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Building the group  |
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~ as "All of the information required to reconstruct the private key is published on the blockchain". But it was not as easy ;-)
Maybe also valid for Satoshi's mined blocks? ~ We can visualize with the blockchain data, how satoshi Mine his coins: |
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Why ? JeanLuck kangaroo keyrate is 1 Giga, I use kangaroo what 55 terra, and I was try BSGS what 500 Peta(but I no so like BSGS, kangaroo I like more), so WanderingPhilospher make his own soft what has exakeys... If you use dead software this not meant what other uses so software 2.
Well you might be right about BSGS (but I have never used that software before so take this with a grain of salt), but I don't think you can hit 55 terakeys per second on kangaroo without a GPU farm and leverage the kangaroo server feature. Maybe they have ... Do you know, if there are already asics like these  not to mine coins but for point addition? |
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yes... but I think 1000 pubkeys in range 100 will be faster then 1 in 120 ?
If you downgrade to range 100, you will have ~1,000,000 pubkeys and only 1 of them is the right one. You would have to calculate far more than the original one in #120. |
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I actually wondered with regards to TTD's 64's pool (I use btw). I am not saying this is happening at all, unlikely but think of the scenario
We mine to a pool like TTD. in the background TTD is receiving back all addresses and private keys. They could potentially have a bloom filter with a list of all addresses with a balance. checking every key against that. Basically making a cracking pool. I know the odds are still astronomical but it could be done. but would it ever be profitable? searching for more keys than just a single puzzle tx?
I am not condoning this btw, I would never steal, it's just a theory.
If they http://www.ttdsales.com/64bit/login.php are trying to solve #64 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN: It is possible Other addresses ~#256: It is impossible  |
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back then privkeys, had no value.
they were not important.
satoshi;s philosophy was if people lost their keys it made the coins left in circulation more scarce
he showed no desire, excitement or effort in hoarding. no mention of wanting/needing to accumulate. he preferred sharing coins for others to use for bug testing
as did others back then. making faucets to give coins away
hoarding/accumulating coin mindsets only begun in 2011
If it's the case why did Satoshi hoard over 1 million bitcoin? Those were mined before 2011. He did not hoard these coins. He had to mine them to let the network run. Without mining these, we wouldn't have Bitcoin today. If more people connected to the network and made the hashrate up, we wouldn't have these coins. How can we know this? He marked these coins so that we can separate them and analyze them. https://bitslog.com/2019/04/16/the-return-of-the-deniers-and-the-revenge-of-patoshi/https://bitcointalk.org/index.php?topic=507458.0That does not mean, that Satoshi isn't owning other coins than these marked. |
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my server has over 40TB of storage space and 3TB of ram, im also using the work split option on the kangaroo clients to keep the ram table low. each work file gets to about 3GB then saves to the harddrive freeing up the ram
That sounds good. But one should always know, that kangaroo is not deterministic. ... Anyway, I'm working on the merger to reduce memory consumption. I added the plot of probability of success:  Hi,
I did a small program to calculate chance of finding the key (without taking in consideration DP overhead) after a certain number or group operation.
Each value are the result of 100.000 trials so we can expect a precision of 0.3%.
0.5*sqrt(N) P=4.606 %
1.0*sqrt(N) P=17.162 %
1.5*sqrt(N) P=34.138 %
2.0*sqrt(N) P=52.314 %
2.5*sqrt(N) P=68.049 %
3.0*sqrt(N) P=80.350 %
3.5*sqrt(N) P=88.846 %
4.0*sqrt(N) P=94.103 %
4.5*sqrt(N) P=97.164 %
5.0*sqrt(N) P=98.746 %
5.5*sqrt(N) P=99.424 %
6.0*sqrt(N) P=99.752 %
I will increase accuracy and number of point and add a nice plot to the README of the Kangaroo program.
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I created a pool for Kangaroo, Puzzle 120(Full Range) DP 22, All kangaroos recorded by server so you can leave and rejoin at anytime.
With DP 22 you would need to store ~300,000,000,000 distinguished points to solve #120 and then comparing them. We (me and zielar) solved #115 after ~2^33.36 DP (DP25) (More than 300GB of DP). I do not know exactly how long the run takes due to unwanted interruption.  #120, DP22 -> ~10TB of distinguished points and then comparing them |
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I created a pool for Kangaroo, Puzzle 120(Full Range) DP 22, All kangaroos recorded by server so you can leave and rejoin at anytime.
