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Proof of work - 51% attack
Proof of work scheme with one and only one correct answerProof of work for standard computersSimple proof of work example?Is there an “additive” proof-of-work?Stateless proof-of-work system with 0-roundtrip timeCan “Approximate Hashing” work in a Proof-of-Work context?Prove that you have $K$ bytes of memoryIs there a proof-of-work scheme vulnerable to length extension attack?Are there Guaranteed Proof-of-work Systems?What Proof of Work does IOTA use?
$begingroup$
In the 51% attack on the Proof Of Work Consensus Mechanism, it's often said that you need more hashing power than the rest of the network combined. Though I can't understand why you need that much.
If we break the blockchain network down in Alice, Bob, Charlie and Eve, and they all currently are honest. They all have 1/4 of the total hashing power. From what I understand, they all compete against eachother to calculate the right hash of a block, and the one winning the race gets its block on the blockchain. This happens one time each 10 minutes. Since they're all competing, the hash-power used to create a block in 10 minutes, is 1/4 of the total hashing power.
Now eve decides to go rogue, and builds her own blockchain. The difference is now, that she don't compete with anyone. That means that she can generate a block every 10 minutes (ish) on her private blockchain, keeping up to speed with the rest of the network. Having just a bit more power, she can then build blocks faster than Alice, Bob and Charlie combined, since they are "competing". At some point Eve's blockchain grows larger than the others, and she broadcasts it, and Alice, Bob and Charlie will per the rules switch to Eve's blockchain.
In this scenario the network was compromised by only using 25%(+1) of the hashing power. So why would Eve need 51%? Doesn't she just need to be faster than the fastest honest miner?
Hope you can tell me why I'm wrong.
proof-of-work
edited Apr 19 at 21:24
Wossname
1033
asked Apr 19 at 10:27
Benjamin LarsenBenjamin Larsen
232
$endgroup$
add a comment |
$begingroup$
In the 51% attack on the Proof Of Work Consensus Mechanism, it's often said that you need more hashing power than the rest of the network combined. Though I can't understand why you need that much.
If we break the blockchain network down in Alice, Bob, Charlie and Eve, and they all currently are honest. They all have 1/4 of the total hashing power. From what I understand, they all compete against eachother to calculate the right hash of a block, and the one winning the race gets its block on the blockchain. This happens one time each 10 minutes. Since they're all competing, the hash-power used to create a block in 10 minutes, is 1/4 of the total hashing power.
Now eve decides to go rogue, and builds her own blockchain. The difference is now, that she don't compete with anyone. That means that she can generate a block every 10 minutes (ish) on her private blockchain, keeping up to speed with the rest of the network. Having just a bit more power, she can then build blocks faster than Alice, Bob and Charlie combined, since they are "competing". At some point Eve's blockchain grows larger than the others, and she broadcasts it, and Alice, Bob and Charlie will per the rules switch to Eve's blockchain.
In this scenario the network was compromised by only using 25%(+1) of the hashing power. So why would Eve need 51%? Doesn't she just need to be faster than the fastest honest miner?
Hope you can tell me why I'm wrong.
proof-of-work
edited Apr 19 at 21:24
Wossname
1033
asked Apr 19 at 10:27
Benjamin LarsenBenjamin Larsen
232
$endgroup$
add a comment |
$begingroup$
In the 51% attack on the Proof Of Work Consensus Mechanism, it's often said that you need more hashing power than the rest of the network combined. Though I can't understand why you need that much.
If we break the blockchain network down in Alice, Bob, Charlie and Eve, and they all currently are honest. They all have 1/4 of the total hashing power. From what I understand, they all compete against eachother to calculate the right hash of a block, and the one winning the race gets its block on the blockchain. This happens one time each 10 minutes. Since they're all competing, the hash-power used to create a block in 10 minutes, is 1/4 of the total hashing power.
Now eve decides to go rogue, and builds her own blockchain. The difference is now, that she don't compete with anyone. That means that she can generate a block every 10 minutes (ish) on her private blockchain, keeping up to speed with the rest of the network. Having just a bit more power, she can then build blocks faster than Alice, Bob and Charlie combined, since they are "competing". At some point Eve's blockchain grows larger than the others, and she broadcasts it, and Alice, Bob and Charlie will per the rules switch to Eve's blockchain.
In this scenario the network was compromised by only using 25%(+1) of the hashing power. So why would Eve need 51%? Doesn't she just need to be faster than the fastest honest miner?
