nordwork313
Newbie
Activity: 3
Merit: 0
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May 12, 2026, 08:44:09 PM |
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I want to clarify the idea more mathematically.
I am not saying that grouping 256 bytes into groups of 50 magically breaks secp256k1 or HASH160.
I agree that if the private key is uniformly random in the full interval, then no grouping method gives an advantage.
For a uniform key space of size N, after testing X candidates, the probability is simply:
[
P_{\text{full}}=\frac{X}{N}
]
That is true.
But my point is about early-hit probability under a structured prior.
If the key was not generated uniformly, but instead has some dictionary-like or byte-structured origin, then the search problem is different.
In that case, we should not search the full interval in blind order first. We should search smaller structured subsets first.
Let:
[
D = 256
]
be the byte dictionary.
Let:
[
g = 50
]
be the size of one active group.
Let:
[
r
]
be the number of required bytes that must be present together in one group before the internal search becomes useful.
The probability that one random group of 50 bytes contains all r required bytes from a dictionary of 256 bytes is:
[
p=\frac{\binom{50}{r}}{\binom{256}{r}}
]
For example, if the key depends on 9 required bytes, then:
[
p=\frac{\binom{50}{9}}{\binom{256}{9}}
]
Numerically:
[
p \approx 2.22 \times 10^{-7}
]
So one group is not enough.
But the idea is not to test only one group.
The idea is to keep generating different regroupings and testing each group independently.
After n independent regrouping attempts, the probability that at least one group contains all required bytes is:
[
P_n = 1-(1-p)^n
]
This is the important part.
The chance grows with every regrouping.
For r=9:
[
p \approx 2.22 \times 10^{-7}
]
If we test 256 independent groups:
[
P_{256}=1-(1-p)^{256}
]
[
P_{256}\approx 0.0000568
]
That is about:
[
0.00568%
]
or roughly:
[
1 \text{ chance in } 17,600
]
That is not a guarantee.
But it is a measurable early-hit probability.
If we test one million different groups:
[
P_{1,000,000}=1-(1-p)^{1,000,000}
]
[
P_{1,000,000}\approx 0.199
]
That is about:
[
19.9%
]
If we test five million different groups:
[
P_{5,000,000}=1-(1-p)^{5,000,000}
]
[
P_{5,000,000}\approx 0.670
]
That is about:
[
67%
]
If we test ten million different groups:
[
P_{10,000,000}=1-(1-p)^{10,000,000}
]
[
P_{10,000,000}\approx 0.891
]
That is about:
[
89.1%
]
So the method is not based on one fixed group.
It is based on repeated regrouping of the dictionary.
The goal is to eventually create a group of 50 bytes that contains all required bytes.
Once this happens, the internal active search space becomes much smaller than the full key interval.
This is where the early-hit advantage comes from.
The full brute-force range can still be searched separately.
But searching the full range blindly gives:
[
P_{\text{full}}=\frac{X}{N}
]
where N is extremely large.
In the structured method, if a successful group G contains the required bytes, then the search is temporarily concentrated inside a much smaller space:
[
S_G \ll N
]
The probability model becomes:
[
P_{\text{structured}} = P(k \in S_G)\cdot \frac{X}{|S_G|}
]
The structured search is better than full blind search when:
[
P(k \in S_G)\cdot \frac{X}{|S_G|} > \frac{X}{N}
]
After cancelling X:
[
P(k \in S_G) > \frac{|S_G|}{N}
]
This is the mathematical condition.
If the subset has more probability mass than its size would have under a uniform distribution, then searching it first is rational.
That is exactly the idea.
The method does not claim that every group is better.
It claims that if the real key has dictionary structure, then some groups have much higher value than random full-range scanning.
The full range treats every key as equally likely:
[
P(k)=\frac{1}{N}
]
The dictionary method assumes that the key may come from a smaller generation process:
[
k = f(b_1,b_2,\ldots,b_r)
]
where:
[
b_i \in D
]
If that assumption is correct, then the effective entropy is not the entropy of the full interval.
It is closer to:
[
H_{\text{eff}} = \log_2 |S|
]
instead of:
[
H_{\text{full}} = \log_2 N
]
That is why this method can have a better chance of an early hit.