With DP 22 you would need to store ~300,000,000,000 distinguished points to solve #120 and then comparing them. |
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Who knows why he left the mined coins from 2009/2010 untouched? It can only be speculation, but remember his motivation -- to create a peer to peer currency. He had to make it look like the new currency was being used, but also not look like an individual couldn't join. He couldn't look like he would hold the majority forever or no one would want to join. He could even have sent those coin off to the void to create the mystery, and drum interest. Perhaps he hid coin behind different riddles and tests to test the resilience of the network. Him mining made it look like there were more users than there were, before the extra nonce tagging him was discovered.
Not only the ExtraNonce. One can say he marked these mined coins intentionally. https://bitslog.com/2019/04/16/the-return-of-the-deniers-and-the-revenge-of-patoshi/ |
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This is a benchmark post with speeds for Kangaroo GPU Solver. All the tests were made with default DP and default grid size (calculated by a program). I guess that some plays with DP and grid size could change (increase or decrease) the speed. If somebody knows the optimal values, please let us know. Card Model Grid size DP Tested speed
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GTX 1050 Ti Grid(12x256) DP 16 115 MKey/sec
GTX 1080 Ti Grid(56x256) DP 15 500 MKey/sec
Tesla T4 16Gb Grid(80x128) DP 14 565 MKey/sec
RTX 2080ti 11Gb Grid(136x128) DP 13 1225 MKey/sec
Tesla V100 32Gb Grid(160x128) DP 13 1420 MKey/sec
---------------------------------------------------------------
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 mean that is x2
what you want is important what is correct location of x1, x2, x3
if you calc wrong, you will never reach at your target
It doesn't matter what your x1 is or x2. Someone will jump from a different point to eg. Point 2 in that example and in this case that point will be Point 1 with (x1, y1) and someone else will jump to eg. Point 6 in that example and in that case that point will be Point 1 with (x1, y1). But both will have as result the point x = 0x991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8 y = 0x14c3c6d1a538e95f34bf05f71d9e90b8ba6195e33f0d6dd7c97695e21a09ca63 as their reference point (the lowest values for x and y in the 6-Point-Group) and will go on with that point. Thereafter they will always have the same reference point and in the case of 'tame' and 'wild' it would lead to a solution. |
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So it works in the full range of 2^256 so what would be the expected operations?
- 1 point addition to have Point (x1, y1)
- 1 subtraction to get y2
- 2 multiplications to get x2 and x3
- comparisions to get lowest x and lowest y
Then you will have all x and y coordinates for all 6 points with the effort of less than 2 point additions, what will increase the speed enormously. |
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So it works in the full range of 2^256 so what would be the expected operations?
- 1 point addition to have Point (x1, y1)
- 1 subtraction to get y2
- 2 multiplications to get x2 and x3
- comparisions to get lowest x and lowest y
Then you will have all x and y coordinates for all 6 points with the effort of less than 2 point additions, what will increase the speed enormously. |
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Pollard's kangaroo / lambda / rho accelerator It will lead to inner loops, but all solvable. Profit: with one point addition, one will cover 6 points. When will you be done with that project? here is pubkey 02991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8 could you tell me first example if its x1 ? x2 ? x3 ? if its x1 then whats x2 and x3 print pubkeys , it will help to vistors for understand about x1 x2 x3 thankx Example: pubkey = 02991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8 This point becomes Point (x1, y1), but we don't know if it is Point 1, 2, 3, 4, 5 or 6. from our offline server: Point 1 (x1, y1)
x1 = 0x991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8
y1 = 0xeb3c392e5ac716a0cb40fa08e2616f47459e6a1cc0f2922836896a1ce5f631cc
Point 2 (x2, y2)
x2 = 0xa673e97568057fb5f41c35d6ed6c88ef97510d71222b3686ef892f4ccc2af536