Hope you can tell me why I'm wrong.
proof-of-work
edited Apr 19 at 21:24
Wossname
1033
asked Apr 19 at 10:27
Benjamin LarsenBenjamin Larsen
232
$endgroup$
In the 51% attack on the Proof Of Work Consensus Mechanism, it's often said that you need more hashing power than the rest of the network combined. Though I can't understand why you need that much.
If we break the blockchain network down in Alice, Bob, Charlie and Eve, and they all currently are honest. They all have 1/4 of the total hashing power. From what I understand, they all compete against eachother to calculate the right hash of a block, and the one winning the race gets its block on the blockchain. This happens one time each 10 minutes. Since they're all competing, the hash-power used to create a block in 10 minutes, is 1/4 of the total hashing power.
Now eve decides to go rogue, and builds her own blockchain. The difference is now, that she don't compete with anyone. That means that she can generate a block every 10 minutes (ish) on her private blockchain, keeping up to speed with the rest of the network. Having just a bit more power, she can then build blocks faster than Alice, Bob and Charlie combined, since they are "competing". At some point Eve's blockchain grows larger than the others, and she broadcasts it, and Alice, Bob and Charlie will per the rules switch to Eve's blockchain.
In this scenario the network was compromised by only using 25%(+1) of the hashing power. So why would Eve need 51%? Doesn't she just need to be faster than the fastest honest miner?
Hope you can tell me why I'm wrong.
proof-of-work
proof-of-work
edited Apr 19 at 21:24
Wossname
1033
asked Apr 19 at 10:27
Benjamin LarsenBenjamin Larsen
232
edited Apr 19 at 21:24
Wossname
1033
asked Apr 19 at 10:27
Benjamin LarsenBenjamin Larsen
232
edited Apr 19 at 21:24
Wossname
1033
edited Apr 19 at 21:24
Wossname
1033
edited Apr 19 at 21:24
Wossname
1033
1033
asked Apr 19 at 10:27
Benjamin LarsenBenjamin Larsen
232
asked Apr 19 at 10:27
Benjamin LarsenBenjamin Larsen
232
asked Apr 19 at 10:27
Benjamin LarsenBenjamin Larsen
232
232
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Since they're all competing, the hash-power used to create a block in 10 minutes, is 1/4 of the total hashing power.
That's not correct.
If the work effort is set to $n$, that is, it takes an expected $2^n$ hashes before someone finds a valid preimage. If Alice, Bob, Charlie and Eve are all working equally hard, then we would expect that each one would end up computing an average of $2^n / 4$ hashes before someone finds a preimage. It might be Alice, who has computed only $2^n / 4$ hashes, but still a total of $2^n$ hashes were computed.
So, if Eve goes off on her own, she'll need to compute her own $2^n$ hashes herself before she finds a preimage; in that time, Alice, Bob and Charlie will end up computing $3 cdot 2^n$ hashes, enough to find an expected 3 links in the hash chain, before Eve can compute one - Eve cannot win this race in the long term.
(Note: this analysis rather assumes that when Alice finds the preimage, she instantly announces it, and the others immediately start searching based for the next preimage. This isn't precisely true; however that isn't nearly a large enough effect to give Eve an advantage when she has only 1/4 of the computing power)
answered Apr 19 at 11:39
ponchoponcho
94.8k2151248
$endgroup$
$begingroup$
That is a really good explanation, thank you :)
$endgroup$
– Benjamin Larsen
Apr 19 at 11:48
$begingroup$
But what if Eve decides to go rogue not necessarily now, but simlpe the next time she is able to find a valid preimage unusually fast (say, within $frac11002^n$ hashes)? This is expected to happen within about 100 rounds, so may be worth waiting for
$endgroup$
– Hagen von Eitzen
Apr 19 at 14:58
$begingroup$
@HagenvonEitzen: however, that doesn't buy her much in the long term - she would get an early start initially, and might be able to find another hash quicker than Alice et al finds two, however in the long term, the greater resources of the honest miners will outrun Eve
$endgroup$
– poncho
Apr 19 at 16:13
add a comment |
Your Answer
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$begingroup$
Since they're all competing, the hash-power used to create a block in 10 minutes, is 1/4 of the total hashing power.
That's not correct.
If the work effort is set to $n$, that is, it takes an expected $2^n$ hashes before someone finds a valid preimage. If Alice, Bob, Charlie and Eve are all working equally hard, then we would expect that each one would end up computing an average of $2^n / 4$ hashes before someone finds a preimage. It might be Alice, who has computed only $2^n / 4$ hashes, but still a total of $2^n$ hashes were computed.
