Hello,

I have now experienced a few times that smaller coins (I've mostly looked at Cryptonote coins in that regard) suffer from sudden jumps and drops in the hashrate, particularly when the hashrate jumps to a 50x+ value for a while - taking the difficulty with it very soon - and then drops again - leaving the difficulty far too high for a prolonged amount of time (which, for me, smells like someone taking advantage of small coins by suddenly launching a huge amount of hashing power to it, making some quick blocks, then when the difficulty adjusts, they go away and to the same to another coin, leaving their victim coin users with multi-hour block times for a long time).

Also, this of course means that a coin suffering this kind of "attack" also gets into a >51% situation with the whale, and on top of that I've also once seen confusion and chain splits because at day 1 of a coin launch with a super low initial difficulty of 100 or so (from a test someone did on a weak machine), people dumped 200KH/s onto it and we got millisecond block times and a big mess.

Is there some kind of counter-strategy for this problem? I was wondering about this for a while and didn't find much info about it.

What about all those small Cryptonote coins, do they just cross their fingers people won't do this to their coin, and if it happens, they sit it out and get 1-hour block times for a few days and then hope it won't happen again, or what?

Thanks!

EDIT: Follow-up question: Would reducing the difficulty window help solving the problem? For example, having a CryptoNote coin which uses the last 8 blocks instead of 720 to calculate difficulty. It would appear that this would help, but it then makes me wonder why most coins I saw didn't have a low setting like this (Monero also has 720). I guess with too low a value, difficulty would be even more unpredictable - but what is too low?

- Make sure the directory in which the wallet is located is not in a path containing non-English characters
- Make sure your Windows username does not have non-English characters. If it has, you need to specify a data directory (without non-English characters in the path) manually instead of using the default one (because the default one will be C:\Users\\AppData\Roaming\Bytecoin, so if your username has non-English characters, it's again problematic). You can do so using the --data-dir parameter.

(Also answered on GitHub)

This does not answer my question.

In fact, in the meantime, I found the solution myself (in case anybody else is interested): For Scrypt coins (not for SHA-256 or X11), the difficulty values used in miner softwares and pools (but not the coin difficulty itself!) are scaled by a factor of 65536 (2^16). Divide the difficulties by that factor if you want to compare them to the block difficulty. Then suddenly everything falls in place again.

Or, said in another way: A share with difficulty 1 is actually difficulty 2^(-16). So you need 65536 more shares than you think in order to find a block (on average, of course).

I posted the full explanation here: https://bitcoin.stackexchange.com/a/61166/62320

Hello,

I am struggling to understand PPLNS (I need to implement it in a pool). I know in general how the system works, but what is a mystery to me is where the N comes from.

I read on many places on the Internet that N = 2 * difficulty is a good value, and that this would mean that on average a miner will earn rewards from two blocks for the same shares.

However, it simply does not add up for me.

Practical example: BitConnect pool. Let's say BitConnect's difficulty is currently ~900,000.

An AntMiner L3+ submits a share at difficulty ~150,000 (which counts as a share value of 150,000 then) every ~15 seconds, so in 3 minutes, that one AntMiner has already filled the whole "N" of 900,000x2=1,800,000. That would, in reverse, mean that I would expect to mine a BitConnect block every 90 seconds on average, with one AntMiner, which is of course not true.

100 AntMiners running on our pool gave a share value of 230,000,000 in the last 5 minutes, which would have been more than 100 times the whole "N" already!

So something can't be right here, and I don't understand what. Should I not multiply the submitted share with the work difficulty? But then a small miner would get the same reward as a large one which makes no sense. Or is the whole Internet wrong about the 2*diff for "N", and that it's based on the assumption that it takes on average a value of shares equal to the difficulty to find a block? Doesn't sound likely.

On average, it takes 2^32 hashes to find a valid share at difficulty 1, right? And on average, it takes X shares to find a share at difficulty X, hence why you increase the difficulty at the pool for your miner if the miner is powerful (it will then less often submit a share), right...? But the last point starts to make less sense already, together with the fact that you need a share at difficulty Y to find a block if the currently coin block difficutly is Y... And the pool software counts a share with difficulty X the same as X shares with difficulty 1. At the end I'm left confused how it all works together and how I can arrive at a value for "N" which *actually* results in a miner getting on average two rewards for the same share...!

Please help me understand what is wrong.

Thanks,
CherryDT

Hello everyone, am I mistaken or is the difficulty adjustment actually every block and not every 2016 blocks?