digaran
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September 01, 2023, 11:47:09 PM |
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I'm bored, nothing interesting is happening around these woods no more. Nobody has new and exciting ideas no more.
Something on mind for some time now, I wonder why 160 bit? Why not keep adding coins to 165, 170 etc?
And is it really feasible brute forcing any key beyond 80 bit? With no pub exposed of course.
I guess from 81 to 159 will not be solved in the next decade, why bother keeping them in there when we can use them as extra incentive for higher ranges, like adding them to 165, 170, 175 etc?
Good luck to all.😴
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james5000
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September 02, 2023, 01:23:35 AM Last edit: September 02, 2023, 09:28:57 PM by Mr. Big |
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Can someone please explain how to divide a point on the curve by 2? I have been working on software for searching by public key, and I can divide by 2 by performing scalar multiplication by the inverse scalar, but this operation takes a long time. If anyone knows a faster way to divide, I can improve my code and release the tool for others to test.
I'm bored, nothing interesting is happening around these woods no more. Nobody has new and exciting ideas no more.
Something on mind for some time now, I wonder why 160 bit? Why not keep adding coins to 165, 170 etc?
And is it really feasible brute forcing any key beyond 80 bit? With no pub exposed of course.
I guess from 81 to 159 will not be solved in the next decade, why bother keeping them in there when we can use them as extra incentive for higher ranges, like adding them to 165, 170, 175 etc?
Good luck to all.
Have you noticed that there are at most 2^160 private keys instead of 2^256? Any key above that will generate a repeated RIPEMD160 hash. That's why there are no keys from 161 to 256 anymore; the creator of the challenge realized it was redundant.
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lordfrs
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September 02, 2023, 08:26:55 AM |
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Can someone please explain how to divide a point on the curve by 2?
I have been working on software for searching by public key, and I can divide by 2 by performing scalar multiplication by the inverse scalar, but this operation takes a long time. If anyone knows a faster way to divide, I can improve my code and release the tool for others to test.
If you're going to divide a point by 2, you multiply it by the inverse of 2 by N.
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digaran
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September 02, 2023, 08:30:46 AM |
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Can someone please explain how to divide a point on the curve by 2?
I have been working on software for searching by public key, and I can divide by 2 by performing scalar multiplication by the inverse scalar, but this operation takes a long time. If anyone knows a faster way to divide, I can improve my code and release the tool for others to test.
There is no such a thing as division in EC math, there is only add, subtract and multiply, whatever you do, at the end you'll have to use * by inverse.
I wish there was a puzzle 161, so kalos could tackle it by finding rmd160 collisions, he seems to be interested in that.😅
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_Counselor
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September 02, 2023, 08:54:16 AM |
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I have been working on software for searching by public key, and I can divide by 2 by performing scalar multiplication by the inverse scalar, but this operation takes a long time. If anyone knows a faster way to divide, I can improve my code and release the tool for others to test.
If you're using division by a fixed number, first you can cache inverse of this number to avoid inversion on each step, second you can decompose inverse into fixed double-add chain, then optimize it with special formulas for point tripling and readditions. But it is will be still expensive opertaion.
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james5000
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September 02, 2023, 09:35:56 AM Last edit: September 02, 2023, 09:27:54 PM by Mr. Big |
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Can someone please explain how to divide a point on the curve by 2?
I have been working on software for searching by public key, and I can divide by 2 by performing scalar multiplication by the inverse scalar, but this operation takes a long time. If anyone knows a faster way to divide, I can improve my code and release the tool for others to test.
If you're going to divide a point by 2, you multiply it by the inverse of 2 by N. I'm currently doing it exactly like this, but it's proving to be very expensive this way.
