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Geniuses who found the key for address #66, can you tell me if the key is between 37396512239913013824 and 51680708854858323072?
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If I would be the creator I would laugh so hard about some of the things discussed here.
Guys, a logarithm is an abstract concept, not some math function.
You get a thing called a "base change". In this case we're dealing with a change of base of an element from some position in a finite field (private keys) to an element in the same position in a finite group (EC public keys) and the problem is to solve for the position without a way to go back from the latter to the first (which is assumed to be hard, but not yet proven). And this in the best case that we even have such an element, and not some fingerprint of it (an address), which makes the problem levels of more absurdly difficult. WTF is with the real numbers field log2 discussion, it makes no sense, we already know the ranges double in size at each step, of course any polynomial regression or whatever is a straight line. Dividing 1 by (2**64) is four levels of magnitude below a double-precision IEEE floating point, so what errors do you expect, they will always be after the 64-th zero decimal digit in reality. Nevermind the fact that there's an infinity of real numbers between any two real numbers, so an infinity of computations. Take 7 as a private key and try to solve back from [1/4, 1/8) interval, mission impossible.
This is not an analytical problem, it's a group theory problem.
log2 - It's just a way of representing numbers in a different way. I'll ask you again, do you know what you're talking about? |
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The concept of a logarithm was invented just a few centuries ago and immediately changed engineering as we know it.
import math
from mpmath import mp
def calculate_log2(decimal_value):
log2_value = mp.log(decimal_value, 2)
return log2_value
def calculate_reverse_log2(log2_value):
decimal_value = mp.power(2, log2_value)
return decimal_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)
]
mp.dps = 20 # Set the high decimal precision
for ordinal, decimal_value in target_numbers:
log2_result = calculate_log2(decimal_value)
reverse_result = calculate_reverse_log2(log2_result)
print(f"Puzzle:{ordinal}: Log(2) for {decimal_value} is approximately {log2_result}. Reverse calculation: {reverse_result}") Look result of this madness: Puzzle:1: Log(2) for 1 is approximately 0.0. Reverse calculation: 1.0 Puzzle:2: Log(2) for 3 is approximately 1.5849625007211561815. Reverse calculation: 3.0 Puzzle:3: Log(2) for 7 is approximately 2.8073549220576041074. Reverse calculation: 7.0 Puzzle:4: Log(2) for 8 is approximately 3.0. Reverse calculation: 8.0 Puzzle:5: Log(2) for 21 is approximately 4.3923174227787602889. Reverse calculation: 21.0 Puzzle:6: Log(2) for 49 is approximately 5.6147098441152082149. Reverse calculation: 49.0 Puzzle:7: Log(2) for 76 is approximately 6.2479275134435854938. Reverse calculation: 76.0 Puzzle:8: Log(2) for 224 is approximately 7.8073549220576041074. Reverse calculation: 224.0 Puzzle:9: Log(2) for 467 is approximately 8.8672787397096619133. Reverse calculation: 467.0 Puzzle:10: Log(2) for 514 is approximately 9.0056245491938781069. Reverse calculation: 514.0 Puzzle:11: Log(2) for 1155 is approximately 10.173677136303419893. Reverse calculation: 1155.0 Puzzle:12: Log(2) for 2683 is approximately 11.389631339260521112. Reverse calculation: 2683.0 Puzzle:13: Log(2) for 5216 is approximately 12.348728154231077553. Reverse calculation: 5216.0 Puzzle:14: Log(2) for 10544 is approximately 13.364134655008051742. Reverse calculation: 10544.0 Puzzle:15: Log(2) for 26867 is approximately 14.713547616912692731. Reverse calculation: 26867.0 Puzzle:16: Log(2) for 51510 is approximately 15.652564919610652675. Reverse calculation: 51510.0 Puzzle:17: Log(2) for 95823 is approximately 16.548084361224413154. Reverse calculation: 95823.0 Puzzle:18: Log(2) for 198669 is approximately 17.600007248708430135. Reverse calculation: 198669.0 Puzzle:19: Log(2) for 357535 is approximately 18.447724952285439321. Reverse calculation: 357535.0 Puzzle:20: Log(2) for 863317 is approximately 19.719530872026151871. Reverse calculation: 863317.00000000000001 Puzzle:21: Log(2) for 1811764 is approximately 20.788963611792287227. Reverse calculation: 1811764.0 Puzzle:22: Log(2) for 3007503 is approximately 21.520134745822105762. Reverse calculation: 3007503.0 Puzzle:23: Log(2) for 5598802 is approximately 22.416686729787820277. Reverse calculation: 5598802.0 Puzzle:24: Log(2) for 14428676 is approximately 23.782435585948494073. Reverse calculation: 14428676.0 Puzzle:25: Log(2) for 33185509 is approximately 24.984050066697330736. Reverse calculation: 33185509.0 Puzzle:26: Log(2) for 54538862 is approximately 25.700781261712878111. Reverse calculation: 54538862.0 Puzzle:27: Log(2) for 111949941 is approximately 26.738278526958637998. Reverse calculation: 111949941.0 Puzzle:28: Log(2) for 227634408 is approximately 27.762143403294801415. Reverse calculation: 227634408.0 Puzzle:29: Log(2) for 400708894 is approximately 28.577979290797464122. Reverse calculation: 400708894.0 Puzzle:30: Log(2) for 1033162084 is approximately 29.944419458082398243. Reverse calculation: 1033162084.0 Puzzle:31: Log(2) for 2102388551 is approximately 30.969382178280594153. Reverse calculation: 2102388551.0 Puzzle:32: Log(2) for 3093472814 is approximately 31.526580209327912218. Reverse calculation: 3093472814.0 Puzzle:33: Log(2) for 7137437912 is approximately 32.732759144627864676. Reverse calculation: 7137437912.0000000001 Puzzle:34: Log(2) for 14133072157 is approximately 33.718356052472843908. Reverse calculation: 14133072157.0 Puzzle:35: Log(2) for 20112871792 is approximately 34.22740003868583903. Reverse calculation: 20112871792.0 Puzzle:36: Log(2) for 42387769980 is approximately 35.302929017096708804. Reverse calculation: 42387769980.000000001 Puzzle:37: Log(2) for 100251560595 is approximately 36.544833738746849477. Reverse calculation: 100251560595.0 Puzzle:38: Log(2) for 146971536592 is approximately 37.096745824716051977. Reverse calculation: 146971536592.0 Puzzle:39: Log(2) for 323724968937 is approximately 38.235977688802476225. Reverse calculation: 323724968937.0 Puzzle:40: Log(2) for 1003651412950 is approximately 39.868395419757349213. Reverse calculation: 1003651412950.0 Puzzle:41: Log(2) for 1458252205147 is approximately 40.407377394423366271. Reverse calculation: 1458252205147.0 Puzzle:42: Log(2) for 2895374552463 is approximately 41.396887129359569265. Reverse calculation: 2895374552463.0 Puzzle:43: Log(2) for 7409811047825 is approximately 42.752573892536879788. Reverse calculation: 7409811047825.0 Puzzle:44: Log(2) for 15404761757071 is approximately 43.808441604030467369. Reverse calculation: 15404761757071.0 Puzzle:45: Log(2) for 19996463086597 is approximately 44.184810076602017917. Reverse calculation: 19996463086597.0 Puzzle:46: Log(2) for 51408670348612 is approximately 45.547076931749783679. Reverse calculation: 51408670348612.0 Puzzle:47: Log(2) for 119666659114170 is approximately 46.766014580697737785. Reverse calculation: 119666659114170.0 Puzzle:48: Log(2) for 191206974700443 is approximately 47.442128478217754077. Reverse calculation: 191206974700443.0 Puzzle:49: Log(2) for 409118905032525 is approximately 48.539513532885657356. Reverse calculation: 409118905032525.00001 Puzzle:50: Log(2) for 611140496167764 is approximately 49.118497410306637905. Reverse calculation: 611140496167764.0 Puzzle:51: Log(2) for 2058769515153876 is approximately 50.870703748687580964. Reverse calculation: 2058769515153876.0 Puzzle:52: Log(2) for 4216495639600700 is approximately 51.904965885818221825. Reverse calculation: 4216495639600700.0 Puzzle:53: Log(2) for 6763683971478124 is approximately 52.586730675690989872. Reverse calculation: 6763683971478124.0 Puzzle:54: Log(2) for 9974455244496707 is approximately 53.147159473916182081. Reverse calculation: 