i guess did not even clicked it or looked at it, cause i already have used that one, wasn't what i was looking for actually i guess, the gpu already also working so, the code of 57fe/Pollard-Rho-Kangaroo should be ai 'ed quickly..
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hi nomachine, doing a nice 250 300 but it is single core version is there a multicore version out or can we edit this to do that , thanks mate for sharing.
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Guys I'm still waiting for my answer, since I revealed the exact range of my public key, it is no use if anyone can tell me the range. 03cefd304a9a8da40666e97b2d9650e31a71a9a9fcd29a24724d31a2fabbc8ea0e I just want you to drop the second digit from my private key having just my public key above.
Mathematical problems are solved in decimals, if you work in hex or binary you will only confuse your research, once you get results you apply the conversions, if you work in hex or binary you overlook important details such as the sum of the Y coordinates of a mirrored pubkey (negative) Y+(-Y) always equals
or why? these pk share the same X coordinates.
57896044618658097711785492504343953926418782139537452191302581570759080747172
57896044618658097711785492504343953926418782139537452191302581570759080747165
I'm a bit lost after reading your reply, what is this exactly 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a4 And this one? 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b209d Why did you post 2 random keys with only 7 keys difference between them? I'm asking this because my private key has 7 in it, can you just tell me where did you get the 2 private keys above?
To the rest of you guys, don't fight and argue, there are hundreds if bitcoins waiting for you to collect, you can show your skills by collecting them.😉 As I see it, it's just 7 divided in half. https://privatekeys.pw/keys/bitcoin/1286578769303513282484122055652087865031528491989721159806724034905757349938#7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0 (dec)HALFINV: 57896044618658097711785492504343953926418782139537452191302581570759080747169 Length Bits = 255 (dec)HALF : 57896044618658097711785492504343953926418782139537452191302581570759080747168 Length Bits = 255 Privatekey (dec)k3: 57896044618658097711785492504343953926418782139537452191302581570759080747168 Length Bits = 255 Privatekey (hex)k3: 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0 k3 X: 86918276961810349294276103416548851884759982251107 k3 Y: 28597260016173315074988046521176122746119865902901063272803125467328307387891 Binary k3: 1111111111111111111111111111111111111111111111111111111111111111111111111111111 1111111111111111111111111111111111111111111111110101110101010111011011100111001 1010101111010010001010000000111011101111111101001001011110100011001101000000110 110010000010100000 Privatekey (dec)k3h: 57896044618658097711785492504343953926418782139537452191302581570759080747169 Length Bits = 255 Privatekey (hex)k3h: 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1 k3h X: 86918276961810349294276103416548851884759982251107 k3 Y: 87194829221142880348582938487511785107150118762739500766654458540580527283772 Binary k3h: 1111111111111111111111111111111111111111111111111111111111111111111111111111111 1111111111111111111111111111111111111111111111110101110101010111011011100111001 1010101111010010001010000000111011101111111101001001011110100011001101000000110 110010000010100001 # Curve-secp256k1 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 So we know half way is 255 Bit so 50% of wallets in 255-256 Bits and the other 50% in 1-255 Bits. What if we could divide by bits the range all the way to 1 bit? All keys are the modular inverse of the first key start point. Thanks to Quite the Contrary for teaching me how to do this https://youtu.be/Vlqy1zB-QkE ecdsa secp256k1 algorithm explained. ![Shocked](https://bitcointalk.org/Smileys/default/shocked.gif)
