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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 13/10/2023, 12:43:58 UTC
@Zahid888 and @nomachine and all other seed lovers, I have something for you.  https://www.bleepingcomputer.com/news/security/bounty-offered-for-secret-nsa-seeds-behind-nist-elliptic-curves-algo/
You guys are really talented and I'm sure you can come up with a script to brute force those seeds, don't forget to share the script.😉
that will not even be the case but the bounty alone is worth it if it would be paid in cryptocurrency but is the resources investment going to be covered with profits made afterwards? that is the real question here
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 12/10/2023, 13:16:37 UTC
If someone will please solve this puzzle for me then I will give you some of the profits I make.

Please let me know when you have a solution for me.  I am willing to pay up to 10% of all profits to you just for solving the puzzle for me.

Thanks.  Wink
I would gladly love to solve the puzzle for you and I accept the offer of your 10% profit. All you need is to load $200,000 in a vast ai GPU rental website in order for me to be able to proceed and I assure you a maximum result time of 3 days. We will be able to solve either puzzle 130 or 66 and share the profit as you've suggested in the sharing ratio of 90:10.
let me know or you can send me an inbox with the next steps. Thank you,
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 07/10/2023, 18:38:42 UTC




I've been receiving many DMs about this script, and I'm not sure what people find interesting in it. I made this script purely for fun.
I want to emphasize that it should not be used for any malicious purposes or to mislead anyone. Its sole purpose is for education, to show how a single seed can create an entire key. Here's the example script.

Code:
import random
from bit import *
import random

List = ["13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so","1BY8GQbnueYofwSuFAT3USAhGjPrkxDdW9",
        "1MVDYgVaSN6iKKEsbzRUAYFrYJadLYZvvZ","19vkiEajfhuZ8bs8Zu2jgmC6oqZbWqhxhG"]
count = 0
found = 0
for i in range (100000000):
    seed = random.getrandbits(128)
    seed_value = int(seed)
    random.seed(seed_value)
    seed = str(seed_value)
    a = random.randrange(2**65,2**66)
    key = Key.from_int(a)
    addr66 = key.address
    hex66 = "%00x" % a
    b = random.randrange(2**66,2**67)
    key = Key.from_int(b)
    addr67 = key.address
    hex67 = "%00x" % b
    c = random.randrange(2**67,2**68)
    key = Key.from_int(c)
    addr68 = key.address
    hex68 = "%00x" % c
    d = random.randrange(2**68,2**69)
    key = Key.from_int(d)
    addr69 = key.address
    hex69 = "%00x" % d
    count+=1
    print('\n\n=============================== Derived From OneSeed =============================\n\nSeeed : {}\nCount : {}\nFound : {}\nAddress 60 : 1Kn5h2qpgw9mWE5jKpk8PP4qvvJ1QVy8su | Private Key : fc07a1825367bbe\nAddress 61 : 1AVJKwzs9AskraJLGHAZPiaZcrpDr1U6AB | Private Key : 13c96a3742f64906\nAddress 62 : 1Me6EfpwZK5kQziBwBfvLiHjaPGxCKLoJi | Private Key : 363d541eb611abee\nAddress 63 : 1NpYjtLira16LfGbGwZJ5JbDPh3ai9bjf4 | Private Key : 7cce5efdaccf6808\nAddress 64 : 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN | Private Key : f7051f27b09112d4\nAddress 65 : 18ZMbwUFLMHoZBbfpCjUJQTCMCbktshgpe | Private Key : 1a838b13505b26867\nAddress 66 : {} | Private Key : {}\nAddress 67 : {} | Private Key : {}\nAddress 68 : {} | Private Key : {}\nAddress 69 : {} | Private Key : {}\nAddress 70 : 19YZECXj3SxEZMoUeJ1yiPsw8xANe7M7QR | Private Key : 349b84b6431a6c4ef1'.format(seed, count, found, '13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so', hex66, '1BY8GQbnueYofwSuFAT3USAhGjPrkxDdW9', hex67, '1MVDYgVaSN6iKKEsbzRUAYFrYJadLYZvvZ', hex68, '19vkiEajfhuZ8bs8Zu2jgmC6oqZbWqhxhG', hex69), end='\r')
   
    if addr66 in List or addr67 in List or addr68 in List or addr69 in List:
        print('\n\n================================ Found ================================\n\n')
        g=open(u"Found.txt","a")
        g.write("From one seed:         " +seed+'\n')
        g.write("\n" +addr66)
        g.write("\n" +addr67)
        g.write("\n" +addr68)
        g.write("\n" +addr69)
        g.write("\n"+seed + "\n"+ "\n")
        g.close()
        found += 1
    break

And yes, with great politeness and simplicity... I have allowed the newbie to DM me just to ensure that the creator doesn't drop any hints accidentally  Grin This doesn't mean that everyone should keep messaging, except the creator Cheesy, He can message me, 24x7 Smiley

i APPRECIATE YOUR NEWBIES, BUT CAN THEY ALSO FUND THIS WALLET "1h7EsfXpgCLftpU4vucjkCbX1Wktop8GM" FOR ABILITY TO CONTINUE HAVING THE SAME FUN EVERYONE ELSE IS HAVING?

NOTHING IS TOO SMALL FROM YOUR NEWBIES. I JUST WANT TO PAY THE ELECTRICAL BILLS ACCRUED AND ALSO BE ABLE TO KEEP THESE MACHINES RUNNING. THANK YOU IN ADVANCE TO YOUR NEWBIES
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 07/10/2023, 17:44:25 UTC
nothing is impossible for God to do.
Only one thing is also impossible for God to do, suicide. Because energy can not be destroyed or created, if he is not the purest kind of energy then what?


Anyhow, your mistake is searching randomly/sequentially, doesn't matter what method you use, 66 bit is very large, you should have at least consult an expert before spending money, or just do a simple calculation like this :

Your speed rate = 5 billion keys per second, how many seconds is 1 year = 31536000 * 5b/2^66 = 467 years to search 66 bit.  If you believe in "luck" and "chance", you should buy lottery tickets to try your luck.

Anyone trying to brute force #66, please stop and first calculate the profit and cost.
WELL Hope, it is the quintessential human delusion, simultaneously the source of our greatest strength, and your greatest weakness.
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 07/10/2023, 16:46:22 UTC
Creator use master keys or child keys? BIP-32 or BIP-39 format? Who can know?

