Problem 12 · Pairing odd-degree vertices · 4 points
Made by AI and not reviewed by a real person.
A robot walks on a 3-by-4 grid of intersections, with streets only between horizontal or vertical neighbors. If it must walk every street and return to its start, what is the fewest street segments it can walk?
- 20
- 24
- 22
- 23
- 21
Show a hint
- On a walk that returns to its start, every arrival at an intersection needs a departure. Which intersections have an odd number of streets?
- Think of the extra walks as paths pairing those intersections. How far apart could the paired intersections be?
Show the answer and solution
Answer: C — 22
The grid has 9 horizontal and 8 vertical streets, for 17 in all. Six intersections have three streets each, so the robot must repeat streets to make their total numbers of visits even. The odd intersections at the middle of the left and right edges are 3 segments apart; if paired with each other, the other four odd intersections need at least 2 more segments. If paired separately, each needs a path of at least 2 segments, and the remaining pair needs at least 1. Either way, at least 5 segments must be repeated. This is possible: repeat all 3 streets across the middle row and the central street in each of the top and bottom rows. Every intersection then has an even number of traversals, allowing a closed walk of 17 + 5 = 22 segments. The tempting 20 counts one repeated segment per pair, overlooking the distances between them.