# Plant 9 trees in 10 rows of 3

"Tree-planting" puzzles are also known as "points and lines" puzzles. The English puzzle author and mathematician Henry Ernest Dudeney was very fond of them. In 1917, Dudeney published a collection of puzzles called "Amusements in Mathematics", which also contains the following classic puzzle:

Sir Isaac Newton's tree-planting puzzle: A farmer wants to plant 9 trees in such a way that there are 10 rows of 3 trees. How does he do this?

Note: the different rows are not allowed to be collinear.

P.S. I know of two different solutions. If you know any more that would be great and if you are able to prove that there aren't any more solutions that would be even greater. I consider two solutions different when there is no graph isomorphism possible between them.

• @HughMeyers the different rows are not allowed to be collinear with each other
– Ivo
Mar 29, 2016 at 14:02
• Funny.. the way English is used by my peers in my region of the USA, "10 rows of 3 trees" means 30 trees, no more, no less. Looking at the highest voted answer and assuming it is at least on the right track, I would have said, "A famer wants to plant 9 trees so that 10 straight lines can be drawn that each pass through exactly 3 trees." Mar 29, 2016 at 15:20
• Are you allowed to warp spacetime?
– Mark
Mar 29, 2016 at 21:25
• @Mark That would be lateral-thinking, I think. Mar 30, 2016 at 0:17
• For those who want to read more about this, there are several external references to the so-called "Orchard-Planting Problem" on Wikipedia and MathWorld. The general problem is still apparently an area of active research, and apparently has some interesting ties to elliptic curves. Mar 30, 2016 at 13:29

A third solution. I've been searching systematically for them (I've eliminated all the combinatorial solutions other than ABC, A14, A25, A36, B15, B26, B34, C16, C24, C35 - it is the only one with geometric solutions).

To see that it is distinct from the others, note that it has a tree with 7 immediate neighbors.

I believe that is all of them, though. Once I double check to make sure I haven't missed something, I try to put something together to show it.

• Nice find! very impressive. Also interesting that unlike the other two solutions there is no line symmetry
– Ivo
Apr 2, 2016 at 19:13
• @IvoBeckers - I would not be surprised if it could be redrawn with symmetry. This form is just what came up by my method of investigation. Apr 2, 2016 at 21:13

I think there might be quite a lot of possible answers, but here is one of them:

Sorry for the bad MS Paint skills ;)

The numbers at the intersections represents the trees, and the lines are added to indicate the 10 "rows" of 3.

EDIT:

I think there might be quite a lot of possible answers, but here is one of them

Let's rephrase that to: "I think there are barely any possible answers, but here is the easiest one."

I tried to come up with another one, but the only one I was able to find eventually was already represented by Marius.

Legend:
1..9 are the trees.
X = empty space.

Solution 1:

1X2
X3X
456
X7X
8X9

Lines:

01. 148
02. 269
03. 136
04. 234
05. 678
06. 479
07. 456
08. 357
09. 159
10. 258

picture:

Solution 2 (built with my awesome ms paint skills):

Lines:

01. 123
02. 165
03. 184
04. 295
05. 278
06. 376
07. 389
08. 345
09. 694
10. 587

• you forgot 456 on your picture Mar 29, 2016 at 16:31
• I didn't add the picture. Some good samaritan addet it. Mar 29, 2016 at 16:41
• Nice, those are the 2 I knew. I'll accept this answer in 24 hours if no one can come with more solutions or a proof that more solutions aren't possible
– Ivo
Mar 29, 2016 at 21:37
• I'm not entirely sure, but I believe your two solutions are the only ones possible. I've tried to come up with a second solution, different from your second one, but every time I either end up with 8 lines of 3, or I need 10 coins in order to create the 9 lines. Pretty frustrating, especially since the solution I posted as answer (the same as your first one) was so easy to find.. Mar 30, 2016 at 11:40
• @KevinCruijssen. I've been trying to find a 3rd way but failed so far. But absence of evidence is not evidence of absence. We either find a mathematical proof that these are the only 2 options or we find a third one. We cannot conclude these are the only 2 because we're not smart enough to find a 3rd one :). Mar 30, 2016 at 11:50

I've completely rewritten this post to hopefully be more coherent.

The geometric puzzle suggests a related combinatorial puzzle:

Given nine objects, how many non-isomorphic collections are there of ten sets of three objects each, such that no two of the sets have more than one object in common?

Where two collections are isomorphic if there is a permutation of the 9 objects (trees) that transforms one collection into the other.

Every solution to the geometric puzzle provides a solution to the combinatorial one. Solutions to the combinatorial puzzle thus restrict the possibilities for solutions to the geometric one, and provide guidance on finding them.