y2 = 0xeb3c392e5ac716a0cb40fa08e2616f47459e6a1cc0f2922836896a1ce5f631cc
Point 3 (x3, y3)
x3 = 0xc06d5d9f69b4cb8d6f720d8f106b442956061673b01e9da1cb0886fe59dd2860
y3 = 0xeb3c392e5ac716a0cb40fa08e2616f47459e6a1cc0f2922836896a1ce5f631cc
Point 4 (x4, y4)
x4 = 0x991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8
y4 = 0x14c3c6d1a538e95f34bf05f71d9e90b8ba6195e33f0d6dd7c97695e21a09ca63
Point 5 (x5, y5)
x5 = 0xa673e97568057fb5f41c35d6ed6c88ef97510d71222b3686ef892f4ccc2af536
y5 = 0x14c3c6d1a538e95f34bf05f71d9e90b8ba6195e33f0d6dd7c97695e21a09ca63
Point 6 (x6, y6)
x6 = 0xc06d5d9f69b4cb8d6f720d8f106b442956061673b01e9da1cb0886fe59dd2860
y6 = 0x14c3c6d1a538e95f34bf05f71d9e90b8ba6195e33f0d6dd7c97695e21a09ca63
(Now we can say that the example point was Point 1, but that is not important.) Remember: x1 = x4 and x2 = x5 and x3 = x6 y1 = y2 = y3 and y4 = y5 = y6 Lowest x = x1 or x = x4 x = 0x991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8 Lowest y = y4 or y = y5 or y = y6 y = 0x14c3c6d1a538e95f34bf05f71d9e90b8ba6195e33f0d6dd7c97695e21a09ca63 That Point (x, y) would be the reference point to go on with. From that point you jump to another Point (x1, y1) according to your kangaroo / rho. It doesn't matter if you jumped to Point 1 or 2 or 3 or 4 or 5 or 6, your reference point would be that Point (x, y) in all cases. That makes kangaroo / rho faster. For example: A 'tame' that jumps to Point 2 will go on with Point 4. A 'wild' that jumps to Point 5 will also go on with Point 4 and we would have a solution. But this only works if you have the full Bitcoin range (1 ... n) like in this project and not in a range like the puzzle #120 (2^119 ... 2^120 - 1). |
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Pollard's kangaroo / lambda / rho accelerator It will lead to inner loops, but all solvable. Profit: with one point addition, one will cover 6 points. When will you be done with that project? here is pubkey 02991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8 could you tell me first example if its x1 ? x2 ? x3 ? if its x1 then whats x2 and x3 print pubkeys , it will help to vistors for understand about x1 x2 x3 thankx Example: pubkey = 02991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8 This point becomes Point (x1, y1), but we don't know if it is Point 1, 2, 3, 4, 5 or 6. from our offline server: Point 1 (x1, y1)
x1 = 0x991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8
y1 = 0xeb3c392e5ac716a0cb40fa08e2616f47459e6a1cc0f2922836896a1ce5f631cc
Point 2 (x2, y2)
x2 = 0xa673e97568057fb5f41c35d6ed6c88ef97510d71222b3686ef892f4ccc2af536
y2 = 0xeb3c392e5ac716a0cb40fa08e2616f47459e6a1cc0f2922836896a1ce5f631cc
Point 3 (x3, y3)
x3 = 0xc06d5d9f69b4cb8d6f720d8f106b442956061673b01e9da1cb0886fe59dd2860
y3 = 0xeb3c392e5ac716a0cb40fa08e2616f47459e6a1cc0f2922836896a1ce5f631cc
Point 4 (x4, y4)
x4 = 0x991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8
y4 = 0x14c3c6d1a538e95f34bf05f71d9e90b8ba6195e33f0d6dd7c97695e21a09ca63
Point 5 (x5, y5)
x5 = 0xa673e97568057fb5f41c35d6ed6c88ef97510d71222b3686ef892f4ccc2af536
y5 = 0x14c3c6d1a538e95f34bf05f71d9e90b8ba6195e33f0d6dd7c97695e21a09ca63
Point 6 (x6, y6)
x6 = 0xc06d5d9f69b4cb8d6f720d8f106b442956061673b01e9da1cb0886fe59dd2860
y6 = 0x14c3c6d1a538e95f34bf05f71d9e90b8ba6195e33f0d6dd7c97695e21a09ca63
(Now we can say that the example point was Point 1, but that is not important.) Remember: x1 = x4 and x2 = x5 and x3 = x6 y1 = y2 = y3 and y4 = y5 = y6 Lowest x = x1 or x = x4 x = 0x991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8 Lowest y = y4 or y = y5 or y = y6 y = 0x14c3c6d1a538e95f34bf05f71d9e90b8ba6195e33f0d6dd7c97695e21a09ca63 That Point (x, y) would be the reference point to go on with. From that point you jump to another Point (x1, y1) according to your kangaroo / rho. It doesn't matter if you jumped to Point 1 or 2 or 3 or 4 or 5 or 6, your reference point would be that Point (x, y) in all cases. That makes kangaroo / rho faster. For example: A 'tame' that jumps to Point 2 will go on with Point 4. A 'wild' that jumps to Point 5 will also go on with Point 4 and we would have a solution. But this only works if you have the full Bitcoin range (1 ... n) like in our project https://bitcointalk.org/index.php?topic=5347791.0 and not in a range like the puzzle #120 (2^119 ... 2^120 - 1). |
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When will you be done with that project?
