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Okay, I think I figured out how the sieving algorithm works.
Suppose that you are looking for 2p-1 chains (eg. 1531, 3061, 6121, 12241, 24481) of length 3 whose origins are multiples of 100 and you want a sieve going up to 7. You need to work out 4 moduli: 2,3,5,7. Let w be the origin, so w+1 is the first prime, 2w+1 is the second prime, 4w+1 is the third prime.
A lot of math from here on is modulo math; if you see me making weird statements like 2*4=3 or 3/2=5, it's because they're true in modular arithmetic under whatever modulus I'm talking about.
Modulo 2:
w+1 != 0, so w != 1
2w+1 != 0, so 2w != 1, which is always true
4w+1 != 0, so 4w != 1, which is always true
Hence, w = 0 (mod 2)
Modulo 3:
w+1 != 0, so w != 2
2w+1 != 0, so 2w != 2, so w != 2/2 = 1
4w+1 != 0, so 4w != 2, so w != 2/4 = 2/1 = 2
Hence, w , ϵ [0,1,2] - [1,2] =
(mod 3)
Modulo 5:
w+1 != 0, so w != 4
2w+1 != 0, so 2w != 4, so w != 4/2 = 2
4w+1 != 0, so 4w != 4, so w != 4/4 = 1
Hence, w , ϵ [0,1,2,3,4] - [1,2,4] = [0,3] (mod 5)
Modulo 7:
w+1 != 0, so w != 6
2w+1 != 0, so 2w != 6, so w != 6/2 = 3
4w+1 != 0, so 4w != 6, so w != 6/4 = 5
Hence, w , ϵ [0,1,2,3,4,5,6] - [3,5,6] = [0,1,2,4] (mod 7)
Now, since we're looking for multiples of 100, we need to convert these modular statements into modular statements about the cofactors. That is to say, given w=100k, we want claims about k.
Base 2: w = 0 (mod 2) is true regardless of k, so this condition happens to be irrelevant here
Base 3: w = 0 (mod 3), so 100k = 0 (mod 3), so k = 0 (mod 3)
Base 5: w ϵ [0,3] (mod 5) is true regardless of k, so once again this condition can be thrown out
Base 7: w ϵ [0,1,2,4] (mod 7), so k ϵ [0/100,1/100,2/100,4/100] = [0/2,1/2,2/2,4/2] = [0,4,1,2] (mod 7)
Thus, we have k = 0 (mod 3), k ϵ (0,1,2,4) (mod 7), so altogether by Chinese Remainder Theorem k ϵ [0,9,15,18] (mod 21).
Let's try it out!
k = 9, w = 900. We have 901, 1801, 3601 . Unfortunately, 901 is divisible by 17 and 53, so fail. But, note that none of these three numbers is divisible by 2,3,5,7, so the sieve did its job.
k = 15, w = 1500, so first number is 1501 - once again not prime. In fact, the first origin that actually works is 26700, followed by 33300 and 54600.
Now, let's try a k outside the sieve.
k = 16, w = 1600, so the chain is 1601, 3201, 6401. As it turns out, 3201 is divisible by three. Once again, the sieve did its job by removing this possibility from consideration. In fact, every k removed by the sieve is guaranteed to fail on either 2,3,5 or 7 (actually just 3 or 7 since failing on 2 or 5 is impossible).
The actual algorithm is somewhat more complex to account for both kinds of Cunningham chains and bitwin chains, and the process is the same.
Now, there's another optimization: primorial numbers. 100 was a nice factor since it removed the possibility of any number in the chain from being divisible by 2 or 5 right off the bat. What if we had used 210 instead? Then, we could have done the sieve for some higher primes, like [11,13,17], instead, and gotten to the goal much faster - indeed, the first origin that works there is 2310, only the eleventh value tested even without any sieving! Primorial numbers are numbers that are divisible by all small primes up to a certain point. If we can shift around nonces in the block until the hash is divisible by at least a reasonably sized primorial, then we can achieve further speedups (although, the Primecoin paper states, not particularly large ones).
In the client, the sieve goes up to 1 million. So, the question is, how much an optimization over the randomness of the Prime Number Theorem (namely, that random numbers n bits long have a 1/n chance of being prime) do Primecoin miners actually get?
