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Topic: Pubkey scaling/subtracting/other tips for reducing search time (Read 2089 times)

full member
Activity: 706
Merit: 111
" I got it down to 104 bits today, but with 32,000 pubkeys; better than the normal 2^16 normally required, but I can't figure out a way to shrink it down to one key... "

for 10 bit down = 1024 pubkeys
for 20 bit down = 1024*1024 = 1048576 pubkeys
for 30 bit down = 1024*1024*1024 = 1073741824 pubkeys

1048576 and 1073741824 pubkeys with each other addition and mutiplication will return you 260 pubkeys apear where 16 pubkeys sure inside 10 bit down from main pubkey
these 260 pubkeys again played for get 30 bit down for 1/720 pubkeys
now you can start to find with above tip



You can only do it by 10 bit, 20, bit, or 30 bit down?
jr. member
Activity: 48
Merit: 11
I'm not sure of the significance of these numbers so maybe a backstory on how you came about these numbers could be useful. Does it make the overall search time faster?  Huh

Those numbers are factors of secp256k1 modified order

n    = 115792089237316195423570985008687907852837564279074904382605163141518161494337 (prime number)
n-1 = 115792089237316195423570985008687907852837564279074904382605163141518161494336 (composite number)

factorisation of n-1 into primes: 2*2*2*2*2*2 * 3 * 149 * 631 * 107361793816595537 * 174723607534414371449 * 341948486974166000522343609283189

18051648 = 2*2*2*2*2*2 * 3 * 149 * 631
9025824   = 2*2*2*2*2 * 3 * 149 * 631
6017216   = 2*2*2*2*2*2 * 149 * 631

and so on

I don't see any special magic here. In fact, BTC's real private key can be divisible by any prime number and also can be prime by itself. Due to the fact that secp256k1 is a cyclic group, we cannot check whether the result of dividing an unknown number (public key) is an integer or a fraction (in real math, outside of cyclic group), so we cannot make any meaningful conclusions from the results obtained.


CAN YOU PLEASE TELL ME HOW YOU ARE SELECTING THOSE PRIME NUMBERS CORRECTLY.


