Author

Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 319. (Read 228650 times)

newbie
Activity: 10
Merit: 0
Andzhig, you should write your native language, because when you write in English I always get an impression that you have undertreated meningitis.

lol.  I'm glad I'm not the only one that was left just scratching my head.

Hey Holy_ship, can you check on the thread about your ndv vanitygen build?  I have a couple questions in there and I can't message you because I'm still a newbie on here.  Smiley
member
Activity: 228
Merit: 10
Hi guys,
In continuation to this thread: https://bitcointalksearch.org/topic/brute-force-on-bitcoin-addresses-video-of-the-action-1305887
While playing around with my bot, I found out this mysterious transaction:
https://blockchain.info/tx/08389f34c98c606322740c0be6a7125d9860bb8d5cb182c02f98461e5fa6cd15
those 32.896 BTC were sent to multiple addresses, all the private keys of those addresses seem to be generated by some kind of formula.
For example:
EDIT: If you find the solution feel free to leave a tip Smiley 1DPUhjHvd2K4ZkycVHEJiN6wba79j5V1u3
This is one of the things that I really love about this mysterious world of Bitcoin and cryptocurrency, you can always find something interesting in it, althought I am not going to resolve this puzzle because I am a dummy Tongue
newbie
Activity: 27
Merit: 0
looks really easy to crack

why would reward be so high
It is not easy to crack the bitcoin address. Bitcoin cryptography is using Elliptic Curve Cryptography (ECC) which it looks like has a pattern but it is not. It looks easy to crack but it is not. So, let them go a head and we see.
jr. member
Activity: 115
Merit: 1
Quote from: Andzhig
phrase ip addresses (sha256) but there is also not a small figure 4294967296.

4294967296 keys to check is job for 80 seconds. OK, if they are not sequential, let it be - 1 hour.  Grin

Andzhig, you should write your native language, because when you write in English I always get an impression that you have undertreated meningitis.
jr. member
Activity: 184
Merit: 3


Andzhig

Пpивeтcтвyю ,интepecнa тeмa взлoмa бтк ,ecть нapaбoтки
нaпиши пoчтy в личкy -oбcyдим

Thank you, from me a cracker like Harry Dunn plumber in the movie "Dumb & Dumber" when he came to the house of a woman. Specially for you there is an idea scan through a phrase ip addresses (sha256) but there is also not a small figure 4294967296.

The LBC project[/url] is currently running at 346.02 Mkeys/s

Actually LBC is still checking the 2^55 range. So even pasting numbers in http://brain.evilbs.com/ by hand is faster than LBC  Grin

You are your "fork" in several 1080, 1070 as yet apparently also did not catch wildfowl. From where to have already gone or left this venture? I'm your program .. 49 .., 57 .., 58 ..., 59 ..., 60..., scanned in scatter, uselessly, can he be between 66-72 (30045390491869460+36028797018963968=66074187510833428 < 72057594037927936) who knows, as in step 611140496167764+1125899906842624=1737040403010388 < 2058769515153876 < 2251799813685248.
jr. member
Activity: 115
Merit: 1
The LBC project[/url] is currently running at 346.02 Mkeys/s

Actually LBC is still checking the 2^55 range. So even pasting numbers in http://brain.evilbs.com/ by hand is faster than LBC  Grin
jr. member
Activity: 115
Merit: 1
Python is not hipsterish enough!
True mama's cryptokiddies choose PHP parser of directory.io
And adding right magic constant to skip "unnecessary" pages is vital!

So, first you should look at you mom's Crystal Ball to find the magic number!
newbie
Activity: 14
Merit: 0


Andzhig

Пpивeтcтвyю ,интepecнa тeмa взлoмa бтк ,ecть нapaбoтки
нaпиши пoчтy в личкy -oбcyдим
jr. member
Activity: 184
Merit: 3
Yes it is clear that compared to LBC this is a louse (but if desired and a louse a lion can gnaw) this is so for luck, suddenly there is a magic step for a random house. There are many options you can think of, and step through a number, sort only even, odd, take each 128,256,512 ... multiply them ... floating scan area, some magical jumps, etc.

Quote
from bitcoin import *
import random
import math

while True:
    f = random.random() + 15.1
    g = f
    b = 1152921504606846976 - ((((((((((g * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128)  - 1143914305352105984) * 128)
    c = math.trunc(b)
    ran = c
    myhex = "%064x" % ran
    myhex = myhex[:64]
    priv = myhex
    pub = privtopub(priv)
    pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed")
    addr = pubtoaddr(pubkey1)
    n = addr
    if n.strip() == "1HAX2n9Uruu9YDt4cqRgYcvtGvZj1rbUyt":
        print ("found!!!",addr,myhex)
        s1 = myhex
        s2 = addr

        f=open(u"C:/a.txt","a")
        f.write(s1)
        f.write(s2)       
        f.close()
        break
    else:
        print ("searching...",f,addr,myhex)
formula 1152921504606846976 - ((((((((((15.9 * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128)  - 1143914305352105984) * 128)

after 15.9xxxxxxxxxxxxxxxxxxxxxxxxxxx a random number of 29, 30 numbers can be sausage immediately at all after 7 steps of the puzzle.


