Search content
Sort by
Showing 19 of 19 results by mckemo
Re: [CHALLENGE] 5 BTC Reward – ECDSA Structured Nonce k Puzzle (1M Signatures)
I’m releasing a cryptographic challenge designed for experts in Bitcoin ECDSA internals, elliptic curve analysis, and nonce pattern vulnerabilities. There is a 5 BTC reward for the first person who solves it. Read the structure carefully.
The Puzzle Overview:
I have posted a file containing 1,000,000 valid signatures for one fixed public key.
Each signature is given as (sig_num, r, s, z, A, rx, ry) where:
r, s, and z are standard ECDSA signature parameters (z is the hashed message)
A is added to the private key to produce the nonce k used for that signature
k = d + A (mod n)
rx, ry are the elliptic curve point coordinates of r
PUBLIC KEY:
04c1c1e912c51061424286bdea075e0a19a96be1869566f4ebc9ea3e565f9c334d1779371fd313e dc2955b14f3eaabf8af027f77a7b3e1e908839d4f7ee81aef28
X = 87638989873003743107580407194345607023493955367007042197569832403610862629709
Y = 10617364466289823353593438673072375587688363537404447133738278709366498193192
Important Details
The nonce values for signatures increase incrementally:
1st signature: k = d + 2
2nd signature: k = d + 3
3rd signature: k = d + 4
... and so forth, continuing this pattern for all 1,000,000 signatures.
In every signature, r == s
Apply for only this dataset,
CSV file download (1 million signatures): https://[Suspicious link removed]/1million_rsz
Edit : Additional 2,000,000 ECDSA Signatures Released for Making this puzzle Solvable
Data Access: https://[Suspicious link removed]/Another1M_RSZ
Data Access: https://[Suspicious link removed]/Another1M_RSZ2
Important: All signatures satisfy r ≠ s.
Each line contains: r, s, z, ry
where r and s are the ECDSA signature components, z is the message hash, and ry is the y-coordinate of the curve point corresponding to r.
Bounty:
Recover any valid k and post it here with the corresponding signature index.
Or recover the private key directly from any subset of signatures.
Post your result here along with your Bitcoin address to receive the bounty.
💰 Prize: 5 BTC
⏱ Paid within 24 hours of verified result.
If you successfully recover any nonce k or the private key d, post your result here along with your BTC address to receive the 5 BTC reward.
Rewards & Rules
5 BTC payout within 24 hours after proof of valid nonce or private key recovery.
Puzzle Purpose:
This cryptographic challenge is an integral step in advancing the development of a novel cryptocurrency protocol inspired by Bitcoin’s UTXO model, yet architected with quantum-resistant cryptographic primitives. By analyzing structured ECDSA nonce patterns and their vulnerabilities, the goal is to rigorously test classical elliptic curve assumptions, improve nonce generation schemes, and inform the design of next-generation signature algorithms resilient against quantum adversaries.
The Puzzle Overview:
I have posted a file containing 1,000,000 valid signatures for one fixed public key.
Each signature is given as (sig_num, r, s, z, A, rx, ry) where:
r, s, and z are standard ECDSA signature parameters (z is the hashed message)
A is added to the private key to produce the nonce k used for that signature
k = d + A (mod n)
rx, ry are the elliptic curve point coordinates of r
PUBLIC KEY:
04c1c1e912c51061424286bdea075e0a19a96be1869566f4ebc9ea3e565f9c334d1779371fd313e dc2955b14f3eaabf8af027f77a7b3e1e908839d4f7ee81aef28
X = 87638989873003743107580407194345607023493955367007042197569832403610862629709
Y = 10617364466289823353593438673072375587688363537404447133738278709366498193192
Important Details
The nonce values for signatures increase incrementally:
1st signature: k = d + 2
2nd signature: k = d + 3
3rd signature: k = d + 4
... and so forth, continuing this pattern for all 1,000,000 signatures.
