COBRAS
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July 28, 2021, 05:45:10 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
Bro, what exact you mean ? 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 ?
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BitcoinADAB
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July 28, 2021, 10:42:21 AM |
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![](https://ip.bitcointalk.org/?u=https%3A%2F%2Flh3.googleusercontent.com%2FwNM-L8bDMPQZvgYpIG6TRDp_Vjn3oxMEuH3vQ_3QzTHC_ZnT1GwWgkBKLSWYuSc4RW1gqDkz3NhVbqc55YYp_AsAbTFSheOxLr_3%3Ds0&t=665&c=58lJZ7NkJLhl3A) 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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COBRAS
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July 28, 2021, 12:00:15 PM |
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![](https://ip.bitcointalk.org/?u=https%3A%2F%2Flh3.googleusercontent.com%2FwNM-L8bDMPQZvgYpIG6TRDp_Vjn3oxMEuH3vQ_3QzTHC_ZnT1GwWgkBKLSWYuSc4RW1gqDkz3NhVbqc55YYp_AsAbTFSheOxLr_3%3Ds0&t=665&c=58lJZ7NkJLhl3A) 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.
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fxsniper
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November 20, 2021, 02:12:13 AM |
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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?
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bigvito19
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November 20, 2021, 12:59:06 PM |
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![](https://ip.bitcointalk.org/?u=https%3A%2F%2Flh3.googleusercontent.com%2FwNM-L8bDMPQZvgYpIG6TRDp_Vjn3oxMEuH3vQ_3QzTHC_ZnT1GwWgkBKLSWYuSc4RW1gqDkz3NhVbqc55YYp_AsAbTFSheOxLr_3%3Ds0&t=665&c=58lJZ7NkJLhl3A) 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?
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mausuv
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November 20, 2021, 03:45:28 PM |
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this tool write subtract in c language https://github.com/WanderingPhilosopher/Windows-KeySubtractor/releases/tag/v1.0yes 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
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ajeev
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July 28, 2022, 04:36:09 AM |
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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
=
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](https://bitcointalk.org/Smileys/default/rolleyes.gif) dec = 104856515000339101452906010972016177983340459646873053037069757574992766929113 hex = E7D2AF2FC9FF9CAF795D2EED256567679873FC547A513AA85A8A2E2776AB88D9 pubkey = 038141a3381c97660163ce69acf22d5a0cc8c09fcbb624fa556ff17629b4b31918 can any one know how he found this. thanks
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NotATether (OP)
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July 28, 2022, 07:03:19 AM |
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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.
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fxsniper
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July 29, 2022, 10:19:29 AM |
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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?
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ajeev
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July 29, 2022, 01:21:07 PM |
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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
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fxsniper
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July 30, 2022, 05:49:07 AM |
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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?
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wedom
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July 30, 2022, 06:59:32 AM |
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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) % N1. Modular Inverse R 2. Q multiply Modular Inverse R 3. Total Modular N4. the result is z
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wedom
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July 30, 2022, 07:02:11 AM |
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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://bitcointalk.org/index.php?topic=5244940.msg60649916#msg60649916
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ajeev
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August 03, 2022, 05:34:38 AM |
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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](https://bitcointalk.org/Smileys/default/huh.gif) 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
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wedom
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August 03, 2022, 12:26:56 PM |
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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](https://bitcointalk.org/Smileys/default/huh.gif) 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: 1 2 3 4 6 8 12 16 24 32 48 64 96 149 192 298 447 596 631 894 1192 1262 1788 1893 2384 2524 3576 3786 4768 5048 7152 7572 9536 10096 14304 15144 20192 28608 30288 40384 60576 94019 121152 188038 282057 376076 564114 752152 1128228 1504304 2256456 3008608 4512912 6017216 9025824 18051648 107361793816595537 214723587633191074 322085381449786611 429447175266382148 644170762899573222 858894350532764296 1288341525799146444 1717788701065528592 2576683051598292888 3435577402131057184 5153366103196585776 6871154804262114368 10306732206393171552 15996907278672735013 20613464412786343104 31993814557345470026 47990721836018205039 63987629114690940052 67745291898271783847 95981443672036410078 127975258229381880104 135490583796543567694 174723607534414371449 191962887344072820156 203235875694815351541 255950516458763760208 270981167593087135388 349447215068828742898 383925774688145640312 406471751389630703082 511901032917527520416 524170822603243114347 541962335186174270776 698894430137657485796 767851549376291280624 812943502779261406164 1023802065835055040832 1048341645206486228694 1083924670372348541552 1397788860275314971592 1535703098752582561248 1625887005558522812328 2096683290412972457388 2167849340744697083104 2795577720550629943184 3071406197505165122496 3251774011117045624656 4193366580825944914776 4335698681489394166208 5591155441101259886368 6503548022234091249312 8386733161651889829552 10094048492842495793203 11182310882202519772736 13007096044468182498624 16773466323303779659104 20188096985684991586406 26033817522627741345901 30282145478527487379609 33546932646607559318208 40376193971369983172812 52067635045255482691802 60564290957054974759218 78101452567883224037703 80752387942739966345624 104135270090510965383604 110250596354215468384319 121128581914109949518436 156202905135766448075406 161504775885479932691248 208270540181021930767208 220501192708430936768638 242257163828219899036872 312405810271532896150812 323009551770959865382496 330751789062646405152957 416541080362043861534416 441002385416861873537276 484514327656439798073744 624811620543065792301624 646019103541919730764992 661503578125292810305914 833082160724087723068832 882004770833723747074552 969028655312879596147488 1249623241086131584603248 1323007156250585620611828 1666164321448175446137664 1764009541667447494149104 1938057310625759192294976 2499246482172263169206496 2646014312501171241223656 3528019083334894988298208 4998492964344526338412992 5292028625002342482447312 7056038166669789976596416 10584057250004684964894624 16427338856778104789263531 21168114500009369929789248 32854677713556209578527062 49282016570334314367790593 65709355427112419157054124 98564033140668628735581186 131418710854224838314108248 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