If I can say anything it'd be for the author of this puzzle to bless me beyond my own imagination
I have tried my best but I am still not going to give up either. you can search or do a lookup of my wallet address. I have definitely invested over $30k in the puzzle 66 alone throughout the whole random scan and my wallet address is 1h7EsfXpgCLftpU4vucjkCbX1Wktop8GM, you can do your search from the private key scanner or 66 bits puzzle
I spend so much resources there but no luck
I am running my search lowkey now but I am almost being drained but still not giving up
A blessing is nothing imaginable or unimaginable but nothing is impossible for God to do.
I just pray for the ability to be able to keep  up as it is so much fun
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 06/10/2023, 17:09:20 UTC
Why do you use linux time? Satoshi was using windows, either vista or seven, maybe even 8, or xp. 😂 we just need to look for all the keys. Lol
You know what we should do? We should use the same technic to search for rmd160 hash patterns and find collisions. 🙃

I can assume that the Satoshi was using some script (errors = ZERO) with Lagrange interpolation, determining the formula that generates the values may require a different approach, such as symbolic regression or other mathematical techniques together with random numbers and time. I am lost in numbers and hypotheses now. I need green grass.. Grin

There is nothing to worry about for now, maybe some could go back in time to see what position Satoshi was taking while throwing the dice or flipping the coins, maybe that way we can know what position we need to also take in order to be able to replicate the same method. How many minutes do you think you need to throw a dice or flip a coin to get puzzle 1 to puzzle 256?
256+254+254+253+252+251+250+249+248+247+246+245+244+243+242+241+240+239+238+237+236+235+234+233+232+231+230+229+228+227+226+225+224+223+222+221+220+219+218+217+216+215+214+213+212+211+210+209+208+207+206+205+204+203+202+201+200+199+198+197+196+195+194+193+192+191+190+189+188+187+186+185+184+183+182+181+180+179+178+177+176+175+174+173+172+171+170+169+168+167+166+165+164+163+162+161+160+159+158+157+156+155+154+153+152+151+150+149+148+147+146+145+144+143+142+141+140+139+138+137+136+135+134+133+132+131+130+129+128+127+126+125+124+123+122+121+120+119+118+117+116+115+114+113+112+111+110+109+108+107+106+105+104+103+102+101+100+99+98+97+96+95+94+93+92+91+90+89+88+87+86+85+84+83+82+81+80+79+78+77+76+75+74+73+72+71+70+69+68+67+66+65+64+63+62+61+60+59+58+57+56+55+54+53+52+51+50+49+48+47+46+45+44+43+42+41+40+39+38+37+36+35+34+33+32+31+30+29+28+27+26+25+24+23+22+21+20+19+18+17+16+15+14+13+12+11+10+9+8+7+6+5+4+3+2+1 = 32895

so you can as well agree with me that the problem here has nothing to do with a pattern or some kind of formula... we have so many ways of getting public keys without making any mistakes. ZERO FREE ERROR VERIFIED METHODS

There is no assumptions here, whenever the person that solved puzzle 120 and 125 is ready he will solve puzzle 130 again. Now for puzzle 66, the highest pool scanned so far is less than 5% so don't expect any results any time soon from these pools. the key range is so vast and the computational resources is so limited We can't complain about wrong bit keys only except there's an evidence that someone successfully scan the full range for any puzzle level and the is not found within that range so please let's stop making assumptions and let's focus on better ways to tackle the puzzle. @Digaran mentioned something earlier and said

Why do you use linux time? Satoshi was using windows, either vista or seven, maybe even 8, or xp. 😂 we just need to look for all the keys. Lol
You know what we should do? We should use the same technic to search for rmd160 hash patterns and find collisions. 🙃
let us just assume that this is possible 160 bit is a whole new world and for RMD160 collision you would find so many SHA256 matching so many RMD160 you won't be limited to finding the private keys for puzzle 66 within just the 66 bit range. there would be an enormous range to cover and you would find perfectly matching private keys matching so many RMD160 hashes. but without further ado, I must say we have a great task at hand and we will stop at nothing to find an end to this puzzle
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 06/10/2023, 06:09:19 UTC
As the "master math guru" of this community 🤥😂  I have always wondered, can God make it so when 2+2 you get 5, I mean he can do anything right? So how could we believe all of the rules of physics, mathematics? If it's possible to change the rules like that? It would be like saying since God can do anything, he should be able to clone himself infinitely, or more importantly, can God kill himself? These questions are taunting and impossible to know the answer for sure, but logic says he shouldn't be able to make 2+2=5, or create clones of himself or self destruct, that means logically even God has limits to his power.
@digaran
So I ran the code and I got 63 pubkeys
I need to ask what would be the target 1 and target 2
and how can I further divide the pubkey of 130 to as low as 100 or 90 or 80 bit range
after hitting a target from the public key results, how would I further calculate the private key to get my target private key
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 05/10/2023, 14:11:18 UTC
Don't you think you are assuming too much, with the little we know about the creator?

The same if for you, Don't you think that you are assuming too much?

Bump!!

You said the he did it manually I said that it was made with and script or program.

WE DON'T HAVE WAY TO KNOW IT.

So we can finished all of our posts with the next sentence.

All that I write here is my personal opinion or speculation there is no proof of anything and all other users are also speculating

I am about to let it all go...
I have tried so may ways to get this to work
maybe we have been looking at the wrong direction the whole time
I need to ask a question, maybe the whole RMD160 or bas58 bruteforce is too slow
but 1 thing is also certain, there is no way to know the public key of number 66 if we dont have to compare the private key with the RMD160 or base58 I could have suggestedfor us to try bruteforcing for the public key maybe it could save us so much time and also boost the speed too.

can we sequentially generate public keys within the range of 66 bit without having anything to do with the private key?
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 05/10/2023, 09:11:54 UTC
It's been months since 125 was solved and they didn't bother to post the private key, we even asked the author to reveal the key after 1 month giving them time to sweep those garbage coins, but still no news, no keys revealed, does that mean this community is only good to contribute but not good enough to fulfill their request?

Not only 125 but 120 is also missing, if we want to solve this "puzzle" we need to have all the pieces which includes 120 and 125 keys, since the author listened to our call and added extra funds to the puzzles, why not do us another favour and reveal 120, 125 keys? It could be sent via PM to a trusted person so they can cash forked coins and give it to someone who really needs it and then they can reveal the emptied keys for us to have a look. 

Maybe I'm expecting too much? Just open your laptop and copy paste the keys in someone's inbox with instructions on what to do. Lol
How much money is in the forked coins anyways, anyone knows?

@digaran we all know you are the mathematics guru in the community, you should have found a way to reduce puzzle 130 to as low as 100 bit range without ending up with so many public keys. hint us through some real work to do as so many of us are no longer willing to keep dying on the puzzle 66. we need a headstart on getting towards puzzle 130 and we got to know for sure what we are doing instead of just trying to bruteforce the whole keys like I am currently doing with puzzle 66 and my electricity bills killing. let's dabble into 130 and turn off these machines. I have so many paper and biro to make some calculations with. NB: I don't mind sharing. if we need to create a pool too but not for 66, as there are so many pools already and all they're scanning is the first 26 bit range which consists of 33554432 possible guesses to find the key.

