alexxino
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April 13, 2025, 12:23:47 PM |
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I have developed a prediction model based on the @HomelessPhD model for alpha values: https://github.com/HomelessPhD/BTC32It targets puzzle private keys with a deviation of 0.0000000001 %. import numpy as np from mpmath import mp import argparse
# Set very high precision mp.dps = 300
# Target numbers (ordinal, value) target_numbers = [ (1, 1), (2, 3), (3, 7), (4, 8), (5, 21), (6, 49), (7, 76), (8, 224), (9, 467), (10, 514), (11, 1155), (12, 2683), (13, 5216), (14, 10544), (15, 26867), (16, 51510), (17, 95823), (18, 198669), (19, 357535), (20, 863317), (21, 1811764), (22, 3007503), (23, 5598802), (24, 14428676), (25, 33185509), (26, 54538862), (27, 111949941), (28, 227634408), (29, 400708894), (30, 1033162084), (31, 2102388551), (32, 3093472814), (33, 7137437912), (34, 14133072157), (35, 20112871792), (36, 42387769980), (37, 100251560595), (38, 146971536592), (39, 323724968937), (40, 1003651412950), (41, 1458252205147), (42, 2895374552463), (43, 7409811047825), (44, 15404761757071), (45, 19996463086597), (46, 51408670348612), (47, 119666659114170), (48, 191206974700443), (49, 409118905032525), (50, 611140496167764), (51, 2058769515153876), (52, 4216495639600700), (53, 6763683971478124), (54, 9974455244496707), (55, 30045390491869460), (56, 44218742292676575), (57, 138245758910846492), (58, 199976667976342049), (59, 525070384258266191), (60, 1135041350219496382), (61, 1425787542618654982), (62, 3908372542507822062), (63, 8993229949524469768), (64, 17799667357578236628), (65, 30568377312064202855), (66, 46346217550346335726), (67, 132656943602386256302), (68, 219898266213316039825), (70, 970436974005023690481) ]
# Provided alpha values alpha_values = { 20: -5.509999999999367, 21: -4.8099999999993575, 22: 6.690000000000806, 23: 4.309999999879222, 24: -5.109999999999362, 25: -2.4933006701008145, 26: 23.19000000000104, 27: 50.01000000165623, 28: 50.01000000165623, 29: 4.490000000000775, 30: -4.30999999999935, 31: -4.8036514600379405, 32: 2.7900000000007505, 33: 10.110000011655663, 34: 8.496609620547714, 35: 1.6905278399975268, 36: 2.2900000000007434, 37: 6.506918969897979, 38: 1.8912248499932935, 39: 2.9099999998792025, 40: -2.809999999999329, 41: 5.2945524899731184, 42: 5.690000000000792, 43: -4.691035100114615, 44: -3.9099999999993447, 45: 2.5900000000007477, 46: 50.01000000165623, 47: -4.50218804004683, 48: 6.909999999879259, 49: 28.510000011655915, 50: 2.495913989964804, 51: -2.5900000001208756, 52: -2.5099999999993248, 53: 50.01000000165623, 54: 2.4900000000007463, 55: -5.00999999999936, 56: 3.8099999998792153, 57: -2.1943843400942242, 58: 8.890000000000837, 59: -3.009999999999332, 60: -2.2099999999993205, 61: 3.1099999998792054, 62: -6.000280060058447, 63: -2.490000000120874, 64: -2.809999999999329, 65: -19.689999988344763, 66: 2.7931615399815364, 67: -4.709999999999356, 68: 8.692996380248756, 70: -24.301635149307526 }
def calculate_prediction(puzzle_number, alpha): # Get all previous data points ordinals = np.array([x[0] for x in target_numbers if x[0] < puzzle_number], dtype=float) values = np.array([x[1] for x in target_numbers if x[0] < puzzle_number], dtype=float) if len(ordinals) < 2: print(f"Not enough data points to make a prediction for puzzle {puzzle_number}") return None, None, None log_values = np.array([float(mp.log(val)) for val in values], dtype=float) weights = np.linspace(0.5, 1, len(ordinals)) coefficients = np.polyfit(ordinals, log_values, 1, w=weights) a = mp.exp(coefficients[1]) b = mp.exp(coefficients[0]) correction = (((2 ** puzzle_number) - 1) - (2 ** (puzzle_number - 1))) / alpha predicted_value = (a * mp.power(b, puzzle_number)) - correction return predicted_value, a, b