In every signature, r == s
Apply for only this dataset,
CSV file download (1 million signatures): https://[Suspicious link removed]/1million_rsz
Edit : Additional 2,000,000 ECDSA Signatures Released for Making this puzzle Solvable
Data Access: https://[Suspicious link removed]/Another1M_RSZ
Data Access: https://[Suspicious link removed]/Another1M_RSZ2
Important: All signatures satisfy r ≠ s.
Each line contains: r, s, z, ry
where r and s are the ECDSA signature components, z is the message hash, and ry is the y-coordinate of the curve point corresponding to r.
Bounty:
Recover any valid k and post it here with the corresponding signature index.
Or recover the private key directly from any subset of signatures.
Post your result here along with your Bitcoin address to receive the bounty.
💰 Prize: 5 BTC
⏱ Paid within 24 hours of verified result.
If you successfully recover any nonce k or the private key d, post your result here along with your BTC address to receive the 5 BTC reward.
Rewards & Rules
5 BTC payout within 24 hours after proof of valid nonce or private key recovery.
Puzzle Purpose:
This cryptographic challenge is an integral step in advancing the development of a novel cryptocurrency protocol inspired by Bitcoin’s UTXO model, yet architected with quantum-resistant cryptographic primitives. By analyzing structured ECDSA nonce patterns and their vulnerabilities, the goal is to rigorously test classical elliptic curve assumptions, improve nonce generation schemes, and inform the design of next-generation signature algorithms resilient against quantum adversaries.
To the person who created this problem...
Please solve this problem first.
There are 13.5 bits.
It is exactly the same, R == S, and Z is created by me.
I can create millions of them, but it's a waste of time, so I will only present two problems in the same format.
Here is the public key:
02145d2611c823a396ef6712ce0f712f09b9b4f3135e3e0aa3230fb9b6d08d1e16
R = 0x4806fcc582332d33610d925fc06afcbb8b141cb4d87fa401effdfb59d4011f99
S = 0x4806fcc582332d33610d925fc06afcbb8b141cb4d87fa401effdfb59d4011f99
Z = 0xa5f9777f19c3d9d3443e09e0a8120b413da3a78bc8b182791bf14cd4a33466a3
R = 0xe12ec23c6d7ee189c248807bac60206a5cf8ea7f899214ca9d0a2a8ddc44d7cf
S = 0xe12ec23c6d7ee189c248807bac60206a5cf8ea7f899214ca9d0a2a8ddc44d7cf
Z = 0x6d17d3bc1305497d13b0bc8f7abe94ccd448a1377ea79d5e7f57da8d4ede8f1
Nice try, this belongs to 135 puzzle xDDD
Re: [CHALLENGE] 5 BTC Reward – ECDSA Structured Nonce k Puzzle (1M Signatures)
⭐ Merited by garlonicon (1)
I call bullshit here.
- r mod 256 Verteilung (Top 5):
Wert 0x8f : 4077 mal
Wert 0x29 : 4065 mal
Wert 0x91 : 4037 mal
Wert 0xd7 : 4034 mal
Wert 0xfc : 4032 mal
- s mod 256 Verteilung (Top 5):
Wert 0xfa : 4057 mal
Wert 0xaf : 4046 mal
Wert 0x82 : 4045 mal
Wert 0xc0 : 4044 mal
Wert 0xe6 : 4039 mal
- (r - s) mod 256 Verteilung (Top 5):
Wert 0xbb : 4077 mal
Wert 0xc5 : 4055 mal
Wert 0xe4 : 4054 mal
Wert 0x73 : 4039 mal
Wert 0x41 : 4038 mal
PS C:\Users\mc_ke\Desktop> python3 comp.py signatures.csv
Verarbeitet 100000 Signaturen...
Verarbeitet 200000 Signaturen...
Verarbeitet 300000 Signaturen...
Verarbeitet 400000 Signaturen...
Verarbeitet 500000 Signaturen...
Verarbeitet 600000 Signaturen...
Verarbeitet 700000 Signaturen...
Verarbeitet 800000 Signaturen...
Verarbeitet 900000 Signaturen...
Verarbeitet 1000000 Signaturen...
