wedom
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July 23, 2021, 03:25:02 PM |
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0XFC07A1825367BBE * inv(0xdb09b0615ad40a0) Interesting constraction ! I now will try to understand why is worked ))) Respect. if a*b=p, then b = p/a = p * a^-1 This is correct, but this 038141a3381c97660163ce69acf22d5a0cc8c09fcbb624fa556ff17629b4b31918 not privkey, this is publick key.Look: https://ibb.co/58B7ccrI did not write that this is a private key. In the formula, you can see that this is exactly the public key. I just pointed out that for this formula the equality A*B <> C, if B = 038141a3381c97660163ce69acf22d5a0cc8c09fcbb624fa556ff17629b4b31918
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COBRAS
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July 23, 2021, 03:40:00 PM Last edit: July 23, 2021, 04:43:38 PM by COBRAS |
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0XFC07A1825367BBE * inv(0xdb09b0615ad40a0) Interesting constraction ! I now will try to understand why is worked ))) Respect. if a*b=p, then b = p/a = p * a^-1 BIG thank you !!!This is correct, but this 038141a3381c97660163ce69acf22d5a0cc8c09fcbb624fa556ff17629b4b31918 not privkey, this is publick key.Look: https://ibb.co/58B7ccrI did not write that this is a private key. In the formula, you can see that this is exactly the public key. I just pointed out that for this formula the equality A*B <> C, if B = 038141a3381c97660163ce69acf22d5a0cc8c09fcbb624fa556ff17629b4b31918 How to find something with known privkey for G(base point of secp256k1) for adding after to our calculus (s omethin like KNOWNx64bytePrivkey*G(originalBP)+x1*G2(not originalBP)+x2*G4(not originalBP)=OURTARGET)? Regard.
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WanderingPhilospher
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Shooters Shoot...
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July 23, 2021, 03:47:49 PM |
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0XFC07A1825367BBE * inv(0xdb09b0615ad40a0) Interesting constraction ! I now will try to understand why is worked ))) Respect. if a*b=p, then b = p/a = p * a^-1 BIG thank you !!!This is correct, but this 038141a3381c97660163ce69acf22d5a0cc8c09fcbb624fa556ff17629b4b31918 not privkey, this is publick key.Look: https://ibb.co/58B7ccrI did not write that this is a private key. In the formula, you can see that this is exactly the public key. I just pointed out that for this formula the equality A*B <> C, if B = 038141a3381c97660163ce69acf22d5a0cc8c09fcbb624fa556ff17629b4b31918 How to find something with known privkey for G(base point of secp256k1) for adding after to our calculus ? Regard. He was telling you that your formula is incorrect. See man, you can't keep posting and asking people to help find x y and z when you really do not understand the math and formulas behind the curve... Finding something with known privkey is easy...you already know the privkey!
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COBRAS
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July 23, 2021, 03:56:46 PM |
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0XFC07A1825367BBE * inv(0xdb09b0615ad40a0) Interesting constraction ! I now will try to understand why is worked ))) Respect. if a*b=p, then b = p/a = p * a^-1 BIG thank you !!!This is correct, but this 038141a3381c97660163ce69acf22d5a0cc8c09fcbb624fa556ff17629b4b31918 not privkey, this is publick key.Look: https://ibb.co/58B7ccrI did not write that this is a private key. In the formula, you can see that this is exactly the public key. I just pointed out that for this formula the equality A*B <> C, if B = 038141a3381c97660163ce69acf22d5a0cc8c09fcbb624fa556ff17629b4b31918 How to find something with known privkey for G(base point of secp256k1) for adding after to our calculus ? Regard. He was telling you that your formula is incorrect. See man, you can't keep posting and asking people to help find x y and z when you really do not understand the math and formulas behind the curve... Finding something with known privkey is easy...you already know the privkey! you already know the privkey!
What are you talk about ?
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WanderingPhilospher
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Shooters Shoot...
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July 23, 2021, 04:03:17 PM |
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How to find something with known privkey for G(base point of secp256k1) for adding after to our calculus ? What are you talking about?
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COBRAS
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July 23, 2021, 04:04:05 PM Last edit: July 23, 2021, 04:16:18 PM by COBRAS |
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Wedom, How to find something with known privkey for G(original base point of secp256k1) for adding after to our calculus (for exaple, something like KNOWNx64bytePrivkey*G(originalBP)+x1*G2(not originalBP)+x2*G4(not originalBP)=OURTARGET)?
What you think about this ?
Regard !!!
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WanderingPhilospher
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July 23, 2021, 04:46:14 PM |
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Wedom, How to find something with known privkey for G(original base point of secp256k1) for adding after to our calculus (for exaple, something like KNOWNx64bytePrivkey*G(originalBP)+x1*G2(not originalBP)+x2*G4(not originalBP)=OURTARGET)?
What you think about this ?
Regard !!!
Still doesn't make sense COBRASso you can take any 64BIT (not byte) known private key and add or multiply G to it and you are only adding 1s. G = 0x1 Known 64bit privkey = 0x8234567890abcdf0 0x8234567890abcdf0 + G = 0x8234567890abcdf1 0x8234567890abcdf0 + G + G = 0x8234567890abcdf2 0x8234567890abcdf0 + G + G + G + G = 0x8234567890abcdf4 Now imagine if you are trying to find YOURTARGET that is in a 2^120 range...how many additions (multiplications) do you have to do?! So again, what are you trying to do? You always want to be shady/sneaky without telling people what you are really trying to do. You give zero examples at lower bit ranges so people can follow what the heck you are talking about.
