What we need are asics
not to mine coins but for more jumps/sec
and
look at this, may be interesting: Calculating Satoshi's coins https://bitcointalksearch.org/topic/calculating-satoshis-coins-5347791
Topic: Pollard's kangaroo ECDLP solver - page 48. (Read 60095 times)
Is there a list with devices/cards and related jumps/sec?
What we need are asics
not to mine coins but for more jumps/sec
and
look at this, may be interesting: Calculating Satoshi's coins https://bitcointalksearch.org/topic/calculating-satoshis-coins-5347791
What we need are asics
not to mine coins but for more jumps/sec
and
look at this, may be interesting: Calculating Satoshi's coins https://bitcointalksearch.org/topic/calculating-satoshis-coins-5347791
Quote
I try sript with 30 and 34. Then 30 I got 109 pubkeys, then 32 - 137 pubkeys. And Yes - script hard to run, requirements are give me strange errors...
If you are using Python 3.x, all you have to change is:pubkey = '03a2efa402fd5268400c77c20e574ba86409ededee7c4020e4b9f0edbee53de0d4'
to whatever pubkey you want
and at the bottom
shiftdown(pubkey, 32)
change the 32 to however many pubkeys you want.
yes, I do it. I try now find a privkey....
edit: I wright understand what 30,32 etc is a power of 2 ? so If we talk about 120 pubkey, then I use this scripts with "30" I will get pubkey diveded by 2^30 ? so result will be "pubkey of 2^90" ?
NotATether, answer please ?
EDIT:
I'm realy not understand how many exact sustracted from pubkey this code:
# This function makes all the downscaled pubkeys obtained from subtracting
# numbers between 0 and divisor, before dividing the pubkeys by divisor.
def shiftdown(pubkey, divisor):
Q = pub2point(pubkey)
# k = 1/divisor
k = pow(divisor, N - 2, N)
for i in range(divisor+1):
P = Q - (i * G)
P = k * P
if (P.y % 2 == 0):
prefix = "02"
else:
prefix = "03"
hx = hex(P.x)[2:].zfill(64)
hy = hex(P.y)[2:].zfill(64)
print(prefix+hx, "04"+hx+hy)
so, if devisor = 5 this mean a just FIVE not any ^ and 2^5 etc.....
Quote
I try sript with 30 and 34. Then 30 I got 109 pubkeys, then 32 - 137 pubkeys. And Yes - script hard to run, requirements are give me strange errors...
If you are using Python 3.x, all you have to change is:pubkey = '03a2efa402fd5268400c77c20e574ba86409ededee7c4020e4b9f0edbee53de0d4'
to whatever pubkey you want
and at the bottom
shiftdown(pubkey, 32)
change the 32 to however many pubkeys you want.
yes, I do it. I try now find a privkey....
edit: I wright understand what 30,32 etc is a power of 2 ? so If we talk about 120 pubkey, then I use this scripts with "30" I will get pubkey diveded by 2^30 ? so result will be "pubkey of 2^90" ?
Quote
I try sript with 30 and 34. Then 30 I got 109 pubkeys, then 32 - 137 pubkeys. And Yes - script hard to run, requirements are give me strange errors...
If you are using Python 3.x, all you have to change is:pubkey = '03a2efa402fd5268400c77c20e574ba86409ededee7c4020e4b9f0edbee53de0d4'
to whatever pubkey you want
and at the bottom
shiftdown(pubkey, 32)
change the 32 to however many pubkeys you want.
yes, I do it. I try now find a privkey....
edit: I wright understand what 30,32 etc is a power of 2 ? so If we talk about 120 pubkey, then I use this scripts with "30" I will get pubkey diveded by 2^30 ? so result will be "pubkey of 2^90" ?
it was just an example (random k generated)
from 120 puzzle we substract the starting range 800000000000000000000000000000 and we obtain the following K 03D943C68C5F958B14524CDBEF68168D6C018E101F4F7074E0C31686384A82F52F
N/2 - above K = 026C25951D1BB97C7DE3020DF0F725487FC33CC1210A155755514B7BF384AACFAD
From this K which is basically N/2 - Puz120 K + START-RANGE we add random K in the interval (0x1 - 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFF)
We compute 10kk points in the range (N/2 - 10kk, N/2), points which cover the following range if we exclude parity (N/2 - 10kk, N/2 + 10kk)
Maybe i will draw something tomorrow for better understanding.