def interpolate_alpha(puzzle_number): """Interpolate alpha for missing puzzle numbers""" known_puzzles = sorted(alpha_values.keys()) if puzzle_number < min(known_puzzles) or puzzle_number > max(known_puzzles): return 0.0 # Default value if outside known range # Find closest lower and higher puzzle numbers lower = max([p for p in known_puzzles if p < puzzle_number]) higher = min([p for p in known_puzzles if p > puzzle_number]) # Linear interpolation alpha_range = alpha_values[higher] - alpha_values[lower] position = (puzzle_number - lower) / (higher - lower) return alpha_values[lower] + (alpha_range * position)
def format_mp(value, decimals=None): """Format mpmath value with specified decimal places""" if decimals is not None: return f"{float(value):,.{decimals}f}".replace(',', '') return mp.nstr(value, mp.dps)
def main(puzzle_number): if puzzle_number == 69: # Special case for puzzle 69 - use interpolated alpha alpha = interpolate_alpha(69) predicted_value, a, b = calculate_prediction(69, alpha) print("\nPuzzle 69 Prediction:") print(f"Using interpolated alpha: {alpha:.6f}") print(f"Base exponential model: {format_mp(a)} * {format_mp(b)}^n") print(f"Predicted Value: {format_mp(predicted_value, 2)}") # Compare with n=70 n70_value = next(x[1] for x in target_numbers if x[0] == 70) growth_rate = n70_value / predicted_value print(f"Growth to n=70: {float(growth_rate):.6f}x") print(f"Actual n=70 value: {format_mp(n70_value)}") return if puzzle_number not in alpha_values: print(f"No alpha value available for puzzle {puzzle_number}") return alpha = alpha_values[puzzle_number] predicted_value, a, b = calculate_prediction(puzzle_number, alpha) if predicted_value is None: return real_value = next((x[1] for x in target_numbers if x[0] == puzzle_number), None) print(f"\nPuzzle {puzzle_number} Calculation:") print(f"Using alpha: {alpha}") print(f"Base exponential model: {format_mp(a)} * {format_mp(b)}^n") if real_value is not None: print(f"Real Value: {format_mp(real_value)}") print(f"Predicted Value: {format_mp(predicted_value, 2)}") if real_value is not None: error_percentage = abs((predicted_value - real_value) / real_value) * 100 print(f"Percentage error: {format_mp(error_percentage, 10)}%")
if __name__ == "__main__": parser = argparse.ArgumentParser(description="Calculate puzzle values using provided alpha values.") parser.add_argument("puzzle_number", type=int, help="Puzzle number to calculate") args = parser.parse_args() main(args.puzzle_number)
If anyone knows what to do next with this - here you go. python3 test.py 68 Puzzle 68 Calculation: Using alpha: 8.692996380248756 Base exponential model: 0.68972090704257297155081468833675824716161023623689914791953650562051467261304 4419972598777231639024182489182413058156841483121232343334326952277688035030318 1311309574260048040167879840662353500733521425500975021869380914054885130184442 40544507079388243203845236809052465256422016436329883051889524096 * 2.00446149587820417640494244738360788449393817055989304597589376020405549833178 4239005365118701510143617670147330576209009738631394722016490414926974514298110 3364613583395133560524755600665108274702675918184495985595413655135405238097981 4864413726673117061462450025066948963321434607944444810622903302^n Real Value: 219898266213316039825 Predicted Value: 219898266215462633472.00 Percentage error: 0.00000000% python3 test.py 69 Puzzle 69 Prediction: Using interpolated alpha: -7.804319 According to this, puzzle 69 starts with 1ba45e..... I don't have the nerves to improve this for the better.  I'm trying to redo your work here! I have tried to get the alphas using Octave but it gives me super different values, none of them negative! Did you make changes in the analysis.m file? I have modified the 'btc32_keys_dec.csv file setting only from 20# to 70# but alphas are way too different than yours. Any clue what I'm doing wrong?
