# The Tree is Growing

I just found this new type of puzzle and I think the rule set is fun! So, I decided to create one.

Rules (taken from the original puzzle):

1. Create non-intersecting non-looping trees such that the trees fill all cells. Trees must sprout from seeds (given triangles) in the direction they point. Tree branches must not pass through black edges. Leaves (tree branch endpoints) from the same tree must not share a row or column. If given, a number on a seed must equal the number of leaves on the final tree.
2. Trees may branch and travel in any direction (except branching on the seed). Different trees must not intersect or share cells. Leaves from different trees may share the same row/column. The seed of a tree is not a leaf, despite it being an endpoint.

The Puzzle:

Kudos to scor for the original puzzle. You also can solve this on Penpa if you prefer online solving. Hope you guys enjoy it!

• @user39583 Yes, you're right and I apologize. I need to be more carefull. Thanks for pointing it out! Hopefully it's unique now.
– Nusi
Nov 4, 2021 at 11:21
• The Penpa needs to be corrected so that the two think lines between the cells (5,1) and (5,2) and the cells (6,9) and (6,10) are covered. Nov 4, 2021 at 13:58

I feel like this is the intended answer but I found two places where lines could go either way unless I misread the instructions (for the green and blue leaves/stars).

UPDATE: I noticed the image in OP had two additional lines added to it, so I updated my solution, which is unique now.

SOLVED:

Steps:

First of all, one thing to note is that no closed 2x2 alcove can have more than 1 leaf. If it had 2, both would need to share the row or column, and since there's only one entrance, they'd be on the same tree. (I'll call this the ALCOVE RULE)

We start by filling some lines and adding some leaves on dead ends. Trees greater than 1 should at least reach a square with a possible intersection, which forces the 3s on both sides to reach longer.

If yellow tree doesn't go into R4C4, the four leaves on the top rows will have to be attached to a 1 or a 3, due to the narrow pathway in C7. Yellow tree needs to pick at least one of them. On the other hand, it can' not go into R5C3 because if it picks all the top leaves two of them will be in the same row (R1C2 and R1C9).

If purple leaf goes in R6C4, R7C4, R8C4, or R9C4, we'll get too many leaves on the bottom central area to join with the orange 3 tree. Thus, purple either ends on R7C5 or the marked leaf in R10C2. In any case, R6C4 must at least extend to its left. (I've also added red borders to indicate trees not joining each other).

R5C7 must join the pink 3 tree; if it didn't, red tree would have 2 leaves, R5C7 and R1C9. Per the ALCOVE RULE, Green tree must reach R4C2 and R5C1. Both sides have space for two leaves at most for this tree (either R5C2 and R2C3, or R8C1 and R9C3). As it needs 3 leaves, we need to go into both areas.

Blue tree must go down C10 due to the ALCOVE RULE too. It can at most fit one leaf by going into R4C9, so it must also extend to split once more (at least), thus reaching R7C10. R5C2 cannot be a leaf of the yellow tree as it'd be in the same column as R1C2. It must be a leaf for green then. This also means that R2C5 and R3C6 must be yellow leaves. If we count yellow leaves, we have: R1C2; one in the R2C2 alcove or in R5C3; R2C5 (we need to extend through R1) and R3C6 (we can't have one in the alcove above the tree as it would collide with R2C5). We also extend green tree at least one space in C1 to make space for the third leaf and have it not colliding with R5C2.

Yellow can't have a leaf in the alcove as it would collide with R2C5 or R3C6. Thus, the leaf in the alcove belongs to green and the last leaf for yellow is R5C3. Green leaf is forced in the alcove to avoid colliding with R5C2.

ALCOVE RULE means that light grey must go down C4, as it can fit at most a leaf in the lower left alcove and a leaf in C1 or C2. For its other two leaves it must reach R10C2 and R8C5 at least. This locks purple leaf too. Purple can't grab both R9C6 and R10C9 as both leaves in R10 would collide. It must then go into R7C3, and R10C9 needs to connect upwards, with the orange tree.

The grey leaf in lower left alcove must be R9C3. This means R9C6 can't be a grey leaf (it must be orange). The fourth grey leaf must go into R7C1 to avoid more collisions. This also fixes brown and green leaves.

Orange must dip into R6C8 because it can't fit a leaf in the alcove below it (due to R9C6). Pink must then connect with R4C8 to make wiggle room for its 3 leaves. Blue needs 3 leaves which must be in different columns. The only place where it could have a leaf in C8 is in R9C8, forcing it to enter the lower right alcove. R6C9 must be blue too, as if it was an orange leaf it would collide.

We're getting close to the end; the third blue leaf must go in C10, so the only place available is R8C10, which also locks the dark grey leaf. This gives us the pink star in R4C9, which means R1C9 can't be pink, it must be red. The places for the last pink leaves are tight, but they fit, giving us the solved grid:

• Oh, I just saw that the image is updated (but the Penpa link isn't). The additional lines disambiguate my solution. Nov 4, 2021 at 13:54
• Correct, good spot. I nearly got to your solution! I hope you can explain the logic a little more in your answer. Nov 4, 2021 at 14:13
• +1. Got the same result! My method was more or less to make greedy branches that ate as much space as possible and then prune them back to make more space as more branches were added.
– Qami
Nov 4, 2021 at 14:51
• @hexomino and Teresa Lisbon - I'll try to make a step-by-step Nov 4, 2021 at 15:18
• @ArturoVialArqueros Thanks a lot and +1. I think your technique is really good and I will try to follow it for future arboretum puzzles. Nov 5, 2021 at 7:23