Pubkey scaling/subtracting/other tips for reducing search time

[wedom]

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/58B7ccr

I 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

[1715105711]

1715105711

[1715105711]

1715105711

[COBRAS]

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/58B7ccr

I 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 (somethin like KNOWNx64bytePrivkey*G(originalBP)+x1*G2(not originalBP)+x2*G4(not originalBP)=OURTARGET)?


Regard.

[WanderingPhilospher]

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/58B7ccr

I 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!

[COBRAS]

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/58B7ccr

I 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 ?

[WanderingPhilospher]

How to find something with known privkey for G(base point of secp256k1) for adding after to our calculus ?

What are you talking about?

[COBRAS]

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 !!!

[WanderingPhilospher]

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 COBRAS


so 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.

[brainless]

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

=

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

[BitcoinADAB]

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.

[COBRAS]

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

=

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 !!!

[bigvito19]

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?

[BitcoinADAB]

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.

[brainless]

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

[COBRAS]

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.

I think this method  in Jacobian coordinates and endomorphism  will be more good.


https://paulmillr.com/posts/noble-secp256k1-fast-ecc/#unsafe-multiplication-for-key-recovery

[BitcoinADAB]

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:

Code:

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).

[bigvito19]

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:

Code:

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?

[BitcoinADAB]

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.

[bigvito19]

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?

[j2002ba2]

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
(aⁱx, b^jy) = cⁱd^j*(k*G)
with
a³ = 1 mod p (matching the chosen value of c)
b² = 1 mod p
c³ = 1 mod n (matching the chosen value of a)
d² = 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

[brainless]

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