So, if Eve goes off on her own, she'll need to compute her own $2^n$ hashes herself before she finds a preimage; in that time, Alice, Bob and Charlie will end up computing $3 cdot 2^n$ hashes, enough to find an expected 3 links in the hash chain, before Eve can compute one - Eve cannot win this race in the long term.
(Note: this analysis rather assumes that when Alice finds the preimage, she instantly announces it, and the others immediately start searching based for the next preimage. This isn't precisely true; however that isn't nearly a large enough effect to give Eve an advantage when she has only 1/4 of the computing power)
answered Apr 19 at 11:39
ponchoponcho
94.8k2151248
$endgroup$
$begingroup$
That is a really good explanation, thank you :)
$endgroup$
– Benjamin Larsen
Apr 19 at 11:48
$begingroup$
But what if Eve decides to go rogue not necessarily now, but simlpe the next time she is able to find a valid preimage unusually fast (say, within $frac11002^n$ hashes)? This is expected to happen within about 100 rounds, so may be worth waiting for
$endgroup$
– Hagen von Eitzen
Apr 19 at 14:58
$begingroup$
@HagenvonEitzen: however, that doesn't buy her much in the long term - she would get an early start initially, and might be able to find another hash quicker than Alice et al finds two, however in the long term, the greater resources of the honest miners will outrun Eve
$endgroup$
– poncho
Apr 19 at 16:13
add a comment |
$begingroup$
Since they're all competing, the hash-power used to create a block in 10 minutes, is 1/4 of the total hashing power.
That's not correct.
If the work effort is set to $n$, that is, it takes an expected $2^n$ hashes before someone finds a valid preimage. If Alice, Bob, Charlie and Eve are all working equally hard, then we would expect that each one would end up computing an average of $2^n / 4$ hashes before someone finds a preimage. It might be Alice, who has computed only $2^n / 4$ hashes, but still a total of $2^n$ hashes were computed.
So, if Eve goes off on her own, she'll need to compute her own $2^n$ hashes herself before she finds a preimage; in that time, Alice, Bob and Charlie will end up computing $3 cdot 2^n$ hashes, enough to find an expected 3 links in the hash chain, before Eve can compute one - Eve cannot win this race in the long term.
(Note: this analysis rather assumes that when Alice finds the preimage, she instantly announces it, and the others immediately start searching based for the next preimage. This isn't precisely true; however that isn't nearly a large enough effect to give Eve an advantage when she has only 1/4 of the computing power)
answered Apr 19 at 11:39
ponchoponcho
94.8k2151248
$endgroup$
$begingroup$
That is a really good explanation, thank you :)
$endgroup$
– Benjamin Larsen
Apr 19 at 11:48
$begingroup$
But what if Eve decides to go rogue not necessarily now, but simlpe the next time she is able to find a valid preimage unusually fast (say, within $frac11002^n$ hashes)? This is expected to happen within about 100 rounds, so may be worth waiting for
$endgroup$
– Hagen von Eitzen
Apr 19 at 14:58
$begingroup$
@HagenvonEitzen: however, that doesn't buy her much in the long term - she would get an early start initially, and might be able to find another hash quicker than Alice et al finds two, however in the long term, the greater resources of the honest miners will outrun Eve
$endgroup$
– poncho
Apr 19 at 16:13
add a comment |
$begingroup$
Since they're all competing, the hash-power used to create a block in 10 minutes, is 1/4 of the total hashing power.
That's not correct.
If the work effort is set to $n$, that is, it takes an expected $2^n$ hashes before someone finds a valid preimage. If Alice, Bob, Charlie and Eve are all working equally hard, then we would expect that each one would end up computing an average of $2^n / 4$ hashes before someone finds a preimage. It might be Alice, who has computed only $2^n / 4$ hashes, but still a total of $2^n$ hashes were computed.
So, if Eve goes off on her own, she'll need to compute her own $2^n$ hashes herself before she finds a preimage; in that time, Alice, Bob and Charlie will end up computing $3 cdot 2^n$ hashes, enough to find an expected 3 links in the hash chain, before Eve can compute one - Eve cannot win this race in the long term.
