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I looked into auction houses and they told me the coins would need to be graded and slabbed before they would consider auctioning them. Mine are not slabbed so there is that.
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I assume this is without the spelling error?
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There is an old lady that lives next door. She must have over 100 cats. Gold mine!
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Thanks for the info, is there a consolidated list of the puzzles that have been solved and the likely method used to solve them?
This is the only transaction I know of that is specifically designed to experimentally determine the security strength of Bitcoin. |
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kangaroos have claimed up to 115 bits.
I think that is a new world record but would have to research it to make sure. If so, the person who claimed the 0.115 BTC should get some fame for that - if they want the fame... |
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Here is an experiment to determine the answer to your question https://bitcointalk.org/index.php?topic=1306983.msg51460251#msg51460251Here my comment from that thread: Well, what is the sacramental meaning of the step through 5? And he just brute force? because, it seems, it is necessary that the address be translated to its public key lit up. Arulbero could not until there is no output, find the number or could?)).
The author of the puzzle did that transaction to spend from every fifth address. My guess is that it is an experiment to see how far people can get when there is an available spend transaction versus addresses with no available spend transaction. He has been known to visit this thread. Perhaps he will explain it. Obviously it is much easier to get the private key when there is a spend transaction on the address. #1 through #61 took a long time whereas #65, #70, #75 and #80 were snatched up pretty soon after the author added the spend transaction to those addresses. I expect #85 will also be snatched up in due time. As discussed #85, #90, #95, #100, #105, #110 are all within the realm of possibility given enough time and resources. It looks as if #115 would be a new world record so someone with enough equipment and motivation can probably get that one. Beyond that it is very iffy. Note that since there is a spend transaction on these addresses we will probably see more of them before we ever see the next address without a spend transaction, #62. This is a HUGE difference in effort. I think one of the take home messages here might be that due to this difference in effort and other factors having to do with privacy and the fungibility of Bitcoin in general: do not reuse Bitcoin addresses. Bitcoin addresses should be used exactly twice: once to fund them and once to spend them - then never used again. As I predicted: https://blockchair.com/bitcoin/outputs?q=transaction_id(56857085)&s=index(asc)#Brute force has claimed the BTC up to 63 bits and the next brute force target address has 64 bits. Kangaroos have claimed up to 115 bits the and next target for the kangaroos is 120 bits. |
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bytcoin, I appreciate your eagerness and willingness to learn about Bitcoin at a technical level and we are trying to help you understand as best we can. What you did was take one signature and make another signature that looks different but, in fact, mathematically it is the same signature. As stated previously: ... is not solvable because the OP crafted the signature from a previous one.is the same signature because: 2*r2*s1 % n = 4d27b04d19bb13563dc45d9d768a13f53f02970024759081601c4bd59aba5f76
r1*s2 % n = 4d27b04d19bb13563dc45d9d768a13f53f02970024759081601c4bd59aba5f76
and
r2*z1 % n = 3d6319695748c424ef2a249456c3a198a6ea34933ba345a8118703ef82d28977
r1*z2 % n = 3d6319695748c424ef2a249456c3a198a6ea34933ba345a8118703ef82d28977
So, do you understand now? Also, think about this just a bit: would Bitcoin be worth about 1 trillion dollars if it was so easy to break? If it was this easy to break it would be broken. If you want to see what an actual attempt to break something similar to Bitcoin looks like then take a look at this paper. It is an attempt to figure out a way to "break" RSA - which is not exactly the same a Bitcoin but for the sake of my point it is close enough: https://eprint.iacr.org/2021/232.pdfNote that most experts in the field do not believe that this "breaks RSA" so don't worry about that. What I want you to notice is the level of math it takes to even attempt to break something like an RSA, ECC, ECDSA or Bitcoin. This is an attempt by a brilliant mathematician and crypto expert. Face facts, you don't have the math for this, I don't have the math for this, the math involve is beyond almost everyone. I encourage you to keep learning but, really, try not to be so melodramatic about it or you will find that everyone here that can help you will just start ignoring you altogether. ASK QUESTIONS and LEARN. For example I though it was interesting when you asked: If you know the mathematical relationship between the k values in two different and valid signatures can you calculate the private key. The answer is yes. The math to show this is not that hard and has been show to you in the other thread (k' = k + n) and in this thread (k' = m * k). This whole "with my very limited mathematical abilities I think that I have found a way to break Bitcoin" is not the way to go about learning about Bitcoin. It is a good way to get ignored by the very people that are here to help you learn. |
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I do not think that address ever had 2562.90203395 that is how much it received over 225 transactions to that address.