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wizard872
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April 13, 2025, 02:50:13 PM |
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I have developed a prediction model based on the @HomelessPhD model for alpha values: https://github.com/HomelessPhD/BTC32It targets puzzle private keys with a deviation of 0.0000000001 %. import numpy as np from mpmath import mp import argparse
# Set very high precision mp.dps = 300
# Target numbers (ordinal, value) target_numbers = [ (1, 1), (2, 3), (3, 7), (4, 8), (5, 21), (6, 49), (7, 76), (8, 224), (9, 467), (10, 514), (11, 1155), (12, 2683), (13, 5216), (14, 10544), (15, 26867), (16, 51510), (17, 95823), (18, 198669), (19, 357535), (20, 863317), (21, 1811764), (22, 3007503), (23, 5598802), (24, 14428676), (25, 33185509), (26, 54538862), (27, 111949941), (28, 227634408), (29, 400708894), (30, 1033162084), (31, 2102388551), (32, 3093472814), (33, 7137437912), (34, 14133072157), (35, 20112871792), (36, 42387769980), (37, 100251560595), (38, 146971536592), (39, 323724968937), (40, 1003651412950), (41, 1458252205147), (42, 2895374552463), (43, 7409811047825), (44, 15404761757071), (45, 19996463086597), (46, 51408670348612), (47, 119666659114170), (48, 191206974700443), (49, 409118905032525), (50, 611140496167764), (51, 2058769515153876), (52, 4216495639600700), (53, 6763683971478124), (54, 9974455244496707), (55, 30045390491869460), (56, 44218742292676575), (57, 138245758910846492), (58, 199976667976342049), (59, 525070384258266191), (60, 1135041350219496382), (61, 1425787542618654982), (62, 3908372542507822062), (63, 8993229949524469768), (64, 17799667357578236628), (65, 30568377312064202855), (66, 46346217550346335726), (67, 132656943602386256302), (68, 219898266213316039825), (70, 970436974005023690481) ]
# Provided alpha values alpha_values = { 20: -5.509999999999367, 21: -4.8099999999993575, 22: 6.690000000000806, 23: 4.309999999879222, 24: -5.109999999999362, 25: -2.4933006701008145, 26: 23.19000000000104, 27: 50.01000000165623, 28: 50.01000000165623, 29: 4.490000000000775, 30: -4.30999999999935, 31: -4.8036514600379405, 32: 2.7900000000007505, 33: 10.110000011655663, 34: 8.496609620547714, 35: 1.6905278399975268, 36: 2.2900000000007434, 37: 6.506918969897979, 38: 1.8912248499932935, 39: 2.9099999998792025, 40: -2.809999999999329, 41: 5.2945524899731184, 42: 5.690000000000792, 43: -4.691035100114615, 44: -3.9099999999993447, 45: 2.5900000000007477, 46: 50.01000000165623, 47: -4.50218804004683, 48: 6.909999999879259, 49: 28.510000011655915, 50: 2.495913989964804, 51: -2.5900000001208756, 52: -2.5099999999993248, 53: 50.01000000165623, 54: 2.4900000000007463, 55: -5.00999999999936, 56: 3.8099999998792153, 57: -2.1943843400942242, 58: 8.890000000000837, 59: -3.009999999999332, 60: -2.2099999999993205, 61: 3.1099999998792054, 62: -6.000280060058447, 63: -2.490000000120874, 64: -2.809999999999329, 65: -19.689999988344763, 66: 2.7931615399815364, 67: -4.709999999999356, 68: 8.692996380248756, 70: -24.301635149307526 }
def calculate_prediction(puzzle_number, alpha): # Get all previous data points ordinals = np.array([x[0] for x in target_numbers if x[0] < puzzle_number], dtype=float) values = np.array([x[1] for x in target_numbers if x[0] < puzzle_number], dtype=float) if len(ordinals) < 2: print(f"Not enough data points to make a prediction for puzzle {puzzle_number}") return None, None, None log_values = np.array([float(mp.log(val)) for val in values], dtype=float) weights = np.linspace(0.5, 1, len(ordinals)) coefficients = np.polyfit(ordinals, log_values, 1, w=weights) a = mp.exp(coefficients[1]) b = mp.exp(coefficients[0]) correction = (((2 ** puzzle_number) - 1) - (2 ** (puzzle_number - 1))) / alpha predicted_value = (a * mp.power(b, puzzle_number)) - correction return predicted_value, a, b
def interpolate_alpha(puzzle_number): """Interpolate alpha for missing puzzle numbers""" known_puzzles = sorted(alpha_values.keys()) if puzzle_number < min(known_puzzles) or puzzle_number > max(known_puzzles): return 0.0 # Default value if outside known range # Find closest lower and higher puzzle numbers lower = max([p for p in known_puzzles if p < puzzle_number]) higher = min([p for p in known_puzzles if p > puzzle_number]) # Linear interpolation alpha_range = alpha_values[higher] - alpha_values[lower] position = (puzzle_number - lower) / (higher - lower) return alpha_values[lower] + (alpha_range * position)