PS: I'm not saying this method will be exponentially "better" but it will double the chances to find k
Why subtract the starting range when we are going to be adding back to it? Is there a specific reason for that?from 120 puzzle we substract the starting range 800000000000000000000000000000 and we obtain the following K 03D943C68C5F958B14524CDBEF68168D6C018E101F4F7074E0C31686384A82F52F
N/2 - above K = 026C25951D1BB97C7DE3020DF0F725487FC33CC1210A155755514B7BF384AACFAD
From this K which is basically N/2 - Puz120 K + START-RANGE we add random K in the interval (0x1 - 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFF)
We compute 10kk points in the range (N/2 - 10kk, N/2), points which cover the following range if we exclude parity (N/2 - 10kk, N/2 + 10kk)
Maybe i will draw something tomorrow for better understanding.
PS: I'm not saying this method will be exponentially "better" but it will double the chances to find k
I'd just add 120 pubkey straight to N/2 and then do incremental additions.
In your opinion is it better to start below N/2 and then pass through (above N/2) to kind of straddle N/2 on both sides?
I guess I don't understand the excluding parity/y you mentioned earlier and how each pub doubles chance to find K.
EDIT: Ok, regardless of start range subtraction (I get it), how does this possibly speed things up? You would still have to search for key from N/2-whatever through N/2 +1, using kangaroo or BSGS or some other key finder, right? I'm interested, I just don't understand how it speeds it up (however small or big), not saying it doesn't, I'm just trying to understand it better. Thanks.
Quote
I try sript with 30 and 34. Then 30 I got 109 pubkeys, then 32 - 137 pubkeys. And Yes - script hard to run, requirements are give me strange errors...
If you are using Python 3.x, all you have to change is:pubkey = '03a2efa402fd5268400c77c20e574ba86409ededee7c4020e4b9f0edbee53de0d4'
to whatever pubkey you want
and at the bottom
shiftdown(pubkey, 32)
change the 32 to however many pubkeys you want.
it was just an example (random k generated)
from 120 puzzle we substract the starting range 800000000000000000000000000000 and we obtain the following K 03D943C68C5F958B14524CDBEF68168D6C018E101F4F7074E0C31686384A82F52F
N/2 - above K = 026C25951D1BB97C7DE3020DF0F725487FC33CC1210A155755514B7BF384AACFAD
From this K which is basically N/2 - Puz120 K + START-RANGE we add random K in the interval (0x1 - 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFF)
We compute 10kk points in the range (N/2 - 10kk, N/2), points which cover the following range if we exclude parity (N/2 - 10kk, N/2 + 10kk)
Maybe i will draw something tomorrow for better understanding.
PS: I'm not saying this method will be exponentially "better" but it will double the chances to find k
from 120 puzzle we substract the starting range 800000000000000000000000000000 and we obtain the following K 03D943C68C5F958B14524CDBEF68168D6C018E101F4F7074E0C31686384A82F52F
N/2 - above K = 026C25951D1BB97C7DE3020DF0F725487FC33CC1210A155755514B7BF384AACFAD
From this K which is basically N/2 - Puz120 K + START-RANGE we add random K in the interval (0x1 - 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFF)
We compute 10kk points in the range (N/2 - 10kk, N/2), points which cover the following range if we exclude parity (N/2 - 10kk, N/2 + 10kk)
Maybe i will draw something tomorrow for better understanding.
