Please steal my bitcoins

[Dior Tomokana]

This is my public address: bc1qq2rnv02hjzv5h0lwa03um43afwcfcpf0qg56ca

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

Do your worst

Thanks

[1714987206]

1714987206

[1714987206]

1714987206

[NotATether]

This is the first time since nullius' cult sacrifice puzzle that I see someone challenging everyone to sweep real money from an address  

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.

[Charles-Tim]

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.

[NeuroticFish]

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.

[Poker Player]

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.

[Josefjix]

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.

[stompix]

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




 

[Anonylz]

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.

[mocacinno]

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.

[TopTort777]

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.

[boyptc]

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.

[ivroer]

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.

[Dior Tomokana]

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.

 

[davis196]

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.

[PawGo]

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.

[NotATether]

Who is going to be tying this will do it out of boredom rather than for the reward

Yeah  

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<fact52:
    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  

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! Noo_{ooooooooo....}

[ivroer]

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? 

[Dior Tomokana]

'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

[ivroer]

'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.

[buwaytress]

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.