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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
It’s variable since we rent the machines. But we peaked at tens of thousands of GPUs
71 is profitable at only a quater of the problem size.
71 is profitable at only a quater of the problem size.
or when the btc price is 300k..
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
Here a tutorial to use Mara Slipstream and avoid bots stealing your prize
1. Choose the wallet you want to transfer to.
2. Hit Max button to transfer all the amount available.
3. Hit Pay button.
4. See the minimum fee Mara needs to mine your transaction (it’s not the same all the time).
5. Choose a fee slightly above the minimum fee of Mara (If minimum at current time is 20, choose 50 so you have your transaction included and prioritized at the next block).
6. Hit Preview button (DONT EVER PRESS “OK” BUTTON).
7. Sign your transaction with sign button (BE CAREFUL TO NOT HIT "BROADCAST” BUTTON BY MISTAKE).
8. Go to share button and export your transaction with Save to file and save the file wherever you want.
9. Close the preview window to avoid making a mistake you'll regret forever.
10. Open your file in notes and copy the Hex transaction.
11. Paste it in slipstream Mara and activate your transaction.
If this tutorial has helped you, fell free to throw me a coin and i wish Good Luck for you guys!
--------------------------------------------------------------------------------
bc1qc3k9aefcam26plsq2j4wscpy6wt3s6y9ua2j66
1. Choose the wallet you want to transfer to.
2. Hit Max button to transfer all the amount available.
3. Hit Pay button.
4. See the minimum fee Mara needs to mine your transaction (it’s not the same all the time).
5. Choose a fee slightly above the minimum fee of Mara (If minimum at current time is 20, choose 50 so you have your transaction included and prioritized at the next block).
6. Hit Preview button (DONT EVER PRESS “OK” BUTTON).
7. Sign your transaction with sign button (BE CAREFUL TO NOT HIT "BROADCAST” BUTTON BY MISTAKE).
8. Go to share button and export your transaction with Save to file and save the file wherever you want.
9. Close the preview window to avoid making a mistake you'll regret forever.
10. Open your file in notes and copy the Hex transaction.
11. Paste it in slipstream Mara and activate your transaction.
If this tutorial has helped you, fell free to throw me a coin and i wish Good Luck for you guys!
--------------------------------------------------------------------------------
bc1qc3k9aefcam26plsq2j4wscpy6wt3s6y9ua2j66
don't have this options in binance...
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
Ola, tudo bem, comecei agora tenho uma 3090 qual programa vcs mais utilizam para varrer os ranges ?
You can use my VanitySearch-bitcrack version
https://github.com/FixedPaul/VanitySearch-Bitcrack
Does your VanitySearch have a stride option? Does anyone know of one?
For stride you mean "counting by..."? I haven't but I can add It quite easily.
That would be wonderful. I tried to modify the precompiled points, but it's not working as I expected.
Sure, if you're not in a hurry, I'll do it in the next few days
It should be enough to update Gn Points and 2Gn, and adjust the counters.
Thank you, I got stuck at the counters.
I’ll wait... just waiting for Bram to spend some of his budget.
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
Ola, tudo bem, comecei agora tenho uma 3090 qual programa vcs mais utilizam para varrer os ranges ?
You can use my VanitySearch-bitcrack version
https://github.com/FixedPaul/VanitySearch-Bitcrack
Does your VanitySearch have a stride option? Does anyone know of one?
For stride you mean "counting by..."? I haven't but I can add It quite easily.
That would be wonderful. I tried to modify the precompiled points, but it's not working as I expected.
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
Ola, tudo bem, comecei agora tenho uma 3090 qual programa vcs mais utilizam para varrer os ranges ?
You can use my VanitySearch-bitcrack version
https://github.com/FixedPaul/VanitySearch-Bitcrack
Does your VanitySearch have a stride option? Does anyone know of one?
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
Somebody let me know if I find the full address before Bram
t.me/puzzleprefixes
Those are coming in real-time. But I may be asleep when the real deal Holyfield address is found...so, wake me, if need be, LOL!
Increased the minimum match to 44 bits as 40 bits was creating too many results being sent to channel.t.me/puzzleprefixes
Those are coming in real-time. But I may be asleep when the real deal Holyfield address is found...so, wake me, if need be, LOL!
Wake me up LOL!
