Author

Topic: Please steal my bitcoins (Read 909 times)

newbie
Activity: 6
Merit: 21
April 16, 2021, 03:19:16 PM
#50
Thanks for your input guys,

it is my calculation that this would be near on impossible but I need to be sure as this is just a test, I will be putting a full bitcoin at risk in future

member
Activity: 77
Merit: 11
April 07, 2021, 01:55:11 AM
#49
I'll give you 0.002Btc close the topic. more valuable time people spend on this topic
newbie
Activity: 6
Merit: 0
April 06, 2021, 11:58:54 AM
#48

What is the probability that this is the private key, from another address? Undecided
[/quote]

If anyone cracks the Pvkey it will be obvious if it is the key to the posted public address.  Also the OP already posted a sig.
   I did a few calculations earlier and figure with my hardware it would take a few hundred years to go through all the permutations...  
legendary
Activity: 1722
Merit: 4711
**In BTC since 2013**
April 06, 2021, 06:38:37 AM
#47
This is my public address: bc1qq2rnv02hjzv5h0lwa03um43afwcfcpf0qg56ca

This is my private key in a randomised order relating to this address: LP5U3KWRLvPwDefz4FVMrAJVFtU4u8pj15w8VSpZF2aaPeY5Qix8

Do your worst

Thanks


Damn this is much harder to win than LOTTO Combination ..


What is the probability that this is the private key, from another address? Undecided
sr. member
Activity: 2618
Merit: 439
April 06, 2021, 06:24:49 AM
#46
This is my public address: bc1qq2rnv02hjzv5h0lwa03um43afwcfcpf0qg56ca

This is my private key in a randomised order relating to this address: LP5U3KWRLvPwDefz4FVMrAJVFtU4u8pj15w8VSpZF2aaPeY5Qix8

Do your worst

Thanks


Damn this is much harder to win than LOTTO Combination ..


Book marking this thread and will check every now and then just to check if there is a winner.

GoodLuck everyone..
newbie
Activity: 6
Merit: 0
April 05, 2021, 07:23:57 PM
#45
Why would Electrum be a clue... ?
legendary
Activity: 1722
Merit: 4711
**In BTC since 2013**
April 05, 2021, 12:12:07 PM
#44
Lance V thanks for all your comments, I have edited the randomised key in the opening post, also to answer your point about the randomisation, the key has been heavily randomised.  I did initially start moving blocks of 2 or more characters around before it dawned on me that that was not the wisest method.  In the end I made a table, as eluded to earlier, and resorted an adjacent random number column multiple times.

If the concern was completely random. So it is extremely difficult to find the right combination..
We are talking about millions of possibilities. It is not impossible, but it takes a very high processing capacity to be able to do this in a timely manner.

If it was done using logic, then it might be easier. Even so, it would be necessary to first discover this logic, which will be equally difficult.

It is still an interesting challenge.
newbie
Activity: 6
Merit: 0
April 04, 2021, 09:25:43 PM
#43
~

I think the reason why my script was so slow is because of Python, maybe if I make a C++ version of it it'll be faster. I just have to implement bitcoi[Suspicious link removed]vtopub myself, however I do it.



I would be interested in seeing C++ code.   I'm an old programmer but have never done any C++
newbie
Activity: 6
Merit: 21
March 31, 2021, 07:52:28 AM
#42
Hi All - thanks for all the great feedback

I have signed the wallet, here are the details:
Message:
Hello this is Dior
Address:
bc1qq2rnv02hjzv5h0lwa03um43afwcfcpf0qg56ca
Signature:
IBsoU2koKJvnjonXZ9O8/AIAAD5ZkgWgMmWOLwArXdV7eJxcKKn3DdDXO2j731mTy5W4vkSdDjyGudaQt7KS+e8=
Screenshot here: https://docs.google.com/drawings/d/1xg05QIzeXNj_1WeXg0uQLRO8qnYBMLjvjGngH8246k8/edit

Hopefully we can all relax and agree that this address belongs to me and only me, and yes I really want you to take them.  On that subject, please note that my use of the word "stealing" was only colourful hyperbole.  If you prefer a disclaimer I totally get that, we don't all live under the same legal system afterall, then here it is.  I guarantee I am the sole owner of the Bitcoin (BTC) recorded at the address stated and I am happy for anyone to move it to any other address and thereby take full and permanent ownership of it.  In fact I welcome it, I will consider it a success if achieved by any means necessary, mathematical or otherwise.

