[Guest]

The professor never took much interest in working the garden. He did, however, like to spend time in the garden constructing puzzles to pose for his fellow puzzle lovers at BrainDen.

There was a particular part of the garden in which the professor wanted to have some trees and bushes planted. Therefore he had arranged for a visit from the local gardener, and here they were – in the garden – discussing the details of the work to be completed.

”So what did you have in mind?”, the gardener asked the professor.

”Well, I want to plant a lot of trees and bushes in this area”, answered the professor.

”Very well!”, said the gardener. ”I know where to buy exactly the type of trees and bushes you should have. Be advised though, that these plants need some spacing to allow sufficient room for the roots. I'd say that no more than 15 plants – that be either trees or bushes – should be planted here”. The gardener continued, ”The trees cost 100 BDB (BrainDen Brainars) each, and the bushes cost 50 BDB each. How many of each do you want?”

The professor thought about this for a little while and said, ”Well, either you plant it exactly as I want it. I will pay for the trees and bushes used, and I'll pay you 400 BDB for the work, or... I will pay you 100 BDB for each row you plant. The rows must follow certain conditions though. A row must consist of at least 3 plants, and you cannot place 2 of the same type next to eachother in a row.”.

The gardener replied: ”You mean, I can make rows of tree-bush-tree, or bush-tree-bush only?”.

”Yes”, the professor replied, ”or more in each row if you wish, but that may add to the total cost of course”.

”Hmmm”. The gardener is still thinking about this.

What do you think he should do? Should he:

- Take 400 BDB as payment, make the garden as the professor wants, while the professor pays for the materials (trees and bushes) used, or

- Design the garden so there are rows of alternating trees/bushes which the gardener must pay for himself, and then charge the professor 100 BDB for each legal row.

[Guest]

The professor thought about this for a little while and said, ”Well, either you plant it exactly as I want it. I will pay for the trees and bushes used, and I'll pay you 400 BDB for the work, or... I will pay you 100 BDB for each row you plant. The rows must follow certain conditions though. A row must consist of at least 3 plants, and you cannot place 2 of the same type next to eachother in a row.”.

What do you think he should do? Should he:

- Take 400 BDB as payment, make the garden as the professor wants, while the professor pays for the materials (trees and bushes) used, or

- Design the garden so there are rows of alternating trees/bushes which the gardener must pay for himself, and then charge the professor 100 BDB for each legal row.

I miss read the problem at first and now I need some thing clarified. When you say rows, do you mean in any direction or is going down a column?

[Guest]

If the professor can do it the way he wants will he still be limited to 15 plants?

If the gardener takes 100/row each row will cost a minimum of 200 (bush-tree-bush), 100 of which will be from the prof and 100 from the gardener.

In the other scenario, the prof pays for ALL the trees plus gives the gardener 400....that doesn't seem like much of a question, unless I missed something.

[Guest]

If the professor can do it the way he wants will he still be limited to 15 plants?

If the gardener takes 100/row each row will cost a minimum of 200 (bush-tree-bush), 100 of which will be from the prof and 100 from the gardener.

In the other scenario, the prof pays for ALL the trees plus gives the gardener 400....that doesn't seem like much of a question, unless I missed something.

Yea, I am questioning that as well.

The only answer is if each plant can be part of another row, for example, a row diagonal, vertical, and horizontal

Edited April 17, 2009 by IDoNotExist

[Guest]

Yea, I am questioning that as well.

The only answer is if each plant can be part of another row, for example, a row diagonal, vertical, and horizontal

The only way to make a profit out of it is to plant it such a way that a few trees count for multiple rows. Seeing how every tree would give you ~$33, each tree would have to count for at least an average of 3 rows to break even.

Not much time now. Will look at it later. Good riddle btw.

[Guest]

I miss read the problem at first and now I need some thing clarified. When you say rows, do you mean in any direction or is going down a column?

In any 2 dimensional direction, yes.

[Guest]

If the professor can do it the way he wants will he still be limited to 15 plants?

If the gardener takes 100/row each row will cost a minimum of 200 (bush-tree-bush), 100 of which will be from the prof and 100 from the gardener.

In the other scenario, the prof pays for ALL the trees plus gives the gardener 400....that doesn't seem like much of a question, unless I missed something.

Well, it's just like the classical plant-9-trees-in-9-rows type of puzzle. So each plant can be part of as many rows as you'd like. But for a row to be counted as valid it can only consist of alternating plant-types (tree-bush-tree-bush... etc).

[Guest]

The only way to make a profit out of it is to plant it such a way that a few trees count for multiple rows. Seeing how every tree would give you ~$33, each tree would have to count for at least an average of 3 rows to break even.

