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Author Topic: Running out of Difficulty?  (Read 1752 times)
ColdHardMetal
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May 21, 2011, 05:51:11 AM
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I had an odd thought the other day that I don't think has been discussed before. This might also be stemming from my imperfect knowledge of the crypto/math/algorithm behind the block generation process, so if it sounds retarded please correct as needed.

My understanding is that in order to increase the difficulty of block generation, the number of leading zeroes in the hashes we are looking for is increased.

For example 000jdhfg is an easier problem to solve than 00000hfs (for some reason).

So my question I guess is that if hashes are of a fixed length (I don't know if they are, or what that length might be), won't there come a time where the hash is all leading zeroes and it cant be made any more difficult?

What happens next?

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FreeMoney
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May 21, 2011, 05:53:48 AM
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I don't know the length of the number, but I know it's not a problem.

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wumpus
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May 21, 2011, 05:54:58 AM
 #3

When the difficulty is that high solving SHA256 will have to be so easy and quick it's probable that SHA256 has been poked full of holes, and it's time to switch to another hashing algorithm.

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Ian Maxwell
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May 21, 2011, 05:58:26 AM
 #4

Yes, there is a maximum possible difficulty as things stand now. It's exactly 26,959,535,291,011,309,493,156,476,344,723,991,336,010,898,738,574,164,086,137,773,096,960.

I would like to live in the world where this is a problem.

Edited to add: Raising the difficulty doesn't mean specifically that the target has more leading zeros, only that the target is a smaller number. If the difficulty increases by something small like 5%, no more leading zeros will probably be required.

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ColdHardMetal
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May 21, 2011, 06:04:52 AM
 #5

So technically "Yes", but practically "No". Got it.

Thanks.

theymos
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May 21, 2011, 06:13:25 AM
 #6

Block hash collisions would be a more serious problem than difficulty limitations if we ever got that far, since only one block can ever be produced at the max difficulty: all other blocks would be seen as duplicates of that block.

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May 21, 2011, 06:18:31 AM
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Block hash collisions would be a more serious problem than difficulty limitations if we ever got that far, since only one block can ever be produced at the max difficulty: all other blocks would be seen as duplicates of that block.
When that happens we have found the Armageddon block!

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May 21, 2011, 06:21:46 AM
 #8

Mine with the large hadron collider to find the god block ?
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May 21, 2011, 07:19:06 AM
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Mine with the large hadron collider to find the god block ?
I need an improbability drive in my 5970

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May 21, 2011, 09:00:34 AM
 #10

If we run out of bits in SHA256 (and I'm not sure if we will, even if we used every particle in the universe for computation), we can just switch to SHA512, or whatever.

Bitcoin has been VERY well considered.  It will take a fundamental change in our knowledge of either physics, or information theory, and probably both, to break it entirely.

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May 21, 2011, 12:31:00 PM
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If we run out of bits in SHA256 (and I'm not sure if we will, even if we used every particle in the universe for computation), we can just switch to SHA512, or whatever.

Bitcoin has been VERY well considered.  It will take a fundamental change in our knowledge of either physics, or information theory, and probably both, to break it entirely.
more likely was switching hash TYPE than size adjust, cuz all SHA family considered unsafe enough to start contest[for replacement] by NSA and their EU colleagues.
"knowledge itself is power' (c)
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May 21, 2011, 01:35:38 PM
 #12

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Block hash collisions would be a more serious problem than difficulty limitations if we ever got that far, since only one block can ever be produced at the max difficulty: all other blocks would be seen as duplicates of that block.

This is a very interesting relationship, because there is a finite number of blocks that can be uniquely hashed at every difficulty. Has anyone done the calculations as to what the theoretical number of unique block are vs the computing power of the network?  Could this be a serious limitation of bitcoins?  If bitcoins were ever accepted widely, you could assume the hash rate of the network would increase much more, thereby increasing difficulty and lowering number of unique block hashes available.

I am assuming non unique hashes are not acceptable, but maybe i am wrong on that?  In the case they are acceptable, this could potentially be exploited for forgery, in the case they aren't acceptable, this presents a "hidden" increase in difficulty thats currently not taken into account in statistics.  I dont know the math, but perhaps its so minuscule as to not matter.
theymos
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May 22, 2011, 02:12:05 AM
 #13

Block collisions would break the system, since people would have different ideas about which blocks are which.

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May 22, 2011, 03:13:38 AM
 #14

Collisions don't have to break the system.  If two blocks had the same hash, it would be trivial to figure out from context when one was which.  Even easier would be to just avoid the issue by making sure a new hash is unique.

2^256 is an absurdly huge number.  It is in the ballpark of the number of particles in the universe.  I don't think we'll ever run out of blocks or hashes.  And if we start getting close, it would be trivial to jump to SHA512 (or whatever).

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Basiley
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May 22, 2011, 03:26:35 AM
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next ? [inimaginably today]stronger hash in use, probably different type of them[not SHA family].
bittrader
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May 22, 2011, 03:52:33 AM
 #16

Yes, there is a maximum possible difficulty as things stand now. It's exactly
26,959,535,291,011,309,493,156,476,344,723,991,336,010,898,738,574,164,086,137,773,096,960.

Hey, Ian, how'd you get this number? I can see why the max difficulty would be 2^224. But 2^224 is 26,959,946,667,150,639,794,667,015,087,019,630,673,637,144,422,540,572,481,103,610,249,216 according to gmp....which is pretty close to your number, but not exactly equal. I'm just curious about where the difference between our numbers comes from.
theymos
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May 22, 2011, 04:09:34 AM
 #17

The minimum target is slightly off from 2224.

A block with a hash equal to the target is also valid, so the real max difficulty would divide by zero.

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