6 ms·

As far as I can tell, you're confusing collisions with cycles. I don't think any of the main cryptographic hashes have problems with short cycles.

paulgb 17y ago
I could well be confusing the two, because I'm not sure if I even know what cycles are :-). I'm guessing from the name that cycles are when h(h(h(... h(a)))) = a? That's not quite what I'm getting at. I'll try explaining my concern a little better. Suppose you have a hash function h. Let p be the probability that h(x) == h(y) for any unique x and y. Now suppose you have some arbitrary inputs a and b. h(a) = h(b) with probability p h(h(a)) = h(h(b)) with probability 2p h(h(h(a))) = h(h(h(b))) with probability 3p etc. The probability at each step increases because a collision is possible at each step, but a collision in a previous step guarantees a collision in the current one. With a good hash function, p should be astronomically low, so I guess that's why it isn't a problem in this case? Or am I completely off-base in my assumptions?
- pmjordan 17y ago
  I see what you mean now. And your intuition on cycles was spot-on. And to answer your question (I'm not a crypto expert, so take with a pinch of salt): With SHA-1, that probability is generally accepted to be 2^(-160). 1000 is roughly 2^10, so the final probability is about 2^(-150). I guess that's still improbable enough... But I think you do make a very valid and interesting point.
  - tptacek 17y ago
    The loop has nothing to do with the probability of a collision. If your key comes out of /usr/share/dict/words, you have a 2^18 search space, not a 2^160 space. If you have a 2^18 search space, you better hope your constant factors are very high. Hence the loop.
    - paulgb 17y ago
      Yeah, I get why the loop is there. But I still think it increases the probability of a collision by nearly 1000x over the probability of a collision without the loop. Maybe a better way to explain my concern is to consider hashing something once versus twice, rather than 1000 times. Let h be a hash function and p be the probability P(h(x) = h(y)) for randomly chosen x and y. Say you have unique inputs a and b. Obviously, P(h(a) = h(b)) = p. Now consider P(h(h(a)) = h(h(b))). h(h(a)) = h(h(b)) if: h(a) = h(b) [probability p] OR h(a) != h(b) [probability 1-p], and h(h(a)) = h(h(b)) [probability p]. So the total probability when we repeat the hash is p + (1-p) * p Since p is small, the probability of a collision has nearly doubled just by applying the hash function again. [note: Higher up in this thread I mistakenly implied that the probability increased linearly. With a low number of iterations and realistic values for p it should be close to linear, but not exact.] [edit:formatting]
      - tptacek 17y ago
        H(m) -> SHA1(SHA1(m)) is not the same hash function as H(m) -> SHA1(m).
        paulgb 17y ago
        Right, but if SHA1(m) = SHA1(n), then SHA1(SHA1(m)) = SHA1(SHA1(n)) also. The converse is not true. There will be cases where SHA1(SHA1(m)) = SHA1(SHA1(n)), but SHA1(m) != SHA1(n). (Or does SHA1 somehow guarantee that this will not happen?) So it seems to follow that the chances that H(a) = H(b) when H(m) -> SHA1(SHA1(m)) must be higher than when H(m) -> SHA1(m). Am I missing something?
        tptacek 17y ago
        I think you're going to need to better explain the logic behind graf 2 before I can answer you. But on the off chance that we can end this thread gracefully, I'll point out that SHA256(SHA(256(m), m) is SHAd256(m), and considered by Ferguson to address security concerns in straight SHA256.

paulgb 17y ago

I could well be confusing the two, because I'm not sure if I even know what cycles are :-). I'm guessing from the name that cycles are when h(h(h(... h(a)))) = a? That's not quite what I'm getting at. I'll try explaining my concern a little better. Suppose you have a hash function h. Let p be the probability that h(x) == h(y) for any unique x and y. Now suppose you have some arbitrary inputs a and b. h(a) = h(b) with probability p h(h(a)) = h(h(b)) with probability 2p h(h(h(a))) = h(h(h(b))) with probability 3p etc. The probability at each step increases because a collision is possible at each step, but a collision in a previous step guarantees a collision in the current one. With a good hash function, p should be astronomically low, so I guess that's why it isn't a problem in this case? Or am I completely off-base in my assumptions?

pmjordan 17y ago
I see what you mean now. And your intuition on cycles was spot-on. And to answer your question (I'm not a crypto expert, so take with a pinch of salt): With SHA-1, that probability is generally accepted to be 2^(-160). 1000 is roughly 2^10, so the final probability is about 2^(-150). I guess that's still improbable enough... But I think you do make a very valid and interesting point.
- tptacek 17y ago
  The loop has nothing to do with the probability of a collision. If your key comes out of /usr/share/dict/words, you have a 2^18 search space, not a 2^160 space. If you have a 2^18 search space, you better hope your constant factors are very high. Hence the loop.
  - paulgb 17y ago
    Yeah, I get why the loop is there. But I still think it increases the probability of a collision by nearly 1000x over the probability of a collision without the loop. Maybe a better way to explain my concern is to consider hashing something once versus twice, rather than 1000 times. Let h be a hash function and p be the probability P(h(x) = h(y)) for randomly chosen x and y. Say you have unique inputs a and b. Obviously, P(h(a) = h(b)) = p. Now consider P(h(h(a)) = h(h(b))). h(h(a)) = h(h(b)) if: h(a) = h(b) [probability p] OR h(a) != h(b) [probability 1-p], and h(h(a)) = h(h(b)) [probability p]. So the total probability when we repeat the hash is p + (1-p) * p Since p is small, the probability of a collision has nearly doubled just by applying the hash function again. [note: Higher up in this thread I mistakenly implied that the probability increased linearly. With a low number of iterations and realistic values for p it should be close to linear, but not exact.] [edit:formatting]
    - tptacek 17y ago
      H(m) -> SHA1(SHA1(m)) is not the same hash function as H(m) -> SHA1(m).
      - paulgb 17y ago
        Right, but if SHA1(m) = SHA1(n), then SHA1(SHA1(m)) = SHA1(SHA1(n)) also. The converse is not true. There will be cases where SHA1(SHA1(m)) = SHA1(SHA1(n)), but SHA1(m) != SHA1(n). (Or does SHA1 somehow guarantee that this will not happen?) So it seems to follow that the chances that H(a) = H(b) when H(m) -> SHA1(SHA1(m)) must be higher than when H(m) -> SHA1(m). Am I missing something?
        tptacek 17y ago
        I think you're going to need to better explain the logic behind graf 2 before I can answer you. But on the off chance that we can end this thread gracefully, I'll point out that SHA256(SHA(256(m), m) is SHAd256(m), and considered by Ferguson to address security concerns in straight SHA256.