Quote
1152921504606846976 - ((((((((((15,9075849271 * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128)  - 1143914305352105984) * 128)
147573952589676412928 - (((((((((((15,9041635330 * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128)* 128)  - 146421031085069565952) * 128)
18889465931478580854784 - ((((((((((((15,9058575788 * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128)* 128)* 128)  - 18741891978888904441856) * 128)
2417851639229258349412352 - (((((((((((((15,9058575788 * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128)* 128)* 128)* 128)  - 2398962173297779768557568) * 128)
309485009821345068724781056 - ((((((((((((((15,9081997978 * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128)* 128)* 128)* 128)* 128)  - 307067158182115810375368704) * 128)
39614081257132168796771975168 - (((((((((((((((15,9081997978 * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)  - 39304596247310823728047194112) * 128)
5070602400912917605986812821504 - ((((((((((((((((15,9095945490 * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)  - 5030988319655785437190040846336) * 128)
649037107316853453566312041152512 - (((((((((((((((((15,9095945490 * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)  - 643966504915940535960325228331008) * 128)
83076749736557242056487941267521536 - ((((((((((((((((((15,90 * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)  - 82427712629240388602921629226369024) * 128)
10633823966279326983230456482242756608 - (((((((((((((((((((15,90 * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)  - 10550747216542769741173968540975235072) * 128)
1361129467683753853853498429727072845824 - ((((((((((((((((((((15,90 * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)  - 1350495643717474526870267973244830089216) * 128)
174224571863520493293247799005065324265472 - (((((((((((((((((((((15,90 * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)  - 172863442395836739439394300575338251419648) * 128)
22300745198530623141535718272648361505980416 - ((((((((((((((((((((((15,90 * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)  - 22126520626667102648242470473643296181714944) * 128)
2854495385411919762116571938898990272765493248 - (((((((((((((((((((((((15,90 * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)  - 2832194640213389138975036220626341911259512832) * 128)
365375409332725729550921208179070754913983135744 - ((((((((((((((((((((((((15,90 * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)  - 362520913947313809788804636240171764641217642496) * 128)
legendary
Activity: 2646
Merit: 1137
All paid signature campaigns should be banned.

 ... some code ...


The LBC project is currently running at 346.02 Mkeys/s and recently had a peak of about 521 Mkeys/s

How many keys per second does your method check?
jr. member
Activity: 184
Merit: 3
Addresses are all compressed in the puzzle?

Quote
if you like to waste your time trying to find a pattern in random data
And you've been guessing for a few years now already all the methods have been tried all sorts, random programs generators... and without the slightest hint of luck. began to study "python" in pieces from Inet assembled a generator with a selectable range of randomization (until 1 step) and a hex output and a compressed key, It remains to teach how to compare the address lines generated with the puzzle addresses next or in the file and stop work.

***

in the general gemoro there full with jumps, who understands to supplement the code, but for now you can duplicate and run the sites ...


Quote
from bitcoin import *
import random

while True:
    ran = random.randrange(36028797018963968,72057594037927936)
    myhex = "%064x" % ran
    myhex = myhex[:64]
    priv = myhex
    pub = privtopub(priv)
    pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed")
    addr = pubtoaddr(pubkey1)
    n = addr
    if n.strip() == "17aPYR1m6pVAacXg1PTDDU7XafvK1dxvhi":
        print ("found!!!",addr,myhex)
        s1 = myhex
        s2 = addr

        f=open(u"C:/a.txt","a")
        f.write(s1)
        f.write(s2)      
        f.close()
        break
    else:
        print ("searching...",addr,myhex)


the heat went off http://meson.ad-l.ink/7Try5DcVV/image.png I have 20 scanners without problems for the PC to run at the same time, how many on amd ryzen  interesting...

Quote
from bitcoin import *
import random

a = int (input("start scan range:"))
b = int (input("end scan range:"))
c = input ("address:")

while True:
    ran = random.randrange(a,b)
    myhex = "%064x" % ran
    myhex = myhex[:64]
    priv = myhex
    pub = privtopub(priv)
    pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed")
    addr = pubtoaddr(pubkey1)
    n = addr
    if n.strip() == c:
        print ("found!!!",addr,myhex)
        s1 = myhex
        s2 = addr

        f=open(u"C:/a.txt","a")
        f.write(s1)
        f.write(s2)       
        f.close()
        break
    else:
        print ("searching...",addr,myhex)

python exists for android... need to somehow gpu to steal, numba...

