Bitcoin Forum

Hey all,

I'm wondering if anyone can help a newbie out and explain the difference between the difficulty of any particular coin, and the 'best share' report (or any share reporting) of cgminer.

My understanding is when it says something such <accepted ..... 56/8> it's saying you achieved a share of 56 when the minimum required for the pool is 8. The best share is simply the highest value so far submitted to the pool.

If the difficulty is say, 7.6 million, what number would I need to have submitted to represent what would have been a successful solo mine? Am I understanding there is a relationship between the two figures?

Cheers.

Yes, you understood that correctly with the 56/8 example. Any share with a difficulty over the 7.6 millino meets the requirements for a block solve, whether it be a pool or solo mining.

Unless I'm mistaken, this seems crazy for something such as Litecoin.

For example, difficulty is at ~200, yet I'm getting over 200 perhaps every 30 mins or so. Surely I'm missing something as there should be no way I'd be so successful.

Litecoin mining uses scrypt, and most GPUs are in the KH/s range, not the MH/s range. So with 1/1000 of the hashrate, it's expected to be 1/1000 of the shares.

Peleus, I'd like to understand this as well. I'm currently mining an altcoin where the difficulty is 0.063387 but to mine a block the share must be greater than 4154. Unfortunately this thread is what comes up when I google it and doesn't yet contain an answer.

Well, I don't understand exactly why, but the relation appears to be 65536. So if the network difficulty is 0.063387, that times 65536 is 4154, so a share has to be greater than 4154 to be a block. Or, from a freshly restarted cgminer, the "best share" has to exceed 4154 to have found a block.

With current Bitcoin difficulty of 2,621,404,453.0646, a miner needs a best share of 1.718e14 to find a block, which would probably be represented as 172T (because it's 172 tera-somethings. Terradifficulties?).

65536 has nothing to do with bitcoin - that's altcoin crapulence. If the diff is 2.6 billion, you need a share of 2.6 billion difficulty with bitcoin.

I might be wrong, but I think if you are mining a SHA256 coin (like BTC, PPC etc) then the difficulty and the share display are the same. (e.g. difficulty is 2.6G, need to difficulty share solve of 2.6G or more)

If you are mining a scrypt coin (Feathercoin, Coino etc) then the difficulty displayed needs to be multipled by 2^16 (or 65536) to give you the comparative difficulty display.

Perhaps someone who knows what they are talking about can confirm?

Cheers
Z

That sounds correct. You would multiple 65536 x the difficulty of that scrypt coin in cgminer and that is the share you must get accepted to block. At least from all my readings that is how I understand.

For example at the time of this writing Dogecoin network difficulty in cgminer shows 1.09K so I would have to find a share of 1,090 x 65,536 = 71,434,240 to find a successful block.

Hello I hope you understand my english. This subject make me some questions.

1. In practice, is that mean more the difficult is high more leading zero must have the hash to build next block ?

2. For example a 1GH/s SHA-256miners can calculate a very high difficulty ?

3. Or with a 1GH/s in solo mining if you force the difficulty to 1 when the coin is like bitcoin, is that possible ? Is that good ?

4. If I mine in a pool, who ask for less leading zero (depending the pool), is that mean sometimes a miner found an hash with more leading zero (acceptable for the coin difficulty) than this hash solve the block ?



Thanks for your help   ;)