Appears I can cross Iota off my list of competing technologies.
Hard to tell what you mean without knowing what you have sacrificed - C, A or P.
For example to deal with the double-spend attack stdset has been discussing, to maintain consistency and access you must do something to "fence off" partitions (branches) and give up P.
The block chain creates one partition which is the longest chain. Thus it can maintain C and A. You will have to give up one of C, A, or P. The promise of DAG was to not give up any of those.
I think you can make DAG work but you will end up with divergent partions that can't be remerged (thus losing global consistency of value), same as what will happens to Bitcoin if it is forked (even by network partitioning).
This is a serious fundamental theorem that all of us designing consensus algorithms face.
I think you can make DAG work but you will end up with divergent partions that can't be remerged
Why? If there are no conflicting tx's, someone can issue a tx referencing 1 tx from partition 1 and 1 tx from partition 2, et voila.
Due the quoted Prisoner's Dilemma that I outlined (for which I believe your response was inadequate for the following reason) in that no one has an incentive to be first to lengthen the tips, the game theory is going to devolve to everyone agreeing to blacklist double-spends without actually abandoning branches containing conflicting transactions where they have transactions.The Prisoner's Dilemma is only solved in favor of lengthening if double-spends won't cause a branch to be illegitimate. Thus the branches will diverge while they lengthen. Your preferred algorithm will not hold over time. CAP's theorem is guidance, and now you just need to model it or put it into the wild and observe.
You might try to formulating some fencing or "longest-path" algorithm, but you are just going to end up back at a block chain (giving up Partition tolerance) once you have solved the Consistency and Access issues.
Any way if I am wrong, then kindly be the first to disprove the CAP theorem. Good luck with that.
Okay what did you give up: C, A, or P? Remember you said I am wrong, so surely you can answer this very quickly right
CfB and mthcl, the onus is on you to prove and demonstrate what you have given up from the CAP theorem. You have not. Period.
Contrary to my initial upthread enthusiasm, I am leaning towards this DAG concept can not work because it appears to attempt to defeat the CAP (Brewer's) Theorem. Before I was thinking the multiple branches are orthogonal, but it becomes clearer from the game theory issues quoted above, that there are complex dependencies. Analogous issues as the following appear to apply to DAG:
Hard to tell what you mean without knowing what you have sacrificed - C, A or P.
For example to deal with the double-spend attack stdset has been discussing, to maintain consistency and access you must do something to "fence off" partitions (branches) and give up P.
The block chain creates one partition which is the longest chain. Thus it can maintain C and A. You will have to give up one of C, A, or P. The promise of DAG was to not give up any of those.
I think you can make DAG work but you will end up with divergent partions that can't be remerged (thus losing global consistency of value), same as what will happens to Bitcoin if it is forked (even by network partitioning).
This is a serious fundamental theorem that all of us designing consensus algorithms face.
I think you can make DAG work but you will end up with divergent partions that can't be remerged
Due the quoted Prisoner's Dilemma that I outlined (for which I believe your response was inadequate for the following reason) in that no one has an incentive to be first to lengthen the tips, the game theory is going to devolve to everyone agreeing to blacklist double-spends without actually abandoning branches containing conflicting transactions where they have transactions.The Prisoner's Dilemma is only solved in favor of lengthening if double-spends won't cause a branch to be illegitimate. Thus the branches will diverge while they lengthen. Your preferred algorithm will not hold over time. CAP's theorem is guidance, and now you just need to model it or put it into the wild and observe.
You might try to formulating some fencing or "longest-path" algorithm, but you are just going to end up back at a block chain (giving up Partition tolerance) once you have solved the Consistency and Access issues.
Any way if I am wrong, then kindly be the first to disprove the CAP theorem. Good luck with that.
It's silly to build a distributed system violating CAP theorem, in no circumstances anyone of us would even attempt that, it's the same as trying to fly faster than speed of light.
Okay what did you give up: C, A, or P? Remember you said I am wrong, so surely you can answer this very quickly right
CfB and mthcl, the onus is on you to prove and demonstrate what you have given up from the CAP theorem. You have not. Period.
The block chain creates one partition which is the longest chain. Thus it can maintain C and A [but gives up on P]. You will have to give up one of C, A, or P.