2

In Bitcoin and Ethereum, the longest chain with the most proof-of-work (POW) invested in it is accepted as the true chain and used to determine balances and history on the network.

Because of this, if an adversary wants to change a block in a blockchain of height at h, he has to solve the POW for all blocks from h up to the current block. This is very difficult.

However, with IOTA, the entire Tangle is not broadcast to the network similar to the way that blocks are mined and broadcast with Bitcoin.

What takes the place of nodes honoring the longest chain?

How does IOTA ensure that the consensus is permanent with an immutable Tangle?

Improve this question
9
  • When you say "unique Blockchain" and "unique Tangle", do you actually mean "unanimous Blockchain" and "unanimous Tangle"?
    – Zauz
    Aug 9, 2018 at 15:05
  • @Austin Powers I mean for example in Bitcoin to achieve a consensus a miner who creates a new block, he broadcasts his new blockchain (as the longest one) to the network and every node after receiving this longest chain will replace it. Thanks
    – Questioner
    Aug 9, 2018 at 15:17
  • @Zauz I mean every node in the network keep the same blockchain at every time interval. In other word, the blockchain that every node maintains is the same. (All local blockchain are equal.) Thanks
    – Questioner
    Aug 9, 2018 at 15:20
  • What do you mean by "removable" i.e. what counts as a "removed" transaction for you?
    – Zauz
    Aug 9, 2018 at 18:59
  • @Zauz I mean one the main features of the blockchain is immutability, meaning that changing and modifying history of transactions is impossible (or very difficult) in case of Bitcoin the reason is that if an adversary wants to remove a transaction from a previous blocks, then he has to calculate again PoW puzzle for all blocks that are after this modified block that is very difficult. Now, consider in IOTA an adversary wants to remove a transaction from Tangle. Has this action "any cost" for the adversary? And in general, is it possible to delete one of previous transactions from Tangle?Thx
    – Questioner
    Aug 9, 2018 at 19:31

1 Answer 1

Reset to default
2

At the moment, we rely on Milestones issued by the Coordinator. In the future we might be able to rely on weight, which is the DAG equivalent of chain length.

I don't know if Iota will need a Coordinator forever. If the honest transaction rate reaches a certain point, I think we could get rid of it entirely, but with current PoW (Proof of Work) and difficulty that point is just too high. Maybe Network bound PoW will change something about this but we know hardly anything about it.

As for immutability, Iota uses the same mechanism as Blockchains:

enter image description here
enter image description here

The block with hash value cof points at the previous block eef. If someone changes the content of eef, the hash value of the block changes as well. It will no longer be eef, in fact eef will no longer exist.

tangle tangle 404

The transactions with hash value PMF and VQT point at the transaction JOE. If someone tries to change the content of JOE, the hash value of the transaction changes as well. It will no longer be JOE, in fact JOE will no longer exist.

Improve this answer
3
  • Thank you, Yes, but as I mentioned in my question, since IOTA PoW takes only milliseconds, so adversary can calculates the PoW for all vertices (tx) again, such that there will be a new Tangle that nobody is capable of distinguishing correct Tangle from malicious Tangle.
    – Questioner
    Aug 13, 2018 at 15:09
  • Once the honest transaction rate gets high enough, Iota could theoretically (!!!) rely on weight only. With the current PoW system, the number of transaction needed for security without COO would be higher than the amount of transactions the network can handle at all. Migrating to network bound PoW (whatever that is) might help. Also, increasing the MWM would work but then it would be more "expensive" to make transactions which we also don't want.
    – Zauz
    Aug 13, 2018 at 15:22
  • As many people already pointed out, your number is arbitrary. But yes, the number would probably also be incredibly high. But not 1.2 billion! Maybe higher, maybe lower, probably higher if ternary hardware works as predicted, but not 1.2 billion.
    – Zauz
    Aug 13, 2018 at 18:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.