Not because it defeats cryptography.
Not because HASH160 gives partial feedback.
Not because a wrong key tells us which byte is correct.
It gives no such feedback.
The advantage exists only in the search ordering.
The full range is blind.
The dictionary-group method gives priority to subsets that may carry higher prior probability if the key was generated with structure.
For a simple analogy, suppose we search for a number from 10 to 99.
If the number is uniform random, there is no better method than testing candidates.
But if we know the number was created from a pattern, for example digits from a smaller set, then testing that smaller set first gives a higher chance of an early hit.
Full range:
[
90 \text{ candidates}
]
Structured subset:
[
9 \text{ candidates}
]
If the real number is likely to be inside those 9 candidates, searching those first is better.
The same logic applies here.
The question is not:
“Does grouping reduce entropy for a uniform key?”
No, it does not.
The real question is:
“Does the puzzle key have any non-uniform generation structure?”
If yes, then grouping and regrouping the byte dictionary can give a better early-hit probability than blind full-range scanning.
If no, then the method becomes ordinary brute force.
So my claim is limited and mathematical:
1. 256 bytes can be regrouped into many 50-byte active subsets.
2. A single group succeeds if it contains all required bytes.
3. The probability for one group is:
[
p=\frac{\binom{50}{r}}{\binom{256}{r}}
]
4. The probability after n regroupings is:
[
P_n=1-(1-p)^n
]
5. This probability increases with n.
6. Once a good group exists, GPU search inside that group is much smaller than blind full-range search.
7. Therefore, if the key has dictionary structure, this method has a higher chance of an early hit.
8. If the key is perfectly uniform, then there is no advantage.
That is the exact mathematical distinction.
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Menowa*
Jr. Member
Activity: 60
Merit: 1
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May 12, 2026, 09:57:38 PM |
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I wrote in the purpose of use that i want to make my Bitcoin transactions private and safe and secure....(I know its sounds stupid but i didn't have anything else ti say😅)..... But they didn't reply to me at all i even tried with 2 different email addresses but nothing happens, can you tell me what did you wrote in the purpose of use box ??
That clearly looks stupid like you have said. I just wrote I wanted to move funds from low entropy wallets to a standard wallet safely. I didn’t mention anything about these puzzles. You just have to have a valid reason rather than “I just want my transactions privately.” |
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mjojo
Newbie
Activity: 92
Merit: 0
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May 12, 2026, 10:37:14 PM |
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https://iili.io/Bb6w41t.md.pngI have been working for several years on deriving a private key from a public one. The only viable solution lies in mathematical formulas. If you can establish a connection between the +7 shift in the denominator of Formula 3 and the exponent n in Formula 4, and derive the underlying dependency, you will find the solution and be able to extract the private key from the public one. Formulas 1 and 2 exhibit such a dependency, but the +7 term disrupts it. Formulas 3 and 4 lack a connection due to the exponent n. I am open to collaboration with anyone who has strong mathematical expertise. Would you elaborate your formula with pubkey of puzzle 66 or another pubkey with known pvkey Private Key (Dec): 46346217550346335726 Private Key (Hex): 000000000000000000000000000000000000000000000002832ed74f2b5e35ee X: 35681022179418186873973468373280332442853574053612577039287936543695937260509 Y: 55769436514148338618765508764324988000805241152238207056637568135859289137720 Compressed Public Key: 024ee2be2d4e9f92d2f5a4a03058617dc45befe22938feed5b7a6b7282dd74cbdd |
