That’s the true nature of brute force.  u'r right. kango, bsgs, lcg, and all other just a force of brute, but with different chance. |
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like AI query "optimize this code"... no one usefull changes at first glance. |
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Hello, guys! Ultra lightweight CUDACyclone is ready, speed is 1.3Gkeys/s on RTX4060. Key feature - extremely low VRAM usage for rented gpu. Less than 500Mb VRAM on RTX4090. It will work even if Vanity or Keyhunt doesn’t start. And also this is a good studying sample for your education (why not)? Total 7 small files. Link: https://github.com/Dookoo2/CUDACyclonechange math with lelicopter's repo for win compatible. |
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If anyone want collaborate let me know please i can drop in 20 rtx 4090
creating new DLP algo like mix BSGS+Kangoo (LCG+BSGS+Kangoo with ptx asm optimization to reduce not needed calculation). its so interesting to test with such GPUs like U have. but not so fast as wannted. may be month later. only for 135,140,145.... 256. |
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to full 160 bit range still 10% to complete (with all other ranges with CircleTraversal.cuh). Hurry to get all rewards. ps: but in these 10% all the awards.  everyone's hope. |
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i'v got Speed: 3078.09 points*/ms *) "point" is a three values: kG, +sG, -sG, at same one ms. with 1650ti/4gb. its a normal or slow result? (cuda12.9/c++20/win64) |
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Cheers There's just one thing left. How do I get the WIF from puzzle 71?  it's easy. |
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Initializing CUDA constants...
Total GPU memory: 4294639616 bytes
Free GPU memory: 3333160960 bytes
Available for tables: 2666528768 bytes
Max half_N: 10416128
Generating 20832256 points (10416128 positive, 10416128 negative)...
Generation time: 1409.00 ms
Speed: 14785.14 points/ms
First positive point:
x = 3130e6f4bcf13065e63b05c3c26da883962d17e1d1a7b83deddfb82798c37fa8
y = ff3191deb7f22286ffc93dbc46f0bb78489fb82fdf9c745f6d07a0f2cdafecaf
First negative point:
x = 50132620dbf5bb47c30c603a60897e8e74873aac7e42f30c9d1ac42f22399410
y = 0644bbbbe2809bcb11f620e540bdbd1366e8fb80d33b1488e85a0d41d6da2240
Generated 20832256 points successfully!
is good result for 1650ti? |
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🎯 REVISED P71 POSITIONING:
✅ Position: 77.3% ( Success Probability 88.0% )
📍 Target Range: 0x7174 area
🔬 Improved Calibration: Enhanced mathematical modeling
📊 Confidence: Higher precision positioning
🎯 Search Strategy: COVERAGE deployment recommended
Reason for sharing this update:
"The mathematical positioning keeps evolving as I refine the φ-based calculations. This has higher success probability."
step count: 40000000 total bits: 135 range step: 544451787073501541541399371890829 range bits: 109 progress is loaded: 40000000/80000000/9315 solve... |
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And every bit at every position has a 50% chance of being either 0 or 1 wrong. value 101010100101010100101010100101010100101010100....101010100 impossible too. have generator by statistic for huge amount of random numbers. its a very nice with python, but very slow for real work. so do a CircleTraversal (up to 25 5 bit range) for GPU, its a best DLP solution (may be soon will publish). |
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Did you know, that if the puzzle creator spent so much money so easily, how many bitcoins does he still have left? (rhetorically)
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In Case You Missed It
Puzzle #69 Found on 30-4-2025
0x0000000000000000000000000000000000000000000000101d83275fb2bc7e0c
I just save a link to Blockchain explorer and the wallet, so you see very quick when a puzzle is solved (-> funds = 0  ) Blockchain explorer didnt show private key. https://privatekeyfinder.io/bitcoin-puzzle/this link show all puzzle's addresses too. |
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You should add Satoshi's 34k public keys to your list, each one with 50BTC. Maybe someone can calculate how many RTX-5090's you'd need to crack just one of those wallets in 24 hours?  how pub keys was gotten, if no one out TX was?! |
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Did U know? Oddly enough, all algorithms - Pollard, Floyd, Shanks - work the same way (little step, oops, huge step = collision). The only difference is in the choice of step and interpretation of values. |
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I think if RBF is disabled in a TX, the loophole with increasing the fee will not be used by bots?
I'm right?
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Did U know... G = 115792089237316195423570985008687907852837564279074904382605163141518161494338 * G
or simply
G = (SECP256k1.order+1) * G
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Wat ?
Here is an example of an improved enumeration of points from 1 to 10. If we take the cost of one inversion as $1, then we save $4: starting from point 3, we make a step of 5 in the cycle, points 1,2,6,7 are formed using the calculated inversion for 4,5,9,10 respectively. $1 $1
---------+-------------------+---------
$0 | $1 $0 | $1
.-------+-------. .-------+-------.
| $0 | $1 | | $0 | $1 |
| .---+---. | | .---+---. |
| | | | | | | | | |
1 2 3 4 5 6 7 8 9 10
funny. but real life not so easy. The wise man asked the king to pay him one grain of wheat for the first square on the chessboard, two for the second, four for the third, and so on, doubling the number of grains for each subsequent square. |
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Did U know, 0.5G point has extremly low X coordinate  Qx: 0x00000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63
Qy: 0xc0c686408d517dfd67c2367651380d00d126e4229631fd03f8ff35eef1a61e3c
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