CAN YOU PROVIDE ME SOME EXAMPLE 447,596


THANKS
are the divisors of the number n-1
n-1 = 115792089237316195423570985008687907852837564279074904382605163141518161494336
447 = 3 * 149
596 = 4 * 149
the number n-1 contains 448 divisors:
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    89876858673395180223533250879507423543395711228642065285744
    134815288010092770335299876319261135315093566842963097928616
    179753717346790360447066501759014847086791422457284130571488
    269630576020185540670599752638522270630187133685926195857232
    359507434693580720894133003518029694173582844914568261142976
    539261152040371081341199505277044541260374267371852391714464
    1078522304080742162682399010554089082520748534743704783428928
    6414488540731361226607730496888035255996436684289152125202372832747357
    12828977081462722453215460993776070511992873368578304250404745665494714
    19243465622194083679823191490664105767989310052867456375607118498242071
    25657954162925444906430921987552141023985746737156608500809491330989428
    38486931244388167359646382981328211535978620105734912751214236996484142
    51315908325850889812861843975104282047971493474313217001618982661978856
    76973862488776334719292765962656423071957240211469825502428473992968284
    102631816651701779625723687950208564095942986948626434003237965323957712
    153947724977552669438585531925312846143914480422939651004856947985936568
    205263633303403559251447375900417128191885973897252868006475930647915424
    307895449955105338877171063850625692287828960845879302009713895971873136
    410527266606807118502894751800834256383771947794505736012951861295830848
    615790899910210677754342127701251384575657921691758604019427791943746272
    955758792568972822764551844036317253143469065959083666655153552079356193
    1231581799820421355508684255402502769151315843383517208038855583887492544
    1911517585137945645529103688072634506286938131918167333310307104158712386
    2867276377706918468293655532108951759430407197877250999965460656238068579
    3823035170275891291058207376145269012573876263836334666620614208317424772
    4047542269201488933989477943536350246533751547786454991002697257463582267
    5734552755413836936587311064217903518860814395754501999930921312476137158
    7646070340551782582116414752290538025147752527672669333241228416634849544
    8095084538402977867978955887072700493067503095572909982005394514927164534
    11469105510827673873174622128435807037721628791509003999861842624952274316
    12142626807604466801968433830609050739601254643359364973008091772390746801
    15292140681103565164232829504581076050295505055345338666482456833269699088
    16190169076805955735957911774145400986135006191145819964010789029854329068
    22938211021655347746349244256871614075443257583018007999723685249904548632
    24285253615208933603936867661218101479202509286718729946016183544781493602
    30584281362207130328465659009162152100591010110690677332964913666539398176
    32380338153611911471915823548290801972270012382291639928021578059708658136
    45876422043310695492698488513743228150886515166036015999447370499809097264
    48570507230417867207873735322436202958405018573437459892032367089562987204
    61168562724414260656931318018324304201182020221381354665929827333078796352
    64760676307223822943831647096581603944540024764583279856043156119417316272
    91752844086621390985396977027486456301773030332072031998894740999618194528
    97141014460835734415747470644872405916810037146874919784064734179125974408
    129521352614447645887663294193163207889080049529166559712086312238834632544
    183505688173242781970793954054972912603546060664144063997789481999236389056
    194282028921671468831494941289744811833620074293749839568129468358251948816
    259042705228895291775326588386326415778160099058333119424172624477669265088
    388564057843342937662989882579489623667240148587499679136258936716503897632
    603083798111021851164432213586916186733528980620181793659401891362073757783
    777128115686685875325979765158979247334480297174999358272517873433007795264
    1206167596222043702328864427173832373467057961240363587318803782724147515566
    1809251394333065553493296640760748560200586941860545380978205674086221273349
    2412335192444087404657728854347664746934115922480727174637607565448295031132
    3618502788666131106986593281521497120401173883721090761956411348172442546698
    4824670384888174809315457708695329493868231844961454349275215130896590062264
    7237005577332262213973186563042994240802347767442181523912822696344885093396
    9649340769776349618630915417390658987736463689922908698550430261793180124528
    14474011154664524427946373126085988481604695534884363047825645392689770186792
    19298681539552699237261830834781317975472927379845817397100860523586360249056
    28948022309329048855892746252171976963209391069768726095651290785379540373584
    38597363079105398474523661669562635950945854759691634794201721047172720498112
    57896044618658097711785492504343953926418782139537452191302581570759080747168
    115792089237316195423570985008687907852837564279074904382605163141518161494336
newbie
Activity: 14
Merit: 0
I'm not sure of the significance of these numbers so maybe a backstory on how you came about these numbers could be useful. Does it make the overall search time faster?  Huh

Those numbers are factors of secp256k1 modified order

n    = 115792089237316195423570985008687907852837564279074904382605163141518161494337 (prime number)
n-1 = 115792089237316195423570985008687907852837564279074904382605163141518161494336 (composite number)

factorisation of n-1 into primes: 2*2*2*2*2*2 * 3 * 149 * 631 * 107361793816595537 * 174723607534414371449 * 341948486974166000522343609283189

18051648 = 2*2*2*2*2*2 * 3 * 149 * 631
9025824   = 2*2*2*2*2 * 3 * 149 * 631
6017216   = 2*2*2*2*2*2 * 149 * 631

and so on

I don't see any special magic here. In fact, BTC's real private key can be divisible by any prime number and also can be prime by itself. Due to the fact that secp256k1 is a cyclic group, we cannot check whether the result of dividing an unknown number (public key) is an integer or a fraction (in real math, outside of cyclic group), so we cannot make any meaningful conclusions from the results obtained.


CAN YOU PLEASE TELL ME HOW YOU ARE SELECTING THOSE PRIME NUMBERS CORRECTLY.


