eggsylacer
Jr. Member

Activity: 57
Merit: 2
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September 03, 2026, 10:55:38 PM |
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This is not my logic, is a math fact.. Scan any range you want sequential and see what i am talking about. Using kangaroo you cannot see the Address...only public key. Use a vanity and scan a /30 full and store local, after that do a statistic to see how many have 33 and 34... You will see aprox 93% of 34 length. You exclude those from generation and you won 7% time. Adding more exclusions like 4 identical letters or numbers in first 7 hex and much more conditions, and you gain up to 40%
I’m telling you the same thing, but within the points on the curve. Following your logic, the complexity of the search can be reduced by scanning the range. But you apparently don’t understand what complexity is.
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puzzle_72_worker
Jr. Member

Activity: 43
Merit: 15
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September 04, 2026, 05:09:10 AM |
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Some of you got in this domain or subject but you have no ideea what you are doing here. What i am saying and what you responde.... I gave the ideea, now if anybody know what i am saying, please check with real facts, if not good luck on your way loosing time and complicate things.
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detechs
Jr. Member

Activity: 90
Merit: 1
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September 04, 2026, 09:09:37 AM |
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This is not my logic, is a math fact.. Scan any range you want sequential and see what i am talking about. Using kangaroo you cannot see the Address...only public key. Use a vanity and scan a /30 full and store local, after that do a statistic to see how many have 33 and 34... You will see aprox 93% of 34 length. You exclude those from generation and you won 7% time. Adding more exclusions like 4 identical letters or numbers in first 7 hex and much more conditions, and you gain up to 40%
I’m telling you the same thing, but within the points on the curve. Following your logic, the complexity of the search can be reduced by scanning the range. But you apparently don’t understand what complexity is. How do you prove the key doesn't contain those filters without scanning them to double check?
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Bitcoin Puzzle Solvers Discord https://discord.gg/3TVn3PGYtu
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eggsylacer
Jr. Member

Activity: 57
Merit: 2
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September 04, 2026, 09:31:51 AM |
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Some of you got in this domain or subject but you have no ideea what you are doing here. What i am saying and what you responde.... I gave the ideea, now if anybody know what i am saying, please check with real facts, if not good luck on your way loosing time and complicate things.
That’s exactly it you don’t even know the definitions of the terms. What can you even talk about? The fact that you simply “discarded” part of the keys does not mean that the computational complexity of the problem has decreased on its own. Since our problem has a condition that each key is in the range from 2^n to 2^(n+1) - 1, the complexity will accordingly be the same. Of course, I don’t deny the fact of “luck,” and you might just happen to get lucky with the key.
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Pac_Man
Newbie

Activity: 9
Merit: 0
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September 04, 2026, 09:34:38 AM |
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This is not my logic, is a math fact.. Scan any range you want sequential and see what i am talking about. Using kangaroo you cannot see the Address...only public key. Use a vanity and scan a /30 full and store local, after that do a statistic to see how many have 33 and 34... You will see aprox 93% of 34 length. You exclude those from generation and you won 7% time. Adding more exclusions like 4 identical letters or numbers in first 7 hex and much more conditions, and you gain up to 40%
I think you're missing something there. You say that excluding 34-digit addresses saves 7% of the time, but you don't say how you gain that 7%. Because when you're analyzing P2PKH keys, you're actually losing that 7%. By excluding those addresses, you'll only gain the time the code takes to verify the generated addresses with the target address already in memory, and you definitely won't gain any 7%. If, on the other hand, verifying the number of digits is the first condition of all the conditions you created, you'll only gain the time the code would take to go through all those conditions you created, but in any case, the entire process would have to be done by the code until you obtain the address that will be analyzed by all the filters you created. Honestly, I don't see any gain in that.
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eggsylacer
Jr. Member