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SecretAdmirere
Newbie
Activity: 19
Merit: 1
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May 12, 2026, 11:07:31 PM |
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Would you elaborate your formula
Formula is a little bit complicated, but it goes smth like this: Magic Mushrooms X * Magic Mushrooms Y + LSD / E = MC² * Methamphetamine = reality where ECDLP doesn't exist |
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0xastraeus
Newbie
Activity: 43
Merit: 0
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May 12, 2026, 11:26:08 PM |
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Here's some math that's relatable to some of yall... import math
def canonicalize_window() -> str:
codes = (
int(math.sqrt(13225)),
int(math.pow(2, 2) * 29),
int(math.sqrt(13689)),
int(7 * math.pow(2, 4)),
int(math.sqrt(11025)),
int(math.pow(10, 2)),
)
return "".join(map(chr, codes))
if __name__ == "__main__":
print(canonicalize_window())
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And24r
Newbie
Activity: 14
Merit: 0
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May 13, 2026, 03:53:14 AM
Last edit: May 13, 2026, 05:32:46 AM by And24r |
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https://iili.io/Bb6w41t.md.pngI have been working for several years on deriving a private key from a public one. The only viable solution lies in mathematical formulas. If you can establish a connection between the +7 shift in the denominator of Formula 3 and the exponent n in Formula 4, and derive the underlying dependency, you will find the solution and be able to extract the private key from the public one. Formulas 1 and 2 exhibit such a dependency, but the +7 term disrupts it. Formulas 3 and 4 lack a connection due to the exponent n. I am open to collaboration with anyone who has strong mathematical expertise. Would you elaborate your formula with pubkey of puzzle 66 or another pubkey with known pvkey Private Key (Dec): 46346217550346335726 Private Key (Hex): 000000000000000000000000000000000000000000000002832ed74f2b5e35ee X: 35681022179418186873973468373280332442853574053612577039287936543695937260509 Y: 55769436514148338618765508764324988000805241152238207056637568135859289137720 Compressed Public Key: 024ee2be2d4e9f92d2f5a4a03058617dc45befe22938feed5b7a6b7282dd74cbdd It’s a long explanation. I’ve provided examples of formulas and set the task goal. If we derive the formula, we’ll work only with the x coordinate — the y coordinate isn’t needed. Following your example: x is the base point GP; x1 is the public key; 2^n is the private key with the number 2 raised to an unknown power. In the first and second examples, you can substitute any value for x — formula 2 will always return x. However, formulas 3 and 4 won’t work in this case. In general, the point is that if you know the degree n, you can easily adjust the private key to match the x-coordinate of the base point GP through sequential increase, using formula 4. |
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OzBtcOz
Newbie
Activity: 17
Merit: 0
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May 13, 2026, 08:21:17 AM |
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I use cuBitCrack, but getting 4GKeys/s from RTX5090 Could you please anyone help me about optimize for speed? Below code provides auto random jumping, when a space scanning finished cat > auto_random.sh << 'EOF'
#!/bin/bash
TARGET="12VVRNPi4SJqUTsp6FmqDqY5sGosDtysn4"
DEVICE=0
BLOCKS=1024
THREADS=256
POINTS=512
while true; do
pkill cuBitCrack 2>/dev/null
sleep 2
RESULT=$(python3 << 'PYEOF'
import random
start = 0x1000000000000000000
end = 0x1ffffffffffffffffff
chunk_size = 2**46
max_start = end - chunk_size
random_start = random.randint(start, max_start)
random_end = random_start + chunk_size
print(f"{random_start:020x} {random_end:020x}")
PYEOF
)
RANDOM_START=$(echo $RESULT | awk '{print $1}')