The combinatorial puzzle admits only 4 solutions:

Theorem: The following are the only solutions to the combinatorial puzzle:

• Objects: $\{A, B, C, 1, 2, 3, 4, 5, 6\}$ Collections: $$ABC, A14, A25, A36, B15, B26, B34, C16, C24, C35$$ $$ABC, A14, A25, A36, B16, B24, B35, C13, C26, C45$$ $$AB1, AC2, BC3, A34, A56, B46, B25, C16, C45, 123$$
• Objects: $\{A, B, C, D, 1, 2, 3, 4, *\}$ Collections: $$AB*, CD*, AC1, AD2, BC3, BD4, A34, B12, C24, D13$$

Proof:

Given such a collection, define an instance is a set $S$ in the collection, and a tree $t$ with $t\in S$. As there are $10$ sets of $3$ trees each, there are $30$ instances in all. Define the degree of a tree to be the number of instances of which it is a part. A tree of degree $n$ is called an $n$-tree. Since there are $9$ trees, the average degree of a tree is $\frac {30}9 = 3\frac13$. Therefore, some trees must have degree $4$ or more.

Lemma:

• The maximum degree for any tree is $4$.
• Any 4-tree shares a common set with every other tree.
• The minimum degree for any tree is at least half the number of 4-trees.

Proof:

Let $n$ be the degree of a tree $T$. Then the $n$ sets containing $T$ consist of $2n$ instances other than those for $T$. If any two of these instances are for the same tree, they cannot be in the same set, and so the two sets they are in have two points in common (the shared point and $T$) which is not allowed. Thus we must have $2n + 1$ distinct trees (including $T$). So $2n + 1 \le 9$, and $n \le 4$. If $n = 4$, then all trees are represented, so the 4-tree shares a set with every other tree. Now for any tree $T$, it must share a set with every 4-tree, so the count $d_4$ of 4-trees satisfies $d_4 \le 2n$, or $n \ge \frac{d_4}2$

QED

Let $d_n$ be the number of trees of degree $n$. By the lemma $n \le 4$. Then the number of instances (30) is the sum of the degrees for all trees, which is $$30 = 4d_4 + 3d_3 + 2d_2 + 1d_1$$ but $$9 = d_4 + d_3 + d_2 + d_1$$ Eliminating $d_4$ gives $$6 = d_3 + 2d_2 + 3d_1$$ So the total number of trees of degree $< 4$ is at most $6$, leaving at least $3$ 4-trees. By the lemma, the smallest degree possible is therefore $2$, so $d_1 = 0$ and $$6 = d_3 + 2d_2$$ In particular $0 \le d_2 \le 3$.

• $d_2 = 0$ gives $d_3 = 6, d_4 = 3$
• $d_2 = 1$ gives $d_3 = 4, d_4 = 4$
• $d_2 = 2$ gives $d_3 = 2, d_4 = 5$, which contradicts the lemma,
• $d_2 = 3$ gives $d_3 = 0, d_4 = 6$, which contradicts the lemma.

Consider the case $d_2 = 1, d_3 = 4, d_4 =4$. Label the 4-trees $A, B, C, D$, the 3-trees $1, 2, 3, 4$, and the 2-tree "$*$". Since the 2-tree shares sets with all $4$ 4-trees, those sets must be $AB* := \{A, B, *\}$ and $CD*$. There cannot be set of just 4-trees, as any such set would share two trees with $AB*$ or $CD*$. Therefore the sets shared by pairs of 4-trees must have a 3-tree as the third member. By relabeling we can take $AC1$ and $AD2$ as two of the sets. The 4th set for $A$ must be $A34$, as all other trees have been paired with $A$. Suppose $BC2$ were also a set. Then $C34$ would also have to be as $C$ has been paired with all other trees. This conflicts with $A34$, so it cannot be that $BC2$ is a set. Therefore, relabeling if needed, we can take $BC3$ as set, and for similar reasons $BD4$. The remaining sets must be $B12, C24, D13$, again as these are the only remaining trees the 4-trees have not been paired with. This completes the list given in the theorem for this case.

For the case $d_3 =6, d_4 = 3$, label the 4-trees $A, B, C$, and the 3-trees $1,2,3,4,5,6$.

1. Suppose that the 4-trees do not share a common set. In this case we can label the trees so that $AB1, AC2, BC3$ are sets. Each of $A, B, C$ have two more sets each, which they share with $2$ 3-trees. By relabeling, three of these are $A34, A56, B46$. This also requires $B25$. $C$ remains to be paired with $1, 4, 5, 6$, but $46$ and $56$ have already occurred, so the two sets must be $C16$ and $C45$. Finally $123$ makes up the last set.
2. When $ABC$ is a set, the remaining $9$ sets must consist of 3 sets each for $A, B, C$ matching them with a pair of 3-trees. This gives us 9 choices of pairs of 3-trees, of which there are ${6\choose2} = 15$ total, chosen so that each tree appears in 3 of the pairs. To find out how many non-isomorphic choices there are, it is easier to examine the $6$ pairs that were not picked. Since each tree appears in 5 pairs overall, it must appear in exactly two pairs of the leftovers. This allows us to form paths. For example, starting at 1, choose one of the two trees paired with it (say 3), then take the other tree paired with 3 (say 6), then the other tree paired with 6, and so on. As there is always exactly one other paired tree, this cannot end until you come back to 1, forming a loop. Each tree lies in such a loop, each loop has length at least $3$, and the sum of all loop lengths is $6$. So there are only two choices: a loop of length $6$, or $2$ loops of length $3$. Any two loops of length $6$ are isomorphic to each other, as are any pairs of loops of length $3$. The remainders for $$ABC, A14, A25, A36, B15, B26, B34, C16, C24, C35$$ form $2$ loops of length $3$, while the remainders for $$ABC, A14, A25, A36, B16, B24, B35, C13, C26, C45$$ form a loop of length $6$. This exhausts all cases, so the theorem is proved.