Activity: 57
Merit: 2
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September 04, 2026, 09:35:29 AM |
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This is not my logic, is a math fact.. Scan any range you want sequential and see what i am talking about. Using kangaroo you cannot see the Address...only public key. Use a vanity and scan a /30 full and store local, after that do a statistic to see how many have 33 and 34... You will see aprox 93% of 34 length. You exclude those from generation and you won 7% time. Adding more exclusions like 4 identical letters or numbers in first 7 hex and much more conditions, and you gain up to 40%
I’m telling you the same thing, but within the points on the curve. Following your logic, the complexity of the search can be reduced by scanning the range. But you apparently don’t understand what complexity is. How do you prove the key doesn't contain those filters without scanning them to double check? How can we prove that the key contains filters? As mentioned earlier, these are just random values. Before claiming that there are “hidden” patterns here, study how generators and pseudo‑randomness work in principle.
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CrabStick1
Newbie

Activity: 3
Merit: 1
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September 04, 2026, 01:36:55 PM Merited by eggsylacer (1) |
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This is not my logic, is a math fact.. Scan any range you want sequential and see what i am talking about. Using kangaroo you cannot see the Address...only public key. Use a vanity and scan a /30 full and store local, after that do a statistic to see how many have 33 and 34... You will see aprox 93% of 34 length. You exclude those from generation and you won 7% time. Adding more exclusions like 4 identical letters or numbers in first 7 hex and much more conditions, and you gain up to 40%
I’m telling you the same thing, but within the points on the curve. Following your logic, the complexity of the search can be reduced by scanning the range. But you apparently don’t understand what complexity is. How do you prove the key doesn't contain those filters without scanning them to double check? How can we prove that the key contains filters? As mentioned earlier, these are just random values. Before claiming that there are “hidden” patterns here, study how generators and pseudo‑randomness work in principle. You are arguing with someone who said that the creator is a fan of iguanas animal and maybe he has one, and have 1-2 kids because has some children stories inside clues + freddy mercury fan.)))))
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puzzle_72_worker
Jr. Member

Activity: 43
Merit: 15
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September 04, 2026, 02:02:22 PM |
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Nobody needs to argue with no one. I gave you a math fact. If you cannot see behind the ideea or you cannot compute, this do not means is not existing. Don't you want to give you also the formula and result of this? Let's be serious... By generating only 34 length, you will see something interesting on how this BTC addresses are generated and how the hex keys compute. I gave you an ideea, the rest is cancan...do what you want with this info. I do not care about haters...if anyone who tried this and saw howis built, will understand, the rest will see this as garbage... feel free to use softwares that others built instead to build your own formulas and softwares...
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eggsylacer
Jr. Member

Activity: 57
Merit: 2
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September 04, 2026, 06:17:16 PM Last edit: September 05, 2026, 10:49:24 AM by Mr. Big |
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This is not my logic, is a math fact.. Scan any range you want sequential and see what i am talking about. Using kangaroo you cannot see the Address...only public key. Use a vanity and scan a /30 full and store local, after that do a statistic to see how many have 33 and 34... You will see aprox 93% of 34 length. You exclude those from generation and you won 7% time. Adding more exclusions like 4 identical letters or numbers in first 7 hex and much more conditions, and you gain up to 40%
I’m telling you the same thing, but within the points on the curve. Following your logic, the complexity of the search can be reduced by scanning the range. But you apparently don’t understand what complexity is. How do you prove the key doesn't contain those filters without scanning them to double check? How can we prove that the key contains filters? As mentioned earlier, these are just random values. Before claiming that there are “hidden” patterns here, study how generators and pseudo‑randomness work in principle. You are arguing with someone who said that the creator is a fan of iguanas animal and maybe he has one, and have 1-2 kids because has some children stories inside clues + freddy mercury fan.))))) I completely agree, it’s like talking to a wall.
Nobody needs to argue with no one. I gave you a math fact. If you cannot see behind the ideea or you cannot compute, this do not means is not existing. Don't you want to give you also the formula and result of this? Let's be serious... By generating only 34 length, you will see something interesting on how this BTC addresses are generated and how the hex keys compute. I gave you an ideea, the rest is cancan...do what you want with this info. I do not care about haters...if anyone who tried this and saw howis built, will understand, the rest will see this as garbage... feel free to use softwares that others built instead to build your own formulas and softwares...
Under no circumstances should you run unverified programs, and always check them for viruses on dedicated websites.
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mjojo
Newbie