RANDOM_END=$(echo $RESULT | awk '{print $2}')
echo "$(date): Yeni dilim $RANDOM_START - $RANDOM_END" >> auto_log.txt
./bin/cuBitCrack -d $DEVICE \
-b $BLOCKS -t $THREADS -p $POINTS \
-c -o Found.txt \
--keyspace ${RANDOM_START}:${RANDOM_END} \
${TARGET} >> output.log 2>&1
if [ -s Found.txt ]; then
echo "$(date): KEY BULUNDU!" >> auto_log.txt
cat Found.txt >> auto_log.txt
break
fi
done
EOF
chmod +x auto_random.sh |
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username666187
Newbie
Activity: 17
Merit: 0
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May 13, 2026, 09:16:13 AM |
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Anyone here know how to code and want to help me solve the puzzles? I created a system from the ground up complete with axioms definitions and procedural operations i just designed my system so that it happens to use the exact same values as secp256k1 im not saying i solved the discrete log thats impossible instead i created a system where the discrete log doesnt even apply at all
If it's still relevant, I can help write the code for you. Bro if you can code we can cook |
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brainless
Member
Activity: 487
Merit: 35
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May 13, 2026, 10:09:08 AM |
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I use cuBitCrack, but getting 4GKeys/s from RTX5090 Could you please anyone help me about optimize for speed? Below code provides auto random jumping, when a space scanning finished cat > auto_random.sh << 'EOF'
#!/bin/bash
TARGET="12VVRNPi4SJqUTsp6FmqDqY5sGosDtysn4"
DEVICE=0
BLOCKS=1024
THREADS=256
POINTS=512
while true; do
pkill cuBitCrack 2>/dev/null
sleep 2
RESULT=$(python3 << 'PYEOF'
import random
start = 0x1000000000000000000
end = 0x1ffffffffffffffffff
chunk_size = 2**46
max_start = end - chunk_size
random_start = random.randint(start, max_start)
random_end = random_start + chunk_size
print(f"{random_start:020x} {random_end:020x}")
PYEOF
)
RANDOM_START=$(echo $RESULT | awk '{print $1}')
RANDOM_END=$(echo $RESULT | awk '{print $2}')
echo "$(date): Yeni dilim $RANDOM_START - $RANDOM_END" >> auto_log.txt
./bin/cuBitCrack -d $DEVICE \
-b $BLOCKS -t $THREADS -p $POINTS \
-c -o Found.txt \
--keyspace ${RANDOM_START}:${RANDOM_END} \
${TARGET} >> output.log 2>&1
if [ -s Found.txt ]; then
echo "$(date): KEY BULUNDU!" >> auto_log.txt
cat Found.txt >> auto_log.txt
break
fi
done
EOF
chmod +x auto_random.sh When running bitcrack, share that screen to guide u best optimal setting |
13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF
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OzBtcOz
Newbie
Activity: 17
Merit: 0
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May 13, 2026, 12:46:51 PM |
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I use cuBitCrack, but getting 4GKeys/s from RTX5090 Could you please anyone help me about optimize for speed? Below code provides auto random jumping, when a space scanning finished cat > auto_random.sh << 'EOF'
#!/bin/bash
TARGET="12VVRNPi4SJqUTsp6FmqDqY5sGosDtysn4"
DEVICE=0
BLOCKS=1024
THREADS=256
POINTS=512
while true; do
pkill cuBitCrack 2>/dev/null
sleep 2
RESULT=$(python3 << 'PYEOF'
import random
start = 0x1000000000000000000
end = 0x1ffffffffffffffffff
chunk_size = 2**46
max_start = end - chunk_size
random_start = random.randint(start, max_start)
random_end = random_start + chunk_size
print(f"{random_start:020x} {random_end:020x}")
PYEOF
)
RANDOM_START=$(echo $RESULT | awk '{print $1}')
RANDOM_END=$(echo $RESULT | awk '{print $2}')
echo "$(date): Yeni dilim $RANDOM_START - $RANDOM_END" >> auto_log.txt
./bin/cuBitCrack -d $DEVICE \
-b $BLOCKS -t $THREADS -p $POINTS \
-c -o Found.txt \
--keyspace ${RANDOM_START}:${RANDOM_END} \
${TARGET} >> output.log 2>&1
if [ -s Found.txt ]; then
echo "$(date): KEY BULUNDU!" >> auto_log.txt
cat Found.txt >> auto_log.txt
break
fi
done
EOF
