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Author Topic: Bitcoin address protection  (Read 1105 times)
satoshy
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July 10, 2012, 08:00:05 PM
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Hi guys. I'm a bitcoin newbie and have some questions about bitcoin security:

How the Bitcoin network make sure a new generated address isn't already in use?

Can't someone just got access to other addresses by generating a new address check if the address was generated and used before, if yes check for the amount that is currently associated to that address and make a transaction with that address? All these steps in a loop of course. May takes some time but sooner or later the attacker should get an address with some bitcoins associated to. Is it really that simple? Or did I miss something?

How can we secure our addresses from attacks like these?
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DeathAndTaxes
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July 10, 2012, 08:20:29 PM
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Simple 2^256 is big.  

If you took over the world and harness all computing power to do nothing but brute force the bitcoin private key space you wouldn't even be able to check a tiny fraction of 1% of the possible private keys in the next billion or so years.  In your lifetime?  With a mere single planet at your command?  Forget about it.

It is in theory possible to generate an existing address by chance however the odds are so small that some have compared it to getting hit by an asteroid or winning the lottery.   Those are actually off by many orders of magnitude.  2^256 is so large that if you built a theoretically perfect planetary sized super computer which consumed the entire energy output of our star and operated at the thermodynamic limit (impossible to more efficient) you wouldn't have enough energy to count to 2^256 (i.e. 0,1,2,3, ... 2^256) before our star burned out.  Note: that isn't generating keypairs, hashing or looking up values.  It is just counting.

So if you want to think about it in terms of asteroids and lotteries. It would be like winning all the lotteries on the planet AND dying from an asteroid strike at the same time. Smiley
satoshy
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July 10, 2012, 08:49:11 PM
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Thanks for your quick reply DeathAndTaxes:
So then it seems to be secure enough. Smiley
Thanks for your explanation.
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