = E_{Length}(H_{sum}) \oplus H_{sum}$

Parallel hash construction example

Is such construction known to be insecure? Or is it secure?

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one acronym ... PMAC –  user1028028 May 31 at 13:37
    
@user1028028: I'm asking about hash, not MAC. –  LightBit Jun 1 at 7:21
1  
Related question: Counter mode secure hash algorithm –  CodesInChaos Jun 1 at 17:02
    
You can build a hash the same way you build a MAC. If you look at how PMAC works you'll see why your construction is wrong and how to fix it –  user1028028 Jun 2 at 7:51
    
@user1028028: PMAC without key is completely insecure. Fix it, how? –  LightBit Jun 2 at 16:18

2 Answers 2

up vote 3 down vote accepted

It is insecure, since the XOR of many ($n$) independent compression function is much more vulnerable to collision attacks. The generalized birthday attack finds a collision with complexity $$ 2^{\frac{\text{block size}}{1+\lg n}}. $$

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Dmitry said that a collision in the xor can be found with a generalized birthday attack.

Actually, it's a lot worse than that; second preimages of sufficiently long messages can be found in polynomial time; hence an enormous block size (which might help against the birthday attack) doesn't help.

Let us call the size of the output of the ideal block cipher $n$. Then, let us consider a message that consists of (at least) $n$ blocks; and for each block, we consider two distinct alternatives, at position $i$, we would have arbitrary selected blocks $M_i$ and $M'_i$.

We would then compute $H_i = E_i( M_i ) \oplus M_i$ and $H'_i = E_i( M'_i ) \oplus M'_i$.

If we have a target value $Z$ for the exclusive or's (which we would have for a second preimage problem), we would then consider what sum of $H_i$ and $H'_i$ would exclusive-or to the target values; this can be viewed as a linear algebra problem, and can be solved with, say, Gaussian elimination in time $O(n^3)$

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