<p>No, this is not a palpable concern.</p>
<p>Multisig is generally implemented by means of Pay to Script Hash (P2SH) addresses. Hereby, the receiver defines a script that they'll use to spend the funds, and forms an address from a hash of the script. Later, when spending, the original script is revealed and must be fulfilled. In the case of a 2-of-3 multisig, this will require two valid signatures corresponding to two of three explicitly stated public keys.</p>
<p>The only way a collision could be found would be if, hypothetically, a script that only requires one key happened to hash to the same address. It is generally considered infeasible to search for such pre-image collisions other than by brute force since cryptographic hash functions are one-way functions, though.</p>

CC BY-SA 4.0
2020-10-26T19:51:22.630Z
2020-10-26T19:51:22.630Z
Murch
<p>Let's say you found a collision for an address on the Bitcoin network. You have its private key.</p>
<p>Will you be able to transfer any money around or will the network prevent it or does it get verified?</p>
<hr />
<p>Here's another way of putting it: Can the address generated by 2-of-3 pv/pk be accessed by a 4th pv(a collision) or the network always looks for 2 of those 3 signing the transaction. Is that clear now?</p>
<p>Is multisig only a way for shareholders of an address, making sure enough of them are present during a transaction, or does the network actually prevents a transfer without all the necessary signatures?</p>

CC BY-SA 4.0
2020-10-26T13:32:06.473Z
Related: https://bitcoin.stackexchange.com/q/68901/5406, https://bitcoin.stackexchange.com/q/8804/5406
Murch
CC BY-SA 4.0
2020-10-26T13:33:41.630Z
Hi Mojtahedi, welcome to Bitcoin Stack Exchange. Could you clarify how your question relates to multisig? It is not clear to me whether you have sufficient keys to spend the funds or not.
Murch
CC BY-SA 4.0
2020-10-26T19:32:48.583Z
It seems to me that part of your message got lost when you wrote the comment. Please edit your question to clarify what you are asking! :)
Murch
CC BY-SA 4.0
2020-10-26T19:36:48.510Z
Sorry. Apparently, I can't edit it anymore. I deleted it. Those links are really helpful, but my question still stands. Here's another way of putting it: Can the address generated by 2-of-3 pv/pk be accessed by a 4th pv(a collision) or the network always looks for 2 of those 3 signing the transaction. Is that clear now?
Mojtahedi
CC BY-SA 4.0
2020-10-26T19:39:33.447Z
Is multisig **only** a way for shareholders of an address, making sure enough of them are present during a transaction, or does the network actually prevents a transfer without all the necessary signatures?
Mojtahedi
CC BY-SA 4.0
2020-10-26T19:41:43.887Z
Oh, I didn't mean that you should edit your comment, but rather edit your question (the original post at the top). The comments may be cleaned up later to remove visual clutter when the clarifications are put in the question—which is much clearer now.
Murch
<transactions><multi-signature><transaction-verification><money-transfer><key-collision>
Will the Bitcoin network discredit a transaction if you do a simple transfer on a multi-sig after you found the collision on that address?
Mojtahedi
# Will the Bitcoin network discredit a transaction if you do a simple transfer on a multi-sig after you found the collision on that address?

(Asked by: username_0 on 2020-10-26)

Let's say you found a collision for an address on the Bitcoin network. You have its private key.

Will you be able to transfer any money around or will the network prevent it or does it get verified?

---

Here's another way of putting it: Can the address generated by 2-of-3 pv/pk be accessed by a 4th pv(a collision) or the network always looks for 2 of those 3 signing the transaction. Is that clear now?

Is multisig only a way for shareholders of an address, making sure enough of them are present during a transaction, or does the network actually prevents a transfer without all the necessary signatures?

**Tags:** transactions, multi-signature, transaction-verification, money-transfer, key-collision

> **Comments:**
> **username_1**: Related: https://bitcoin.stackexchange.com/q/68901/5406, https://bitcoin.stackexchange.com/q/8804/5406  
> **username_1**: Hi username_0, welcome to Bitcoin Stack Exchange. Could you clarify how your question relates to multisig? It is not clear to me whether you have sufficient keys to spend the funds or not.  
> **username_1**: It seems to me that part of your message got lost when you wrote the comment. Please edit your question to clarify what you are asking! :)  
> **username_0**: Sorry. Apparently, I can't edit it anymore. I deleted it. Those links are really helpful, but my question still stands. Here's another way of putting it: Can the address generated by 2-of-3 pv/pk be accessed by a 4th pv(a collision) or the network always looks for 2 of those 3 signing the transaction. Is that clear now?  
> **username_0**: Is multisig **only** a way for shareholders of an address, making sure enough of them are present during a transaction, or does the network actually prevents a transfer without all the necessary signatures?  
> **username_1**: Oh, I didn't mean that you should edit your comment, but rather edit your question (the original post at the top). The comments may be cleaned up later to remove visual clutter when the clarifications are put in the question—which is much clearer now.  

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## 1 answers

### Answer by username_1 on 2020-10-26. Score: 1 | Accepted answer

No, this is not a palpable concern.

Multisig is generally implemented by means of Pay to Script Hash (P2SH) addresses. Hereby, the receiver defines a script that they'll use to spend the funds, and forms an address from a hash of the script. Later, when spending, the original script is revealed and must be fulfilled. In the case of a 2-of-3 multisig, this will require two valid signatures corresponding to two of three explicitly stated public keys.

The only way a collision could be found would be if, hypothetically, a script that only requires one key happened to hash to the same address. It is generally considered infeasible to search for such pre-image collisions other than by brute force since cryptographic hash functions are one-way functions, though.


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