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Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 30. (Read 192523 times)

hero member
Activity: 630
Merit: 731
Bitcoin g33k
...and hope whichever miner that finds it will return it to you...

if you want to rely on hope Smiley Well, just with HOPE you could try to execute the normal transaction and hope the block will be mined quickly afterward so it'll reach your address before anyone else replaces the TX  Tongue
full member
Activity: 1078
Merit: 219
Shooters Shoot...
Would that work?

Yes, but if you are the minner you can mine your own trasactions without broadcast it publicly. The mined block is only broadcasted if you found the solution for the block, in this case if you are luck enough and no other miner mine the same block height at the same time you will be able to redeem it without problem.

So what needs to be done is this:

If you have the key to #66, rent a crap ton, and I mean a crap ton of hash. Solve a BTC block, then include your transaction within the block you solved Smiley

Ol' Brainless always has a way to do things, but never shares Smiley  I wouldn't doubt he has some way of doing it, he is one who definitely thinks outside of any boxes placed upon us Smiley
hero member
Activity: 862
Merit: 662
Thank you for answer albert0bsd, I have 2 3070 gpu, how can i use them both in keyhunt for bsgs mode?

if you have GPU use kangaroo.

My keyhunt program actually doesn't support GPU yet
newbie
Activity: 25
Merit: 0
By which program you cracked that? and how to find thats in what range?

I use keyhunt the program that i develop:

https://bitcointalksearch.org/topic/keyhunt-development-requests-bug-reports-5322040
https://github.com/albertobsd/keyhunt

But other programs can also be used like Kangaroo

https://github.com/JeanLucPons/Kangaroo

There is no way to know the range of an address.

But since we are talking here of puzzle 66 i did the test in the bit 66 just to check if that address was in that range, and actually the key is of that address is in that range!!

Other address besides of low bit puzzles aren't vulnerable to this.

As soon as the person who finds puzzle 66 sends the transaction to the network, pubkey will appear without requiring network approval. Once Pubkey appears it will take 1 second to find the private key. Then, someone else will spend again because the network approval has not yet occurred. The person with the highest network approval will own the bitcoins. Even if you send with a high transaction fee, the person who receives a higher transaction fee and network approval than you will win. For this reason, Low puzzles are problematic.

Exactly all the non-confirmed transacctions are public avaible in the mempool of each node.

Also there is sites to check them https://mempool.space/ also they offer some api to check for some values

https://mempool.space/docs/api/rest#get-address-transactions-mempool

Once that you get the TX id, you need to download the Raw transaction, decode it and extract the publickey and that is easy to do if you know what are you doing.


If find the Pkey by pubkey is so easy like this, then why puzzle 130 didnt solve yet?

Well for puzzle  66 is really easy (some minutes/seconds) but since the complexity is exponential it will take a lot of time do that for those bits ranges, months or years depending of hardware.

Here we are talking of low bit puzzles less than 80 bits are solvables by GPU almost in less than 10 minutes (in avarage the time needed to mine a block)



Thank you for answer albert0bsd, I have 2 3070 gpu, how can i use them both in keyhunt for bsgs mode?
member
Activity: 260
Merit: 19
the right steps towerds the goal
if someone found 66 PK, can share with me, for withdraw, i know secure way to transfer, other you have seen above test, and definatly, slowly slowly lot of people would have 66 PK, but no one take step for attempt withdraw Smiley
For a drowning person, even a straw is a great support. I will definitely contact you.

When everything is dependent on fate, it's best that we only utilize this code. There's no need for undue force with the computer, no burden on the machine, no excessive worry about electricity costs, and most importantly, without any concern for double spending. Undecided

2^129 - 2^130 | PUB : 0383af736b4eefaa2f697e63b4530129ff34ff945be353ee498b0473238ea91ef4 | PVK : 0x2264f44bd23a25b4a5b8390e650cf9cae
2^134 - 2^135 | PUB : 03a987392d57dd82a9ee630f99a75b3925ca5417869bbdfc76b6f59997e558ee73 | PVK : 0x7a5330f4174f496ca6c4364888e9aef5db
2^139 - 2^140 | PUB : 0389d6d9b49a09988a70238b4c79639ef2881c686134075a485843e4814e36320d | PVK : 0xdffa0bf08d3584c46a48904da5df25016af
2^144 - 2^145 | PUB : 0226d47bfd7e231e9c123683137bf0f34522c03a8604fee9a9f626f0334daf9c0b | PVK : 0x167b8d6bec4db640ca2560f120f0a072bf658
2^149 - 2^150 | PUB : 0363abdd6ae3bc62c6ccf3d34d22b44d90f710fa99931d566a6a8c0da68acca933 | PVK : 0x3e1954a5af437841b085ba454c8b9e1149268a
2^154 - 2^155 | PUB : 035bebccb841e7b812fcfbc5fe86d9f9549e68d3d3da1bffd613e8a3960ca26ddc | PVK : 0x7a7179d2027c342b336fb8fd92347d50d937d1e
2^159 - 2^160 | PUB : 022cc56adb8c5385aefbb3cd38dfc2294560c4f73ad4562f68916535d6ab8ebf66 | PVK : 0xc09735b95a670ac005324cb8ddcf3576df7b3f1e


