paniker this is meaning that you can change the n's
modular elliptic curve
Total of all the wallets n is the last number. n= 115792089237316195423570985008687907852837564279074904382605163141518161494337 (In Dec)
n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141 (In HEX)
Half way of n n//2 = 57896044618658097711785492504343953926418782139537452191302581570759080747169
57896044618658097711785492504343953926418782139537452191302581570759080747169 Lenght Bits = 255
very nice and thanks to boris.. you still here..
yes I am still here, this was the only thing I have found so far. Half way of n n//2 is wrong, check in above posts, mention clearly formula for div thankx |
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How come no one has attempted to put a bloom filter on kangaroo. Kangaroo hasn't been updated of modified.
JeanLucPons commented on May 7, 2020
Yes, I added load/save/work
The multi key will be done with the help of theses new features.
The idea is to allow to pre compute large tame kangaroo file and to solve multiple key using this file.
Then when a key is solved, all the wild will become tame and more keys are solved more chance to solve the others....
JeanLucPons commented on May 8, 2020
Yes,
I will add some note about this on the readme. It is a bit tricky.
Multi key support is not yet supported, for this you will need first to create a large tame array for a given range and then attack keys with it.
maybe he created and testing, but not for public, as last found puzzle 115 by joint vent.. and those version were not publicly available... |
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start the scrypt with double click .. simply to use in Python... You need: faster "bit" there is a library from "ice" https://github.com/iceland2k14/secp256k1 there it is necessary to throw its libraries into the folder with the script. donate BTC: 1DonateZNR9BUaCqJTgXCoyyCpRSosFujRyou dont have GPU ? any GPU ? This code is only for CPU, working with two Threads... Sure, I use GPU too, but I don't can coding for Cuda applications like, vanitygen, Kangaroo, etc... i will sugggest same working your code at c and gpu, above link use git scripts, working 10005 better then you python code , ttoal same function, even if your cpu have gpu inside, like i3-6100 process have gpu, its also work on this rotar, and auto use gpu at cpu based command, https://github.com/phrutis/Rotor-Cuda, for fully random at gpu like bitcrack is BitCrack2, use that, you will forget python dont forget merit me....  |
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start the scrypt with double click .. simply to use in Python... You need: faster "bit" there is a library from "ice" https://github.com/iceland2k14/secp256k1 there it is necessary to throw its libraries into the folder with the script. donate BTC: 1DonateZNR9BUaCqJTgXCoyyCpRSosFujRyou dont have GPU ? any GPU ? |
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Please write mi in words how many years it is:
1.055e+06y
The simpler method is to just move the decimal point depending on the value of " e+number". From your example: | 1.055 | (e+06 = move by 6 decimal places) | | 10.55 | = move by 1 place | | 105.5 | = move by 2 places | | 1055. | = move by 3 places | | 10550. | = move by 4 places | | 105500. | = move by 5 places | | 1055000. | = move by 6 places |
Other note: You can use your forum signature for off-topic messages. You can access it in: " Profile->Forum Profile Information". Thanks!  Am I correct with 6.02311e+07y?6 .02311 (e+07 = move by 7 decimal places) 60 .2311 = move by 1 place 602 .311 = move by 2 places 6023 .11 = move by 3 places 60231 .1 = move by 4 places 602311 . = move by 5 places 6023110 . = move by 6 places 60231100 . = move by 7 places use big calc https://www.calculator.net/big-number-calculator.html |
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Which version are you running? It seems salvagewallet was removed from 0.21.0, so if you're running a version pre 0.21.0, you could try the initial command given by BitMaxz.
EDIT: i just logged on to my system, and had a look for myself (-help). Seems like the command is:
./bitcoin-wallet salvage --"wallet.dat"
At least, that how it works for me..
On windows, that would make
bitcoin-wallet.exe salvage --"c:\\path to\\wallet.dat"
(at least, i think it'll need double slashes)
So basically, if you're running >=0.21.0, you could try to switch the salvage command and the path...
i do I:\Bitcoin\bitcoin-qt.exe -salvagewallet I:\bit\wallets\wallet.dat and I:\Bitcoin\bitcoin-qt.exe -salvagewallet I:\\bit\\wallets\\wallet.dat the same effect. Flashes for a second. But nothing starts and nothing happens -salvage does not work in my version at all. writes that such command is unknown I really hope that my key that I received after processing the information from the first post by the script will somehow help "YmM63dDbt...+....." Did you look at the command i posted vs your command? executable -salvagewallet --"path" is not the same as executable salvage --"path" But first things first: which bitcoin core version are you using? 0.20.1 when you finish your last option for bad db inside wallet.dat recovery, then can try to give me dat file, i will recover db inside dat file manually, and make sure no demand from me, if recovered, will fw back to you as it is |
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lets guess P1 is Even P2 is Odd post your guess and lets get answer in private key from creator it was first what i checked. It is not correlated ( PK is 96, but i try to test how much participant , no one interested, as they loose hopes, maximum newbie apear and start trying brute force or kanagroo, vantisearch etc Enjoy your time |
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In reality, neither coordinate is negative. Both y coordinates are points on the secp256k1 curve, which is defined modulo p, and so there are no negative coordinates, as anything negative simply loops round mod p and becomes positive again. And so we use even and odd to differentiate the two y coordinates.