9974455244496706.9998 Puzzle:55: Log(2) for 30045390491869460 is approximately 54.737993190511333013. Reverse calculation: 30045390491869460.0 Puzzle:56: Log(2) for 44218742292676575 is approximately 55.295507509568065689. Reverse calculation: 44218742292676575.0 Puzzle:57: Log(2) for 138245758910846492 is approximately 56.940012835374135824. Reverse calculation: 138245758910846492.0 Puzzle:58: Log(2) for 199976667976342049 is approximately 57.472609298293031082. Reverse calculation: 199976667976342049.0 Puzzle:59: Log(2) for 525070384258266191 is approximately 58.865288438176815787. Reverse calculation: 525070384258266190.99 Puzzle:60: Log(2) for 1135041350219496382 is approximately 59.977450564669282481. Reverse calculation: 1135041350219496382.0 Puzzle:61: Log(2) for 1425787542618654982 is approximately 60.306464728992728608. Reverse calculation: 1425787542618654982.0 Puzzle:62: Log(2) for 3908372542507822062 is approximately 61.761273698209320329. Reverse calculation: 3908372542507822061.9 Puzzle:63: Log(2) for 8993229949524469768 is approximately 62.963545065677060031. Reverse calculation: 8993229949524469767.8 Puzzle:64: Log(2) for 17799667357578236628 is approximately 63.948484083037149251. Reverse calculation: 17799667357578236628.0 Puzzle:65: Log(2) for 30568377312064202855 is approximately 64.728673773273428832. Reverse calculation: 30568377312064202855.0 So, puzzle 66 private key is from
65.000000000000000000 log(2)
to
66.000000000000000000 log(2)You can test puzzle 15 (or any) with this formula : from mpmath import mp
import random
import secp256k1 as ice
import sys
def calculate_reverse_log2(log2_value):
decimal_value = mp.power(2, log2_value)
return decimal_value
mp.dps = 20 # Set the decimal precision to a sufficiently high value
target = "1QCbW9HWnwQWiQqVo5exhAnmfqKRrCRsvW"
while True:
random_log2_value = mp.mpf(random.uniform(14.0, 15.0))
reverse_result = calculate_reverse_log2(random_log2_value)
HEX = "%064x" % int(reverse_result)
dec = int(HEX, 16)
caddr = ice.privatekey_to_address(0, True, dec)
message = "[+] {}".format(dec);messages = [];messages.append(message);output = ''.join(messages) + "\r";sys.stdout.write(output);sys.stdout.flush()
if target in caddr:
wifc = ice.btc_pvk_to_wif(HEX)
print(f"\n\033[32m[+] PUZZLE SOLVED: {wifc} \033[0m")
break Result is instant. I can not see pattern here... Can you see ??  You can even do regression analysis  I'm lost between the size of the numbers and the precision required here . . . There is no pattern. But it's not random either according to the polynomial analysis. There is an exact math formula for making this puzzle with some script, errors = ZERO. With high decimal precision (mp.dps = 20 at least) And the formula is in the creator's mind. Your polynomial regression plot suggests that the source of this generation is the same. OMG, sorry if their any scientist here, but this IS THE NATURE OF THE PUZZLE !!!  It IS BUILDING on powers of 2 ranges of hex. puzzle 65 - 2^64 to 2^65 puzzle 64 - 2^63 to 2^64 puzzle 63 - 2^62 to 2^63 and so on... so its OBVIOUS will appear a power of 2 graph. Now I invited you to the the standard deviation and realize the size of problem... This is NOT what I pointed out... The reality is: when you do a HD wallet with the child ( there are 213 childs from the original) is very hard to create on deterministic range of hex. The fact you need the parent extended public key, is because you need the chain code (32 bytes) to do the math. Its not a simple factor.... That's why remembered me a lot how profanity works But is crucial to understand first how HD wallets works btw This puzzle is very strange. If it's for measuring the world's brute forcing capacity, 161-256 are just a waste (RIPEMD160 entropy is filled by 160, and by all of P2PKH Bitcoin). The puzzle creator could improve the puzzle's utility without bringing in any extra funds from outside - just spend 161-256 across to the unsolved portion 51-160, and roughly treble the puzzle's content density. If on the other hand there's a pattern to find... well... that's awfully open-ended... can we have a hint or two?  