Lesson notes
These notes explain the problem and current implementation. They are a reading companion, not a verbatim transcript of the recording.
Determine whether a robot can reach any exit from the upper-left cell of a rectangular room. Open cells contain 0, blocked cells contain 1, and exits contain 2. The robot can move only down or right. It does not move up or left, so this is a directed reachability problem rather than a general four-direction maze search.
Define the state as a cell
Let the answer for (row, column) mean whether an exit is reachable starting there. A position outside the room is false, an exit is true, and a blocked cell is false. An open cell succeeds if either the cell below it or the cell to its right succeeds.
The same cell can be reached by several paths. Without memoization, the recursion repeatedly solves identical suffix regions and can grow exponentially. Store each completed Boolean answer in a memo table. Its entries need three states: unknown, known false, and known true. Treating false as unknown would repeat all unsuccessful searches.
Trace a shared dead end
Consider [[0, 0, 1], [0, 0, 1], [1, 0, 2]]. From the start, moving down or right can eventually reach the center cell. From there the downward route reaches the bottom-middle cell and then the exit. The right neighbor is blocked. Memoizing the center's true result lets a later path reuse it immediately.
Now replace the bottom-middle cell with a wall. The center becomes false. Caching that false is just as useful as caching a success: both incoming paths can stop without exploring the same blocked suffix again. The memo describes reachability from a cell, independent of the path used to arrive there.
Explain why no visiting marker is needed
Every recursive move increases row or column. A path can never return to an earlier cell, so the dependency graph has no cycle. Each state depends only on states farther down or right, and completed answers can be reused safely. General four-direction movement would require a cycle-aware visited policy instead of this argument.
The recurrence checks exactly the legal next moves. If either next state reaches an exit, prefixing the current open cell gives a valid path. If both fail, every legal continuation fails. The out-of-bounds, blocked, and exit base cases complete the correctness argument.
Read the table's absence representation
Java can use boxed Boolean entries so null means unknown. Other languages use optional Booleans, a numeric sentinel, or a map that distinguishes an absent key from a stored false value. The distinction must survive lookup and comparison. A truthiness check alone often collapses unknown and false.
Each of r × c cells has its result computed at most once, with constant work per transition. Time and memo storage are O(r × c). Recursive depth is O(r + c) because a path advances only down and right. The room itself need not be modified. A bottom-up table can remove recursion by processing dependencies from the bottom-right toward the top-left.
Test the movement rule
Use an exit at the start, a blocked start, no exit, a single row, a single column, and a room with several routes to one suffix. Include an arrangement whose only geometric route would require moving left or up; it must fail under this contract. An empty room has no starting cell and returns false.
For an independent oracle, propagate reachable open cells forward from the start and report whether any reachable cell is an exit. This evaluates the directed graph in the opposite direction from the recursive suffix question. A follow-up can count paths, which changes Boolean OR into addition and requires a numeric overflow policy for large rooms.
Continue with the implementation
Read the JavaScript walkthrough for the current code and links to the other programming languages. Use the original lesson for practice.