Gesamtanzahl Signaturen: 1000000
Anzahl r == s: 0 (0.00%)
- r mod 256 Verteilung (Top 5):
Wert 0x8f : 4077 mal
Wert 0x29 : 4065 mal
Wert 0x91 : 4037 mal
Wert 0xd7 : 4034 mal
Wert 0xfc : 4032 mal
- s mod 256 Verteilung (Top 5):
Wert 0xfa : 4057 mal
Wert 0xaf : 4046 mal
Wert 0x82 : 4045 mal
Wert 0xc0 : 4044 mal
Wert 0xe6 : 4039 mal
- (r - s) mod 256 Verteilung (Top 5):
Wert 0xbb : 4077 mal
Wert 0xc5 : 4055 mal
Wert 0xe4 : 4054 mal
Wert 0x73 : 4039 mal
Wert 0x41 : 4038 mal
- r mod 65536 Verteilung (Top 5):
Wert 0xc5b1 : 36 mal
Wert 0x1da8 : 34 mal
Wert 0x2d4b : 34 mal
Wert 0x7420 : 33 mal
Wert 0xad74 : 32 mal
- s mod 65536 Verteilung (Top 5):
Wert 0x7ff6 : 35 mal
Wert 0x5f03 : 33 mal
Wert 0xd24d : 33 mal
Wert 0x1262 : 32 mal
Wert 0x3be4 : 32 mal
PS C:\Users\mc_ke\Desktop> python3 comp.py signatures.csv
Anzahl Signaturen geladen: 1000000
--- Mod 256 Verteilungen (Top 5) ---
r mod 256: [(143, 4077), (41, 4065), (145, 4037), (215, 4034), (252, 4032)]
s mod 256: [(250, 4057), (175, 4046), (130, 4045), (192, 4044), (230, 4039)]
(r - s) mod 256: [(187, 4077), (197, 4055), (228, 4054), (115, 4039), (65, 4038)]
--- Mod 65536 Verteilungen (Top 5) ---
r mod 65536: [(50609, 36), (7592, 34), (11595, 34), (29728, 33), (44404, 32)]
s mod 65536: [(32758, 35), (24323, 33), (53837, 33), (4706, 32), (15332, 32)]
--- Bitlängen Histogramme ---
r bit length: [(256, 499930), (255, 250110), (254, 124566), (253, 62505), (252, 31318), (251, 15894), (250, 7844), (249, 3956), (248, 1983), (247, 965)]
s bit length: [(256, 500272), (255, 250244), (254, 124687), (253, 62349), (252, 31290), (251, 15578), (250, 7839), (249, 3862), (248, 1955), (247, 967)]
z bit length: [(256, 499691), (255, 249950), (254, 125122), (253, 62729), (252, 31026), (251, 15892), (250, 7690), (249, 4027), (248, 1980), (247, 974)]
--- Ry mod 256 Verteilung (Top 5) ---
[(249, 4111), (219, 4083), (120, 4078), (191, 4064), (27, 4051)]
--- Korrelationen mod 256 ---
r vs s: 0.0021
r vs z: -0.0006
s vs z: -0.0007
r vs Ry: -0.0009
s vs Ry: -0.0001
--- Suche nach Signaturpaaren mit gleichem r (für Schlüssel-/Nonce-Recovery) ---
Keine Paare mit gleichem r gefunden, versuche weitere Analysen...