Lance V thanks for all your comments, I have edited the randomised key in the opening post, also to answer your point about the randomisation, the key has been heavily randomised.  I did initially start moving blocks of 2 or more characters around before it dawned on me that that was not the wisest method.  In the end I made a table, as eluded to earlier, and resorted an adjacent random number column multiple times.

So far, my money is quite literally on NotaTether  Grin

I guess what this boils down to is how easy is it to rearrange the private key and successfully use it.  If we agree it is near impossible, great I will leave them there until quantum computing or BTC dollar appreciation changes that.  If someone takes them very easily, great but less so as I will need to rethink things.  If we think it is hard but not worth the effort then that's fine too, but can we quantify how hard that is, that would be useful information.

Thanks and good luck




sr. member
Activity: 1764
Merit: 260
Binance #SWGT and CERTIK Audited
March 31, 2021, 07:36:21 AM
#41
There are lots of doubt and negative comment. If you will check the address which sent money to the address mentioned @op, you'll notice that it has big balance on it. So it probably was sent to create this "game" for fun, or maybe for some security purposes Smiley
If you don't want to play in on it, then don't haha. Not all people does something like this for money or any reward.
legendary
Activity: 3290
Merit: 16489
Thick-Skinned Gang Leader and Golden Feather 2021
March 31, 2021, 06:56:25 AM
#40
A single GPU can go through 2^52 combinations of stuff quite quickly so it's not unreasonable to assume that it can also cycle through 52! permutations as well.
No it can't: 52!/2^52=17909712639806203109773682781470098163210004180908203.125
legendary
Activity: 1568
Merit: 6660
bitcoincleanup.com / bitmixlist.org
March 31, 2021, 06:44:21 AM
#39
This is my opinion:

WE the members here should not attempt to do this UNLESS the OP will here and now "sign" the address in the original post.  This and this ALONE proves that the address is owned/controlled by them.

OP ---- why would you be unwilling to sign this address if in fact it is in your control?

The coins were moved to that address 50 minutes before OP's topic was made. So my guns are likely on that she owns them.

When people go through all the combinations, 0.002 BTC will be worth more than all the gold in the universe.

Not really. A single GPU can go through 2^52 combinations of stuff quite quickly so it's not unreasonable to assume that it can also cycle through 52! permutations as well.
copper member
Activity: 1554
Merit: 489
Stop the war!
March 31, 2021, 05:58:23 AM
#38
If my math is right then we have 2*(51!) ~ 3*10^66 combinations. That's quite a big number.
And for each of those combinations the key has to be converter to address and checked.

Maybe somebody can further (greatly) reduce the numbers, but I don't know how.

I believe that people are not going to waste their time for 0.002B. If someone has a system that reduces the combinations (I doubt it) then maybe they will try it but it is a lot of effort for so little reward.

When people go through all the combinations, 0.002 BTC will be worth more than all the gold in the universe.
legendary
Activity: 3290
Merit: 16489
Thick-Skinned Gang Leader and Golden Feather 2021
March 31, 2021, 05:40:48 AM
#37
I like what you're doing Smiley But as you can see, many people are sceptical, and we've seen many Newbies over the years who waste people's time. Some, however, are honest. So I have a proposal to prove you're honest at this moment: can you move the 0.002BTC to another address, wait for it to confirm, and post the private key to the current address? If you do: don't overpay on transaction fees (your last transaction could have been a lot cheaper).
After that, you can continue your quest by randomizing the new private key and posting it.

It's a bit of a hassle, but I'd really appreciate you doing this. If you do, I believe your post is worth reading so I'll Merit it.

Don't worry everyone, the address is funded and it will remain funded forever, I'm not that sort of girl.
I realize my proposal means you'll have to break this promise. It's up to you Smiley

If my math is right then we have 2*(51!) ~ 3*10^66 combinations. That's quite a big number.
And for each of those combinations the key has to be converter to address and checked.

Maybe somebody can further (greatly) reduce the numbers, but I don't know how.
If you account for duplicate characters, the number of possibilities drops significantly:
Code:
     3 V
      3 P
      3 F
      3 8
      3 5
      2 w
      2 p
      2 e
      2 a
      2 U
      2 L
      2 4
      1 z
      1 x
      1 v
      1 u
      1 t
      1 r
      1 j
      1 i
      1 f
      1 Z
      1 Y
      1 W
      1 S
      1 R
      1 Q
      1 M
      1 K
      1 J
      1 D
      1 A
      1 3
      1 2
      1 1
It may even be "easier" to brute-force a random funded address. So indeed OP makes a good point with this topic.