We used to throw 100 BTC at a gambling site in those days. I remember paying 25 BTC for a couple of cases of Girl Scout cookies on a whim just to support the "Bitcoin economy". I spent or lost a lot of BTC in my time.
I have some left, not 2000 BTC, but I managed to keep a few.
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If you have two valid signatures using the same private key where k' = 2k then:
From each message we can derive the z value (hash of the message) so:
First message and signature (m, r, s, z)
Second message and signature (m', r', s', z')
Therefore: ks = z + rdA and k's' = z' + r'dA
Therefore: (sk - z)/r = (s'k' - z')/r'
But in this case k' = 2k so:
(sk - z)/r = (2s'k - z')/r'
So all you have to do is solve for k. All the other values: s, z, r, s', z', and r' are all known.
rr'(sk - z)/r = rr'(2s'k - z')/r'
r'(sk - z) = r(2s'k - z')
r'sk - r'z = 2rs'k - rz'
r'sk - r'z +rz' = 2rs'k
k = (r'sk - r'z +rz')/2rs'
Once you know k you can simply calculate the private key, dA = (sk - z)/r
I still do not see what your two signatures have to do with the BTC at 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx
These two things: how to solve for the private key when you know k' = 2k and the BTC stored at 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx seem to be unrelated, right?
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The prize was offered in this thread: https://bitcointalk.org/index.php?topic=5321454.0Unfortunately, he locked the thread before I could claim the prize  but I have the solution for the prize !!! From each message we can derive the z value (hash of the message) so: First message and signature (m, r, s, z) Second message and signature (m', r', s', z') Therefore: ks = z + rd A and k's' = z' + r'd ATherefore: (sk - z)/r = (s'k' - z')/r' But in this case k' = 2k so: (sk - z)/r = (2s'k - z')/r' So all you have to do is solve for k. All the other values: s, z, r, s', z', and r' are all known. rr'(sk - z)/r = rr'(2s'k - z')/r' r'(sk - z) = r(2s'k - z') r'sk - r'z = 2rs'k - rz' r'sk - r'z +rz' = 2rs'k k = (r'sk - r'z +rz')/2rs' Once you know k you can simply calculate the private key, d A = (sk - z)/r Please send my 3,350 bitcoins to 1BurtWEejbnKeBRsvcydJvsNztB1bXV5iQ That was easy! Thanks! Easiest $170,000,000.00 I have ever earned. Still waiting on payment. |
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Yes, as we have all told you before:
If there is a known mathematical relationship between the two nonce used (k1 = k0 + n OR k1 = m * k0) then you can derive the private key from the two signatures.
This is why the k value is picked at random (hopefully from a secure random number generator).
The key phrase here is "known mathematical relationship between the two nonce"
We have shown you the math for this - several times. You should be able to do the math yourself at this point.
What does the address 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx have to do with anything?
Are you really claiming to know that there are two signatures in the block chain using the private key for 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx that were constructed with k1 = 2 * k0?
How would you know this?
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Anyhow, the halving will take care of some of this, the revenue will drop by 75% in 7 years so even if the price will hit 200k will be at the same levels as now, the rest will probably be solved if we stop with all those renewables that only work with subsidies and go full nuclear, constant power no matter what time of the year and of the day.
The next halving may have less impact than you think. Currently the miners are getting about 5 BTC/hour from fees and 6(50/2 3) = 37.5 BTC/hour from the subsidy. Assuming steady growth in transaction demand until the next halving this could easily grow to 10 BTC/hour in fees. So at the next halving we could have 10 BTC/hour in fees and 6(50/2 4) = 18.75 BTC/hour from the subsidy, all at a higher price. We will have to see how it all plays out... |
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Your calculations are straightforward, but the usage is not proportional to the price of a bitcoin because the price of electricity is not inelastic. As the usage goes up, the cost of electricity will increase more quickly.