def format_mp(value, decimals=None): """Format mpmath value with specified decimal places""" if decimals is not None: return f"{float(value):,.{decimals}f}".replace(',', '') return mp.nstr(value, mp.dps)
def main(puzzle_number): if puzzle_number == 69: # Special case for puzzle 69 - use interpolated alpha alpha = interpolate_alpha(69) predicted_value, a, b = calculate_prediction(69, alpha) print("\nPuzzle 69 Prediction:") print(f"Using interpolated alpha: {alpha:.6f}") print(f"Base exponential model: {format_mp(a)} * {format_mp(b)}^n") print(f"Predicted Value: {format_mp(predicted_value, 2)}") # Compare with n=70 n70_value = next(x[1] for x in target_numbers if x[0] == 70) growth_rate = n70_value / predicted_value print(f"Growth to n=70: {float(growth_rate):.6f}x") print(f"Actual n=70 value: {format_mp(n70_value)}") return if puzzle_number not in alpha_values: print(f"No alpha value available for puzzle {puzzle_number}") return alpha = alpha_values[puzzle_number] predicted_value, a, b = calculate_prediction(puzzle_number, alpha) if predicted_value is None: return real_value = next((x[1] for x in target_numbers if x[0] == puzzle_number), None) print(f"\nPuzzle {puzzle_number} Calculation:") print(f"Using alpha: {alpha}") print(f"Base exponential model: {format_mp(a)} * {format_mp(b)}^n") if real_value is not None: print(f"Real Value: {format_mp(real_value)}") print(f"Predicted Value: {format_mp(predicted_value, 2)}") if real_value is not None: error_percentage = abs((predicted_value - real_value) / real_value) * 100 print(f"Percentage error: {format_mp(error_percentage, 10)}%")
if __name__ == "__main__": parser = argparse.ArgumentParser(description="Calculate puzzle values using provided alpha values.") parser.add_argument("puzzle_number", type=int, help="Puzzle number to calculate") args = parser.parse_args() main(args.puzzle_number)
If anyone knows what to do next with this - here you go. python3 test.py 68 Puzzle 68 Calculation: Using alpha: 8.692996380248756 Base exponential model: 0.68972090704257297155081468833675824716161023623689914791953650562051467261304 4419972598777231639024182489182413058156841483121232343334326952277688035030318 1311309574260048040167879840662353500733521425500975021869380914054885130184442 40544507079388243203845236809052465256422016436329883051889524096 * 2.00446149587820417640494244738360788449393817055989304597589376020405549833178 4239005365118701510143617670147330576209009738631394722016490414926974514298110 3364613583395133560524755600665108274702675918184495985595413655135405238097981 4864413726673117061462450025066948963321434607944444810622903302^n Real Value: 219898266213316039825 Predicted Value: 219898266215462633472.00 Percentage error: 0.00000000% python3 test.py 69 Puzzle 69 Prediction: Using interpolated alpha: -7.804319 According to this, puzzle 69 starts with 1ba45e..... I don't have the nerves to improve this for the better.  i tried found key in range [ 1ba45e000000000000 ~ 1ba45effffffffffff ], but not match key. fail
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AlanJohnson
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April 13, 2025, 03:08:16 PM |
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you should already be rich enough to participate in this puzzle.
So, honestly speaking, this puzzle was made by a rich person for rich people—the rich will get richer, and the poor will get poorer. Right?  Yep... Or maybe you manage to use vast.ai for free by hacking their system—exploiting 3,000 GPUs, stolen credit cards, or even the puzzle creator himself, as you’ve already written somewhere. The problem is, you could end up in prison. It’s better to just go fishing.  Do you really think these puzzles are solved fairly? With savings? Who even has that much in savings? Come on!  Let's clear few things up: 1) This whole thing isn't any form of lottery or puzzle or any other "play" for fun 2) This whole thing isn't meant to make someone happy or make someone poor rich 3) Nobody has the obligation to do anything with it 4) This whole thing is made as an indicator of how safe bitcoin is at current time (some sort of measurement tool for how much cracking power is available out there to attack bitcoin cryptograhy) You are here long enough to know that i think... Yes, it's frustrating to see "free" money sitting there but we can't grab them but the bitter truth is ... it's over for small players. I will still be watchinge the progress but it's better to do something else than wasting energy and time on this at this point of difficulty.
