wedom
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September 28, 2021, 02:35:37 PM |
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by the way dont brain (mind) brainless but you got me man~ now before die i really want to see script which you are using to reduce number of keys . consider it as my last wish ![Grin](https://bitcointalk.org/Smileys/default/grin.gif) Giving a ready-made script is very easy. I would hope for hints, and get to the result myself.
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WanderingPhilospher
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September 28, 2021, 03:51:20 PM |
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@WanderingPhilosopher If you store it in a file, you can lookup the position. But is there a formula for shifting it up by knowing the position of the pubkey?
@fxsniper
Do you read? It is a memory tradeoff! If you bitshift a 256 bit key 128 bit you will have 2^128 keys to check in a range of 1-2^128. 2^128 is how many petabytes???
Please inform yourself about ECDLP. I invested last few weeks in my freetime to understand how it works and what to do.
Etar showed example: And this pubkey is 03d01115d548e7561b15c38f004d734633687cf4419620095bc5b0f47070afe85a with privkey 0xC To produce real privkey need multiply privkey by divisor 0xC*0x8 = 0x60 After this need add to result founded public key number (7) Totaly privekey = 0x60 +0x7=0x67
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WanderingPhilospher
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September 28, 2021, 03:53:40 PM |
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Here is no magic, here is script to shiftdown pubkey: import random import math import hashlib import base58 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 isoncurve(pt,p): """ returns True when pt is on the secp256k1 curve """ (x,y)= pt return (y**2 - x**3 - 7)%p == 0
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 uncompressed 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
def hash160(hex_str): sha = hashlib.sha256() rip = hashlib.new('ripemd160') sha.update(hex_str) rip.update( sha.digest() ) return rip.hexdigest() # .hexdigest() is hex ASCII def getbtcaddr(pubkeyst): hex_str = bytearray.fromhex(pubkeyst) # Obtain key: key_hash = '00' + hash160(hex_str)
# Obtain signature:
sha = hashlib.sha256() sha.update( bytearray.fromhex(key_hash) ) checksum = sha.digest() sha = hashlib.sha256() sha.update(checksum) checksum = sha.hexdigest()[0:8]
return (base58.b58encode( bytes(bytearray.fromhex(key_hash + checksum)) )).decode('utf-8')
def checkpub(realpub, temppub, id): localpt = ptmul(temppub, 1024, p) localaddpt = ptmul(g, id, p) respub= addpt(localpt,localaddpt, p) print ("respub-> ", compresspub(respub)) #secp256k1 constants Gx=0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798 Gy=0x483ADA7726A3C4655DA4FBFC0E1108A8FD17B448A68554199C47D08FFB10D4B8 n=0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141 p = 2**256 - 2**32 - 977 g= (Gx,Gy)
compressed_key='0234c1fd04d301be89b31c0442d3e6ac24883928b45a9340781867d4232ec2dbdf' point=getuncompressedpub(compressed_key)
print(getbtcaddr("04%064x%064x"%point)) print(getbtcaddr(compressed_key)) divisor = 2**3 newpub=ptdiv(point,divisor,p,n)
(partGx,partGy)=ptdiv(g,divisor,p,n) print ("1 Fraction part-> (%x,%064x)" % (partGx,partGy))
with open('pub.txt', 'w') as f: f.write("04%064x%064x"%newpub) f.write('\n') print ("Compressed NewPUB (",0,")-> ", compresspub(newpub),"addr",getbtcaddr(compresspub(newpub))) i=1 (pointx,pointy)=(partGx,partGy) while i<divisor: (newpubtempx,newpubtempy) = addpt(newpub,(pointx,p-pointy), p) f.write("04%064x%064x"%(newpubtempx,newpubtempy)) f.write('\n') print ("Compressed NewPUB (",i,")-> ", compresspub((newpubtempx,newpubtempy)),"addr",getbtcaddr(compresspub((newpubtempx,newpubtempy)))) (pointx,pointy) = addpt((pointx,pointy),(partGx,partGy), p) i=i+1
In this example i use pubkey 0234c1fd04d301be89b31c0442d3e6ac24883928b45a9340781867d4232ec2dbdf, privkey is 0x67 and upper range is 2^7 Divisor is 2^3, so new upper range is 2^7-2^3=2^4 In file pub.txt you will find all pubkeys and their number is equil to divisor. if you try to find all this keys in range 0x1:0xf you will see that only one pubkey will be lie in range And this pubkey is 03d01115d548e7561b15c38f004d734633687cf4419620095bc5b0f47070afe85a with privkey 0xC To produce real privkey need multiply privkey by divisor 0xC*0x8 = 0x60 After this need add to result founded public key number (7) Totaly privekey = 0x60 +0x7=0x67 same old stuff with different script :p Etar said, no magic...Etar has probably lost more knowledge on curve math, programming, etc. then most of us have ever known. Super smart, and helpful.