Activity: 97
Merit: 0
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September 04, 2026, 11:10:22 PM |
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~
As a reminder Gx of secp256k1 is also a valid point on secQ256k1, maybe it would be better for the author to use that curve and generate Qs of puzzle exposed Qs and somehow share them. Let's try breaking the ECC before the machines do it for us. Gx = 53718550993811904772965658690407829053653678808745171666022356150019200052646 the same x point of N/2 in secp256k1 and secq256k1 N/2,x = 86918276961810349294276103416548851884759982251107 how many you found the same x point on both curve?
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eggsylacer
Jr. Member

Activity: 57
Merit: 2
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September 04, 2026, 11:17:38 PM |
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As a reminder Gx of secp256k1 is also a valid point on secQ256k1, maybe it would be better for the author to use that curve and generate Qs of puzzle exposed Qs and somehow share them. Let's try breaking the ECC before the machines do it for us. Gx = 53718550993811904772965658690407829053653678808745171666022356150019200052646 the same x point of N/2 in secp256k1 and secq256k1 N/2,x = 86918276961810349294276103416548851884759982251107 how many you found the same x point on both curve? There is no division on the curve; only addition and multiplication based on addition.
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brainless
Member


Activity: 505
Merit: 46
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September 05, 2026, 06:38:56 AM |
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Cyclone app with function f for load list ranges works on cpu, any other app for gpu ?
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13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF
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andyfibe
Newbie

Activity: 14
Merit: 0
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September 05, 2026, 06:50:34 AM Last edit: September 06, 2026, 04:34:34 AM by andyfibe |
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~
As a reminder Gx of secp256k1 is also a valid point on secQ256k1, maybe it would be better for the author to use that curve and generate Qs of puzzle exposed Qs and somehow share them. Let's try breaking the ECC before the machines do it for us. b1.global Fast-Paced AI Matches to Monetize Every Second Gx = 53718550993811904772965658690407829053653678808745171666022356150019200052646 the same x point of N/2 in secp256k1 and secq256k1 N/2,x = 86918276961810349294276103416548851884759982251107 how many you found the same x point on both curve? There is no division on the curve; only addition and multiplication based on addition. Spot on about addition and multiplication Even though there's no real division on the curve, multiplying by the inverse scalar ($k^{-1}$) pretty much do the same job
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fecell
Jr. Member

Activity: 196
Merit: 2
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September 05, 2026, 06:53:15 AM |
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multiplying by the inverse scalar (...) pretty much do the same job exactly!
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eggsylacer
Jr. Member

Activity: 57
Merit: 2
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September 05, 2026, 08:14:31 AM Last edit: September 05, 2026, 08:28:32 AM by eggsylacer |
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It says right in the text "multiplying Q by the modular multiplicative inverse of a modulo n". You don't even understand what you're reading, lol... if there was a division operation on the curve, then the curve would have been hacked long ago... That is, there would be no discrete logarithm problem.
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puzzle_72_worker
Jr. Member

Activity: 43
Merit: 15
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September 05, 2026, 08:49:16 AM |
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@eggsylacer as i was saying to others, are you a PHD, double PHD or nobel prize winner? That's the problem, here o forum who is arguing have no ideea about the crypto domain...you follow like sheeps the software built of others, use AI and you think that all the things AI is giving you is perfect... 30 years ago you had no chance to go in a domain similar to this one, and now using AI in 2-3 minutes you think you are GOD... If anyone is smart enough, built your own algorithm, own softwares and don't give advices to others if you cannot prove your real world identity or what study, academic or Nobel Prize.... Give it a break...this forum is letting anyone who can use AI thinking and comporting like Einstein.... If you cannot prove that, or invest at least 1.000.000Euro in research, you are just a child with a laptop that use AI and think he is a hacker.... Also using multiple accounts from fake email addresses....
This is reality...
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fecell
Jr. Member

Activity: 196
Merit: 2
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September 05, 2026, 09:00:41 AM |
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That is, there would be no discrete logarithm problem. of'coz. ECDLP and div(G, d) different casuality.
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eggsylacer
Jr. Member

