Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 336. (Read 242853 times)

Maybe somehow the word (passphrase) tear out of the private key. Or there only one way, you can select 64 characters hexadecimal system and this will be a private key, you can get the same address through the word but through the address we can not see this word. As the author of a puzzle in general he scattered keys in space If he did not select the address manually. Or he has a hardware wallet (deterministic). How can we repeat the principle of his puzzle just like that scattered at the same addresses. Or was it an isolated case in the history of the universe? 

WOw that is really tiresome puzzle i would recommend you to take some photos and pictures on how you came up with that given so that we can fully understand on how are we going to crack that puzzle. I know that would lead to an account holder with huge money on it. Well let me help cracking that one but i need more data regarding the puzzle.

1. There is no true randomness in computers.
2. Can you count? 30 keys from 55 were in second half.

But OK, I've counted myself just now and was expecting a bit more than 30/55 chances

I do not understand all this, If I'm smart enough to do this, it would be nice to do it. Would the author be a miner? This also raises the question whether someone else rather than the author did cash out

Unless and until they don't.  The selection of key is totally random within the range with all values equally likely.  Seeing or creating "trends" in random data is a very human thing to do but unfortunately for your theory the private keys for each range were created by having a computer create a 256 bit secure random number and then masking the result down to the desired key length.

If you search the top half of the range for any key there is exactly a 50% chance of the key being in the top half of the range.

Not exactly. He just mentioned that values are random. But randomness can be very different.
If you look at the table with keys 1-55 you'll see that keys tend to sit in second part of range.

Theory is explained here (in short):
https://lbc.cryptoguru.org/about

and here (a lot of reading):
https://bitcointalk.org/index.php?topic=1877935.260

Software is available in at least two options:

LBC
or
https://bitcointalksearch.org/topic/faster-and-autonomous-large-bitcoin-collider-fork-3102823

I've read most of this thread and am absolutely mesmerized by everyone's knowledge of the inner workings of things. I would love to be up to par with everyone on this. so who wants to tutor me and/or provide a link to something that explains more in-depth?

you are wasting your time.  The person who created the puzzle has already posted exactly how he did it.For step n you do not need to search for any possible solutions from step n-1

Approx search space for puzzle56 is 36028797018963968-72057594037927936 

Only issue now is to compile Vanitygen in Windows.I know this as a benchmark for my own calculations. we'll see soon Who will succeed. I'm not likely.

really interesting!!!
I want toparticipate too!!!!
ahahahahahah
I like things like this one.

To use the baby-step giant-step algorithm you need the actual public key, the address (P2PKH) is not enough. The public key is exposed when someone (e.g. the owner) makes a transfer out of the account.

everything is already done https://bitcointalksearch.org/topic/m.13424809 https://bitcointalksearch.org/topic/m.13463670

here are the other pvk decimal values I was able to find:

Address 15: 26867
Address 16: 51510
Address 17: 95823
Address 18: 198669
Address 19: 357535
Address 20: ?

Well and how he from a transaction the public key received? sent a bit bitcoin to the address? What prevents him from doing the same thing with 56? I gave him an approximate search space.

                                              puzzle step difference                       2^difference
44        15404761757071     120349701227 >          62460552416       8796093022208        8733632469792
45        19996463086597     156222367864 >          35872666637     17592186044416       17556313377779
46        51408670348612     401630237098 >        245407869234      35184372088832      34938964219598
47      119666659114170     934895774329 >        533265537231      70368744177664      69835478640433
48      191206974700443    1493804489847 >       558908715518     140737488355328     140178579639810
49      409118905032525    3196241445566 >      1702436955719    281474976710656     279772539754937
50      611140496167764    4774535126310 >      1578293680744    562949953421312     561371659740568
51    2058769515153876   16084136837139 >    11309601710829   1125899906842624   1114590305131795
52    4216495639600700   32941372184380 >    16857235347241   2251799813685248   2234942578338007
53    6763683971478124   52841281027172 >    19899908842792  4503599627370496   4483699718527704  
54    9974455244496708   77925431597630 >    25084150570458   9007199254740992   8982115104170534
55  30045390491869460  234729613217730 >  156804181620100 18014398509481984  17857594327861884  
                                                           >                           36028797018963968  358*************


8140<->8156 in this range multiplied by 128 6 times should be 56 (if we proceed from the previous steps not 357 and not 359).

36028797018963968-8140<->8156+234729613217730x128              scanning intervals (dec)                

35800098600386560 35844079065497600                       59318788069777664 53689288535564544
35804496646897664 35848477112008704                       58755838116356352 53126338582143232
35808894693408768 35852875158519808                       58192888162935040 52563388628721920
35813292739919872 35857273205030912                       57629938209513728 52000438675300608
35817690786430976 35861671251542016                       57066988256092416 51437488721879296
35822088832942080 35866069298053120                       56504038302671104 50874538768457984
35826486879453184 35870467344564224                       55941088349249792 50311588815036672
35830884925964288 35874865391075328                       55378138395828480 49748638861615360
35835282972475392                                                   54815188442407168
35839681018986496                                                   54252238488985856  

You talk some crap as usual, there are no spends on p56, so public address is unknown.

I think that a novice programmer can find 56. Here that there will be further through 10-20 steps...

Arulbero looking for 56 between 00000000000000000000000000000000000000000000000000B2BE1F9B67E100 00000000000000000000000000000000000000000000000000D2BE1F9B67E100

This "trick" is explained here and in LBC thread.
If someone spends PART of puzzle56, then Arulbero (or anyone who is capable of) can spend the rest.

I guess, this doesn't prove regular BTC wallets to be vulnerable, as their location in 2^159 address space is unknown.

Anyway huge respect to arulbero. He definitely didn't sleep at math classes.

When do you calculate a private address for a 56bit puzzle?

Prema