1. Mallory, Alice and Bob found block on their farm.
  2. They disconnect from internet, simulating split and change all of nodes in their segment to malicious.
  3. Malicious nodes decrease complication of block founding, and dramatically outrace whole of world.
  4. After a week, they restore original clients on their nodes and plug the network back.
  5. Their chain bigger than world's and they gather total week's bounty 25,200 btc, more than 2,520,000 $.

Is it possible?

What part of code protect us? Raw code is appreciated.

Related question:

Is the Bitcoin protocol secure if two parts of the network have no connection?

Scenario: disaster splits the Internet into dozens unconnected fragments

What would happen if a portion of the Bitcoin network was separated from the rest of the network?

What's the impact of net-splits on block generation... over a long period of time?


2 Answers 2


No, this attack is not possible.

The chosen branch is the one representing the highest total difficulty, not the highest number of blocks.

Unless the attacker has more hashrate than the honest network combined, he will not be able to find a chain with higher total difficulty, even if in its chain the number of blocks is higher.


Is it possible?

Yes and no.

The behaviour you are describing a normal feature of the functioning network, that is, the most difficult to produce (often but not necessarily the longest) chain will always win.

Malicious nodes decrease complication of block founding, and dramatically outrace whole of world.

When the two networks rejoin, the second one having a lower difficulty, the new blocks would be instantly rejected. This is due to bitcoin looking for the most difficult chain to produce, not the ultimate highest block. A shorter chain with a higher cumulative difficulty is preferred over a longer, easier to product once.

Moreover, it would take a very large amount of computing power even just to ride the difficulty down. A single entity or group solving just a single block at difficulty 112M is unlikely, and they would need to solve up to 2015 of them to reach the next adjustment period. If they had enough hashpower to do that, they would be a lot better off financially just mining on the real network rather than attempting to attack it.

  • Even though checkpoints currently exist to protect to chain up to some hardcoded points, they only exist to enable a performance optimization (dropping signature verification before the last checkpoint). A new block synchronization mechanism that is being developed will perhaps allow dropping those checkpoints enitely. In any case, they don't exist as a security mechanism on themself - they're way too infrequently updated for that, and relying on them would be strong centralization. Commented Sep 22, 2013 at 18:06
  • Alright, I've removed that sentence from the answer. I mainly added that as an afterthought as the difficulty-not-height renders an attack like that impossible anyway.
    – Anonymous
    Commented Sep 22, 2013 at 22:12

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