Post
Topic
Board Development & Technical Discussion
Merits 5 from 4 users
Re: Proof-of-Approval: A Better Blockchain Consensus Protocol
by
monsterer2
on 20/03/2018, 10:01:37 UTC
⭐ Merited by DarkStar_ (2) ,Welsh (1) ,LeGaulois (1) ,TheBeardedBaby (1)
Stakes (or stake changes) in any block affect decision process in forks containing that block only, not sibling forks.

A - B0 - C0 - D0
  \ B1 - C1

Assuming B0/C0/D0 is the "honest" fork and B1/C1 is an adversarial attempt to a long-range revision attack, any stake changes in B1 and C1 do not affect decision process on the fork B0/C0 etc. Stakes for the decision process are calculated just before each block for that specific fork. More details are in Section 3.3.1 of the paper.

Hi Shunsai,

In the following example, say I had a majority of stake at block A. Then at B0, C0 all the way up to head block H0, I sell off my majority of stake, understanding that there is an upper limit on stake transfer.

Code:
A - B0 - C0 - D0 - H0
  \ B1 - C1 - D1 - H1

Then, I create a double spend of my stake using the historical private keys, which I still own, thus sending it to myself in a fork continuing fork B1 - C1 - D1 that I create and approve myself (majority stakeholder).

This is not a 50% attack, because I don't own the value held in the private keys anymore, the cost for doing this is nothing.

Why doesn't the network just accept this fork as canon, given that its the same length (or longer if I chose) than the genuine fork?
  
Quote
In another scenario, some stakeholders may only approve their own blocks in an attempt to tilt the winning chances in their own favor. This is a real possibility. A possible mechanism to deter such an "attack" is to simply "blacklist" such stakeholders temporarily. An honest stakeholder can tell within a few blocks with certainty if another stakeholder is not approving his/her valid blocks.

My question is, what does the stakeholder have to gain by approving anyone's blocks but his own? If he has an edge by not doing this, there is a very real problem.