PS: I'm not saying this method will be exponentially "better" but it will double the chances to find k
Sure,
I will just give a simple example.
Lets say you get the key close to the N half, lets take for example a random point 7fffffffffffffffffffffffffffffff5d1777e13d618b66b81ee5de17cf885a (we pushed for example 120 puzzle to this k) which results in
K = 0269da924844fcc5852ef6808f0e0bc9e851bd7b6b23d6e55d4b152ab0101fedfd
The diff between N/2 and our point is 3ff6921a42c4b727ca4968504b9846
With a big enough hash table which we need to compute as we want (random/incremental), lets say we compute from 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681A20A0 to 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681B20A0 (small example) then our table will contain 0x10000 public keys, but in reality it is 0x20000 (double).
If we discard Y or parity from our K computation when we generate random steps (increments) we have a double change to get it right.
If our random increment difference was bigger than normal hash table size (but not bigger than double) than we will have a match and we can easily compute the right k from the known K.
Dont know if i was clear enough, if not i will try to illustrate it
Made need the illustration now I will just give a simple example.
Lets say you get the key close to the N half, lets take for example a random point 7fffffffffffffffffffffffffffffff5d1777e13d618b66b81ee5de17cf885a (we pushed for example 120 puzzle to this k) which results in
K = 0269da924844fcc5852ef6808f0e0bc9e851bd7b6b23d6e55d4b152ab0101fedfd
The diff between N/2 and our point is 3ff6921a42c4b727ca4968504b9846
With a big enough hash table which we need to compute as we want (random/incremental), lets say we compute from 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681A20A0 to 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681B20A0 (small example) then our table will contain 0x10000 public keys, but in reality it is 0x20000 (double).
If we discard Y or parity from our K computation when we generate random steps (increments) we have a double change to get it right.
If our random increment difference was bigger than normal hash table size (but not bigger than double) than we will have a match and we can easily compute the right k from the known K.
Dont know if i was clear enough, if not i will try to illustrate it
But one quick question, if we push as you said and the diff between N/2 and our point is 3ff6921a42c4b727ca4968504b9846, would that be big enough to solve since 120 starting range is
800000000000000000000000000000? 120s pubkey added to 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681B20A0 should at least be a difference of 800000000000000000000000000000, right?
Sure,
I will just give a simple example.
Lets say you get the key close to the N half, lets take for example a random point 7fffffffffffffffffffffffffffffff5d1777e13d618b66b81ee5de17cf885a (we pushed for example 120 puzzle to this k) which results in
K = 0269da924844fcc5852ef6808f0e0bc9e851bd7b6b23d6e55d4b152ab0101fedfd
The diff between N/2 and our point is 3ff6921a42c4b727ca4968504b9846
With a big enough hash table which we need to compute as we want (random/incremental), lets say we compute from 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681A20A0 to 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681B20A0 (small example) then our table will contain 0x10000 public keys, but in reality it is 0x20000 (double).
If we discard Y or parity from our K computation when we generate random steps (increments) we have a double change to get it right.
If our random increment difference was bigger than normal hash table size (but not bigger than double) than we will have a match and we can easily compute the right k from the known K.
Dont know if i was clear enough, if not i will try to illustrate it
I will just give a simple example.
Lets say you get the key close to the N half, lets take for example a random point 7fffffffffffffffffffffffffffffff5d1777e13d618b66b81ee5de17cf885a (we pushed for example 120 puzzle to this k) which results in
K = 0269da924844fcc5852ef6808f0e0bc9e851bd7b6b23d6e55d4b152ab0101fedfd
The diff between N/2 and our point is 3ff6921a42c4b727ca4968504b9846
With a big enough hash table which we need to compute as we want (random/incremental), lets say we compute from 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681A20A0 to 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681B20A0 (small example) then our table will contain 0x10000 public keys, but in reality it is 0x20000 (double).
If we discard Y or parity from our K computation when we generate random steps (increments) we have a double change to get it right.
