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So absolute truth on the blockchain is held by majority rule?

c0achmcguirk 10y ago
Yes, in a sense. The longest blockchain is the "truth." But with independently acting miners and mining pools there isn't a "majority" in the sense of a group of actors bullying the minority. The 51% problem describes this but it probably won't happen. The miners don't want to be a part of a pool that achieves over 50% hashing power because it would undermine bitcoin and sabotage their efforts. Currently the biggest mining pool is either AntPool or F2Pool. They both have about 15% of the total hashing power [1]. - [1] https://blockchain.info/pools https://blockchain.info/pools
Taek 10y ago
Not quite. Bitcoin has very specific rules that transactions must follow. For example, there is a very strict way that you can create new money. Existing money can only be spent by the owner, etc. So, if the longest chain violates one of the core rules, that chain is ignored. You only ever follow the longest chain that also follows all of the rules.
- runeks 10y ago
  It's not the longest chain (most blocks) that is preferred, it's the chain with the most cumulative work. Ie. the valid chain whose sum of block difficulties is the greatest.
bshanks 10y ago
Yes, if you imagine that the number of "votes" that each computer gets is proportional to its computational power ("hashing power" or hashpower). (if instead, each node got one vote, then it would be too easy for someone to create a million fake nodes to game the vote) Also as Taek points out, the protocol also contains some rules that constrain the content of the chain. No honest node will accept a chain violating these rules even if a majority of the hashpower supports such a chain. Note that this describes a "proof-of-work" chain. There are other forms of blockchains, such as "proof-of-stake", in which the number of "votes" is proportional to something other than computational power.

Yes, in a sense. The longest blockchain is the "truth." But with independently acting miners and mining pools there isn't a "majority" in the sense of a group of actors bullying the minority. The 51% problem describes this but it probably won't happen. The miners don't want to be a part of a pool that achieves over 50% hashing power because it would undermine bitcoin and sabotage their efforts. Currently the biggest mining pool is either AntPool or F2Pool. They both have about 15% of the total hashing power [1]. - [1] https://blockchain.info/pools https://blockchain.info/pools

Taek 10y ago

Not quite. Bitcoin has very specific rules that transactions must follow. For example, there is a very strict way that you can create new money. Existing money can only be spent by the owner, etc. So, if the longest chain violates one of the core rules, that chain is ignored. You only ever follow the longest chain that also follows all of the rules.

runeks 10y ago
It's not the longest chain (most blocks) that is preferred, it's the chain with the most cumulative work. Ie. the valid chain whose sum of block difficulties is the greatest.

bshanks 10y ago

Yes, if you imagine that the number of "votes" that each computer gets is proportional to its computational power ("hashing power" or hashpower). (if instead, each node got one vote, then it would be too easy for someone to create a million fake nodes to game the vote) Also as Taek points out, the protocol also contains some rules that constrain the content of the chain. No honest node will accept a chain violating these rules even if a majority of the hashpower supports such a chain. Note that this describes a "proof-of-work" chain. There are other forms of blockchains, such as "proof-of-stake", in which the number of "votes" is proportional to something other than computational power.