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kTimesG
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April 13, 2025, 03:39:06 PM |
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I have developed a prediction model based on the @HomelessPhD model for alpha values:
According to this, puzzle 69 starts with 1ba45e.....
i tried found key in range [ 1ba45e000000000000 ~ 1ba45effffffffffff ], but not match key. fail What did you expect? I hope it wasn't an expensive lesson to learn. Only thing here that should be clearly stated is that a model you are using for prognosis should have a number of parameters less than a number of observations - simple polynomial approximation with an order of n will ideally fit the dataset of n points (it will have exactly n roots) just memorizing the data instead of retrieving the logic behind the data - same correct for complex models like SVM or neural networks - they could just memorize (overfit) the data but in a less obvious way - so be carefull using them or you will lose your precious time for mathematical madness.
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Off the grid, training pigeons to broadcast signed messages.
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farou9
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April 13, 2025, 04:54:04 PM |
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i tried found key in range [ 1ba45e000000000000 ~ 1ba45effffffffffff ], but not match key. fail silly question but you made shore to check the compressed addresses
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denyAKA BLACKANGEL
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April 13, 2025, 09:03:00 PM |
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i hoppe this help,if you have a succes you can write
1f3010a2b8793d347f 19vkiEaebeLmrjJkuoYtLetpobkjWLcv9D 177590b696ee793041 19vkiEah776mXf2m8nXRBFSyKeffPDiwpL 1f4eb0b57715269183 19vkiEaJV6PWrWTGhqBhfVvZqJ1Fo9NFTD 16833c184839cc383a 19vkiEaNn9TRKZ4Y7gyNNob4WTEnc5D5FU 1ae324e2c108b33737 19vkiEakZw2EengwkWRBikpXAbL2TzFJHW 17716a73fe1e9f233b 19vkiEatqjCiA4qzVYmVTTgvZVm1499Yqg 1873f9c83bdd2ddc5b 19vkiEaq6wcVd9VzxGu3wPLcqPVmAY44b2 17ecea3fd10f9309a1 19vkiEa49TuRsfmk5gokx2uQQFzmyDH3sN 17ecea3fd123280cdc 19vkiEaWGUW6Phcnr6KLGbKNN3q2P9PwjW 1e5b4754a58d3fc59d 19vkiEaSFApUaWLea6NCtUWwYZkyD9vGM8 19b0b29424564ba65e 19vkiEaBi5VkRbADHVH3gNSJTrynNwEjZS 1e857a4c954bc9896a 19vkiEaBGF19bq7nhKUAM91WR7p5LQL68 1f171b6772f3daf7b3 19vkiEafXjcmpggNZijSULAZ6XT9M8j6MX 1f1568dbd4f8378a26 19vkiEaGNcbwBf8QVPh5cy3LmyENBYcuuM 177590b696dc4dedc2 19vkiEaXNynCbJteXbTPdsE9x9cDcd88bq 13871339bbcf9f2539 19vkiEaZmk6Jwf7Ep4oYviwht3brThkaE1 15f74cc42bac0c9077 19vkiE1v5PagT8TPkP3JWN9MGiH5ooQiQ
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papiro08
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April 13, 2025, 09:30:15 PM |
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I found this haha it's a joke  Hash160: 61eb8a50c8673cd3fd9d73ee90c733ff24b6e64f BTC Address: 19vkiEajf1jD23km6G6DdSQXH2VhWP3kHG
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denyAKA BLACKANGEL
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April 13, 2025, 09:42:13 PM |
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ahhahha you need more not one and when you have luck you have a match
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pbies
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April 13, 2025, 10:00:43 PM |
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Since few weeks ago I know for now that we won't hit puzzle 69 without broader talk about maths in this case.
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BTC: bc1qmrexlspd24kevspp42uvjg7sjwm8xcf9w86h5k
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GoldTiger69
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April 13, 2025, 10:20:00 PM |
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I found this haha it's a joke  Hash160: 61eb8a50c8673cd3fd9d73ee90c733ff24b6e64f BTC Address: 19vkiEajf1jD23km6G6DdSQXH2VhWP3kHG I found this one: PK: 19665F6056877D8B5B Public Key : 02502C2518C06231BDDACA06491E0B790DB9478B56B8DCD2E255CA7E7380E97267 Found Hash160 : 61eb8a50c8673c376c4bdb96d179afecaadbd140 I'm very curious about the distance between the 2 addresses.
