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The well-ordered orchard



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5 horizontal rows

4 vertical columns

4 diagonals with slope 1 (ROL, NKH, MJG, IFC)

4 diagonals with slope -1 (SNI, TOJE, PKFA, LGB)

2 knight's move diagonals with slope -2 (AJS, BKT)

Nice, and extremely well ordered.

But the gardener was more frugal. He used fewer trees.

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Wow! His name wasn't Martin, was it?

Please correct me if I'm wrong: a line of 4 counts as one row of 4, rather than two rows of three, right?

Sometimes it's Martin, a semi-idol of mine, but not this time.

It's gardener, lower case, with two "e"s.

Correct. A row of four is not also two (or four) rows of three.

A row is a row, and its size is the number of trees aligned with it.


If the rows of length greater than three are excluded, there are (only) eighteen rows.

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Assuming rows are straight lines, no two different rows may share two trees. Since a row consists of at least 3 trees, there are

3C2 = 3 combinations of tree pairs in each row.
Thus for 19 rows we need at least 19*3 = 57 pairs of trees. The minimum number that allows for that many is 12 trees. (12C2 = 66.)
A tree shares a row with other two trees. It cannot share another row with any of those two trees. If maximum rows in which a tree is a participant is x and total number of trees is N,

then 2x + 1 = N. Solving for N=12, x=5.5.

Since no fractional trees are allowed, each tree can participate in at the most 5 rows. 12*5 = 60. However, since each row has 3 trees, 60/3 = 20.
And so it appears a gardener could construct 20 rows with 12 trees.

Now all gardener has left to do is to use these limits as a guide for planting his garden.

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plasmid, on 15 Mar 2013 - 00:21, said:

I've been stuck on this one. Maybe it would help to know whether or not the following figure is a subset of the final answer, since I keep coming back to it and it seems to at least be a local minimum (or whatever the appropriate term would be)

attachicon.gifnine dots.jpg

You're on the right track.

  • you can clearly keep the corners that define the rectangle.
  • you can keep the others, with slight horizontal adjustments
  • add three points

Two of the added points are not on one of the three present horizontal lines (possibly extended).

Also, review the clue in post 19.

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