Does anyone else see the zcash in the first 5 puzzle addresses?
what do you mean ?
https://imgur.com/a/4vdcD11
Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 47. (Read 253699 times)
what do you mean ?
https://imgur.com/a/4vdcD11
Does anyone else see the zcash in the first 5 puzzle addresses?
Puzzle 1, PVK=1, ZEC p2pkh address = t1UYsZVJkLPeMjxEtACvSxfWuNmddpWfxzs
Puzzle 2, PVK=3, ZEC p2pkh address = t1VLyEX9gpXZdZeVXeuAvqPRPxj8u8qiVHL
Puzzle 3, PVK=7, ZEC p2pkh address = t1SSFwcYTiLApC5RXEZvNWsLdTPeWQxduZU
Puzzle 4, PVK=8, ZEC p2pkh address = t1XaScJtVuFehnJP2dEMDyRVWTkjQXJ4PU7
Puzzle 5, PVK=15, ZEC p2pkh address = t1Wxyub9LgLu6gdrRcZGYmRYM5nG51YqEqL
...
Puzzle 66, PVK=2832ED74F2B5E35EE, ZEC p2pkh address = t1LsC22pjUpfCd5ATPzbWA5Aq3HZTZXJWuS
what do you mean ?
Does anyone else see the zcash in the first 5 puzzle addresses?
On Nov. 1st 2024 at 0:00 AM UTC I will post here a BTC P2PKH address (compressed) associated with an 80-bits secure private key.
The address owns 0.005 BTC (~340 $)
Starting from that time, at some point during the next 24 hours an outgoing transaction, containing unspent outputs from that address, will be broadcasted on the BTC network (public mempool, not Mara).
Let's see who breaks it. I only ask one thing if that happens: what method was used to break the key.
Together with the target address I will also provide the range of the search interval.
This will be in the form (just an example):
The Hamming length of the range will therefore be 80 contiguous bits, but they may start anywhere.
If your first instinct is to yell "this is not a 80-bits secure private key", please note this competition is NOT for you, do not expect a response from me.
Get your tools ready boys.
Don't mind KtimesGThe address owns 0.005 BTC (~340 $)
Starting from that time, at some point during the next 24 hours an outgoing transaction, containing unspent outputs from that address, will be broadcasted on the BTC network (public mempool, not Mara).
Let's see who breaks it. I only ask one thing if that happens: what method was used to break the key.
Together with the target address I will also provide the range of the search interval.
This will be in the form (just an example):
Code:
minKey = 0xf2e542b46066c4e6f91abc80000000000000000000185e689447431d74c5b133
maxKey = 0xf2e542b46066c4e6f91abcbfffffffffffffffffffd85e689447431d74c5b133
maxKey = 0xf2e542b46066c4e6f91abcbfffffffffffffffffffd85e689447431d74c5b133
The Hamming length of the range will therefore be 80 contiguous bits, but they may start anywhere.
If your first instinct is to yell "this is not a 80-bits secure private key", please note this competition is NOT for you, do not expect a response from me.
Get your tools ready boys.
If I think about your mention line
"Let's see who breaks it. I only ask one thing if that happens: what method was used to break the key.
"
If some one break, there will no money till they explain you about breaking method, then you will release money
If it's Yes
May I think
It's same old strategy, like happen at telegram groups, some one post challenge with 10k, one guy pick message and post at discord groups for 200 bucks, upon solution finding, result some one break, and person ask write him for post method in private message, for release his award
Maybe this is similar game
The only strategy is the one explained in my post, there are no hidden tricks, it just needs to be read very very carefully - all the people here failed to 100% correctly see that, I will explain why after it is over. If someone cracks the key, replaces the TX, wins the competition. It would be nice though to let me/us know what method he used (I only care about the algorithm).
One last hint: secp256k1 is a modular group. Don't expect that all existing tools take this into consideration fully.
5 second for math and max 15 min at gpu
On Nov. 1st 2024 at 0:00 AM UTC I will post here a BTC P2PKH address (compressed) associated with an 80-bits secure private key.
