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