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Author Topic: Why do hashes begin with 0000000...?  (Read 1419 times)
TradeFortress
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October 16, 2012, 07:02:53 AM
 #1

For example, this page: http://blockchain.info/blocks/BTC%20Guild

Is there any reason for the hash being like that?
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October 16, 2012, 07:07:27 AM
 #2

For example, this page: http://blockchain.info/blocks/BTC%20Guild

Is there any reason for the hash being like that?

Yes, this is how the difficulty works. For a miner to successfully record the blocks, they have to have the transactions and block header all work out to have a hash less than a certain number. So the beginning will always have a bunch of zeros because the hash has to work out less than a certain target.

See the article on difficulty for more details.
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October 16, 2012, 07:10:03 AM
 #3

The SHA256 "solution" hash for a block requires x number of zeros at the start as the way of controlling the "difficulty" to find such a hash by continually trying to hash the block header with a different "nonce" until such a solution is found.

This is the "proof of work" that is fundamental to the Bitcoin block chain method to prevent double spends.

With CIYAM anyone can create 100% generated C++ web applications in literally minutes.

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October 16, 2012, 07:42:26 AM
 #4

You will know that we have reached the economic, technological and cosmological singularity once we've reached the 00000000000000000000000000000000000000000000000000000000000000000 hash.

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October 16, 2012, 08:37:15 AM
 #5

You will know that we have reached the economic, technological and cosmological singularity once we've reached the 00000000000000000000000000000000000000000000000000000000000000000 hash.

And when that happens, division by zero will became possible and universe will cease to exist.  Wink
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October 16, 2012, 08:52:44 AM
 #6

You will know that we have reached the economic, technological and cosmological singularity once we've reached the 00000000000000000000000000000000000000000000000000000000000000000 hash.

And when that happens, division by zero will became possible and universe will cease to exist.  Wink

To find "00000000000000000000000000000000000000000000000000000000000000000" hash is not harder than to find "458264a810936934867095f02578963498570298345702946587165c845698074" one.
Meni Rosenfeld
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October 16, 2012, 10:03:41 AM
 #7

You will know that we have reached the economic, technological and cosmological singularity once we've reached the 00000000000000000000000000000000000000000000000000000000000000000 hash.

And when that happens, division by zero will became possible and universe will cease to exist.  Wink

To find "00000000000000000000000000000000000000000000000000000000000000000" hash is not harder than to find "458264a810936934867095f02578963498570298345702946587165c845698074" one.
"00000000000000000000000000000000000000000000000000000000000000000" has a lower Kolmogorov complexity, making finding it a more impressive achievement than "458264a810936934867095f02578963498570298345702946587165c845698074".

Of course, all it would mean is that SHA-256 has been sufficiently broken.

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TangibleCryptography
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October 16, 2012, 11:15:04 AM
 #8

You will know that we have reached the economic, technological and cosmological singularity once we've reached the 00000000000000000000000000000000000000000000000000000000000000000 hash.

And when that happens, division by zero will became possible and universe will cease to exist.  Wink

To find "00000000000000000000000000000000000000000000000000000000000000000" hash is not harder than to find "458264a810936934867095f02578963498570298345702946587165c845698074" one.
"00000000000000000000000000000000000000000000000000000000000000000" has a lower Kolmogorov complexity, making finding it a more impressive achievement than "458264a810936934867095f02578963498570298345702946587165c845698074".

Of course, all it would mean is that SHA-256 has been sufficiently broken.

It could mean that however it could simply be pure luck.  A solution for a block (any block) today could be found that produces an all zero hash.
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October 16, 2012, 11:32:34 AM
 #9

It could mean that however it could simply be pure luck.  A solution for a block (any block) today could be found that produces an all zero hash.

Doubtful. Who will calc 2255 hashes in average to find such a solution?
TangibleCryptography
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October 16, 2012, 11:48:21 AM
 #10

It could mean that however it could simply be pure luck.  A solution for a block (any block) today could be found that produces an all zero hash.

Doubtful. Who will calc 2255 hashes in average to find such a solution?

Well that is the thing about probability.  While on AVERAGE (over an infinite number of universes) it will take 2^255 attempts to find that hash the same thing can be said of any individual hash.

The odds of finding this exact hash:
000000000000006472e36fb3dc1f77b846feabe001b2b86fc78fe8110cd6ce4f  (solution for the most recent block #203545)
is also one in 2^256 and if you were looking for this specific hash using random inputs it would take on average 2^255 attempts.


There isn't anything special or harder about an all zero hash.  Difficulty simply reduces the set of valid hashes relative to the set of all possible hashes and thus controls how many attempts on average will be necessary to find a valid one.
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October 16, 2012, 11:54:36 AM
 #11

I see. Btw, is there any proof that output of SHA256 is distributed uniformly? I suspect that some outputs r met more often than others.
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October 16, 2012, 12:02:35 PM
 #12

I see. Btw, is there any proof that output of SHA256 is distributed uniformly? I suspect that some outputs r met more often than others.

No hashing algorithm is perfectly distributed.  That goal isn't even realistic or attempted.  The theoretical "perfect" hashing function would be a Random Oracle.  All practical implementations fall short of this ideal.  SHA256 has been studied extensively to ensure its distribution is (for lack of a better term) "randomly distributed enough" to make using any imperfections in distribution an ineffective attack vector.


Here is a paper on forcing a collision attack (a method of exploiting non-random distribution) on a round reduced version of SHA256.
http://eprint.iacr.org/2008/131.pdf

By round reduced it means that this "attack" is only theoretical at best because it involves a shortened version of the SHA256 algorithm.  In theory similar attacks could eventually be found on longer versions and possibly someday even the full length version.
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October 16, 2012, 12:14:16 PM
 #13

There isn't anything special or harder about an all zero hash.
If you ask me, "will you bet your life on the fact that no SHA-256 hash will be found by 2020 with a value of 00000000000000000000000000000000000000000000000000000000000000000, without SHA-256 being broken?", I'll say "sure", because this has a probability of about 1 in 10^50, and will thus not happen.

If you ask me the same about b0867d52439eedac556187087c69bbe6fdb3c47f266780775931068f18fae223, I will refuse. Why? Because I don't know how you came up with b0867d52439eedac556187087c69bbe6fdb3c47f266780775931068f18fae223. Maybe you generated it truly randomly (and then the answer is the same). But maybe you came up with it by calculating some SHA-256 hash, and then I will lose even before the bet began.

What makes me so confident in the 00000000000000000000000000000000000000000000000000000000000000000 case, and not in the b0867d52439eedac556187087c69bbe6fdb3c47f266780775931068f18fae223 case, is that 0 has a much lower Kolmogorov complexity (and the describing program has no reference to anything remotely relevant to SHA-256).

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