If our random increment difference was bigger than normal hash table size (but not bigger than double) than we will have a match and we can easily compute the right k from the known K.
Dont know if i was clear enough, if not i will try to illustrate it
Another method instead of pubkey reducing is actually increasing it.
Lets not forget about curve properties and how it can actually help us solve a puzzle a little faster.
if you manage to push the K to the right limit in the curve "middle" 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681B20A0 than it will be twice faster.
What i mean with twice faster is that we can try to match a K without parity
If half = 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681B20A0 than all other points after will generate same X but with inverse parity.
Basically it's easier to push the key to the middle and increment it until a match is found because of this property.
Just my 2 cents
I'm tracking, but can you give an example? Lets not forget about curve properties and how it can actually help us solve a puzzle a little faster.
if you manage to push the K to the right limit in the curve "middle" 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681B20A0 than it will be twice faster.
What i mean with twice faster is that we can try to match a K without parity
If half = 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681B20A0 than all other points after will generate same X but with inverse parity.
Basically it's easier to push the key to the middle and increment it until a match is found because of this property.
Just my 2 cents
Another method instead of pubkey reducing is actually increasing it.
Lets not forget about curve properties and how it can actually help us solve a puzzle a little faster.
if you manage to push the K to the right limit in the curve "middle" 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681B20A0 than it will be twice faster.
What i mean with twice faster is that we can try to match a K without parity
If half = 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681B20A0 than all other points after will generate same X but with inverse parity.
Basically it's easier to push the key to the middle and increment it until a match is found because of this property.
Just my 2 cents
Lets not forget about curve properties and how it can actually help us solve a puzzle a little faster.
if you manage to push the K to the right limit in the curve "middle" 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681B20A0 than it will be twice faster.
What i mean with twice faster is that we can try to match a K without parity
If half = 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681B20A0 than all other points after will generate same X but with inverse parity.
Basically it's easier to push the key to the middle and increment it until a match is found because of this property.
Just my 2 cents
Quote
in my view , we are talking off topic here , as my posts were deleted in past at this forum, and admin/mod and creator of this thread, reported, offtopic discus not allowed, maybe we get this warning again, till that some one wakeup, seems at all thread , bitcrack, 100 btc puzzle, 32 btc puzzle, all people sleeping, not fruit full posts, rear time post only left with newbie for just asking, how to install bitcrack, whats speed getting , Q etc
love to stay silent, before some one again report and post delete
Start a new thread...or I canlove to stay silent, before some one again report and post delete
Too late.... I already did.
Quote
in my view , we are talking off topic here , as my posts were deleted in past at this forum, and admin/mod and creator of this thread, reported, offtopic discus not allowed, maybe we get this warning again, till that some one wakeup, seems at all thread , bitcrack, 100 btc puzzle, 32 btc puzzle, all people sleeping, not fruit full posts, rear time post only left with newbie for just asking, how to install bitcrack, whats speed getting , Q etc
love to stay silent, before some one again report and post delete
Start a new thread...or I canlove to stay silent, before some one again report and post delete
Quote
in my view , we are talking off topic here , as my posts were deleted in past at this forum, and admin/mod and creator of this thread, reported, offtopic discus not allowed, maybe we get this warning again, till that some one wakeup, seems at all thread , bitcrack, 100 btc puzzle, 32 btc puzzle, all people sleeping, not fruit full posts, rear time post only left with newbie for just asking, how to install bitcrack, whats speed getting , Q etc
love to stay silent, before some one again report and post delete
Start a new thread...or I can love to stay silent, before some one again report and post delete
Quote
btw i can offer 115bit range only 1 key, for make this key i need 0.75btc for buying 3090rtx gpu's for my calc fast to within 7 days Smiley, but here no one have this for work of months to days Smiley
For me, to even consider it, I'd have to know your method/how you cut it to 115 with only 1 key. I don't doubt you can, just never seen it.Best I did once was to cut it down to 108 bit, with 1 out of 2048 keys being in the range.
