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Showing 13 of 13 results by zbyszek2
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Re: The Legend of Satoshi Nakamato, FINAL STEP PUBLISHED.... 4.87 BTC GRAND PRIZE!
by
zbyszek2
on 03/02/2018, 17:20:31 UTC
there is a bug in the article: "red (0) or yellow (1)"  should be yellow (0) or red (1).
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Re: The Legend of Satoshi Nakamato, FINAL STEP PUBLISHED.... 4.87 BTC GRAND PRIZE!
by
zbyszek2
on 28/01/2018, 00:57:40 UTC

Exactly. I did research that as well for itod: https://bitcointalk.org/index.php?topic=766000.msg26795620#msg26795620

Like you said, when dealing with chaos, to receive X bits of your choice, you may very likely only have to change X bits in the input, regardless of the intermediate steps. In my research it's a loop of SHA256. The Private->public key computation serves just as well, hence the private key could be like 200 zeros followed by 56 or less bits of actual random data. It could be also Bacon26-encoded 10 random words from the English dictionary, whatever meets the entropic-I/O-shift requirements.

Yes, one of your entries helped me to realize my mistake, but as mentioned by Pan Troglodytes, the correlations between different bit streams are still  there
(amongst them correlation between green/blue and slim/fat)
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Re: The Legend of Satoshi Nakamato, FINAL STEP PUBLISHED.... 4.87 BTC GRAND PRIZE!
by
zbyszek2
on 27/01/2018, 00:25:00 UTC

Zbyszek2 sampled WIF keys and performed analysis and proved in his post earlier, that it is impossible to see such a correlation in real WIF key.

In short - you discard blob on a weak assumption that it is not random. All the data is not random and should be thus discarded, don't you agree?

If the data were perfectly random you would not be able to read it into WIF key.
.....

Guys I would like to apologize for a mistake I made in my previews analysis.
I assumed that the private WIF key is random (private/public key pair was generated with a vanitygen like tool without modifications)
so my conclusion was that the "yellow/red", "green/blue" bit stream can not encode a  private key.

Unfortunately my assumption might not nesesery be true,
in fact there only needs to be about length("FLAMEN6")*6bits= 7*6bits = 42bits  random bits in a stream to generate a public key with "1FLAMEN6" prefix,
so not only the "yellow/red", "green/blue" stream may contain a valid private key, but also only the "short/long" stream is enough to create a valid private key with a  "1FLAMEN6" prefix.

Even more, it's probably possible to modify the original vanitygen source code so, that the time it would take to find such a private key is the same as now (on my laptop the estimation is ~5 days). The only change would be to feed vanitygen not with random numbers but with a desired bit pattern + the 42 random bits.

If this is the case in this puzzle, you have to decide for yourself.
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Board Bitcoin Discussion
Re: The Legend of Satoshi Nakamato, FINAL STEP PUBLISHED.... 4.87 BTC GRAND PRIZE!
by
zbyszek2
on 21/01/2018, 21:45:33 UTC
Yet another idea:

some leaves are pointing inside chess squares,
what if they indicate the reading order of  flames sides and the direction.
Direction could be imposed by the corner spirals:
if a leaf is pointing  left we go CCW, otherwise we go CW.

https://imgur.com/oLzQaRP

unfortunately I'm not sure if the green color is correct (might also be "inner top" flames)
and if the black color should be taken into account, if not we would have this order (based on leaf marked yellow):
1. inner right CCW
2. outer bottom CW
3. outer top CW
4. inner left CW
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Board Bitcoin Discussion
Re: The Legend of Satoshi Nakamato, FINAL STEP PUBLISHED.... 4.87 BTC GRAND PRIZE!
by
zbyszek2
on 09/01/2018, 01:09:49 UTC
I'm having a difficult time following your path. The eye points to the (#2) leaf, but then what? The leaf points down & right. Wouldn't we go to the next leaf?

If the idea is to follow the vine that the leaf points to, then that's hopeless. The vine disappears almost immediately and doesn't give any indication of direction.

The leaf with the #3 on it points back up into the flames. But there is no number 4? And yet there is a number 5?
hm... yes I've probably missed some numbers Cheesy

here you go, the full path:
https://imgur.com/gallery/rI1x1

1. follow  leafs
2. follow vine branches (cyan color)
section 1, cyan path, do something with:
3. blue flame  
section 2, cyan path, do something with:
4. white square (probably position)  
5. white queen/bishop
6. blue knight
7. rays (?)
8. white square (probably position)
9. go to  section 3
section 3, yellow path, do something with:
(and so one..., unfortunately we are still in section 1)
(leafs marked green are probably associated with leaf 12)

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Board Bitcoin Discussion
Re: The Legend of Satoshi Nakamato, FINAL STEP PUBLISHED.... 4.87 BTC GRAND PRIZE!
by
zbyszek2
on 08/01/2018, 23:57:24 UTC
I would like to share an idea with you, it was probably mentioned before but with no much response,
idea: The branches and leafs show a path to the solution

The beginning of this puzzle is (quote from "The Phoenix and the Turtle" by WILLIAM SHAKESPEARE)
So between them love did shine
That the Turtle saw his right
Flaming in the Phoenix' sight:
Either was the other's mine.  

here is a picture of the idea:
https://imgur.com/gallery/XeMgS

the puzzle is divided into "sections". The path, what to do next, is shown by the branches (sometimes partially obscured by other objects) and leafs.
According to this theory we are in section 1, arrow 1. :-)

some observations: there is the phoenix hair, a brunch "skipping" 1 or 2 branches below, a K (or R or B) and a leaf pointing at the small blue flame
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Re: The Legend of Satoshi Nakamato, FINAL STEP PUBLISHED.... 4.87 BTC GRAND PRIZE!
by
zbyszek2
on 07/01/2018, 04:01:26 UTC
I'm having trouble reproducing the length XOR decode. My understanding from crax0r's breakdown is: you concat the length bits in this order:
....

a python (2.7) script for you (not the prettiest but it works):
Code:
#!/usr/bin/python

def splitString(s, size):
    return [s[i:i+size] for i  in range(0, len(s), size)]

bac = "ABCDEFGHIJKLMNOPQRSTUVWXYZ_______________________________________________________";
#      12345678901234567890123456

inner_short_long = ('0110110110100010110110110011','110111110110110111110010','100011100010110111110111110111110','0111001101101')
outer_short_long = ('001101100111001101','001111000101','0011111011010001','01101101')

(it_l2r, ir_t2b, ib_r2l, il_b2t) = inner_short_long
(ot_l2r, or_t2b, ob_r2l, ol_b2t) = outer_short_long
ib_l2r = ib_r2l[::-1]
il_t2b = il_b2t[::-1]
it_r2l = it_l2r[::-1]
ot_r2l = ot_l2r[::-1]
ob_l2r = ob_r2l[::-1]

rbn_l2r = '011010'

encrypted_track = "".join([ib_l2r, il_t2b, ir_t2b, it_r2l, ot_r2l, ob_l2r]);

print 'encrypted_track = ', encrypted_track
xor = (rbn_l2r * 22) [:-1]
print 'xor             =  ', xor

decrypted_track = ("{0:b}").format(int(encrypted_track, 2) ^ int(xor, 2))
print 'decrypted_track =  ', decrypted_track
indexed_track = (decrypted_track*5)[::5]
print 'indexed_track   =  ', indexed_track
splited_track = splitString(indexed_track,5)
print 'splited_trak    = ' , splited_track

print 'decoded bacon   = ' , [ bac[int(x,2)] for x in splited_track]

output:
Code:
encrypted_track =  011111011111011111011010001110001101101100111011011111011011011111001011001101101101000101101101101011001110011011001000101101111100
xor             =   01101001101001101001101001101001101001101001101001101001101001101001101001101001101001101001101001101001101001101001101001101001101
decrypted_track =   10010010010010010010111000011000000100001110110010010010110010010000110000000100000001000100000100110000011010110000101100000110001
indexed_track   =   10011001110010000101011001101000000101001000101000100100101000100110000010101000010110010010001000000000001001000101100000000011101
splited_trak    =  ['10011', '00111', '00100', '00101', '01100', '11010', '00000', '10100', '10001', '01000', '10010', '01010', '00100', '11000', '00101', '01000', '01011', '00100', '10001', '00000', '00000', '01001', '00010', '11000', '00000', '01110', '1']
decoded bacon   =  ['T', 'H', 'E', 'F', 'M', '_', 'A', 'U', 'R', 'I', 'S', 'K', 'E', 'Y', 'F', 'I', 'L', 'E', 'R', 'A', 'A', 'J', 'C', 'Y', 'A', 'O', 'B']


@RealOnTheMF - I'm impressed!

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Board Bitcoin Discussion
Re: The Legend of Satoshi Nakamato, FINAL STEP PUBLISHED.... 4.87 BTC GRAND PRIZE!
by
zbyszek2
on 06/01/2018, 00:23:24 UTC
If someone is interested in symbols:

https://imgur.com/gallery/F22X4

of course they are very subjective Wink
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Re: The Legend of Satoshi Nakamato, FINAL STEP PUBLISHED.... 4.87 BTC GRAND PRIZE!
by
zbyszek2
on 05/01/2018, 17:11:18 UTC
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Board Bitcoin Discussion
Re: The Legend of Satoshi Nakamato, FINAL STEP PUBLISHED.... 4.87 BTC GRAND PRIZE!
by
zbyszek2
on 05/01/2018, 16:07:29 UTC
A face is the most universally recognized object that the human brain can decode from any contour. ...

speaking of faces:

https://imgur.com/a/uJR0s

image was rotated CCW 90 deg, added to its mirror with 50% alpha,

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Re: The Legend of Satoshi Nakamato, FINAL STEP PUBLISHED.... 4.87 BTC GRAND PRIZE!
by
zbyszek2
on 04/01/2018, 23:28:45 UTC

EDIT: still, they can be a mapping into more random data (a board has been mentioned)

Yes, that's what I'm suspecting too
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Re: The Legend of Satoshi Nakamato, FINAL STEP PUBLISHED.... 4.87 BTC GRAND PRIZE!
by
zbyszek2
on 04/01/2018, 22:56:11 UTC
I would like to share some statistics I've made,
the conclusion: It is not possible that the "yellow/red", "green/blue" bits (only) encode a  private key
(or at least very unlikely)
(EDIT: see bottom comment)

looking at the distribution of  1's in the "yellow/red", "green/blue" bit streams it looks like the stream is random
Code:
yellow/red  00010101001011011100101100000000101110111000111100001010111000011100010111111101011100111101001100101001101100110001111000110000101110101100011101110010
green/blue  10001101001010111100001110110111110110010000101101000000110011011100011000101010011110010000001110100100010111111111011000011111101001011011010010101000
we have 78 - 1's in "yellow/red" stream and 77 -1's in "green/blue" stream (nearly 50% of all 152x2 bits)
but if we take only bits if a flame is long:
Code:
short/long  01101101101000101101101100111101111101101101111100101000111000101101111101111101111100111001101101101101101000101101111100101000111100101100111001101100
yellow/red   00 01 10 1   0 11 01 11  0000 01011 01 10 01111  0 1   111   0 11 00101 11111 10111  111  10 11 01 10 11 1   1 00 11110  1 0   1011  1 11  011  11 00
green/blue   00 11 10 1   1 11 00 11  1101 11101 00 00 01011  0 0   110   0 11 00110 01010 00111  010  00 11 01 01 00 0   1 11 10110  0 1   1010  0 10  010  01 10  
we get:
P(yellow/red = 1 | long) = 0,6559139785
P(green/blue = 1 | long) = 0,5161290323
this is very strange (assuming the streams are random) but still possible, so let just check every second bit and cut out only bits according to 011011 pattern:
Code:
pattern     0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0
yellow/red    0 0   0 1   0 1   1 1   0 0   1 1   1 1   1 1   0 1   1 1   0 1   0 0   1 1   0 1   1 1   0 1   1 1   1 1   1 0   1 1   1 0   1 1   1 1   0 1   1 0  
green/blue    0 1   0 1   1 1   0 1   1 0   1 0   0 0   1 1   0 0   1 0   0 1   0 1   1 1   0 1   0 0   0 1   1 0   0 0   1 1   0 1   0 1   1 1   0 1   0 0   1 1  

P(yellow/red = 1 | '0 1 1' = 1) = 0,68
P(green/blue = 1 | '0 1 1' = 1) = 0,52
this is very unlikely to happen for a private key, how unlikely?
I've tested 1000 WIF private keys generated with vanitygen in base64, base58 and binary coding,  the highest P(private key = 1| '0 1 1' = 1) was ~0.62

Here are my statistics:
1. average P for a private key generated with vanitygen (1000 samples)
Code:
vanity key  bin            base64          base58         ascii 7bit     ascii 8bit
avr         0,4910042715   0,4928240574    0,4593443252  0,538281408 0,470996232
2. P for short/long, yellow/red, green/blue and slim/fat bit streams with cutouts (columns)
https://imgur.com/gallery/JRZFs
Code:
all '10101010’ '01010101’ 1 1 0 1 1 0 0 1 0 0 short/long 1 short/long 0 yellow/red 1 yellow/red 0 green/blue 0 green/blue 1 slim/fat 1 slim/fat 0
short/long 0,6118421053 0,6578947368 0,5657894737 1 0 1 0 0,7820512821 0,4324324324 0,6233766234 0,6 0,7333333333 0,4935064935
yellow/red 0,5131578947 0,5263157895 0,5 0,68 0,2307692308 0,6559139785 0,2881355932 1 0 0,5584415584 0,4666666667 0,56 0,4675324675
green/blue 0,5065789474 0,5263157895 0,4868421053 0,52 0,5384615385 0,5161290323 0,4915254237 0,5512820513 0,4594594595 1 0 0,6266666667 0,3896103896
slim/fat 0,4934210526 0,4210526316 0,5657894737 0,52 0,2307692308 0,5913978495 0,3389830508 0,5384615385 0,4459459459 0,6103896104 0,3733333333 1 0
Bit count 152 76 76 50 26 93 59 74 78 77 75 75 77


in case of 2. we can clearly see there is a correlation between  "short/long" -  "yellow/red" streams and "green/blue" - "slim/fat" streams


EDIT:
In this analysis I assumed that the private key is random, but that is not necessarily true,
in fact we only need 42bits of "randomnes" to generate a public key with a "1FLAMEN6" prefix,
so after all my main conclusion is probably wrong
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Re: The Legend of Satoshi Nakamato, FINAL STEP PUBLISHED.... 4.87 BTC GRAND PRIZE!
by
zbyszek2
on 04/01/2018, 20:55:57 UTC
Have you tried using the inner shape as the third bit (skinny line or blob of paint)?


you mean something like this:
Code:
short/long  01101101101000101101101100111101111101101101111100101000111000101101111101111101111100111001101101101101101000101101111100101000111100101100111001101100
yellow/red  00010101001011011100101100000000101110111000111100001010111000011100010111111101011100111101001100101001101100110001111000110000101110101100011101110010
green/blue  10001101001010111100001110110111110110010000101101000000110011011100011000101010011110010000001110100100010111111111011000011111101001011011010010101000
slim/fat    01001100001100011101011101011100001110001101011100010000111111000110011000100011010000011001001101001100011000101001011100011111110100111011011000111100

it is a possibility, in some cases the blobs (slim/fat) look like there were corrected.