Begin with a finite sequence of blocks in a row, each in one of 3 colors: red, blue, yellow.
Below each pair of neighboring blocks place a new block with the color rule: If the blocks are the same color use that color but if they are different use the third color.
Example:
r b y y b
y r y r
b b b
b b
b
How can the color of the last block be easily predicted from the top row?
Note: I don't know the full answer but can solve special cases.
Relabel the colours 0,1,2 in any order. Then, if a_1 is the daughter of a_2 & a_3, we have
a_1 = 2(a_2+a_3)mod 3 (Eq1)
Consider the case of a 4 row triangle. The entry at the base we call a_1, and the top row is a_7,..,a_10. Iterate the mod equation 3 time gives
a_1 = 2(a_7 + a_10)mod 3 (Eq2)
Suppose there are N rows, where N = 3M + R, 0 <= R <3 Use induction on Eq2 to express a_1 in terms of elements in row 3M. Then apply Eq1 to express those elements in row 3M in terms of elements in row N.
The index of the leftmost element of row 3M is K = (3M+1)(3M+2)/2. Then induction gives:
a_1 = 2^K Sum over J M!/J!(M-J)! a_(K+3J) mod 3 (Eq3)
=(-1)^K Sum over J M!/J!(M-J)! a_(K+3J) mod 3
where J runs from 0 to M.
For example, with M= 3, we get K = 55 and
a_1 = 8( a_55 +3a_58 +3a_61 +a_64) mod 3 = -a_55 -a_64 mod 3
Edited on December 25, 2018, 10:11 pm
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Posted by FrankM
on 2018-12-21 09:18:14 |
The good parts (sketch of proof)