@OP: did you really use a random order, or did you manually shuffle some (blocks of) characters? The latter would greatly increase the chance of ever finding this key.

This is my public address: bc1qq2rnv02hjzv5h0lwa03um43afwcfcpf0qg56ca
the correct randomised key is: LP5U3KWRLvPwDefz4FVMrAJVFtU4u8pj15w8VSpZF2aaPeY5Qix8
Quoting for future reference Smiley You should probably update the incorrect randomised key at the start of the topic.

many probably feel bad about the word 'stealing' and doing anything related to it.
It's like saying: "my Bitcoin is in this list. Good luck stealing it".

If there were any way to confirm that the private key thing is legit and solvable (by referring to a trusted forum member to confirm this, for instance), it could make the whole thing pretty popular here on Bitcointalk.
I thought of that, but decided it's better if OP proves once that he's currently honest, without trusting anyone else with the funded private key. Trusting someone else with the private key kinda goes against what Bitcoin stands for.

sr. member
Activity: 1820
Merit: 418
Telegram: @worldofcoinss
March 31, 2021, 02:53:51 AM
#36
Only the comment I can leave here is that private key must not be revealed to only one, if known to anyone, the person will be able steal the bitcoin the private key can unlock on blockchain.

Read carefully, the OP wants you to to take away his bitcoins. Thats the sole reason of private key to be there.
full member
Activity: 1442
Merit: 106
March 30, 2021, 06:45:07 PM
#35
The idea of trying to trace the wallet balance and then cracking the private key is the last thing that should come to someone's mind. Imagine the complexity that is in the occassion of validating a transaction and then you want to try out in scrambling the alphanumeric combination to get a private key to unlock the wallet, it will be possible but not in the ner future as you will so get frustrated.

This is my public address: bc1qq2rnv02hjzv5h0lwa03um43afwcfcpf0qg56ca

This is my private key in a randomised order relating to this address: LP5U3KWRLvPwDefz4FVMrAJVFtU4u8pj15w8VSpZF2aaPeY5QIx8

Do your worst

Thanks


hero member
Activity: 761
Merit: 606
March 30, 2021, 06:21:58 PM
#34
This is my opinion:

WE the members here should not attempt to do this UNLESS the OP will here and now "sign" the address in the original post.  This and this ALONE proves that the address is owned/controlled by them.

OP ---- why would you be unwilling to sign this address if in fact it is in your control?
full member
Activity: 1736
Merit: 116
March 30, 2021, 05:52:32 PM
#33
This is an interesting thing for me, because as long as I am active in this forum most people are looking for a safe way so that their Bitcoin is not stolen.
But the opening post did the unexpected by inviting people to steal his Bitcoin. I don't know what the real purpose the opening post is doing this,
but it would be very interesting if there were members of this forum who managed to crack the private key. I agree that people who have
programming skills want to solve this puzzle not because of the reward, because the number of Bitcoins contained at the address is not large.
But maybe he want to test its ability to hack the wallet. Hopefully someone will succeed, I will continue to monitor this topic.
hero member
Activity: 1876
Merit: 524
March 30, 2021, 05:43:16 PM
#32
I don't think much has to be dealt with for 0.002 btc. When I look at the alphabetic and numerical numbers, I don't think it's worth the trouble at all. It is worth the effort if there is a large amount in it, but I do not recommend anyone to bother.
legendary
Activity: 2576
Merit: 1252
Leading Crypto Sports Betting & Casino Platform
March 30, 2021, 05:13:44 PM
#31
It is really strange and not popular to think that someone is barely giving out his information and telling everybody to steal whatever amount is contained by that account. At first look, it seems like I got to think the account is not yours because you are daring everybody to try stealing it if they can and that might be a crime to commit for it might belong to some innocent individual and have just stored that amount for better use in future. If that will be really yours giving out a randomized private key even after knowing the amount of Bitcoin you have in your account will still not push people in here to hack such because it will just be a hassle to be done so barely better keep that as long as you can or better leave it forgotten if you are really a hopeless one.
legendary
Activity: 952
Merit: 1385
March 30, 2021, 03:50:57 PM
#30
Brute-force algorithm would be:
1) Let's say WIF starts with K (later - repeat with L) - remove K from the set of letters
2) Calculate combinations from the rest of the letters (51) taking 45 letters (do not use 6 letters for checksum) (18009460)
3) Calculate permutations for each of 45 letters
4) For each permutation try to decode WIF: K + calculated result + fake checksum, which should be ignored (just extract privkey from WIF).