chmod +x auto_random.sh When running bitcrack, share that screen to guide u best optimal setting NVIDIA GeForce R 12799 / 32109MB | 1 target 3891.43 MKey/s (1,593,567,084,544 total) [00:06:47][2026-05-13.11:37:57] [Info] Compression: compressed [2026-05-13.11:37:57] [Info] Starting at: 0000000000000000000000000000000000000000000001EE119455F12CEE4EF4 [2026-05-13.11:37:57] [Info] Ending at: 0000000000000000000000000000000000000000000001EE119475F12CEE4EF4 [2026-05-13.11:37:57] [Info] Counting by: 0000000000000000000000000000000000000000000000000000000000000001 [2026-05-13.11:37:57] [Info] Initializing NVIDIA GeForce RTX 5090 [2026-05-13.11:37:57] [Info] Generating 134,217,728 starting points (5120.0MB) [2026-05-13.11:38:07] [Info] 10.0% [2026-05-13.11:38:08] [Info] 20.0% [2026-05-13.11:38:08] [Info] 30.0% [2026-05-13.11:38:08] [Info] 40.0% [2026-05-13.11:38:08] [Info] 50.0% [2026-05-13.11:38:09] [Info] 60.0% [2026-05-13.11:38:09] [Info] 70.0% [2026-05-13.11:38:09] [Info] 80.0% [2026-05-13.11:38:09] [Info] 90.0% [2026-05-13.11:38:09] [Info] 100.0% [2026-05-13.11:38:09] [Info] Done NVIDIA GeForce R 12799 / 32109MB | 1 target 3864.52 MKey/s (15,749,779,292,160 total) [01:07:27] Any Above Value of 134,217,728 starting points (5120.0MB), I get out of memory error. And CUDA version is 13.0 |
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Niekko
Member
Activity: 111
Merit: 26
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May 13, 2026, 02:10:45 PM |
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NVIDIA GeForce R 12799 / 32109MB | 1 target 3891.43 MKey/s (1,593,567,084,544 total) [00:06:47][2026-05-13.11:37:57] [Info] Compression: compressed
[2026-05-13.11:37:57] [Info] Starting at: 0000000000000000000000000000000000000000000001EE119455F12CEE4EF4
[2026-05-13.11:37:57] [Info] Ending at: 0000000000000000000000000000000000000000000001EE119475F12CEE4EF4
[2026-05-13.11:37:57] [Info] Counting by: 0000000000000000000000000000000000000000000000000000000000000001
[2026-05-13.11:37:57] [Info] Initializing NVIDIA GeForce RTX 5090
[2026-05-13.11:37:57] [Info] Generating 134,217,728 starting points (5120.0MB)
[2026-05-13.11:38:07] [Info] 10.0%
[2026-05-13.11:38:08] [Info] 20.0%
[2026-05-13.11:38:08] [Info] 30.0%
[2026-05-13.11:38:08] [Info] 40.0%
[2026-05-13.11:38:08] [Info] 50.0%
[2026-05-13.11:38:09] [Info] 60.0%
[2026-05-13.11:38:09] [Info] 70.0%
[2026-05-13.11:38:09] [Info] 80.0%
[2026-05-13.11:38:09] [Info] 90.0%
[2026-05-13.11:38:09] [Info] 100.0%
[2026-05-13.11:38:09] [Info] Done
NVIDIA GeForce R 12799 / 32109MB | 1 target 3864.52 MKey/s (15,749,779,292,160 total) [01:07:27]
Any Above Value of 134,217,728 starting points (5120.0MB), I get out of memory error. And CUDA version is 13.0
it's not very much bad, think that BitCrack it's a real bruteforce, and you step at 0x01. With my 5090 i going at 4750 MKey/s, but I've modded it a little. You will never going at speed like BSGS, because it's not a real speed, yours it's real. |
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0xastraeus
Newbie
Activity: 43
Merit: 0
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May 13, 2026, 02:14:09 PM |
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NVIDIA GeForce R 12799 / 32109MB | 1 target 3891.43 MKey/s (1,593,567,084,544 total) [00:06:47][2026-05-13.11:37:57] [Info] Compression: compressed
[2026-05-13.11:37:57] [Info] Starting at: 0000000000000000000000000000000000000000000001EE119455F12CEE4EF4
[2026-05-13.11:37:57] [Info] Ending at: 0000000000000000000000000000000000000000000001EE119475F12CEE4EF4
[2026-05-13.11:37:57] [Info] Counting by: 0000000000000000000000000000000000000000000000000000000000000001
[2026-05-13.11:37:57] [Info] Initializing NVIDIA GeForce RTX 5090
[2026-05-13.11:37:57] [Info] Generating 134,217,728 starting points (5120.0MB)
[2026-05-13.11:38:07] [Info] 10.0%
[2026-05-13.11:38:08] [Info] 20.0%
[2026-05-13.11:38:08] [Info] 30.0%
[2026-05-13.11:38:08] [Info] 40.0%
[2026-05-13.11:38:08] [Info] 50.0%
[2026-05-13.11:38:09] [Info] 60.0%
[2026-05-13.11:38:09] [Info] 70.0%
[2026-05-13.11:38:09] [Info] 80.0%
[2026-05-13.11:38:09] [Info] 90.0%
