Topic: Blockchain reorganization attack (Read 232 times)
Ok, thank you all.
The explanation listed here as reorganization attack does not require 51%? I don't understand.
https://en.bitcoin.it/wiki/Irreversible_Transactions#Blockchain_reorganization_attack
https://en.bitcoin.it/wiki/Irreversible_Transactions#Blockchain_reorganization_attack
Quote
Also called alternative history attack. This attack has a chance to work even if the merchant waits for some confirmations, but requires relatively high hashrate and risk of significant expense in wasted electricity to the attacking miner.
The reason for mentioning the 51% rate is that in order for this attack to succeed, you send your coins to a third party, withdraw your money, and then reverse the transaction history and for this scenario to succeed, you need at least 3 confirmations, meaning that you will mine 3 consecutive blocks faster than the rest of the network Something that will not happen unless it controls more than half of the network hashrate.
Explanation better ----> https://en.bitcoin.it/wiki/Majority_attack
The explanation listed here as reorganization attack does not require 51%? I don't understand.
https://en.bitcoin.it/wiki/Irreversible_Transactions#Blockchain_reorganization_attack
https://en.bitcoin.it/wiki/Irreversible_Transactions#Blockchain_reorganization_attack
In *theory* the same way in theory I can go out to the local 7-11 buy 3 lottery tickets and hit the jackpot on all of them.
ANYONE can get insanely lucky and find a block with not a lot of hashpower. IF you have ignored the previous block you can then broadcast your own in it's place. BUT all the other nodes have already SEEN the block you ignored so they will not accept yours. You then have to get insanely lucky a 2nd time to find a 2nd block that you would build on top of the other one you found and broadcast that. And do all of that BEFORE some other mining pool found a block on top of the proper chain.
You might be able to do it with less then the accepted amount of hashpower needed. You would be better off buying lottery tickets.
-Dave
The explanation listed here as reorganization attack does not require 51%? I don't understand.
https://en.bitcoin.it/wiki/Irreversible_Transactions#Blockchain_reorganization_attack
https://en.bitcoin.it/wiki/Irreversible_Transactions#Blockchain_reorganization_attack
Are a blockchain reorganization attack and a 51% attack the same thing?
Slightly different. The former refer to type of the attack, while the latter refer to how to perform an attack.
Is 51% of the hashing power necessary for a reorg attack?
No. But 51% of hashrate ensure success of your attack.
An attacker is not required to have 51% in order to force a reorg, but the probability of success drops very quickly with lower hash rate and more blocks. This is why it is recommended to wait for additional confirmations if there is a risk due to a reorg.
They could force a reorg if they own most of the networking infrastructure, since Bitcoin Core packets are unencrypted (unless you use Tor to connect to it), they will be able to read any packet they want and drop some of them, preventing the relaying to a large portion of nodes of mined blocks that the operator don't like until other blocks are mined.
Jam enough blocks, and you can cause a mempool congestion, possibly even dropping transactions if the total size is over 300MB.
Even Hetzner could (theoretically) do this as they own a large percentage of nodes. If it fails, it would cause a chain split and would be even more destructive than a few reorged blocks (as transactions inside would go back into the mempool).
I'm not even sure Hetzner could theoretically do that when,
1. Most pool have it's own infrastructure, especially for fast block propagation (e.g. FIBRE and Falcon).
2. Decent amount of node (which accept incoming connection) use Tor.
3. Only 6.1% (which accept incoming connection) Hetzner.
4. Bitcoin Core connect to 10 other nodes and try to diversify it based on IP block.
P.S. I use https://bitnodes.io/dashboard/ as primary reference.
Are a blockchain reorganization attack and a 51% attack the same thing? Is 51% of the hashing power necessary for a reorg attack?
Yes, those are essentially the same thing. You can do reorg attacks with less than 51% as well, such as happens with selfish mining. The exact percentage needed depends on various network assumptions and behaviour of other miners.There's a chance not all 51% attacks are reorg ones though as some may try to change other things
Those other things also require reorgs. Avoiding reorgs means always extending the longest (most worked) chain, which is the honest behaviour and thus cannot be called an attack, even if one were to have > 50% of hashpower.With 51%, an attacker can do these things:
1. Reorg previous blocks, removing (and optionally replacing) transactions that have already been confirmed.
2. Censor any or all transactions.
3. Claim all of the block rewards.
1. Reorg previous blocks, removing (and optionally replacing) transactions that have already been confirmed.
2. Censor any or all transactions.
3. Claim all of the block rewards.
2 requires re-orging other blocks that include the txs one wants to censor.
3 requires re-orging ALL other blocks.
An attacker is not required to have 51% in order to force a reorg, but the probability of success drops very quickly with lower hash rate and more blocks. This is why it is recommended to wait for additional confirmations if there is a risk due to a reorg.
They could force a reorg if they own most of the networking infrastructure, since Bitcoin Core packets are unencrypted (unless you use Tor to connect to it), they will be able to read any packet they want and drop some of them, preventing the relaying to a large portion of nodes of mined blocks that the operator don't like until other blocks are mined.
Jam enough blocks, and you can cause a mempool congestion, possibly even dropping transactions if the total size is over 300MB.
Even Hetzner could (theoretically) do this as they own a large percentage of nodes. If it fails, it would cause a chain split and would be even more destructive than a few reorged blocks (as transactions inside would go back into the mempool).
Are a blockchain reorganization attack and a 51% attack the same thing? Is 51% of the hashing power necessary for a reorg attack?
With 51%, an attacker can do these things:
1. Reorg previous blocks, removing (and optionally replacing) transactions that have already been confirmed.
2. Censor any or all transactions.
3. Claim all of the block rewards.
An attacker is not required to have 51% in order to force a reorg, but the probability of success drops very quickly with lower hash rate and more blocks. This is why it is recommended to wait for additional confirmations if there is a risk due to a reorg.
I'd say a "reorg attack" is a type of 51% attack (perhaps a big subset of attacks that could occur once someone controls more than 50% of hashing power).
There's a chance not all 51% attacks are reorg ones though as some may try to change other things (such as enforcing a malicious consensus on the protocol - for example to credit themselves a certain amount of btc). It might also be argued it becomes a denial of service attack once the reorg is broadcast while the hashing power majority remains with the attacker.
There's a chance not all 51% attacks are reorg ones though as some may try to change other things (such as enforcing a malicious consensus on the protocol - for example to credit themselves a certain amount of btc). It might also be argued it becomes a denial of service attack once the reorg is broadcast while the hashing power majority remains with the attacker.
Are a blockchain reorganization attack and a 51% attack the same thing? Is 51% of the hashing power necessary for a reorg attack?
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