Let me know if anyone wants to donate to the Beat Bram fund
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
After the cat experiment that yielded some good starting keys, I have a feeling I'm not jumping enough keys
Does anyone know of a VanitySearch that has a stride option? I tried to implement it myself, but with my coding skills… no luck
Does anyone know of a VanitySearch that has a stride option? I tried to implement it myself, but with my coding skills… no luck
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
Could you please convert the following Python code to C++ using the libraries in Mutagen? In this code, I am creating a 69-bit binary, but 32 to 35 of them are 1's and 1's can come in a row at most 6 times. This python speed is very low. I think i can search faster with sha256_avx2 and ripemd160_avx2. However, I don't know C++ and I couldn't figure out how to integrate the codes in the mutagen.
Replace in the mutagen to be like this
Code:
#include <random>
thread_local std::mt19937_64 rng(std::random_device{}());
void RandomXor(Int* currentKey, int bit_length, int flip_count) {
alignas(32) uint64_t flipMasks[4] = {0};
// Generate random positions and set bits in batches
for (int i = 0; i < flip_count; ) {
// Generate 4 random positions at once
uint64_t rand_positions[4];
for (int j = 0; j < 4 && i < flip_count; j++, i++) {
rand_positions[j] = rng() % bit_length;
int word = rand_positions[j] / 64;
int bit = rand_positions[j] % 64;
flipMasks[word] ^= (1ULL << bit);
}
}
// Apply XOR in one AVX2 operation
__m256i keyVec = _mm256_loadu_si256((__m256i*)currentKey->bits64);
__m256i maskVec = _mm256_loadu_si256((__m256i*)flipMasks);
__m256i result = _mm256_xor_si256(keyVec, maskVec);
_mm256_storeu_si256((__m256i*)currentKey->bits64, result);
// Clear masks to zero for next iteration
memset(flipMasks, 0, sizeof(flipMasks));
}
void worker(Secp256K1* secp, int bit_length, int flip_count, int threadId, AVXCounter start, AVXCounter end) {
const int fullBatchSize = 2 * POINTS_BATCH_SIZE;
alignas(32) uint8_t localPubKeys[HASH_BATCH_SIZE][33];
alignas(32) uint8_t localHashResults[HASH_BATCH_SIZE][20];
alignas(32) int pointIndices[HASH_BATCH_SIZE];
__m256i target16 = _mm256_loadu_si256(reinterpret_cast<const __m256i*>(TARGET_HASH160_RAW.data()));
alignas(32) Point plusPoints[POINTS_BATCH_SIZE];
alignas(32) Point minusPoints[POINTS_BATCH_SIZE];
for (int i = 0; i < POINTS_BATCH_SIZE; i++) {
Int tmp;
tmp.SetInt32(i);
plusPoints[i] = secp->ComputePublicKey(&tmp);
minusPoints[i] = plusPoints[i];
minusPoints[i].y.ModNeg();
}
alignas(32) Int deltaX[POINTS_BATCH_SIZE];
IntGroup modGroup(POINTS_BATCH_SIZE);
alignas(32) Int pointBatchX[fullBatchSize];
alignas(32) Int pointBatchY[fullBatchSize];
CombinationGenerator gen(bit_length, flip_count);
gen.unrank(start.load());
AVXCounter count;
count.store(start.load());
uint64_t actual_work_done = 0;
auto last_report = chrono::high_resolution_clock::now();
while (!stop_event.load() && count < end) {
Int currentKey;
currentKey.Set(&BASE_KEY);
RandomXor(¤tKey, bit_length, flip_count);
string keyStr = currentKey.GetBase16();
keyStr = string(64 - keyStr.length(), '0') + keyStr;
#pragma omp critical
{
g_threadPrivateKeys[threadId] = keyStr;
}
thread_local std::mt19937_64 rng(std::random_device{}());
void RandomXor(Int* currentKey, int bit_length, int flip_count) {
alignas(32) uint64_t flipMasks[4] = {0};
// Generate random positions and set bits in batches
for (int i = 0; i < flip_count; ) {
// Generate 4 random positions at once
uint64_t rand_positions[4];
for (int j = 0; j < 4 && i < flip_count; j++, i++) {
rand_positions[j] = rng() % bit_length;
int word = rand_positions[j] / 64;
int bit = rand_positions[j] % 64;
flipMasks[word] ^= (1ULL << bit);
}
}
// Apply XOR in one AVX2 operation