I am the creator. You are quite right, 161-256 are silly. I honestly just did not think of this. What is especially embarrassing, is this did not occur to me once, in two years. By way of excuse, I was not really thinking much about the puzzle at all. I will make up for two years of stupidity. I will spend from 161-256 to the unsolved parts, as you suggest. In addition, I intend to add further funds. My aim is to boost the density by a factor of 10, from 0.001*length(key) to 0.01*length(key). Probably in the next few weeks. At any rate, when I next have an extended period of quiet and calm, to construct the new transaction carefully. A few words about the puzzle. There is no pattern. It is just consecutive keys from a deterministic wallet (masked with leading 000...0001 to set difficulty). It is simply a crude measuring instrument, of the cracking strength of the community. Finally, I wish to express appreciation of the efforts of all developers of new cracking tools and technology. The "large bitcoin collider" is especially innovative and interesting! It's not a puzzle in the sense of the word puzzle. It is a "It is just consecutive keys from a deterministic wallet (masked with leading 000...0001 to set difficulty)." Since the function on the graph of Polynomial Regression of Log(2) behaves as linear, we can assume that the sequence generated by this function has one source. P.S. If you don't know what I'm talking about, there's no point in opening a dispute. |
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I formulated the question: is it possible to create a number system in which:
1=1
2=3
3=7
4=8
5=21
6=49
.....
Also I have a question, how did the creator access the addresses above puzl#160 repeatedly? I.e. he was transferring btc from these addresses to smaller addresses. Did he really save every key to the address and "manually" (using an algorithm) spell them out?
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wish me luck, current speed is 100K keys per sec.
100kk/s is very low. You're better off using keyhunt by alberto You may get +1Mk/s even on a potato CPU Perhaps his algorithm works exponentially? That is, the time to find the key decreases exponentially. |
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hi all i read the post since one year sorry for my english i found a interesting think but i does takes me further. in every elliptic curve like y^2= x^3+7 there is something interesting like :
if P(1,y1) - k times--> Q(-29/3 ,y2)
P(2,y3) --k times--> Q(-3,y2)
so on there is a simple math here where k is always too know independent from whcih curve we work .
i don't want to give more information this operation is 10 times faster then k*G= and find the x value
I think to solve the equation we need to introduce a third Z axis |
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The concept of a logarithm was invented just a few centuries ago and immediately changed engineering as we know it.
import math
from mpmath import mp
def calculate_log2(decimal_value):
log2_value = mp.log(decimal_value, 2)
return log2_value
def calculate_reverse_log2(log2_value):
decimal_value = mp.power(2, log2_value)
return decimal_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)
]
mp.dps = 20 # Set the high decimal precision
for ordinal, decimal_value in target_numbers:
log2_result = calculate_log2(decimal_value)
reverse_result = calculate_reverse_log2(log2_result)