The address owns 0.005 BTC (~340 $)
Starting from that time, at some point during the next 24 hours an outgoing transaction, containing unspent outputs from that address, will be broadcasted on the BTC network (public mempool, not Mara).
Let's see who breaks it. I only ask one thing if that happens: what method was used to break the key.
Together with the target address I will also provide the range of the search interval.
This will be in the form (just an example):
The Hamming length of the range will therefore be 80 contiguous bits, but they may start anywhere.
If your first instinct is to yell "this is not a 80-bits secure private key", please note this competition is NOT for you, do not expect a response from me.
Get your tools ready boys.
Don't mind KtimesGThe address owns 0.005 BTC (~340 $)
Starting from that time, at some point during the next 24 hours an outgoing transaction, containing unspent outputs from that address, will be broadcasted on the BTC network (public mempool, not Mara).
Let's see who breaks it. I only ask one thing if that happens: what method was used to break the key.
Together with the target address I will also provide the range of the search interval.
This will be in the form (just an example):
Code:
minKey = 0xf2e542b46066c4e6f91abc80000000000000000000185e689447431d74c5b133
maxKey = 0xf2e542b46066c4e6f91abcbfffffffffffffffffffd85e689447431d74c5b133
maxKey = 0xf2e542b46066c4e6f91abcbfffffffffffffffffffd85e689447431d74c5b133
The Hamming length of the range will therefore be 80 contiguous bits, but they may start anywhere.
If your first instinct is to yell "this is not a 80-bits secure private key", please note this competition is NOT for you, do not expect a response from me.
Get your tools ready boys.
If I think about your mention line
"Let's see who breaks it. I only ask one thing if that happens: what method was used to break the key.
"
If some one break, there will no money till they explain you about breaking method, then you will release money
If it's Yes
May I think
It's same old strategy, like happen at telegram groups, some one post challenge with 10k, one guy pick message and post at discord groups for 200 bucks, upon solution finding, result some one break, and person ask write him for post method in private message, for release his award
Maybe this is similar game
The only strategy is the one explained in my post, there are no hidden tricks, it just needs to be read very very carefully - all the people here failed to 100% correctly see that, I will explain why after it is over. If someone cracks the key, replaces the TX, wins the competition. It would be nice though to let me/us know what method he used (I only care about the algorithm).
One last hint: secp256k1 is a modular group. Don't expect that all existing tools take this into consideration fully.
Code:
N = 115792089237316195423570985008687907852837564279074904382605163141518161494337
def inv(v):
return pow(v, N-2, N)
def divnum(a, b):
return ( (a * inv(b) ) % N )
x = 0x3b050b7264187e2bcf8b2d50f5feb5 # - 0x262794
y = 0x3b050b7264187e2bcf8b2d50f5feb5 # - 0x262794
# Specify the value of i for which you want to get the result
target_i = 99990000000000 # Replace 5000 with the desired value
# Initial value of X
X_initial = divnum(x, 99990000000000)
# Final value of X after target_i steps
X_final = (X_initial - (target_i * divnum(1, 1)) % N) % N
# Sum of all X values over target_i steps
S = divnum((target_i * (X_initial + X_final%N) %N), 2 )% N
# Final value of y
y_final = (y - S) % N
if y_final <= 2**190:
print("input:")
print(hex(0x3971621b0ac11b09e7741edd106f916e5))
print("y", hex(y_final), target_i)
print("X", hex(X_final* 99990000000000 %N))