Ideas for optimization:
- check which letters are duplicated
- check min/max possible WIF (min: KwDiB..., max: L5o...)


Other remarks:
Possible prefix & rest of letters:
Code:
"Kw" "LP5U3WRLvPwDefz4FVMrAJVFtU4u8pj158VSpZF2aaPeY5Qix8",
"Kx"  "LP5U3WRLvPwDefz4FVMrAJVFtU4u8pj15w8VSpZF2aaPeY5Qi8",
"Kz"  "LP5U3WRLvPwDef4FVMrAJVFtU4u8pj15w8VSpZF2aaPeY5Qix8",
"L1"  "P5U3KWRLvPwDefz4FVMrAJVFtU4u8pj5w8VSpZF2aaPeY5Qix8",
"L2"  "P5U3KWRLvPwDefz4FVMrAJVFtU4u8pj15w8VSpZFaaPeY5Qix8",
"L3"  "P5UKWRLvPwDefz4FVMrAJVFtU4u8pj15w8VSpZF2aaPeY5Qix8",
"L4"  "P5U3KWRLvPwDefzFVMrAJVFtU4u8pj15w8VSpZF2aaPeY5Qix8",
"L5"  "PU3KWRLvPwDefz4FVMrAJVFtU4u8pj15w8VSpZF2aaPeY5Qix8"

Additionally, because of https://bitcointalksearch.org/topic/m.56676121 I think we may assume that "i" is not the last character.
member
Activity: 224
Merit: 36
March 30, 2021, 03:04:16 PM
#29


I think the reason why my script was so slow is because of Python, maybe if I make a C++ version of it it'll be faster. I just have to implement bitcoin.prvtopub myself, however I do it.



There might be tools to help with that. I've just googled something called "Cython".

However, in my experience, those sort of translators never fully work and there is some tweaking to do.
sr. member
Activity: 1666
Merit: 426
March 30, 2021, 10:05:49 AM
#28
Do you really want them to steal your bitcoins? Anyways I don't know the purpose of posting that here in this forum because, to be honest, it is only now I've seen a person who wants to steal his/her own bitcoin by anybody, is it a challenge or not? But first of all, let us just confirm if this wallet has a bitcoin or he is simply trolling us all of a sudden. If this was really a challenge or an event then we are excited to participate here mate.
legendary
Activity: 3248
Merit: 1402
Join the world-leading crypto sportsbook NOW!
March 30, 2021, 09:58:12 AM
#27
This could be an interesting game and attract more participants if it were branded differently. You're asking people to steal the coins, but many probably feel bad about the word 'stealing' and doing anything related to it. Also, it's not great that you're a newbie, so there isn't a good reason to trust you that this thing is legit, especially after and capitalized I controversy. But you could make it a challenge with a 0.002 reward, write some conditions and the description of it as a challenge, not as a post asking people to steal the coins. If there were any way to confirm that the private key thing is legit and solvable (by referring to a trusted forum member to confirm this, for instance), it could make the whole thing pretty popular here on Bitcointalk.
legendary
Activity: 2716
Merit: 1225
Once a man, twice a child!
March 30, 2021, 09:16:55 AM
#26
Only the comment I can leave here is that private key must not be revealed to only one, if known to anyone, the person will be able steal the bitcoin the private key can unlock on blockchain.
That's the essence of OP's post. The OP is actually daring someone to crack the PKey combination and steal the funds there.

I believe the OP mixed and scrambled the figures and that's going to be a hard nut for anyone to crack fixing them up.
sr. member
Activity: 1106
Merit: 391
March 30, 2021, 09:01:35 AM
#25
Hilarious! has anyone checked to see if the wallet actually contains anything or the op is just here to troll, to be honest, i have never seen where anyone wants to have their asset stolen voluntarily, it has always been the opposite, i can already tell there is nothing that will worth the effort.


op must make fun of us by changing a little bit of the private key he gave or he forgot his password or something else.

but I saw that the contents in the wallet were very little and not so valuable with the difficulties faced to solve it. Cheesy
newbie
Activity: 6
Merit: 21
March 30, 2021, 08:56:26 AM
#24
Good luck NotATether, hope my error didn't cost you too much time.

I'll definitely share the reason for this experiment and I'll definitely do it here first, but not just yet.  I need this to run for a bit first to prove how secure or otherwise a completely exposed but randomised key is

Anyone saying it's easy and so redundant, great help yourself to the BTC and prove how easy this was all along.

