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

[itod]

Also, the repeatling pattern alphabetcanary found is most definitely 100% there. It's going one direction in the inner ring and the opposite direction in the outer. So at the very least, the lengths of the even flames do not encode information. The colors still could, but that changes your count..., something to consider.

I agree that pattern alphabetcanary found changes everything. Since we are looking for vanity address, the pattern could not be part of the key unless we imagine that author developed special software that would make a key that hashes to 1FLAME, and in the same time puts a pattern inside the key, which is a borderline impossible regarding the length of the pattern. Conclusion is that pattern breaks the reading of the flames, and understanding the pattern is necessary to understand how the flames should be read. Or the pattern could be the actual reading and everything else in the flames is not important, which doesn't seem likely but is possible. In any case IMHO the flames are unusable until we understand why the pattern is there and what does it represent.

[feedo]

Wow!

That is great, meaningul and logical post. Great reading, thank you!

though i believe that sini is an honest member trying to share something useful but his long post is completely useless! why?,  simply because he thinks in a logical way!

this puzzle is not difficult..(difficult means it requires time, effort, thinking in a logical way, working in teams, solving in order steps 1 leads to 2 then 3,4,5...etc
but this is not the case at all in this puzzle (in fact it's a big troll) and all those who have thought in a logical way failed and still they are failing after 3 years! why?!

because simply if you think in a logical way you are highly probable wrong and will fail and waste your time, not because it's difficult but those trolls who created this puzzle embedded things in an extremely highly unpredictable crazy way! so that its impossible to see those embedded things.

you must think in a highly crazy unlogical way to get only one fraction of the solution!
turn your brain on and think about that, you see that white piece on the board and your brain tells you that this is queen, right? wrong! as this piece is simply Alice and you need that letter A and you must think in crazy way and you should be crazy lucky to find those characters ea oem 011 and arrange them correctly to get only a small fraction of this puzzle!
you see those green things and your brain tells you that those are leaves and you start to count them, right? again you are wrong! those are not leaves!

your eyes see those colored beautiful ordered flames and your brain tells you to count them and arrange them according to their lengths, colors... etc ... and you are on the wrong way!!
the flames don't hold the solution (maybe they only represents 10% of the puzzle) and you don't need all of them! and again you have to think in a crazy fucking way that only those trolls who created this puzzle know!
even the colors.. who told you that there is only blue,red,green, yellow,orange?
don't you see that each of those colors has many gradients (light green, dark green, dark blue, purple...etc)

i am not sad because i can't solve this puzzle and take the money, i'll be so happy if anyone deserves the prize win, but i am sad because people still wasting their precious time thinking in a logical wrong way trying to solve this biggest troll after almost 3 years!

the only one who benefits from that is coin artist (or better Coin Troll)

[sini]

What the fuck did I just read

[itod]

i am not sad because i can't solve this puzzle and take the money, i'll be so happy if anyone deserves the prize win, but i am sad because people still wasting their precious time thinking in a logical wrong way trying to solve this biggest troll after almost 3 years!

The only way for you not to be sad is to leave us here to waste our time, and not waste your precious time on this at all. Do you consider us idiots so you know better what should we do with our time then ourselves?

Please stop this conspiracy theories, either try to contribute something useful or leave if you do not enjoy the puzzle. You may find it surprising, but some nerds actually enjoy hard, almost unsolvable things. The harder it is the more enjoyable is time spent here.

[sini]

I'm used to walls of frustration. It comes with the hobby. This puzzle at least has produced interesting conclusions within a week. Some puzzles I've looked at, have produced nothing for years.

[kn0w0n3]

You may find it surprising, but some nerds actually enjoy hard, almost unsolvable things. The harder it is the more enjoyable is time spent here.

This

[feedo]

And my personal opinion after some thinking. This puzzle won't be solved. I don't know if it is on purpose or simple mistake, but think about it. Only one person benefiting from it it is creator and we are all wasting time. 3 years!!! Really?!

+1

[alphabetacanary]

The two missing flames is a good theory and deserves to be tried.  I'm going to share the path I attempted which assumes there is no missing data.

