Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 205 / 4 / a / Solution

The graph has edges $1\to 2\leftarrow 3$, $1\to 4$, $2\to 4$, $3\to 5$, $4\to 5$, $4\to 6$, and $6\to 5$. Any D-separating set for 1 and 6 must contain 4 because of the directed path $1\to 4\to 6$. Conditioning on 4 activates the collider $1\to 4\leftarrow 2$ and, through its descendant, the collider $1\to 2\leftarrow 3$. The remaining route through $3\to 5\leftarrow 6$ is open exactly when 5 is conditioned on and 3 is not. Hence all separating sets, among the nonendpoint vertices, are

For the second graph, $2\perp \perp 5\mid 1,6$ forces colliders on the unblocked two-edge paths: $2\to 3\leftarrow 5$ and $2\to 4\leftarrow 5$. Acyclicity then forces $1\to 4$ and $6\to 4$. Thus its edges are

Solved by gpt-5.6-sol high.