(a) Let f(n) count the number of different perfect covers of a 2 -by- n chessboard by dominoes.

step 1 All right, so here we have a checkerboard that is 2 by n long. Okay, so we have a checkerboard a

(a) Let f(n) count the number of different perfect covers of...

Key Concepts

Perfect Covering

Perfect covering refers to a configuration where every square in the board is covered exactly once by a domino, ensuring that there are no unused squares or overlaps. This concept is essential for setting up the problem correctly and avoiding partial or redundant tilings in the enumeration process.

Combinatorial Enumeration

Combinatorial enumeration is the process of counting the number of ways certain patterns or configurations can be arranged. In tiling problems, it involves determining the number of perfect covers that exist for a given board size, which is a fundamental aspect of combinatorial analysis and often requires recursive or dynamic programming strategies.

Domino Tiling

Domino tiling involves covering a given board completely using domino pieces (rectangles covering two adjacent squares) without overlaps or gaps. This concept is a classical example in combinatorics and geometry that examines how simple shapes can be arranged in various configurations to perfectly cover a region.

Recurrence Relations

Recurrence relations are equations that express a sequence in terms of its preceding values. In tiling problems, these relations are used to relate the number of tilings for a board of a particular size to the number of tilings for smaller boards, enabling efficient computation and revealing underlying patterns in the counts.

Fibonacci Sequence

The Fibonacci sequence is a well-known numerical series where each term is the sum of the two preceding ones. In the context of domino tilings of a 2-by-n board, the number of tiling configurations often follows a Fibonacci-like recurrence relation, highlighting a direct connection between a geometric covering problem and this famous sequence.

Transcript

00:01 All right, so here we have a checkerboard that is 2 by n long.

00:09 Okay, so we have a checkerboard and it's n long.

00:15 We have two options here.

00:17 So first we can either place our domino vertically or we could place it horizontally.

00:26 And when we place this first domino, either horizontally or vertically, that's all the possible cases, right? so if a sub n is the number of ways to cover the board, then it's going to equal the number of ways you cover it by placing it first horizontally, plus the number of ways when you would place it first vertically.

00:58 So let's consider those cases.

01:01 So case one is when we place the nominal vertically.

01:10 Okay, well, we've placed it vertically, and we've eliminated this rightmost space, right? and this was n long.

01:18 So that means we have a checkerboard that is now n minus one long that still needs to be filled.

01:27 And so how many ways is that? well, that's going to be the number of ways to cover a board of length n minus one.

01:38 Okay, and the other possibility is horizontal.

01:40 So if we place our first one horizontally here in the green, we're sort of obligated to put a horizontal one below it, right? there's not space to put a vertical one there.

01:57 So now the length that we're working with is n minus 2 instead of the n minus 1.

02:06 So the number of ways in which we can arrange the rest of the tiles on this one is going to be the number of ways we can arrange them on a board of length.

02:16 Length n minus 2.

02:17 So the total number of ways, then a sub n, is going to be equal to the sum of these two cases.

02:25 So that's just a sub n minus 1 plus a sub n minus 2.

02:33 Okay, so that is the solution.

02:38 So then we're looking for the initial conditions.

02:44 So what happens if n is equal to 1? so then we would have a 2 by 1 checkerboard.

02:55 And well, there is exactly one way to place tiles on this, right? it's to place one tile on it vertically.

03:03 So a of 1 is going to be equal to 1.

03:08 Now, if n is equal to 2, then we have a 2 by 2 option.

03:15 So we could place them two horizontally...