Code:
from bit import *
import random

L = [
'03633cbe3ec02b9401c5effa144c5b4d22f87940259634858fc7e59b1c09937852',
'02145d2611c823a396ef6712ce0f712f09b9b4f3135e3e0aa3230fb9b6d08d1e16',
'031f6a332d3c5c4f2de2378c012f429cd109ba07d69690c6c701b6bb87860d6640',
'03afdda497369e219a2c1c369954a930e4d3740968e5e4352475bcffce3140dae5',
'03137807790ea7dc6e97901c2bc87411f45ed74a5629315c4e4b03a0a102250c49',
'035cd1854cae45391ca4ec428cc7e6c7d9984424b954209a8eea197b9e364c05f6',
'02e0a8b039282faf6fe0fd769cfbc4b6b4cf8758ba68220eac420e32b91ddfa673'
]

while True:
    seed = random.getrandbits(161)
    random.seed(seed)
    for exp in range(129, 164, 5):
        a = random.randrange(2**exp, 2**(exp+1))
        pk = Key.from_int(a)
        key = Key.from_int(a)
        pub = pk._pk.public_key.format(compressed=True)
        addrpub = pub.hex()
        if addrpub in L:
            print('================found=====================\n')
            f=open(u"Key_Found.txt","a")
            f.write("Seed : " + str(seed) + "\n")
            f.write("Prvk : " + hex(a) + "\n")
            f.close()
            input('================Exit=====================\n')
            exit()
        #if addrpub.startswith('03633cbe3') or addrpub.startswith('02145d26') or addrpub.startswith('031f6a33') or addrpub.startswith('03afdda4') or addrpub.startswith('03137807') or addrpub.startswith('035cd185') or addrpub.startswith('02e0a8b0'):
        print(f"2^{exp} - 2^{exp+1} | PUB : {addrpub} | PVK : {hex(a)}", end = '\n')
       

If fortune is on your side, a single core of a computer is sufficient for you..  Good...Luck...Luck...And....Only....Luck
hero member
Activity: 862
Merit: 662
Would that work?

Yes, but if you are the minner you can mine your own trasactions without broadcast it publicly. The mined block is only broadcasted if you found the solution for the block, in this case if you are luck enough and no other miner mine the same block height at the same time you will be able to redeem it without problem.
donator
Activity: 1055
Merit: 1020
If there are thousands of bots watching the 66 address, which I think there are, they will all watch the mempool as well and keep increasing the fee against each other.  Maybe the answer is to send the transaction first with the entire amount as the fee minus 1 sat to your address and hope whichever miner that finds it will return it to you.  This way no one can increase the fee any higher and your timestamp is first.  Would that work?
hero member
Activity: 862
Merit: 662
By which program you cracked that? and how to find thats in what range?

I use keyhunt the program that i develop:

https://bitcointalksearch.org/topic/keyhunt-development-requests-bug-reports-5322040
https://github.com/albertobsd/keyhunt

But other programs can also be used like Kangaroo

https://github.com/JeanLucPons/Kangaroo

There is no way to know the range of an address.

But since we are talking here of puzzle 66 i did the test in the bit 66 just to check if that address was in that range, and actually the key is of that address is in that range!!

Other address besides of low bit puzzles aren't vulnerable to this.

As soon as the person who finds puzzle 66 sends the transaction to the network, pubkey will appear without requiring network approval. Once Pubkey appears it will take 1 second to find the private key. Then, someone else will spend again because the network approval has not yet occurred. The person with the highest network approval will own the bitcoins. Even if you send with a high transaction fee, the person who receives a higher transaction fee and network approval than you will win. For this reason, Low puzzles are problematic.

Exactly all the non-confirmed transacctions are public avaible in the mempool of each node.

Also there is sites to check them https://mempool.space/ also they offer some api to check for some values

https://mempool.space/docs/api/rest#get-address-transactions-mempool

Once that you get the TX id, you need to download the Raw transaction, decode it and extract the publickey and that is easy to do if you know what are you doing.


If find the Pkey by pubkey is so easy like this, then why puzzle 130 didnt solve yet?