Let's be creative since the puzzle creator has already boosted our morale with the 10x increment. creativity is the key to success
Let's think out of the box. I saw someone trying to paint a picture with the RMD160 too, these are creative ideas. puzzle 125 and 130 pks have nothing to do with puzzle 130 just as puzzle 1 to 65 has nothing to do with puzzle 66. we just have to start coming to the realization that these thing called RANDOM is not a joke, the bit range is also not a thing to play with. and with the exposed pubkeys we have an upper edge to calculate something rather than just bruteforcing throughout
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Merits 1 from 1 user
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 01/10/2023, 07:37:09 UTC
⭐ Merited by citb0in (1)
Hi all,

to what extent can the performance and the hit probability of these two tools be compared and what significance do they then have?

For example:

- we run Kangaroo in multi-GPU mode on several modern GPUs and achieve a displayed speed of 15 GKey/s

- we run keyhunt in BSGS mode on 8 CPU threads with about 16GB RAM allocation and get about 80 PKey/s.

Is Kangaroo with multi-GPU usage more advantageous and more likely to get a hit, even though keyhunt with CPU usage in BSGS can process 80 PKey/s?

From this numerical example, one would think that keyhunt with the 80 PKey/s would have to be worlds better. But is that the case or is it not possible to compare this at all and the given speed rates are completely useless for the comparison?


the easiest example I can give to you for this question is a "web" and a "gunshot"
now picture both having the same momentum
Kangaroo would be considered a "web" moving at the same momentum as the "gunshot" from BSGS
and what do webs do? it spreads, covering so much area and what do gunshots do? it moves on a straight line
Now the preference is you choice and your available resources
they're both super fast like I've explained earlier.
let's say you have a single 4090 and 16 cpu threads and you combined both and you got the speed of 5GKey/s
and you choose not to use the 4090 but you rather use the 16 cpu thread on your machine and you have 128GB RAM and you chose -k 7000 and you achieved the speed of 1 EKey/s you got so much momentum on both processes but based on my calculations, you would be 90% slower using the latter compared to the former because they are both having ETAs which can easily be calculated
Mind you this has nothing to do with what I am saying, it has everything to do with your preference and resources
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 30/09/2023, 14:33:04 UTC
Based on the information provided, it appears that the pattern for generating the private key (PVK) values for these Bitcoin addresses involves incrementing the PVK values by a specific sequence of integers. Let's calculate the PVK value for Address 15 and beyond based on the observed pattern:

Address 15: PVK value = Address 14's PVK value + 27
Address 15: PVK value = 10544 + 27
Address 15: PVK value = 10571

So, the PVK value for Address 15 is 10571.

You can continue this pattern to calculate the PVK values for addresses beyond Address 15 by adding the next integer in the sequence to the PVK value of the previous address.

What  is for 66 then ?  Grin

I can even help you further how to program this

import math

# Given list of numbers
numbers = [
    1, 3, 7, 8, 21, 49, 76, 224, 467, 514, 1155, 2683, 5216, 10544, 26867, 51510,
    95823, 198669, 357535, 863317, 1811764, 3007503, 5598802, 14428676, 33185509,
    54538862, 111949941, 227634408, 400708894, 1033162084, 2102388551, 3093472814,
    7137437912, 14133072157, 20112871792, 42387769980, 100251560595, 146971536592,
    323724968937, 1003651412950, 1458252205147, 2895374552463, 7409811047825,
    15404761757071, 19996463086597, 51408670348612, 119666659114170, 191206974700443,
    409118905032525, 611140496167764, 2058769515153876, 4216495639600700,
    6763683971478124, 9974455244496707, 30045390491869460, 44218742292676575,
    138245758910846492, 199976667976342049, 525070384258266191, 1135041350219496382,
    1425787542618654982, 3908372542507822062, 8993229949524469768,
    17799667357578236628, 30568377312064202855
]


Give me math formula that can calculate pattern for WIF 66 here ?
Whoever succeeds in doing this will receive a greater prize than the Nobel Prize.

p.s.
It doesn't even have to be an exact number.
That it is accurate to approximately +/- 8 decimal places.

you must be a very big joker if you think there is a pattern in the numbers you see up there, the only pattern you can ever find is the puzzle numbers 1 to 65, anything apart from that then you must be joking. if you like turn it to ASCII or WIF or HEX or DEC or Base 9 or Base 2 or any format you want. a Random number will forever remain a random number

the only problem we are having here is the fact that we cannot generate a defined public key range (e.g 66 bit range public keys) without having to involve the private key. what would be the usage of the curve if we could do that then?
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 28/09/2023, 13:45:55 UTC
I have made a new discovery today, ( at least for me is new). You all know about lambda right?  Here :
5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72

Multiplying any point by lambda will *metamorphose our x but not our y coordinates. I just realized if we take for example :
04bcace2e99da01887ab0102b696902325872844067f15e98da7bba04400b88fcb483ada7726a3c 4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8

04c994b69768832bcbff5e9ab39ae8d1d3763bbf1e531bed98fe51de5ee84f50fb483ada7726a3c 4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8

0479be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798483ada7726a3c 4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8

And dividing all three keys above by 2, will give us :
041b7b24bfec17f17cfad4ed5d26f3ba3f2ecde9516100d16dbfa3986c27223893c0c686408d517 dfd67c2367651380d00d126e4229631fd03f8ff35eef1a61e3c

04e484db4013e80e83052b1267603def814791291a8a098469934ed0bc5f7e2739c0c686408d517 dfd67c2367651380d00d126e4229631fd03f8ff35eef1a61e3c

0400000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63c0c686408d517 dfd67c2367651380d00d126e4229631fd03f8ff35eef1a61e3c

If we divide the keys above by any number, all their y coordinates will be identical, EC is like a giant web of interconnected triangles, by any chance is there any EC calculation which only y coordinates are used for an operation?


* I was reminded of Nevermore in dota. 🙃



Since I was nominated as the master mathematician, I invented this number  1073741824, there should be that many puzzle #66 identical addresses in 66 bit range?  Or maybe there are only 4 identical addresses?


if there are 4 identical addresses but our targets are the first 10 characters of puzzle 66, then there are 1073741824 addresses having similar first 10 characters within the 66 bit range

but we need to also note that if there are 4 identical addresses within the 66 bit range then our chances of hitting 1 should be 9223372036854775808 which still falls within 63 bit range which defnitely won't be an easy task to fin in any way because that is about 58 years on my 4090.

we need a better algorithm
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 27/09/2023, 20:20:17 UTC
Look closely, I am working with artistic binary, not with base58, Ripemd160, decimal, or hexadecimal. and just print output in any format, If you have any output result even in any format, please show us. 100% of people are trying randomly even in 130 also, and I am also trying randomly, but with some logic. You are a master of mathematics, so you can solve 130 directly even with your mobile. All the best!
[/quote]

I don't know if this will help you in any way:

the problem is I have almost 2 thousand of these which will be posted as a code below, in order to maintain orderliness and less consumption of space

Code:

13zb1hQbUMfsukCbQ68xfiRUKmQsjnuN4i - 48373146411643185729 - 20d45a6a761982f4c90f65d426f7f529e296d53b - 29F4FF05E95F42A41 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZhbhBQjwQVHnkvfNJTv
13zb1hQbUNQk542NaQsTARn4DHMJyNeHyF - 56079894187336241810 - 20d45a6a761994c64363d754ac40d31b1956e6ff - 30A43CEA3ADAB9292 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZrWNidYDipR4MKTPUez
13zb1hQbUS3MinKVRsfLv7dnWJvaxGEkp2 - 52723525194923494754 - 20d45a6a7619ec52e73033af2735e1d20805de55 - 2DBAF96D2D5C5CD62 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZndMYToFaPpKxu88cjD
13zb1hQbUSkr7XJdPxdsTbj1pktPYL7tV9 - 51637160598513618584 - 20d45a6a7619fd9206ab6d54e7420173783299af - 2CC9C0BFA9C338E98 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZmNXEdH4k77QPgCFjWK
13zb1hQbUTKNyCRK2TVAf5hwkLa9Uu645s - 46307419374150351508 - 20d45a6a761a0b17ee9d1607a1ba6707925baade - 282A50092D545FA94 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZfDCe8fYcijS1VJFjb8
13zb1hQbUTYE5CZ5rgM3484Cu7VjUvWS5j - 45530387498103924700 - 20d45a6a761a106f025a34d6c08ad869bafe27f7 - 277DC6E32FE5EE7DC - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZeK7AgYYCCD7RPSEKwY
13zb1hQbUUKpzmjCTNBRDreH8mPKJwSk3y - 40783378601333411922 - 20d45a6a761a2363c7081e895a67f2fe4358ebab - 235FBAAC220BED052 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZYprWu8QCpVmZF5bxfM
13zb1hQbUXTKFHSYSy9DqKxRo4AQDGRsTM - 50710785841633655170 - 20d45a6a761a6ed4cb272462c99e21757b4f056b - 2BFC0E70BB37B7982 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZkJR3bamqyuJkR4yET8
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13zb1hQbYfCh4tLsFigCmhazz3epDuCbNQ - 48319096639085843099 - 20d45a6a76310316b604f954c8bbdda490518a7f - 29E8FEA63407B929B - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZhY51Ts4Eg8JAFiaDAM
13zb1hQbYgwZkxQnjCRHojfBPmbKwNcDH8 - 72323466770253861848 - 20d45a6a76312d05410f3f2c035a368ed438f90e - 3EBB091D765BFD3D8 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qaBHPXttYdEA91nSsfCB
13zb1hQbYi7eVzTRTMbhQXLW9Lf4tjK9y9 - 69223021477850718522 - 20d45a6a7631495233503d0a28b2cda6f8c5643f - 3C0A993465A87213A - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qa7hXWgFetfdV3mbkhjz
13zb1hQbYicmXEraYfreCSJJb2nPgPy1Y8 - 50739863439266470124 - 20d45a6a7631556d21157878f342c8884789d650 - 2C02834F7794ED8EC - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZkLN7XEA4eXuDGToCFM
13zb1hQbYkE8CSJE9Dczmvvzmhm1CUdPmN - 44916764761681479414 - 20d45a6a76317c3bb1d3c4402f91b6356d3e8c5f - 26F5867924FCEC6F6 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZdby7CbVSJPg5QiNYZt
13zb1hQbYmXUp2kp1MnkCfm28Yccr1JDZJ - 49186179685162308609 - 20d45a6a76319b8eab500d387da1a1d19c453596 - 2AA9869D2F8A25401 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZiYCeKgWkgpm4GACVjy
13zb1hQbYnjxtCUX5nnowUK6p81Cy4RTNR - 40063177750128182863 - 20d45a6a7631b8db402cd786858d5cdb1fd3f9c4 - 22BFCFFE661B9464F - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZXza2f2MZdRcVaxsm7Q
13zb1hQbYr65coaa35fwkmrDh1SiUa7U2U - 47556563278343707287 - 20d45a6a7632098c30e3686402166a0d082be821 - 293FADA5610101A97 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZgewutd5K1rpGSj9fx5
13zb1hQbYt7Dm4FycZoLJkp6KbZGH4Nszc - 53059098480127906699 - 20d45a6a76323a3dbacc7605f964f61d668faf7d - 2E057C8F2D27E138B - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZo1rQw3fabbtbMnVitQ
13zb1hQbYta7n82gaH6kKBiZdXtEKet4Kz - 51379696827744687905 - 20d45a6a7632456bf85e5687ad15ad81f32e4177 - 2C9095A0B24FC9721 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZm5G6LjvzssU17Deiac
13zb1hQbYtqamAxgXWrvaivH1LR2hFbnyW - 52248453983132922421 - 20d45a6a76324bd9b585e3c0f4f794e020ac8f2a - 2D517CC12CFFB3235 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZn5WEmL3oJCR9jofCUs
13zb1hQbYwABKw7N2QkN2sBsapkiNb5QKm - 44302072124455547399 - 20d45a6a763283ccc9ba6f710d88eee18aa7901b - 266D093DFEFE87207 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZctktRSFm8kPC6B7s2W
13zb1hQbYyKFrt8MJqABBc9qYhjQqor4vf - 67626845178234534785 - 20d45a6a7632b7cb0759ce42c4b8c6a46204a7e2 - 3AA82D1632101FB81 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qa5rWp4AWSeQ629A81T4
13zb1hQbYzjxDxHtxLizTcc4Y4JruUYPe7 - 44246694740411011399 - 20d45a6a7632da2b258158e4308d35933543b999 - 2660BD66FF757D147 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZcq3Z4GsW4XYrfzMu9j
13zb1hQbZ2m1fWDppGDU1vZrXD96gU6s7s - 41185244817163056292 - 20d45a6a76330ad40f0423d58207bc26473d864e - 23B8F61F3D28074A4 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZZHoAY9Y9bGszExd6zx
13zb1hQbZ7drTvxYRWwmswQWW4hgZwMjph - 42019911220101952898 - 20d45a6a76338066bb9a3ec519124f00fdc4163b - 24724B69FA7A0B582 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZaFkkQmqZUpxiEX3SXa
13zb1hQbZ7h14f2RR3E8he367D9DBe3ajX - 42340799662575025650 - 20d45a6a763381b5c164488fbb3b392970cfdad0 - 24B98BCF641C37DF2 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZadGWypcrEnCP4jX6UB
13zb1hQbZ7uFRvwiNetS4mWeiN1CQPN5d9 - 73298960546708798659 - 20d45a6a763386cd1977367af813a8872852b1f0 - 3F93A38433FF33CC3 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qaCQniq2fg3BqiSD75WK
13zb1hQbZ8Q3gsemMLnkQitsGdJrPjQVi1 - 41108554716948110979 - 20d45a6a763392c599ed3914bdfa3b5c217a9527 - 23A7EECB66D669E83 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZZCexUaW44fFd714JEx
13zb1hQbZAoFwgfFDZMYnqpDshGZjJr6Dq - 47446912354920464392 - 20d45a6a7633cca3d326c7f9f020a8e17390d7d1 - 292754B66FC208008 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZgXbY7XLYDuDrp26UzB
13zb1hQbZAzTuretB5jaNU7ConrLPwEoPr - 46688318888855073721 - 20d45a6a7633d14c5a69331557f1c12fcd082300 - 287EE3AAA025DF3B9 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZfejm7SuigQwfWsgn68