CAN YOU PROVIDE ME SOME EXAMPLE 447,596


THANKS
jr. member
Activity: 48
Merit: 11
from fastecdsa import curve
from fastecdsa.point import Point
import bit

G = curve.secp256k1.G
N = curve.secp256k1.q
DIV = '02AE3482B19E840288CC9B302AD9F5DC017AB796D3690CC8029017A8AF3503BE8E'
pubkey = '03ec0f4d728d248698a59d3a50a0469da06fdb8019700dfc5de9eae2dd93fc2bc8'

def pub2point(pub_hex):
    x = int(pub_hex[2:66], 16)
    if len(pub_hex) < 70:
        y = bit.format.x_to_y(x, int(pub_hex[:2], 16) % 2)
    else:
        y = int(pub_hex[66:], 16)
    return Point(x, y, curve=curve.secp256k1)


Q = pub2point(pubkey)
R = pub2point(DIV)
z= Q / R

print(z)


-----------------------------------------
>>>    z= Q / R
TypeError: unsupported operand type(s) for /: 'Point' and 'Point'




CAN ANYONE HELP,HOW TO CORRECT THIS

THANKS
I answer here
https://bitcointalksearch.org/topic/m.60649916
jr. member
Activity: 48
Merit: 11

z= Q / R



I think the operation is


z = Q * modInv(R, N) % N

1.  Modular Inverse R
2.  Q multiply Modular Inverse R
3.  Total Modular N
4.  the result is z

My understanding divided in the Elliptic Curve is not mean  10/2=5 is not divided by normal math
divided in Elliptic Curve = multiply the value with Inverse value  and the result is Modular with N

It is correct or not?
Did I understand wrong?


z = Q * modInv(R, N) % N

1.  Modular Inverse R
2.  Q multiply Modular Inverse R
3.  Total Modular N
4.  the result is z
member
Activity: 406
Merit: 45

z= Q / R



I think the operation is


z = Q * modInv(R, N) % N

1.  Modular Inverse R
2.  Q multiply Modular Inverse R
3.  Total Modular N
4.  the result is z

My understanding divided in the Elliptic Curve is not mean  10/2=5 is not divided by normal math
divided in Elliptic Curve = multiply the value with Inverse value  and the result is Modular with N

It is correct or not?
Did I understand wrong?
hero member
Activity: 501
Merit: 3855
TypeError: unsupported operand type(s) for /: 'Point' and 'Point'

CAN ANYONE HELP,HOW TO CORRECT THIS

THANKS

You cannot divide a 'Point' by another 'Point' for the same reason that you cannot multiply a 'Point' by another 'Point'.

See here: https://crypto.stackexchange.com/questions/88214/how-do-i-multiply-two-points-on-an-elliptic-curve
newbie
Activity: 14
Merit: 0
from fastecdsa import curve
from fastecdsa.point import Point
import bit

G = curve.secp256k1.G
N = curve.secp256k1.q
DIV = '02AE3482B19E840288CC9B302AD9F5DC017AB796D3690CC8029017A8AF3503BE8E'
pubkey = '03ec0f4d728d248698a59d3a50a0469da06fdb8019700dfc5de9eae2dd93fc2bc8'

def pub2point(pub_hex):
    x = int(pub_hex[2:66], 16)
    if len(pub_hex) < 70:
        y = bit.format.x_to_y(x, int(pub_hex[:2], 16) % 2)
    else:
        y = int(pub_hex[66:], 16)
    return Point(x, y, curve=curve.secp256k1)


Q = pub2point(pubkey)
R = pub2point(DIV)
z= Q / R

print(z)


-----------------------------------------
>>>    z= Q / R
TypeError: unsupported operand type(s) for /: 'Point' and 'Point'




CAN ANYONE HELP,HOW TO CORRECT THIS

THANKS
member
Activity: 406
Merit: 45
I just follow this topic about subtracting

how can I use subtracting to find some point that adds a point then the result is pubkey 120?
subtracting is not real subtracting like   9-1=8  and I can add 8+1=9  like this right
this subtracting will be subtracted to another point right
So, pubkey can not do reverse roll back one step before right?