--- Häufige Kollisionswerte der unteren 16 Bits ---
r:
Häufigste Werte der unteren 16 Bits:
Wert 0xc5b1 : 36 mal
Wert 0x1da8 : 34 mal
Wert 0x2d4b : 34 mal
Wert 0x7420 : 33 mal
Wert 0xad74 : 32 mal
Wert 0x154d : 32 mal
Wert 0xfc1c : 32 mal
Wert 0x883a : 31 mal
Wert 0xc9fc : 31 mal
Wert 0x6296 : 31 mal
s:
Häufigste Werte der unteren 16 Bits:
Wert 0x7ff6 : 35 mal
Wert 0x5f03 : 33 mal
Wert 0xd24d : 33 mal
Wert 0x1262 : 32 mal
Wert 0x3be4 : 32 mal
Wert 0xf292 : 32 mal
Wert 0x4712 : 32 mal
Wert 0x87b3 : 31 mal
Wert 0x9574 : 31 mal
Wert 0xf1d7 : 31 mal
z:
Häufigste Werte der unteren 16 Bits:
Wert 0x24dd : 35 mal
Wert 0x20f0 : 33 mal
Wert 0xe731 : 33 mal
Wert 0x3275 : 33 mal
Wert 0x3725 : 32 mal
Wert 0xc70f : 32 mal
Wert 0x8f93 : 32 mal
Wert 0x2d6c : 32 mal
Wert 0x25a3 : 32 mal
Wert 0xeedd : 31 mal
- r mod 256 Verteilung (Top 5):
Wert 0x8f : 4077 mal
Wert 0x29 : 4065 mal
Wert 0x91 : 4037 mal
Wert 0xd7 : 4034 mal
Wert 0xfc : 4032 mal
- s mod 256 Verteilung (Top 5):
Wert 0xfa : 4057 mal
Wert 0xaf : 4046 mal
Wert 0x82 : 4045 mal
Wert 0xc0 : 4044 mal
Wert 0xe6 : 4039 mal
- (r - s) mod 256 Verteilung (Top 5):
Wert 0xbb : 4077 mal
Wert 0xc5 : 4055 mal
Wert 0xe4 : 4054 mal
Wert 0x73 : 4039 mal
Wert 0x41 : 4038 mal
PS C:\Users\mc_ke\Desktop> python3 comp.py signatures.csv
Verarbeitet 100000 Signaturen...
Verarbeitet 200000 Signaturen...
Verarbeitet 300000 Signaturen...
Verarbeitet 400000 Signaturen...
Verarbeitet 500000 Signaturen...
Verarbeitet 600000 Signaturen...
Verarbeitet 700000 Signaturen...
Verarbeitet 800000 Signaturen...
Verarbeitet 900000 Signaturen...
Verarbeitet 1000000 Signaturen...
Gesamtanzahl Signaturen: 1000000
Anzahl r == s: 0 (0.00%)
- r mod 256 Verteilung (Top 5):
Wert 0x8f : 4077 mal
Wert 0x29 : 4065 mal
Wert 0x91 : 4037 mal
Wert 0xd7 : 4034 mal
Wert 0xfc : 4032 mal
- s mod 256 Verteilung (Top 5):
Wert 0xfa : 4057 mal
Wert 0xaf : 4046 mal
Wert 0x82 : 4045 mal
Wert 0xc0 : 4044 mal
Wert 0xe6 : 4039 mal
- (r - s) mod 256 Verteilung (Top 5):
Wert 0xbb : 4077 mal
Wert 0xc5 : 4055 mal
Wert 0xe4 : 4054 mal
Wert 0x73 : 4039 mal
Wert 0x41 : 4038 mal
- r mod 65536 Verteilung (Top 5):
Wert 0xc5b1 : 36 mal
Wert 0x1da8 : 34 mal
Wert 0x2d4b : 34 mal
Wert 0x7420 : 33 mal
Wert 0xad74 : 32 mal
- s mod 65536 Verteilung (Top 5):
Wert 0x7ff6 : 35 mal
Wert 0x5f03 : 33 mal
Wert 0xd24d : 33 mal
Wert 0x1262 : 32 mal
Wert 0x3be4 : 32 mal
PS C:\Users\mc_ke\Desktop> python3 comp.py signatures.csv
Anzahl Signaturen geladen: 1000000
--- Mod 256 Verteilungen (Top 5) ---
r mod 256: [(143, 4077), (41, 4065), (145, 4037), (215, 4034), (252, 4032)]
s mod 256: [(250, 4057), (175, 4046), (130, 4045), (192, 4044), (230, 4039)]
(r - s) mod 256: [(187, 4077), (197, 4055), (228, 4054), (115, 4039), (65, 4038)]
--- Mod 65536 Verteilungen (Top 5) ---