Activity: 57
Merit: 2
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September 05, 2026, 09:03:00 AM Last edit: September 05, 2026, 10:48:56 AM by Mr. Big |
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@eggsylacer as i was saying to others, are you a PHD, double PHD or nobel prize winner? That's the problem, here o forum who is arguing have no ideea about the crypto domain...you follow like sheeps the software built of others, use AI and you think that all the things AI is giving you is perfect... 30 years ago you had no chance to go in a domain similar to this one, and now using AI in 2-3 minutes you think you are GOD... If anyone is smart enough, built your own algorithm, own softwares and don't give advices to others if you cannot prove your real world identity or what study, academic or Nobel Prize.... Give it a break...this forum is letting anyone who can use AI thinking and comporting like Einstein.... If you cannot prove that, or invest at least 1.000.000Euro in research, you are just a child with a laptop that use AI and think he is a hacker.... Also using multiple accounts from fake email addresses....
This is reality...
That's right, people like you, after reading the answers of neural networks, believe that you have acquired intimate knowledge in areas where, at least, there is no information in the public domain. As proof, the person above dropped a link to the neural network's response and at the same time does not even understand what is written there... This is how it used to be: people cited Google answers as proof and said that this was an unassailable truth. And now it’s the same thing, only with AI. If you think that asking at AI "how to hack bitcoin" will give you the answer, then I have bad news for you.
That is, there would be no discrete logarithm problem. of'coz. ECDLP and div(G, d) different casuality. What about not using AI to communicate with other people? Otherwise, why do I need you as a barrier between me and the AI?? As already understood above, you don’t understand what you’re talking about.
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fecell
Jr. Member

Activity: 196
Merit: 2
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September 05, 2026, 09:13:42 AM |
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That is, there would be no discrete logarithm problem. of'coz. ECDLP and div(G, d) different casuality. What about not using AI to communicate with other people? Otherwise, what do you need for, as a the strip between me and the AI? As already understood above, you don’t understand what you’re talking about. C:\2>solver_kav0m_v6-3.exe 145 [+] Loaded puzzle 145 from key_list.py [!] Memory budget clamps table M: 500000000 -> 246771126 (pinned in auto_m6_P145.txt) total bits: 145 range bits: 124 table scalars M: 246,771,126 base threads: 2097152 probes to scan: 5,248,241,427,689,071,100,347,925,727,234,088,107,000,201,216 expected probes: 7.681e+34
[+] Resuming search from saved counter: 7,121 [!] Initializing GPU secp256k1... Initializing secp256k1 for GPU... secp256k1 constants initialized on GPU Initializing precomputed table for base point G... Table size: 256 entries (0.02 MB) Precomputed table initialized successfully in global memory secp256k1 GPU initialization complete!
=== GPU Device Information === Device Name: NVIDIA GeForce GTX 1650 Ti Compute Capability: 7.5 CUDA Driver Version: 13.3 CUDA Runtime Version: 13.3 Total Global Memory: 4.00 GB Shared Memory per Block: 48.00 KB Max Threads per Block: 1024 Max Block Dimensions: 1024 x 1024 x 64 Max Grid Dimensions: 2147483647 x 65535 x 65535 Warp Size: 32 Number of Multiprocessors: 16 Memory Bus Width: 128 bits Free Memory at startup: 3.21 GB Total Memory (system view): 4.00 GB ===============================
[!] Allocating GPU memory... [+] Loading table cache: chk6_P145_step246771126.dat [!] Building sorted index (CUB radix sort on GPU)... [+] index built in 8.7 s (246771126 entries) [+] Loading bloom filter: blm6_P145_step246771126.dat [+] Tables built: 246771126 entries (3.03 GB keys+idx+bloom). Free VRAM: 0.18 GB [19:11:18] [+] [3,824,131,506,176 / 22,300,745,198,530,623,141,535,718,272,648,361,505,980,416] S: 108604118.3 pr/s, E: 09:46:51, ETA: 2.241e+19 y, progress: 0.0000% [19:11:27] [+] [3,825,205,248,000 / 22,300,745,198,530,623,141,535,718,272,648,361,505,980,416] S: 108606131.1 pr/s, E: 09:47:00, ETA: 2.241e+19 y, progress: 0.0000%
forgot AI. its just a 108606131.1 probes/s at low video card. and its not over....speeeedup are waiting.
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