Author

Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 325. (Read 229288 times)

jr. member
Activity: 115
Merit: 1
actually 134.58 bits

Are you on topic, man? 134.58 is just an AVERAGE distance in 2^256 between addresses. And nothing more.
jr. member
Activity: 30
Merit: 1
For those interested:

Puzzle
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Key
1
3
7
8
21
49
76
224
467
514
1155
2683
5216
10544
26867
51510
95823
198669
357535
863317
1811764
3007503
5598802
14428676
33185509
54538862
111949941
227634408
400708894
1033162084
2102388551
3093472814
7137437912
14133072157
20112871792
42387769980
100251560595
146971536592
323724968937
1003651412950
1458252205147
2895374552463
7409811047825
15404761757071
19996463086597
51408670348612
119666659114170
191206974700443
409118905032525
611140496167764
2058769515153876
4216495639600700
6763683971478124
9974455244496710
Low
0
2
4
8
16
32
64
128
256
512
1024
2048
4096
8192
16384
32768
65536
131072
262144
524288
1048576
2097152
4194304
8388608
16777216
33554432
67108864
134217728
268435456
536870912
1073741824
2147483648
4294967296
8589934592
17179869184
34359738368
68719476736
137438953472
274877906944
549755813888
1099511627776
2199023255552
4398046511104
8796093022208
17592186044416
35184372088832
70368744177664
140737488355328
281474976710656
562949953421312
1125899906842620
2251799813685250
4503599627370500
9007199254741000
High
1
3
7
15
31
63
127
255
511
1023
2047
4095
8191
16383
32767
65535
131071
262143
524287
1048575
2097151
4194303
8388607
16777215
33554431
67108863
134217727
268435455
536870911
1073741823
2147483647
4294967295
8589934591
17179869183
34359738367
68719476735
137438953471
274877906943
549755813887
1099511627775
2199023255551
4398046511103
8796093022207
17592186044415
35184372088831
70368744177663
140737488355327
281474976710655
562949953421311
1125899906842620
2251799813685250
4503599627370490
9007199254740981
18014398509482000
% In the space
100.00%
100.00%
100.00%
0.00%
33.33%
54.84%
19.05%
75.59%
82.75%
0.39%
12.81%
31.02%
27.35%
28.71%
63.99%
57.20%
46.21%
51.57%
36.39%
64.66%
72.78%
43.41%
33.49%
72.00%
97.80%
62.54%
66.82%
69.60%
49.28%
92.44%
95.80%
44.05%
66.18%
64.53%
17.07%
23.36%
45.89%
6.94%
17.77%
82.56%
32.63%
31.67%
68.48%
75.13%
13.67%
46.11%
70.06%
35.86%
45.35%
8.56%
82.86%
87.25%
50.18%
10.74%
full member
Activity: 706
Merit: 111
So, LBC is going through first 160 bits of 256? But Rico says it is not.


https://lbc.cryptoguru.org/stats

actually 134.58 bits

jr. member
Activity: 115
Merit: 1
So, LBC is going through first 160 bits of 256? But Rico says it is not.
full member
Activity: 706
Merit: 111
It exhausts each sequential search space, its not brute forcing the whole range

Thanks for the answer! And what's the magic of extracting 2^160 from 2^256? That's totally unclear Huh

adr1 = ripemd160(sha256(pubkey(rand(2^256-2^160)+2^160)))

this line makes no sense to me  Huh

2^256 is from 1 to 115792089237316195423570985008687907853269984665640564039457584007913129639936.

2^160 is from 1 to 1461501637330902918203684832716283019655932542976

But the sha256-hash of LONG==1 is almost in the end of 2^256
48635463943209834798109814161294753926839975257569795305637098542720658922315

LONG==1461501637330902918203684832716283019659932542976 (around 2^161 ) is also near hash of 1.

So, if LBC is not going sequentally from PK == 1 to 2^256, does it really solve this puzzle?!  Huh



Example it cracks keys like this:


00001
000011
0000111
00001000
000010101
0000110001
00001001100
000011100000
0000111010011
00001000000010
000010010000011
0000101001111011
00001010001100000
000010100100110000
0000110100011110011
00001100100100110110
000010111011001001111
0000110000100000001101
00001010111010010011111
000011010010110001010101
0000110111010010100110100
00001011011110010000001111
000010101010110111001010010
0000110111000010101000000100


and so on
jr. member
Activity: 115
Merit: 1
It exhausts each sequential search space, its not brute forcing the whole range

Thanks for the answer! And what's the magic of extracting 2^160 from 2^256? That's totally unclear Huh

adr1 = ripemd160(sha256(pubkey(rand(2^256-2^160)+2^160)))

this line makes no sense to me  Huh

2^256 is from 1 to 115792089237316195423570985008687907853269984665640564039457584007913129639936.

2^160 is from 1 to 1461501637330902918203684832716283019655932542976

But the sha256-hash of LONG==1 is almost in the end of 2^256
48635463943209834798109814161294753926839975257569795305637098542720658922315

LONG==1461501637330902918203684832716283019659932542976 (around 2^161 ) is also near hash of 1.

So, if LBC is not going sequentally from PK == 1 to 2^256, does it really solve this puzzle?!  Huh
full member
Activity: 706
Merit: 111
Hey, guys!

Please, anyone can explain how do you predict when LBC will find the key?

I've read (lbc.cryptoguru.org/man/theory) several times, but do not understand that pseudo code (I'm programmer, but noob in crypto)
The point I do not get completely: rico claims that they do not brute 2^256 range, because it is too large and do some hack to make range 2^159.

I of course understand that after sha256 we get a compression of range. But shouldn't sets 2^256 and 2^159 overlap chaotically? So how LBC chooses the right keys from 2^256 that map to unique addresses in 2^159 ?

For solving this puzzle shouldn't LBC brute 2^256 range ?!


It exhausts each sequential search space, its not brute forcing the whole range
hero member
Activity: 812
Merit: 500
This is still going on.

GL
jr. member
Activity: 115
Merit: 1
Hey, guys!

Please, anyone can explain how do you predict when LBC will find the key?

I've read (lbc.cryptoguru.org/man/theory) several times, but do not understand that pseudo code (I'm programmer, but noob in crypto)
The point I do not get completely: rico claims that they do not brute 2^256 range, because it is too large and do some hack to make range 2^159.

I of course understand that after sha256 we get a compression of range. But shouldn't sets 2^256 and 2^159 overlap chaotically? So how LBC chooses the right keys from 2^256 that map to unique addresses in 2^159 ?

For solving this puzzle shouldn't LBC brute 2^256 range ?!
member
Activity: 322
Merit: 10
Don't fully understand but ok.
full member
Activity: 706
Merit: 111
LBC is going to start finding the 55th key in a few hours.

With 3GH/s should take around ~37 days

0.55 BTC plus BCH plus BTG yields around

~0.66 BTC = $8,500

Basically about 2x as profitable as regular ETH mining with a RX470/570 if you can find it in 37 days and not closer to the end around 74 days.



is it going to find the 55th key or start finding it in a few hours?
legendary
Activity: 3808
Merit: 1723
LBC is going to start finding the 55th key in a few hours.

With 3GH/s should take around ~37 days

0.55 BTC plus BCH plus BTG yields around

~0.66 BTC = $8,500

Basically about 2x as profitable as regular ETH mining with a RX470/570 if you can find it in 37 days and not closer to the end around 74 days.

full member
Activity: 706
Merit: 111
Y'all gave up yet.....
newbie
Activity: 9
Merit: 0
legendary
Activity: 2646
Merit: 1137
All paid signature campaigns should be banned.
You are wasting your time.  The person who created the puzzle has already posted exactly how he did it.  It is random.  You will not find a pattern to predict the private keys.
newbie
Activity: 1
Merit: 0
We should find a solution instead of brute forcing. What do you think?

The newest approach is very promising. As a basis we use the binaries of the private keys in a descending order and a special cellular automaton:

██░░█░░█░░██░██░     
██░█░░░████░░██   
█░█░░█░░██░░░░     
█░█░░░██░░░░░     
█░█░░████░██     
█░░█░░░░░██     
█░░░░░░░█░           
███░█░░██     
███░░░░░     
█░░██░░     
██░░░█     
█░█░█           
█░░░               
███           
██           
█           

Cellular Automaton:
Input:    ██░  █░█  ░█░  █░░  ░░░  ░░█  ░██  ███  ░░█  ░██
Output: ██░  █░█  ░█░  █░░  ░░░  ░░█  ░██  ███  ░░█  ░██

The automaton speciality is that there isn't one static rule for the output, but the output depends on how often the rule has been executed before (same column). One big hint that someone has come up with this is: All the rules in- and outputs are the same!!

Example (See beginning of first row -> second row):

██░   
  █

██░ => █    because the function is executed the (3 % x = 1) th time

then second -> third:

██░ => ░    because the function is executed the (3 % x = 2) th time

██░
  ░


When you double check the results with 6 digit US patents ( US137821, US497622 !!! ) you see where this leads to ...

Keep up the good work!

...

https://1trash.github.io/
legendary
Activity: 1120
Merit: 1037
฿ → ∞
I can sweep for you keys in the future only from the 32btc contest. So for the 55bit key and so forth. I can't sweep any other address due to the fact that it might belong to someone.

Its really not that difficult. For Bitcore you just download their wallet with the included blockchain (300mb) and it syncs quick since their blockchain is mostly empty and just go to Debug and put in "importprivkey XXXXX" and you are all set. You can sell on Crytopia.

For BitcoinGold I just used the Coinomi app from the Android app store. Just build a new wallet and there is an option there to sweep a BitcoinGold address.

Thanks for letting me keep the funds, I will keep half and donate the other half to your LBC Pot. As soon as either Poloniex or Bitfinex allows BitcoinGold deposits.

Sure - all of what you said goes d'accord with our stab at it - limited to the puzzle transactions it shall be.

It's just that we do only care about the BTC as that is where the project started off. So transferring the funds to a custodial address then publishing the privkey refers to BTC only and means the forks become at the time of publishing the privkey accessible to everybody else.

NOT OUR PROBLEM. I might add.

So by transferring these also to custodial addresses, you would actually save them for their rightful owners, because we do not care.
Also we certainly know technically how to sweep the funds, but the hassle of another blockchain and another exchange-account seems not worth it.

legendary
Activity: 3808
Merit: 1723
I sweeped both your BitCore and BitcoinGold funds before somebody else got to it. Post a BitCore and BitcoinGold address in this thread and I will send them back to you.

Here are those transaction ids

https://chainz.cryptoid.info/btx/tx.dws?172551.htm

http://btgexp.com/tx/f78617a5e6bf438c7a310639f6d8fe10c194a74be6ac98b2c329024a4d1df745

EDIT: Please post your addresses where you want the coins sent in the 32btc contest thread since its unmoderated.

Thank you for the noble offer, but it seems not worth the hassle to go for every BTC fork. (We didn't even claim Bitcoin Cash).

If you want to support the project - which would be greatly appreciated - you can keep both BitCore and BitcoinGold funds you sweeped and transfer whatever you deem appropriate to the LBC Pot (https://blockchain.info/address/1LBCPotwPzBvBcTtd7ADGzCWPXXsZE19j6).

Thank you.

If this "mode of operation" is ok for you, feel free to send me an email address via PM, we could inform you in advance of private keys found, so you could see if you can process the BitCore and BitcoinGold attached to them.


I can sweep for you keys in the future only from the 32btc contest. So for the 55bit key and so forth. I can't sweep any other address due to the fact that it might belong to someone.

Its really not that difficult. For Bitcore you just download their wallet with the included blockchain (300mb) and it syncs quick since their blockchain is mostly empty and just go to Debug and put in "importprivkey XXXXX" and you are all set. You can sell on Crytopia.

For BitcoinGold I just used the Coinomi app from the Android app store. Just build a new wallet and there is an option there to sweep a BitcoinGold address.

Thanks for letting me keep the funds, I will keep half and donate the other half to your LBC Pot. As soon as either Poloniex or Bitfinex allows BitcoinGold deposits.


legendary
Activity: 1120
Merit: 1037
฿ → ∞
I sweeped both your BitCore and BitcoinGold funds before somebody else got to it. Post a BitCore and BitcoinGold address in this thread and I will send them back to you.

Here are those transaction ids

https://chainz.cryptoid.info/btx/tx.dws?172551.htm

http://btgexp.com/tx/f78617a5e6bf438c7a310639f6d8fe10c194a74be6ac98b2c329024a4d1df745

EDIT: Please post your addresses where you want the coins sent in the 32btc contest thread since its unmoderated.

Thank you for the noble offer, but it seems not worth the hassle to go for every BTC fork. (We didn't even claim Bitcoin Cash).

If you want to support the project - which would be greatly appreciated - you can keep both BitCore and BitcoinGold funds you sweeped and transfer whatever you deem appropriate to the LBC Pot (https://blockchain.info/address/1LBCPotwPzBvBcTtd7ADGzCWPXXsZE19j6).

Thank you.

If this "mode of operation" is ok for you, feel free to send me an email address via PM, we could inform you in advance of private keys found, so you could see if you can process the BitCore and BitcoinGold attached to them.
legendary
Activity: 3808
Merit: 1723
Cross posting this



Congrats on finding the private key. I don't think you know but you can also claim BitCore and Bitcoin GOLD with that private key. I see you or somebody already claimed Bitcoin Cash.

I sweeped both your BitCore and BitcoinGold funds before somebody else got to it. Post a BitCore and BitcoinGold address in this thread and I will send them back to you.

Here are those transaction ids

https://chainz.cryptoid.info/btx/tx.dws?172551.htm

http://btgexp.com/tx/f78617a5e6bf438c7a310639f6d8fe10c194a74be6ac98b2c329024a4d1df745


EDIT: Please post your addresses where you want the coins sent in the 32btc contest thread since its unmoderated.
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