***

Quote
Key one has a random one bit key, key two has a random two bit key, key three has a random three bit key, etc.
deterministic random? most likely generated by time in windows.
legendary
Activity: 2646
Merit: 1137
All paid signature campaigns should be banned.
After considering a bunch of pvks found by the bots I noticed that many addresses were on the same transaction, after a tight check I saw what kind of pattern the sequence was. One thing I noticed perhaps this transaction was made by the Bitcoin developers. Some GPU bots would be best to violate it, as shorena has said. Is anyone out there?
Yes, ALL the Bitcoins in the puzzle transaction were placed there in a single transaction.  The author of the puzzle transaction came into this thread and explained the whole thing.  Yes, there is a well defined pattern to the puzzle transaction.  All the private keys are random but each key uses an increasing bit length.  Key one has a random one bit key, key two has a random two bit key, key three has a random three bit key, etc.  All of this work trying to find out a pattern to the random private keys is a complete and total waste of time.  "Fun" maybe, if you like to waste your time trying to find a pattern in random data.  If you bothered to read the thread you would know all this and would not have to guess.
newbie
Activity: 32
Merit: 0
After considering a bunch of pvks found by the bots I noticed that many addresses were on the same transaction, after a tight check I saw what kind of pattern the sequence was. One thing I noticed perhaps this transaction was made by the Bitcoin developers. Some GPU bots would be best to violate it, as shorena has said. Is anyone out there?
legendary
Activity: 1974
Merit: 1077
^ Will code for Bitcoins
Maybe if we learn a bit more from how the numbers were generated it could greatly optimize the program. Grin

We know exactly how the numbers were generated because the author of the puzzle came into this thread and told us.

Read back and find it if you are interested.

Where is the fun in knowing something? Isn't it much more interesting to throw around half-baked theories and draw maps where X marks the hidden treasure chest?
legendary
Activity: 2646
Merit: 1137
All paid signature campaigns should be banned.
Maybe if we learn a bit more from how the numbers were generated it could greatly optimize the program. Grin

We know exactly how the numbers were generated because the author of the puzzle came into this thread and told us.

Read back and find it if you are interested.
newbie
Activity: 177
Merit: 0
I am incompetent to solve a problem like this, I know its trading at Bittrex and Binance.

but hopefully at the end ny code like this can be solved and I will congratulate the person who broke the code. yuhuuu
jr. member
Activity: 184
Merit: 3
fast ... last 55 as I understand the author of the puzzle specially flashed the public key (threw some bitcoin and deducing). to see how it works baby step gigant step.

they all lie side by side (in its macrocosm)), The scanner would be with a random (127,7-16 signs randomisation) and a step up-down for a couple million.

                 56                                                                  49                                                          42                                    
127,*************************   127     127,27325883491766944644041359425  127,34166804622509516775608062744  
63,583036446150909604124734642028  63,5    63,660348182749885381781496107578    63,668431836413219571113586425781  
31,861576626618332619411688710898  31,75   31,787429311634468831471167504787   31,771796089393319562077522277832
15,90613502904471082688431238239   15,875  15,908679856821891007712110877037   15,926393464003922417759895324707
7,941484368221046585833988729064   7,9375  7,9644791460323176579549908638        7,966582541557727381587028503418

                35
127,414638392627239227294921875  127
63,588673464953899383544921875  
31,7922732196748256683349609375
15,90996809303760528564453125  
7,938812442123889923095703125  

  
                28
127,151996135711669921875           127
63,58295440673828125    
31,7968273162841796875  
15,876374721527099609375
7,9462490081787109375  

 
                21
127,1361083984375                         127
63,58837890625
31,82952880859375
15,9052734375
7,954345703125
 

                14
127,359375                                     127      

                 7
127,40625

you can try to search 7,9 she meets frequently for the next step 576460752303423488 - (((((((((7,94910898 * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128)  - 571957152676052992) * 128)
next step 576460752303423488 * 128 = 73786976294838206464 - 576460752303423488 = 73210515542534782976 formula 73786976294838206464 - ((((((((((7,94910898 * 128) * 128)  * 128) * 128)* 128)  * 128) * 128)* 128) * 128)  - 73210515542534782976) * 128)

if you look at the beginning

0               1           2          
1               3           4          
2               7           8            
3               8          16        
4              21          32          
5              49          64          
6              76         128          
7             224         256          
8             467         512            
9             514        1024 < 1018,15625
10           1155       2048
11           2683       4096
12           5216       8192
13           10544    16384  
14           26867    32768
15           51510    65536
16           95823   131072 < 130324 < 748
17          198669   262144  
18          357535   524288
19          863317  1048576
20         1811764  2097152
21         3007503  4194304
22         5598802  8388608
23        14428676 16777216  < 16664492 112724 (16777216-112724=16664492/128/128/128)
24        33185509 33554432  
25        54538862 67108864  
26       111949941 134217728    
27       227634408 268435456