Well for puzzle  66 is really easy (some minutes/seconds) but since the complexity is exponential it will take a lot of time do that for those bits ranges, months or years depending of hardware.

Here we are talking of low bit puzzles less than 80 bits are solvables by GPU almost in less than 10 minutes (in avarage the time needed to mine a block)
newbie
Activity: 25
Merit: 0
thousands of bots are drooling waiting for your transaction, your hard earned bitcoins will evaporate like steam over a pot
now no one will scan 66 address ddd

How can bots get ahead of me? Let's say I solved the puzzle without using the site and only I know the private key at the time of sending the transaction. please explain how this will happen because I apparently don’t understand something


As soon as the person who finds puzzle 66 sends the transaction to the network, pubkey will appear without requiring network approval. Once Pubkey appears it will take 1 second to find the private key. Then, someone else will spend again because the network approval has not yet occurred. The person with the highest network approval will own the bitcoins. Even if you send with a high transaction fee, the person who receives a higher transaction fee and network approval than you will win. For this reason, Low puzzles are problematic.


If find the Pkey by pubkey is so easy like this, then why puzzle 130 didnt solve yet?
jr. member
Activity: 37
Merit: 1
34Sf4DnMt3z6XKKoWmZRw2nGyfGkDgNJZZ
thousands of bots are drooling waiting for your transaction, your hard earned bitcoins will evaporate like steam over a pot
now no one will scan 66 address ddd

How can bots get ahead of me? Let's say I solved the puzzle without using the site and only I know the private key at the time of sending the transaction. please explain how this will happen because I apparently don’t understand something


As soon as the person who finds puzzle 66 sends the transaction to the network, pubkey will appear without requiring network approval. Once Pubkey appears it will take 1 second to find the private key. Then, someone else will spend again because the network approval has not yet occurred. The person with the highest network approval will own the bitcoins. Even if you send with a high transaction fee, the person who receives a higher transaction fee and network approval than you will win. For this reason, Low puzzles are problematic.
member
Activity: 316
Merit: 34
Can you crack and post the PVK of that signed address??

Code:
-----BEGIN BITCOIN SIGNED MESSAGE-----
I also own this address
-----BEGIN SIGNATURE-----
13zb1hQbWVuYdZoAkztVrNrm65aReL2pYD
H8SlTYJ7a/Mp5cra9VzqgDFMGRQUYfA5NLrSCb0GkwbeEDqx8vKWGYWX3YmiqIU8nl6THdprDK/k34y1GQrFFDk=
-----END BITCOIN SIGNED MESSAGE-----

It took 5 minutes in my laptop and less of 1 minute in main computer

WTF, I'm utterly astounded to realize that the smaller keys are not secure by any means, and searching for the larger ones is proving to be an insurmountable challenge for us. Given the magnitude of this predicament, it seems wise for us to disengage from this futile endeavor. The journey thus far has been both eye-opening and taxing.  Angry Sad Cry
if someone found 66 PK, can share with me, for withdraw, i know secure way to transfer, other you have seen above test, and definatly, slowly slowly lot of people would have 66 PK, but no one take step for attempt withdraw Smiley
sr. member
Activity: 1429
Merit: 264
"Do not disturb my kangaroos." --  Archimedes

 Grin

"It's Vogon poetry to me" -- Me.

So it's here you keep all the bright heads. Stuck on a riddle.
newbie
Activity: 25
Merit: 0
Can you crack and post the PVK of that signed address??

Code:
-----BEGIN BITCOIN SIGNED MESSAGE-----
I also own this address
-----BEGIN SIGNATURE-----
13zb1hQbWVuYdZoAkztVrNrm65aReL2pYD
H8SlTYJ7a/Mp5cra9VzqgDFMGRQUYfA5NLrSCb0GkwbeEDqx8vKWGYWX3YmiqIU8nl6THdprDK/k34y1GQrFFDk=
-----END BITCOIN SIGNED MESSAGE-----

It took 5 minutes in my laptop and less of 1 minute in main computer

the advantage is that no one knows exactly when the 66 puzzle will be solved, it could be a day, a month, a year or 100 years
I think that no one will monitor the 66 puzzle address for that long

That is cheap to do that, Actually I am doing it.




By which program you cracked that? and how to find thats in what range?
member
Activity: 286
Merit: 15
"Do not disturb my kangaroos." --  Archimedes

 Grin
member
Activity: 260
Merit: 19
the right steps towerds the goal
Can you crack and post the PVK of that signed address??