print("Xfin",((X_initial- X_final %N)%N))
code not work with pubkeys, this is little modifications need. Next time maybe, sorry
after run scrypt y = 0x1027136fb927635998880000
put y to x and y in scrypt,
target_i = 99990
Code:
N = 115792089237316195423570985008687907852837564279074904382605163141518161494337
def inv(v):
return pow(v, N-2, N)
def divnum(a, b):
return ( (a * inv(b) ) % N )
x = 0x1027136fb927635998880000 # - 0x262794
y = 0x1027136fb927635998880000 # - 0x262794
# Specify the value of i for which you want to get the result
target_i = 99990#000000000 # Replace 5000 with the desired value
# Initial value of X
X_initial = divnum(x, 99990000000000)
# Final value of X after target_i steps
X_final = (X_initial - (target_i * divnum(1, 1)) % N) % N
# Sum of all X values over target_i steps
S = divnum((target_i * (X_initial + X_final%N) %N), 2 )% N
# Final value of y
y_final = (y - S) % N
if y_final <= 2**190:
print("input:")
print(hex(0x3971621b0ac11b09e7741edd106f916e5))
print("y", hex(y_final), target_i)
print("X", hex(X_final* 99990000000000 %N))
print("Xfin",((X_initial- X_final %N)%N))
result y = 0x1027136f73c75f4b7e15bbf2
brute y, frecover priv after
another variant:
Code:
x = 0x1027136fb927635998880000 # - 0x262794
y = 0x1027136fb927635998880000 # - 0x262794
# Specify the value of i for which you want to get the result
target_i = 99990000000000
target_i = 9999000000000#0
resukt :
0x1027136fb927635998880000
y 0xeb2eb4982170b0c69860000 9999000000000
X 0xcec0f8c941f82ae13a00000
Xfin 9999000000000
put new y to x and y in scrypt result:
input:
0xeb2eb4982170b0c69860000
y 0xd63fa5a5088552d58d10000 9999000000000
X 0xb77e7665d0f2a60e49e0000
Xfin 9999000000000
[Program finished]
etc
On Nov. 1st 2024 at 0:00 AM UTC I will post here a BTC P2PKH address (compressed) associated with an 80-bits secure private key.
The address owns 0.005 BTC (~340 $)
Starting from that time, at some point during the next 24 hours an outgoing transaction, containing unspent outputs from that address, will be broadcasted on the BTC network (public mempool, not Mara).
Let's see who breaks it. I only ask one thing if that happens: what method was used to break the key.
Together with the target address I will also provide the range of the search interval.
This will be in the form (just an example):
The Hamming length of the range will therefore be 80 contiguous bits, but they may start anywhere.
If your first instinct is to yell "this is not a 80-bits secure private key", please note this competition is NOT for you, do not expect a response from me.
Get your tools ready boys.
Don't mind KtimesGThe address owns 0.005 BTC (~340 $)
Starting from that time, at some point during the next 24 hours an outgoing transaction, containing unspent outputs from that address, will be broadcasted on the BTC network (public mempool, not Mara).
Let's see who breaks it. I only ask one thing if that happens: what method was used to break the key.
Together with the target address I will also provide the range of the search interval.
This will be in the form (just an example):
Code:
minKey = 0xf2e542b46066c4e6f91abc80000000000000000000185e689447431d74c5b133
maxKey = 0xf2e542b46066c4e6f91abcbfffffffffffffffffffd85e689447431d74c5b133
maxKey = 0xf2e542b46066c4e6f91abcbfffffffffffffffffffd85e689447431d74c5b133
The Hamming length of the range will therefore be 80 contiguous bits, but they may start anywhere.
If your first instinct is to yell "this is not a 80-bits secure private key", please note this competition is NOT for you, do not expect a response from me.
Get your tools ready boys.
If I think about your mention line
"Let's see who breaks it. I only ask one thing if that happens: what method was used to break the key.
