Source: TSTST-2019-P3
Category: Existence and Construction

Problem:
On an infinite square grid we place finitely many cars, which each occupy a single cell and face in one of the four cardinal directions. Cars may never occupy the same cell. It is given that the cell immediately in front of each car is empty, and moreover no two cars face towards each other (no right‑facing car is to the left of a left‑facing car within a row, etc.). In a move, one chooses a car and shifts it one cell forward to a vacant cell. Prove that there exists an infinite sequence of valid moves using each car infinitely many times.

Solution:
Let $S$ be any rectangle containing all the cars. Partition $S$ into horizontal strips of height $1$, and color them red and green in an alternating fashion. It is enough to prove all the cars may exit $S$.

To do so, we outline a five-stage plan for the cars:

1. All vertical cars in a green cell may advance one cell into a red cell (or exit $S$ altogether), by the given condition. (This is the only place where the hypothesis about the cell immediately in front of each car being empty is used!)
2. All horizontal cars on green cells may exit $S$, as no vertical cars occupy green cells.
3. All vertical cars in a red cell may advance one cell into a green cell (or exit $S$ altogether), as all green cells are empty.
4. All horizontal cars within red cells may exit $S$, as no vertical cars occupy red cells.
5. The remaining cars exit $S$, as they are all vertical. 

The solution is complete.


Grading guidelines:
(Partial) 1. Introduced a bounding rectangle S containing all cars, and reduced the problem to showing that all cars can eventually be moved out of S (then continue moving cars forward forever). 2. Used a striping/coloring of S into alternating horizontal strips (or an equivalent invariant/partition) and stated a staged plan separating vertical moves from horizontal moves based on the colors. 3. Correctly justified one key stage of the plan, e.g. that all vertical cars on one color can advance one step (using the “front cell is empty” condition), or that horizontal cars on a given color can exit because no opposing-direction blockage can occur within a row. (Almost) 1. Presented the full five-stage schedule (move vertical cars from one color to the other, then let horizontal cars on that color exit; repeat with the other color; finish with remaining vertical cars) and proved that after these stages every car exits S, but omitted a localized case check for cars on the boundary of S when “exiting S”. 2. Carried out the coloring/strip argument and proved that each stage is legal and decreases the number of cars inside S, but left a minor gap in explaining why stage 2 (or stage 4) is always possible for all horizontal cars (e.g. why a horizontal car is never blocked by another car during its march to the boundary). 3. Solution is almost complete, but made minor mistakes which are not negligible.