Two Moving Objects on a 5x5 Grid
Consider a 5×5 grid. Rows are numbered 1 to 5 from top to bottom and columns 1 to 5 from left to right, so cell (1,1) is the top-left corner. Two objects move independently. Object D is a domino covering two adjacent cells. At step 1 it covers (1,3) and (2,3), so its inner end touches the centre cell (3,3) from above. At each step D rotates 90° clockwise about the centre while staying two cells long and keeping its inner end adjacent to the centre: step 2 covers (3,4) and (3,5), step 3 covers (4,3) and (5,3), step 4 covers (3,1) and (3,2), and step 5 repeats step 1. Object P is a single dot. At step 1 it is at (1,1). At each step it moves one cell clockwise around the outer perimeter of the same grid. It moves through (1,2), (1,3) and so on, turning the corner at (1,5). The two objects are allowed to share a cell.
1. At which step, if any, do D and P first share a cell? Show your tracking.
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Track both cycles step by step.
| Step | D cells | P cell | | 1 | (1,3),(2,3) | (1,1) | | 2 | (3,4),(3,5) | (1,2) | | 3 | (4,3),(5,3) | (1,3) | | 4 | (3,1),(3,2) | (1,4) | | 5 | (1,3),(2,3) | (1,5) | | 6 | (3,4),(3,5) | (2,5) | | 7 | (4,3),(5,3) | (3,5) | | 8 | (3,1),(3,2) | (4,5) | | 9 | (1,3),(2,3) | (5,5) | | 10 | (3,4),(3,5) | (5,4) | | 11 | (4,3),(5,3) | (5,3) |
First shared cell is (5,3) at step 11. Earlier near-misses are step 3, when P is at (1,3), but D has already rotated to the bottom; step 7, when P reaches (3,5) but D is at the bottom; and step 9, when P reaches (5,5) but D has returned to the top.
2. State the position of each object at step 20 and explain how you used periodicity.
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Use the periods. D repeats every 4 steps: 20 ÷ 4 = 5 remainder 0, so step 20 has the same orientation as step 4, left: (3,1) and (3,2). P repeats every 16 steps because the 5×5 perimeter has 16 cells: 20 ÷ 16 = 1 remainder 4, so step 20 is the same position as step 4, which is (1,4). They do not overlap at step 20. The shortcut is valid only because each object's motion is exactly periodic; mixing the two periods without reducing them would give the right answer here only by coincidence.
3. Suppose the grid were enlarged to 7×7 with the same rules, and D started above the centre. State the cells D covers at step 3 and the period of P.
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The centre of a 7×7 grid is (4,4). D's inner end must be adjacent to the centre, so at step 1 it covers (2,4) and (3,4) (above), at step 2 (4,5) and (4,6), at step 3 (5,4) and (6,4), at step 4 (4,3) and (4,2). The four-step rotation period is unchanged; only the distances from the centre change. P's perimeter grows from 16 to 24 cells, so its period becomes 24. The two periods are now 4 and 24, with least common multiple 24, so the combined pattern repeats every 24 steps instead of every 16. The first meeting step must be recalculated; it is not simply the old step 11.