My new software is for public keys, inspired by BSGS but a little different, using division and subtraction. Currently, I'm solving 35 bits in 3 seconds; this time is due to the delay in dividing a point. However, I believe that with the correct configuration, I will soon be able to break larger keys. I have some theories that have worked out very well in my tests, and I will be trying out more things to confirm. For example, for a 35-bit key, I reduce the challenge #35 to 3 bits using my code. You input the public key and the 3-bit key '111,' and it returns 0x4aed21170, which is the equivalent private key of the public key. key: 111 found: 00000000000000000000000000000000000000000000000000000004aed21170 Total time: 0 h, 0 m, 2 s "111101" would be the key for #38. key: 111101 found: 00000000000000000000000000000000000000000000000000000022382facd0 Total time: 0 h, 0 m, 13 s
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ing1996
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September 02, 2023, 10:15:26 AM |
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So any update on the progress of finding this key? First offset = 03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710 Half of above = 0291001b0dc6e5a2628cb4698eb00a6fb7dbd276dc2b214795f2fe52e61243aa9b Half of 130? 0337374e00a32eaf009e9946035c0e69085627b60a844637d2b958dd83bcfa4383 The following is the subtracted key from #130 Second offset = 03d99bb89e8db75d20b882f13f8086fb39221858fa211de0346c926a93ae259b3a Half of above? 03a3dc00bf5f7e7eec691569c7f67a15d3cdbb3a9994c9a5ec1430cffdb622cf9f
Now subtract half of first offset from half of #130 to get half of second offset.
Second offset is known, we need to work on first offset's half, use -1 divide by 2 script to reduce 18 bits from it, you'll have millions of new offsets and one of them is the target, now divide the #130 range by 2, subtract 18 bits from it and use the new range as your search range, input those millions offset keys and search the range.
Don't just try blind searching.😉
Hi! Can you show me how you do it?, to remove on video. I can't understand how you make the -1 divide by 2 scenario, and how you get the millionth offsets.
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digaran
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September 02, 2023, 10:45:02 AM Last edit: September 02, 2023, 11:06:11 AM by digaran |
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So any update on the progress of finding this key? First offset = 03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710 Half of above = 0291001b0dc6e5a2628cb4698eb00a6fb7dbd276dc2b214795f2fe52e61243aa9b Half of 130? 0337374e00a32eaf009e9946035c0e69085627b60a844637d2b958dd83bcfa4383 The following is the subtracted key from #130 Second offset = 03d99bb89e8db75d20b882f13f8086fb39221858fa211de0346c926a93ae259b3a Half of above? 03a3dc00bf5f7e7eec691569c7f67a15d3cdbb3a9994c9a5ec1430cffdb622cf9f
Now subtract half of first offset from half of #130 to get half of second offset.
Second offset is known, we need to work on first offset's half, use -1 divide by 2 script to reduce 18 bits from it, you'll have millions of new offsets and one of them is the target, now divide the #130 range by 2, subtract 18 bits from it and use the new range as your search range, input those millions offset keys and search the range.
Don't just try blind searching.
Hi! Can you show me how you do it?, to remove on video. I can't understand how you make the -1 divide by 2 scenario, and how you get the millionth offsets. Hi there, I can't make a video but here is the script you can use to reduce bits. from sympy import mod_inverse import secp256k1 as ice pub=ice.pub2upub('Here Compressed Public Key') N=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
k=mod_inverse(2,N) neg1=ice.point_negation(ice.scalar_multiplication(1))
def ters(Qx,Scalar): ScalarBin = str(bin(Scalar))[2:] le=len(ScalarBin) for i in range (1,le+1): if ScalarBin[le-i] == "0": Qx=ice.point_multiplication(k,Qx) else: Qx=ice.point_addition(Qx,neg1) Qx=ice.point_multiplication(k,Qx) return ice.point_to_cpub(Qx)
for x in range(1,65536): f= (ters(pub,x)) data= open(“halfpub.txt”,”a”) data.write(f+”\n”) data.close() Note this is where you decide how many bits should be reduced for x in range(1,65536): For example reducing 26 bits requires 67108864 to be generated, 1 of them is the correct 26 bit reduced key.
My new software is for public keys, inspired by BSGS but a little different, using division and subtraction. Currently, I'm solving 35 bits in 3 seconds; this time is due to the delay in dividing a point. However, I believe that with the correct configuration, I will soon be able to break larger keys. I have some theories that have worked out very well in my tests, and I will be trying out more things to confirm. For example, for a 35-bit key, I reduce the challenge #35 to 3 bits using my code. You input the public key and the 3-bit key '111,' and it returns 0x4aed21170, which is the equivalent private key of the public key. key: 111 found: 00000000000000000000000000000000000000000000000000000004aed21170 Total time: 0 h, 0 m, 2 s "111101" would be the key for #38. key: 111101 found: 00000000000000000000000000000000000000000000000000000022382facd0 Total time: 0 h, 0 m, 13 s Will you release for public to use?
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mcdouglasx
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September 02, 2023, 10:51:23 AM |
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I'm bored, nothing interesting is happening around these woods no more. Nobody has new and exciting ideas no more.