You can try to print the number first example take the second step 7,954345703125 * 128 = < 1018,15625 * 128 = < 130324 ; 131072-130324= < 748 a prime number to find 131072-748=130324/128/128 = 7,954345703125 formula 131072 - ((((((((7,954345703125  * 128) * 128)  - 130048) * 128) = 95744 slightly askew because 95823/128=748,6171875 ; 131072 - 748,6171875 = 130323,3828125/128/128= 7,954308032989501953125 with this value 131072 - ((((((((7,954308032989501953125  * 128) * 128)  - 130048) * 128) = 95823  here you can try 749

take the previous step 7,9462490081787109375 * 128 = 1017,119873046875 * 128 = < 130191,34375 on the second multiplication, the tail does not disappear * 128 = < 16664492 (etc with each puzzle...)

you can take a number from the first step or from the second (any) and choose a number for the formula, for example last thing 7,941484368221046585833988729064*128*128=130113,27988893362726230407133698 on the idea should be enough for the first 8 characters (meaning scanning). 131072 - 130113 = 959 and let's go 959,2 959,3 959,4.. 959,23 959,28... 959,50 959,59.. or 959,random... 960...  

or all can be much simpler to take the number and multiply by 128,  749*128=95872 (95823), (880,656494140625) 881 * 128* 128 = 14434304 (14428676)... (958,72011106637273769592866301537) 959*128*128*128*128*128*128=4217726604148736 (4216495639600700)... what number to multiply 7 times to get 56.

or the program that the whole numbers searched for and multiplied by the comparator compared with the key.
newbie
Activity: 55
Merit: 0
Nice attempt DannyHamilton! There must be a way to make it much more efficient as the last three addresses were found so fast it couldn't have been by brute forcing.

Maybe if we learn a bit more from how the numbers were generated it could greatly optimize the program. Grin
jr. member
Activity: 184
Merit: 3
As here more precisely... formula 36028797018963968 - (36028797018963968:128) = 35747322042253312 ; 36028797018963968-(30045390491869460:128) = 35794067405746237,84375 all this is divided further into 128




71681994167562459          71494644084506624                                        
35794067405746237,84375 35747322042253312
17936473077884353,46875 17873661021126656
8954357973713819,15625  8936830510563328
4470658255186115,53125  4468415255281664


  127,332801540126                            127    
  63,583036446150909604124734642028 63,5  
  31,861576626618332619411688710898 31,75  
  15,90613502904471082688431238239  15,875
   7,941484368221046585833988729064  7,9375


63,5:128... etc left column like always more...
0,00388080056434026547876737882337  0,003875732421875


127:128:128 = 0,00775146484375 means left 0,00776... 0,00776x128=0,993...


36028797018963968 - ((((((((63,583036446150909604124734642028 * 128) * 128) * 128) * 128) * 128) * 128) * 128) - 35747322042253312) * 128)  
72057594037927936 - ((((((((127,16607289230181920824946928406 * 128) * 128) * 128) * 128) * 128) * 128) * 128) - 71494644084506624) * 128)


formula 360... 56 presumably begins 0,993 (+ 6 some numbers = 1000000 combinations for scanning)
72057594037927936 - (((((((((0,993123456 * 128) * 128) * 128) * 128) * 128) * 128) * 128) * 128)  - 71494644084506624) * 128)

substitute the formula in https://qalculate.github.io/downloads.html generate 1000000 digits of 6 digits add after 0,993 you do not need to sort everything out, you can start at 0,993623456 or 0,993723456

6 digits are taken from the least error compensated by scanning, you can take 127,1660728 but there are already 7 digits so that the more the better, but the error grows in case of wrong.

in the next step, the left side can is starting to become less, who knows... 127x2 = 254x2 = 508

Can there is a variant of search of a key as the whole number (in any python or something like that).

***

Although this can be done with the formula 358))

(((180143985 - 178575943) * 100000000) + 77925431597630) * 128
(((360287970 - 358123456) * 100000000) + 234729613217730) * 128

***

If collisions located 0 - 2^160 (and addresses themselves in the area +- 2^250-255) then where is the collision puzzle?
jr. member
Activity: 32
Merit: 2
Not checked if you are right with p1-55
but half of 2^55 is still a job for gf1080 GPU for several years

Try to narrow range a bit more, please Cheesy

Hey buddy,

we search for an partner, maybe you are interested ? I tried send you a PN but it was not possible.

Best Regards
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