Topic: The Legend of Satoshi Nakamato, FINAL STEP PUBLISHED.... 4.87 BTC GRAND PRIZE! - page 46. (Read 108519 times)

LOL seriously? more than 2 and half years and you consider that a hurry!!
i think you stand for coin artist because she doesn't want us to solve it to become more famous as long as time passes and nobody solve it and the prize value get bigger, she is clever thats why she doesn't give any clue or even useless fake one to waste time

Why are you in such a hurry to have your puzzle solved then?

It may not have been solved in 2 years due to lack of exposure. I did not know about this puzzle until fairly recently.., It has been great reading through this and past ARGs from the same creator/creators though. Put a little mystery back into the world..., Which is why I don't want to see it end prematurely. If you really are the creator of this puzzle, You've put together a challenge that one doesn't come across to often in their life. I hope you'll let it play out..., Also, if you really did make this.., Thank you for the challenge.

@Feedo

I believe she stated that the book was not a hint.., Although, even if it was a hint. That doesn't necessarily mean you'll understand what the hint is simply from reading the book.

without the rabbit help we need another 1000 years to solve it, even coin artist didn't give any help or useful clue,, i wasted 2 hours reading that useless hint that she posted (Martyrs Love book) that turns to be nothing more than wasting of time and distraction of attention.

I am here just staring at the JPG ...thinking about "lossy compression"

It's like when someone gives you the solution to a riddle. The fun is in trying to solve the thing, not in being told how to do it.

If you are indeed the creator, you can do as you please with your puzzle. I just enjoy the mystery in it..,

@feedo

The face you circled is what I originally thought was a joker.., I don't think it is a face though.

I see your sv now, but I'm not convinced by that.

Personally, I hope that the rabbit is not actually giving out clues. A large part of the reason I enjoy working on this puzzle, is because it has not been solved in so long. I enjoy the mystery behind it.., I also like seeing the ideas other people come up with..., IDK, I just feel like hints of this nature (If they really are hints) kind of ruin the whole thing..., Actually, the disappoint I experienced seeing someone posting hints is what prompted me to make an account in the first place.

Any who.., It seems as though the rabbit is actually giving plausible hints.., but please stop rabbit.

It makes sense that the picture should be oriented so you can see the rabbit. It also makes sense that the information we need to extract is in the whole picture, as apposed to just in the flames. I also think it is a pretty crazy that the portion of the keys containing ribbons that could encode A1, fall into the A1 tile. If we number the tiles as follows:

9 09  A9
8 08  A8
7 07  A7
6 06  A6
5 05  A5
4 04  A4
3 03  A3
2 02  A2
1 01  A1  B1  C1  D1  E1  F1
   0    A    B    C    D    E    F


I also have a suspicion (that if this hint is actually a hint) some of these tiles need to be flipped.


As far as the rabbits black and white picture, I don't see any face, I thought I saw the face of a joker in it the very first time it was posted, but upon closer examination i think it was just my tired mind.


I've tried a lot of different things, and haven't really discovered much that hasn't been said already, at least nothing I want to share at the moment.


One thing I did notice that I haven't seen mentioned, that may or may not mean anything, is that all of the tracks of flames are divisible by 4, except for three of them.



EDIT:

Since I posted, I might as well ask.... If you flip the picture so you can see the rabbit. It appears as though there is an arrow pointing to the flame in the bottom left corner. If you zoom into the arrow it looks like it is just a brush stroke, and so I dismissed it (for the most part). I'm curious if anyone thinks that this might actually be purposeful, or if the general consensus is that it is indeed just a brush stroke.

Even with image edit i cant see something ! What do you think?

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

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

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?

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.

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.

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

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


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.

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:
 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:


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 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).


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

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:


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:

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):

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.


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?

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

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 ....)