13zb1hQbZByoUcmMqU1BbvTpV4uGZtBAC8 - 39970028619426507997 - 20d45a6a7633e921ed2ceed9ee482e0e5c6f8600 - 22AB21142531D2CDD - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZXtKpZNbG3Tv94hUzgm
13zb1hQbZCPBSKqh1W5H1nJcR6XrCr1PQG - 56712421402939884764 - 20d45a6a7633f2d9c5af02e0d3207f4517adb0d3 - 3130AFECA6A7E04DC - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZsEnHg7Wx9pfhx8uQVd
13zb1hQbZDED5RSGFTEwceLsPcSV1gmknC - 55623916971135547896 - 20d45a6a7634073b22575e019fa15535dab17c6f - 303EFD9C6F05B39F8 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZqyofEjKVs6ATaF6QyN
13zb1hQbZHz22BoPi6WomdnDqv9S7W4eWb - 60830966206306538975 - 20d45a6a763479e17eb2e70f300082c6475e311a - 34C330177E6B155DF - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZwzuAn3JWb9UvvLd4L9
13zb1hQbZKJphfDfXNinY3a7c7jgBn9qEL - 71760461253069144487 - 20d45a6a763499ceb603a9bad2f958d507b154ce - 3E3E05F4E735E81A7 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qaAdeJD7EiUs1W6aN9nd
13zb1hQbZKwipzYgbZxf6ojD6489Uvp7ND - 63135733581907222313 - 20d45a6a7634a9254db8f6b39c5633484ca42596 - 36C2F2F2F6F47DF29 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZzfRDGXP36kfd7MTgPh
13zb1hQbZNiYCx1iLvGmqNeFZUy8UF2KFQ - 59109423803581444657 - 20d45a6a7634ebfeb13fc7d054c498aa948ebe77 - 3344EDBC6327DB231 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZv1Uz6yGEnbCr65UUBY
13zb1hQbZPWNeHTyw3AaG5ryiKYjPxyAdg - 46316994679194156202 - 20d45a6a7634ff0c40fff2287dfc6fe357ee6c23 - 282C7054345067CAA - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZfDqsfn6q9nRGAkb1Cg
13zb1hQbZPwbRxZQjysh6kw1GdMP7mgY6s - 51484651588548046457 - 20d45a6a7635098819aa599e708e55167515208b - 2CA7E39D7A8282E79 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZmCJCz3nuzxYW3ZoQLG
13zb1hQbZQxB2Uo9vStqyySebae1nWznZ4 - 62707459898691393621 - 20d45a6a763521e1d4c37892ef85cd8e24cd03e9 - 3663DA68A1C8D9855 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZzAhsuRrGmyN336Qre4
13zb1hQbZYKCT3ZseAEzMeknCrxHdRGN4B - 56318207850861952808 - 20d45a6a7635d363cf974abfabb40a0ad0c393a2 - 30D9277A7593DFF28 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZrnMPxRkBxgdj5rWYep
13zb1hQbZgqJpWZEKh5rKodYgG4tEAuiUm - 42146233785976033142 - 20d45a6a7636a0c8b26b880bdaa3be75271e4ce3 - 248E58053F9AB0F76 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZaQDxB6hqYfP92bj98w
13zb1hQbZhH3MqugYnYJHrJSuvxMhDPzq4 - 39901960523427979431 - 20d45a6a7636ab7b21d1a57d594103ddf086dcbc - 229C03DAFB895A0A7 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZXom92w4keNmMX7aden
13zb1hQbZkEAMBdGCHzhsoYyVpQc9idTVx - 37928688135156917422 - 20d45a6a7636f29cfc16e30973ae5ea6b4793a8c - 20E5DC4E77499E4AE - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZVXU7E6qRdnfVsuDjvn
13zb1hQbZnEmeMRKk4dWoRvS3c3UjhupHM - 67915966895582247278 - 20d45a6a76372315eda44276e018e77c32962769 - 3AE85FC0EAC5E796E - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qa6Bu4DVuoZbytWGnmRs
13zb1hQbZtXa2N9tZDWWnrwv1ecyyaPViW - 37818162477213485699 - 20d45a6a7637baba38d216d75440a4158a510661 - 20CD51A6760510A83 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZVQ4LAjTkUfwa8ZLyWE
13zb1hQbZu5DpBkwxV8xVMQKiu172ZnXZD - 37332141779891591579 - 20d45a6a7637c7e26f1679739d23c9f59f2032ed - 20616692796F5199B - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZUqUT1hobdzF69RsaFZ
13zb1hQbZv8cU8RZ7Jsih7jywZFmUjDfhg - 59503748966032306180 - 20d45a6a7637e167564113280fd1d1f5b06301b9 - 339C7C86B945F7004 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZvTvJzccvf78k8Cbxuf
13zb1hQbZvApRnT7VfASyyaitsiS6hQ8mS - 70197752828550488997 - 20d45a6a7637e2521c4a4b53dcc785bb5a04f359 - 3CE308446206CA3A5 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qa8psjfDazNuCghxVMxQ
13zb1hQbZwHvxzBBNejGtsAn9EP3GcdnzY - 63014719132667385149 - 20d45a6a7637fd631b1858820308c1843888f919 - 36A81412F53A6953D - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZzXJeesJ2Xi4hsvVqYi
13zb1hQbZwaBjkgF2pDKL8cj6vGGoYQ3kM - 69858139391242271699 - 20d45a6a76380424dac4e91b8e5082e06819a046 - 3C979F7A6ECB7FBD3 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qa8S7A1MPgs9JKEgSQJt
13zb1hQbZx5pDaycRAJKCewQwoapXYhqhS - 59939320530308382581 - 20d45a6a76381075d4feb1f6b7760f99bf6da2d5 - 33FD33E733F4F3F75 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZvy82GBicciCVBUWQzy
13zb1hQbZxJoWv4Tg5BEp2Wh6jrHJ3izUQ - 38151524982449520153 - 20d45a6a763815dbf2060244293d14de1cb66192 - 2117571D60A583E19 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZVnQc36yxjhrUDV2PAf
13zb1hQbZzQTg82osBGLewsYb577Hk4MWk - 71636956086031639744 - 20d45a6a7638486e37de7cbf87b2aca95faeb7d4 - 3E2299802EA2C64C0 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qaAVN3rkT1YBYJ9nq29X
13zb1hQba3haK5kdfNjuep2K8vggsX881g - 71692370163003147489 - 20d45a6a763897dfbbdbc937335fcc3e6513d945 - 3E2EE76D21F9DC4E1 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qaAZ5XVtMsEJekK415ia
13zb1hQba4sqbb6QZwYQx5DNVm9UVNFjEe - 69023330083310371733 - 20d45a6a7638b44005ec24ed38bcb2655c0166af - 3BDE421026E3F8395 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qa7U91uQci5qGdZTdp5G
13zb1hQba6j84GRShkeWwU1utueP1Y1nvj - 71238881664467444713 - 20d45a6a7638e0d8ab3a8cf0e7738513deca18c8 - 3DCA3596F1EABA3E9 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qaA2g9PcgZ4vPLqVoCAf
13zb1hQba6yjaeZR74QhL5UWuVTmBBLmVk - 43560962407148933968 - 20d45a6a7638e6eba955cc0fb83a1607331539ba - 25C87A0837041D350 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZc356ZdF921RRbwTwkG
13zb1hQbaMSop4NgeJ6svfvsURwjWvRvmi - 50873275498753923907 - 20d45a6a763a43b5ee0d67822081babbea68342f - 2C2022E8727CF9B43 - KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZkVJtC9HWnC9WwrzJmA
 

