Added a pull request for "-purgetx" option.
Sorry I still don't know what a 'pull request' is although I've seen the term floating around this forum. Is this -purgetx likely to end up in the official client any time soon or should I find it elsewhere? I'm in the process right now of learning a little more linux; just downloaded ubuntu livecd. Pull request is a github-ism. It is essentially a patch or fork off the code tree.
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kjj, how'd u find the block?
I just clicked back through the chain until I saw a 9000 -> 8999 + 1 transaction around the right date.
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Hmm. Krugman isn't usually this eloquent. Not usually this brief either. I think that quote is from an impostor.
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If a decent implementation of Shor's algorithm ever shows up, it will be the end of all cryptography, not just bitcoin. At that point, we will literally be carrying discrete milligrams and micrograms of gold around in our pockets, because nothing else will be trusted.
Not true. There are still plenty of encryption algorithms that are not weakened by Shor. You are right. I was thinking this morning that it had been many years since I looked into it, so I did some more checking, and found this paper on Quantum Resistant Public Key Cryptography.
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I know i was i imagine anyone else running in the 300 and under mhash range is done as well if they are paying for their power. It basically cut my daily profits in half. I wish others luck but i feel like you remove the hobbyist and small timers so early in its development you remove a lot of interest in the coins just how this works i guess.
I'm sorry you missed out on the goldrush, you can't get a bunch of free bitcoins, and you are bitter about it. Don't let the door hit your ass on the way out.
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So why we can't mining without network connection, if next bock is unique, we just need to submit the result to internet when next block is done caculation Is that true? and it often take more then a month ( difficult may changes durring caculation, what will happen if you don't join a pool and difficult changes durring you caculation on one block?)
Because each block contains the header of the previous block. Right now we are on block # 127114. No one cares if you come up with an acceptable replacement for block # 123000, which is what you'd get if you spent a month mining disconnected. We've moved on. We don't care any more.
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If a decent implementation of Shor's algorithm ever shows up, it will be the end of all cryptography, not just bitcoin. At that point, we will literally be carrying discrete milligrams and micrograms of gold around in our pockets, because nothing else will be trusted.
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6870s yield terrible MHash/$. Well I shouldn't say terrible, but it is quite poor, if you buy new it's one of the few reasonable options left I suppose. Buying used is a far superior option if you can wrangle it though.
Opinions vary. When I did the math, it gave the best Mhash/sec to $ ratio, and by a considerable margin. Also, it was in stock almost everywhere. Anyone that is thinking about mining needs to make a spreadhseet and list all of the options along with their hash values. Then shop a bunch of sites, looking for the best price on each card. From there it is a simple matter to find the best Mhash/sec per $ value. Adding electrical costs (if such apply to you) to that spreadsheet is trivial.
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kjj, I have two questions about the situation you're describing:
1.) So if two blocks are found at the same time, on opposite sides of the world, you're saying eventually when it all settles down, one person is going to lose their 50btc?
(also what do you mean by "will be put back into the pool?" )
2.) I completely understand why with the current extreme difficulty this rarely happens, however back when the difficulty was much easier, and GPU miners had just came out so people were hitting jackpots all the time, would you say this happened much more often back then?
1) Yes. That is actually the reason that no node accepts a coinbase transaction as an input until it has been in the chain for 100 blocks. As for your parenthetical question, let me back up and answer PRCman first. wait, so If I have 2 cards in my computer, working as 2 separate workers, are they basically competing against each other? That is, are they doing the exact calculations at the exact same time, or will they be doing different/random calculations in an attempt to find the block?
There's no competition in the sense you're thinking. Within one difficulty period, your mining will result in the same number of bitcoins regardless of whether there are 1 or 10,000,000 other miners on the network. Mining is competitive only because the network periodically adjusts itself by increasing the difficulty of the hashing work to be done, based on how quickly it has been done in the past. The more miners there are, the faster the work will get done. The faster the work gets done, the harder it will become in the future. I don't understand, why in computing period, there is no compation, won't 2 miners caculate several same has values while they are mining? This is a key concept that you must understand: There is not one true next block that everyone is working on. Each miner is allowed to create their own block, including their own coinbase transaction, and their own subsets of pending transactions that they wish to include in it. Each miner is working on their own version of the next block.The nodes communicate transactions amongst themselves. Each one can make their own decision as to which ones are included in the block they are working on. This is where the transaction fee comes in, in case you've been wondering about that little detail. It doesn't come up much now, because generally each block contains all or most outstanding transactions in the global pool, but that isn't necessary. Different nodes can pick different ones. Say there are 7 transactions pending when block A is completed, numbered 1 to 7. Say the node that creates block B includes transactions 1-5 in the new block, and the node that creates transaction X includes 1-3, and 7 in their new block. When blocks B and X are done, the nodes will have different ideas of which transactions are done, and which ones are still pending. The node working on B will think that transactions # 6 and #7 need to be put into the next next block (block C), while the X node thinks that transactions #4, #5 and #6 are still outstanding. When block "C" comes around, the X node then rolls back all transactions, #1, #2, #3, and #7. But the new chain includes #1, #2, and #3, so it can discard them. It also contains #7, so it can discard that one too. And now it re-broadcasts transactions #4, #5 and #6 as valid transactions. The rest of the netowrk will pick them up (but probably knew them already) and work at including them in the next next next block. As for question 2) from PcChip, I wasn't around back then, but the answer is no. The validity of a block is based on the difficulty of that block. If the earlier blocks were easier, it was because there was less computational power being applied to the system back then, so it balances out. The odds of a block race are the same now (give or take a bunch for difficulty lag) as they were back then.
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Yawn.
These stories come out every couple of months, and have for years, and years, and years.
The dollar has some big problems, as does the US in general, but there is simply no viable alternative, and no prospects for an alternative. Also, nearly every man, woman and child on the face of the planet has a vested interest in preventing a full blown collapse.
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You didn't search before you posted this, did you? Also, there is no danger of hashing power getting too concentrated. Anyone that has the ability to command that much computing power totally has better things to do with it.
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Why would you want to do that?
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More likely, the site is wildly inaccurate about the aggregate hashing speed, as it always is after a difficulty change, and it just subtracts the published hashing speeds of the various known pools from the horribly inflated global estimate and concludes that "other" is doing a lot of work.
I bet it'll look normal again in a day or two.
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Looks like we got a second invalid block.
Any thought of rolling the shares from the invalid round into the next round?
Correct me if I'm wrong in this, but wouldn't that defeat the concept of the invalid insurance/reward those of us that donate 2.5% or more get? It would diminish our reward a bit, but not eliminate it.
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Maybe. Not all superconducting ring reduction systems are quantum in nature. See this IEEE article. The good stuff is on page 2.
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Looks like we got a second invalid block.
Any thought of rolling the shares from the invalid round into the next round?
Block 151 has already been solved. So far it's not showing invalid to me. Oh, heh. I wasn't suggesting that we had consecutive invalid blocks. There was one before, round 8.
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Normally, yes, but you can set it in the conf file.
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If that is all you want, you just need to edit your bitcoin.conf, and make sure the rpc port is available from the outside (forward the port on your router).
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Looks like we got a second invalid block.
Any thought of rolling the shares from the invalid round into the next round?
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