print(f"Puzzle:{ordinal}: Log(2) for {decimal_value} is approximately {log2_result}. Reverse calculation: {reverse_result}") Look result of this madness: Puzzle:1: Log(2) for 1 is approximately 0.0. Reverse calculation: 1.0 Puzzle:2: Log(2) for 3 is approximately 1.5849625007211561815. Reverse calculation: 3.0 Puzzle:3: Log(2) for 7 is approximately 2.8073549220576041074. Reverse calculation: 7.0 Puzzle:4: Log(2) for 8 is approximately 3.0. Reverse calculation: 8.0 Puzzle:5: Log(2) for 21 is approximately 4.3923174227787602889. Reverse calculation: 21.0 Puzzle:6: Log(2) for 49 is approximately 5.6147098441152082149. Reverse calculation: 49.0 Puzzle:7: Log(2) for 76 is approximately 6.2479275134435854938. Reverse calculation: 76.0 Puzzle:8: Log(2) for 224 is approximately 7.8073549220576041074. Reverse calculation: 224.0 Puzzle:9: Log(2) for 467 is approximately 8.8672787397096619133. Reverse calculation: 467.0 Puzzle:10: Log(2) for 514 is approximately 9.0056245491938781069. Reverse calculation: 514.0 Puzzle:11: Log(2) for 1155 is approximately 10.173677136303419893. Reverse calculation: 1155.0 Puzzle:12: Log(2) for 2683 is approximately 11.389631339260521112. Reverse calculation: 2683.0 Puzzle:13: Log(2) for 5216 is approximately 12.348728154231077553. Reverse calculation: 5216.0 Puzzle:14: Log(2) for 10544 is approximately 13.364134655008051742. Reverse calculation: 10544.0 Puzzle:15: Log(2) for 26867 is approximately 14.713547616912692731. Reverse calculation: 26867.0 Puzzle:16: Log(2) for 51510 is approximately 15.652564919610652675. Reverse calculation: 51510.0 Puzzle:17: Log(2) for 95823 is approximately 16.548084361224413154. Reverse calculation: 95823.0 Puzzle:18: Log(2) for 198669 is approximately 17.600007248708430135. Reverse calculation: 198669.0 Puzzle:19: Log(2) for 357535 is approximately 18.447724952285439321. Reverse calculation: 357535.0 Puzzle:20: Log(2) for 863317 is approximately 19.719530872026151871. Reverse calculation: 863317.00000000000001 Puzzle:21: Log(2) for 1811764 is approximately 20.788963611792287227. Reverse calculation: 1811764.0 Puzzle:22: Log(2) for 3007503 is approximately 21.520134745822105762. Reverse calculation: 3007503.0 Puzzle:23: Log(2) for 5598802 is approximately 22.416686729787820277. Reverse calculation: 5598802.0 Puzzle:24: Log(2) for 14428676 is approximately 23.782435585948494073. Reverse calculation: 14428676.0 Puzzle:25: Log(2) for 33185509 is approximately 24.984050066697330736. Reverse calculation: 33185509.0 Puzzle:26: Log(2) for 54538862 is approximately 25.700781261712878111. Reverse calculation: 54538862.0 Puzzle:27: Log(2) for 111949941 is approximately 26.738278526958637998. Reverse calculation: 111949941.0 Puzzle:28: Log(2) for 227634408 is approximately 27.762143403294801415. Reverse calculation: 227634408.0 Puzzle:29: Log(2) for 400708894 is approximately 28.577979290797464122. Reverse calculation: 400708894.0 Puzzle:30: Log(2) for 1033162084 is approximately 29.944419458082398243. Reverse calculation: 1033162084.0 Puzzle:31: Log(2) for 2102388551 is approximately 30.969382178280594153. Reverse calculation: 2102388551.0 Puzzle:32: Log(2) for 3093472814 is approximately 31.526580209327912218. Reverse calculation: 3093472814.0 Puzzle:33: Log(2) for 7137437912 is approximately 32.732759144627864676. Reverse calculation: 7137437912.0000000001 Puzzle:34: Log(2) for 14133072157 is approximately 33.718356052472843908. Reverse calculation: 14133072157.0 Puzzle:35: Log(2) for 20112871792 is approximately 34.22740003868583903. Reverse calculation: 20112871792.0 Puzzle:36: Log(2) for 42387769980 is approximately 35.302929017096708804. Reverse calculation: 42387769980.000000001 Puzzle:37: Log(2) for 100251560595 is approximately 36.544833738746849477. Reverse calculation: 100251560595.0 Puzzle:38: Log(2) for 146971536592 is approximately 37.096745824716051977. Reverse calculation: 146971536592.0 Puzzle:39: Log(2) for 323724968937 is approximately 38.235977688802476225. Reverse calculation: 323724968937.0 Puzzle:40: Log(2) for 1003651412950 is approximately 39.868395419757349213. Reverse calculation: 1003651412950.0 Puzzle:41: Log(2) for 1458252205147 is approximately 40.407377394423366271. Reverse calculation: 1458252205147.0 Puzzle:42: Log(2) for 2895374552463 is approximately 41.396887129359569265. Reverse calculation: 2895374552463.0 Puzzle:43: Log(2) for 7409811047825 is approximately 42.752573892536879788. Reverse calculation: 7409811047825.0 Puzzle:44: Log(2) for 15404761757071 is approximately 43.808441604030467369. Reverse calculation: 15404761757071.0 