Let's say you know the colour of the first square, what colours can you figure out next?

Colour it like a chessboard, but there's more than one possible chessboard!

Brute force the color of the first square!

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Consider the contribution of each pillar to the answer.

How does doing the operation on a pillar change the contribution to the answer?

You only want to do an operation on a pillar if the top cell contributes 2 to the answer.

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Let's say you know whether you flipped the first cell or not, which new columns must be flipped?

Create a graph of $$$n$$$ nodes, draw an edge between two values if they both exist in the same column, what structure does this graph have?

It is a bunch of disjoint cycles (since each node has an in-degree of $$$2$$$).

Each cycle is independent, and there are $$$2$$$ ways to flip each cycle.

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All trees are bipartite.

Find a $$$2$$$-coloring of a tree in one query.

You can find all nodes adjacent to one node in one query.

The min(number of black, number of white) nodes is $$$\leq \left \lfloor{\frac{n}{2}}\right \rfloor$$$.

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Try to solve this problem for $$$n \leq 15$$$.

Try BFSing on the bitmasks.

Do we really care about what elements are in the bitmask?

Only the number of nodes in each mask matters.

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Make a directed graph on the sand.

The graph should satisfy the following property: if a node $$$a$$$ can reach another node $$$b$$$ by a sequence of edges, disturbing $$$a$$$ will disturb $$$b$$$.

Compress scc's in the graph.

You must disturb all nodes with in-degree $$$0$$$ in the new graph.

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Continued from $$$F1$$$, read that first.

The $$$a_i^\texttt{th}$$$ nodes from the bottom of each column must eventually be disturbed (call them special).

If a special node can reach another special node, you can delete that node.

Each node will remove some contiguous subsegment of the remaining special nodes' columns.

Find a greedy strategy to cover all of the columns.

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