Just check is the 1 in the set.
One can snuggle pictures to each other and to edge of stand.
Let's join regular triangles to three edges of hexagon (to 1st, 3rd and 5th edges).
One can check if lexicographically smallest string wich is equals to first string in statement the same as second.
Let's paint bottom-right cell to black color. Then let's calculate number of ways to came to each black cell avoiding previous black cells via dp.
Let's remember Pick's theorem and let's consider all potencial sides of polygon separately. Do we really need to consider them all?
Lightened part of path is union of disjoint segments. Hence if we know what segments of path each continious subsequence of spotlights can lighten then we can solve the problem with simple (for such problem munber) dp.
How one can to dicover this information?
When we starts with some spotlight and will add spotlights one-by-one, we can't store all information about it process because we have union of disjoint segments of each step. Union of disjoint segments have too much parameters to store all possible unions. But if we interested of possibility of segment, it is important to know only leftmost and rightmost points of union and leftmoost unlightened point beween them.