I'm offline till the morning so I'm afraid I'll be quiet until then but happy to respond to any feedback when I get back

legendary
Activity: 952
Merit: 1385
March 30, 2021, 08:51:17 AM
#23
Brute-force algorithm would be:
1) Let's say WIF starts with K (later - repeat with L) - remove K from the set of letters
2) Calculate combinations from the rest of the letters (51) taking 45 letters (do not use 6 letters for checksum) (18009460)
3) Calculate permutations for each of 45 letters
4) For each permutation try to decode WIF: K + calculated result + fake checksum, which should be ignored (just extract privkey from WIF).

Ideas for optimization:
- check which letters are duplicated
- check min/max possible WIF (min: KwDiB..., max: L5o...)
legendary
Activity: 1568
Merit: 6660
bitcoincleanup.com / bitmixlist.org
March 30, 2021, 08:42:44 AM
#22
~

I think the reason why my script was so slow is because of Python, maybe if I make a C++ version of it it'll be faster. I just have to implement bitcoin.prvtopub myself, however I do it.



Good luck NotATether, hope my error didn't cost you too much time.

I think Python had caused me more lost time than this omission in fact.
legendary
Activity: 2912
Merit: 6403
Blackjack.fun
March 30, 2021, 07:49:19 AM
#21
Besides all you have to do is keep an eye on my address.

That would be just a stop loss situation, if you decide to move the coins in 7 days from now on it it would simply mean someone who has actually tried to do it would have wasted his time. That again, it's just 100$, I would be far more suspicious if the sum would have been 10-100x higher.

I know some of you have said I could be trolling and given an inaccurate key but that would ruin my experiment and my experiment is worth far more than an unfunny joke.

What would be the purpose of the experiment? It's just a question of math and quantifying reward per effort.
If the funds don't get moved in two hundred years it means it's safe to randomize your key and let it sit on three hundred forums and nobody will touch your coins? If the funds get swiped in one month what does this prove?
I'm more interested in your method of interpreting and quantifying the results than in the actual result.

'I' on the 50th place - wrong character, not from Base58 set.

Nice one!

legendary
Activity: 2968
Merit: 3684
Join the world-leading crypto sportsbook NOW!
March 30, 2021, 07:33:55 AM
#20
The thing with all this is that you need to go through all this AND trust that:
- the letters and numbers are indeed real and the author hasn't also changed a few of them for trolling
- the author won't move the coins tomorrow or in the next hours, making all your effort a waste

Doing this for 100$  based on a post created by a newbie?
Who is going to be tying this will do it out of boredom rather than for the reward

Most valid comment I have seen in a while.

I don't knock people, and it's nice if it's $10 or $100, but you have no idea if this is a troll, so yeah, whoever's gonna do it won't be doing it for the reward, just for the pure sake of being bored. There's no bragging rights either as there's no puzzle to solve, no IQ to claim.

Good luck still, hope someone has fun.
member
Activity: 89
Merit: 14
March 30, 2021, 07:17:00 AM
#19
Quote
'I' on the 50th place - wrong character, not from Base58 set.

When I just repeated this I noticed that after I add a comma after the small case i, Excel changes it to I


Seems to be a plausible explanation, MS Office is notorious for being overzealous with auto correction.
newbie
Activity: 6
Merit: 21
March 30, 2021, 07:08:41 AM
#18
Quote
'I' on the 50th place - wrong character, not from Base58 set.
Guy's I'm so so sorry the "I" became capitilised it should be an i

I was horrified when I saw this and I have just retraced my steps as follows

In order to randomise the key I extracted it from Electrum (free clue)
I pasted in Excel
in order to transpose so that I could create an adjacent =rand() column, I had to use text to columns as otherwise the cell key is just a single cell
I had to delimit and the easiest way was to introduce a comma between each character
When I just repeated this I noticed that after I add a comma after the small case i, Excel changes it to I
i is included in Base 58, I is not

I'm totally devastated and sorry and I understand if that makes me look dishonest, I'm real sorry.

I have just rechecked each character one by one to make sure no other errors crept in and I confirm that 100% each character is correct with the correct capitalisation. The 50th character i was the only one affected so the correct randomised key is: LP5U3KWRLvPwDefz4FVMrAJVFtU4u8pj15w8VSpZF2aaPeY5Qix8

Once again I'm really really sorry
member
Activity: 89
Merit: 14
March 30, 2021, 07:07:23 AM
#17
This is my private key in a randomised order relating to this address: LP5U3KWRLvPwDefz4FVMrAJVFtU4u8pj15w8VSpZF2aaPeY5QIx8

'I' on the 50th place - wrong character, not from Base58 set.

