<p>A 51% attack does not allow the attacker to spend coins they do not own; the usual rules of validity still apply for transactions and blocks. If those rules are broken, the block/transaction will be ignored by the rest of the network, no matter how much hashpower the miner making invalid blocks has. </p>

<p><a href="https://bitcoin.stackexchange.com/q/658">This question</a> has a more complete explanation of what a 51% attacker can, and can not do. </p>

CC BY-SA 4.0
2019-07-21T03:22:33.023Z
2019-07-21T03:22:33.023Z
chytrik
<p>In the paper by Nakamoto, miner can accept the mined block only if all transaction are valid. </p>

<p>Assume that there are 5 miners in the world. 4 of 5 are honest that have 10% of 100% hash power and 1 of 5 is malicious that has 90% of all hash power.</p>

<p>Let's say group A consists of honest nodes (4 of 5).
Let's say group B consists of malicious node (1 of 5).</p>

<p>Group B wants to create a block which has a fraud transaction (e.g. Alice wants to spend 1000 BTC but she actually does not have it).
Group B is successively creating and propagating some blocks with huge hash power, but Group A does not accept these blocks because transactions in blocks are invalid.
Yes, 51% attack is not accepted with only hash power. So, length of chain of group A is short but group B is long. But group A does not accept chain of group B.</p>

<p>So my conclusion is that 51% attack makes a mess in the network, not coin control.</p>

<p>I'm confused. Please help me understand where I am wrong.</p>

<blockchain>
How can a 51% attack be successful?
Hyunsoo
# How can a 51% attack be successful?

(Asked by: username_0 on 2019-07-21)

In the paper by Nakamoto, miner can accept the mined block only if all transaction are valid.

Assume that there are 5 miners in the world. 4 of 5 are honest that have 10% of 100% hash power and 1 of 5 is malicious that has 90% of all hash power.

Let's say group A consists of honest nodes (4 of 5).
Let's say group B consists of malicious node (1 of 5).

Group B wants to create a block which has a fraud transaction (e.g. Alice wants to spend 1000 BTC but she actually does not have it).
Group B is successively creating and propagating some blocks with huge hash power, but Group A does not accept these blocks because transactions in blocks are invalid.
Yes, 51% attack is not accepted with only hash power. So, length of chain of group A is short but group B is long. But group A does not accept chain of group B.

So my conclusion is that 51% attack makes a mess in the network, not coin control.

I'm confused. Please help me understand where I am wrong.

**Tags:** blockchain


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## 1 answers

### Answer by username_1 on 2019-07-21. Score: 2 | Accepted answer

A 51% attack does not allow the attacker to spend coins they do not own; the usual rules of validity still apply for transactions and blocks. If those rules are broken, the block/transaction will be ignored by the rest of the network, no matter how much hashpower the miner making invalid blocks has.

[This question](https://bitcoin.stackexchange.com/q/658) has a more complete explanation of what a 51% attacker can, and can not do.


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