Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 272. (Read 208030 times)

@arulbero
I have an idea about #58. Can I message you the details?

American or British system ? -> https://www.quadrillion.com/number.shtm

quadrillion = 10^15 -> 10 quadrillion = 10^16?  10^16 = 2^53  -->  2 lists of 2^26 keys = 2 * 2^2 = 8 seconds  

quadrillion = 10^24 -> 10 quadrillion = 10^25?  10^25 = 2^83  --> 2 lists of 2^41 keys ? NO! Too much for my RAM!


Let's say that the max list size (for my RAM) is 2^28 keys.
 
To retrieve the #59, I need to compute 2 list of 2^29, I have to split the first in 2 * 2^28, then generate the 2^29 keys of the second list 2 times against each half of the first,  tot: 2*2^28 + 2*2^29 = 3*2^29 instead of 2*2^29 (96 seconds instead of 64 seconds).

To retrieve the #61, I need to compute 2 list of 2^30, I have to split the first in 4 * 2^28, then generate the 2^30 keys of the second list 4 times against each quarter of the first list,  tot: 4*2^28 + 4*2^30 = 5*2^30 instead of 2*2^30 (320 seconds instead of 128 seconds).

To retrieve the #63, I need to compute 2 list of 2^31, I have to split the first in 8 * 2^28, then generate the 2^31 keys of the second list 8 times against each part of the first list,  tot: 8*2^28 + 8*2^31 = 9*2^31 instead of 2*2^31 (1152 seconds instead of 256 seconds).

To retrieve the #65, I need to compute 2 list of 2^32, I have to split the first in 16 * 2^28, then generate the 2^32 keys of the second list 16 times against each part of the first list,  tot: 16*2^28 + 16*2^32 = 17*2^32 instead of 2*2^32 (4352 seconds instead of 512 seconds).

Time is for the #(57+x) = (2^(x/2)+1) lists of 2^((57+x-1)/2) keys instead of only 2 lists of 2^(57+x-1)/2 keys


For the #61 = #(57+4) = (2^2+1) * (2^30) = (2^2  + 1) lists of 2^30  -> 5 * 2^6 seconds = about 320 seconds = 5 min 20 sec

For the #71 = #(57+14) = (2^7+1) * (2^35) = (2^7  + 1) lists of 2^35  -> 2^7 * 2^11 seconds = about 2^18 seconds = 73 hours

For the #83 = #(57+26) = (2^13+1) * (2^82/2) = (2^13  + 1) lists of 2^41  -> 2^13 * 2^17 seconds = about 2^30 seconds = 34 years  (if I had enough RAM, 2 * 2^41 = 2^42 keys = 2^18 seconds = 73 hours !!! I would need a RAM:  2^13  = 8000 times 32 GB)

anyone notice the other transactions on #57? how on earth did they do that? Those are hellafied vanity addresses.
edit: Sorry pre coffee me.. someone just sent to those addresses .. had me freaking out for a sec.. like oh man if they
can generate vanity addresses that long that fast they pretty much own all the btc... *panic mode engaged*
Then *doh! eureka moment*  NOW it's funny.

Could you please show us how to change the code so it can search only range 2^56 to 2^57 - 1?

Have you tried to search a key in space of 10 quadrillion? How many hours you think it can finish that large space?

Nope
Somehow Mr  1AqEgLuT4V2XL2yQ3cCzjMtu1mXtJLVvww has given some BTC (0.1) to 1LqJ9cHPKxPXDRia4tteTJdLXnisnfHsof   I wonder why

https://www.blockchain.com/btc/tx/c1c8a3c8c9dccc3dbbc470e294a0610a2bc2e22c41f9a0c2a06ef5dc1d6d4480

Of course.
Many blocks mined by Satoshi have txs with "pay to public key" script (P2PK) instead of "pay to public key hash" script (P2PKH, pay to address).

The public keys are known, but the btc are still there:

block #100

https://www.blockchain.com/it/btc/tx/2d05f0c9c3e1c226e63b5fac240137687544cf631cd616fd34fd188fc9020866

PUSHDATA(65)[04e70a02f5af48a1989bf630d92523c9d14c45c75f7d1b998e962bff6ff9995fc5bdb44f1793b3749 5d80324acba7c8f537caaf8432b8d47987313060cc82d8a93] CHECKSIG

@arulbero

If the public key is revealed it is still safe from bruteforce if the attacker don't know the range of bits to search for?

Is this puzzle still available?and remain unsolved thats why the answer keeps on coming even this thread has been posted three years ago?
Anyway this has been a big prize if the 32bitcoin remain as the bounty for the puzzle,no wonder how much the value for now of bitcoin

That works only if you have enough RAM to store 2^28 keys. Otherwise that program cannot retrieve #57.

Besides if you want to retrieve only #57 and you don't modify the code, it starts always from 1 to 2^57 - 1 (instead from 2^56 to 2^57 - 1)

I you try this code (https://gist.github.com/jhoenicke/2e39b3c6c49b1d7b216b8626197e4b89) and want to get all puzzle private keys up to #57 , you need to change
 giant steps up to  2^28

#define GSTEP (1<<28)

otherwise it stops at #52

@arulbero thanks mate!

Because 2^57 has 58 digits (in binary form '1' + 57 '0'). Then 2^57 could be the key #58, not #57

key 1 -> 1 bit :  1  ( from 2^0 to 2^1 - 1)

key 2 -> 2 bit :  from 2 to 3 (from 2^1 to 2^2 - 1) : 10 or 11

key 3 -> 3 bit : from 4 to 7 (from 2^2 to 2^3 - 1) : 100 or 101 or 110 or 111

key 57 -> 57 bit : from 2^56 to 2^57 - 1

key 58 -> 58 bit : from 2^57 to 2^58 - 1

If you look at

key #3
.....0000000000000000000000000000000000000000000000000000000000000111 (to address 19ZewH8Kk1PDbSNdJ97FP4EiCjTRaZMZQA)

and at the #4
.....0000000000000000000000000000000000000000000000000000000000001000 (to address 1EhqbyUMvvs7BfL8goY6qcPbD6YKfPqb7e)

key #3 = 2^3 - 1 = 7
key #4 = 2^3      = 8

Btw why -1?

I don't know, did you test it?

Look at this thread:
https://bitcointalksearch.org/topic/m.47210876
https://bitcointalksearch.org/topic/m.47263144

Wow faster than libsecp256k1

Well it ain't newbie in fact he ranked up to member. Anyway, I do not see something relevant with the data presented. It maybe some sort of coincidence. Even if so BTC won't be decrypted that easily and for that the reward is nothing. I bet no one could manage to predict on or cracked the things with BTC.

About 16.7 MKeys/s, 1 GKeys/min.
 
Let's say 2^24 Keys/s.
To compute 2 lists of 2^28 keys, it takes 2 * 2^4 seconds, about 30 seconds. That is the time it took to retrieve the key #57.

In your code how many keys can you generate per minute?

Right!


You have to use a library tailored to elliptic computations.

The most famous is the libsecp256k1 written by Pieter Wuille (it is in Bitcoin Core).

These programs https://gist.github.com/jhoenicke/2e39b3c6c49b1d7b216b8626197e4b89 and https://github.com/klynastor/supervanitygen use this library.

My code instead uses a different library written by myself for the LBC project.