Something on mind for some time now, I wonder why 160 bit? Why not keep adding coins to 165, 170 etc?
And is it really feasible brute forcing any key beyond 80 bit? With no pub exposed of course.
I guess from 81 to 159 will not be solved in the next decade, why bother keeping them in there when we can use them as extra incentive for higher ranges, like adding them to 165, 170, 175 etc?
Good luck to all.😴
It's not that we don't have new ideas, it's that we don't share them for our own interest, I thought about sharing how I reduce the bits from 130 to 107, in a range of 1000 possible pubkey, but I haven't done it, because the The time when you did something important and people contributed to the developer. I don't have how to update my PC, so bit 100-107 is still too much for me. An example of what I am saying is keyhunt, it is a tool used by thousands or millions and has only received a little more than 10k in donations. If I get to unlock a puzzle, I'll share it, while not because nowadays people don't even post the pk of the last puzzles solved.
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BTC bc1qxs47ttydl8tmdv8vtygp7dy76lvayz3r6rdahu
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digaran
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September 02, 2023, 11:01:51 AM |
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I'm bored, nothing interesting is happening around these woods no more. Nobody has new and exciting ideas no more.
Something on mind for some time now, I wonder why 160 bit? Why not keep adding coins to 165, 170 etc?
And is it really feasible brute forcing any key beyond 80 bit? With no pub exposed of course.
I guess from 81 to 159 will not be solved in the next decade, why bother keeping them in there when we can use them as extra incentive for higher ranges, like adding them to 165, 170, 175 etc?
Good luck to all.😴
It's not that we don't have new ideas, it's that we don't share them for our own interest, I thought about sharing how I reduce the bits from 130 to 107, in a range of 1000 possible pubkey, but I haven't done it, because the The time when you did something important and people contributed to the developer. I don't have how to update my PC, so bit 100-107 is still too much for me. An example of what I am saying is keyhunt, it is a tool used by thousands or millions and has only received a little more than 10k in donations. If I get to unlock a puzzle, I'll share it, while not because nowadays people don't even post the pk of the last puzzles solved. If you can solve a puzzle with your idea, delete all social media accounts, remove yourself from online world, use a new pc with tor+vpn enabled to transfer the funds and never ever come back to this forum to post or share anything, and only solve the next puzzle after 4, 6 month. If you like to live long. This is my advice to anyone. However if you first share your idea and then puzzles start getting solved, nobody would look at your direction but if your idea could be used to crack any 256 bit key, then cryptocurrencies no more!
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kalos15btc
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September 02, 2023, 11:28:34 AM |
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I'm bored, nothing interesting is happening around these woods no more. Nobody has new and exciting ideas no more.
Something on mind for some time now, I wonder why 160 bit? Why not keep adding coins to 165, 170 etc?
And is it really feasible brute forcing any key beyond 80 bit? With no pub exposed of course.
I guess from 81 to 159 will not be solved in the next decade, why bother keeping them in there when we can use them as extra incentive for higher ranges, like adding them to 165, 170, 175 etc?
Good luck to all.😴
i already find two puzzle with my methode of hash160 lowering the range of searching,and still im going to 130 135 my methode is working its just lowering and find the range in subed pub key to lower bit range 2 times yes when you find same 14 digits of hash160 than means that adress is in that range, but the search is not for one adress, i get a lot of addresses im trying , and no im not going to share the pk of any address i work hard for this, i advise you to try on 120 my methode, and get your bitcoin cash prize, ps: ill leave 50 satoshi on 130 for you, thank you
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digaran
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September 02, 2023, 11:59:18 AM Last edit: September 02, 2023, 01:09:02 PM by digaran |
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i already find two puzzle with my methode of hash160 lowering the range of searching,and still im going to 130 135 my methode is working its just lowering and find the range in subed pub key to lower bit range 2 times yes when you find same 14 digits of hash160 than means that adress is in that range, but the search is not for one adress, i get a lot of addresses im trying , and no im not going to share the pk of any address i work hard for this, i advise you to try on 120 my methode, and get your bitcoin cash prize, ps: ill leave 50 satoshi on 130 for you, thank you
So you already solved 120 and 125? Can we see their private keys? Since you have worked hard for them, and don't want to share them, you could sign a message with them without revealing them. I advise you to grind all the puzzles with no mercy, don't even leave a single sat for anyone. Kalos the merciless.😅 Ps, keep your pk ( whatever that means ) for your collection.
