PLEASE NOTE! A different problem that uses this same ruleset that can likely only be solved with brute force and a computer program has been posted to math.se, viewable HERE. I was unaware of the crossposting etiquette on stackexchange, my bad comrades!



  • The game takes place on a board with a square grid. For the purposes of this puzzle, you can assume the board size is infinite.
  • The board's grid squares are all in a "blank state" at the start of the game.
  • Clicking (or tapping, etc.) a blank grid square places a tier1 tile.
  • When three tier1 tiles are found to be adjacent in any of the 4 cardinal directions, a tier2 tile is placed at the site of the tier1 tile that was most recently placed. The other two tier1 tiles that formed the chain of three are reverted to blank grid squares.
  • An extension of the previous rule: three adjacent tier2 tiles create a tier3 tile, etc. Tier5 tiles are the greatest tile possible (see the postulates section for the reason why).
  • The edges of the game board (not relevant in this problem) and all non-blank grid squares are considered "walls". No tile can exist on top of or inside a wall; this game exists on only one "gameObject layer" (shout-out to people whom've used Unity for game making!).


  • Tier2 tiles cannot be created adjacent to 4 walls. Only tier1 tiles have the property of being able to be placed on any blank grid square.
  • Tier3 tiles cannot be created adjacent to 3 walls.
  • Tier4 tiles cannot be created adjacent to 2 walls.
  • Tier5 tiles cannot be created adjacent to any walls. Tier6 tiles are impossible to create because of this: due to the rule that all non-blank grid squares are considered walls, a Tier5 tile cannot be created adjacent to any other tile, and therefore cannot be arranged into a chain of three adjacent tiles to reach the next tier.

Important to note!

  • In this puzzle, what we care about are your unique grid square clicks. Because the least recent two tiles that formed a higher tier tile are reverted back to blank grid tiles, it's possible to click a specific grid square multiple times in the creation process of any tile tier3 and above. Unique grid square clicks are defined as the complete set of board squares that you clicked at least once, even if they end up as blank tiles at the end of your solution.


Earlier today I posted this question on the mathematics stack exchange. While it hasn't received a proper answer, StackExchange user Servaes offered this formula for calculating the minimum number of unique grid square clicks needed to create a tile of a given tier:

"The minimum amount of grid squares required to reach a $T_n$ tile is . . . $T_n=2n-1$"

According to this proposed formula, a tier4 tile can be made with 7 unique grid square clicks (27 clicks total), and a tier5 tile can be made with 9 unique grid square clicks (81 clicks total). While I agree it's possible to create a tier4 tile in 7 unique grid square clicks, I disagree that it's possible to make a tier5 in 9.

And so I pose this conundrum to you:

What is the fewest amount of unique grid square clicks needed to create a tier5 tile? Please show your work: I want to understand the shape formed by your unique grid square clicks, and I want to know the chronological order and coordinates of your clicks (81 clicks is a lot to give coords for, so feel free to only give the order and coords of your t4 tiles (3 total ordered coords), t3 tiles (9 total ordered coords), or t2 tiles if you're feeling spicy (27 total ordered coords)).


While it's explaining the separate problem I linked earlier in this post, and while it also contains potentially confusing terms that are not relevant to this specific puzzle, I have created a YouTube video that may help you grok the game I've described here. Click this link to check it out!


2 Answers 2


I can do it with



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Start by making a T4 at the bottom. Make another one above in the same way. Make a T3 in the top right corner, and use the space in the bottom right to make one below it. The rest is simple.

About the minimum required

You need at least 2 of each lower tier plus an empty space, which is adjacent to one of each tier. This takes 9 spaces, but it's impossible to do without having a T4 in a corner, which we can't do. So the minimum is between 10 and 12. I highly doubt 10 is possible, but 11 might be.

  • $\begingroup$ howdy! i just learned (sorry, new) that crossposting is frowned upon, but I wanted to let you know that (1.) i edited the post for this problem to disclaim that a different problem that uses this exact tile ruleset was posted to math.se, and (2.) someone over there found a solution to THIS problem that uses 11 unique tile clicks, down from 12!!!! next up: the hunt for 10? ...also: (3.) p.s. if you wanna check that 11 unique click solution out, there's a link to the math.se problem post at the top of this one in the PLEASE NOTE! section $\endgroup$ Commented May 4, 2019 at 3:42

I think it can be done in 25 unique grid clicks.

The steps are as follows:

I get a 5*5 square as the grid

. . 2 2 2 - . . 3 . . - . . 4 . .

. . 2 2 2 - . . 3 . . - . . 4 . .

. . 2 2 2 - . . 3 . . - . . 4 . .

. . 2 2 2 - . . 3 . . - . . 4 . .

. . 2 2 2 - . . 3 . . - . . 4 . .

And you get 5 in the center of the grid.

  • $\begingroup$ A t5 tile is definitely creatable using 25 unique grid square clicks (5*5 = 25)!! But unfortunately this is not the correct answer, because I was just able to create a t5 tile using 13 unique grid square clicks. Now the question has shifted a little! Is 13 the least possible amount of unique grid square clicks?? If you can prove that, you've solved the question!!! $\endgroup$ Commented May 3, 2019 at 16:49

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