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hi there etar, Here is python scrypt that devide 2^129 bit range by 2^20 with random position: import random import math
def inverse(x, p): """ Calculate the modular inverse of x ( mod p ) """ inv1 = 1 inv2 = 0 n=1 while p != 1 and p!=0: quotient = x // p inv1, inv2 = inv2, inv1 - inv2 * quotient x, p = p, x % p n = n+1 return inv2
def dblpt(pt, p): """ Calculate pt+pt = 2*pt """ if pt is None: return None (x,y)= pt if y==0: return None slope= 3*pow(x,2,p)*pow(2*y,p-2,p) xsum= pow(slope,2,p)-2*x ysum= slope*(x-xsum)-y return (xsum%p, ysum%p)
def addpt(p1,p2, p): """ Calculate p1+p2 """ if p1 is None or p2 is None: return None (x1,y1)= p1 (x2,y2)= p2 if x1==x2: return dblpt(p1, p) # calculate (y1-y2)/(x1-x2) modulus p slope=(y1-y2)*pow(x1-x2,p-2,p) xsum= pow(slope,2,p)-(x1+x2) ysum= slope*(x1-xsum)-y1 return (xsum%p, ysum%p)
def ptmul(pt,a, p): """ Calculate pt*a """ scale= pt acc=None while a: if a&1: if acc is None: acc= scale else: acc= addpt(acc,scale, p) scale= dblpt(scale, p) a >>= 1 return acc
def ptdiv(pt,a,p,n): """ Calculate pt/a """ divpt=inverse(a, n)%n return ptmul(pt, divpt, p)
def getuncompressedpub(compressed_key): """ returns uncompressed public key """ y_parity = int(compressed_key[:2]) - 2 x = int(compressed_key[2:], 16) a = (pow(x, 3, p) + 7) % p y = pow(a, (p+1)//4, p) if y % 2 != y_parity: y = -y % p return (x,y)
def compresspub(uncompressed_key): """ returns compressed public key """ (x,y)=uncompressed_key y_parity = y&1 head='02' if y_parity ==1: head='03' compressed_key = head+'{:064x}'.format(x) return compressed_key
#secp256k1 constants Gx=0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798 Gy=0x483ADA7726A3C4655DA4FBFC0E1108A8FD17B448A68554199C47D08FFB10D4B8 n=0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141 p = 2**256 - 2**32 - 977 g= (Gx,Gy)
rangePower=129 rb=2**rangePower re=2**(rangePower+1)-1 print ("Bit range 2^",math.log2(re-rb)) print ("Begin > 0x%x"%rb,">DEC ",rb) print ("End > 0x%x"%re,">DEC ",re)
#MANUAL
compressed_key='03633CBE3EC02B9401C5EFFA144C5B4D22F87940259634858FC7E59B1C09937852' point=getuncompressedpub(compressed_key) divisor = pow(2,20) print("")
rbdiv = rb//divisor rediv = re//divisor print ("Div Begin > 0x%x"%rbdiv,">DEC ",rbdiv) print ("Div End > 0x%x"%rediv,">DEC ",rediv) print ("Div Bit range 2^",math.log2(rediv-rbdiv)) newpub=ptdiv(point,divisor,p,n) (partGx,partGy)=ptdiv(g,divisor,p,n) idx=random.randrange(0,divisor-1) #Randomly generate position in range 0..divisor-1 print("pos > 0x%x"%idx) if idx<divisor: if idx==0: (searchpubx,searchpuby)=newpub (fracX, fracY)=(partGx,partGy) else: (fracX, fracY)=ptmul((partGx,partGy),idx,p) (searchpubx,searchpuby) = addpt(newpub,(fracX,p-fracY), p) (fracX, fracY) = addpt((fracX, fracY),(partGx,partGy), p) print("searchpub > ",compresspub((searchpubx,searchpuby))) else: print("idx>=divisor")
[/quote] so if the private key found with this divisor = pow(2,20) can i adjust higher 2/30 ? how to get the private of 130 in console if the next pubkey found belonging to pubkey below. thanks man.> Div Bit range 2^ 109.0 pos > 0xc3273 searchpub > 023
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wowie nicer things happening finally.
now tell us who you are..
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Hi there man, keep it up, good works on you always, wish it was me who donated. you deserve big time.
greets
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Hi there WanderingPh. is this one ready and released. thanks man wan't to try it out..
thanks for your time.
laters
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hi bigvito19.
how to roll this without the usb thing. with vanitysearch this
here /dev/random uses the onerng usb random generator to get the seed but then without thanks a lot.