If you interpret the data stream 6 bits at a time, with the traversal path I outlined in a previous post, you get strings that look like this:

Code:

UTCUGYUYBFCYYYFUUTCYBGCFU2Mnsq3?cqkarYvtusft8c30rxBEB?FxwHn1tVRjXHFdQ?RYiUKh 
UTCUGYUYBFCYYYFUUTCYBGCFU9ukq?YS3u4Erbiz10JY?a?enBZNCVm1oy?ByVQ?V?DimryqomWs 
UTCUGYUYBFCYYYFUUTCYBGCFU2Mnsq3?cqkarYvtusft8c30rxAsv4r9ApK?DcgNaprUh1gZPdnQ 
EgEG6Gn6nngGppn86GnppEn6nd?cQHrytSLCtk?cav4rxH4LnBZNCTQyU?heqAzMR?D3uApnq?Kr 
deudqZdZvruZZZrddeuZvqurdrMnsq3?cqkarYvtusft8c30rxAsv4r9ApK?DcgNaprUh1gZPdnQ 
deudqZdZvruZZZrddeuZvqurdxCMG5Zf?C?sFWP7?6oZ3X2TKvVNCVm1oy?ByVQ?V?DimryqomWs 

(Here, I've concatenated all the data starting with lengths, then inner, then outer colors in the order of the pattern. This is just ONE of a many ways of concatenating the data. Just for demonstration purposes, I used a base 58 alphabet here and use ? when the value goes beyond 58.  By no means is this a complete list).

Now, due to the pattern in the lengths, the beginning part sticks out like a sore thumb because there are actually only 8 possible combinations for 6 bits when you fix 3 of them (0.1.1.).  Turns out only 7 are actually used.  That's why the characters repeat. So let's split the strings where the repeated characters end:

Code:

EgEG6Gn6nngGppn86GnppEn6n d?cQHrytSLCtk?cav4rxH4LnBZNCTQyU?heqAzMR?D3uApnq?Kr
UTCUGYUYBFCYYYFUUTCYBGCFU 2Mnsq3?cqkarYvtusft8c30rxBEB?FxwHn1tVRjXHFdQ?RYiUKh
UTCUGYUYBFCYYYFUUTCYBGCFU 9ukq?YS3u4Erbiz10JY?a?enBZNCVm1oy?ByVQ?V?DimryqomWs
UTCUGYUYBFCYYYFUUTCYBGCFU 2Mnsq3?cqkarYvtusft8c30rxAsv4r9ApK?DcgNaprUh1gZPdnQ
deudqZdZvruZZZrddeuZvqurd rMnsq3?cqkarYvtusft8c30rxAsv4r9ApK?DcgNaprUh1gZPdnQ
deudqZdZvruZZZrddeuZvqurd xCMG5Zf?C?sFWP7?6oZ3X2TKvVNCVm1oy?ByVQ?V?DimryqomWs

Oh look, you end up with exactly 51 characters left over after the pattern ends.  (Note: This actually means that you start taking 'key' bits while two bits away from the end of the lengths but I don't think that really matters. That could be the tricky part. There's no reason the key needs to start exactly on a boundary between lengths and colors, for example.)

So my theory is that the first part is supposed to be used as a key to deciphering the second part.  Of course, if you leave the lengths to the end, then the whole thing gets reversed but you end up with the same situation.  Basically, 51 *ciphered* characters representing the bitcoin key, then another key to deciphering THAT key. If you try this, you will of course have to keep the values those '?' represent in my examples so you can apply math operations on them.  I think once you do the right math, everything will fall within 0-57 as expected.

So in short, my theory is:

Flames = FlameEncode(cipherkey + Cipher(bitcoinpk, cipherkey))
or
Flames = FlameEncode(Cipher(bitcoinpk, cipherkey) + cipherkey)

I've tried many operations trying to use those 25 characters as a deciphering key for the 2nd part with no luck (add, subtract, xor, bit shifting, even hill cipher 5x5 matrix).

Good luck!

[edit : This theory means the cipher key has no meaning. It doesn't have to have any meaning. It was chosen to be conspicuous so you can separate it from the random data which is the bitcoin key]

[sini]

The outer ring, if you count orange flames, has almost the exact same number of flames as the inner ring. The thing is having a clear definition of what is an 'orange' flame and a colored one.

