On a background 545 years an error per 0,5 years to make all any +/- 0,1 percents
Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 300. (Read 240757 times)
If I calculated everything correctly, then the space from 2000000000000001 to 4000000000000001 can be scanned
in about 545 years if the average key generation rate is about 134 Mkey / s
Confirm or disprove my calculations.
Looks like something is wrong in your calculation.in about 545 years if the average key generation rate is about 134 Mkey / s
Confirm or disprove my calculations.
I think should be like this:
134000000 keys/s * 60 sec * 60 min * 24 hours = 11577600000000 keys/day
(4000000000000001 - 2000000000000001) / 11577600000000 = 172.75 days
Same, correct me pls if I'm wrong.
The space is hexadecimal!
2000000000000000 hex = 2305843009213693952 decimal
134000000 Keys/ sec
199164.1626255609065782 days
545.6552400700298810362 years
Then yes, you are right, it will take just only ~545 years.
But there is also a good news: imagine what will be the price of BTC after 545 years!!!
If I calculated everything correctly, then the space from 2000000000000001 to 4000000000000001 can be scanned
in about 545 years if the average key generation rate is about 134 Mkey / s
Confirm or disprove my calculations.
Looks like something is wrong in your calculation.in about 545 years if the average key generation rate is about 134 Mkey / s
Confirm or disprove my calculations.
I think should be like this:
134000000 keys/s * 60 sec * 60 min * 24 hours = 11577600000000 keys/day
(4000000000000001 - 2000000000000001) / 11577600000000 = 172.75 days
Same, correct me pls if I'm wrong.
The space is hexadecimal!
2000000000000000 hex = 2305843009213693952 decimal
134000000 Keys/ sec
199164.1626255609065782 days
545.6552400700298810362 years
If I calculated everything correctly, then the space from 2000000000000001 to 4000000000000001 can be scanned
in about 545 years if the average key generation rate is about 134 Mkey / s
Confirm or disprove my calculations.
Looks like something is wrong in your calculation.in about 545 years if the average key generation rate is about 134 Mkey / s
Confirm or disprove my calculations.
I think should be like this:
134000000 keys/s * 60 sec * 60 min * 24 hours = 11577600000000 keys/day
(4000000000000001 - 2000000000000001) / 11577600000000 = 172.75 days
Same, correct me pls if I'm wrong.
If I calculated everything correctly, then the space from 2000000000000001 to 4000000000000001 can be scanned
in about 545 years if the average key generation rate is about 134 Mkey / s
Confirm or disprove my calculations.
in about 545 years if the average key generation rate is about 134 Mkey / s
Confirm or disprove my calculations.
Ten years ago, Satoshi and Craig created a website where just like that from donations distributed 5 bitcoins to all those in need,and now some 0.5 bitcoins would solve my almost all problems.
I join my friend and say if God in the world let him drip on my purse incentive prize for the promotion of cryptocurrency to the masses!!! 1HqqeH1VWDY9vJq8nF4C7BHQ3Yvzxak48V
Just so you know: begging is not allowed on this forum and you can actually be banned for begging. Not sure if this is begging but be careful. I join my friend and say if God in the world let him drip on my purse incentive prize for the promotion of cryptocurrency to the masses!!! 1HqqeH1VWDY9vJq8nF4C7BHQ3Yvzxak48V
...the puzzle above 70 in the next fifty years will not be solved...
We have to try continue here https://cloud.google.com/tpu/ Why did 70, 75 and 80 move?
And 85 too ....
Go back a few posts ...
Why did 70, 75 and 80 move?
Sometimes, when I look at clouds, I see that they look kind of like animals. I have seen a cloud that looks like a rabbit. One time I saw this cloud that looked very much like an elephant. I have seen many various animals in clouds. I believe that if you look hard enough at any cloud you will see that it looks like some sort of animal.
You too?! Oh mate, I thought that I'm the only one who see it . Just yesterday I saw a horse-cloud, that was beautiful!
I decided to drive the statistics.
Calculate the partial repetition of a part of a public address, starting from the end or from the beginning. Bit repetitions.
I.e
10101010111......010101010111010101010
01011010100......010101001011010101010
Match the end to 11 bits.
1101011011011010101010111......010101010110101010101
1101011011011001011010100......010101001010010101010
Match from the beginning to 14 bits.
Private keys get over in a row. And looking for repetition in a public address. When repetition is found, the difference in the private key is calculated.
At the moment, found such a minimum range.
Matching 19 or less bits are found with ranges of less than 128 keys.
Of course, a jump (pass) to 380k keys, at a search speed of 1M per second, is nothing. But finding large ranges, you can make big jumps.
Sometimes, when I look at clouds, I see that they look kind of like animals. I have seen a cloud that looks like a rabbit. One time I saw this cloud that looked very much like an elephant. I have seen many various animals in clouds. I believe that if you look hard enough at any cloud you will see that it looks like some sort of animal. Calculate the partial repetition of a part of a public address, starting from the end or from the beginning. Bit repetitions.
I.e
10101010111......010101010111010101010
01011010100......010101001011010101010
Match the end to 11 bits.
1101011011011010101010111......010101010110101010101
1101011011011001011010100......010101001010010101010
Match from the beginning to 14 bits.