Well spotted... (or validated if you were tinkering with your WIF recovery tool?)

This should be interesting... will OP be amending and/or explaining?  Undecided
legendary
Activity: 1568
Merit: 6660
bitcoincleanup.com / bitmixlist.org
March 30, 2021, 07:01:31 AM
#16
Who is going to be tying this will do it out of boredom rather than for the reward

Yeah  Lips sealed

Some multithreaded script I was working on the last few hours:

Code:
from bitcoin import *
import threading
import time
import math
import sys

pub = "bc1qq2rnv02hjzv5h0lwa03um43afwcfcpf0qg56ca"
randprv = "LP5U3KWRLvPwDefz4FVMrAJVFtU4u8pj15w8VSpZF2aaPeY5Qix8"
_1M=1024**2
nthreads = 48

def index2permarray(k, size):
  if size <= 1:
    return [0]
  arr = []
  while size > 1:
    multiplier = math.factorial(size-1)
    digit = math.floor(k/multiplier)
    arr += [digit]
    k = k % multiplier
    size = size-1
  return arr

fact52array = []
for i in range(0,53):
  fact52array += [math.factorial(i)]

def index2permutation(i,n,p):
  arr=""
  for k in range(1, n + 1): # k goes from 1 to n
    f = fact52array[n - k]  # compute factorial once per iteration
    d = i // f            # use integer division (like division + floor)
    arr = arr + str(p[d])          # print permuted number with trailing space
    p = p[:d] + p[d+1:]        # delete p[d] from p
    i = i % f             # reduce i to its remainder
  return arr

class checkKey (threading.Thread):
   def __init__(self, threadID, low, high):
      threading.Thread.__init__(self)
      self.threadID = threadID
      self.low = low
      self.high = high
   def run(self):
     for i in range(self.low, self.high+1):
        if (self.high-self.low+1 - i) % 100 == 0:
          print("Thread {}: {}%".format(self.threadID, i/(self.high-self.low+1)))
        prv = index2permutation(i,52,randprv)
        try:
          testpub = prvtopub(prv)
          if testpub == pub:
             printf("Match: " + prv)
             sys.exit(0)
        except Exception as s:
          pass

fact52 = math.factorial(52)
def main():
  threadcount=0
  k=0
  while _1M*k    while (threading.activeCount() <= nthreads):
      checkKey(threading.activeCount(), _1M*k, _1M*(k+1)).start()
      threadcount += 1
      k += 1
    time.sleep(0.01)
    #print(str(_1M*k/fact52*100) + "% ", end="", flush=True)

main()

Basically what it does is that it spawns a couple of threads and maps a specific number to a permutation, then takes hundreds of thousands of those permutations in one shot. The first batch of 48 threads don't even finish their work in a reasonable amount of time  Undecided

Some more unrelated rubbish that I copied and pasted from the internet which some people may find useful:

Code:
# Python 3 implementation of the approach 

# Given the array arr[] of N elements and a permutation array P[], the task is to permute the given array arr[] based on the permutation array P[].
# Function to permute the the given
# array based on the given conditions
def permute(A, P, n):
      
    # For each element of P
    for i in range(n):
        next = i
  
        # Check if it is already
        # considered in cycle
        while (P[next] >= 0):
              
            # Swap the current element according
            # to the permutation in P
            t = A[i]
            A[i] = A[P[next]]
            A[P[next]] = t
              
            temp = P[next]
  
            # Subtract n from an entry in P
            # to make it negative which indicates
            # the corresponding move
            # has been performed
            P[next] -= n
            next = temp

def int_from_code(code):
    """
    :type code: list
    :rtype: int
    """
    num = 0
    for i, v in enumerate(reversed(code), 1):
        num *= i
        num += v

    return num

def code_from_int(size, num):
    """
    :type size: int
    :type num: int
    :rtype: list
    """
    code = []
    for i in range(size):
        num, j = divmod(num, size - i)
        code.append(j)

    return code


def perm_from_code(base, code, pick=None):
    """
    :type base: list
    :type code: list
    :rtype: list
    """
    if pick:
        return _perm_from_code_pick(base, code)