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One thing I can't quite figure out with Primecoin is exactly what the relationship is between the main measure of mining output, "primes per second", and how long it actually takes to find a block. Right now chains need to be 7 primes long, and each prime must (usually) be at least 256 bits long, so naively (assuming primes are random):
P(chain of 7 primes) = P(number is a prime)^7 = 1/256^7 (~1/10^17) by the Prime Number Theorem
Since every number we're dealing with is odd, we can improve that bound to 1/128^7 (1/10^15), but the client seems to be using an advanced sieving algorithm (it looks like much more than just the basic Sieve of Erastothenes) to cut even more corners, so I would not be surprised if the number of primes needed to scan through to get a block is far less than this (it must be, as otherwise there's no way miners with even as much as 100 pps would be getting blocks so frequently). So, the question is, does anyone know what the ratio is, whether mathematically or empirically?
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I'm predicting bottom at $0.20.
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Yay!
{
"blocks" : 787,
"currentblocksize" : 1000,
"currentblocktx" : 0,
"errors" : "",
"generate" : true,
"genproclimit" : -1,
"primespersec" : 1,
"pooledtx" : 0,
"testnet" : false
}
Mining on laptops is fun.
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It looks as if some have missed the idea behind this article...
Comparing mid-90s media FUD regarding the internet to current media FUD regarding bitcoin.
Do you guys really think Bitcoin Magazine would seriously suggest people don't use the internet?
Our website is free, print edition $8.88. Which do you think we want you to use?  |
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Critical question: is the encryption and decryption client side or server side? If it's the former, this is huge. If it's the latter, interesting but not that much. |
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BIPS charges a 2.5% fee on the consumer side. I was told this in person; it caught me by surprise too, as I had thought their whole 0% deal was some kind of loss leader.
Clarification: the mis-charges were an error. BIPS charges a 0% fee for BTC-to-BTC transfer, and charges 2.5% if you convert the bitcoins to fiat through their service in the form of its below-market exchange rate. Optionally, merchants can check a checkbox (visible only when you have a verified bank account) to take on the fee themselves, in which case they can sell at full price. |
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BIPS charges a 2.5% fee on the consumer side. I was told this in person; it caught me by surprise too, as I had thought their whole 0% deal was some kind of loss leader.
Clarification: the mis-charges were an error. BIPS charges a 0% fee for BTC-to-BTC transfer, and charges 2.5% if you convert the bitcoins to fiat through their service in the form of its below-market exchange rate. Optionally, merchants can check a checkbox (visible only when you have a verified bank account) to take on the fee themselves, in which case they can sell at full price.
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I agree with one thing, coinlab isn't getting 75M, they'll be lucky if mtgox agrees to settle for 1M if mtgox was in breach of contract, probably even less.
But if a contract was made between coinlab and mtgox, there will be break up fees for sure.
The point of initially citing figures like 75M is never to actually get 75M. It's actually quite a bit more devious than that. Excerpting William Poundstone's Priceless: The Myth of Fair Value: Psychologists Gretchen Chapman and Brian Bornstein tested this idea in a 1996 experiment, when Liebeck v. McDonald’s was much in the news. They presented eighty University of Illinois students with the hypothetical case of a young woman who said she contracted ovarian cancer from birth control pills and was suing her HMO. Four groups each heard a different demand for damages: $100; $20,000; $5 million; and $1 billion. The mock jurors were asked to give compensatory damages only. Anyone who wants to believe in the jury system must find the results astonishing. | Demand................. | Award (average) | | $100 | $990 | | $20,000 | $36,000 | | $5 million | $440,000 | | $1 billion | $490,000 |
S. Reed Morgan, of the McDonald’s coffee lawsuit, has described attorneys such as himself as “entrepreneurs.” By seeking liability suit jackpots, professional litigators provide incentives for big companies to worry about the safety of their products. Less sympathetic observers dismiss this as “lottery litigation.” Either way, attorneys facing the legal wheel of fortune sometimes refrain from asking jurors for a specific amount. They fear that a reasonable figure might preempt a windfall, and a high-end figure could boomerang. Chapman and Bornstein’s experiment suggests otherwise. The title of their paper says it all: “The More You Ask For, the More You Get.” |
