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>posted the private key before

When you post a transaction, the public key is in the transaction (inside the field "sigscript") . With the public key known you only need 2^(66/2) checks (instead of 2^66), which can be done really fast.

So some bot watched the address, obtained the public key, computed the private key from it, and front-ran the original submitter probably with a deal from a mining pool to make sure his transaction is enforced.



Edit: I see it's because I'm this instance there was less entropy, I guess a normal transaction has a lot more bits to guess

Why doesn't this happen with every large transaction then? Someone tries to move 10 BTC, instantly stolen?

Basically you're saying that every single Bitcoin transaction can be stolen "really fast".

Am I missing a step here?


It is based on the fact that the upper range limit of the private key used in the puzzle is known. A securely generated private key would not be vulnerable even if its public key is known.

The second post on this thread[0] has a helpful chart that makes it easier to understand.

0: https://bitcointalk.org/index.php?topic=5218972.0


Thanks


> When you post a transaction, the public key is in the transaction (inside the field "sigscript")

Is that true for every single Bitcoin transaction?

> With the public key known you only need 2^(66/2) checks (instead of 2^66), which can be done really fast.

Then how comes not all Bitcoin transactions are front-ran like that and Bitcoin is not worth zero already? 2^33 is indeed nothing: 8 billion (so I understand this can be easily cracked).


>Is that true for every single Bitcoin transaction

I think so, for outgoing transaction (aka to remove from the address), it's kind of needed to verify the signature.

The 2^66 is only for this game where only 66 bits were left unknown. In the general case obtaining the private key from the public key is much longer.


Ah gotcha, that's what I missed. Thanks for your explanation. For a regular address, even with the public key, if there are 256 unknown bits it'd be 2^128, which is statistically unlikely to be solvable.




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