lets guess P1 is Even P2 is Odd post your guess and lets get answer in private key from creator |
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sorry to say, link was https://github.com/AdamISZ/ConfidentialTransactionsDoc/blob/master/essayonCT.texremoving part of these from my post btw my monkey brain was thinking, first thread created about scrypt buy/sell, some time later that show was ended then all start discusion at " Let test my scrypt for find a privkey " and actual title should be " Let test my scrypt for generate R S Z " similar... and i see that show was ended too, could you please update last post whats new thread for new show, that will help us to come and enjoy those next shows, Thankx |
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30240 came from astro
360 days in year
7 day in week
12 month in year
360 x 7 x 12 = 30240
Full ver
2520
360 x 7 = 2520
For complete numerology 30240 will work
2520 will make u stuck in some part of calc
2 447 7572 564114 3 596 9536 752152 4 631 10096 1128228 6 894 14304 1504304 8 1192 15144 2256456 12 1262 20192 3008608 16 1788 28608 4512912 24 1893 30288 6017216 32 2384 40384 9025824 48 2524 60576 18051648 64 3576 94019 96 3786 121152 149 4768 188038 192 5048 282057 298 7152 376076 30288 what other guys call 30240 as i call this shit a 30288WTF ?  dude you are going against your magic numbers , somebody please hand over me tomato ketchup i will assume it soda and drink it today where the hell i left my .75 btc? sorry i forgot  for kids in math, first learn word and meaning 1. ecc secp256k1 number = 1bit to 256bit 2. numberology = 0 to infinite i post both side magic numbers, , unfortunatly you all unable to understand, further no details from my side, extreme knowledge for your hightech brains brainless posted, definatly i am thousands year back living in A.D, as i dont know 360 or 365 in a year, here are muslim , christ, and jews, chinese all have there own year calender, and can better understand why 360 Enjoy your time |
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30240 came from astro
360 days in year
7 day in week
12 month in year
360 x 7 x 12 = 30240
Full ver
2520
360 x 7 = 2520
For complete numerology 30240 will work
2520 will make u stuck in some part of calc
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Giving you all one more tip , in total numerology, only 30240 is is dividable from 1 to 10, mean 5 even 5 odd, at same time, and no floating result i gurantee, you never see 30240 secrets 30240.0000 10.0000 3024.0000 30240.0000 9.0000 3360.0000 30240.0000 8.0000 3780.0000 30240.0000 7.0000 4320.0000 30240.0000 6.0000 5040.0000 30240.0000 5.0000 6048.0000 30240.0000 4.0000 7560.0000 30240.0000 3.0000 10080.0000 30240.0000 2.0000 15120.0000 30240.0000 1.0000 30240.0000 if you multuply 30240 to any numbers, and result could also div by 1 to 10, and in result no floating point 30240 * 777 = 23496480.0000 23496480.0000 10.0000 2349648.0000 23496480.0000 9.0000 2610720.0000 23496480.0000 8.0000 2937060.0000 23496480.0000 7.0000 3356640.0000 23496480.0000 6.0000 3916080.0000 23496480.0000 5.0000 4699296.0000 23496480.0000 4.0000 5874120.0000 23496480.0000 3.0000 7832160.0000 23496480.0000 2.0000 11748240.0000 23496480.0000 1.0000 23496480.0000 hope like in my preivios post for magic numbers, you could not understand, same hope you will not understand secrets behind this only 1 magic numberology |
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How about I give you a 120 bit public key and you give me exactly one 115 bit public key.
