Some analytical sets give you a small network — nodes connected by channels that behave differently — and then hang several questions off the same picture. The official HAT sample paper spends six of its ten analytical questions on exactly this shape: seven supervisors, a one-way message circuit, a two-way radio, and a stream of "can this message reach that supervisor through exactly one intermediary?" items.
That is the highest-leverage set on the paper. You read the setup once and answer six questions from one diagram. Candidates who skip drawing the diagram answer those six questions by re-reading the paragraph each time, lose the clock, and guess. Candidates who draw it once collect six marks in under four minutes.
Model the network as two kinds of line on one page, and never mix them up.
The model omits capacity and collisions: the real questions never ask how many messages fit, only whether a route exists and how long it is.
The reading order that works. (1) Copy every channel onto your diagram, arrows for the circuit, plain lines for the radio. (2) Mark which nodes have no radio at all — they can only be reached or left by the circuit. (3) Only then read question one.
Four question shapes, one skill each.
| Shape | What it asks | Method |
|---|---|---|
| Exactly k intermediaries | A path of length k + 1 hops | Walk candidates; count the middle nodes |
| Minimum intermediaries | The shortest legal path | Breadth-first: try 1 hop, then 2, then 3 |
| Requires both channels | No all-circuit or all-radio path exists | Prove the single-channel routes fail |
| Forced waypoint | Every path goes through node X | Remove X; if the target becomes unreachable, X is forced |
Direction is the trap, not arithmetic. On a one-way circuit the reverse hop simply does not exist: if the circuit only runs C → D, then a message from D to C must find another channel or another way round the loop. Radio hops are symmetric, so it is almost always the circuit's direction that decides a question.
Use this network throughout — it has the same two-channel shape as the sample paper:
One-way circuit: A -> B -> C -> D -> E -> A Two-way radio: B - F, F - G, D - G (Any node may act as an intermediary.)
Question 1 — exactly one intermediary from B to D? Walk the circuit: B → C → D. The middle node is C, and no two-hop all-radio path exists because B and D share no radio line. Answer: C.
Question 2 — shortest route from F to A? F's radio exits are B and G. Via B the only onward move is the circuit B → C → D → E → A, giving F-B-C-D-E-A with four intermediaries. Via G: G's radio reaches D, and D continues on the circuit D → E → A, giving F-G-D-E-A with three intermediaries — G, D and E. Answer: 3 intermediaries, the path F-G-D-E-A.
Question 3 — which pair needs both channels? A to F: every move from A is on the circuit (A → B → C → D → E → A), and F hangs only off the radio, so the path A-B-F uses the circuit A → B and then the radio B - F. Two channels, one intermediary. Answer: A to F.
A message from B to D uses exactly one intermediary. Which node is it?
What is the minimum number of intermediaries for a message from F to A?
Which pair requires using both the circuit and the radio?
A message travels from G to B through the fewest possible intermediaries. Which node does it pass through?
Without looking back: what does "exactly two intermediaries" mean in hops, and which channel type is directional?
The answer is: a four-node path sender-X-Y-receiver, three hops with two middle nodes; only the one-way circuit is directional — the radio works either way.
On the one-way circuit, traffic may travel only along the , while the two-way radio works in directions. An intermediary is any node strictly the sender and the receiver, and the endpoints are never counted. A path with exactly one intermediary has hops.
Show that a message from E to F requires both channels, and give a shortest route with its intermediary count.
E and F sit on different sides of the network: E is on the one-way circuit only, F on the radio only. So any path must include at least one circuit hop and at least one radio hop — both channels are mandatory.
Shortest route: E → A → B then B - F gives E-A-B-F, with intermediaries A and B — two of them. The radio hops B-F and F-G never shorten this, because from G you would still need G-D and then D-E-A-B to start from the wrong side.
Check the reverse direction while you are here: F to E is the same three hops read backwards on the radio legs, but the circuit legs must still run with the arrows (D to E, not E to D), which is why path-finding on these sets is directional even when the endpoints look symmetric.
You can now draw a two-channel network once and answer every question hung on it: exact hop counts, minimum intermediaries, both-channel proofs and forced waypoints. The next lesson returns to verbal deduction — syllogisms — where the diagram is Venn circles instead of routing lines, but the habit is the same: draw the structure before reading the options.