[2026-05-13.11:38:09] [Info] 100.0%
[2026-05-13.11:38:09] [Info] Done
NVIDIA GeForce R 12799 / 32109MB | 1 target 3864.52 MKey/s (15,749,779,292,160 total) [01:07:27]
Any Above Value of 134,217,728 starting points (5120.0MB), I get out of memory error. And CUDA version is 13.0
Did you compile correctly for 5090? try lowering blocks and higher points like -b 256 -p 4096 or something see how that goes. Play around with it. |
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brainless
Member
Activity: 487
Merit: 35
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May 13, 2026, 02:33:53 PM |
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I use cuBitCrack, but getting 4GKeys/s from RTX5090 Could you please anyone help me about optimize for speed? Below code provides auto random jumping, when a space scanning finished cat > auto_random.sh << 'EOF'
#!/bin/bash
TARGET="12VVRNPi4SJqUTsp6FmqDqY5sGosDtysn4"
DEVICE=0
BLOCKS=1024
THREADS=256
POINTS=512
while true; do
pkill cuBitCrack 2>/dev/null
sleep 2
RESULT=$(python3 << 'PYEOF'
import random
start = 0x1000000000000000000
end = 0x1ffffffffffffffffff
chunk_size = 2**46
max_start = end - chunk_size
random_start = random.randint(start, max_start)
random_end = random_start + chunk_size
print(f"{random_start:020x} {random_end:020x}")
PYEOF
)
RANDOM_START=$(echo $RESULT | awk '{print $1}')
RANDOM_END=$(echo $RESULT | awk '{print $2}')
echo "$(date): Yeni dilim $RANDOM_START - $RANDOM_END" >> auto_log.txt
./bin/cuBitCrack -d $DEVICE \
-b $BLOCKS -t $THREADS -p $POINTS \
-c -o Found.txt \
--keyspace ${RANDOM_START}:${RANDOM_END} \
${TARGET} >> output.log 2>&1
if [ -s Found.txt ]; then
echo "$(date): KEY BULUNDU!" >> auto_log.txt
cat Found.txt >> auto_log.txt
break
fi
done
EOF
chmod +x auto_random.sh When running bitcrack, share that screen to guide u best optimal setting NVIDIA GeForce R 12799 / 32109MB | 1 target 3891.43 MKey/s (1,593,567,084,544 total) [00:06:47][2026-05-13.11:37:57] [Info] Compression: compressed [2026-05-13.11:37:57] [Info] Starting at: 0000000000000000000000000000000000000000000001EE119455F12CEE4EF4 [2026-05-13.11:37:57] [Info] Ending at: 0000000000000000000000000000000000000000000001EE119475F12CEE4EF4 [2026-05-13.11:37:57] [Info] Counting by: 0000000000000000000000000000000000000000000000000000000000000001 [2026-05-13.11:37:57] [Info] Initializing NVIDIA GeForce RTX 5090 [2026-05-13.11:37:57] [Info] Generating 134,217,728 starting points (5120.0MB) [2026-05-13.11:38:07] [Info] 10.0% [2026-05-13.11:38:08] [Info] 20.0% [2026-05-13.11:38:08] [Info] 30.0% [2026-05-13.11:38:08] [Info] 40.0% [2026-05-13.11:38:08] [Info] 50.0% [2026-05-13.11:38:09] [Info] 60.0% [2026-05-13.11:38:09] [Info] 70.0% [2026-05-13.11:38:09] [Info] 80.0% [2026-05-13.11:38:09] [Info] 90.0% [2026-05-13.11:38:09] [Info] 100.0% [2026-05-13.11:38:09] [Info] Done NVIDIA GeForce R 12799 / 32109MB | 1 target 3864.52 MKey/s (15,749,779,292,160 total) [01:07:27] Any Above Value of 134,217,728 starting points (5120.0MB), I get out of memory error. And CUDA version is 13.0 BLOCKS=170 THREADS=512 POINTS=8192 Now need to optimize -t -p, mostly point up and down with make set When ever need to adjust t and p Use 32 +- in above mention , report here screen shot for see what's going on |
13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF
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OzBtcOz
Newbie
Activity: 17
Merit: 0
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May 13, 2026, 02:35:55 PM |
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NVIDIA GeForce R 12799 / 32109MB | 1 target 3891.43 MKey/s (1,593,567,084,544 total) [00:06:47][2026-05-13.11:37:57] [Info] Compression: compressed
[2026-05-13.11:37:57] [Info] Starting at: 0000000000000000000000000000000000000000000001EE119455F12CEE4EF4
[2026-05-13.11:37:57] [Info] Ending at: 0000000000000000000000000000000000000000000001EE119475F12CEE4EF4
[2026-05-13.11:37:57] [Info] Counting by: 0000000000000000000000000000000000000000000000000000000000000001