    perm = base.copy()
    for i in range(len(base) - 1):
        j = code[i]
        perm[i], perm[i+j] = perm[i+j], perm[i]

    return perm

def perm_from_int(base, num, pick=None):
    """
    :type base: list
    :type num: int
    :rtype: list
    """
    code = code_from_int(len(base), num)
    return perm_from_code(base, code, pick=pick)

def code_from_perm(base, perm, pick=None):
    """
    :type base: list
    :type perm: list
    :rtype: list
    """
    if pick:
        _code_from_perm_pick(base, perm)

    p = base.copy()
    n = len(base)
    pos_map = {v: i for i, v in enumerate(base)}

    w = []
    for i in range(n):
        print(pos_map, perm)
        d = pos_map[perm[i]] - i
        w.append(d)

        if not d:
            continue
        t = pos_map[perm[i]]
        pos_map[p[i]], pos_map[p[t]] = pos_map[p[t]], pos_map[p[i]]
        p[i], p[t] = p[t], p[i]

    return w

def int_from_perm(base, perm, pick=None):
    """
    :type base: list
    :type perm: list
    :rtype: int
    """
    code = code_from_perm(base, perm, pick=pick)
    return int_from_code(code)



This is my private key in a randomised order relating to this address: LP5U3KWRLvPwDefz4FVMrAJVFtU4u8pj15w8VSpZF2aaPeY5QIx8

'I' on the 50th place - wrong character, not from Base58 set.

Don't tell me I just wasted my time descrambling an invalid WIF! Nooooooooooo....
legendary
Activity: 952
Merit: 1385
March 30, 2021, 06:31:37 AM
#15
This is my private key in a randomised order relating to this address: LP5U3KWRLvPwDefz4FVMrAJVFtU4u8pj15w8VSpZF2aaPeY5QIx8

'I' on the 50th place - wrong character, not from Base58 set.
hero member
Activity: 3150
Merit: 937
March 30, 2021, 06:21:27 AM
#14
Quote
P.S I know some of you have said I could be trolling and given an inaccurate key but that would ruin my experiment and my experiment is worth far more than an unfunny joke.

What is your experiment and what do you want to prove with this?
If you wanna conduct an experiment,you could use test Bitcoins,which are very cheap.
What's the point of wasting 0.002 BTC?
By the way,are you a girl or a boy?You say "I'm not that sort of girl".
If you are a girl,then welcome to Bitcoin community.We really need more girls. Grin
newbie
Activity: 6
Merit: 21
March 30, 2021, 06:11:05 AM
#13
Don't worry everyone, the address is funded and it will remain funded forever, I'm not that sort of girl.  Besides all you have to do is keep an eye on my address.  Maybe one day it will be worth those permutations.

I may have made a mistake providing my public address as well as the randomised private key, that may have made this task simple enough for a knowledgeable person to quickly grab them, if so I hope that someone does take them, if not for the BTC then for the theoretical proof.

Good luck

P.S I know some of you have said I could be trolling and given an inaccurate key but that would ruin my experiment and my experiment is worth far more than an unfunny joke.

 


member
Activity: 89
Merit: 14
March 30, 2021, 06:07:21 AM
#12
Agree with most people here who are saying the monetary value is not the "prize" , so it's really just a "for fun" puzzle if someone feels motivated for a technical challenge.

Starting points for someone trying to crack this would be to look at what can be excluded

There's only 35 unique symbols (of 58) used, meaning 23 can be excluded from combination/permutations.

It appears to be a compressed public key format because it's 52 characters in length, so your pattern with start with: L or K (not a 5 for uncompressed)

You can reduce the target length by a few characters because you don't need to "crack" the base58 checksum, this should reduce the search complexity by a few orders of magnitude... but it is still a pretty big one at that!

Good luck, if you solve it you should post back here and share any "shortcuts" you took in attempting to solve.
hero member
Activity: 3024
Merit: 680
★Bitvest.io★ Play Plinko or Invest!
March 30, 2021, 06:07:02 AM
#11
Hilarious! has anyone checked to see if the wallet actually contains anything or the op is just here to troll
He's not trolling about the address, it has 0.002BTC which you can check through the explorer.