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brainless
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July 26, 2021, 03:21:36 PM |
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you just explained that it has already been spend found. whe testing that key or what, what addition to add if found to get the right pvk. address. thanks
This is another good example with privrange ( ((2^60) - 986458768829923488 ) pubkey: 0x db09b0615ad40a0 * 038141a3381c97660163ce69acf22d5a0cc8c09fcbb624fa556ff17629b4b31918
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0348e843dc5b1bd246e6309b4924b81543d02b16c8083df973a89ce2c7eb89a10d(this is has a range 2^60 (address - 1Kn5h2qpgw9mWE5jKpk8PP4qvvJ1QVy8su))This is a part of privkey - 986458768829923488 in hex db09b0615ad40a0.No one can't find a privkey of 038141a3381c97660163ce69acf22d5a0cc8c09fcbb624fa556ff17629b4b31918 ? dec = 104856515000339101452906010972016177983340459646873053037069757574992766929113 hex = E7D2AF2FC9FF9CAF795D2EED256567679873FC547A513AA85A8A2E2776AB88D9 pubkey = 038141a3381c97660163ce69acf22d5a0cc8c09fcbb624fa556ff17629b4b31918
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13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF
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BitcoinADAB
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July 26, 2021, 08:28:29 PM Last edit: July 26, 2021, 08:56:49 PM by BitcoinADAB |
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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.
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COBRAS
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July 26, 2021, 08:31:25 PM Last edit: July 26, 2021, 08:57:02 PM by COBRAS |
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you just explained that it has already been spend found. whe testing that key or what, what addition to add if found to get the right pvk. address. thanks
This is another good example with privrange ( ((2^60) - 986458768829923488 ) pubkey: 0x db09b0615ad40a0 * 038141a3381c97660163ce69acf22d5a0cc8c09fcbb624fa556ff17629b4b31918
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0348e843dc5b1bd246e6309b4924b81543d02b16c8083df973a89ce2c7eb89a10d(this is has a range 2^60 (address - 1Kn5h2qpgw9mWE5jKpk8PP4qvvJ1QVy8su))This is a part of privkey - 986458768829923488 in hex db09b0615ad40a0.No one can't find a privkey of 038141a3381c97660163ce69acf22d5a0cc8c09fcbb624fa556ff17629b4b31918 ? dec = 104856515000339101452906010972016177983340459646873053037069757574992766929113 hex = E7D2AF2FC9FF9CAF795D2EED256567679873FC547A513AA85A8A2E2776AB88D9 pubkey = 038141a3381c97660163ce69acf22d5a0cc8c09fcbb624fa556ff17629b4b31918 Bro !!! It worked !!!!! E7D2AF2FC9FF9CAF795D2EED256567679873FC547A513AA85A8A2E2776AB88D9 * db09b0615ad40a0 babahtBamBamBamtratTatatattaTam ))) - fc07a1825367bbe p.s. I was send you 1 merit for "proof of work" Big thank you Brainless !!!
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bigvito19
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July 27, 2021, 12:07:44 PM |
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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?
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BitcoinADAB
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July 27, 2021, 05:32:30 PM |
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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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brainless
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July 27, 2021, 06:41:18 PM |
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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
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13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF
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BitcoinADAB
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July 27, 2021, 09:17:24 PM Last edit: July 27, 2021, 10:20:36 PM by BitcoinADAB |
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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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bigvito19
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July 27, 2021, 11:00:10 PM |
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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). So it works in the full range of 2^256 so what would be the expected operations?
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BitcoinADAB
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July 28, 2021, 12:21:46 AM |
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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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bigvito19
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July 28, 2021, 01:45:41 AM |
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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. Expected operations in time as in how long would it take to solve a key?
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j2002ba2
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July 28, 2021, 02:04:53 AM |
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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. - 1 point addition to have Point (x1, y1)
- 2 subtractions to get y2 and the corresponding private key
- 4 multiplications to get x2 and x3, and their corresponding private keys
- 3 comparisions to get lowest x and lowest y
The speedup works only on Pollard Rho, at most sqrt(6) = 2.44 times. For Kangaroo only the negation (y) is applicable, with speedup at most 1.41 times (and bigger variance?) - AFAIK Jean-Luc uses it already. All this is well known: if we have a point (x,y) = k*G, the 6 points are (a ix, b jy) = c id j*(k*G) with a 3 = 1 mod p (matching the chosen value of c) b 2 = 1 mod p c 3 = 1 mod n (matching the chosen value of a) d 2 = 1 mod n i∈{0,1,2} j∈{0,1}. One can calculate the numbers by finding the primitive roots mod p and n I.E. r p = 77643668876891235360856744073230947502707792537156648322526682022085734511405 r n = 106331823171076060141872636901030920105366729272408102113527681246281393517969 a = (r p(p-1)/3) 2 = 55594575648329892869085402983802832744385952214688224221778511981742606582254 b = r p(p-1)/2 = 115792089237316195423570985008687907853269984665640564039457584007908834671662 = -1 c = r n(n-1)/3 = 37718080363155996902926221483475020450927657555482586988616620542887997980018 d = r n(n-1)/2 = 115792089237316195423570985008687907852837564279074904382605163141518161494336 = -1
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brainless
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July 28, 2021, 04:06:33 AM |
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in short when you create x1, x2, x3 pubkey will be break in raw, and back to reconstruct, and you will get proper series of x1, x2, x3 above my pubkey proper x1, x2, x3 is here x1 = a673e97568057fb5f41c35d6ed6c88ef97510d71222b3686ef892f4ccc2af536 x2 = 991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8 x3 = c06d5d9f69b4cb8d6f720d8f106b442956061673b01e9da1cb0886fe59dd2860
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
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13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF
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