Not much time now. Will look at it later. Good riddle btw.

Exactly right, scsw. And thanks. It took quite some time to construct it, so I'm glad it pleases :-)

Now I just hope it wont prove too easy

[Guest]

Exactly right, scsw. And thanks. It took quite some time to construct it, so I'm glad it pleases :-)

Now I just hope it wont prove too easy

It's not too easy. I've been thinking for a while and I've found ways for the gardener to earn 100 and 200 Brainars, and a way to earn 400 Brainars but he has to use 16 plants. Haven't found the right solution yet, but it could come to me at any moment.

[plasmid]

T B T B T
B T B T B

9 rows of three going East-West (there are three three-element rows within each of the five-thingy rows)
5 rows going North-South
2 rows going diagonally from corner to corner (that alternate B-T-B)
(one East-Northeast to West-Southwest)
(one West-Northwest to East-Southeast)
gross income: $1600 bux
gross cost: 8xB + 7xT = 400 + 700 = $1100 bux
profit: $500 bux
More than he'd make from the professor even if he didn't order anything

B T B T B

Edit: just realized I might be able to make some more cash...

Can I also count 2 rows of four within each of the horizontal rows, and one row of five for each of the horizontal rows, to bring the total profit up to 1400?

Edited April 17, 2009 by plasmid

[Guest]

or more in each row if you wish, but that may add to the total cost of course

Except this quote makes it sound like a row with >3 plants is still just 1 row.

[Guest]

Except this quote makes it sound like a row with >3 plants is still just 1 row.

Ya exactly, that's why the puzzle is hard.

[Guest]

Edit: just realized I might be able to make some more cash...

Can I also count 2 rows of four within each of the horizontal rows, and one row of five for each of the horizontal rows, to bring the total profit up to 1400?

T B T B T
B T B T B

9 rows of three going East-West (there are three three-element rows within each of the five-thingy rows)
5 rows going North-South
2 rows going diagonally from corner to corner (that alternate B-T-B)
(one East-Northeast to West-Southwest)
(one West-Northwest to East-Southeast)
gross income: $1600 bux
gross cost: 8xB + 7xT = 400 + 700 = $1100 bux
profit: $500 bux
More than he'd make from the professor even if he didn't order anything

B T B T B

Back. Whew. A lot of work today. I have about 10 minutes.

Let's see. Can you double/triple count a row? Rereading the OP;

”Yes”, the professor replied, ”or more in each row if you wish, but that may add to the total cost of course”.

This is a tab bit ambiguous.

EDIT: Ah, just saw jbriddler's post.

Edited April 17, 2009 by scsw

[plasmid]

Except this quote makes it sound like a row with >3 plants is still just 1 row.

Good point everyone.

Now yer forcin' me to get creative...

All rows run diagonally. Five rows through each of the trees on the sides. Six rows through the center tree.

TTT
B B B B B B

16 rows = $1600
12xB + 3xT = 600 + 300 = $900
Profit = $700

B B B B B B

[Guest]

Edit: just realized I might be able to make some more cash...

Can I also count 2 rows of four within each of the horizontal rows, and one row of five for each of the horizontal rows, to bring the total profit up to 1400?

B T B T B
T B T B T
B T B T B

9 rows of three going East-West (there are three three-element rows within each of the five-thingy rows)
5 rows going North-South
2 rows going diagonally from corner to corner (that alternate B-T-B)
(one East-Northeast to West-Southwest)
(one West-Northwest to East-Southeast)
gross income: $1600 bux
gross cost: 8xB + 7xT = 400 + 700 = $1100 bux
profit: $500 bux
More than he'd make from the professor even if he didn't order anything

It's a good suggestion, but I'm sorry to say that the professor was very specific in that a row could consist of more than 3 plants - but that is still only 1 row then. So this solution has a total of 8 rows.

EDIT: Ahhh, I see this was pointed out already.

Edited April 17, 2009 by uhre

[Guest]

Good point everyone.

Now yer forcin' me to get creative...

All rows run diagonally. Five rows through each of the trees on the sides. Six rows through the center tree.

TTT
B B B B B B

16 rows = $1600
12xB + 3xT = 600 + 300 = $900
Profit = $700

B B B B B B

Oh, I hadn't noticed this reply. That is very good, Plasmid. That is even one more row than I found, and a lower cost. Well done.

[Guest]

Well, Plasmid did it. Improve it if you can, but I doubt that it's possible:

My original solution for 15 legal rows, and a total profit of 450 BDB:

Plasmids solution for 16 legal rows, and a total profit of 600 BDB:

Edited April 18, 2009 by uhre