Can someone help to make this script working? class EllipticCurve: def __init__(self, p, a, b, g_x, g_y, n): self.p = p self.a = a self.b = b self.g_x = g_x self.g_y = g_y self.n = n
def add_points(self, p1, p2): # Point addition logic here pass
def subtract_points(self, p1, p2): # Point subtraction logic here pass
def scalar_multiply(self, point, scalar): # Scalar multiplication logic here pass
# Replace these values with your desired parameters N = ... P = ... G = (..., ...) # (x-coordinate, y-coordinate)
# Create an instance of the EllipticCurve class with your parameters curve = EllipticCurve(P, 0, 7, G[0], G[1], N)
# Replace with your target public key and desired number of subtractions target_public_key = "..." num = 100
# Example of point subtraction loop subtract_point = G # Initialize subtract_point with base point G
for t in range(num + 1): # Perform point subtraction result = curve.subtract_points(subtract_point, G)
# Convert the result to hexadecimal representation h = (result[0], result[1])
# Print or store the result as needed print("Subtraction result for iteration {}: {}".format(t, h))
# Update subtract_point for the next iteration subtract_point = result
# Note: You need to implement the add_points and subtract_points methods # with the actual point addition and subtraction logic for your curve.
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lordfrs
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September 03, 2023, 09:10:50 AM |
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from sympy import mod_inverse import secp256k1 as ice pub=ice.pub2upub('Here Compressed Public Key') N=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
k=mod_inverse(2,N) neg1=ice.point_negation(ice.scalar_multiplication(1))
def ters(Qx,Scalar): ScalarBin = str(bin(Scalar))[2:] le=len(ScalarBin) for i in range (1,le+1): if ScalarBin[le-i] == "0": Qx=ice.point_multiplication(k,Qx) else: Qx=ice.point_addition(Qx,neg1) Qx=ice.point_multiplication(k,Qx) return ice.point_to_cpub(Qx)
for x in range(1,65536): f= (ters(pub,x)) data= open(“halfpub.txt”,”a”) data.write(f+”\n”) data.close() Note this is where you decide how many bits should be reduced for x in range(1,65536): For example reducing 26 bits requires 67108864 to be generated, 1 of them is the correct 26 bit reduced key. Will you release for public to use? Yes this is my code, I can write bit reduction code in another way if you want. If you want, you can reduce bits by subtraction. I'm developing a new algorithm for downgrading from a known bit range to the bit range I want. When I complete it by giving only 1 correct pubkey, I will share it here.
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lordfrs
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September 03, 2023, 09:35:08 AM |
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i already find two puzzle with my methode of hash160 lowering the range of searching,and still im going to 130 135 my methode is working its just lowering and find the range in subed pub key to lower bit range 2 times yes when you find same 14 digits of hash160 than means that adress is in that range, but the search is not for one adress, i get a lot of addresses im trying , and no im not going to share the pk of any address i work hard for this, i advise you to try on 120 my methode, and get your bitcoin cash prize, ps: ill leave 50 satoshi on 130 for you, thank you
So you already solved 120 and 125? Can we see their private keys? Since you have worked hard for them, and don't want to share them, you could sign a message with them without revealing them. I advise you to grind all the puzzles with no mercy, don't even leave a single sat for anyone. Kalos the merciless.😅 Ps, keep your pk ( whatever that means ) for your collection.
