One possible solution is to have competitor 0 play competitor 2 (competitor indices are 0-based) at intersection (0,1). Then have competitor 1 play competitor 3 at intersection (2,1). Then schedule all the other games at intersection (1,1).
Here's one possible ordering:
Game 1 (Competitor 0 vs. Competitor 2)
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Competitor 0 goes from (0,0) to (0,1): 1
Competitor 2 goes from (0,2) to (0,1): 1
Game 2 (Competitor 1 vs. Competitor 3)
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Competitor 1 goes from (2,0) to (2,1): 1
Competitor 3 goes from (2,2) to (2,1): 1
Game 3 (Competitor 0 vs. Competitor 1)
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Competitor 0 goes from (0,1) to (1,1): 1
Competitor 1 goes from (2,1) to (1,1): 1
Game 4 (Competitor 2 vs. Competitor 3)
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Competitor 2 goes from (0,1) to (1,1): 1
Competitor 3 goes from (2,1) to (1,1): 1
At this point, all the competitors are at (1,1) so no further travel is necessary for the remaining games. The sum of all the distances is 8. |