Strimko Special Number?

Probability Level 5

Like the game of Sudoku , Strimko involves filling out all distinct numbers for all rows, columns, and equal-length strings. As shown, each string consists of adjacent cells, connected either diagonally, vertically, or horizontally. In addition, it is also possible for multiple strings to cross each other.

For this problem, consider the following set of $3 \times 3$ empty, unconnected cells. Given the conditions of creating a Strimko puzzle, how many unique arrangements of 3 strings are there, such that an empty puzzle can have at least a valid solution?

Clarification: Multiple distinct constructions, where the whole setup can all be reoriented (via rotation or reflection) to look like each other, are all counted as one.

This original problem has been inspired by Grabarchuk family's puzzles featured in Wall Street Journal .

8 or less 9 10 11 12 13 14 15 or more

1 solution

Michael Huang
Oct 2, 2018

By the rules of Strimko , color the empty set of $3 \times 3$ cells as shown:

Figure 1. Set of 3-by-3 cells colored by the rules of sudoku Figure 1. Set of 3-by-3 cells colored by the rules of sudoku

where neither of the cells within rows and columns share the same colors. Since $2 \times 2$ square within $3 \times 3$ square consists of diagonal cells of the same colors, no strings can cross each other. Thereby, the connected strings can be constructed as follow:

Figure 2. Three cases, each illustrating distinct possibility of constructing a game of Strimko Figure 2. Three cases, each illustrating distinct possibility of constructing a game of Strimko

where the dashed-line groups indicate multiple possibilities for the construction. Without loss of generality, positioning a straight line in any other way gives the same or similar results. Thus, for each of the cases, we consider the given constructions.

Case I: Since the form is symmetric, we count $1$ unique string arrangement. Rotating it gives the similar construction.

Case II: Since the construction is entirely asymmetric, the number of unique string arrangements is $3^2 = 9$ , which is the total number of ways to connect two sets of three triangular vertices.

Case III: For one of the triangular groups, since two possible trees of common forms are reflective, then there are $3 - 1 = 2$ distinct ways to create an initial tree (as denoted by the orange-colored lines from Figure 3):

Figure 3. Two distinct cases for two-strings setups. Figure 3. Two distinct cases for two-strings setups.

Firstly, for the right-side diagram, since the formed tree is unique out of the three possible trees, there are also $3 - 1 = 2$ possible trees for the bottom-right triangular set - one identical to the orange tree and another distinct.
Secondly, for the left-side diagram, since the L-shaped tree from the right-side diagram is already taken, this leaves the empty triangular set with $3 - 1 = 2$ possible trees.

Both of these cases altogether show that there are $2^2 = 4$ string arrangements.

Thus, there are $1^2 + 2^2 + 3^2 = \boxed{14}$ unique constructions.

Interestingly, the number of unique arrangements is combinatorially the total number of squares of any size within $3 \times 3$ chessboard. This follows the topic on combinatorial proofs of powers sum from the article I read.

Strimko Special Number?

1 solution

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