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?




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.

  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.