(Note: this analysis rather assumes that when Alice finds the preimage, she instantly announces it, and the others immediately start searching based for the next preimage. This isn't precisely true; however that isn't nearly a large enough effect to give Eve an advantage when she has only 1/4 of the computing power)
answered Apr 19 at 11:39
ponchoponcho
94.8k2151248
$endgroup$
Since they're all competing, the hash-power used to create a block in 10 minutes, is 1/4 of the total hashing power.
That's not correct.
If the work effort is set to $n$, that is, it takes an expected $2^n$ hashes before someone finds a valid preimage. If Alice, Bob, Charlie and Eve are all working equally hard, then we would expect that each one would end up computing an average of $2^n / 4$ hashes before someone finds a preimage. It might be Alice, who has computed only $2^n / 4$ hashes, but still a total of $2^n$ hashes were computed.
So, if Eve goes off on her own, she'll need to compute her own $2^n$ hashes herself before she finds a preimage; in that time, Alice, Bob and Charlie will end up computing $3 cdot 2^n$ hashes, enough to find an expected 3 links in the hash chain, before Eve can compute one - Eve cannot win this race in the long term.
(Note: this analysis rather assumes that when Alice finds the preimage, she instantly announces it, and the others immediately start searching based for the next preimage. This isn't precisely true; however that isn't nearly a large enough effect to give Eve an advantage when she has only 1/4 of the computing power)
answered Apr 19 at 11:39
ponchoponcho
94.8k2151248
answered Apr 19 at 11:39
ponchoponcho
94.8k2151248
answered Apr 19 at 11:39
ponchoponcho
94.8k2151248
answered Apr 19 at 11:39
ponchoponcho
94.8k2151248
94.8k2151248
$begingroup$
That is a really good explanation, thank you :)
$endgroup$
– Benjamin Larsen
Apr 19 at 11:48
$begingroup$
But what if Eve decides to go rogue not necessarily now, but simlpe the next time she is able to find a valid preimage unusually fast (say, within $frac11002^n$ hashes)? This is expected to happen within about 100 rounds, so may be worth waiting for
$endgroup$
– Hagen von Eitzen
Apr 19 at 14:58
$begingroup$
@HagenvonEitzen: however, that doesn't buy her much in the long term - she would get an early start initially, and might be able to find another hash quicker than Alice et al finds two, however in the long term, the greater resources of the honest miners will outrun Eve
$endgroup$
– poncho
Apr 19 at 16:13
add a comment |
$begingroup$
That is a really good explanation, thank you :)
$endgroup$
– Benjamin Larsen
Apr 19 at 11:48
$begingroup$
But what if Eve decides to go rogue not necessarily now, but simlpe the next time she is able to find a valid preimage unusually fast (say, within $frac11002^n$ hashes)? This is expected to happen within about 100 rounds, so may be worth waiting for
$endgroup$
– Hagen von Eitzen
Apr 19 at 14:58
$begingroup$
@HagenvonEitzen: however, that doesn't buy her much in the long term - she would get an early start initially, and might be able to find another hash quicker than Alice et al finds two, however in the long term, the greater resources of the honest miners will outrun Eve
$endgroup$
– poncho
Apr 19 at 16:13
$begingroup$
That is a really good explanation, thank you :)
$endgroup$
– Benjamin Larsen
Apr 19 at 11:48
$begingroup$
That is a really good explanation, thank you :)
$endgroup$
– Benjamin Larsen
Apr 19 at 11:48
$begingroup$
But what if Eve decides to go rogue not necessarily now, but simlpe the next time she is able to find a valid preimage unusually fast (say, within $frac11002^n$ hashes)? This is expected to happen within about 100 rounds, so may be worth waiting for
$endgroup$
– Hagen von Eitzen
Apr 19 at 14:58
$begingroup$
But what if Eve decides to go rogue not necessarily now, but simlpe the next time she is able to find a valid preimage unusually fast (say, within $frac11002^n$ hashes)? This is expected to happen within about 100 rounds, so may be worth waiting for
$endgroup$
– Hagen von Eitzen
Apr 19 at 14:58
$begingroup$
@HagenvonEitzen: however, that doesn't buy her much in the long term - she would get an early start initially, and might be able to find another hash quicker than Alice et al finds two, however in the long term, the greater resources of the honest miners will outrun Eve
$endgroup$
– poncho
Apr 19 at 16:13
$begingroup$
@HagenvonEitzen: however, that doesn't buy her much in the long term - she would get an early start initially, and might be able to find another hash quicker than Alice et al finds two, however in the long term, the greater resources of the honest miners will outrun Eve
$endgroup$
– poncho
Apr 19 at 16:13
add a comment |
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