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 Gardener's Woe (Posted on 2006-06-24)
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.

 See The Solution Submitted by Jer Rating: 4.0000 (1 votes)

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 'Arcs' for part 2 | Comment 9 of 26 |
"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 33.19 units.

 Posted by brianjn on 2006-06-24 21:54:21

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