[2026-05-13.11:37:57] [Info] Initializing NVIDIA GeForce RTX 5090
[2026-05-13.11:37:57] [Info] Generating 134,217,728 starting points (5120.0MB)
[2026-05-13.11:38:07] [Info] 10.0%
[2026-05-13.11:38:08] [Info] 20.0%
[2026-05-13.11:38:08] [Info] 30.0%
[2026-05-13.11:38:08] [Info] 40.0%
[2026-05-13.11:38:08] [Info] 50.0%
[2026-05-13.11:38:09] [Info] 60.0%
[2026-05-13.11:38:09] [Info] 70.0%
[2026-05-13.11:38:09] [Info] 80.0%
[2026-05-13.11:38:09] [Info] 90.0%
[2026-05-13.11:38:09] [Info] 100.0%
[2026-05-13.11:38:09] [Info] Done
NVIDIA GeForce R 12799 / 32109MB | 1 target 3864.52 MKey/s (15,749,779,292,160 total) [01:07:27]
Any Above Value of 134,217,728 starting points (5120.0MB), I get out of memory error. And CUDA version is 13.0
it's not very much bad, think that BitCrack it's a real bruteforce, and you step at 0x01. With my 5090 i going at 4750 MKey/s, but I've modded it a little. You will never going at speed like BSGS, because it's not a real speed, yours it's real. Thanks your answer, but when I checked other forums, they presented their benchmarks up to 8 GKeys/s. So, I thouht that there is something wrong with my GPU or compiling or etc. |
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Niekko
Member
Activity: 111
Merit: 26
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May 13, 2026, 03:06:59 PM |
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it's not very much bad, think that BitCrack it's a real bruteforce, and you step at 0x01. With my 5090 i going at 4750 MKey/s, but I've modded it a little.
You will never going at speed like BSGS, because it's not a real speed, yours it's real.
Thanks your answer, but when I checked other forums, they presented their benchmarks up to 8 GKeys/s. So, I thouht that there is something wrong with my GPU or compiling or etc. with BitCrack ? My 5090 was never going at 8GKeys/s. |
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Grzegorz2022
Newbie
Activity: 24
Merit: 0
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May 13, 2026, 03:16:33 PM |
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NVIDIA GeForce R 12799 / 32109MB | 1 target 3891.43 MKey/s (1,593,567,084,544 total) [00:06:47][2026-05-13.11:37:57] [Info] Compression: compressed
[2026-05-13.11:37:57] [Info] Starting at: 0000000000000000000000000000000000000000000001EE119455F12CEE4EF4
[2026-05-13.11:37:57] [Info] Ending at: 0000000000000000000000000000000000000000000001EE119475F12CEE4EF4
[2026-05-13.11:37:57] [Info] Counting by: 0000000000000000000000000000000000000000000000000000000000000001
[2026-05-13.11:37:57] [Info] Initializing NVIDIA GeForce RTX 5090
[2026-05-13.11:37:57] [Info] Generating 134,217,728 starting points (5120.0MB)
[2026-05-13.11:38:07] [Info] 10.0%
[2026-05-13.11:38:08] [Info] 20.0%
[2026-05-13.11:38:08] [Info] 30.0%
[2026-05-13.11:38:08] [Info] 40.0%
[2026-05-13.11:38:08] [Info] 50.0%
[2026-05-13.11:38:09] [Info] 60.0%
[2026-05-13.11:38:09] [Info] 70.0%
[2026-05-13.11:38:09] [Info] 80.0%
[2026-05-13.11:38:09] [Info] 90.0%
[2026-05-13.11:38:09] [Info] 100.0%
[2026-05-13.11:38:09] [Info] Done
NVIDIA GeForce R 12799 / 32109MB | 1 target 3864.52 MKey/s (15,749,779,292,160 total) [01:07:27]
Any Above Value of 134,217,728 starting points (5120.0MB), I get out of memory error. And CUDA version is 13.0
it's not very much bad, think that BitCrack it's a real bruteforce, and you step at 0x01. With my 5090 i going at 4750 MKey/s, but I've modded it a little. You will never going at speed like BSGS, because it's not a real speed, yours it's real. Thanks your answer, but when I checked other forums, they presented their benchmarks up to 8 GKeys/s. So, I thouht that there is something wrong with my GPU or compiling or etc. how long does it take you to find the key to puzzle 70 |
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Divaytis
Newbie
Activity: 3
Merit: 0
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May 13, 2026, 04:08:55 PM |