QED

There is not a 1-1 correspondence between combinatorial solutions and geometric solutions. Indeed, the two geometric solutions found so far both correspond to $$ABC, A14, A25, A36, B15, B26, B34, C16, C24, C35$$.

In fact,

Theorem: There is no geometric solution that corresponds to $$AB*, CD*, AC1, AD2, BC3, BD4, A34, B12, C24, D13$$.

Proof: Suppose there is. Then the rows corresponding to $AB*$ and $CD*$ form either a V, T, or X shape, with the intersection at $*$. Assuming no lines are parallel, the 3-trees must lie somewhere on the lines shown. Rows formed by the 3-tree on a line and a 4-tree not on it must also pass through a 3-tree on another line. So the location of the first 3-tree determine the location of the second 3-tree. Placing the 3-tree from the $AD$ line in varying regions of that line determines regions for the 3-trees on the $AC$ and $BD$ lines, which in turn each require regions for the 3-tree on the $BC$ line. But in all cases, the required regions on the $BC$ line do not overlap, making it impossible to place this tree. The case when some of the lines are parallel can be considered in the limit, and thus fail as well. It is impossible to construct a geometric solution for this combitorial case.

$$\text{ V - Shape}$$ $$\begin{array}{c|cc|cc} AD & AC & DB & BC(AC) & BC(BD)\\\hline AD1 & AC2 & BD4 & BC2 & BC3\\ AD2 & AC2 & BD1 & BC2 & BC3\\ AD3 & AC3 & BD2 & BC3 & BC4\\ AD4 & AC4 & BD3 & BC1/4 & BC2\\ AD5 & AC1 & BD3 & BC1 & BC2\\ AD6 & AC2 & BD4 & BC2 & BC3\end{array}$$

$$\text{ T - Shape}$$ $$\begin{array}{c|cc|cc} AD & AC & DB & BC(AC) & BC(BD)\\\hline AD1 & AC4 & BD4 & BC1/4 & BC3\\ AD2 & AC4 & BD1 & BC1/4 & BC3\\ AD3 & AC1 & BD1 & BC1 & BC3\\ AD4 & AC2 & BD2 & BC2 & BC1/4\\ AD5 & AC3 & BD3 & BC3 & BC2\\ AD6 & AC4 & BD4 & BC1/4 & BC3\end{array}$$

$$\text{ X - Shape}$$ $$\begin{array}{c|cc|cc} AD & AC & DB & BC(AC) & BC(BD)\\\hline AD1 & AC2 & BD4 & BC1/4 & BC2\\ AD2 & AC2 & BD1 & BC1/4 & BC2\\ AD3 & AC3 & BD2 & BC3 & BC1/4\\ AD4 & AC4 & BD3 & BC2 & BC3\\ AD5 & AC1 & BD3 & BC2 & BC3\\ AD6 & AC2 & BD4 & BC1/4 & BC2\end{array}$$

QED

Paul Sinclair has showed that there're only 4 potential combinations. He has showed that there're no geometry solution for the last one.

Since projective transformation keep lines, and by using projective transformation, we could assign the coordinates of any four points (no three of them in same line) arbitrary.

So for the 2nd combination, we could map A to infinity point at x-axis (1:0:0), and C to infinity point at y-axis (0:1:0), 1->(0,0) and 4->(1,0) and it is easy to show that 3->(0,1) and 5->(1,t) (that's point 5 in line x=1) as picture below. It is required also that the line through point 2 and 6 should be parallel to y-axis since there's line C26. But Geogebra shows that even if we moving point 5 in line x=1 in any way, line 26 always pass point x:(0,1/2), which shows that line 26 does not parallel to y-axis so that the 2nd combination has not geometry solution.

Similarly for 3rd combination, by setting A(0,0), B infinity point at x-axis, C infinity point at y-axis, we could get picture below where point 5 could move in line y=1, and 3 is infinity point at line 12. It is also required that point 4 must in line A3 so that line A4 should be parallel to line 12. Using Geogebra to draw the trajectory of point 4 as point 5 moving. We could find the trajectory doesn't intersect with line A3 so that it is not a valid geometry solution too.

So the 1st combination is the unique valid solution under projective transformation.