the problem about trying to find a collision is very difficult...

I haven't seen anyone with the first 10 characters of puzzle 66 wallet address "13zb1hQbWV" maybe within this range, the first 10 character might be the solution to the puzzle itself

I have waited over 10 days straight on a 4090 going at 5 billion per second looking for the first 10 character "13zb1hQbWV" but all I always get is very close so the actual range has nothing to do with the characters of the wallet address or the characters of the rmd160 has there is not a single relationship between these

I even went as far as getting the closest difference between each rmd160 first 10 characters I got and I wote this code

Code:
import subprocess
import time
import random
import string
import signal
import sys
import os
import logging
import multiprocessing

# Define a function to save the current state (current_key) to a checkpoint file
def save_checkpoint(current_key, checkpoint_filename):
    with open(checkpoint_filename, 'w') as checkpoint_file:
        checkpoint_file.write(current_key)

# Define a function to load the current state (current_key) from a checkpoint file
def load_checkpoint(checkpoint_filename):
    if os.path.exists(checkpoint_filename):
        with open(checkpoint_filename, 'r') as checkpoint_file:
            return checkpoint_file.read()
    else:
        return None

def generate_next_key(current_key):
    # Convert the current_key to an integer, add 2385407605810, and convert back to hexadecimal
    next_key = hex(int(current_key, 16) + 2385407605810)[2:].zfill(18)
    return next_key

def key_search_worker(start_key, end_key):
    try:
        command = f"KeyHunt-Cuda.exe -g --gpux 2048,512 -m address --coin BTC --range {start_key}:{end_key} 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so"
        subprocess.run(command, shell=True, check=True)
    except subprocess.CalledProcessError as e:
        logging.error(f"Error in subprocess: {e}")

def main():
    checkpoint_filename = "checkpoint.txt"  # Specify the checkpoint filename
    num_processes = 4  # Adjust as needed

    # Load the last saved checkpoint (if any)
    current_key = load_checkpoint(checkpoint_filename)
    if current_key is None:
        current_key = "2058E1E06903F1DCC"  # Specify your desired starting key here

    try:
        while True:
            processes = []
            for _ in range(num_processes):
                # Generate the next key and update current_key
                next_key = generate_next_key(current_key)

                # Split the key range for each process
                start_key = current_key
                end_key = next_key

                # Save the current state (current_key) to the checkpoint file at regular intervals
                if _ % 10 == 0:
                    save_checkpoint(current_key, checkpoint_filename)

                # Create a process to search for keys in the specified range
                process = multiprocessing.Process(target=key_search_worker, args=(start_key, end_key))
                processes.append(process)
                process.start()

                # Update current_key
                current_key = next_key

            # Wait for all processes to complete
            for process in processes:
                process.join()

    except KeyboardInterrupt:
        for process in processes:
            process.terminate()
            process.join()

        print("Script terminated by user.")
        try:
            input("Press Ctrl+C again to exit the script.")
        except KeyboardInterrupt:
            print("Exiting script.")
            sys.exit(0)

if __name__ == "__main__":
    logging.basicConfig(filename="key_search.log", level=logging.ERROR)
    main()

it adds the difference to the current key to get the next key and it scans it for the specified time before terminating the code and adding the difference again to the current key but no luck still

so its obvious the private key to puzzle 66 has nothing to do with the distance between the characters of the rmd160 or wallet addresses. it is what it is
Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 21/09/2023, 01:48:00 UTC

I really appreciate your advice, yeah i was just wondering if i could try my luck with all the unsolved addresses at once lmao.

If anyone can correct me with this:
Since Puzzle 64 was found 2022-09-10
does that mean it's been a year now 2023-09-20 and still the search for Puzzle 66 on going?!

if you do your math well
the normal time for scanning the 66 bit range on an average cpu would take nothing less than 500 years
now we have pools that are scanning and they're barely 4% gone on the 66 bit range...
so this is to tell you that you should expect at least another 1 or 2 more years for the 66 bit range to be defeated successfully
but the pubkey ranges might be the appropriate range of choice for you as these gets defeated almost every 4 to 6 months
Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 16/09/2023, 07:59:50 UTC
but it is excellent, even better than C for testing. of new ideas,

I agree with this, personally i think that testing and proof of concept are the only thing that python is good for. But once that the proof of concept is proven to be useful or good then is time to move to another language. If speed is determining for the algorithm


the fact that you think that there are no new things is simply the result of your own mental fatigue

What F.. mental Fatigue? I already shared here a lot of programs, ideas, users share with me their ideas on telegram and there is nothing new...

What i am tryting to said is if you have a vague idea without math background or without enough testing, then OPEN a new thread discuss their idea there and if the idea was proven to be good, then link that new topic here sharing a brief sumary of it and how to use.

I have this code already which is not an idea for now but a work in progress but we might be needing to know if it's a good idea

Code:
#include "SECP256k1.h"
#include "Point.h"
#include "sha256.h"
#include "Int.h"
#include "ripemd160.h"
#include <iostream>
#include <iomanip>
#include <cstring>
#include <chrono>
#include <thread>
#include <mutex>
#include <atomic>
#include <vector>
#include <fstream>
#include <sstream>
#include <ctime>
#include <boost/multiprecision/cpp_int.hpp>
#include <gmpxx.h>

constexpr uint64_t START_VALUE = 576565752303423488ULL;
constexpr uint64_t END_VALUE = 900'000'000'000'000'000ULL;
constexpr uint64_t INCREMENT = 1ULL;

