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Author Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it  (Read 80796 times)
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January 06, 2016, 04:09:27 PM
 #161

Its an easy way to check whether the first bit is always 1 for a given step, which it is. Thus you can limit the search space. For step n you do not need to search for any possible solutions from step n-1
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January 06, 2016, 04:13:09 PM
 #162

Its an easy way to check whether the first bit is always 1 for a given step, which it is. Thus you can limit the search space. For step n you do not need to search for any possible solutions from step n-1
This was known from page 1 of this thread.

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January 06, 2016, 04:18:26 PM
 #163

EDIT: this formula doesn't make any sense I know, just playing around Smiley

Just for fun the average of this:

Code:
1.00000000
1.50000000
1.75000000
1.00000000
1.31250000
1.53125000
1.18750000
1.75000000
1.82421875
1.00390625
1.12792969
1.31005859
1.27343750
1.28710938
1.63983154
1.57196045
1.46214294
1.51572418
1.36388779
1.64664650
1.72783279
1.43408918
1.33485842
1.72003222
1.97801048
1.62538475
1.66818412
1.69600850
1.49275696
1.92441434
1.95800192
1.44051053
1.66181426
1.64530614
1.17072322
1.23364647
1.45885221
1.06935867
1.17770458
1.82563129

Is 1.482530615

So starting at 1.482530615 into the next range and doing a butterfly search might be a tiny bit faster.
 

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January 06, 2016, 04:22:56 PM
 #164

Maybe these random-like numbers came from hashing operations? For example: n-th key = truncate(SHA256(f((n-1)-th key))). It will be still hopeless if the process involves a strong passphrase though.
I think this would be a good way to do it.  Then the author just needs to remember the seed and algorithm instead of remembering 256 private keys.

Although keeping track of 256 private keys is not that hard in the first place.

The algorithm would need to "skip over" any private keys in the generated sequence that had the undesirable result of a 0 in the most significant bit after the masking operation.

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January 06, 2016, 04:44:46 PM
 #165

So starting at 1.482530615 into the next range and doing a butterfly search might be a tiny bit faster.
butterfly search... so the numbers were generated using chaos theory  Wink

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January 06, 2016, 04:56:57 PM
 #166

I noticed that the results for x increase then reset every 3 steps, 0-2, 3-5, 6-8, etc...except for 15-21, this pattern continues when n > 21.  Just a simple observation, hopefully helpful, probably not though.

Here are the results for the values we know given those 2 mentioned formulas:

y = 2^n * x

and so x = y / 2^n

AND

y = 2^n + x

and so x = y - 2^n


n     Known Results (y)     x = y / 2^n     x = y - 2^n
011.000000000
131.500000001
271.750000003
381.000000000
4211.312500005
5491.5312500017
6761.1875000012
72241.7500000096
84671.82421875211
95141.003906252
1011551.12792969131
1126831.31005859635
1252161.273437501120
13105441.287109382352
14268671.6398315410483
15515101.5719604518742
16958231.4621429430287
171986691.5157241867597
183575351.3638877995391
198633171.64664650339029
2018117641.72783279763188
2130075031.43408918910351
2255988021.334858421404498
23144286761.720032226040068
24331855091.9780104816408293
25545388621.6253847520984430
261119499411.6681841244841077
272276344081.6960085093416680
284007088941.49275696132273438
2910331620841.92441434496291172
3021023885511.958001921028646727
3130934728141.44051053945989166
3271374379121.661814262842470616
33141330721571.645306145543137565
34201128717921.170723222933002608
35423877699801.233646478028031612
361002515605951.4588522131532083859
371469715365921.069358679532583120
383237249689371.1777045848847061993
3910036514129501.82563129453895599062


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January 06, 2016, 07:55:04 PM
 #167

I noticed that the results for x increase then reset every 3 steps, 0-2, 3-5, 6-8, etc...except for 15-21, this pattern continues when n > 21.  Just a simple observation, hopefully helpful, probably not though.

Here are the results for the values we know given those 2 mentioned formulas:

y = 2^n * x

and so x = y / 2^n

AND

y = 2^n + x

and so x = y - 2^n


n     Known Results (y)     x = y / 2^n     x = y - 2^n
011.000000000
131.500000001
271.750000003
381.000000000
4211.312500005
5491.5312500017
6761.1875000012
72241.7500000096
84671.82421875211
95141.003906252
1011551.12792969131
1126831.31005859635
1252161.273437501120
13105441.287109382352
14268671.6398315410483
15515101.5719604518742
16958231.4621429430287
171986691.5157241867597
183575351.3638877995391
198633171.64664650339029
2018117641.72783279763188
2130075031.43408918910351
2255988021.334858421404498
23144286761.720032226040068
24331855091.9780104816408293
25545388621.6253847520984430
261119499411.6681841244841077
272276344081.6960085093416680
284007088941.49275696132273438
2910331620841.92441434496291172
3021023885511.958001921028646727
3130934728141.44051053945989166
3271374379121.661814262842470616
33141330721571.645306145543137565
34201128717921.170723222933002608
35423877699801.233646478028031612
361002515605951.4588522131532083859
371469715365921.069358679532583120
383237249689371.1777045848847061993
3910036514129501.82563129453895599062


Very interesting observation!!! Thank you for sharing

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January 06, 2016, 08:02:28 PM
 #168

Nice colletive work y'all! Gonna try to use matlab to make some intergers experiments.
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January 06, 2016, 08:05:27 PM
 #169

31-33 doesn't follow the pattern either.  I missed that.

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January 06, 2016, 08:36:47 PM
Last edit: January 06, 2016, 08:51:04 PM by Bulista
 #170

31-33 doesn't follow the pattern either.  I missed that.

Yes, I see, but it was well observed.

Here is a chart of those values with the ups and downs for x = y / 2^n:



EDIT: and btw,

this is the sequence:



and this is x = y - 2^n:

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January 06, 2016, 09:00:38 PM
 #171


-snip-

I think that  what you have done is to pretty much prove it is random and there is no predictive formula.

I'm not sure.

I got some infos that are getting me to believe that there is a possible formula behind it.

I give you an example. I'm playing around by creating random formulas, and I get pretty much similar results. Yet they are predictable with a formula.

All I use for inputs are 2 arrays, one with the current position and another with the sequential list of prime numbers.

For example:

Consider n = count, p = prime numbers, and y = sequence based on the formula 2^n + (n mod p) * Log(n+1, 2) <--- Random formula I invented.

y / 2^p + 1 and y-2^p *-1 are similar formulas to what was shown before for the var x in the real sequence, their results also appear to be random...

n     p     y = 2^n + (n mod p) * Log(n+1, 2)     y / 2^p + 1     y-2^p *-1
0     2              1     1.25000000     3
1     3              3     1.37500000     5
2     5              7     1.22406016     25
3     7              14     1.10937500     114
4     11              25     1.01234752     2023
5     13              45     1.00548399     8147
6     17              81     1.00061679     130991

Yet this sequence is breakable with a simple formula.

EDIT: this formula doesn't make any sense I know, just playing around Smiley

Am I the only one noticing that this guy makes a random formula using prime numbers and his first 3 results are exactly the ones as the sequence for this puzzle?

It seems pretty obvious that there is some formula using prime numbers behind this, no? And possibly the formula he posted is not so far from the real one.

Keep up the good work all!

Yea, I was actually surprised by those 3 first results since I was genuinely trying random stuff.

Maybe with some tweaks we can get to the right formula, if there is indeed one Grin

One thing I'm sure if we stop looking we won't know.
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January 06, 2016, 09:23:55 PM
 #172

-snip-
EDIT: and btw,

this is the sequence:



and this is x = y - 2^n:



Given the numbers we are dealing with you should probably use a log scale.
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January 06, 2016, 11:20:22 PM
 #173

Is there any way to tell what time an address was generated?  I'm asking because the results for x = y / 2^n could be converted to a time of day.

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January 07, 2016, 12:42:32 AM
 #174

Is it possible that the entity who created this is seeking employees capable of cracking it for its endeavor? I recall the NSA (or some other) putting out a contest, with the reward for solvers to become employees. Just a thought.
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January 07, 2016, 01:20:37 AM
 #175

The transactions seem to originate from address 173ujrhEVGqaZvPHXLqwXiSmPVMo225cqT  which had a Total received amount of   56,457.80848111 BTC and a Final Balance of    312.04932734 BTC

Definitely a very big player of some kind.

a player of one special word

its called bs

their is no sequence i dont think

spam thread

why would u try to solve a sequence of random numbers

found in a transaction

does the question even makes sense to anyone

it may be some kind of idiot test

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January 07, 2016, 06:10:03 AM
 #176

why would u try to solve a sequence of random numbers
Because everyone reads only the subject of this topic  Grin

Quote
it may be some kind of idiot test
This board? The whole bct.o? Yes.
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January 07, 2016, 08:02:30 AM
 #177

The transactions seem to originate from address 173ujrhEVGqaZvPHXLqwXiSmPVMo225cqT  which had a Total received amount of   56,457.80848111 BTC and a Final Balance of    312.04932734 BTC

Definitely a very big player of some kind.
...
their is no sequence i dont think
...

if you dont know if there is sequence or not, why are you talking shit?
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January 07, 2016, 08:05:31 AM
 #178

if you dont know if there is sequence or not, why are you talking shit?
Because this is public resource and anyone can say whatever he wants
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January 07, 2016, 09:23:49 AM
 #179

if you dont know if there is sequence or not, why are you talking shit?
Because this is public resource and anyone can say whatever he wants

Good point  Cheesy

Is there any way to tell what time an address was generated?  I'm asking because the results for x = y / 2^n could be converted to a time of day.

That's not possible
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January 07, 2016, 12:50:37 PM
 #180

-snip-
EDIT: and btw,

this is the sequence:



and this is x = y - 2^n:



Given the numbers we are dealing with you should probably use a log scale.

Log 2:

http://imgur.com/8aW8tj6
http://imgur.com/dGguJcK
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