bonanova Posted February 15, 2013 Report Share Posted February 15, 2013 The gardener planted his orange trees in a superbly ordered manner. Not counting pairs of trees as a "row", he created nineteen rows of trees. How many trees was it necessary for him to plant? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted April 8, 2013 Author Report Share Posted April 8, 2013 Time to plant the trees. Assuming interest on this puzzle is spent, let's close it out. Honorable mention to plasmid, whose last post was clearly on the right track. Quote Link to comment Share on other sites More sharing options...
0 CaptainEd Posted February 15, 2013 Report Share Posted February 15, 2013 (edited) a 5x5 square gives me at least 32 "rows", so I'm looking for something smaller. Also, a 4 x 5 square gives me at least 21 "rows". Still searching... Edited February 15, 2013 by CaptainEd Quote Link to comment Share on other sites More sharing options...
0 CaptainEd Posted February 15, 2013 Report Share Posted February 15, 2013 (edited) A B C E F G H I J K L M N O P R S T 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) Edited February 15, 2013 by CaptainEd Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 15, 2013 Author Report Share Posted February 15, 2013 A B C E F G H I J K L M N O P R S T 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. Quote Link to comment Share on other sites More sharing options...
0 CaptainEd Posted February 15, 2013 Report Share Posted February 15, 2013 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? Quote Link to comment Share on other sites More sharing options...
0 CaptainEd Posted February 15, 2013 Report Share Posted February 15, 2013 (edited) I see your point A B D F G H I J L M O P Q R S DFG HIJL MOP QRS ADH BIM FJQ GOR LPS BFP AJOS DIR HMQ AIQ QOL AFL BJR MJG DJP Edit: from "more frugal" to "better still" that meets the OP with even fewer trees Edited February 15, 2013 by CaptainEd Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 15, 2013 Author Report Share Posted February 15, 2013 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. Clue: If the rows of length greater than three are excluded, there are (only) eighteen rows. Quote Link to comment Share on other sites More sharing options...
0 k-man Posted February 15, 2013 Report Share Posted February 15, 2013 18 rows formed by 13 trees. I can get 19 and even more with 14 trees. Is this close to your solution, Bonanova? Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted February 15, 2013 Report Share Posted February 15, 2013 13t20r? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 15, 2013 Author Report Share Posted February 15, 2013 18 rows formed by 13 trees. I can get 19 and even more with 14 trees. Is this close to your solution, Bonanova? Yes. Very close. It's as close as you can come and not be the minimum. That was a hint at the answer. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 15, 2013 Author Report Share Posted February 15, 2013 orchard.jpg 13t20r? OK that counts as over-achievement I guess. It's not the fewest trees that can make eighteen rows, but it probably is the fewest that can make twenty. Honorable mention! Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 26, 2013 Author Report Share Posted February 26, 2013 So the answer is 12 trees making 18 rows. Can you draw the layout? Quote Link to comment Share on other sites More sharing options...
0 augustinus Posted February 26, 2013 Report Share Posted February 26, 2013 (edited) X B C X D E F G H I J K X M N X 4 horizontal 4 vertical 4 outer diagonal parts (BD, CG,HM, NK) 6 inner diagonal parts (BFK,EJ, DIN,CEH,FI, GJM) X's are empty spots Edited February 26, 2013 by augustinus Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 26, 2013 Author Report Share Posted February 26, 2013 On the right track. We need at least three trees per row. Quote Link to comment Share on other sites More sharing options...
0 tammie Posted February 28, 2013 Report Share Posted February 28, 2013 i thought it was 19 rows. 18 of three and one of more right? total using 12 trees? still don`t have the layout but b4 i spend the time trying i want to make sure i have the right information So the answer is 12 trees making 18 rows. Can you draw the layout? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 1, 2013 Author Report Share Posted March 1, 2013 Hi tammie, and welcome, if no one else has said it, to the Den. Yes, you have it right. There are 12 trees. Eighteen rows have 3 and only 3 trees in them. There is a "bonus" row, if you want to call it that, and it has 4 trees. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 10, 2013 Author Report Share Posted March 10, 2013 Shamelessly bumping this puzzle. It's worth the effort to solve, IMO. 12 trees making 18 rows of 3 trees each. Quote Link to comment Share on other sites More sharing options...
0 Prime Posted March 11, 2013 Report Share Posted March 11, 2013 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. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 11, 2013 Author Report Share Posted March 11, 2013 The orchard occupied a rectangular plot of land. To permit optimal sun exposure and air circulation, seven of the trees were planted along its boundary. The other five were scattered in its interior. Quote Link to comment Share on other sites More sharing options...
0 plasmid Posted March 15, 2013 Report Share Posted March 15, 2013 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) Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 15, 2013 Author Report Share Posted March 15, 2013 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) nine 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. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 18, 2013 Author Report Share Posted March 18, 2013 Almost time to harvest the oranges. Anyone still working on this, or are there millions wanting to know how it's possible? More time if you like, otherwise I'll post the layout at week's end. Quote Link to comment Share on other sites More sharing options...
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bonanova
The gardener planted his orange trees in a superbly ordered manner.
Not counting pairs of trees as a "row", he created nineteen rows of trees.
How many trees was it necessary for him to plant?
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