// Define a range of factors
constexpr double MIN_FACTOR = 64.0;
constexpr double MAX_FACTOR = 1028.0;
constexpr double FACTOR_INCREMENT = 1.00;

std::atomic<uint64_t> currentValue(START_VALUE);
std::atomic<uint64_t> totalKeys(0);
std::mutex printMutex;
std::mutex resultMutex;
std::vector<std::array<uint8_t, 20>> ripemd160Hashes;

std::chrono::time_point<std::chrono::system_clock> startTime;
std::atomic<bool> matchFound(false);
std::string currentHexPrivateKey;

void loadRIPEMD160Hashes() {
    std::ifstream file("wallets.txt");
    if (file.is_open()) {
        std::string hexHash;
        while (file >> hexHash) {
            if (hexHash.length() != 40) {
                std::cerr << "Invalid RIPEMD160 hash length: " << hexHash.length() << std::endl;
                continue; // Skip invalid hashes
            }

            std::array<uint8_t, 20> hash;
            for (int i = 0; i < 20; ++i) {
                std::string byteHex = hexHash.substr(2 * i, 2);
                hash[i] = static_cast<uint8_t>(std::stoi(byteHex, nullptr, 16));
            }

            ripemd160Hashes.push_back(hash);
        }
        file.close();
        std::cout << "Loaded " << ripemd160Hashes.size() << " RIPEMD160 hashes from file." << std::endl;
    }
}

std::string hexBytesToHexString(const uint8_t* bytes, size_t length) {
    std::stringstream ss;
    ss << std::hex << std::setfill('0');
    for (size_t i = 0; i < length; ++i) {
        ss << std::setw(2) << static_cast<int>(bytes[i]);
    }
    return ss.str();
}

bool hasMinimumMatchingCharacters(const uint8_t (&hash)[20]) {
    for (const std::array<uint8_t, 20>& loadedHash : ripemd160Hashes) {
        bool isMatch = true;
        for (int j = 0; j < 19; ++j) {  // Loop through the first 5 bytes (40 bits)
            if (hash[j] != loadedHash[j]) {
                isMatch = false;
                break; // If any character doesn't match, stop checking
            }
        }
        if (isMatch) {
            return true;
        }
    }
    return false;
}

void printProgress() {
    startTime = std::chrono::system_clock::now();

    while (!matchFound.load(std::memory_order_relaxed)) {
        {
            std::lock_guard<std::mutex> lock(resultMutex);

            auto now = std::chrono::system_clock::now();
            auto elapsed = std::chrono::duration_cast<std::chrono::seconds>(now - startTime);
            double keysPerSecond = static_cast<double>(totalKeys.load()) / elapsed.count();

            std::cout << "\rTime: " << elapsed.count() << "s, Keys: " << totalKeys.load()
                      << ", Keys/s: " << std::fixed << std::setprecision(5) << keysPerSecond
                      << ", Current: " << currentValue.load()
                      << ", Priv Key: " << currentHexPrivateKey << std::flush;
        }
        std::this_thread::sleep_for(std::chrono::seconds(5));
    }
}

void counterWorker(int threadId, Secp256K1& secp256k1, int numThreads) {
    mpz_class current(START_VALUE + threadId * INCREMENT);

    while (current <= END_VALUE) {
        for (double factor = MIN_FACTOR; factor <= MAX_FACTOR; factor += FACTOR_INCREMENT) {
            mpz_class result = current * static_cast<uint64_t>(factor);

            std::string hexPrivateKey = result.get_str(16); // Get hex representation directly
            currentHexPrivateKey = hexPrivateKey;

            Int privateKey;
            privateKey.SetBase16(hexPrivateKey.c_str());

            Point publicKey = secp256k1.ComputePublicKey(&privateKey);

            uint8_t compressedPublicKey[33];
            secp256k1.GetPublicKeyRaw(true, publicKey, reinterpret_cast<char*>(compressedPublicKey));

            uint8_t publicKeyHash[32];
            sha256_33(compressedPublicKey, publicKeyHash);

            uint8_t ripemd160Hash[20];
            ripemd160(publicKeyHash, 32, ripemd160Hash);

            if (hasMinimumMatchingCharacters(ripemd160Hash)) {
                std::string matchedPrivateKey = hexPrivateKey; // Store the private key for printing
                std::string matchedRipemd160 = hexBytesToHexString(ripemd160Hash, 20); // Convert RIPEMD160 to hex string

                {
                    std::lock_guard<std::mutex> lock(printMutex);
                    std::cout << "\nMatching RIPEMD160 hash found. Private Key: " << matchedPrivateKey << ", RIPEMD160: " << matchedRipemd160 << std::endl;
                }

                {
                    std::ofstream foundFile("found.txt", std::ios::app); // Append mode
                    if (foundFile.is_open()) {
                        foundFile << "Matched Private Key: " << matchedPrivateKey << ", RIPEMD160: " << matchedRipemd160 << std::endl;
                        foundFile.close();
                    }
                }
            }
        }

        totalKeys.fetch_add(1);

        // Convert the mpz_class to uint64_t and update the currentValue atomically
        currentValue.store(static_cast<uint64_t>(current.get_ui()), std::memory_order_relaxed);

        current += INCREMENT * numThreads;
    }

    // Signal that this thread has completed its work
    matchFound.store(true, std::memory_order_relaxed);
}

int main() {
    loadRIPEMD160Hashes();

    Secp256K1 secp256k1;
    secp256k1.Init();

    std::vector<std::thread> threads;

    // Start the progress printing thread
    std::thread progressThread(printProgress);

    const int numThreads = std::thread::hardware_concurrency();
    for (int i = 0; i < numThreads; ++i) {
        threads.emplace_back(counterWorker, i, std::ref(secp256k1), numThreads);
    }

    for (auto& thread : threads) {
        thread.join();
    }

    // Wait for the progress thread to complete
    progressThread.join();

    std::cout << std::endl;

    return 0;
}

now I get 1.5million/s if I have 1 factor on my 16 threaded pc which I normally get nothing less than 55 million/s on keyhunt compress and exhomorphism
the more the factor to be multiplied the less the keys/s
and come to think of it, so many private keys are just numbers and these numbers have factors except it's a prime number
Now I can solve puzzle 65 without the public key in just a few seconds if you have the right factors in place
we need to see if we can get this code working on GPU which will also make so much momentum and also if we can get it faster on cpu
it's just an idea in progress and \i don't know if it's a good idea that's why I have decided to put it on here
Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 09/09/2023, 12:51:24 UTC
Kangaroo can find a key in any range, regardless of its size. Even if you specify the wrong range, kangaroo will still find the key, but it will take much longer.

JLP's kangaroo implementation does have a problem with small ranges because by default there are too many kangaroos in a small range. You can adjust CPU/GPU group size to fix it, search with cpu only, or search with less threads/grid size.

