Quote from: alphabetacanary on January 04, 2018, 05:58:06 PM

(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)