"
If some one break, there will no money till they explain you about breaking method, then you will release money
If it's Yes
May I think
It's same old strategy, like happen at telegram groups, some one post challenge with 10k, one guy pick message and post at discord groups for 200 bucks, upon solution finding, result some one break, and person ask write him for post method in private message, for release his award
Maybe this is similar game
Code:
N = 115792089237316195423570985008687907852837564279074904382605163141518161494337
def inv(v):
return pow(v, N-2, N)
def divnum(a, b):
return ( (a * inv(b) ) % N )
x = 0x3b050b7264187e2bcf8b2d50f5feb5 # - 0x262794
y = 0x3b050b7264187e2bcf8b2d50f5feb5 # - 0x262794
# Specify the value of i for which you want to get the result
target_i = 99990000000000 # Replace 5000 with the desired value
# Initial value of X
X_initial = divnum(x, 99990000000000)
# Final value of X after target_i steps
X_final = (X_initial - (target_i * divnum(1, 1)) % N) % N
# Sum of all X values over target_i steps
S = divnum((target_i * (X_initial + X_final%N) %N), 2 )% N
# Final value of y
y_final = (y - S) % N
if y_final <= 2**190:
print("input:")
print(hex(0x3971621b0ac11b09e7741edd106f916e5))
print("y", hex(y_final), target_i)
print("X", hex(X_final* 99990000000000 %N))
print("Xfin",((X_initial- X_final %N)%N))
code not work with pubkeys, this is little modifications need. Next time maybe, sorry
after run scrypt y = 0x1027136fb927635998880000
put y to x and y in scrypt,
target_i = 99990
Code:
N = 115792089237316195423570985008687907852837564279074904382605163141518161494337
def inv(v):
return pow(v, N-2, N)
def divnum(a, b):
return ( (a * inv(b) ) % N )
x = 0x1027136fb927635998880000 # - 0x262794
y = 0x1027136fb927635998880000 # - 0x262794
# Specify the value of i for which you want to get the result
target_i = 99990#000000000 # Replace 5000 with the desired value
# Initial value of X
X_initial = divnum(x, 99990000000000)
# Final value of X after target_i steps
X_final = (X_initial - (target_i * divnum(1, 1)) % N) % N
# Sum of all X values over target_i steps
S = divnum((target_i * (X_initial + X_final%N) %N), 2 )% N
# Final value of y
y_final = (y - S) % N
if y_final <= 2**190:
print("input:")
print(hex(0x3971621b0ac11b09e7741edd106f916e5))
print("y", hex(y_final), target_i)
print("X", hex(X_final* 99990000000000 %N))
print("Xfin",((X_initial- X_final %N)%N))
result y = 0x1027136f73c75f4b7e15bbf2
brute y, frecover priv after
another variant:
Code:
x = 0x1027136fb927635998880000 # - 0x262794
y = 0x1027136fb927635998880000 # - 0x262794
# Specify the value of i for which you want to get the result
target_i = 99990000000000
target_i = 9999000000000#0
resukt :
0x1027136fb927635998880000
y 0xeb2eb4982170b0c69860000 9999000000000
X 0xcec0f8c941f82ae13a00000
Xfin 9999000000000
put new y to x and y in scrypt result:
input:
0xeb2eb4982170b0c69860000
y 0xd63fa5a5088552d58d10000 9999000000000
X 0xb77e7665d0f2a60e49e0000
Xfin 9999000000000
[Program finished]
etc
https://www.techrxiv.org/users/837126/articles/1228731-dual-vanity-address-attack-dvaa
someone is researching new ways:)
someone is researching new ways:)
The "new way" is much slower than the privKey-to-ripemd attack which is used for address-only puzzles. Basically brute-force on steroids, nothing "advanced" as the paper claims.
Quote
The process of generating dual vanity addresses is computationally intensive, requiring massive numbers of hash calculations. If many users or attackers engage in generating such addresses, it could lead
to increased demand on network resources. Although the Bitcoin network is designed to handle high
volumes of transactions and computations, a significant surge in address generation activities could
strain node resources, potentially affecting network performance and transaction processing times.
to increased demand on network resources. Although the Bitcoin network is designed to handle high
volumes of transactions and computations, a significant surge in address generation activities could
strain node resources, potentially affecting network performance and transaction processing times.
https://www.techrxiv.org/users/837126/articles/1228731-dual-vanity-address-attack-dvaa
someone is researching new ways:)
someone is researching new ways:)
What do you all think #67 hex starts with?
#66 I don't believe in randomization, no one can be that lucky! also brute force from the start would of required around 700 billion keys per sec to find in 2years
I think the finder knew exactly the hex started with 2832 or had some mathematical way of working out a starting point of 283 for 000000000000000000000000000000000000000000000002832ed74f2b5e35ee
How do you think they found it?