read back my post i need for process that required 3090 gpu's, i dont have, for base on cpu its required multiple days, and i am systems are busy already to process some other projects btw for all you 1 Q, do you know those digits which can be divide without floating point in results from 1 to 10 thos digit multiplywhat ever, result also could be div without float from 1 to 10 remember i am brainless but my talks are not useless, in past i post magic numbers for division and multiply you all never make any tests on use of those magic numbers, and talking like higly math experts about div/mul above all my post have lot of clue for your new thinking levels, but i see you all still bound in basics waiting your research for those numbers who div from 1 to 10 without float in results |
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guys big lol here plus some explanation from my side. the script shared by NotATether is just doing opposite of divisor it is increasing 5 bit and giving you last key 5 bit uper range. only difference is while doing divisor you will get 1 key from down 5 bit and other all from defined uper bits range on exact same distance from reference value of this formula N=115792089237316195423570985008687907852837564279074904382605163141518161494337
for i in range(32):
print(i, hex((((N-1)//32)*i+1)%N))
lets say we have key 10 and we do divisor of 10 so what we get is like this (just example) upper reference range 210 , 200 , 190 , 180 .................. until 10 assume after divisor each key will have 5 bit down from reference ranges based of exact same distance 205 , 195 , 185 , 175 ................. until 5 here is problem , position in 32 keys list is not always same so you cant guess which key is in which range at which position. but if you guess one key correctly which range that key is then you can find easily which is correct 5 bit down key at what position as all other keys ranges are always in sequence. if you do increment of one key all the keys will get increment as well and perhaps they will switch position also in divisor output. now talk about NotATether script,the script he posted is doing mod inverses and it is just multiplying value until reach 5 uper bit. (no one can get 120 how can they will get 125 lolololo) exapmle key is in 120 range and you used his script output will be 120, 121 , 122, 123, 124, 125 thats it . all 32 keys will be from 120 to 125 range and some range will have 2 or more keys. but guaranteed all keys will be between 120 ~ 125 range at known position as 125 bit will be at 32 position and divisor will have only one key from lower 5 bit guaranteed (not known position) and all other keys from exact same distance with upper reference range. hope you get the point. now talk about brainless theory - NotATether and brainless are misunderstanding each other brainless maybe joked that he reduced keys 720 by doing multiplication , addition and subtraction bla bla bla until 90 or 100 bit but NotATether is insisting what he explained inside his posts is not a way & there is also no way to achieve that and perhaps he never achieved that one and just keep lying. now what i think is brainless have to explain this to community " 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 as how he claimed this one and plus dont forget he claimed before that he found the 120 key but no plan to cash it but same time he asked .75 bitcoin to provide 115 range one key to buy 3090 (WTF) well i think i got headache now time to drink coffee  .75 bitcoin to provide 115 range one key to buy 3090 this statment is about 4july2021 to 5july2021 and claim for have key could be very next minut, but not same time btw you are writing a very interesting topics, and defiantly not boring and same time you are getting headache and need coffee, you are writing seems upset read your above post, and if you feel more headache, then next time dont try coffee, try some wine, special red wine, dont mind stay happy and healthy and wealthy |
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Your Q that you have just linked to says this: guys lets say 32 divisors will return 32 keys , one of will be guaranteed for lower range but what about other all keys , i guess they all are also valid ~~ so my question is all other keys are random in 256 range or what?
from here, my answer were no random, and exact in all 256bit range, and what is exact location, i explain about how to calc exact range, simple, bro if you mislead by my those answers and explanation, i love to say SORRY |
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I think a lot of people in this thread have tried to solve puzzle 120 with a kangaroo. But since bitcoins are still in place, it means no one has solved it. Brainless somehow found a way to reduce the number of keys to search and if I understand the translation correctly, solved the puzzle 120, but did not withdraw funds. I could be wrong.
I am telling you, brainless' method is baloney. (I reverse engineered his script and I can share it right here if you want - it is useless). Brainless did not manage to reduce a 120 bit key to a smaller range (and he definitely did not solve #120). He made 2^20 zones, each of which are 2^20th the size of the 2^256 range. (He said that himself, read his posts again carefully). Also see: little bit more example
dec 33
hex 21
pubkey 021697ffa6fd9de627c077e3d2fe541084ce13300b0bec1146f95ae57f0d0bd6a5
use your program to get 32 keys of above pubkey
you will see 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 in your list
one step ahead pubkey is 1.03125 ( 03bb2228d3ea32cb3c1eb160cc824a4ba8115f9a7f415d18ddcaac8193defc2c47 + 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 )
and in private key F7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFEC4D965FF79CE5B39E1D3CB9869B48F38
029eda0cebe3c594b59add6dccbff3347f06ad09e83e0b9279dd821cc94284c5d0
hope you will test each zone for your understand
Sorry to rain on this, but this method is useless. I managed to decode how to split multiple keys using this method (I even made a script that reproduces your example, but it's private at the request of someone else, not that the results would be useful/practical to anyone here), and it turns out you're just using mod(32)-sized ranges. Now as we all know, when you modinv a small number on secp256k1 you get a very large number (and modinv a "decimal number" is really just modinv(denominator) * numerator so this still holds for that) This effectively means that the range becomes humungous, way bigger than 2^120 where we started.