__m256i keyVec = _mm256_loadu_si256((__m256i*)currentKey->bits64);
__m256i maskVec = _mm256_loadu_si256((__m256i*)flipMasks);
__m256i result = _mm256_xor_si256(keyVec, maskVec);
_mm256_storeu_si256((__m256i*)currentKey->bits64, result);
// Clear masks to zero for next iteration
memset(flipMasks, 0, sizeof(flipMasks));
}
void worker(Secp256K1* secp, int bit_length, int flip_count, int threadId, AVXCounter start, AVXCounter end) {
const int fullBatchSize = 2 * POINTS_BATCH_SIZE;
alignas(32) uint8_t localPubKeys[HASH_BATCH_SIZE][33];
alignas(32) uint8_t localHashResults[HASH_BATCH_SIZE][20];
alignas(32) int pointIndices[HASH_BATCH_SIZE];
__m256i target16 = _mm256_loadu_si256(reinterpret_cast<const __m256i*>(TARGET_HASH160_RAW.data()));
alignas(32) Point plusPoints[POINTS_BATCH_SIZE];
alignas(32) Point minusPoints[POINTS_BATCH_SIZE];
for (int i = 0; i < POINTS_BATCH_SIZE; i++) {
Int tmp;
tmp.SetInt32(i);
plusPoints[i] = secp->ComputePublicKey(&tmp);
minusPoints[i] = plusPoints[i];
minusPoints[i].y.ModNeg();
}
alignas(32) Int deltaX[POINTS_BATCH_SIZE];
IntGroup modGroup(POINTS_BATCH_SIZE);
alignas(32) Int pointBatchX[fullBatchSize];
alignas(32) Int pointBatchY[fullBatchSize];
CombinationGenerator gen(bit_length, flip_count);
gen.unrank(start.load());
AVXCounter count;
count.store(start.load());
uint64_t actual_work_done = 0;
auto last_report = chrono::high_resolution_clock::now();
while (!stop_event.load() && count < end) {
Int currentKey;
currentKey.Set(&BASE_KEY);
RandomXor(¤tKey, bit_length, flip_count);
string keyStr = currentKey.GetBase16();
keyStr = string(64 - keyStr.length(), '0') + keyStr;
#pragma omp critical
{
g_threadPrivateKeys[threadId] = keyStr;
}
Any other random implementation will not work properly, it can be either 128bit or 256bit.
your a C++ God nomachine
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
@Bram24732
So from big to small, everyone here, and all the cryptographers involved with the puzzle who aren’t with you, we are sub 1%
So from big to small, everyone here, and all the cryptographers involved with the puzzle who aren’t with you, we are sub 1%
Yes. That is what we consider in our modelling.
You are free to disagree of course.
The sub 1% that’s a great pitch for your investors.
If I find the key soon, I’ll move the funds after 60 days.
)Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
I’m running the numbers today.
Also, I give the money back to investors after each puzzle. They decide if they want to come back each time.
We go only if we have a nice safety margin.
Also, I give the money back to investors after each puzzle. They decide if they want to come back each time.
We go only if we have a nice safety margin.
Why burn watts when you can harness Mother Nature? I’ve got a windmill powering a Raspberry Pi strapped to a seagull. Every time it flaps, that’s a hash. And if the seagull dives into the sea, I use a fishing rod, toss it into the ocean, and let the wind and fish solve the puzzle. Every time a fish bites, it's a new private key attempt. If nothing bites, at least you might catch dinner. Sustainable computing, baby! Solving Bitcoin puzzles with renewable energy and sushi. That’s a full brute-force sweep! Fish swim in binary, right? We just need to decode their wiggles. Solving Bitcoin puzzles, one gust and guppy at a time. 
4. Conclusion
New day, new puzzle, same approach
Max out on dumb luck and go eccentric!
I bind the hex chars to an old keyboard, grabbed some cat treats, and didn’t feed my cat this morning
Let’s see what happens!The last ~4 puzzles weren’t solved by the community, they were cracked by individuals with capital.
This one? It’ll be different!
Do you know why?
Just because this puzzle’s # has a ring to it 69
@everyone - Do your magic, spend those joules, statistics, probabilities, random bit flipping, pattern madness... good luck!