Code:
-----BEGIN BITCOIN SIGNED MESSAGE-----
I also own this address
-----BEGIN SIGNATURE-----
13zb1hQbWVuYdZoAkztVrNrm65aReL2pYD
H8SlTYJ7a/Mp5cra9VzqgDFMGRQUYfA5NLrSCb0GkwbeEDqx8vKWGYWX3YmiqIU8nl6THdprDK/k34y1GQrFFDk=
-----END BITCOIN SIGNED MESSAGE-----

It took 5 minutes in my laptop and less of 1 minute in main computer

WTF, I'm utterly astounded to realize that the smaller keys are not secure by any means, and searching for the larger ones is proving to be an insurmountable challenge for us. Given the magnitude of this predicament, it seems wise for us to disengage from this futile endeavor. The journey thus far has been both eye-opening and taxing.  Angry Sad Cry
member
Activity: 76
Merit: 16
Lol, yeah, I do not think you understand the Kangaroo algo.

It's all laid out for you in various readings/papers.

I never said 9 billion kangaroos. Do you understand the algo? When I say "find" x amount of tames and wilds, it is referring to the points/distances found by each type of kangaroo. You store tame and wild points (Based on DP used) and distances, that are generated from the tame and wild kangaroos, hopping all around.
2^66.05 - 2^32 (DP size that I stated) = 2^34.05 stored DPs. 2^33.05 tames and 2^33.05 wilds. 2^33.05 = 8,892,857,981; so roughly 9 billion points and distances stored (tames and wilds) to solve, on average. Could be a little higher, could be a little lower. So no, I was not "kidding". And yes, my times are based on math, and the space complexity is what I said, roughly 9 billion points & distances per tame and wild, to solve. I can't give you exact amount of GBs required because each Kangaroo program stores points differently, different amount of bytes and different formats, binary vs plain text. One would need to calculate it based on their DP and how the points/distances are stored.

But yes, you could set out 2 kangaroos, 1 tame, and 1 wild, and eventually solve, in many many years, or you could get lucky and solve within minutes, hours, days.

I doubt whoever solved 120/125, if they used the kangaroo algo, set a DP of less than 28. They would have an enormous amount of DP overhead, that JLP explains well in his github:
Code:
DP overhead according to the range size (N), DP mask size (dpBit) and number of kangaroos running in paralell (nbKangaroo).

110 and 115 were both solved with DP 25. I know that during the 115 run, the grid sizes for the GPUs were choked down and another part of the code was reduced, to prevent a massive DP overhead. And when finally solved, I do believe total DPs stored (points w distances) was a smidge over the expected total of 2^33.55
So your estimated times include the DP overhead?
Because I did a lot of simulations on lower puzzles to get min/avg/max jumps and stored footprints, and, as you can guess, when the DP criteria kicks in, storage goes down, number of jumps until a collision goes up. And there's a lot of ways to improve either space or time, but not both.

So it is not fair to still call the time complexity anywhere near O(sqrt(n)) when you're using 2*sqrt(sqrt(n)) stored items instead of the average 2*sqrt(n).

Steps until a collision can range between expected average / 10, and 3 * the expected average.

And JLP kangaroo is not some godly reference, I did not use it at all, for one I don't even agree with the way jumps and hashmap keys are used, but that's another story.

No one says you need to have 50/50 tames and kangs. And no one says you even need to store the entire traveled distance. There are various techniques.
hero member
Activity: 862
Merit: 662
Can you crack and post the PVK of that signed address??

Code:
-----BEGIN BITCOIN SIGNED MESSAGE-----
I also own this address
-----BEGIN SIGNATURE-----
13zb1hQbWVuYdZoAkztVrNrm65aReL2pYD
H8SlTYJ7a/Mp5cra9VzqgDFMGRQUYfA5NLrSCb0GkwbeEDqx8vKWGYWX3YmiqIU8nl6THdprDK/k34y1GQrFFDk=
-----END BITCOIN SIGNED MESSAGE-----

It took 5 minutes in my laptop and less of 1 minute in main computer

the advantage is that no one knows exactly when the 66 puzzle will be solved, it could be a day, a month, a year or 100 years
I think that no one will monitor the 66 puzzle address for that long

That is cheap to do that, Actually I am doing it.
jr. member
Activity: 42
Merit: 1
the advantage is that no one knows exactly when the 66 puzzle will be solved, it could be a day, a month, a year or 100 years
I think that no one will monitor the 66 puzzle address for that long
hero member
Activity: 862
Merit: 662
Can you crack and post the PVK of that signed address??

Well, the puzzle 66 is expected to be in the 66 bit key space, is your key also in that range?

If that is true, then anyone would be able to crack it
jr. member
Activity: 42
Merit: 1
Just kidding, but the risk here is if the hacker set a high transaction fee, the hackers transaction may be confirmed before yours
now there is a lot of money at stake, I think that there are a lot of bots attached to the 66 address puzzle
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