1
$\begingroup$

The question:

I want to pave a $2\times n$ rectangle with blocks of two types, A and B, as illustrated http://image.prntscr.com/image/4254aa94b979415b895e65cd90f700ac.png

long edges are length $2$ and short edges are length $1$. I want to know in how many ways this can be done. Reflections of combinations DO count as separate combinations.

(a) Find a linear recursive equation for $X_n$, the number of pavings of a $2\times n$ rectangle.

I have no idea what to do except start by placing blocks on the left-side of the $2\times n$ rectangle, but I don't know what else I can do.

  • 0
    When you say reflections of combinations DO count, does that mean you can use the blocks $A$ and $B$ in any orientation? I.e. could I lay $A$ flat with its long side on the bottome? – 2017-01-26
  • 0
    Yes, you can. Any rotation that is a 90 degree multiple. – 2017-01-26

1 Answers 1

2

You have $X_n$ as the number of ways to pave a $2 \times n$ rectangle. At the right hand end you might have a type B piece, a vertical type A piece, or two horizontal type A pieces. If you take off the A piece(s) you are left with a tiled rectangle. If you take off a B piece you are left with a tiled rectangle with one extra square, so define $Y_n$ at the number of ways to tile a $2 \times n$ rectangle plus the top square of the next column, which is the same as the number of ways to tile the rectangle plus the bottom square of the next column. One of these rectangles plus square can either have a B piece attached to a rectangle, or a horizontal A piece sticking out. This should suggest a set of coupled recurrences for $X_n,Y_n$.

  • 0
    So would you describe the overall recursion as: Xn = Xn-1 + Xn-2 + Yn-2 ? Implying that the number of ways to tile a 2xn rectangle is equal to the number of ways to tile it by starting with either an A, two rotated As or a B on one side – 2017-01-26
  • 0
    Not quite. Because I defined the $Y$ with one orientation and there are two orientations for the B tile you should have $2Y_{n-2}$ at the end. Now you need to write a recurrence for $Y_n$ as well. Then you can either eliminate one variable between the two recurrences or look at a vector of $X_n,Y_n$ and the updating being a matrix multiply. – 2017-01-26
  • 0
    How would you build a recurrence for Yn in terms of Xn? I'm sorry if this sounds silly, this question is just outside of the realm of my knowledge. – 2017-01-26
  • 0
    You can either have a B attached to a rectangle or have a horizontal A attached to a rectangle plus square to make a rectangle plus square. There are two terms on the right here. – 2017-01-26