It depends on the people who will join the project. First they have to understand it and think that it is possible. |
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Hi, is there a list with devices/cards and related jumps/sec for Pollard's kangaroo (point addition)?
Thanks
Here is something... Thanks to DaveF. Has anyone put together (or started to put together) a list of CPUs / Video Cards & the speed you can get out of them.
I know it's a newer project and Jean_Luc is working VERY VERY hard on it so getting accurate numbers is going to be a moving target. But for now all we can do is look through the thread and see who is running what to get a general idea.
So far I have pulled from this thread:
GPU: GPU #0 GeForce GTX 1080 Ti (28x128 cores) Grid(224x128)
914.418 MK/s (GPU 896.216 MK/s)
GPU: GPU #0 GeForce GTX 1050 Ti (6x128 cores) Grid(48x128)
220.180 MK/s (GPU 220.180 MK/s)
GPU: GPU #0 GeForce GT 520M (1x48 cores) Grid(8x128)
10.233 MK/s (GPU 7.026 MK/s)
GPU: GPU #0 GeForce RTX 2070 (36x64 cores) Grid(288x128)
1535.880 MK/s (GPU 1470.257 MK/s)
Added 30-April-2019
GPU: GPU #0 GeForce GTX 1060 3GB (9x128 cores) Grid(72x128)
321.929 MK/s (GPU 321.929 MK/s)
GPU: GPU #0 GeForce GTX 1080 (20x128 cores) Grid(160x128)
672.062 MK/s (GPU 672.062 MK/s)
Added 1-May-2019
GPU: GPU #0 Tesla V100-SXM2-16GB (80x64 cores) Grid(640x128)
GPU: GPU #3 Tesla V100-SXM2-16GB (80x64 cores) Grid(640x128)
GPU: GPU #2 Tesla V100-SXM2-16GB (80x64 cores) Grid(640x128)
GPU: GPU #1 Tesla V100-SXM2-16GB (80x64 cores) Grid(640x128)
7260.449 MK/s (GPU 7212.931 MK/s)
So 7260 / 4 = 1815 MK/s
GPU: GPU #0 GeForce GTX 750 (4x128 cores) Grid(32x128)
104.960 MK/s (GPU 94.405 MK/s) (2^32.12)
Added 3-May-2019
i7-7700K CPU Number of CPU thread: 8
22.092 MK/s (GPU 0.000 MK/s)
With -t 7
Number of CPU thread: 7
21.609 MK/s
Added 8-May-2019
EVGA RTX 2080 XC ULTRA
1427.967 MK/s (GPU 1424.946 MK/s)
Added 23-May-2019
GPU: GPU #0 GeForce GTX 1660 Ti
961.319 MK/s (GPU 961.319 MK/s)
GPU: GPU #0 GeForce RTX 2080 Ti (68x64 cores) Grid(544x128)
GPU: GPU #1 GeForce RTX 2080 Ti (68x64 cores) Grid(544x128)
5128.213 MK/s (GPU 5128.213 MK/s)
So 5128 / 2 = 2564 MK/s
Added 8-June-2019
GPU: GPU #0 GeForce GTX 960M (5x128 cores) Grid(40x128)
117.802 MK/s (GPU 117.802 MK/s)
Added 23-July-2019
GPU: GPU #0 GeForce GTX 1660 (22x64 cores) Grid(176x128)
839.061 MK/s (GPU 839.061 MK/s)
Added 25-July-2019
GPU: GPU #0 GeForce GTX 1650 (14x64 cores) Grid(112x128)
511.906 MK/s (GPU 511.906 MK/s) (2^36.97)
Added 21-Nov-2019
GPU: GPU #0 GeForce GTX 970 (13x128 cores) Grid(104x128)
360.322 MK/s (GPU 331.442 MK/s) (2^32.77)
Added 25-Nov-2019
GPU: GPU #0 GeForce GTX 980 (16x128 cores) Grid(128x128)
375.384 MK/s (GPU 375.384 MK/s)
GPU: GPU #0 GeForce RTX 2060 SUPER (34x64 cores) Grid(272x256)
[1361.71 Mkey/s][GPU 1361.71 Mkey/s]
GPU: GPU #0 GeForce RTX 2080 SUPER (48x64 cores) Grid(384x256)
[2001.52 Mkey/s][GPU 2001.52 Mkey/s]
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Pollard's kangaroo / lambda / rho accelerator for our project It leads to inner loops, but all solvable. Profit: with one point addition, one will cover 6 points. |
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