Find repeating patterns in the algorithm, normalize it into some code statements. You do not need to go through all N squares and check whether each of them have been visited upon every iteration.

From 2, try to solve all possible triangles with bigger and bigger length, by hand. Until one point you will find it. GLHF.

Since A & A = 1 and logical OR is True when at least one operand is True, the response is "1 OR 0 OR 0" = 1.
(Logical AND has a higher priority than OR)

Damn, I didn't think it was possible to do it in one line.
Nice job!

Take a look at the last 2 columns, the 3s and 5s.

Not an issue. This kata is BF-specific.

Its first arg (enclosed in ()) is actually a generator expression. You may find enlightening this reading about iterators :)

By 'take two of different color each day' it means you can't pick the same color twice, but you CAN pick for example from 1 & 3 on day1, and 1 & 3 on day2.

This is why for solve([8,1,4]) you can pick 1 & 2, then 1 & 3 (4 times), and only the first color will be remaining. For solve([8,2,8]), it's 1 & 2, 2 & 3 and then 7x 1 & 3.

Does that make it clearer?