Path puzzles: find bottlenecks before drawing a route

Translate the board into connections

When a puzzle asks you to connect locations, first distinguish the picture from the connections it permits. Two objects that look close may not have a legal link. Two distant objects may connect through a portal. A route plan should use the actual allowed moves.

Our network has six nodes: A, B, C, D, E, and F. Its undirected edges are A–B, B–C, C–A, C–D, D–E, E–F, and F–D. Each edge can be crossed in either direction. A triangle on the left meets a triangle on the right through the single edge C–D. The basic goal is to get from A to F.

A-----B       E
 \   /       / \
   C---------D---F

Edges: AB, BC, CA, CD, DE, EF, FD

The connector is unavoidable

Every route from A to F must cross C–D. Remove that edge and A, B, C are disconnected from D, E, F. That makes C–D a bridge in this particular network. The triangles contain alternate local routes; the connector does not.

A shortest route is A, C, D, F, which uses three edges. Taking A, B, C, D, E, F also reaches the exit, but uses five. If the goal is simply to arrive using as few edges as possible, there is no reason to visit B or E. If the goal awards collectibles there, the objective has changed and the longer route may be necessary.

Changing the goal changes the solution

Suppose the objective is now to visit every node exactly once, starting at A and finishing at F. A, B, C, D, E, F is valid: every consecutive pair has an edge, and no node repeats. The three-edge route fails because it misses B and E.

Why finish the left triangle before crossing? Once you cross C–D, returning to an unvisited B would require revisiting C or D. Under the no-repeat rule that return is impossible. The connector therefore imposes an order: complete the left component, cross once, then complete the right component.

A return-to-start requirement can make the task impossible

Now ask for a loop that starts at A, visits all six nodes, and returns to A without repeating any other node. Such a loop cannot exist in this network. Reaching the right triangle requires crossing C–D, and returning requires crossing the same connection again. That repeats its endpoints.

This impossibility follows from the listed edges, not from the drawing style. Adding a second connector such as B–E changes the situation: A, B, E, F, D, C, A is then a valid loop. Before repeatedly retrying a route, check whether an overlooked link or a different rule is required.

Use the idea without applying it to the wrong mechanic

For a wire-connection or maze board, inspect narrow corridors that join larger areas. If a one-way door replaces C–D, direction becomes a separate constraint. If switches open connections, the network changes with state; a static drawing no longer tells the whole story.

The graph model does not establish whether a drawn bridge can support weight or whether a laser reflects correctly. Those puzzles need additional rules. For a connectivity puzzle, the practical sequence is to write the goal precisely, list legal neighbors, locate unavoidable connectors, and only then optimize the route.