Private keys get over in a row. And looking for repetition in a public address. When repetition is found, the difference in the private key is calculated.
At the moment, found such a minimum range.
Code:
24 bits: 2970 * 128 = 380160 (+-128) pkeys
23 bits: 330 * 128 = 42240 (+-128) pkeys
22 bits: 33 * 128 = 4244 (+-128) pkeys
21 bits: 11 * 128 = 1408 (+-128) pkeys
20 bits: 10 * 128 = 1280 (+-128) pkeys
Of course, a jump (pass) to 380k keys, at a search speed of 1M per second, is nothing. But finding large ranges, you can make big jumps.
I decided to drive the statistics.
Calculate the partial repetition of a part of a public address, starting from the end or from the beginning. Bit repetitions.
I.e
10101010111......010101010111010101010
01011010100......010101001011010101010
Match the end to 11 bits.
1101011011011010101010111......010101010110101010101
1101011011011001011010100......010101001010010101010
Match from the beginning to 14 bits.
Private keys get over in a row. And looking for repetition in a public address. When repetition is found, the difference in the private key is calculated.
At the moment, found such a minimum range.
Matching 19 or less bits are found with ranges of less than 128 keys.
Of course, a jump (pass) to 380k keys, at a search speed of 1M per second, is nothing. But finding large ranges, you can make big jumps.
Calculate the partial repetition of a part of a public address, starting from the end or from the beginning. Bit repetitions.
I.e
10101010111......010101010111010101010
01011010100......010101001011010101010
Match the end to 11 bits.
1101011011011010101010111......010101010110101010101
1101011011011001011010100......010101001010010101010
Match from the beginning to 14 bits.
Private keys get over in a row. And looking for repetition in a public address. When repetition is found, the difference in the private key is calculated.
At the moment, found such a minimum range.
Code:
24 bits: 2970 * 128 = 380160 (+-128) pkeys
23 bits: 330 * 128 = 42240 (+-128) pkeys
22 bits: 33 * 128 = 4244 (+-128) pkeys
21 bits: 11 * 128 = 1408 (+-128) pkeys
20 bits: 10 * 128 = 1280 (+-128) pkeys
Of course, a jump (pass) to 380k keys, at a search speed of 1M per second, is nothing. But finding large ranges, you can make big jumps.
Hello to everyone,
62. Wallet Hex Code Start: Starting with 2F (B, D, E) ......
62. Wallet Hex Code Start: Starting with 2F (B, D, E) ......
Based on what?
Hello to everyone,
62. Wallet Hex Code Start: Starting with 2F (B, D, E) ......
62. Wallet Hex Code Start: Starting with 2F (B, D, E) ......
I can assure you at this stage that you missed the right range :-)
Cracking bitcoin private key would be near to impossible. The wallets are unique and if it is a cold storage though one can still have the difficulties in accessing it for bitcoin has stored in that hardware or cold storage where one could not access if it is not doing online activities. So, forget that reward OP and start earning btc in a decent way like joining a bounty campaign or investing btc with your hard earn money.
Hello to everyone,
62. Wallet Hex Code Start: Starting with 2F (B, D, E) ......
62. Wallet Hex Code Start: Starting with 2F (B, D, E) ......
Why ? How ?
...the puzzle above 70 in the next fifty years will not be solved...
We have to try continue here https://cloud.google.com/tpu/ I will answer in Russian.
Кaк бyдeм дeлить нe yбитoгo кaбaнчикa?
Bceм пo 0.1BTC Кaк бyдeм дeлить нe yбитoгo кaбaнчикa?
On first machine (600GB RAM):
On second machine (80GB RAM):
... and nothing changes from two days on both machines.
Something is wrong?
Am I to understand that no more changes are expected in these processes?
Can someone modify the code in such a way that the process is visible (along with the percentage)?
for machine's memory is 600GBCode:
Check bit = 90 only, pubkey is:
035c38bd9ae4b10e8a250857006f3cfd98ab15a6196d9f4dfd25bc7ecc77d788d5
Build Hash, MEM size = 384GB
On second machine (80GB RAM):
Code:
Check bit = 90 only, pubkey is:
035c38bd9ae4b10e8a250857006f3cfd98ab15a6196d9f4dfd25bc7ecc77d788d5
Build Hash, MEM size = 64GB
Search bits = 90
Search Keys.... from 20000000000000000000000 to 3ffffffffffffffffffffff
... and nothing changes from two days on both machines.
Something is wrong?
Am I to understand that no more changes are expected in these processes?
Can someone modify the code in such a way that the process is visible (along with the percentage)?
at least this code need update:
uint32_t entry = xst.n[0] & (HASH_SIZE-1);
to
uint64_t entry = xst.n[0] & (HASH_SIZE-1);
and update
typedef struct hashtable_entry {
uint32_t x;
uint32_t exponent;
} hashtable_entry;
to
typedef struct hashtable_entry {
uint32_t x;
uint64_t exponent; //this will take more memory
} hashtable_entry;
For 5TB RAM changes are that same how for 600GB? :-)
Кaк ? ктo нaйдeт тoт и зaбepeт нeмнoгo ocтaлocь
нeмнoгo ocтaлocь Jump to: