The Legend of Satoshi Nakamato, FINAL STEP PUBLISHED.... 4.87 BTC GRAND PRIZE!

[Neantis]

More I think about it more I think it's broken (maybe because of painting technique?)  (that is why she posted new picture - and she says that nowhere she said it was a clue - maybe she gave up on this but it's bad PR to said that people have followed broken puzzle for 3 years).

There isn't no single point to catch first, even 1flamen6 is no good. I'm quite annoyed and tired, probably like a lot of people before me.

If it will not be solved I hope she will leave solution in her last will  (I wish her long and happy life, don't get me wrong! )

That is all for now, I'm going to bed. I have already trouble in work because of this %$#@#$ puzzle, so tomorrow only read some or post but no quality analysis. Fortunatelly I switched off computer during weekend and manage to have a great time with kids, it was worth it.

- I also think that there is too much guessing but that's maybe because we haven't seen the obvious clue.

- The new picture (high-rez) is not a clue but helped many people. We stopped working on wrong path (jpeg glitch which showed as text in the painting)

- With the new picture (.tif) you can see brush strokes, you can see how she painted it (the order) and even though I'm not sure you can spot some mistake she made and fixed (see the first flame, bottom right, inner track => Looks like it was a purple flame first and changed to a red one).

- It's not bad PR to have an unsolved puzzle. The most famous puzzles were unsolved for a long time.

- Don't burn out on the puzzle, if you don't see anything take a break, come back to it later.

[alphabetacanary]

1) We have 17 leaves, 17x3=51 characters of WIF. WIF is encoded in Base58 (26 letters capital and small and numbers without O, 0, l, I, +, / - so it's smaller Base64).
Could anyone tell me what's the lowest bit number to encode it ? Is 6 bits enough or we need at least 7 ?

Yes, 6 bits is sufficient to cover Base58.  But if you come across 6 bits that represent a value >= 58, you have to decide how to handle that as it would not address a B58 character.  You can wrap around or consider it invalid.

[regisek]

Thanks for replies and @Neantis for good word and encouragement

I want to clarify - when I was writing about new picture I have meant this from 15th of December

https://twitter.com/coin_artist/status/941452262978748417

http://pbs.twimg.com/media/DRC10xTVoAAjR9w.jpg

[Mr.Smithers]

regisek

Why you don't follow me?

Your thinking is good

Im was say !there no queen!

Yari Shogi

1234567890ABCDEF

One from best clue

Actually this is contrary to what they say on Yari Shogi webpage, that the fields of the chessboard (or yarishogiboard, if you will) should be numbered from 1 to 7 and then from A to I. But anyway it hardly can be considered a good clue, as it was pointed out in this thread many times that 7+9 = 16 = number of hex symbols. So try harder and try something less obvious this time

You see only six ribbons on the key?

Follow me! Hey!

In easy say... Keyhole and keyhole is way to the key

Now you see 00? Hey!

no, actually I do not

[Pan Troglodytes]

More I think about it more I think it's broken (maybe because of painting technique?)  (that is why she posted new picture - and she says that nowhere she said it was a clue - maybe she gave up on this but it's bad PR to said that people have followed broken puzzle for 3 years).

this bit I don't get. Are you saying that the new picture actually is a hint or not?

I have few observations and questions, maybe it will help someone.

1) We have 17 leaves, 17x3=51 characters of WIF. WIF is encoded in Base58 (26 letters capital and small and numbers without O, 0, l, I, +, / - so it's smaller Base64).
Could anyone tell me what's the lowest bit number to encode it ? Is 6 bits enough or we need at least 7 ?

2) Notice that queen is not transparent (you can't see chessboard under) except base, where you can clearly see red chess tile (so that you don't mase mistake and assume other colour). Maybe it's important?

3) You say that orange flames are not important. I disagree. If they were not important, she could paint colorful flames more fat? For one thing, they determine the exact place of our 3-coloured friends.

4) What about 6 ribbons on key? Maybe they are some sort of checksum? Like you know, a in Base64 or Q in Base58, when you invented some sort of method to decode chessboard tiles and you stumble apon tile with ribbons on it you can verify?

5) also notice that keyhole along with roots of branches of vines (where they punch through chessboard) are never inside tile, always on borders. Interesting.

That is all for now, I'm going to bed. I have already trouble in work because of this %$#@#$ puzzle, so tomorrow only read some or post but no quality analysis. Fortunatelly I switched off computer during weekend and manage to have a great time with kids, it was worth it.

re 1) I was thinking that maybe the leafs they point to some interesting places: some of them seem to point to some flames, and then some of them point to some chessboard squares, but then again some of them point to nowhere or it is not clear where they point.
re 1) - and there is order to the leafs, at lest some of them clearly grow from the same vine

re 2) note that yari shogi squares have no color assigned

re 4) the red ribbons seem to be coding 011010 bin = 26 dec = 1A hex and 1A may indicate the rightmost upper square. Or not. 26 may be the number of letters A-Z

[asher1101]

I will not be answering one of those

[Pan Troglodytes]

Still you count flame and ignore stupid white rabbit?

http://www.chessvariants.com/ms.dir/yarishogi.html

There no G -just start before A with 0

And thanks to that numbering suggestion the key and the ribbons end up to be on the A1 square, the same number that is encoded in the ribbons. It seems that maybe  the ribbons were intended as the indication on how to count squares, but since nobody figured it out in 2 years .... white rabbit showed us the solution
 

Im will stop show all of you the way, why you say about me -the rabbit - troll?

Its late its late, its no many time to solve

thanks, no reason to get offended, the language you are using is a little odd -  maybe trollish and rabbitish they are similar? and that is why they made the connection

[alphabetacanary]

I find it interesting that this book, entitled "The Shakespearean CIphers Examined" shows the Turtle and Phoenix poem on p.123 as evidence of Francis Bacon's signature.  The claim was made by William Stone Booth whose work is now regarded as 'bunk'.  However, Francis Bacon is credited for inventing a cipher that essentially boils down to a binary encoding.  Probably means nothing but I just found it interesting.



http://marshallfoundation.org/library/wp-content/uploads/sites/16/2014/06/Shakespearean-Ciphers_II_watermark.pdf

[Neantis]

Bacon's biliteral cipher was the main lead some time ago.

I tried some stuff with it, nothing interesting I found but that's because I'm missing something I'm sure.

This picture is nice and explain the concept very well

http://www.cabinetmagazine.org/issues/40/FriedmanNYPL2_FINAL.jpg

Maybe it's something similar to this but with more than just binary. I saw a cypher using a pig where 6 parts of the pig are either 1 or 0 (eye open/close, tail big/short, mouth open/close ....)

[feedo]

Still you count flame and ignore stupid white rabbit?

http://www.chessvariants.com/ms.dir/yarishogi.html

There no G -just start before A with 0

Im will stop show all of you the way, why you say about me -the rabbit - troll?

Its late its late, its no many time to solve

But if we start before A with 0 then we count 0ABCDEF and 012345678 vertically?
and we have two keyhole not just one and the two holes is 00 so we have 00011010?
1from me 1follow me may be we have to start from the rabbit head which points to left upward with his eyes or we should counts toward him ?!!

and that head of a person in the white image that you posted then deleted, who is that person? there is person in the image, i see him i am sure, but is he important?!

the rabbit is the solution, but i am lost... i am lost... i am lost

[680cfbb909]

I'm a bit late to the party but done some experiements with the flames on the border.

The datasets provided by other users seem to be consistent with a number of 152 flames. But no one knows in which order or direction you have to read them.
So what I did is the hard way: I wrote a little script, which tries every possible combination you can read the flames in.

The dataset I used is:

Code:

outer_top_left_to_right = "001101100111001101"
outer_right_top_to_bottom = "001111000101"
outer_bottom_right_to_left = "0011111011010001"
outer_left_bottom_to_top = "01101101"

inner_top_left_to_right = "0110110110100010110110110011"
inner_right_top_to_bottom = "110111110110110111110010"
inner_bottom_right_to_left = "100011100010110111110111110111110"
inner_left_bottom_to_top = "0111001101101"

Of course I tried them forwards and backwards, so I don't miss the right combination.
What I did than, is use the binary strings (152 bit long each) and parse them to ASCII. So I got lots of strings (around 200 MB as a txt file with 10339728 lines)

Example:

Code:

m>Ñ}÷Ú8ž*}·Þm¢ÛYÍ–Î
m>Ñ}÷Ú8ž*}·Þm¢Û[:Îl
m>Ñ}÷Ú8ž*}·Ý³¬æÌÛE¶
m>Ñ}÷Ú8ž*}·Ý³³mÚÎl
m>Ñ}÷Ú8ž-³¬æÌÛE¶O¶û
m>Ñ}÷Ú8ž-³¬æÄûo¼ÛE¶
m>Ñ}÷Ú8ž-³³mÚÎlO¶û
m>Ñ}÷Ú8ž-³³mÙ>ÛîÎl
m>Ñ}÷Ú8ž-³“í¾ìæÌÛE¶
m>Ñ}÷Ú8ž-³“í¾ómÚÎl
m>Ñ}÷Ú8æÚ-µœØx©ößvÎ
m>Ñ}÷Ú8æÚ-µœØx¶ÎO¶û
m>Ñ}÷Ú8æÚ-µœØŸmöx¶Î
m>Ñ}÷Ú8æÚ-µœØŸm÷lãÅ
m>Ñ}÷Ú8æÚ-µœÙlãÅO¶û
m>Ñ}÷Ú8æÚ-µœÙläûo³Å

As expected, there are a lot of non printable characters in there, so I stripped all of them and took a look at my file.
What is left looks like this (just a few lines of the output left after filtering):

Code:

Ž-÷ß9µ·}·ÈñO´[h¶ÎÎl
Ž-÷ß9µ·}·Èñ[h¶ÎÎl>Ñ
Ž-÷ß9µ·}·Èñ[h¶ÌûFÎl
Ž-÷ß9µ·}·ÈûFÎl<VÚ-³
Ž-÷ß9µ·}·ÈûFÎlm¢Û3Å
Ž-÷ß9µ·}·ÈûDñlæÆÚ-³
Ž-÷ß9µ·}·ÈûDñ[h¶ÎÎl
Ž-÷ß9µ·}·ÈûE¶‹lìæÃÅ

What I did know is look for words ("you", "www", "dot", "found", etc) in this file. But nothing can be found.

So what I can say is:
- if my dataset is right and the encoding is in the length of the flames (long = 1, short = 0) you need to find the right starting point
- this starting point is not on a edge
- maybe we need to decode the flames not in ASCII

I hear you guys say: "But 680cfbb909, what about that Bacon cipher we told you about?"
Don't worry, I got you covered!

Bacon cipher uses 5 bit (A/B) to encode one letter. We have 152 flames. So either there are 2 flames too much, or we need to find 3 more to
make the cipher work right.

But I'm a scientist and we do stupid things all the the time, so I wrote my binary strings (152 bit each) to a file and tried to decode them with
that bacon-dude's cipher.

What I found are some words in the strings, mostly "YOU" but to be honest: I decoeded 1,7 GB of binary strings. If you decode such a huge
dataset, it is possible to find a lot of words by coincidence. Of course I checked the lines strings that have a "YOU" in it after I threw that
bacon cipher on to them. But none of those lines made any sense after the "YOU". There's only gibberish.

Conclusion:
Bacon Cipher and ASCII encoding doesn't work on the flames if we don't have the right starting point or use some other encoding before
we convert them to ASCII or use bacon to decrypt them.


So how to find a starting point? TBH, I don't have the slightest idea.
Let's look at the poem.

Code:

Phoenix and the Turtle fled 
In a mutual flame from hence. 
...
Two distincts, division none: 
Number there in love was slain. 
...
Flaming in the Phoenix' sight: 
Either was the other's mine.

What I think is important here is "phoenix' sight". So take a look at the phoenix' eye.
It is the only element, which is aligned with the chess fields and their borders.
https://imgur.com/a/UqNcp
If you follow this line to the top, it ends right on a leaf which points to a spot that
has no flame in it.

Maybe this can be the starting point? But what about the ribbons on the key? They aren't there for nothing. And in which direction should we read the flames?
Maybe those greek-key-thingies tell us something?

[Pan Troglodytes]

Still you count flame and ignore stupid white rabbit?

http://www.chessvariants.com/ms.dir/yarishogi.html

There no G -just start before A with 0

Im will stop show all of you the way, why you say about me -the rabbit - troll?

Its late its late, its no many time to solve

But if we start before A with 0 then we count 0ABCDEF and 012345678 vertically?

I was thinking that the chessvariants page together with the description from our friend white rabbit suggest to count 0ABCDEF and 123456789 vertically (the picture should be rotated 90 degrees clockwise so the rabbit is visible instead of the dove).

and we have two keyhole not just one and the two holes is 00 so we have 00011010?

I was looking for that 00 in the picture and I found it in the hole in the key (not a keyhole in the lock, but a hole in the key), it looks kind of as double O or double 0. But maybe it is like catching at straws

[crax0r]

I'm a bit late to the party but done some experiements with the flames on the border.

The datasets provided by other users seem to be consistent with a number of 152 flames. But no one knows in which order or direction you have to read them.
So what I did is the hard way: I wrote a little script, which tries every possible combination you can read the flames in.

The dataset I used is:

Code:

outer_top_left_to_right = "001101100111001101"
outer_right_top_to_bottom = "001111000101"
outer_bottom_right_to_left = "0011111011010001"
outer_left_bottom_to_top = "01101101"

inner_top_left_to_right = "0110110110100010110110110011"
inner_right_top_to_bottom = "110111110110110111110010"
inner_bottom_right_to_left = "100011100010110111110111110111110"
inner_left_bottom_to_top = "0111001101101"

Of course I tried them forwards and backwards, so I don't miss the right combination.
What I did than, is use the binary strings (152 bit long each) and parse them to ASCII. So I got lots of strings (around 200 MB as a txt file with 10339728 lines)
(...)

You seem to be forgetting that the pattern of 0 1 1 is VERY consistent across the flame-bits.
If you read the flames from inner top, right, bottom, left then at the top of the left go to outer and read down left, then bottom, right, and the top, you would end up right above the point where you started reading from.
(PS. You can also start from outer top and end in inner top, pattern 0 1 1 also prevails.)
(PPS. If you look at the 0 1 1 pattern, then the only safe-transition from inner to outer is on left-side's top.)

Now, if you read every even-numbered flame-length in that stream (first flame is flame-0, lets index from 0), you will get repeated pattern of 0 1 1. Since it's repeated then:
   A) it cannot encode any data or
   B) we should read by skipping some colors, only then such lengths can encode some data. For example, if we read only 78 of the outer-red-colored flames, we still could treat inner color as a bit of data and flame height as a bit of data, regardless if its a part of 011 pattern or not.


Lets focus on:
A) So, if we assume pattern 0 1 1 is there to guide us where to start and end, lets take Inner-top as a starting point.
It starts with short-length flame number 0, so its a pattern-bit-0, that flame still has 2 color-bits, the next flame is uneven, so we take full 3 bits out of it, then the third flame is even again, its long, so its pattern-bit-1, 2 bits again, next flame is uneven again, so 3 bits etc etc.

So, from 152 flames, we get 76 flames that encode 1 bit of 011 pattern and 2 bits of data AND 76 flames that encode full 3 bits. That's 76*2+76*3=380 bits.

It's basically a very solid proof that you can only read the bits in general 2 ways:
1) inner top to outer top (all around the painting as described above)
2) outer top to inner top (all around the painting as described above)

On top of that, since you have 3 data-types, you can choose to read them:
1) In pairs of 2 flames, so 76 * (2+3) bits
2) SplitA: even flames first, 76*2 then uneven ones, 76*3
3) 6 permutations on how to order the data-types in 3bit-flames: height=H, innerColor=I, outerColor=O, then HIO, HOI, OIH, OHI, IHO, IOH
    2 permutations on how to order the data-types in 2bit-flames: innerColor=I, outerColor=O, then IO, OI
    In total: 12 permutations
4) Read each data-type separately, then combine using 6 permutations listed above. So we first make stream of 76bits of Heights, then stream of 152 bits of InnerColor, then 152bits of OuterColor
5) Possibly, 2 out of 5 bits have to be combined into 1 (like Phoenix and the Dove), making it 1+3 bits of data. I like this one cause 76*4=304, an exact length of a WIF-compressed private key.
6) All the 380bits, read in one of many possible "ways" listed above can be:
  a) WIF key (296/304 bits)
  b) base64 key (44*6bits)
  c) pure 32byte long priv key (256bits)
  c) QR 25x25 raw data (skipping statics, to encode 32 bytes in binary mode you need QR Version2, with L-Level ECC == 389bits, missing 9 bits from the rabbits hat? ribbons?)
      For example:

Code:

rs32 389 01101001000111010101111011111000010000101011110001111000101000001101100010101101001010000101001100000101111000011000000010001111110111001010010100000111010001110111110001111000110110100110001000010000101001110110001101100100111010111001111000001101000001001001011001101101011101110001110101001011100110000011100101110111001110000000000110110101100000101110110111010101001001010100010100111

██████████████    ████  ██    ██    ██████████████
██          ██      ██████  ██  ██  ██          ██
██  ██████  ██    ████████  ██████  ██  ██████  ██
██  ██████  ██  ████        ██      ██  ██████  ██
██  ██████  ██      ██  ██  ██████  ██  ██████  ██
██          ██  ██      ████████    ██          ██
██████████████  ██  ██  ██  ██  ██  ██████████████
                    ██  ██                        
  ████  ██████      ██  ██  ████  ██    ██  ██    
    ██  ██      ████          ██  ████████        
████        ██      ██      ████████████  ██████  
  ██  ██      ██  ██          ██████  ██      ████
██  ████████████      ████████      ████  ████  ██
    ████        ██        ██        ██  ██    ████
██  ████    ██  ████  ████    ██    ██████  ██  ██
████    ████  ████          ████  ██          ██  
  ██    ██  ██████    ████  ██████████████  ██  ██
                ██████  ██████  ██      ██    ████
██████████████  ██  ██  ██    ████  ██  ██  ██████
██          ██      ████        ██      ██  ██████
██  ██████  ██      ██  ██████  ████████████████  
██  ██████  ██    ██████                    ████  
██  ██████  ██  ████  ██  ████          ██  ██████
██          ██    ████  ██████  ██  ██  ██    ██  
██████████████    ██  ██  ██      ██  ██    ██████
389
6022d55444dc4f15f3a4306fb979b49a2bf66488a594ab7f806aab278b0e83804ed197

 d) some Bacon-Ciphered text
  e) ascii url

Bonus, funnily, if you read 18 flames from inner top, skipping the flame-0 (which might be there to "tell" us what zero-values are: short, blue, yellow), then you get:

Code:

101 (2) 5 10 (2)
000 (0) 0 00 (0)
110 (2) 6 10 (2)
110 (2) 6 10 (2)
001 (1) 1 01 (1)
001 (1) 1 01 (1)
000 (0) 0 00 (0)
010 (1) 2 01 (1)
110 (2) 6 10 (2)

Format is: databits (1-bitcount) datavalue sumbits (sumvalue)
Note: databits were masked with 011
As you see, by some strange coincidence, the first 9-pairs (so 18 flames) that encode 3+2 bits are somehow connected via this logic. I said it's funny cause im 99% sure its just a coincidence (chance is 1:4^9 so 1 in 262144) (unless Coin_Artist can tip otherwise).

[colonel_koval]

You seem to be forgetting that the pattern of 0 1 1 is VERY consistent across the flame-bits.
(PPS. If you look at the 0 1 1 pattern, then the only safe-transition from inner to outer is on left-side's top.)

I confirm, I made the same observation. So basically we can have only one way to go through the flames - or maybe the second way if we read them backwards. EDIT: Times 152 as we can start anyplace

Now, if you read every even-numbered flame-length in that stream (first flame is flame-0, lets index from 0), you will get repeated pattern of 0 1 1. Since it's repeated then:
   A) it cannot encode any data or
   B) we should read by skipping some colors, only then such lengths can encode some data. For example, if we read only 78 of the outer-red-colored flames, we still could treat inner color as a bit of data and flame height as a bit of data, regardless if its a part of 011 pattern or not.

OR: C) it can be a form of indices into the inner part of the picture, and then the repeating pattern is essential and cannot be discarded. The inner part of the picture would contain the randomized information if C) option is correct.

[BTG-BTC]

I am the only one who does not understand the subject?

[Pan Troglodytes]

Bonus, funnily, if you read 18 flames from inner top, skipping the flame-0 (which might be there to "tell" us what zero-values are: short, blue, yellow), then you get:

Code:

101 (2) 5 10 (2)
000 (0) 0 00 (0)
110 (2) 6 10 (2)
110 (2) 6 10 (2)
001 (1) 1 01 (1)
001 (1) 1 01 (1)
000 (0) 0 00 (0)
010 (1) 2 01 (1)
110 (2) 6 10 (2)

Format is: databits (1-bitcount) datavalue sumbits (sumvalue)
Note: databits were masked with 011
As you see, by some strange coincidence, the first 9-pairs (so 18 flames) that encode 3+2 bits are somehow connected via this logic. I said it's funny cause im 99% sure its just a coincidence (chance is 1:4^9 so 1 in 262144) (unless Coin_Artist can tip otherwise).

I didn't get that part, can you please elaborate? You are using some column names that I am sure you understand very well but they are at best criptic for other people . Also, I didn't catch the regularity in the databits (how you call them) you pasted above, only that there seem to be 9 flames not 18 as you claim.

[crax0r]

Bonus, funnily, if you read 18 flames from inner top, skipping the flame-0 (which might be there to "tell" us what zero-values are: short, blue, yellow), then you get:

Code:

101 (2) 5 10 (2)
000 (0) 0 00 (0)
110 (2) 6 10 (2)
110 (2) 6 10 (2)
001 (1) 1 01 (1)
001 (1) 1 01 (1)
000 (0) 0 00 (0)
010 (1) 2 01 (1)
110 (2) 6 10 (2)

Format is: databits (1-bitcount) datavalue sumbits (sumvalue)
Note: databits were masked with 011
As you see, by some strange coincidence, the first 9-pairs (so 18 flames) that encode 3+2 bits are somehow connected via this logic. I said it's funny cause im 99% sure its just a coincidence (chance is 1:4^9 so 1 in 262144) (unless Coin_Artist can tip otherwise).

I didn't get that part, can you please elaborate? You are using some column names that I am sure you understand very well but they are at best criptic for other people . Also, I didn't catch the regularity in the databits (how you call them) you pasted above, only that there seem to be 9 flames not 18 as you claim.

