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Author Topic: How exactly 51% attack would work?  (Read 1217 times)
twolifeinexile (OP)
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January 17, 2013, 10:46:07 PM
 #1

What happened exactly??
51%er double spend or spend bitcoins with a zero balance account then accept in their block, since their block is faster enough that their block chain grows faster then honest block calcs (which will reject fake spend) and eventually people have to accept?
What if people realize it and even they think they will not reverse the action, since that fake coin is still being able to be traced, and since the probability of reverse is not zero, people may accept the blockchain , but they do not have to accept fake coins as a payment (you pay me with fake coin, I reject to give you back services, since there is a possibility it is reversed.)

So not sure how 51% attack work.
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DannyHamilton
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January 17, 2013, 11:04:29 PM
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There are a couple of 51% attacks.  Here is one possible example.

Imagine the following:

Long blockchain exists.
The current completed block is block number 216942.

I create a transaction sending 5 BTC to you and publish it on the honest bitcoin network.

I start working on creating a block 216943a.  In my block I include a transaction that spends that same 5 BTC to me instead of you.

Someone on the network completes block 216943 and includes the transaction that sent the 5 BTC to you in that block.

You now have 1 confirmation and give me whatever I purchased from you.

I complete my own bock 216943a that has the transaction sending the 5 BTC to me instead of you.

Everyone else is now working on block 216944 which builds on top of 216943.
I am working on block 216944a that builds on top of my block 216943a.

Since I have more hashing power than everyone else combined, I get lucky and find my block first.

I publish my 2 blocks 216943a and 216944a.

The entire bitcoin system accepts my blocks as the real ones, block 216943 is abandoned (replaced with 216943a) and all miners start working on block 216945a.

The transaction sending the 5 BTC to you vanishes, and is replaced with the transaction sending 5 BTC to me instead.

Now I have whatever I purchased from you, AND I still have my 5 BTC.

I stop my 51% attack, and from now on, everybody continues to build on top of my 216943a giving me more and more confirmations.

twolifeinexile (OP)
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January 19, 2013, 03:56:43 PM
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There are a couple of 51% attacks.  Here is one possible example.

Imagine the following:

Long blockchain exists.
The current completed block is block number 216942.
......


Thank you very much for explaining this,

then later if by any chance, if people realize you are paying yourself in the main chain and pay me in the orphaned chain, though they may not abandon the main chain, each one at least can choose not receiving your bitcoins from the "tainted" wallet( by ways of treating payment from tainted wallet as invalid -- they are doing this may not because they want to punish you or help me, but may only due to trying avoid involving the dispute?

Is that possible?
DannyHamilton
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January 19, 2013, 04:17:13 PM
 #4

There are a couple of 51% attacks.  Here is one possible example.

Imagine the following:

Long blockchain exists.
The current completed block is block number 216942.
......


Thank you very much for explaining this,

then later if by any chance, if people realize you are paying yourself in the main chain and pay me in the orphaned chain, though they may not abandon the main chain, each one at least can choose not receiving your bitcoins from the "tainted" wallet( by ways of treating payment from tainted wallet as invalid -- they are doing this may not because they want to punish you or help me, but may only due to trying avoid involving the dispute?

Is that possible?
If I'm smart about my scamming, I won't re-use any address.  I'll use a new address for EVERY transaction.  Then you can't identify which addresses are mine, so you won't know which coins to avoid.  There is no way for anybody (except me and the original recipient) to know which transaction is the "right one".  Blocks get orphaned every day, it is a normal process required for bitcoin to work properly.  51% attack just takes advantage of this normal process to steal/scam.
quanium
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January 19, 2013, 09:03:29 PM
 #5

This is sort of freaky but I figure due to the nature and physics of computing, no single group or entity will ever achieve that much control of the network. (unless perhaps via quantum computing?)
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January 19, 2013, 09:21:14 PM
 #6

Actually it would take like 10 millions $ for a 51% attack...

DannyHamilton
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January 19, 2013, 11:07:49 PM
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Actually it would take like 10 millions $ for a 51% attack...
I suspect you number is a bit low, and that certain limitations would make it quite a bit more difficult than dropping $10 million on a table and saying "Set up a 51% miner for me."
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January 19, 2013, 11:10:30 PM
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Actually 10 millions $ should be more than enough not only to buy enough computers and gpu but to get the required buildings and ppl to operate the rigs. Not 10? Then 20, and we have more than enough to also pay some nice holidays  Wink

DannyHamilton
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January 20, 2013, 12:33:37 AM
 #9

It's an interesting proposition, but I'm not convinced.
mikerbiker6
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April 13, 2014, 07:30:23 PM
 #10

But having 51% of the total hashing power is not a guarantee that you discover a block faster. You are more likely to find the next block but it is not certain.
On the other hand, you could also outrun someone with only 1% by getting very lucky right?

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Deadly7
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April 14, 2014, 05:15:43 AM
 #11

Thanks for the explanation.
Mobius7
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April 14, 2014, 08:39:42 AM
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This is sort of freaky but I figure due to the nature and physics of computing, no single group or entity will ever achieve that much control of the network.

Once upon a time (In 2011), deepbit was close to having 50% of total hashrate.

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