The key knowledge required is: the number of non-intersecting paths between a source and destination in an unweighted, directed graph is same as the maxflow between them.

To apply this to the problem, first count the number of empty cells at the edges. If it is not 4 then the answer is no. If it is 4, we need to check if we can choose 2 of the edge points as entries (sources) and the rest 2 as exits (sinks). Let us try all possible combinations of entries (there are 4C2 of them).

To check if a entry/exit assignment is valid we have two approaches based on the key knowledge stated above:

1. Convert the grid to a directed graph with a source and sink. Add an edge from source to each entry, add an edge from each exit to the sink. For each cell, make two nodes in the transformed graph (in_node and out_node). Add an edge from in_node to out_node. Now for each valid (empty) neighbouring cell add an edge from out_node of current cell to in_node of the neighbouring cell. All edges have capacity 1. Now if the max-flow in this graph is 2 that means it is secure
2. Another approach is — note that the above approach is essentially finding two augmenting paths between source and node. That can be implemented without the transformed graph by finding a path from a source to sink and trying from another source but keeping track of the previous path and only travelling in reverse direction in this path (similar to residual network). This requires careful implementation.

One optimization is — note that we can always swap the entries with their corresponding exits. So, we can always fix the first point as an entry and try each of the other point as an entry. This cuts run time by 2 as we need to try only 3 possibilities instead of 6.

Since maxflow is only 2, Edmond Karp should work. Even Dinic's will work. The approach 2 above worked for larger n*m but we decided to keep it small to allow standard flow approaches to pass too.