Code:

databits (1-bitcount) datavalue sumbits (sumvalue)
101      (2)          5         10      (2)

See, 101 are 3-databits from flame 1, its got 2 number 1's, hence 1-bitcount of it is 2. Similarly 110 and 011 would be 2 as well.
Next, there are 2-sumbits from flame 2, they, as a value are number 2.
Why it works: in 3 bits maximum of bits set to 1 is 3 -  "111". With 2 bits you can encode numbers 0,1,2,3, hence 2 bits can be used to count 3bit "1bitcount" perfectly. ECC sort of.

[feedo]

i don't know why you guys are considering only the flames, if its all about the flames and their order then this could be solved 2 years ago. we should focus on the hidden things (the rabbit) instead of the dove and phoenix. i feel like coin artist don't want us to solve the puzzle.. 2 years and she have'nt yet give any clue.. those high quality images and the poem things that she posted early is nothing!.. just to distract our attention.. this way maybe she want to become famous and get more attention as time passes and no one can solve it and bitcoin get higher in value and thus the prize itself.

the useful clues is from the rabbit, just focus on them.. he posted 8 times but there are 2 posts out of them deleted, the first post maybe deleted by the mod but the second one deleted by the rabbit himself, i don't know why?! maybe it hides important clue.

my question was if we start counting the yari shogi 9*7 board as 0ABCDE Horizontally and thus we get 012345678 vertically, in this case there is no G and no 9 !!
he said in one post

Yari Shogi  1234567890ABCDEF  but this is 7*10 board so where we put 9? and if there is 0 before A like what he said then the keyhole in the first line is on A0 not A1

we have to arrange those numbers on the board correctly.
there are two keyholes not only one and the two form 00 and with the ribbons becomes 00011010 (he put this code in his first deleted post)

Keyhole and keyhole is way to the key
1from me is solution (maybe he refers to direction or the starting position should be from the rabbit)

the rabbit in the image is very important (Still you count flame and ignore stupid white rabbit?)

in the second deleted post he posted this modified image but i don't know who is in the image? ( there is head and two eyes in the middle).. and why he deleted that post?


anyone here can tell me how i upload image to post instead of inserting link?

[dimouj]

i don't know why you guys are considering only the flames, if its all about the flames and their order then this could be solved 2 years ago. we should focus on the hidden things (the rabbit) instead of the dove and phoenix. i feel like coin artist don't want us to solve the puzzle.. 2 years and she have'nt yet give any clue.. those high quality images and the poem things that she posted early is nothing!.. just to distract our attention.. this way maybe she want to become famous and get more attention as time passes and no one can solve it and bitcoin get higher in value and thus the prize itself.

the useful clues is from the rabbit, just focus on them.. he posted 8 times but there are 2 posts out of them deleted, the first post maybe deleted by the mod but the second one deleted by the rabbit himself, i don't know why?! maybe it hides important clue.

my question was if we start counting the yari shogi 9*7 board as 0ABCDE Horizontally and thus we get 012345678 vertically, in this case there is no G and no 9 !!
he said in one post

Yari Shogi  1234567890ABCDEF  but this is 7*10 board so where we put 9? and if there is 0 before A like what he said then the keyhole in the first line is on A0 not A1

we have to arrange those numbers on the board correctly.
there are two keyholes not only one and the two form 00 and with the ribbons becomes 00011010 (he put this code in his first deleted post)

Keyhole and keyhole is way to the key
1from me is solution (maybe he refers to direction or the starting position should be from the rabbit)

the rabbit in the image is very important (Still you count flame and ignore stupid white rabbit?)

in the second deleted post he posted this modified image but i don't know who is in the image? ( there is head and two eyes in the middle).. and why he deleted that post?


anyone here can tell me how i upload image to post instead of inserting link?

The key with ribbons is at the place A1 as the binary hides behind so left from it its the 01 .

Check the board again its actually 7×9 not 7×10

Upload an image to imgur and provide the link!
Greetings from Greece

[feedo]

https://www.youtube.com/watch?v=yvJtwh_jcpk

Real flame?

Remember thats stupid rabitt say

Thank you Rabbit!, without you we can't advance and solve this puzzle not even in 200 years!