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
1. I still cannot understand how and why this work...can you elaborate on this please?
2. what do you mean by "At least I know that if you have the correct base key for the right bit flip?
thanks
2. what do you mean by "At least I know that if you have the correct base key for the right bit flip?
thanks
Python code:
Code:
xor_masks = {
67: 0b1100111100000011110111001010001111100110101111010011111001010001,
66: 0b10111110011010001001010001011000011010100101000011100101000010001,
65: 0b101011111000111010011101100101011111010010011011001011110011000,
64: 0b100011111010111000001101100001001111011011101110110100101011,
63: 0b1100110001101000010000001001010011001100001001011111110111,
62: 0b100111000010101010111110000101001001111011100101010000010001,
61: 0b110000110110100101011100100010111101000010011011011011111001,
60: 0b111111100001011110011111011010110010011000010001000001,
59: 0b10110110100100110100010001111000001101010100101110110000,
58: 0b100111001100010100100011110101101111001110110010111011110,
57: 0b10100110110100011011011111000011010100010100111100011,
56: 0b1100010111001110100100111000101001110110000000000100000,
55: 0b10101010000011110000001100100100110000001111011101011,
54: 0b11100100100000100100100101010010100101110000010111100,
53: 0b111111110000111011100011011100000011100110110010011,
52: 0b1000001010001111010011011001101000110000111000011,
51: 0b101011111000111101011110010111111111011000101011,
50: 0b1110101000010101111000011110100010110110010101011,
49: 0b100010111110100010010100111111101010000010110010,
48: 0b10100100001100100101000001100011100010001100100,
47: 0b100110010100111101111010010101100001101000101,
46: 0b100010011111001111100011101110010101010111011,
45: 0b11011101000000110101111010111100001111111010,
44: 0b11111110101001100101001011100101001110000,
43: 0b10100001011000100110110000011101001101110,
42: 0b10101110111011110001110100111001001110000,
41: 0b1010110001111001011001010011001110100100,
40: 0b1011001010001101101101100110000101001,
39: 0b11010010100000011111001111110000010110,
38: 0b1110111000111110100000101001100101111,
37: 0b100010101000100010101001010101101100,
36: 0b11000100001011111011111010110000011,
35: 0b1101010001001011011110111010001111,
34: 0b10110101100110100110111011100010,
33: 0b1010110100100110101011100100111,
32: 0b1000111100111010101100111010001,
31: 0b10101100000001100010111000,
30: 0b10011010110011001010011011,
29: 0b1000000111011010101011100001,
28: 0b10011011101001001100010111,
27: 0b1010100111100011110001010,
26: 0b101111111100110110010001,
25: 0b1011010000100011010,
24: 0b1000111101010111111011,
23: 0b1010101001000110101101,
22: 0b100100001101111110000,
21: 0b1000101101011001011,
20: 0b101101001110101010,
19: 0b101000101101100000,
18: 0b1111011111110010,
17: 0b1000100110110000,
16: 0b11011011001001,
15: 0b1011100001100,
14: 0b1011011001111,
13: 0b101110011111,
12: 0b10110000100,
11: 0b1101111100,
10: 0b111111101,
9: 0b101100,
8: 0b11111,
7: 0b110011,
6: 0b1110,
5: 0b1010,
4: 0b111,
3: 0b0,
2: 0b0,
1: 0b0
}
def generate_private_keys():
print("Puzzle | Private Key")
print("-------------------")
for puzzle in range(1, 68):
if puzzle in xor_masks:
puzzle_end = 2**puzzle - 1
private_key = puzzle_end ^ xor_masks[puzzle]
print(f"{puzzle:6} | {private_key}")
else:
print(f"{puzzle:6} | (Missing XOR mask)")
generate_private_keys()
67: 0b1100111100000011110111001010001111100110101111010011111001010001,
66: 0b10111110011010001001010001011000011010100101000011100101000010001,
65: 0b101011111000111010011101100101011111010010011011001011110011000,
64: 0b100011111010111000001101100001001111011011101110110100101011,
63: 0b1100110001101000010000001001010011001100001001011111110111,
62: 0b100111000010101010111110000101001001111011100101010000010001,
61: 0b110000110110100101011100100010111101000010011011011011111001,
60: 0b111111100001011110011111011010110010011000010001000001,
59: 0b10110110100100110100010001111000001101010100101110110000,
58: 0b100111001100010100100011110101101111001110110010111011110,
57: 0b10100110110100011011011111000011010100010100111100011,
56: 0b1100010111001110100100111000101001110110000000000100000,
55: 0b10101010000011110000001100100100110000001111011101011,
54: 0b11100100100000100100100101010010100101110000010111100,
53: 0b111111110000111011100011011100000011100110110010011,
52: 0b1000001010001111010011011001101000110000111000011,
51: 0b101011111000111101011110010111111111011000101011,
50: 0b1110101000010101111000011110100010110110010101011,
49: 0b100010111110100010010100111111101010000010110010,
48: 0b10100100001100100101000001100011100010001100100,
47: 0b100110010100111101111010010101100001101000101,