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NotATether
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September 28, 2021, 04:11:58 PM Merited by studyroom1 (1) |
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This makes actually no sense, because the underlying principle is different. ~snip So we are halving the public key 5 times. This means, that when we do this on key 11, we get 32 keys, from where one of them has the correct endsequence of the private key. So we have to check all keys if they are in the range of 2^6. If we get one, than we know how the first 6 bits of the privatekey are. We can from here determine the last 5 bits easily. I am not interested in the other ranges, as i have now a total valid privatekey
Had a freak day today, so my brain's (no pun intended) not very attentive right now. Will need multiple drinks of jolt cola before I can go over what you did. I recently did a little research on private key generation via /dev/urandom. More than 100 million keys were used for the test. The bit allocation was about the same = 0.5 Maybe the old versions of OpenSSL, which were probably used in the puzzle creation, had some vulnerabilities and gave some deviation?
It's possible, but also remember a) OpenSSL uses multiple entropy sources - one of which is urandom (another is the entropy it stores in a file somewhere) and 2) the real-life sample size is very small - there were at least 256 256-bit random numbers generated in the process of making the puzzle, and possibly more as several of the numbers were probably not in the appropriate range. 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)
Well all I did is reverse engineer the algorithm until it completely matched the examples that brainless provided so congratulations on explaining in simple terms why the algorithm is so absurd. 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 720 keys means you've only reduced it by some factor between 2^109 and 2^110, how do you even begin shifting a key down that far while not making the number of group ops explode as well?
Guys fyi 1024 is not div point in ecc, posting here, divideable magic digits for ecc, these will help you to decide bitrange to divide, use these magic ecc div numbers, for pollard kanagroo or other manual div
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
What the hell is a div point? This is just a dump of x-points that are on the secp256k1 curve (x=5 is not on the curve IIRC). Again though, I could pick any x-point on the curve to divide by, so this dump doesn't answer the question - How is creating multiple zones supposed to shift the range down, when all it's doing is shifting it up (massively)? ![Roll Eyes](https://bitcointalk.org/Smileys/default/rolleyes.gif)
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)
0.75 BTC at the time wouldn't buy you just a 3090, it would buy you an entire fucking 8Pack OrionX (almost).
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WanderingPhilospher
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September 28, 2021, 11:35:39 PM |
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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 In brainless' defense concerning this specific convo, what he says is true. The locations aren't random and you can calculate exact range in the example he gave. I went through and found each key in their respective range. Sometimes we have to remember that not everyone speaks this language or that language and that can lead to confusion...sometimes. If you go back to the convo, he does walk you through how to find exact range/location for the example he provided. No magic, no range/key shrinkage, just ranges on the curve.