My theory stands as well that the outer ring and inner ring are needed to decode one another. I believe the outer ring is more likely the key, and the inner ring the cipher text. But the rest is a mystery.

[feedo]

The outer ring, if you count orange flames, has almost the exact same number of flames as the inner ring. The thing is having a clear definition of what is an 'orange' flame and a colored one.

My theory stands as well that the outer ring and inner ring are needed to decode one another. I believe the outer ring is more likely the key, and the inner ring the cipher text. But the rest is a mystery.

According to your great theory we need only 10% of the image to solve the puzzle! what an amazing theory!  it sounds like coin-artist is stupid 

https://imgur.com/a/91nrE

[sini]

Why are you getting salty. You're literally oozing cognitive dissonance. The turtle dove and phoenix were important. The chess board, not as much.

[feedo]

The chess board, not as much.

LOL , so you don't have another great theory for the board?

[feedo]

Oh, I know how not to be a troll

"Count, the number of flames, there is hidden, she hid, it must be, there is no other solution, only flames, any other solution is bad"

Now it fits and I'm not a troll anymore?

Every normal person will think twice about what they count at all and if their puzzles are worth it

I miss you 

[smracer]

(Here, I've concatenated all the data starting with lengths, then inner, then outer colors in the order of the pattern. This is just ONE of a many ways of concatenating the data. Just for demonstration purposes, I used a base 58 alphabet here and use ? when the value goes beyond 58.  By no means is this a complete list).

Have you tried using the inner shape as the third bit (skinny line or blob of paint)?

Each inner flame is either a blob or a line.

Inner Color, Outer Color, (Line/Blob)

8 variations of flames taken in pairs.  76 pairs of 64 variations.

I still think we are looking for a BIP38 encrypted key starting with 6PR.

[CompNeuro]

(Here, I've concatenated all the data starting with lengths, then inner, then outer colors in the order of the pattern. This is just ONE of a many ways of concatenating the data. Just for demonstration purposes, I used a base 58 alphabet here and use ? when the value goes beyond 58.  By no means is this a complete list).

Have you tried using the inner shape as the third bit (skinny line or blob of paint)?

Each inner flame is either a blob or a line.

Inner Color, Outer Color, (Line/Blob)

8 variations of flames taken in pairs.  76 pairs of 64 variations.

I still think we are looking for a BIP38 encrypted key starting with 6PR.

Can I ask why you think that's the case?

[Pan Troglodytes]

5). We begin to look at the requirements for the output. We see that this is closest to the WIF format. Given we know WIF is base 58 we can do a calculation to know the minimum amount of information needed to express one character:

ceil(log2(58)) = 6 bits

6 bits * 51 characters = 306 bits

So given, if the flames were used to encode a WIF string you'd need 2 flames for each character.

Actually WIF requires 304 bits which is (so it happens) 38 bytes which is 32 bytes for the 64hex private key format+1 byte (x80 header)+1 byte (x01 footer) + 4 bytes checksum. This is why 152-flames-coding-two-bits-only theory (with bits coded by inner&outer color and using length only as a pacemaker) was so appealing to many.

- 2 missing symbols represents the death of the phoenix and turtle dove.

I am sorry to have written the end to your theory of two missing flames symbolizing two loving souls. It was beautiful but unsound.

I'll give a fairly big hint. There's two missing flames in the inner ring. Beyond that, the way things are encoded is a mystery. But that should be enough, if you can reason with the theme of the poem, to catch up with what he's likely talking about.

Are those the two flames supposedly representing the death of the phoenix and turtle dove?

One more point. You published a nice google document with some information on missing flames in the outer border. The document has been subsequently edited with the relevant information removed. What is really your group's purpose ?

I used 6 bits assuming that each character was encoded using a fixed length (see Bacon Cipher). I edited it (I only have editing permissions) because I did not have conclusive evidence to keep it in the document. The document is my personal platform for notes, what I see as distracting or misleading I remove.

