An unlucky gardener planted a 10x10 square array of 100 old seeds out in the garden. Only 5 of these seeds have germinated including one at the southwest corner (0,0) where a slug is currently reducing it to ground level.
When it finishes it will head directly to the next closest doomed plant. After it eats that one it will again leave a slime trail to the closest remaining plant and so on until the garden is no more.
Where are the 4 remaining seedlings if the path crawled by the slug is the longest possible and it never has to choose between two equidistant snacks?
Note: Although the slug will never have to choose between two equidistant seedlings, this doesn't imply that no two are equidistant.
Next find the locations if 6 seedlings had germinated instead of 5.
"Arc' for part 1 provides part of the solution
I shall not go through my reasoning here except to suggest that I looked for a symmetry to optimise both sides of the garden but came up with a rotation.
I have 4 plants at each corner of the garden, I discovered that I could not have the 2 internal plants on x = 3 to 6.
My solution therefore was (0,0), (2,6), (0,9), (9,9), (7,3) and (9,0) being a distance of ≈28.86 units.
And two invalid solutions under the rules:If my internal plants were at (3,6) and (6,3) the slug would have to make a choice. If by going via (3,6) to (0,9) the path is
≈30.9 units whereas by going via (3,6),(6,3) and (9,0) the path is
Posted by brianjn
on 2006-06-24 21:54:21