r mod 65536: [(50609, 36), (7592, 34), (11595, 34), (29728, 33), (44404, 32)]
s mod 65536: [(32758, 35), (24323, 33), (53837, 33), (4706, 32), (15332, 32)]
--- Bitlängen Histogramme ---
r bit length: [(256, 499930), (255, 250110), (254, 124566), (253, 62505), (252, 31318), (251, 15894), (250, 7844), (249, 3956), (248, 1983), (247, 965)]
s bit length: [(256, 500272), (255, 250244), (254, 124687), (253, 62349), (252, 31290), (251, 15578), (250, 7839), (249, 3862), (248, 1955), (247, 967)]
z bit length: [(256, 499691), (255, 249950), (254, 125122), (253, 62729), (252, 31026), (251, 15892), (250, 7690), (249, 4027), (248, 1980), (247, 974)]
--- Ry mod 256 Verteilung (Top 5) ---
[(249, 4111), (219, 4083), (120, 4078), (191, 4064), (27, 4051)]
--- Korrelationen mod 256 ---
r vs s: 0.0021
r vs z: -0.0006
s vs z: -0.0007
r vs Ry: -0.0009
s vs Ry: -0.0001
--- Suche nach Signaturpaaren mit gleichem r (für Schlüssel-/Nonce-Recovery) ---
Keine Paare mit gleichem r gefunden, versuche weitere Analysen...
--- Häufige Kollisionswerte der unteren 16 Bits ---
r:
Häufigste Werte der unteren 16 Bits:
Wert 0xc5b1 : 36 mal
Wert 0x1da8 : 34 mal
Wert 0x2d4b : 34 mal
Wert 0x7420 : 33 mal
Wert 0xad74 : 32 mal
Wert 0x154d : 32 mal
Wert 0xfc1c : 32 mal
Wert 0x883a : 31 mal
Wert 0xc9fc : 31 mal
Wert 0x6296 : 31 mal
s:
Häufigste Werte der unteren 16 Bits:
Wert 0x7ff6 : 35 mal
Wert 0x5f03 : 33 mal
Wert 0xd24d : 33 mal
Wert 0x1262 : 32 mal
Wert 0x3be4 : 32 mal
Wert 0xf292 : 32 mal
Wert 0x4712 : 32 mal
Wert 0x87b3 : 31 mal
Wert 0x9574 : 31 mal
Wert 0xf1d7 : 31 mal
z:
Häufigste Werte der unteren 16 Bits:
Wert 0x24dd : 35 mal
Wert 0x20f0 : 33 mal
Wert 0xe731 : 33 mal
Wert 0x3275 : 33 mal
Wert 0x3725 : 32 mal
Wert 0xc70f : 32 mal
Wert 0x8f93 : 32 mal
Wert 0x2d6c : 32 mal
Wert 0x25a3 : 32 mal
Wert 0xeedd : 31 mal
Re: Presenting a new version of Pollard's kangaroo secp256k1 ECDLP solver
I just want to update that the code now supports 254-bit range and multiple GPUs (on linux only).
Can not compile it:
Merge.cpp:10:21: error: 'filesystem' is not a namespace-name
10 | namespace fs = std::filesystem;
| ^~~~~~~~~~
Merge.cpp: In function 'bool isRegularFile(const string&)':
Merge.cpp:20:12: error: 'fs' has not been declared
20 | return fs::is_regular_file(filePath);
| ^~
Merge.cpp: In member function 'void Kangaroo::MergeDir(std::string&, std::string&)':
Merge.cpp:77:30: error: 'fs' has not been declared
77 | for (const auto& entry : fs::directory_iterator(dirName)) {
| ^~
Merge.cpp:78:13: error: 'fs' has not been declared
78 | if (fs::is_regular_file(entry.path())) {
| ^~
make: *** [Makefile:60: obj/Merge.o] Error 1
If the pubkey is revealed,may someone use BSGS algorithm to crack the privkey and double spend the coins? 