The ranges look something like this (P = public key): 1*modinv(32) 2*modinv(32) 1P 2*modinv(32) 3*modinv(32) 2P ... 32*modinv(32) 33*modinv(32) (32*modinv(32) = 1) 32P The ranges are much bigger than when you divide-and-split them down by a number. Now of course you can replace 32 with an extremely huge number to make the ranges manageable/smaller but what's the point? Now you have an unmanageable amount of public keys. Please read Q and A and my explaination related to that Q start from here https://bitcointalk.org/index.php?topic=5244940.msg57818498#msg57818498 |
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is there any specific way to recognize which exactly is the divisor key which will lead us 5 bit down?
i guess not but brainless can you confirm it ;D
i mean if i do divisor this key ~ 0348e843dc5b1bd246e6309b4924b81543d02b16c8083df973a89ce2c7eb89a10d will get this output 0313c264941c883aabdb8c8edb51ebd682c2ac07fa433b3d315b0133cc8170450c
02af206b94d6b8af8880b1fc9a95d2d34c5470dcac1de4b73f34e3d9dd898ebccb
02c010afde4f60bbd61f4252d81ec2e6ea3a80e409fb7f0427f802d6684341859b
025e110e2b5204cdfd6822e7f89a7aae7144ca1793ee99125519afe69a14819a74
033d6c769f853f6d42bec9db3fb654c6f24a64483813505e1516fac0f09a9efbbd
03073cf28f3fe95ee3d4cfc445bbccb2f28331c83a24e425ed99546e5b95b15de1
028275bb253a4a0690513e1129fae1a4a2f76d3e44c1b7eddcd83ae07a58862799
0255769cff42221f9a0a62967d4dfa011d44fad98560db79be23e5ea6f7f8029a4
03ac77a409287008dd8b11bc081b8477055bfb481a3101ed87e6ef05b00c9a0640
0327192559eb626d0c533aa5194ebbee2c18ac235ae46fce84cc960324595adda2
02f939484b1ee08a8fcbd7b5903c6998c990b73b3b97a0ed540c95b65878e21527
029db274aeb17416dba16b940e267e76a1e15ed7d56f7e6e90b63b2e87d567903f
028c43a1fd26ca6c24451b9d47e04c8aa4827ac8a9cb452ccb05009e45d7e89d96
02e8f6094a5fa52287654fb930226c1f56d22e6bb32df66bed8a4e7cad7024703e
02ce502e8e73ce24592814254787f71b64cefed0a98779245331613552b6ff0a92
03475697cfab1926d8698e361200ae9acbbd627910293b54dba9ed8e1eaf76ff56
03b114ce9210cf1a31152b55d635e5a51aadc0cbbf686c5e03fabce4dc32fa65b7
0268ea2fb0120ef4060b86ef556bc9914496933346e7efaca77d9a6cce73a07ced
0336547f7008e93e117f0f4cae806d553ef4418fd5dd087854553355256c25c267
02abdaa757d845d3cafbb32a7a3f179f1f5c9e9bc702ee58840574036bfc1e40c9
03fb447f682b32a32a01ffdd10473ea61eed4228267dbdb8b4bb3567d55948a6a8
03e32401503b65f3b9438774bf5fe2f714fecbe5f53828180a3f388b4a84d46f32
020674c090056a6aedcd52ea3ee92d403d7426e6cd1d5217d622d54e60afa89383
02410388a9508f7b547d715dfe5d127ce7dc828d1611675e6440535a81c936ee05
02daaeea93f97a224cc7de6e665d7d8cb5a08a6ed8712db8067d6e85838379b9e1
03c1d907646ff1bc0219718a6aed820d87bc4c03fb07a4da0fd83f94cf192b29a7
034da2824d16a6d1a6c9ed58df015f2d303f91edfe0427fdfc6accea6fb774c0c3
033929e93f62ad6bd54c9e81f5775ed59f173cd4d7eaef14b444509d285074a7e4
03e4ed13c6328577e37da6cc7cbf3671ce8baada7593fba038734139a8b1783d37
029b413c07d1efb4848e3828ee9f370964f0fb04169e0053c3485e7a98dba27d4b
0335f2e3d043050272c3185f50fbea2321a9b63954c19928ff05a879faf62847a5
032819ab50e1387dc69437b4c24ce4b178169eaa8295696e5395f8c491542ab62e
021d118f1578dcf147ae29be89af46ddeeb5fb50b08edbb6441564897ef01022a7 where 0335f2e3d043050272c3185f50fbea2321a9b63954c19928ff05a879faf62847a5 is the only key leading to the target so my question & my opinion it is not necessary to jump to the target from only 5 bit down specific key , i guess we can jump from any key listed above if we know the private key? @brainless dont use your brain less , tell me this is possible or not? Everything is possible |
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