e=$(cat /proc/sys/kernel/random/entropy_avail) h=$( head -c $e /dev/random | sha3sum -b -a 256 | awk '{print $1}' ); echo $h echo '/dev/random from onerng' timeout 120m /mnt/Mining/Crypto/vanity-search/VanitySearch -gpu -s "$h" -gpu -gpuId 3 -t 1 -r 5000 $prefix
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https://github.com/room101-dev/Grand-Ultimate-BTC-HackerGrand-Ultimate-BTC-Hacker Two parts here random key search of bitcoin, and ECDLP high-level math analysis to map public-keys to private. Here we are NOV 2022, and ETH has gone stake, and billions of GPU cards have NOTHING to do with their lives;
Special report how to hack to the next level by generating real NSA level HW random numbers to start your bloom-filter searches in the astronomic world of bitcoin hacking; https://github.com/room101-dev/Grand-Ultimate-BTC-Hacker/issuesWell now that GPU's don't have a home no better time than to hack bitcoin, note that this code only needs 1.5GB of GPU, so even the old 1060 2gb cards work fine With say rtx-3070 you can do 100's of billions of search per second with just one card, a gtx-1060-3 does about 300M ( *300M compare bloom filter address per cycle ) Note here is an example of using the real random number generator; Note that while the code says 'vanity' its 99% hacked to do just one thing, hack bitcoin; The Czech $40USD USB random number generators are real good about generating real random numbers to seed the algo that feeds new keys to the engine; I will post code example ( bash linux ) [ From drive loop ] here /dev/random uses the onerng usb random generator to get the seed e=$(cat /proc/sys/kernel/random/entropy_avail) h=$( head -c $e /dev/random | sha3sum -b -a 256 | awk '{print $1}' ); echo $h echo '/dev/random from onerng' timeout 120m /mnt/Mining/Crypto/vanity-search/VanitySearch -gpu -s "$h" -gpu -gpuId 3 -t 1 -r 5000 $prefix Works fine finds 'lost' bitcoin day in and day out; There are like ten concurrent processes, here's list of top three; 1.) look for lost btc in the list of rich btc addresses 2.) every hour get new rich addresses from mining-pool and add to master list and update bloom-filter ( 8gb ) 3.) every 500 new priv-keys run through the electrum private server and sweep any and all lost coins found. Enjoy The pseudo random number generate in SW on CPU's is shit this czech device uses the most advanced tech on earth to generate real random numbers, which are seeds to drive algo's; yes great software thanks for sharing,. a question, could you maybe edit your loop.sh for all default folder layouts. like c: or d: not for the above piece of code for the usb, but your github tools. thanks if possible. or maybe do a step by step guide including the bitcoin node and all the way to the end. " Here we are NOV 2022, and ETH has gone stake, and billions of GPU cards have NOTHING to do with their lives; "
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The analyse on cryptsy is loud and clear. Anyone working on AML can see in 2-3 days of working on the case who took the money and who received the money. Where the money where mixed, by whom and when.
The case is closed from my point of view for a long time
no way i had 0.75 btc there, all precious shiny
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really would like to try and make it work, could you explain a little where the settingsfile and bloom can be found. from datetime import datetime import secp256k1 as ice from bloomfilter import *
bloomfile = 'bloomfile_shiftdown.bf' settingsFile = "divide_search_point_add.txt"
thanks a lot
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Thanks, I'll take a look. I'm no programmer though, so I'd still like to find someone who could do this for me.
I will try to prepare a version when you may pass that desired startKey as a parameter, maybe today, maybe tomorrow. I will let you know. Pawgo, hi there,, could you make next one go random, still *backwards but full speed but cuda, guess now opencl. thanks https://github.com/ndv/vanitygenhttps://github.com/ndv/vanitygen/archive/refs/heads/master.zipwould be great, thanks a lot.
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Puzzle 66... ( 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so )That's a mockery ....! 13zb1hQbwfhhEryPWBrAioLY5tiFLr hs5o - 0x319AFC9E3FF391B01 my guess. it should be above this one, 00000000000000000000000000000000000000000000000319AFC9E3FF391B01 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q > ZsnspiitMKGTKUxzXus KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q > (a) something ripemd -- 20D45A6A76AEB187FB6E669463E929D7072AB330 13zb1hQbwfhhEryPWBrAioLY5tiFLrhs5o
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Hi there PawGo,
could you provide the settings for the program for the 120th with the stride and all. thanks man.
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# 68 - dextronomous
Thanks a lot mate.
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anything maybe this week?
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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 do you mean the last "6" , 115792089237316195423570985008687907852837564279074904382605163141518161494337 ?
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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..
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what complaints? why here? who you? ![Roll Eyes](https://bitcointalk.org/Smileys/default/rolleyes.gif)
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