Both random finding and/or mathematics all ok
So no, the pool did not know the key started with 2832, or 283, or 28, nor even a 2. It was purely, random.
What do you all think #67 hex starts with?
#66 I don't believe in randomization, no one can be that lucky! also brute force from the start would of required around 700 billion keys per sec to find in 2years
I think the finder knew exactly the hex started with 2832 or had some mathematical way of working out a starting point of 283 for 000000000000000000000000000000000000000000000002832ed74f2b5e35ee
How do you think they found it?
Both random finding and/or mathematics all ok
What do you all think #67 hex starts with?
#66 I don't believe in randomization, no one can be that lucky! also brute force from the start would of required around 700 billion keys per sec to find in 2years
I think the finder knew exactly the hex started with 2832 or had some mathematical way of working out a starting point of 283 for 000000000000000000000000000000000000000000000002832ed74f2b5e35ee
How do you think they found it?
... the usual non-sense ...
Can you explain more here or send me a PM please?
where did you got these numbers?
puzzle 120 Public Key is 02ceb6cbbcdbdf5ef7150682150f4ce2c6f4807b349827dcdbdd1f2efa885a2630
can you say what operations you did to come up with this numbers?
because I checked and 0x1d8285b9320c3f15e7c596a87aff5c is not the hex of 108945 or 10894500, without more details nobody can help
you had another post where you posted 9 lines where you need to find i to get the private key of puzzle 130 but you deleted the post...
If anyone else have other ideas I am open to discussions in PM, brainstorming, tests, scripts and team work.
Why would you think a disfunctional illiterate and reported scammer has broken ECDLP? All the code he ever posted is pure bullshit, and everyone ignores him so that maybe he will stop spamming with all his dementia about splitting the public key and etc.
Now, think about it: if YOU were to be split in two, or four, or whatever, does any of the parts contain everything about the entire you, and I mean at the lowest possible detail? There's no such thing as "reducing" information to simplify a problem, all you get in return is having to double up on the effort to recover the information that you lost (by checking all the alternatives that you just dismissed).
We have the forum for brainstorming, unless you have some desire to create a KuKluxKlan for breaking secp256k1 or something, and we're not invited.
Who buy me a coffee
input:
0x3b050b7264187e2bcf8b2d50f5feb8
0x20ec559e1efc0dfed6d56c7e38ed2e2ceb863be1a65de099e3cdbbed62bb7ee7
y 0x1d8285b9320c3f15e7c596a87aff5c 108945
y - xivided pub 120
108945 is i, - "trick", is a number what known and for many(mybe any) privkes / pubkeys before brute. Possible not brute fool range
.y 0x1d8285b9320c3f15e7c596a87aff5c 108945
additional demo:
y 0x1621e44ae5892f506dd430fe5c3f85 8170875
input:
0x3b050b7264187e2bcf8b2d50f5feb8
0x2bd9674194ca33570744b1b9950c9c9ef402e8b645dca52a6a915712a1d2c73e
y 0x1d8285b9320c3f15e7c596a87aff5c 10894500
input:
0x3b050b7264187e2bcf8b2d50f5feb8
0x80b5a45a8c795201045c6aa28ca12ac5437ae3fcd66e3ccaf311c076b460f80c
y 0x24e327277e8f4edb61b6fc5299bf33 13618125
someone can calculate what is divisor ?
p.sI know answer.
@brainless, you steal think what if divide only one integer answer can be ?