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Anyone here know how to code and want to help me solve the puzzles? I created a system from the ground up complete with axioms definitions and procedural operations i just designed my system so that it happens to use the exact same values as secp256k1 im not saying i solved the discrete log thats impossible instead i created a system where the discrete log doesnt even apply at all
If it's still relevant, I can help write the code for you. Bro if you can code we can cook I can help with writing the code without any problems, how can I contact you to discuss everything? |
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speed_user_113
Newbie
Activity: 8
Merit: 0
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May 13, 2026, 05:10:18 PM |
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As i see there is GPU crisys and you cannot find anymore on a good price 4090 or 5090 to rent. I want to see the ones that had scripts and GPU farms rental to steal now the key  (unless they have their own GPU farm) |
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SecretAdmirere
Newbie
Activity: 19
Merit: 1
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May 13, 2026, 07:40:41 PM |
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Thanks your answer, but when I checked other forums, they presented their benchmarks up to 8 GKeys/s. So, I thouht that there is something wrong with my GPU or compiling or etc.
Don't know what and from who, you are getting your BitCrack code, but 5090 is easily capable of 11b+ checked hash160 addresses per second, if not 12b+, but for those numbers you need a custom code beacuse none of those publicly avaliable can achive it (they are not that very well "optimised").. You should get the https://github.com/FixedPaul/VanitySearch-Bitcrack, that one is the fastest publicly avaliable, and it would run at 8b+ at stock speeds and power draw, and if you want to go beyond that well, you need to drasticly change the hashing and FE functions. how long does it take you to find the key to puzzle 70
You can't simply compare "how long it takes to find 2⁷⁰ private key", that is a mode where it hashes pubkeys and checks hash160 target to the computed hash160, totally different space where your bsgs is being used. You should compile and benchmark your FE mul and compare it against kTimesG or RetiredCoder (115/140 Gmul/s on 4090) and get a rough estimate where your performance is at beacuse that one is by far the most impactful one. |
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username666187
Newbie
Activity: 17
Merit: 0
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May 13, 2026, 09:24:40 PM
Last edit: May 14, 2026, 11:40:08 AM by Mr. Big |
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Anyone here know how to code and want to help me solve the puzzles? I created a system from the ground up complete with axioms definitions and procedural operations i just designed my system so that it happens to use the exact same values as secp256k1 im not saying i solved the discrete log thats impossible instead i created a system where the discrete log doesnt even apply at all
If it's still relevant, I can help write the code for you. Bro if you can code we can cook I can help with writing the code without any problems, how can I contact you to discuss everything? Dude allow new members to send you message im not trying to dox my phone number and email address
Anyone here know how to code and want to help me solve the puzzles? I created a system from the ground up complete with axioms definitions and procedural operations i just designed my system so that it happens to use the exact same values as secp256k1 im not saying i solved the discrete log thats impossible instead i created a system where the discrete log doesnt even apply at all
If it's still relevant, I can help write the code for you. Bro if you can code we can cook Allow messages from newbies on your profile so i can pm you Same goes for you "And24u" |
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