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denyAKA BLACKANGEL
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April 13, 2025, 11:17:10 PM |
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i can provide more address with 16 exact caracter but not today 
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nightdreams
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April 14, 2025, 02:09:50 AM |
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i hoppe this help,if you have a succes you can write
1f3010a2b8793d347f 19vkiEaebeLmrjJkuoYtLetpobkjWLcv9D 177590b696ee793041 19vkiEah776mXf2m8nXRBFSyKeffPDiwpL 1f4eb0b57715269183 19vkiEaJV6PWrWTGhqBhfVvZqJ1Fo9NFTD 16833c184839cc383a 19vkiEaNn9TRKZ4Y7gyNNob4WTEnc5D5FU 1ae324e2c108b33737 19vkiEakZw2EengwkWRBikpXAbL2TzFJHW 17716a73fe1e9f233b 19vkiEatqjCiA4qzVYmVTTgvZVm1499Yqg 1873f9c83bdd2ddc5b 19vkiEaq6wcVd9VzxGu3wPLcqPVmAY44b2 17ecea3fd10f9309a1 19vkiEa49TuRsfmk5gokx2uQQFzmyDH3sN 17ecea3fd123280cdc 19vkiEaWGUW6Phcnr6KLGbKNN3q2P9PwjW 1e5b4754a58d3fc59d 19vkiEaSFApUaWLea6NCtUWwYZkyD9vGM8 19b0b29424564ba65e 19vkiEaBi5VkRbADHVH3gNSJTrynNwEjZS 1e857a4c954bc9896a 19vkiEaBGF19bq7nhKUAM91WR7p5LQL68 1f171b6772f3daf7b3 19vkiEafXjcmpggNZijSULAZ6XT9M8j6MX 1f1568dbd4f8378a26 19vkiEaGNcbwBf8QVPh5cy3LmyENBYcuuM 177590b696dc4dedc2 19vkiEaXNynCbJteXbTPdsE9x9cDcd88bq 13871339bbcf9f2539 19vkiEaZmk6Jwf7Ep4oYviwht3brThkaE1 15f74cc42bac0c9077 19vkiE1v5PagT8TPkP3JWN9MGiH5ooQiQ
What program you used to find the prefix new to here
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bitcoinpuzzles621
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April 14, 2025, 03:02:42 AM |
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Oh wow, alot has happened since I stopped checking the forum posts for almost 2 months again. So is there really no sense in trying to solve any of the remaining puzzles now because bram has a superior method thats fast enough to solve all of them? Can anyone tell me as just confirmation? Or maybe I understood some of the previous discussions wrong. I might still keep trying though if anyone thinks that the higher numbered puzzles will still take a very long time to solve no matter what. My method is slow though and I dont have too many GPUs, thats why im asking here again to make sure its even worth the time to keep trying to do any of the remaining unsolved puzzles.
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E36cat
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April 14, 2025, 04:55:19 AM |
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Oh wow, alot has happened since I stopped checking the forum posts for almost 2 months again. So is there really no sense in trying to solve any of the remaining puzzles now because bram has a superior method thats fast enough to solve all of them? Can anyone tell me as just confirmation? Or maybe I understood some of the previous discussions wrong. I might still keep trying though if anyone thinks that the higher numbered puzzles will still take a very long time to solve no matter what. My method is slow though and I dont have too many GPUs, thats why im asking here again to make sure its even worth the time to keep trying to do any of the remaining unsolved puzzles.
go fishing
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wizard872
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April 14, 2025, 05:22:11 AM |
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I have developed a prediction model based on the @HomelessPhD model for alpha values:
According to this, puzzle 69 starts with 1ba45e.....
i tried found key in range [ 1ba45e000000000000 ~ 1ba45effffffffffff ], but not match key. fail What did you expect? I hope it wasn't an expensive lesson to learn. Only thing here that should be clearly stated is that a model you are using for prognosis should have a number of parameters less than a number of observations - simple polynomial approximation with an order of n will ideally fit the dataset of n points (it will have exactly n roots) just memorizing the data instead of retrieving the logic behind the data - same correct for complex models like SVM or neural networks - they could just memorize (overfit) the data but in a less obvious way - so be carefull using them or you will lose your precious time for mathematical madness.