Can someone help to make this script working? class EllipticCurve: def __init__(self, p, a, b, g_x, g_y, n): self.p = p self.a = a self.b = b self.g_x = g_x self.g_y = g_y self.n = n
def add_points(self, p1, p2): # Point addition logic here pass
def subtract_points(self, p1, p2): # Point subtraction logic here pass
def scalar_multiply(self, point, scalar): # Scalar multiplication logic here pass
# Replace these values with your desired parameters N = ... P = ... G = (..., ...) # (x-coordinate, y-coordinate)
# Create an instance of the EllipticCurve class with your parameters curve = EllipticCurve(P, 0, 7, G[0], G[1], N)
# Replace with your target public key and desired number of subtractions target_public_key = "..." num = 100
# Example of point subtraction loop subtract_point = G # Initialize subtract_point with base point G
for t in range(num + 1): # Perform point subtraction result = curve.subtract_points(subtract_point, G)
# Convert the result to hexadecimal representation h = (result[0], result[1])
# Print or store the result as needed print("Subtraction result for iteration {}: {}".format(t, h))
# Update subtract_point for the next iteration subtract_point = result
# Note: You need to implement the add_points and subtract_points methods # with the actual point addition and subtraction logic for your curve. Pcurve = 2**256 - 2**32 - 2**9 - 2**8 - 2**7 - 2**6 - 2**4 -1 # The proven prime N=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141 # Number of points in the field Acurve = 0; Bcurve = 7 # This defines the curve. y^2 = x^3 + Acurve * x + Bcurve Gx = 55066263022277343669578718895168534326250603453777594175500187360389116729240 Gy = 32670510020758816978083085130507043184471273380659243275938904335757337482424
def ECsub(point1x,point1y,point2x,point2y): neg_yq = Pcurve - point2y return ECadd(point1x,point1y,point2x,neg_yq) # point1-point2
you can use ready ice library, it will be faster Ice.pointnegatition finds the inverse of the point. It will be extracted with ice.pointadd The simple logic of the subtraction is to subtract the cord y from the total point point
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james5000
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September 03, 2023, 11:04:51 AM |
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Is there a way to determine if a point on the curve is even or odd?
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albert0bsd
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September 03, 2023, 11:15:41 AM |
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Is there a way to determine if a point on the curve is even or odd?
if i have a satoshi each time that i see that question... Sadly there is no way to determine if a point in the curve is odd or even, also there is no way to determine if bit in any position is 1 o 0 that would totally break ECDSAI already try a lot of things to try to anwser that question, but none of them works and actually i really give up that way, my recomendation is not lose time in that topic (Unless you are full cryptographer with wide knowledge in that field) Welcome to the club by the way.
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ing1996
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September 03, 2023, 11:39:09 AM |
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Is there a way to determine if a point on the curve is even or odd?
if i have a satoshi each time that i see that question... Sadly there is no way to determine if a point in the curve is odd or even, also there is no way to determine if bit in any position is 1 o 0 that would totally break ECDSAI already try a lot of things to try to anwser that question, but none of them works and actually i really give up that way, my recomendation is not lose time in that topic (Unless you are full cryptographer with wide knowledge in that field) Welcome to the club by the way. Hello 👋 all. If we divide an odd number X by 2, as a result we will have a remainder of X.5, is it also impossible to determine whether there is a remainder at this point?.
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yoshimitsu777
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September 03, 2023, 12:13:28 PM |
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hello.For private key 2 the x and y values of pubkey are
c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5 1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a
i change generator points of elliptic curve to the values above. let us assume we want to bruteforce scan the puzzles 1 to 20. end range of puzzle20 is 0x100000 and normaly we would scan 1:100000 but as i changed G I divide the range by 2 therefore new reduced scan range is 0x1 to 0x80000
they keys found are: 4 26 70 101 A30 1498 649B
now we multiply by 2 and get correct prvkeys for seven puzzles 4 7 8 10 13 14 16
we miss thirteen keys and i understand that it didnt find them because they are and our scan used even.
someone please show what are the exact steps to find also rest keys within that range?
please show example with private key 4 for understanding
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Denis_Hitov
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September 03, 2023, 12:56:05 PM |
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hello.For private key 2 the x and y values of pubkey are
c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5 1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a
i change generator points of elliptic curve to the values above. let us assume we want to bruteforce scan the puzzles 1 to 20. end range of puzzle20 is 0x100000 and normaly we would scan 1:100000 but as i changed G I divide the range by 2 therefore new reduced scan range is 0x1 to 0x80000
they keys found are: 4 26 70 101 A30 1498 649B
now we multiply by 2 and get correct prvkeys for seven puzzles 4 7 8 10 13 14 16
we miss thirteen keys and i understand that it didnt find them because they are and our scan used even.
someone please show what are the exact steps to find also rest keys within that range?
please show example with private key 4 for understanding
Please post the code so it's more clear what you're talking about.
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citb0in
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September 03, 2023, 04:18:48 PM |
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a mathematical curiosity that maybe could help the puzzle: all numbers even that respects this succession 4,10,16,22,28,34,40,46.....To infinity divided by 3 plus the sum of +1 to the same number divided by 3, results in an integer, odd number. target = 100 target_2 = 100+1 #= 101
t1= target//3 #= 33.333333333333336 t2= target_2//3 #= 33.666666666666664
r= t1+t2 # = 67 ---snipp--- t1 = t2 = 33 r will not result in 67 as you said, r=66
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