And couldn't do it for less than .50 btc
I guess we could see a mini-business appearing in the future of people selling reduced puzzle transaction keys, but this is obviously a very risky business too in the same way that bought wallet.dat brute-forcing is also high risk (and usually downright scam).
for 20 bit down = 1024*1024 = 1048576 pubkeys
~
1048576 and 1073741824 pubkeys with each other addition and mutiplication will return you 260 pubkeys apear where 16 pubkeys sure inside 10 bit down from main pubkey
these 260 pubkeys again played for get 30 bit down for 1/720 pubkeys
~
1048576 and 1073741824 pubkeys with each other addition and mutiplication will return you 260 pubkeys apear where 16 pubkeys sure inside 10 bit down from main pubkey
these 260 pubkeys again played for get 30 bit down for 1/720 pubkeys
I don't get this tip. When I tried to shift down #120 by 20 bit I was looking at 2^20 total pubkeys generated from this. How do you manage to make do with only 260 or 720 of them? That's even less than the 1024 pubkeys I obtained from shifting 10 bits down.
With brainless, who knows, he is a wizard at pubkey/range reduction.
love to stay silent, before some one again report and post delete
Quote
btw i can offer 115bit range only 1 key, for make this key i need 0.75btc for buying 3090rtx gpu's for my calc fast to within 7 days Smiley, but here no one have this for work of months to days Smiley
For me, to even consider it, I'd have to know your method/how you cut it to 115 with only 1 key. I don't doubt you can, just never seen it.Best I did once was to cut it down to 108 bit, with 1 out of 2048 keys being in the range.
And couldn't do it for less than .50 btc
I guess we could see a mini-business appearing in the future of people selling reduced puzzle transaction keys, but this is obviously a very risky business too in the same way that bought wallet.dat brute-forcing is also high risk (and usually downright scam).
for 20 bit down = 1024*1024 = 1048576 pubkeys
~
1048576 and 1073741824 pubkeys with each other addition and mutiplication will return you 260 pubkeys apear where 16 pubkeys sure inside 10 bit down from main pubkey
these 260 pubkeys again played for get 30 bit down for 1/720 pubkeys
~
1048576 and 1073741824 pubkeys with each other addition and mutiplication will return you 260 pubkeys apear where 16 pubkeys sure inside 10 bit down from main pubkey
these 260 pubkeys again played for get 30 bit down for 1/720 pubkeys
I don't get this tip. When I tried to shift down #120 by 20 bit I was looking at 2^20 total pubkeys generated from this. How do you manage to make do with only 260 or 720 of them? That's even less than the 1024 pubkeys I obtained from shifting 10 bits down.
With brainless, who knows, he is a wizard at pubkey/range reduction.
" I got it down to 104 bits today, but with 32,000 pubkeys; better than the normal 2^16 normally required, but I can't figure out a way to shrink it down to one key... "
for 10 bit down = 1024 pubkeys
for 20 bit down = 1024*1024 = 1048576 pubkeys
for 30 bit down = 1024*1024*1024 = 1073741824 pubkeys
1048576 and 1073741824 pubkeys with each other addition and mutiplication will return you 260 pubkeys apear where 16 pubkeys sure inside 10 bit down from main pubkey
these 260 pubkeys again played for get 30 bit down for 1/720 pubkeys
now you can start to find with above tip
I'll try to digest and understand what you are saying/doing to shrink the range with limited pubkeys...intersting for 10 bit down = 1024 pubkeys
for 20 bit down = 1024*1024 = 1048576 pubkeys
for 30 bit down = 1024*1024*1024 = 1073741824 pubkeys
1048576 and 1073741824 pubkeys with each other addition and mutiplication will return you 260 pubkeys apear where 16 pubkeys sure inside 10 bit down from main pubkey
these 260 pubkeys again played for get 30 bit down for 1/720 pubkeys
now you can start to find with above tip
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