Edit: Are you assuming compressed private key?

it really does not matter. The compressed and noncompressed differ only in one byte (the x01 footer). Please refer to https://en.bitcoin.it/wiki/Wallet_import_format

The relevant data is 32 bytes for a bare private key, 37 bytes for noncompressed wif (no x01 footer), 38 bytes for compressed wif. The base58 is just another representation for those 37 or 38 bytes. Because the first byte is x80 and base58 codeing is not byte-per-byte, it starts always with 5 or K(L) depending if it is compressed or not. Whatever it is, the bitwise contents will be same (regarding the middle part i.e. the bare private key) for all representations.

@sini: Please reshare your outer border missing flames analysis

[zbyszek2]

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.

[Pan Troglodytes]

(Here, I've concatenated all the data starting with lengths, then inner, then outer colors in the order of the pattern. This is just ONE of a many ways of concatenating the data. Just for demonstration purposes, I used a base 58 alphabet here and use ? when the value goes beyond 58.  By no means is this a complete list).

Now, due to the pattern in the lengths, the beginning part sticks out like a sore thumb because there are actually only 8 possible combinations for 6 bits when you fix 3 of them (0.1.1.).  Turns out only 7 are actually used.  That's why the characters repeat. So let's split the strings where the repeated characters end:

Code:

EgEG6Gn6nngGppn86GnppEn6n d?cQHrytSLCtk?cav4rxH4LnBZNCTQyU?heqAzMR?D3uApnq?Kr
UTCUGYUYBFCYYYFUUTCYBGCFU 2Mnsq3?cqkarYvtusft8c30rxBEB?FxwHn1tVRjXHFdQ?RYiUKh
UTCUGYUYBFCYYYFUUTCYBGCFU 9ukq?YS3u4Erbiz10JY?a?enBZNCVm1oy?ByVQ?V?DimryqomWs
UTCUGYUYBFCYYYFUUTCYBGCFU 2Mnsq3?cqkarYvtusft8c30rxAsv4r9ApK?DcgNaprUh1gZPdnQ
deudqZdZvruZZZrddeuZvqurd rMnsq3?cqkarYvtusft8c30rxAsv4r9ApK?DcgNaprUh1gZPdnQ
deudqZdZvruZZZrddeuZvqurd xCMG5Zf?C?sFWP7?6oZ3X2TKvVNCVm1oy?ByVQ?V?DimryqomWs

Oh look, you end up with exactly 51 characters left over after the pattern ends.  (Note: This actually means that you start taking 'key' bits while two bits away from the end of the lengths but I don't think that really matters. That could be the tricky part. There's no reason the key needs to start exactly on a boundary between lengths and colors, for example.)

Of course it does leave 51 chars in base58 encoding. It is by no means surprising. As I said above, 152 flames, if you assume a flame codes only 2 bits i. e. two colors, inner and outer (with length bits EXCLUDED, exactly how you did exclude length information but using a different method), code 304 bits which is exactly the WIF length and as I said above, this is exactly the reason this theory was so appealing to many forum members (around page 29 of this thread I think)

[alphabetacanary]

Have you tried using the inner shape as the third bit (skinny line or blob of paint)?

No.  I noticed those too but just felt it was not consistent enough to count as data.  6PR theory might still be possible but I haven't gone down that road.

[alphabetacanary]

Of course it does leave 51 chars in base58 encoding. It is by no means surprising. As I said above, 152 flames, if you assume a flame codes only 2 bits i. e. two colors, inner and outer (with length bits EXCLUDED, exactly how you did exclude length information but using a different method), code 304 bits which is exactly the WIF length and as I said above, this is exactly the reason this theory was so appealing to many forum members (around page 29 of this thread I think)

Just to clarify terminology, WIF format is not 304 bits (my mistake on a previous post too, BTW). 

WIF implies a base58 encoding because it's the format you import into a wallet.  In this case, it would be either 306 bits for a non-compressed version and 312 for compressed version assuming 6 bits per char.

304 bits = 38 bytes which is enough space for a compressed 'raw' key (not base58, but the bytes before it gets base58 encoded) which fits just the inner/outer color informatin.

I think the subtle difference here is I don't exclude ALL length information when making the key.  I include 2 bits to get up to that 306 number so I can make a base58 string.  But I agree, this is not that surprising since I outlined all this around page 25 of this thread.