Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 259. (Read 229433 times)

No it can not. The random number generator gathers environmental noise from device drivers and other sources into an entropy pool. That's the beauty of the entropy pool, values are not calculated, instead they are measured, emulating what hardware random number generators do, best and expensive hardware generators measure radiation sources which are totally random. Since you do not hardware random generator in normal PC/server, usually temperature variations are used instead.

I have created a pool of sorts to collectively solve bit 64 of this challenge.

if you goto http://www.ttdsales.com/64bit/

create an account and you will get access to the user and stats page.  From there you can download the client software. 

It works with bitcrack and the server to hand out work and keep track of what has been checked.

Each range work load is 000000000 - FFFFFFFFF or 68,719,476,735 keys.  I selected this range because it does not complete to quickly on top hardware and is still easily completed in less then an hour with slower GPUs.

8000000 - FFFFFFF is generated by the server randomly.  There is a btc address I have generated for all 134,217,728 ranges in the DB.  The private key for these addresses are used as proof of work for the range work loads.  When the client software finishes a range and submits this proof the DB marks that range as checked and dispenses another.  Before being dispensed the range is checked against the DB to ensure it hasn't already been checked. What I am calling a random collision.

All work completed is kept track of for each user and displayed on your user stats page.

If/when we find the solution the bitcoin stored in 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN will be split between the contributing users equally according to work completed just like a bitcoin mining pool would for a mined block.  This percentage is displayed on your stats page.

Please check it out and pm me with any thoughts comments questions or ideas for improvements.

It is a work in progress but seems to be functioning nicely for my 6 machines that have been running on it for a couple of weeks now.  Currently running ~1.37 Billion Keys /s.

Can entropy formula use for calculate position zone of puzzle

He almost certainly used some env with good entropy, he obviously knew what he was doing. Zero chance that these random numbers have a weakness on PRNG side.

Recently mentioned this shit https://bitcointalksearch.org/topic/collection-of-18509-found-and-used-brainwallets-4768828 , sha256 brainwallet brut... take any information https://github.com/minimaxir/textgenrnn convert in sha... or any other hashiers (including several times) and in search of treasures. On mobile can used "Pydroid 3 - IDE for Python 3" https://play.google.com/store/apps/details?id=ru.iiec.pydroid3&hl=en_IN but you have to look at the possibility of the device. To appeal, save it is necessary to write there "/sdcard/..."
 

I found hack page or may be webmaster hide page (may be scam)

this page random bitcoin address generator

www.50pansa.info/random.html

so, have any brute force address on java script

I not sure this script is clean or not, just fine to safety file on github

(however it is lowly)

Can possible anybody help to modify to brute force ?
it can run on mobile easy

import random
from bit import Key
import time
import itertools

for X in range(1,10000000000000):
    R=X
    #print(R)
    random.seed(R)

    list = ["16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN","13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so","1BY8GQbnueYofwSuFAT3USAhGjPrkxDdW9","18ZMbwUFLMHoZBbfpCjUJQTCMCbktshgpe"]



            #1       2        3       4       5       6       7       8       9      10

    a0=   [(0, 0)]                                                                         # 1
    a1=   [(1, 0), (0, 1)]                                                                 # 2
    a2=   [(2, 0), (1, 1), (0, 2)]                                                         # 3
    a3=   [(3, 0), (2, 1), (1, 2), (0, 3)]                                                 # 4
    a4=   [(4, 0), (3, 1), (2, 2), (1, 3), (0, 4)]                                         # 5
    a5=   [(5, 0), (4, 1), (3, 2), (2, 3), (1, 4), (0, 5)]                                 # 6
    a6=   [(6, 0), (5, 1), (4, 2), (3, 3), (2, 4), (1, 5), (0, 6)]                         # 7
    a7=   [(7, 0), (6, 1), (5, 2), (4, 3), (3, 4), (2, 5), (1, 6), (0, 7)]                 # 8
    a8=   [(8, 0), (7, 1), (6, 2), (5, 3), (4, 4), (3, 5), (2, 6), (1, 7), (0, 8 )]         # 9
    a9=   [(9, 0), (8, 1), (7, 2), (6, 3), (5, 4), (4, 5), (3, 6), (2, 7), (1, 8 ), (0, 9)] # 10
    a10=  [(9, 1), (8, 2), (7, 3), (6, 4), (5, 5), (4, 6), (3, 7), (2, 8 ), (1, 9)]         # 9
    a11=  [(9, 2), (8, 3), (7, 4), (6, 5), (5, 6), (4, 7), (3, 8 ), (2, 9)]                 # 8
    a12=  [(9, 3), (8, 4), (7, 5), (6, 6), (5, 7), (4, 8 ), (3, 9)]                         # 7
    a13=  [(9, 4), (8, 5), (7, 6), (6, 7), (5, 8 ), (4, 9)]                                 # 6
    a14=  [(9, 5), (8, 6), (7, 7), (6, 8 ), (5, 9)]                                         # 5
    a15=  [(9, 6), (8, 7), (7, 8 ), (6, 9)]                                                 # 4
    a16=  [(9, 7), (8, 8 ), (7, 9)]                                                         # 3
    a17=  [(9, 8 ), (8, 9)]                                                                 # 2
    a18=  [(9, 9)]                                                                         # 1


    while True:

        v1 = random.choice([a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15,a16,a17,a18]) #[a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15,a16,a17,a18]
        v2 = random.choice([a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15,a16,a17,a18])
        v3 = random.choice([a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15,a16,a17,a18])
        v4 = random.choice([a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15,a16,a17,a18])
        v5 = random.choice([a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15,a16,a17,a18])
        v6 = random.choice([a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15,a16,a17,a18])
        v7 = random.choice([a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15,a16,a17,a18])
        v8 = random.choice([a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15,a16,a17,a18])
        v9 = random.choice([a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15,a16,a17,a18])
        v10= random.choice([a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15,a16,a17,a18])

        if v1 == a3:
            if v2 == a11:
                if v3 == a11:
                    if v4 == a14:
                        if v5 == a4:
                            if v6 == a2:
                                if v7 == a10:
                                    if v8 == a2:
                                        if v9 == a10:
                                            if v10 == a10:

                                                print(len(v1),v1)
                                                print(len(v2),v2)
                                                print(len(v3),v3)
                                                print(len(v4),v4)
                                                print(len(v5),v5)
                                                print(len(v6),v6)
                                                print(len(v7),v7)
                                                print(len(v8),v8)
                                                print(len(v9),v9)
                                                print(len(v10),v10)

                                                ur = len(v1)*len(v2)*len(v3)*len(v4)*len(v5)*len(v6)*len(v7)*len(v8)*len(v9)*len(v10)
                                                print(len(v1),len(v2),len(v3),len(v4),len(v5),len(v6),len(v7),len(v8),len(v9),len(v10),"=",ur)
                                                time.sleep(3.0)
                                                count = 0
                                                
                                                if ur >=       1000000:
                                                    if ur <= 150000000:
                                                        
                                                        for x in itertools.product(v1, v2, v3, v4, v5, v6, v7, v8, v9, v10):
                                                            count += 1
                                                            d = ("".join(map(str, (item for sublist in x for item in sublist))))

                                                            ran = int(d)
                                                            key1 = Key.from_int(ran)
                                                            addr1 = key1.address
                                                                                                                                    
                                                            if addr1 in list:

                                                                print (ran,"found!!!")

                                                                s5 = str(ran)
                                                                f=open(u"C:/a.txt","a")
                                                                f.write(s5 + '\n')
                                                                f.close()

                                                                break

                                                            else:
                                                                                                                                            
                                                                #pass
                                                                print(R,len(v1),len(v2),len(v3),len(v4),len(v5),len(v6),len(v7),len(v8),len(v9),len(v10),"=",ur,count,ran,len(d),addr1) #("all steps",lenn,count,ran,len(d),addr1)

        break


 

How to save in a text file all the pubkey passed until reaching the pubkey of origin?

Start: 8000000000000000
End:   ffffffffffffffff

Pubkey: 0330de2c8bc2010aaebbb647c5bac00eb8028f78d795f2cd4532bc6c504c0e01e7

how to save all MK / s FOUND to a txt file?

check out the image:

How to save in a text file all the pubkey passed until reaching the pubkey of origin?

Start: 8000000000000000
End:   ffffffffffffffff

Pubkey: 0330de2c8bc2010aaebbb647c5bac00eb8028f78d795f2cd4532bc6c504c0e01e7

how to save all MK / s FOUND to a txt file?

What for. You still do not have time to catch anything faster others with rigs. All stay at 120 bits.

***

Continuation of the blind blindbirthday...

Choose (random.choice 2x) from a set (00-99) with a fixed seed()

Find

seed(3435)    0000
seed(108918) 0001
seed(4327)    0002
etc..

from 0000 to 9999 10000 similarly 10000 different seed 3435, 108918, 4327, 20723, 9991, 10068, 8175, 10495, 23048, 5511, 3085 etc... seed [2532, 679, 4652, 2530, 620] pz 3056 8377 3120 6420 2855

But we have an unlimited number of numbers for fixed results


seed 2532 pz 3056
seed 3801 pz 3056
seed 4476 pz 3056
seed 6363 pz 3056
etc...
seed 1000000000974 pz 3056
seed 1000000006200 pz 3056
seed 1000000072946 pz 3056
seed 1000000074588 pz 3056
etc...

Are there patterns in the units of steps seed or rand?...

But as in the case of a black hole (2^...)

Here is a wormhole

When increasing the number pz will grow and seed

seed            3056
seeed          30568
seeeeed       305683
seeeeeeed    3056837
seeeeeeeed  30568377 etc

and these seed for numbers 10 or 20 digits long too repeated.

If we choose 2 samples for 4 len num by 1 seed, we can choose by 1 seed from 10 num len set 0000000000-9999999999 for pz 20 len (the question is how much it will occupy the ram of the system resources)

Please if it is possible for us to release the pubkeys from the other addresses that are missing. We are grateful! United We Are One!

Donate: 1C6wmXbktMA1peHMDbDRShAhJ6Wd8Mypw8

I am the creator.

You are quite right, 161-256 are silly.  I honestly just did not think of this.  What is especially embarrassing, is this did not occur to me once, in two years.  By way of excuse, I was not really thinking much about the puzzle at all.

I will make up for two years of stupidity.  I will spend from 161-256 to the unsolved parts, as you suggest.  In addition, I intend to add further funds.  My aim is to boost the density by a factor of 10, from 0.001*length(key) to 0.01*length(key).  Probably in the next few weeks.  At any rate, when I next have an extended period of quiet and calm, to construct the new transaction carefully.

A few words about the puzzle.  There is no pattern.  It is just consecutive keys from a deterministic wallet (masked with leading 000...0001 to set difficulty).  It is simply a crude measuring instrument, of the cracking strength of the community.

Finally, I wish to express appreciation of the efforts of all developers of new cracking tools and technology.  The "large bitcoin collider" is especially innovative and interesting!

The puzzle was solved, wasn't it? Did someone get a prize?

i found this blindbirthday thingie is this a bit what you meant andzhig