The public key is already known information, but that is not enough to crack the private key, and if someone get's access to the private key first they need to spend the coins before be able to double spend it. But looks like you are lost in the topic, what op means is.
Let's say someone finds the private key from puzzles 66 and 67, if they spend the coins and make public the private key, som users could use that private key to make a double spend from the same coins and if they use bigger fees then they could steal those coins. But the way to avoid that is to not make public the private key until it has more than 1 confirmation.
if the public key is known,the bsgs will solve it in seconds. i get 48 exahashes per second.
Hello People.
I need somebody who can help me, building a new chain based on 0.1.0 with same settings and stuff. Like a reset on a private network.
Do you think it is possible? Same genesis-block. same settings (not the network settings since i will only run on one client and try to mine)..
do you think it is possible?
I need somebody who can help me, building a new chain based on 0.1.0 with same settings and stuff. Like a reset on a private network.
Do you think it is possible? Same genesis-block. same settings (not the network settings since i will only run on one client and try to mine)..
do you think it is possible?
blocktime is 10 mins - do you think it is possible, that you e.g. wait 9 mins, set the reward extremly high (like 0.09 btc) and distribute it ?
The 10 minutes is the average time. There are enough cases in which no block has fallen for almost an hour and other cases in which several blocks have fallen within a minute. So there is no guarantee that you will send the transaction to the net just before 10 minutes have elapsed and the next block will fall shortly afterwards. That would be a huge amount of luck that you would also need. The only way out of this dilemma, as already mentioned several times, is to find a large miner who will include your transaction in the next block to be calculated. This large miner must have so much power that he regularly mines blocks and the probability is in your favor that he will ever mine a block. And of course there must be an incentive for this miner, perhaps in the form of a reasonable commission, to comply with your request and make the extra effort to do so. Very unlikely, but not impossible.
All other dubious and questionable methods mentioned here are not target-oriented but misleading and inspire false confidence, just like the last hopeless appeal of a desperate user. Even if you were to find out the key and provide some form of proof here in the forum that you know the key, it would not change the fact that you have no legal right to open the vault and obtain the contents. To make this clear, let's take a simple example: imagine that Alice has known the key from #66 for over a year. This fact alone does not make her the owner of the contents of the safe from puzzle #66. If Mallory finds out the key 3 years later and grabs the coins from #66, they belong to Mallory. That's the way the game is, that's how it's defined in Bitcoin and there is no "law" or "rules". The mighty power is written in Bitcoin protocol and everything is defined in there.
like there is no max or smth. or a security. they should have integrated, that coins are locked in some way as long as the transaction is ongoing and that you can not change the destination address.
What if someone found the key and spent the btc to another wallet ( i know, pubkey...) but with a super high tx fee, like extremely high.
Which key? puzzle key? 66?
If the 66 key is spent it may enter in some bot fight each one will increase the fees compared to other until there is nothing left to spent but the final result is unknown
Not always the higher fee wins, chek my previous reply to this:
Just to mention that when the nodes have many FullRBF transacions not always win that one with more fee, here some examples, Dot with Green margin was mined, some are Testnet and other are mainnet
https://talkimg.com/images/2024/03/19/JJJn9.png
Here was mined a TX with 1 sat/vB instead of a 37 sat/vB
https://talkimg.com/images/2024/03/19/JJVvN.png
Here was mined a TX with 22 sat/vB instead of a 44 sat/vB
https://talkimg.com/images/2024/03/19/JJjia.png
Here was mined a TX with 106 sat/vB instead of a 1032 sat/vB
https://talkimg.com/images/2024/03/19/JJJn9.png
Here was mined a TX with 1 sat/vB instead of a 37 sat/vB
https://talkimg.com/images/2024/03/19/JJVvN.png
Here was mined a TX with 22 sat/vB instead of a 44 sat/vB
https://talkimg.com/images/2024/03/19/JJjia.png
Here was mined a TX with 106 sat/vB instead of a 1032 sat/vB
All depens of what is the transaction that the miner selected to work with his hardware.
I tried to import a private key into bitcoin core but i got this error :
Only legacy wallets are supported by this command (code -4)
what should i do now?
Only legacy wallets are supported by this command (code -4)
what should i do now?
If you don't show what command do you used, then there is no way to know what is wrong.
What kind of address are you trying to import? it looks like it is not a legacy one.
Wow, i just learned that this features exists. so even having the key for 66, you are not safe because of bots just waiting to get a signal from the network and getting the privkey within seconds via bsgs /kangaroo.
blocktime is 10 mins - do you think it is possible, that you e.g. wait 9 mins, set the reward extremly high (like 0.09 btc) and distribute it ?
there is no pattern.
i can exactly tell you why:
The creator of the puzzle posted already, that he created a wallet and masked the keys to match in the key range with 0
Therefore 256 adresses have been generated and he just put as many 000 to fit in. So it is kinda random.
No rythm, no nothing involved.
i can exactly tell you why:
The creator of the puzzle posted already, that he created a wallet and masked the keys to match in the key range with 0
Therefore 256 adresses have been generated and he just put as many 000 to fit in. So it is kinda random.
No rythm, no nothing involved.
Re: Bitcoin’s, Altcoin’s Private Keys are all Compromised. Second proof.
But why do you only point to Alamy, there are lots of images that appear in the result. There's facebook, X, github, and Pinterest and there's even tiktok. Does it mean those keys are also on those websites?
I doubt there is truth to this post. I have kept my private key somewhere safe, I created a wallet that has never been used but only deposit transactions I'm sure there will be nothing of public key that is exposed by it. How will the wallet keys be exposed with that security in mind? This is just something for that meduim page, fud for the crypto holders.
i dont get a single alamy site. it depends on your locationI doubt there is truth to this post. I have kept my private key somewhere safe, I created a wallet that has never been used but only deposit transactions I'm sure there will be nothing of public key that is exposed by it. How will the wallet keys be exposed with that security in mind? This is just something for that meduim page, fud for the crypto holders.
Thats the problem i got here.
The tool von iceland is indeed working. It reduces to a lower keyspace precisely.
But i don´t understand what exactly it is doing.
And thats the point-> I can not reverse anything what i don´t understand what has been done.
Thats what i am asking for. Can somebody help me to understand, what iceland exactly has done, and how to reverse it onto the found privkey?
Keymath won´t help here, because i don´t know, what i got to enter..
i don´t know the divisor or how to even get it..
It seems like it is subtracting with Generatorpoint G, but.. what about the privkey?!
The tool von iceland is indeed working. It reduces to a lower keyspace precisely.
But i don´t understand what exactly it is doing.
And thats the point-> I can not reverse anything what i don´t understand what has been done.
Thats what i am asking for. Can somebody help me to understand, what iceland exactly has done, and how to reverse it onto the found privkey?
Keymath won´t help here, because i don´t know, what i got to enter..
i don´t know the divisor or how to even get it..
It seems like it is subtracting with Generatorpoint G, but.. what about the privkey?!
Thank you for the reply.
But i think this is the wrong turn.
Lets say i use the pubkey from puzzle 35 02f6a8148a62320e149cb15c544fe8a25ab483a0095d2280d03b8a00a7feada13d
And i let the program run and i want to reduce by 10->
it generates me 1024 keys.
one of the keys will be in range 25.
This works like a charm - >
The key 028490315e76ba204a01059765fadf4a4b5c73cc90e86dee99bc7f378a2b8f3e01 is found. -> Privkey 12bb484
i asked chatgpt to put in, what has been calculated and it added "Derived by subtracting 368 times G"
i dont think this is true..
what math do i have to apply, to get the correct privkey (4aed21170) now?
i can´t get it to work
But i think this is the wrong turn.
Lets say i use the pubkey from puzzle 35 02f6a8148a62320e149cb15c544fe8a25ab483a0095d2280d03b8a00a7feada13d
And i let the program run and i want to reduce by 10->
it generates me 1024 keys.
one of the keys will be in range 25.