Puzzle:45: Log(2) for 19996463086597 is approximately 44.184810076602017917. Reverse calculation: 19996463086597.0 Puzzle:46: Log(2) for 51408670348612 is approximately 45.547076931749783679. Reverse calculation: 51408670348612.0 Puzzle:47: Log(2) for 119666659114170 is approximately 46.766014580697737785. Reverse calculation: 119666659114170.0 Puzzle:48: Log(2) for 191206974700443 is approximately 47.442128478217754077. Reverse calculation: 191206974700443.0 Puzzle:49: Log(2) for 409118905032525 is approximately 48.539513532885657356. Reverse calculation: 409118905032525.00001 Puzzle:50: Log(2) for 611140496167764 is approximately 49.118497410306637905. Reverse calculation: 611140496167764.0 Puzzle:51: Log(2) for 2058769515153876 is approximately 50.870703748687580964. Reverse calculation: 2058769515153876.0 Puzzle:52: Log(2) for 4216495639600700 is approximately 51.904965885818221825. Reverse calculation: 4216495639600700.0 Puzzle:53: Log(2) for 6763683971478124 is approximately 52.586730675690989872. Reverse calculation: 6763683971478124.0 Puzzle:54: Log(2) for 9974455244496707 is approximately 53.147159473916182081. Reverse calculation: 9974455244496706.9998 Puzzle:55: Log(2) for 30045390491869460 is approximately 54.737993190511333013. Reverse calculation: 30045390491869460.0 Puzzle:56: Log(2) for 44218742292676575 is approximately 55.295507509568065689. Reverse calculation: 44218742292676575.0 Puzzle:57: Log(2) for 138245758910846492 is approximately 56.940012835374135824. Reverse calculation: 138245758910846492.0 Puzzle:58: Log(2) for 199976667976342049 is approximately 57.472609298293031082. Reverse calculation: 199976667976342049.0 Puzzle:59: Log(2) for 525070384258266191 is approximately 58.865288438176815787. Reverse calculation: 525070384258266190.99 Puzzle:60: Log(2) for 1135041350219496382 is approximately 59.977450564669282481. Reverse calculation: 1135041350219496382.0 Puzzle:61: Log(2) for 1425787542618654982 is approximately 60.306464728992728608. Reverse calculation: 1425787542618654982.0 Puzzle:62: Log(2) for 3908372542507822062 is approximately 61.761273698209320329. Reverse calculation: 3908372542507822061.9 Puzzle:63: Log(2) for 8993229949524469768 is approximately 62.963545065677060031. Reverse calculation: 8993229949524469767.8 Puzzle:64: Log(2) for 17799667357578236628 is approximately 63.948484083037149251. Reverse calculation: 17799667357578236628.0 Puzzle:65: Log(2) for 30568377312064202855 is approximately 64.728673773273428832. Reverse calculation: 30568377312064202855.0 So, puzzle 66 private key is from
65.000000000000000000 log(2)
to
66.000000000000000000 log(2)You can test puzzle 15 (or any) with this formula : from mpmath import mp
import random
import secp256k1 as ice
import sys
def calculate_reverse_log2(log2_value):
decimal_value = mp.power(2, log2_value)
return decimal_value
mp.dps = 20 # Set the decimal precision to a sufficiently high value
target = "1QCbW9HWnwQWiQqVo5exhAnmfqKRrCRsvW"
while True:
random_log2_value = mp.mpf(random.uniform(14.0, 15.0))
reverse_result = calculate_reverse_log2(random_log2_value)
HEX = "%064x" % int(reverse_result)
dec = int(HEX, 16)
caddr = ice.privatekey_to_address(0, True, dec)
message = "[+] {}".format(dec);messages = [];messages.append(message);output = ''.join(messages) + "\r";sys.stdout.write(output);sys.stdout.flush()
if target in caddr:
wifc = ice.btc_pvk_to_wif(HEX)
print(f"\n\033[32m[+] PUZZLE SOLVED: {wifc} \033[0m")
break Result is instant. I can not see pattern here... Can you see ?? You can even do regression analysis I'm lost between the size of the numbers and the precision required here . . . There is no pattern. But it's not random either according to the polynomial analysis. There is an exact math formula for making this puzzle with some script, errors = ZERO. With high decimal precision (mp.dps = 20 at least) And the formula is in the creator's mind. Your polynomial regression plot suggests that the source of this generation is the same. |
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