just searching! i running 2 RTX4090 in my home
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denyAKA BLACKANGEL
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April 14, 2025, 05:56:27 AM |
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i help a little to all,evry time what you find you save and you can see the frequenz and distanz of address f = 1/T this can help a loot you dont need a loot of GPU more it's a luck and knowledg i use python i write a self,not vanity just random address generate and some algorithm 
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kTimesG
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April 14, 2025, 06:35:19 AM |
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i help a little to all,evry time what you find you save and you can see the frequenz and distanz of address f = 1/T this can help a loot you dont need a loot of GPU more it's a luck and knowledg i use python i write a self,not vanity just random address generate and some algorithm  thx for sharing the path to success! There are soooooo many wrong things with it (literally one wrong thing every few words). I have a question for all of ya prefix hunters who stubbornly think there is a single seed of truth in that theory. On April 1st I posted here a joke which in fact, was not a joke at all. I intentionally searched for the wrong H160 of the puzzle 68 prefix. For who's not a math expert: because the puzzle 68 address has the first byte of its H160 larger than 58, then we can divide by 58 and, voila, obtain addresses that start with the same address prefix, but an address which is one less in length. Nobody observed the flaw until I hinted at the issue!After scanning less than 500 trillion keys, a 53-bit prefix was found. Now, here's what another experiment ended up at: After scanning 9000 trillion keys for the actual address (the good hash) of Puzzle 69, there was still no 53 bits or larger prefix found. Question: why did I find a 53 bit prefix of some hash after 500 trillion keys, and no 53 bit prefix after 9000 trillion keys (and counting)? So, never forget that just as well as you might get lucky, you never, ever, ever, have the guarantee to find anything remotely closer than what you expect. Oh, and just for fun, the bad luck scenario was actually done under the theoretical base that prefixes should be spaced apart rather than being close. The magnitude between the min and max distance does a great job of making this concept useless. No idea what you people see, I think you only see what you want to see.
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Off the grid, training pigeons to broadcast signed messages.
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E36cat
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Activity: 64
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April 14, 2025, 07:27:58 AM |
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i help a little to all,evry time what you find you save and you can see the frequenz and distanz of address f = 1/T this can help a loot you dont need a loot of GPU more it's a luck and knowledg i use python i write a self,not vanity just random address generate and some algorithm  thx for sharing the path to success! There are soooooo many wrong things with it (literally one wrong thing every few words). I have a question for all of ya prefix hunters who stubbornly think there is a single seed of truth in that theory. On April 1st I posted here a joke which in fact, was not a joke at all. I intentionally searched for the wrong H160 of the puzzle 68 prefix. For who's not a math expert: because the puzzle 68 address has the first byte of its H160 larger than 58, then we can divide by 58 and, voila, obtain addresses that start with the same address prefix, but an address which is one less in length. Nobody observed the flaw until I hinted at the issue!After scanning less than 500 trillion keys, a 53-bit prefix was found. Now, here's what another experiment ended up at: After scanning 9000 trillion keys for the actual address (the good hash) of Puzzle 69, there was still no 53 bits or larger prefix found. Question: why did I find a 53 bit prefix of some hash after 500 trillion keys, and no 53 bit prefix after 9000 trillion keys (and counting)? So, never forget that just as well as you might get lucky, you never, ever, ever, have the guarantee to find anything remotely closer than what you expect. Oh, and just for fun, the bad luck scenario was actually done under the theoretical base that prefixes should be spaced apart rather than being close. The magnitude between the min and max distance does a great job of making this concept useless. No idea what you people see, I think you only see what you want to see. They like the program`s output, it just feels like a zero.zero win =)
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denyAKA BLACKANGEL
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April 14, 2025, 07:50:57 AM Last edit: April 14, 2025, 08:04:36 AM by denyAKA BLACKANGEL |
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you need a more knowledge even a more inteligent, smart ,dont be negativ you dont know me,be a positiv me frend.i write a more 2000+ brute vanity and experiment code just for bitcoin i try evry algorithm for mathe... 4 years i just watch this forum, there are very intelligent people here for math and IT specialists. I hope you're not the one waiting for someone to post a new code that you can try. Be grateful to those who write code and use it here so that others can also try their luck. Just say thank you.i wish you luck,me frend, just keep search
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Akito S. M. Hosana
Jr. Member
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Merit: 8
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April 14, 2025, 08:32:23 AM |
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Mind tricks don't work on this puzzle—only money does. 
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