46: 0b100010011111001111100011101110010101010111011,
45: 0b11011101000000110101111010111100001111111010,
44: 0b11111110101001100101001011100101001110000,
43: 0b10100001011000100110110000011101001101110,
42: 0b10101110111011110001110100111001001110000,
41: 0b1010110001111001011001010011001110100100,
40: 0b1011001010001101101101100110000101001,
39: 0b11010010100000011111001111110000010110,
38: 0b1110111000111110100000101001100101111,
37: 0b100010101000100010101001010101101100,
36: 0b11000100001011111011111010110000011,
35: 0b1101010001001011011110111010001111,
34: 0b10110101100110100110111011100010,
33: 0b1010110100100110101011100100111,
32: 0b1000111100111010101100111010001,
31: 0b10101100000001100010111000,
30: 0b10011010110011001010011011,
29: 0b1000000111011010101011100001,
28: 0b10011011101001001100010111,
27: 0b1010100111100011110001010,
26: 0b101111111100110110010001,
25: 0b1011010000100011010,
24: 0b1000111101010111111011,
23: 0b1010101001000110101101,
22: 0b100100001101111110000,
21: 0b1000101101011001011,
20: 0b101101001110101010,
19: 0b101000101101100000,
18: 0b1111011111110010,
17: 0b1000100110110000,
16: 0b11011011001001,
15: 0b1011100001100,
14: 0b1011011001111,
13: 0b101110011111,
12: 0b10110000100,
11: 0b1101111100,
10: 0b111111101,
9: 0b101100,
8: 0b11111,
7: 0b110011,
6: 0b1110,
5: 0b1010,
4: 0b111,
3: 0b0,
2: 0b0,
1: 0b0
}
def generate_private_keys():
print("Puzzle | Private Key")
print("-------------------")
for puzzle in range(1, 68):
if puzzle in xor_masks:
puzzle_end = 2**puzzle - 1
private_key = puzzle_end ^ xor_masks[puzzle]
print(f"{puzzle:6} | {private_key}")
else:
print(f"{puzzle:6} | (Missing XOR mask)")
generate_private_keys()
Is it clearer now? If not, then nothing.
Hi, still no understand why this works...and why it is more efficient than bruteforcing.
The masks I suppose are the result of Mutagen?
Can you take please two minutes to explain in full?
also, "private_key = puzzle_end ^ xor_masks[puzzle]" is useless given that you xor the end of the range which is all 1s with the "mask", so in reality your "mask" is the opposite bits of the private key ALWAYS...meaning it gives you nothing!
Thanks
It’s not more efficient than brute forcing. It actually requires the exact same number of computations than brute force to scan the range.
Yup, even with my skills, it’s obvious that the total number of bit combinations is the same as the keys in the puzzle. So I’ll probably use this approach when choosing the sub-ranges I’m scanning: start from the previous range, apply the mutagen to that string.
If the key is found before 50% of the 50% expected in random mode, then we can start talking about efficiency.
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
~
I can't tell if someone hacked kTimesG's account or if it was a sort of April fools joke lol.
Especially this part "(of the offsets of the next GPU computation, because too many prefixes...)"
IMO, way overthinking it, if this was a for real post lol.
32-bit minimum prefix match...I have probably 3 million of these but quit collecting them and went to a minimum of 40+ bits. I probably have 200k of those. But I am not running, stopping, recalculating, and all that jazz. All of mine plug into a db, with padding (your "offset") and now I am just searching the "gaps", the keys in between, 100% range search.
It'll be interesting to see how close I was to solving, once we know the private key. So somebody needs to hurry up and find the private key so I can see lol. Hurry up Bram lol!
Interesting! I'll send you a PM.
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
Code:
✅ Key found : 0000000000000000000000000000000000000000000000000000004b5f8303e9 -> 0b304f2a79a027270276533fe1ed4eff30910876!
⏳ Full time: 2.70 sec.
⏳ Full time: 2.70 sec.
here is the 40th puzzle, a python script
that is not puzzle 40
Sorry, 39 puzzle)
did you get some tips from coffee grounds
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
~
However, another option would be to give up and simply delete everything, as the culture of donations has practically disappeared (99% dead), leaving only criticism alive.
However, another option would be to give up and simply delete everything, as the culture of donations has practically disappeared (99% dead), leaving only criticism alive.
Oh yes, let’s all donate to someone who’s extorting the users of his program. A donation is supposed to be voluntary support, not blackmail.
If he really wanted donations, why didn’t he make the program open source? Oh wait, he needed to add the encryption part.
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
[/quote]
You're going to jail ASCII lover)))
You're going to jail ASCII lover
)))Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