Regarding searching for multiple keys, the kangaroo algorithm only works with one key at a time, so such a search is only possible sequentially. If it were possible to search for several keys simultaneously without loss of speed, then the puzzle's addresses with revealed public keys would have been solved long ago.

so the problem still lies in choice
we have none to make but to keep trying
we are all victims of causality
let's give the solver of 120 and 125 some more time, they'll solve all the puzzles with the exposed public key before anyone is able to solve puzzle 66
even the solo pool is only 3% gone from the search range despite the fact that we scan in bit ranges randomly
the truth is the only way to go is the puzzles with the private key and so much investment to go all in
and if anyone could stack up 100,000 4090s in a single machine which I don't think is possible. then the search would take approximately 20.5 hrs using KeyHunt Cuda (36893488147419103232/100000/5000000000/86400) but if you do the maths towards how much renting 100,000 4090s would turn out to be $830,000 (20.5*0.4*100000) using a standard rate of $0.4/hr for a single 4090. let's just imagine that the key would definitely not be at the last part so we can say 50/50 chance of getting it sooner before the last part which still leaves us with an investment of $415,000
now the target is how much again?
How much profit would you be making?
It's practically impossible to invest any amount of capital in the search for puzzles without the exposed pubkeys while still expecting to make any profit
and mind you "even if you have the private key for the rmd tahts have exactly 39 characters as that of the puzzle 66" for example "20d45a6a762535700ce9e0b216e31994335db8a4" and you thinking searching that range would get you closer to the range of puzzle 66 then you definitely don't understand how sha256 and ripemd160 works
Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 08/09/2023, 16:53:10 UTC
For keys with no available public key, the only possible solution is an especial hardware like an ASIC miner to grind sha256 and rmd160 hashes at a rate of trillions/s for the cheapest ASIC and thousands of T/s for advanced ASICs. Then you could wait 10 years and see if development can increase the hash per second for your ASIC or not.

So, other than engineers, a factory with billions in equipment and a few million to spend on prototypes, what else we don't have?

Oh and that's just for keys from 66 up to 80.
However if you know the public key, there is at least a chance that you could work out things mathematically.

Regarding unexposed keys, of course there is a mathematical solution which is finding hash 160 and then hash 256 collisions, but I haven't seen any tool doing that, and it's not discussed here at all.

Ps, I have some ideas about finding collisions, it is related to sha256 checksum operation, if I could change the checksum requirement from 8 first characters of the second hash, by making it 32 character, I could work on finding collisions, in fact if we could replace sha256 with rmd160 and code something to extract rmd160 double hash checksum by using the first 20 characters of either first or second hash, again we could start working on collisions.

But that's all just an idea, I don't know if it works or not.

quote "that's the idea that has been running through my mind all along"

I have been trying to code some kind of collision where it'd only workout sha256 and rmd 160 while retaining the pubkey
But even if it worked out as I've been trying we'd only be stuck with probably 118 if we can successfully extract the pubkey from sha256 rmd collisions

but that'd definitely still give us an edge over the current situation where puzzle 66 is grinding the hell out of me
Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 08/09/2023, 16:47:53 UTC
I've tried keyhunt random on CPU very slow.
And cubitcrack counting random ranges I give it time to time gave me about 10x speed of keyhunt, but without the random option, I don't like it.

I've tried to find any relation or pattern between the previous puzzles and got absolutely nothing!

I'd like to have that random option on GPU! is there any tool for that please?

Good thing I've learned some Python and found it's actually slow  Grin
I started to learn C++ and CUDA hopefully to create a better tool, is there any libraries you can recommend to work with?

Looking at the puzzles that are still unsolved and how big those ranges are! there's no way in our life time! But it's fun to play!

vanbitcracken2 or KeyHuntCuda

but mind you, random option resets your chance to 100 times the initial time to find the key you are looking for withing the specified 66 bit range
for example
If it'd take you 300 years to scan the range
expect 30,000 years on a random search and you will still not stand the chance either, except you're able to program the random to note the range it had already scanned which is impossible because why would it be called random then?

rrandom option is the least thought anyone should be thinking about right now

only except you'd want to scan a specific 50 bit range randomly which you'd have to not down just like the puzzle 66 pool.
so if you want to consent to Random search for the puzzle 66
It's best to resort to the specific bit range search where you'd have to note every defeated range in order for you not to multiscan the range you've already scanned
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Merits 1 from 1 user
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 23/08/2023, 15:08:00 UTC
⭐ Merited by citb0in (1)
Alright, after numerous calculations, I've come to the point where I need someone who is interested in puzzle 66 and capable of counting 48 bits within a few minutes. I have a total of 588 ranges for them to work on, with the opportunity to claim the prize of 6.6 BTC. Alternatively, if the author of the puzzle – the one who created this challenge to test the difficulty of increasing bits – has come across my post, I challenge them to assist me in improving my counting speed. I am confident in my ability to crack puzzle 66 in a matter of days.

what would the scanner get for wasting time and resources if the keys needed don't fall within these ranges?

I hope you know that addresses don't reflect on other addresses, they have no business whatsoever... if we are talking about pubkeys then It is pretty understandable, and I suggest you have the public key for number 66 to be able to have such confidence about the range to be scanned and if I were you and I had the pubkey then I wouldn't hesitate using kangaroo or keyhunt bsgs where it wont take up to 10 minutes to get your key
Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
bestie1549
on 23/08/2023, 14:22:38 UTC
Alright, after numerous calculations, I've come to the point where I need someone who is interested in puzzle 66 and capable of counting 48 bits within a few minutes. I have a total of 588 ranges for them to work on, with the opportunity to claim the prize of 6.6 BTC. Alternatively, if the author of the puzzle – the one who created this challenge to test the difficulty of increasing bits – has come across my post, I challenge them to assist me in improving my counting speed. I am confident in my ability to crack puzzle 66 in a matter of days.


im ready to work with you

As previously mentioned, I have a total of 588 ranges, where is a private key of puzzle 66. Each range requires a 48-bit computation (I am referring to GPU computation). and I am seeking an individual who is capable of performing these computations within few minutes or at least an hour.

As for our collaboration approach, I am currently developing a script to scan all 588 ranges. Once the private key is obtained, the script will automatically perform the following actions: 2% of the amount will be deducted as fees, 59% will be sent to the individual performing the computations, and 39% will be sent to me. my bitcoin address and the computation performer's bitcoin address will be pre-entered into the script. The reason for allocating 59% to the computation performer is due to their resource usage, whether through GPU rental or electricity consumption. I intend to share the entire script with the individual to ensure transparency, while keeping the ranges known only to me. Once the script initiates, neither of us will have any access to it. But when scanning of one range completed, it will be saved along with the proof of work. Certainly, it will take me some time to perform all these processes.

So why not continue searching for that person in the meantime?

(This range is an example of 588 ranges: 3b07a000000000000:3b07affffffffffff, and this address is for proof of work: 1GYpdGZBaqRDKkpFWbaRezrVeGy6g6UNfE, 1Ngdkik5LLGzmzYUWKiPYDQ6CRRGWCPo5L, 19XbiMqyqeaJLDSvHLzhJkxngDYsBPXTQo, 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so)

The person who will perform the counting of this range needs to be scanned and to post here with proof of work, or they can personally DM me as well.  In this interaction, I will assess their operational speed in completing this 48-bit range. For scanning, they can use any software, but my suggestion is to use 'VanBitCrackenS v1.0' because here they can take advantage of multi-GPU. If the puzzle address falls within this range, then it will be yours. 👍

what would the scanner get for wasting time and resources if the keys needed don't fall within these ranges?