-1
$\begingroup$

Suppose I had a hash function which produced a 256-bits output for any given input (such as SHA-3-256). Now suppose i did an exhaustive mapping from every possible 512-bits input to a given 256-bits output, and stored it in a table along with the original 512-bits input.

Now, if I later wanted to find the original 512-bits input based on a 256-bits hash, What is the probability of finding a collision rather than the original input?

It seems that in the case of exhaustive mapping from input to output, the risk of collision get "unacceptable" very fast. Would I be correct to assume that in this case there exists $2^{512}$ inputs, but the hash function is only able to supply $\approx 2^{256}$ (minus unnecessary collisions caused by imperfections in the hash function), so the there must exist $\approx 2^{256}$ collisions?

$\endgroup$
  • 2
    $\begingroup$ 2^256 / 2^512 is not 0.5, but rather 1/2^256. $\endgroup$ – SleuthEye Jun 1 '16 at 21:57
  • $\begingroup$ Just a question : assume that due to some quantum theory, using multiple dimensions and other Sci-fi stuff, you can store each of the pair of $2^{512}$ inputs to their respective output on 1 bit (i.e. 8 per byte). How many Terabytes of data would you need ? :) $\endgroup$ – Biv Jun 1 '16 at 22:03
  • $\begingroup$ @Biv hehe, right, but it is a theoretical question that works just as well on 4 bit output, 8 bit input, which would be easily storable :) $\endgroup$ – Daniel Valland Jun 1 '16 at 22:05
3
$\begingroup$

As pointed out by @kodlu, given an input sequence of 512 bits ($2^{512}$ possibilities), and given a hash function which spreads out the resulting hash approximately evenly, you would wind up with approximately $\frac{2^{512}}{2^{256}} = 2^{256}$ input sequences in your table which store the result of your exhaustive mapping for a given 256 bits hash (provided you could physically store such a large table).

Of course only one of those $2^{256}$ sequences would be the actual original input, and so the probability that you'd pick the correct original sequence from the table by choosing at random one of those $2^{256}$ sequences would be:

$$ P_{\mbox{original}} \approx \frac{1}{2^{256}} \approx 8\times 10^{-78} $$

Correspondingly, the probability that you'd pick a collision would be:

$$ \begin{align} P_{\mbox{collision}} &= 1 - P_{\mbox{original}}\\ &\approx 1 - \frac{1}{2^{256}} \\ &\approx 1 \end{align} $$

In other words it would be nearly impossible to obtain the original sequence and one would almost surely obtain a collision by randomly choosing a sequence from your large exhaustive mapping. This is provided you do not have additional information which would allow you to choose one of the sequence from your table in a better way than a simple random selection.

$\endgroup$
3
$\begingroup$

If you're tossing $n=2^{512}$ balls into $m=2^{256}$ bins, as a model of a good hash function, so $n=m^2$, the average load will be $n/m=m$ so the typical output will have $2^{256}-$fold collisions.

See paper here, last case of theorem 1, whereby the maximum load won't be too far away from average load in a multiplicative sense, in your case.

What about the output with least number of collisions? Probability a fixed bin is missed is $$ (1-(1/m))^n\approx \exp(-n/m)=\exp(-m) $$ so the expected number of missed bins is essentially zero, approximately $m \exp(-m)$.

A poisson approximation can show that with probability almost one the minimum load bin has at least a $$O(m/\log m)-fold ~~~~(1)$$collision. So, many collisions will occur even for the least popular output.

Edit: To put this in the context of @SleuthEye's answer, even if the actual input led to the least popular output your probability of finding the actual input would be something like 1 in $2^{248}$ due to the estimate in (1).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.