legendary
Activity: 1568
Merit: 6660
bitcoincleanup.com / bitmixlist.org
dec = 104856515000339101452906010972016177983340459646873053037069757574992766929113

hex = E7D2AF2FC9FF9CAF795D2EED256567679873FC547A513AA85A8A2E2776AB88D9

pubkey = 038141a3381c97660163ce69acf22d5a0cc8c09fcbb624fa556ff17629b4b31918

can any one know how he found this.

thanks

Well seeing I'm the one who made this thread, they appear to be multiplying the privkey with the pubkey to get a different pubkey/address:

G*p*p = (G*p)*p = P*p, not that this helps with computing any private key. I imagine it would be more useful in generating vanity addresses , provided that two different values are used, not just two p's.
newbie
Activity: 14
Merit: 0
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 ?  Roll Eyes



dec = 104856515000339101452906010972016177983340459646873053037069757574992766929113

hex = E7D2AF2FC9FF9CAF795D2EED256567679873FC547A513AA85A8A2E2776AB88D9

pubkey = 038141a3381c97660163ce69acf22d5a0cc8c09fcbb624fa556ff17629b4b31918

can any one know how he found this.

thanks
jr. member
Activity: 70
Merit: 1
this tool write subtract in c language https://github.com/WanderingPhilosopher/Windows-KeySubtractor/releases/tag/v1.0


yes i know this git code
i am not understand c language so,
                         
input two point
privatekey    3     -    2      =   1
                 point1 - point2 = point3
                                               X                                                                                                                         Y
point1 f9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9      388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672
point2 c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5   1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a


i need 3rd point subtract value

please write easy understand make python or explain mathematicaly
full member
Activity: 706
Merit: 111


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.

Whatever happened to this?
member
Activity: 406
Merit: 45
This thread subtracting stop to talk meaning it is not works right?
if it is possible work can possible explain to new arrival to know again
thread is continue from talk on kangaroo thread page 8x -9x
seem need to understand knowledge  about bit and ECDLP to understand right?
member
Activity: 808
Merit: 20
$$P2P BTC BRUTE.JOIN NOW ! https://uclck.me/SQPJk


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.



Hello.

Maybe is more simple get lovest or highest point from yours screen, and substract her from pubkey and brute after ?

Regard.
copper member
Activity: 75
Merit: 11


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.
member
Activity: 808
Merit: 20
$$P2P BTC BRUTE.JOIN NOW ! https://uclck.me/SQPJk
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



Bro, what exact you mean ?

Quote
Compressed
Address
15wJjXvfQzo3SXqoWGbWZmNYND1Si4siqV
Public Key (hex)
020000000000000000000000000000000000000000000000000000000000000000
P


040000000000000000000000000000000000000000000000000000000000000000218f8534c5bf3 c23b4c1c32b6295484c1473ce4bd628ae258acbb955a0fcb75e 1FPjjJiE8XN6uG2NUmkWK4Au3uSnyKGr5p


If for example any public P(x,Y) key without transaction and without balance, but another point P1(0(zero),Y with ballance and transaction it miant what P(x,Y) on the curve ?  and any other P what have no transactions and no ballance not on the curve and if use this points success will be zero ?
member
Activity: 310
Merit: 34
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
full member
Activity: 204
Merit: 437
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, bjy) = cidj*(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.
rp = 77643668876891235360856744073230947502707792537156648322526682022085734511405
rn = 106331823171076060141872636901030920105366729272408102113527681246281393517969
a = (rp(p-1)/3)2 = 55594575648329892869085402983802832744385952214688224221778511981742606582254
b = rp(p-1)/2 = 115792089237316195423570985008687907853269984665640564039457584007908834671662 = -1
c = rn(n-1)/3 = 37718080363155996902926221483475020450927657555482586988616620542887997980018
d = rn(n-1)/2 = 115792089237316195423570985008687907852837564279074904382605163141518161494336 = -1
full member
Activity: 706
Merit: 111
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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