Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 83. (Read 215611 times)
You have to limit yourself to only searches for #130
Can somebody who say double-spend is possible with RBF-disabled make a video to actually proof this. This is a simple test, which I tried several times on my own addresses and I have always negative results.
If we found a private key then can we import that into bitcoin core?
Opt-in RBF is still default on the current version, so there is still hope lol! 
Well not all the miners but some of them yes that depend of the version and custom configuration, the exact number is unknown
So this is a new feature - Full RBF <> RBF. Are all the miners now set up for Full RBF instead of BIP125?
Well not all the miners but some of them yes that depend of the version and custom configuration, the exact number is unknown
So this is a new feature - Full RBF <> RBF. Are all the miners now set up for Full RBF instead of BIP125?
Well not all the miners but some of them yes that depend of the version and custom configuration, the exact number is unknown
So this is a new feature - Full RBF <> RBF. Are all the miners now set up for Full RBF instead of BIP125?
It is a Node configuration and it may vary depending of the Bitcoin Core Version.
The full name is Full RBF
Check this link:
Replace-by-fee (RBF)
Also this link:
https://en.bitcoin.it/wiki/Transaction_replacement
Can you explain why you feel that the miners will completely ignore the RBF flag if it is set to disable? I know it is up to the miners but wouldn't they mostly stick to the intent of the feature? I know the incentive would be to earn more money but there are many cases where the miner returned excess fee as well.
It is a Node configuration and it may vary depending of the Bitcoin Core Version.
The full name is Full RBF
Check this link:
Replace-by-fee (RBF)
Quote
Replace-By-Fee (RBF) is a node policy that allows an unconfirmed transaction in a mempool to be replaced with a different transaction that spends at least one of the same inputs and which pays a higher transaction fee.
Different node software can use different RBF rules, so there have been several variations. The most widely-used form of RBF today is BIP125 opt-in RBF as implemented in Bitcoin Core 0.12.0 and subsequent versions; this allows the creator of a transaction to signal that they’re willing to allow it to be replaced by a higher-paying version. An alternative form of RBF is full-RBF that allows any transaction to be replaced whether or not it signals BIP125 replaceability.
Different node software can use different RBF rules, so there have been several variations. The most widely-used form of RBF today is BIP125 opt-in RBF as implemented in Bitcoin Core 0.12.0 and subsequent versions; this allows the creator of a transaction to signal that they’re willing to allow it to be replaced by a higher-paying version. An alternative form of RBF is full-RBF that allows any transaction to be replaced whether or not it signals BIP125 replaceability.
Also this link:
https://en.bitcoin.it/wiki/Transaction_replacement
Can you explain why you feel that the miners will completely ignore the RBF flag if it is set to disable? I know it is up to the miners but wouldn't they mostly stick to the intent of the feature? I know the incentive would be to earn more money but there are many cases where the miner returned excess fee as well.
It is a Node configuration and it may vary depending of the Bitcoin Core Version.
The full name is Full RBF
Check this link:
Replace-by-fee (RBF)
Quote
Replace-By-Fee (RBF) is a node policy that allows an unconfirmed transaction in a mempool to be replaced with a different transaction that spends at least one of the same inputs and which pays a higher transaction fee.
Different node software can use different RBF rules, so there have been several variations. The most widely-used form of RBF today is BIP125 opt-in RBF as implemented in Bitcoin Core 0.12.0 and subsequent versions; this allows the creator of a transaction to signal that they’re willing to allow it to be replaced by a higher-paying version. An alternative form of RBF is full-RBF that allows any transaction to be replaced whether or not it signals BIP125 replaceability.
Different node software can use different RBF rules, so there have been several variations. The most widely-used form of RBF today is BIP125 opt-in RBF as implemented in Bitcoin Core 0.12.0 and subsequent versions; this allows the creator of a transaction to signal that they’re willing to allow it to be replaced by a higher-paying version. An alternative form of RBF is full-RBF that allows any transaction to be replaced whether or not it signals BIP125 replaceability.
Also this link:
https://en.bitcoin.it/wiki/Transaction_replacement
Can you explain why you feel that the miners will completely ignore the RBF flag if it is set to disable? I know it is up to the miners but wouldn't they mostly stick to the intent of the feature? I know the incentive would be to earn more money but there are many cases where the miner returned excess fee as well.
Just to mention that when the nodes have many FullRBF transacions not always win that one with more fee, here some examples, Dot with Green margin was mined, some are Testnet and other are mainnet
Just to mention that when the nodes have many FullRBF transacions not always win that one with more fee, here some examples, Dot with Green margin was mined, some are Testnet and other are mainnet
Here was mined a TX with 1 sat/vB instead of a 37 sat/vB
Here was mined a TX with 22 sat/vB instead of a 44 sat/vB
Here was mined a TX with 106 sat/vB instead of a 1032 sat/vB
Here was mined a TX with 1 sat/vB instead of a 37 sat/vB
Here was mined a TX with 22 sat/vB instead of a 44 sat/vB
Here was mined a TX with 106 sat/vB instead of a 1032 sat/vB
Dear Alberto, there is a lot of speculation going around in the group. Thank you for the information you provided.
We want to look for inferior puzzles, but if they are found and someone else steals them, all our efforts will be in vain.
Full information needs to be provided on this issue.
Bitcoin address and Message signature will Leak public key 
?
?Only Signed messages leak the public key. The address alone doesn't leak anything
By which program you cracked that? and how to find thats in what range?
Bitcoin address and Message signature will Leak public key
?
Just to mention that when the nodes have many FullRBF transacions not always win that one with more fee, here some examples, Dot with Green margin was mined, some are Testnet and other are mainnet
Here was mined a TX with 1 sat/vB instead of a 37 sat/vB
Here was mined a TX with 22 sat/vB instead of a 44 sat/vB
Here was mined a TX with 106 sat/vB instead of a 1032 sat/vB
Here was mined a TX with 1 sat/vB instead of a 37 sat/vB
Here was mined a TX with 22 sat/vB instead of a 44 sat/vB
Here was mined a TX with 106 sat/vB instead of a 1032 sat/vB
...and hope whichever miner that finds it will return it to you...
if you want to rely on hope
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 
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
Ol' Brainless always has a way to do things, but never shares
I wouldn't doubt he has some way of doing it, he is one who definitely thinks outside of any boxes placed upon us 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
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?
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
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.
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')
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
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.
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?
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)
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