This works like a charm - >
The key 028490315e76ba204a01059765fadf4a4b5c73cc90e86dee99bc7f378a2b8f3e01 is found. -> Privkey 12bb484
i asked chatgpt to put in, what has been calculated and it added "Derived by subtracting 368 times G"
i dont think this is true..
what math do i have to apply, to get the correct privkey (4aed21170) now?
i can´t get it to work
Hi.
i divided a pubkey with Iceland2k´s PubDiv.
It works and reduced the range as promised.
But how can i now calculate the Privkey back?
https://github.com/iceland2k14/quick
Somebody got an idea?
i divided a pubkey with Iceland2k´s PubDiv.
It works and reduced the range as promised.
But how can i now calculate the Privkey back?
https://github.com/iceland2k14/quick
Somebody got an idea?
Re: [ANN] Dogethereum | Fork 10 000:1 of DOGE [Snapshot: 30.03.2018]
You quoted it already.... but i will Show you again in green. Anyway, they want you to Import the wallet.dat from you dogecoin wallet. so be carefull because then they got your private key.
who is actually responsible for the announcements from the devs
fuck to hell........................
Announcement Twitter
#DogeCoin gets its first fork, congratulations! Snapshot block of #Dogethereum is 2157611 03/30/2018
04:50 - 25. März 2018
30/3/2018
http://dogethereum.net/category/news/
DogeCoin snapshot 03/26/2018 (~2151868 block), everyone whos had DogeCoins before that date
What Date, what Block??
unforgiving if the announcements do not match
stupid mistakes must not occur, who else should take you seriously
fuck to hell........................
Announcement Twitter
#DogeCoin gets its first fork, congratulations! Snapshot block of #Dogethereum is 2157611 03/30/2018
04:50 - 25. März 2018
30/3/2018
http://dogethereum.net/category/news/
DogeCoin snapshot 03/26/2018 (~2151868 block), everyone whos had DogeCoins before that date
What Date, what Block
??Product roadmap
◆ Establish secure stable network - DONE
◆ Warm up and prepare network for public access - DONE
◆ Fork of all coins DOGE 10.000:1 DOGX 30/03/2018
◆ Public advertisement and promo - after release
◆ Listing on crypto Exchanges - after release
◆ Research and development of faster algorithm with better scalability - 2018-2019
◆ TO THE MOON!
Technical details
◆ ICO - NO
◆ Premine - NO
◆ ASIC - NO (only GPU PoW)
◆ Algorithm - Ethash
◆ Block Time - 13 seconds
◆ Block Reward - 3 DOGX
Release date 30/03/2018
Snapshot Block - 2157611
◆ Establish secure stable network - DONE
◆ Warm up and prepare network for public access - DONE
◆ Fork of all coins DOGE 10.000:1 DOGX 30/03/2018
◆ Public advertisement and promo - after release
◆ Listing on crypto Exchanges - after release
◆ Research and development of faster algorithm with better scalability - 2018-2019
◆ TO THE MOON!
Technical details
◆ ICO - NO
◆ Premine - NO
◆ ASIC - NO (only GPU PoW)
◆ Algorithm - Ethash
◆ Block Time - 13 seconds
◆ Block Reward - 3 DOGX
Release date 30/03/2018
Snapshot Block - 2157611
unforgiving if the announcements do not match
stupid mistakes must not occur, who else should take you seriously
Re: [ANN] Dogethereum | Fork 10 000:1 of DOGE [Snapshot: 30.03.2018]
Of Course they ran into Problems.
They want to take a coin and convert it into a token.
U can´t mine TOKENS..
so they take 10000 doge which is 400000 Satoshi for one of their fake shitcoins. it is not a fork at all, you can call it airdrop if you want.
They want to take a coin and convert it into a token.
U can´t mine TOKENS..
so they take 10000 doge which is 400000 Satoshi for one of their fake shitcoins. it is not a fork at all, you can call it airdrop if you want.