Winning Positions

In this week's post, we learn about Winning Positions for Combinatorial Games.

How would you use Winning Positions to solve the following problem? >

In Worked Example 2, how many winning positions are there on a k×n k \times n grid?

#Combinatorics #MathProblem #Math

Note by Calvin Lin
7 years, 10 months ago

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Comments

Let T(k,n)T(k,n) denote the number of winning positions on a k×nk \times n grid. The top row always ends in … W W L W W L W W L, and whether this row starts with a W or with an L depends on nn.

  • If n≡0(mod3)n \equiv 0 \pmod{3}, then the top row is W W L W W L […] W W L, so the last position in the row beneath is an L (because the only possible move brings the opponent in a winning position) and the pattern continues. It's easy to see that 23\frac{2}{3} of all positions in this grid are winning, so,

T(k,n)=23⋅k⋅n(if n≡0(mod3)). T(k,n) = \frac{2}{3} \cdot k \cdot n \quad \left( \text{if }n \equiv 0 \pmod{3} \right).

  • If n≡2(mod3)n \equiv 2 \pmod{3}, then the top row is W L W W L […] W W L, so the last position in the row beneath is an L and the pattern continues. In each row, the number of winning positions is 23⋅(n−2)+1\frac{2}{3}\cdot (n-2) + 1, so,

T(k,n)=k(2(n−2)3+1)(if n≡2(mod3)). T(k,n) = k\left(\frac{2(n-2)}{3}+ 1 \right) \quad \left( \text{if }n \equiv 2 \pmod{3} \right).

  • If n≡1(mod3)n \equiv 1 \pmod{3}, things are getting real interesting. The top row is L W W L […] W W L, so the entire second row is winning, because you can put your opponent in a losing position by moving to the leftmost square of the top row! But that means the last position in the third row is losing, and the third row is exactly the same as the first row, so the first position in the third row ls L, so the entire fourth row is winning. More generally,

the t-th row is{L W W L W W L […] W W L( if t is odd),W W W W W W […] W W W( if t is even). \text{the }t\text{-th row is} \begin{cases} \begin{aligned} \text{L W W L W W L […] W W L}&\quad \left( \text{ if }t\text{ is odd}\right) ,\\ \text{W W W W W W […] W W W}&\quad \left( \text{ if }t\text{ is even}\right) . \end{aligned} \end{cases}

In all even rows, there are nn winning positions. In all odd rows, there are 23⋅(n−1)\frac{2}{3}\cdot(n-1) winning positions. So,

T(k,n)=⌊k2⌋⋅n+⌈k2⌉⋅2(n−1)3(if n≡1(mod3)). T(k,n) = \left \lfloor \frac{k}{2} \right \rfloor \cdot n + \left \lceil \frac{k}{2} \right \rceil \cdot \frac{2(n-1)}{3} \quad \left( \text{if }n \equiv 1 \pmod{3} \right).

Whew.

Tim Vermeulen - 7 years, 10 months ago
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