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Bitcoin multisig has nothing to do with cryptographic secret sharing schemes. Traditional secret sharing relies on fancy mathematical tricks like representing some secret as a polynomial and handing out, say, 10 points on that polynomial, such that any 5 of them are enough to determine what the original polynomial is. Bitcoin does not need that because its job isn't hiding secrets - rather, it's verification. In the case of Bitcoin, you simply have everyone make their signatures and then have miners directly check if at least 5 of the 10 signatures are valid. So the Bitcoin protocol can theoretically be extended to support any kind of hierarchical, anarchical, pseudorandom or whatever other ruleset for transaction validation (that's right, we could implement gambling right in the blockchain!) provided that no human judgement is required to determine validity (so, "redeemable by the owner of 178gb... if <insert name here> is deceased" won't work without a trusted authority). As for what is possible right now, I'm looking at https://en.bitcoin.it/wiki/Script, and it looks like there's a lot of opcodes that nobody seems to be taking advantage of yet. Might something like (sig) (pubkey) OP_CHECKSIG (sig) (pubkey) OP_CHECKSIG (sig) (pubkey) OP_CHECKSIG OP_ADD OP_ADD 2 OP_GREATERTHANOREQUAL (sig) (pubkey) OP_CHECKSIG (sig) (pubkey) OP_CHECKSIG (sig) (pubkey) OP_CHECKSIG OP_ADD OP_ADD 2 OP_GREATERTHANOREQUAL (sig) (pubkey) OP_CHECKSIG (sig) (pubkey) OP_CHECKSIG (sig) (pubkey) OP_CHECKSIG OP_ADD OP_ADD 2 OP_GREATERTHANOREQUAL OP_ADD OP_ADD 2 OP_GREATERTHANOREQUAL for a "2-of-3 per group, 2 groups out of 3" transaction work? |
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What about OKPay? Have they gotten their money back or been compensated?
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[/quote]haha like me you dont wanna spend bitcoin wanna hold bitcoin because value of bitcoin go up only spending is no good idea only if you wanna regret later you sold bitcoin
[/quote]
Here's an idea. Well, actually three ideas.
1. Buy more bitcoins as soon as you spend them.
2. Convince your employer to pay you part of your salary in BTC and spend that.
3. If you're so confident about the future direction of the price put the bitcoins you already have on leverage.
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I have kind of mixed feelings about this. On the one hand I would be very happy if Bitcoinstore managed to reach their goal. On the other hand, I feel they should be able to do this "on their own" without extra community efforts.
Anyway, for me, being an international customer, the problem is after including shipping, VAT and import tax, my domestic alternatives are usually more competitive.
But I really hope that they succeed in their $850k goal, as that would be a significant milestone for Bitcoin!
The $850k in a month is only a temporary thing. If they survive, from here on it will only be 850k every 3 months. |
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I can offer up Bitcoin Magazine's pages on wallet options, common misconceptions and terminology ( part 2) for the site - we have a pretty detailed description of the various wallet options out there, as well as vocabulary lists. Just give credit/link back to us on the page and you can copy over and modify how you see fit. You can also just link to these pages as well as WeUseCoins, the wiki and other stuff in some kind of "Further Reading" section. |
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rJ2gXg1xcYW6756fBj393Dd9ZekJP45odZ
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Back before I joined Bitcoin Magazine, I was writing (really, the sole writer) for a site called Bitcoin Weekly. When the owner of the site (kiba) ran out of money to maintain the site and pay contributors, I came up with an interesting business model. Every week I would write two articles, and kiba would release summaries of the articles together with a Bitcoin address and "ransom". When the ransom was paid, he would release the content. The mechanism actually worked pretty decently, earning me about $3-$6 per hour for my work. There are actually lots of cool and untested business models that Bitcoin, especially together with Bitcointip, can make possible. Just go out and try it!
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Currency de-valuation means de-valued relative to other currencies. It doesn't affect people who deal only in that currency beyond the price of imports going up.
That doesn't sound quite right. Taking the idea to its logical conclusion, if there was a single one-world currency its issuer could simply print trillions of it every day and it wouldn't affect the people one bit because they're all inside the currency bubble and there are no imports. Devaluation can also mean relative to commodities, in which case you can argue that since marginal productivity stays the same in real terms wages would stay the same in real terms, and so it basically would be equivalent to a tax on savings plus partial debt jubilee. In reality, it's usually somewhere in between the two scenarios. |
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I'll be happy to accept this as true once I see some the slightest shred of evidence. If of these "zero worlds" proponents could jump out of a window and decide not to fall, or stand in front of a moving bus and decide not to be hit by it using only the power of their mind that would be pretty compelling.
Only to themselves though. From the perspective of your own mind (alternative QM translation: over all possible universes which you are likely to survive as an observer of), you're overwhelmingly likely to see them splattered on the pavement in more fragments than a Windows XP hard drive partition. |
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