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I'm preparing for a lecture on blockchain, where I'm going to briefly explain the proof-of-work mechanism and the hashcash challenge.

To illustrate the difficulty, I want to give the maximum amount of numbers one has to try in order to find a solution, given a specific difficulty target (at this time, 18,670,168,558,400).

I believe it's a simple formula, but I can't find it again. Can someone please refresh my memory?

Thanks!

R.

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There's no maximum. Instead you can find the formula linking the target and the required number of initial zeroes. Then the average number of hashes you need to calculate is 2^(target number of initial bits). Then you could input the current target to find the average number of hashes.

I don't know the formula for that but for example the block 661657, a recent one, had the target bits as many as about 76, which means it took about 2^76 hashes.

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  • Ah, yes, that's it! That gives me a nice illustration of the computing difficulty of the proof of work! Thanks! – mrgou Dec 16 '20 at 20:46

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