BTW, you can do exponents like this: 2e
Agreed. That is why I said "trend to" perhaps "try to use" that much power would be better since, as you point out, prices would adjust. Thanks. Fixed the typos in all my exponents. |
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Bitcoin Network Power Consumption Estimate
First, note that Bitcoin mining efficiency does not matter when estimating the trend of the power consumption of the entire Bitcoin network.
The power consumption of the entire network depends on five things:
x = the exchange rate [USD/BTC]
e = the era [0..32]
f = the average fees per hour [BTC/hour]
c = the average cost of energy [USD/kWh]
r = the average percentage of income miners spend on energy [unit less ratio]
From the era we can calculate the average hourly BTC subsidy rate:
s = 6(50/2e) [BTC/hour]
And the average amount of BTC all the miners in the world make per hour:
b = s + f [BTC/hour]
From this we can calculate the amount of USD per hour all the miners in the world make:
u = bx [USD/hour]
Given the worldwide average percentage of income miners spend on energy the amount spent worldwide on energy is:
ur [USD/hour]
And finally, the worldwide power consumption is given by:
P = ur/c [kW]
= bxr/c [kW]
= (s + f)xr/c [kW]
= (6(50/2e) + f)xr/c [kW]
Notice that mining efficiency does not enter into this equation and does not matter.
You do not need to know or estimate the average overall efficiency of the mining network unless you want to calculate the difficulty and/or hash rate.
Let’s put in some numbers:
x = $50,000; the exchange rate [USD/BTC]
e = 3; the era [0..32]
f = 5; the average fees per hour [BTC/hour]
c = $0.03; the average cost of energy [USD/kWh]
r = 0.8; the average percentage of income miners spend on energy [unit less ratio]
P = (6(50/2e) + f)xr/c [kW]
= (6(50/23) + 5) 50000 ( 0.8 ) / 0.03 = 56,666,666 [kW] = 57 Gigawatts
World power production/consumption is about 15,000 Gigawatts.
Bitcoin mining will trend toward 57/15,000 = 0.38 % of world power production given these values.
This scales by BTC price so:
BTC at $500,000 means power consumption would trend to 3.8% of worldwide power.
BTC at $5,000,000 means power consumption would trend to 38% of worldwide power.
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On an offline computer, I used a copy of bitaddress.org to enter a brainwallet passphrase of 12 words and the private key and public key were displayed. I printed that offline, and keep the info safe until now. I'm confident I have the public and private key. Just trying to move the coins safely to coinbase.
So: 1: get a reliable wallet like those mentioned in this thread and sweep the private key into that wallet. 2: now, simply transfer the coins from that wallet to coinbase 3: sell the coins |
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The 2009 must be a typo. 2019 perhaps?
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OK, everyone probably knows about this site but it is the coolest thing I have found in a long time. It visually shows the mem pool statistics, size, the fees people are paying, etc. over time. Click here and see the beautiful color graphics for the last three months: https://jochen-hoenicke.de/queue/#BTC,3mYes, look at the third graph down, mem pool size, and notice how high it is (225 MB). This large amount of transactions has to be squeezed into the limited size blocks - therefore the amount you need to pay to get your transaction in the blocks goes up. You can adjust the time window. You can also see the fees being paid on other crypto (ETH, BCH, etc.) to see how they compare to BTC. So cool. Donating to the site author now... EDIT: I will, of course, donate something other than BTC, probably BCH, since a donation of BTC would cost me too much to send right now  |
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Five words for you: Magic The Gathering Online eXchange
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OK, everyone probably knows about this site but it is the coolest thing I have found in a long time. It visually shows the mem pool statistics, size, the fees people are paying, etc. over time. Click here and see the beautiful color graphics for the last three months: https://jochen-hoenicke.de/queue/#BTC,3mDonating to the site author now... |
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