--> https://blockchair.com/bitcoin/address/bc1qq2rnv02hjzv5h0lwa03um43afwcfcpf0qg56ca

But for the private key, it's mixed in order based from op.
legendary
Activity: 2478
Merit: 1492
March 30, 2021, 06:01:59 AM
#10
You say your private key is randomized. How can we be sure that this is not just random letters and numbers? How can we be sure that this wallet is yours and not someone else? Can you sign a message from it? Otherwise you are just a freshly registered account that pushes bitcointalk users to commit a crime by hacking someone else account.
legendary
Activity: 3584
Merit: 5243
https://merel.mobi => buy facemasks with BTC/LTC
March 30, 2021, 05:30:32 AM
#9
You're holding just $116 worth of BTC and changing people to steal it. Please learn to protect your small funds now so that you can do the same for bigger ones.

My best guess is that that's what the OP is doing: seeing if somebody steals those $116. If somebody does, he has learned something about private key(s), and he might be less likely to get his real wallet robbed because of his newfound knowledge.


Hilarious! has anyone checked to see if the wallet actually contains anything or the op is just here to troll, to be honest, i have never seen where anyone wants to have their asset stolen voluntarily, it has always been the opposite, i can already tell there is nothing that will worth the effort.



The address quoted by the OP is indeed funded... However, like said by somebody else: it's not certain that the private key is correct... Might aswell be a wild goose chase.
hero member
Activity: 2562
Merit: 577
March 30, 2021, 05:25:07 AM
#8
Hilarious! has anyone checked to see if the wallet actually contains anything or the op is just here to troll, to be honest, i have never seen where anyone wants to have their asset stolen voluntarily, it has always been the opposite, i can already tell there is nothing that will worth the effort.

legendary
Activity: 2912
Merit: 6403
Blackjack.fun
March 30, 2021, 05:15:53 AM
#7
If my math is right then we have 2*(51!) ~ 3*10^66 combinations
~

I believe that people are not going to waste their time for 0.002B. If someone has a system that reduces the combinations (I doubt it) then maybe they will try it but it is a lot of effort for so little reward.

The thing with all this is that you need to go through all this AND trust that:
- the letters and numbers are indeed real and the author hasn't also changed a few of them for trolling
- the author won't move the coins tomorrow or in the next hours, making all your effort a waste

Doing this for 100$  based on a post created by a newbie?
Who is going to be tying this will do it out of boredom rather than for the reward




 
sr. member
Activity: 1288
Merit: 305
yes
March 30, 2021, 05:04:59 AM
#6
You're holding just $116 worth of BTC and changing people to steal it. Please learn to protect your small funds now so that you can do the same for bigger ones.
legendary
Activity: 1372
Merit: 2017
March 30, 2021, 04:37:30 AM
#5
If my math is right then we have 2*(51!) ~ 3*10^66 combinations. That's quite a big number.
And for each of those combinations the key has to be converter to address and checked.

Maybe somebody can further (greatly) reduce the numbers, but I don't know how.

I believe that people are not going to waste their time for 0.002B. If someone has a system that reduces the combinations (I doubt it) then maybe they will try it but it is a lot of effort for so little reward.
legendary
Activity: 3668
Merit: 6382
Looking for campaign manager? Contact icopress!
March 30, 2021, 04:25:25 AM
#4
If my math is right then we have 2*(51!) ~ 3*10^66 combinations. That's quite a big number.
And for each of those combinations the key has to be converter to address and checked.

Maybe somebody can further (greatly) reduce the numbers, but I don't know how.
legendary
Activity: 1512
Merit: 4795
Leading Crypto Sports Betting & Casino Platform
March 30, 2021, 03:56:05 AM
#3
Only the comment I can leave here is that private key must not be revealed to anyone, if known to anyone, the person will be able steal the bitcoin the private key can unlock on blockchain.
legendary
Activity: 1568
Merit: 6660
bitcoincleanup.com / bitmixlist.org
March 30, 2021, 03:40:26 AM
#2
This is the first time since nullius' cult sacrifice puzzle that I see someone challenging everyone to sweep real money from an address  Shocked

Anyways shouldn't be too hard to do, you gave us the public key anyway along with the private key characters themselves so nothing that a little combination program can't do.

(Ninja edit) Also it's uncompressed since it begins with an "L". (But there is a "5" and a "K" too maybe it is compressed).

There's 0.002BTC inside according to block explorers.
newbie
Activity: 6
Merit: 21
March 30, 2021, 03:35:10 AM
#1
This is my public address: bc1qq2rnv02hjzv5h0lwa03um43afwcfcpf0qg56ca

This is my private key in a randomised order relating to this address: LP5U3KWRLvPwDefz4FVMrAJVFtU4u8pj15w8VSpZF2aaPeY5Qix8

Do your worst

Thanks

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