108945! btw is so hard to understand you sentence, check your grammars
input:
0x3b050b7264187e2bcf8b2d50f5feb8
0x20ec559e1efc0dfed6d56c7e38ed2e2ceb863be1a65de099e3cdbbed62bb7ee7
y 0x1d8285b9320c3f15e7c596a87aff5c 108945
y - xivided pub 120
108945 is i, - "trick", is a number what known and for many(mybe any) privkes / pubkeys before brute. Possible not brute fool range
.y 0x1d8285b9320c3f15e7c596a87aff5c 108945
additional demo:
y 0x1621e44ae5892f506dd430fe5c3f85 8170875
input:
0x3b050b7264187e2bcf8b2d50f5feb8
0x2bd9674194ca33570744b1b9950c9c9ef402e8b645dca52a6a915712a1d2c73e
y 0x1d8285b9320c3f15e7c596a87aff5c 10894500
input:
0x3b050b7264187e2bcf8b2d50f5feb8
0x80b5a45a8c795201045c6aa28ca12ac5437ae3fcd66e3ccaf311c076b460f80c
y 0x24e327277e8f4edb61b6fc5299bf33 13618125
someone can calculate what is divisor ?
p.sI know answer.
@brainless, you steal think what if divide only one integer answer can be ?
Can you explain more here or send me a PM please?
where did you got these numbers?
puzzle 120 Public Key is 02ceb6cbbcdbdf5ef7150682150f4ce2c6f4807b349827dcdbdd1f2efa885a2630
can you say what operations you did to come up with this numbers?
because I checked and 0x1d8285b9320c3f15e7c596a87aff5c is not the hex of 108945 or 10894500, without more details nobody can help
you had another post where you posted 9 lines where you need to find i to get the private key of puzzle 130 but you deleted the post...
If anyone else have other ideas I am open to discussions in PM, brainstorming, tests, scripts and team work.
Who buy me a coffee
input:
0x3b050b7264187e2bcf8b2d50f5feb8
0x20ec559e1efc0dfed6d56c7e38ed2e2ceb863be1a65de099e3cdbbed62bb7ee7
y 0x1d8285b9320c3f15e7c596a87aff5c 108945
y - xivided pub 120
108945 is i, - "trick", is a number what known and for many(mybe any) privkes / pubkeys before brute. Possible not brute fool range
.y 0x1d8285b9320c3f15e7c596a87aff5c 108945
additional demo:
y 0x1621e44ae5892f506dd430fe5c3f85 8170875
input:
0x3b050b7264187e2bcf8b2d50f5feb8
0x2bd9674194ca33570744b1b9950c9c9ef402e8b645dca52a6a915712a1d2c73e
y 0x1d8285b9320c3f15e7c596a87aff5c 10894500
input:
0x3b050b7264187e2bcf8b2d50f5feb8
0x80b5a45a8c795201045c6aa28ca12ac5437ae3fcd66e3ccaf311c076b460f80c
y 0x24e327277e8f4edb61b6fc5299bf33 13618125
someone can calculate what is divisor ?
p.sI know answer.
@brainless, you steal think what if divide only one integer answer can be ?
Priv#130？？？？
Priv#125？？？？
Who know？
Priv#125？？？？
Who know？
Code:
Google Colab 2 CpU modify r += 7
Private Key: 0xade6d7ce3b9b
Ops: 1068032 Stored: 66716
Speed: 84669 ops/s
Finished in 12.6 s
That's just dumb luck, just as well as randomly replacing a strategic "+ 1" (a jump distance needs to be added) by "+ 7" (which has no rational explanation). So the average jump distance is 7x larger, the jump table is five times smaller (never mind that its computation is screwed up anyway), and still, puzzle 48 is found 10 times faster.
However if you try to use this "r += 7" tweak in general (or even for other 48 bit puzzles), then on average the time to solve grows many several times than expected. Some limited brains here may not agree, because of the 48-bit puzzle counter example, but it's actually expected that the time to solve actually grows, and this will always happen as soon as the average jump distance is messed around with (either making it lower or higher than the optimal value for the case of the 2-kangaroo algorithm).
You can always find some weird setup that solves a particular public key in very few steps, but it only work for that single case, while on average case the number of steps will increase. Math and probabilities can't be cheated. Even when sticking to the rules, there will always be puzzles that will be solved much faster, and puzzles that will get solved much slower, and this applies to any algorithm. We only know the average though.
Who buy me a coffee
bump lol
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