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WanderingPhilospher
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September 29, 2021, 03:07:17 AM |
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The cool thing, well, neat to me; is using brainless' "math" I was able to find private keys to x,y and x,-y, or the private keys to both 03 and 02 public keys. The positive and negative Ys that belong to the same X...like I said, I have never found both so it was neat to me lol
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brainless
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September 29, 2021, 03:50:24 AM |
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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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13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF
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wedom
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September 29, 2021, 05:15:27 AM Last edit: September 29, 2021, 05:31:28 AM by wedom |
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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 Thanks for the number hint. I hope to figure it out and solve this problem. In the previous post, I thought the other way, that we need to find such a divisor and such a multiplier that would get the same division residuals. For example, below are the results of division and multiplication using these and other numbers: 2281607788513008375014137388606100914 2281607788513008375014137388605979764 2281607788513008375014137388605858614 2281607788513008375014137388605737464 2281607788513008375014137388605616314 2281607788513008375014137388605495164 2281607788513008375014137388605374014 2281607788513008375014137388605252864 2281607788513008375014137388605131714 2281607788513008375014137388605010564 2281607788513008375014137388604889414 2281607788513008375014137388604768264 2281607788513008375014137388604647114 2281607788513008375014137388604525964 2281607788513008375014137388604404814
don't look at the specific values of the numbers, this is for an example. All that matters is that they end in one digit and the second digit alternates.
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WanderingPhilospher
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September 29, 2021, 05:41:12 AM |
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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 Maybe I am misunderstanding you and what you mean, but any multiple of 2520 would have the same result...
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a.a
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September 29, 2021, 05:58:07 AM |
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What about 2520? 2520 = 2³ * 3² * 5 * 7 2520 / 10 = 252 2520 / 9 = 280 2520 / 8 = 315 2520 / 7 = 360 2520 / 6 = 420 2520 / 5 = 504 2520 / 4 = 630 2520 / 3 = 830 2520 / 2 = 1260 2520 / 1 = 2520 Only difference is, that dividing by 8 will get you a odd number. If you want that it is even when dividing by 8, then just double 2520. So 5040 is much smaller than 30240. So why is according to you only 30240 dividable from 1 to 10? If it is even relevant as 30240 is not a divisor of N-1. References: https://en.wikipedia.org/wiki/Highly_composite_numberhttps://mrob.com/pub/math/numbers-14.html#lc5040
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WanderingPhilospher
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September 29, 2021, 06:04:17 AM |
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What about 2520? 2520 = 2³ * 3² * 5 * 7 2520 / 10 = 252 2520 / 9 = 280 2520 / 8 = 315 2520 / 7 = 360 2520 / 6 = 420 2520 / 5 = 504 2520 / 4 = 630 2520 / 3 = 830 2520 / 2 = 1260 2520 / 1 = 2520 Only difference is, that dividing by 8 will get you a odd number. If you want that it is even when dividing by 8, then just double 2520. So 5040 is much smaller than 30240. So why is according to you only 30240 dividable from 1 to 10? If it is even relevant as 30240 is not a divisor of N-1. References: https://en.wikipedia.org/wiki/Highly_composite_numberhttps://mrob.com/pub/math/numbers-14.html#lc5040 Yeah I already asked about 2520; I did not see where he stated it had to be even or odd, just no float. It could be beneficial, I will have to run more tests tomorrow. But as of now, I can find any key in any range with any divisor. So I did learn something from all of this discussion. Example, if you took a 40 bit public key and divided it by 33, I could find every public key that is generated and ultimately each pub key found in any/every range will lead me back to private key of original pub key, with a click of a button. I ran several tests with divisor of 33 and a few with a divisor of 192. Find one pubkey in any range and you have the private key of original pub key.
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a.a
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September 29, 2021, 06:16:36 AM |
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I saw that you posted before I could post. I had already invested about 20 minutes for the post and was like: "Well, it is already said, but not from everyone" and so I posted it anyway.
Also the 33 division makes sense as the distance between each point will be the mod inverse of 33 or so. Its like cutting the whole N into ranges, as mentioned before. Only problem is, that you are only getting one in the lower bruteforcable range and then can jump from that one and can determine the other 32 private keys.
And yes, you just have to multiply the privatekey times your divisor and then mod N to get the right result.
