With wslconfig I didn't really understand in which folder to make the file, with what name of the folder, of the file and with what extension.
C:\Users\<UserName>\.wslconfig The contents of the file: [wsl2]
#Limits VM memory to use no more than 4 GB, this can be set as whole numbers using GB or MB
memory=192GB More information: https://learn.microsoft.com/en-us/windows/wsl/wsl-config |
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FinderOuter has similar functionality. https://github.com/Coding-Enthusiast/FinderOuterMissing Base16
PrivateKey: 00000000000000000000000000000000000000000000000000******a7b90de4
PubKey: 029588f9aace0310751f37d130afa1b792e70b542b29d373f64a56a50bcd90ac3f
The given key is missing 6 characters.
Total number of permutations to check: 16 777 216
Running in parallel.
Found the key: 00000000000000000000000000000000000000000000000000397f5aa7b90de4
Elapsed time: 00:00:37.3199093
k/s= 453 438 |
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I can't figure out how to calculate the ccap=20 parameter? I have an RTX 4050 Laptop. Help please. ./deviceQuery
./deviceQuery Starting...
CUDA Device Query (Runtime API) version (CUDART static linking)
Detected 1 CUDA Capable device(s)
Device 0: "NVIDIA GeForce RTX 4050 Laptop GPU"
CUDA Driver Version / Runtime Version 12.3 / 12.3
CUDA Capability Major/Minor version number: 8.9
Total amount of global memory: 6140 MBytes (6438780928 bytes)
(020) Multiprocessors, (128) CUDA Cores/MP: 2560 CUDA Cores
GPU Max Clock rate: 2055 MHz (2.06 GHz)
Memory Clock rate: 8001 Mhz
Memory Bus Width: 96-bit
L2 Cache Size: 25165824 bytes
Maximum Texture Dimension Size (x,y,z) 1D=(131072), 2D=(131072, 65536), 3D=(16384, 16384, 16384)
Maximum Layered 1D Texture Size, (num) layers 1D=(32768), 2048 layers
Maximum Layered 2D Texture Size, (num) layers 2D=(32768, 32768), 2048 layers
Total amount of constant memory: 65536 bytes
Total amount of shared memory per block: 49152 bytes
Total shared memory per multiprocessor: 102400 bytes
Total number of registers available per block: 65536
Warp size: 32
Maximum number of threads per multiprocessor: 1536
Maximum number of threads per block: 1024
Max dimension size of a thread block (x,y,z): (1024, 1024, 64)
Max dimension size of a grid size (x,y,z): (2147483647, 65535, 65535)
Maximum memory pitch: 2147483647 bytes
Texture alignment: 512 bytes
Concurrent copy and kernel execution: Yes with 1 copy engine(s)
Run time limit on kernels: Yes
Integrated GPU sharing Host Memory: No
Support host page-locked memory mapping: Yes
Alignment requirement for Surfaces: Yes
Device has ECC support: Disabled
Device supports Unified Addressing (UVA): Yes
Device supports Managed Memory: Yes
Device supports Compute Preemption: Yes
Supports Cooperative Kernel Launch: Yes
Supports MultiDevice Co-op Kernel Launch: No
Device PCI Domain ID / Bus ID / location ID: 0 / 1 / 0
Compute Mode:
< Default (multiple host threads can use ::cudaSetDevice() with device simultaneously) >
deviceQuery, CUDA Driver = CUDART, CUDA Driver Version = 12.3, CUDA Runtime Version = 12.3, NumDevs = 1
Result = PASS |
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So no, if you find any keys with funds, unless it's a puzzle key, you should not touch them because they're not yours, so hands off the merch.😉
Yes, you urgently need to give these keys to me and continue to live in poverty, mama's son  |
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Is it possible to use the same pb table for 50 bit range and for 70 bit range?
Or under each range and a public key the new table is necessary?
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It seems to me that the author of the topic has already made it clear that this is a fake and not a real challenge.
He himself has already abandoned this topic. All he wanted to achieve was to sell his NFT
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What calculator do you use?
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I am also ready to become the CEO of Twitter, with payments in bitcoins.
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Sorry @nc50lc but it does'nt matter it is "out of range". see: import hashlib
g=(0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798, 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8)
p = ZZ( '0xFFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE FFFFFC2F'.replace( ' ', '' ) )
n = ZZ( '0xFFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE BAAEDCE6 AF48A03B BFD25E8C D0364141'.replace( ' ', '' ) )
E = EllipticCurve(GF(p), [0, 7])
G = E.point( g )
def egcd(a, b):
if a == 0:
return (b, 0, 1)
else:
g, y, x = egcd(b % a, a)
return (g, x - (b // a) * y, y)
def modinv(a, m):
g, x, y = egcd(a, m)
if g != 1:
raise Exception('modular inverse does not exist')
else:
return x % m
def verify(r, s,z,public_key):
w = modinv(s, n)
u1 = (z * w) % n
u2 = (r * w) % n
D=u1*G + u2*public_key
x,y=D.xy()
x=int(x)
if (r % n) == (x % n):
print( "signature matches")
else:
print("invalid signature")
r= 111175281461482630465516451385666215051004681245013976528598462758289754744929
s= 70043377187322970975383334126537096260470471254635274932605589652196963378161
z= 1
x1=65484586321995029360829397682915368247978476961863225607803717802088249892660
y1=72074870721525551148484769172216378998698581912792399280515952501346465251009
P=E.point((x1,y1))
x2=40909554126419277592724504966829837604137845573578049527014144934973709534933
y2=87404510172103350666497040794028294741242353586809580318994867241148928032959
P2=E.point((x2,y2))
verify(r,s,z,P)
verify(r,s,z,P2) as you see two differents pubkey are valid for the same transactions. what that means -> need finds "additional" pubkey for valid transactions for addres "0" or "n", then you can spend coins. realy good mathematician can do. Traceback (most recent call last): File "2key.py", line 6, in <module> p = ZZ( '0xFFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE FFFFFC2F'.replace( ' ', '' ) ) NameError: name 'ZZ' is not defined what am I doing wrong? |
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I got the R,S,Z values from ~1000 transactions from one address.
How to find a nonce?
What information can be obtained from a large number of R,S,Z?
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phrutis has closed all its github directories if anyone is interested, I have Fialka, Lostcoins fork of Rotor-cuda.
Post it, it might help. |
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Which card is faster in computing RTX 2060 Super 8GB or RTX 3060 12GB?
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How can knowledge of an even or odd point help in hacking secp256k1?
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If "r" starts with 000 - can this be considered a weak upper bit?
How to determine from a transaction that it has weak bits?
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And if you want to burn some bitcoin, far better to do it to an OP_RETURN output than a burner address.
Can you explain how to do this? Do I understand correctly that there are transactions that cannot be spent even knowing the private key? How to define such transactions? |
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If I understand correctly from the answers on github, the creator does not want to change his code.
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I guess it's always possible to use something like this for money laundering.
I.e.: (1) Send 1BTC to address whose private key is publicly known.
(2) See how your BTC was collected by someone else's bot. (3) Cry because you are a loser  |
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Hello friends;
I think I found a clue.
66. The probability that the wallet is in these ranges is over 90%
3B320044788A17DA0 - 3BC0ABFC3654BAE80
3475FBD690F7B3880 - 34B85C9495BA98840
30389466D233BCDE0 - 305E58789566D2EA0
2E1325EAAC24E8A40 - 2E6917C76BF8F5340
Can you elaborate on what these assumptions are based on? Given that all previous keys were near the end of the range, I would only consider 3B and above |
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If you set up a time server on a local PC and register a redirect from the specified domains, will your capsule open?
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