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ssxb
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September 29, 2021, 06:22:09 AM |
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What about 2520? 2520 = 2³ * 3² * 5 * 7 2520 / 10 = 252 2520 / 9 = 280 2520 / 8 = 315 2520 / 7 = 360 2520 / 6 = 420 2520 / 5 = 504 2520 / 4 = 630 2520 / 3 = 830 2520 / 2 = 1260 2520 / 1 = 2520 Only difference is, that dividing by 8 will get you a odd number. If you want that it is even when dividing by 8, then just double 2520. So 5040 is much smaller than 30240. So why is according to you only 30240 dividable from 1 to 10? If it is even relevant as 30240 is not a divisor of N-1. References: https://en.wikipedia.org/wiki/Highly_composite_numberhttps://mrob.com/pub/math/numbers-14.html#lc5040 Yeah I already asked about 2520; I did not see where he stated it had to be even or odd, just no float. It could be beneficial, I will have to run more tests tomorrow. But as of now, I can find any key in any range with any divisor. So I did learn something from all of this discussion. Example, if you took a 40 bit public key and divided it by 33, I could find every public key that is generated and ultimately each pub key found in any/every range will lead me back to private key of original pub key, with a click of a button. I ran several tests with divisor of 33 and a few with a divisor of 192. Find one pubkey in any range and you have the private key of original pub key. Find one pubkey in any range and you have the private key of original pub keythat what i said before ![Grin](https://bitcointalk.org/Smileys/default/grin.gif)
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ssxb
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September 29, 2021, 06:25:12 AM |
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I saw that you posted before I could post. I had already invested about 20 minutes for the post and was like: "Well, it is already said, but not from everyone" and so I posted it anyway.
Also the 33 division makes sense as the distance between each point will be the mod inverse of 33 or so. Its like cutting the whole N into ranges, as mentioned before. Only problem is, that you are only getting one in the lower bruteforcable range and then can jump from that one and can determine the other 32 private keys.
And yes, you just have to multiply the privatekey times your divisor and then mod N to get the right result.
that what i said before ![Grin](https://bitcointalk.org/Smileys/default/grin.gif)
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_Counselor
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September 29, 2021, 06:26:14 AM |
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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 don't know what you mean by "numerology", but 30240 is not the only such number (some people already told about it) if you multuply 30240 to any numbers, and result could also div by 1 to 10, and in result no floating point
If you multiply any X by any Y, that resulting number will be divisible without remainder by all factors of Y (and X). There is no secrets. That is basics. Fundamental theorem of arithmetic.
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ssxb
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September 29, 2021, 06:31:32 AM |
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i am drinking soda right now and watching screen (no red wine as i am Muslim ![Tongue](https://bitcointalk.org/Smileys/default/tongue.gif) ) 32 keys are in front of me telling me that we are relative to each other and no matter from where you will jump toward us we will be always on distance of 1 with each other. if you knows the private keys you will find these keys are lying but hell yaa man when you work blindly on public keys without knowing private keys you will know they are having relation on curve and on 1 key distance from each other. fck math is beautiful . curse you elliptic curve sorry glass slipped from my hand i need to clean the table ![Roll Eyes](https://bitcointalk.org/Smileys/default/rolleyes.gif)
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a.a
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September 29, 2021, 06:48:14 AM |
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@WP I take a 255 bit pubkey, divide it by 33. Now I only have to search the 2^255/33 ~ 2^252 range to find the key. What a reduction.
@ssxb
Well but what is the new knowledge?
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ssxb
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September 29, 2021, 07:20:39 AM |
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@WP I take a 255 bit pubkey, divide it by 33. Now I only have to search the 2^255/33 ~ 2^252 range to find the key. What a reduction.
@ssxb
Well but what is the new knowledge?
Well but what is the new knowledge? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 hope you get the point ~ congratulation to me today i learnt how to generate numbers in series from 1 to 32 with python. such a genius (wile-e-coyote) now i am telling you guys dont follow me you can hardly program this. (My brainless theory) watching wile e coyote vs bugs bunny
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a.a
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September 29, 2021, 07:24:56 AM |
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So no new knowledge ![Cool](https://bitcointalk.org/Smileys/default/cool.gif)
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brainless
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September 29, 2021, 07:26:34 AM |
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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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13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF
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