MATH 4707 Homework 1
نویسنده
چکیده
the purpose of using it in this homework. Namely, that the Fibonacci numbers, fn, count the tilings on a 2× (n− 1) board. First, suppose that n = 1. Then we have a 2× 0 board, and there is exactly 1 way to tile it. Suppose we have n = 2. Then, we have a 2× 1 board, and again there is exactly one way to tile it: with a single vertical domino. This provides the basis, where f1 = f2 = 1. Now, we need to show the recurrence relation. Suppose that fn−1 and fn−2 count the tilings on a 2 × (n − 2) and 2 × (n − 3) board respectively. Now, consider a 2 × (n − 1) board. For each tiling, we can either have a vertical domino in the first column, or a horizontal domino that covers the first and second columns. In the first case, we just have a 2 × (n − 2) board remaining. In the second case, we have a 2 × (n − 3) board remaining. Thus, the number of ways to tile the 2 × (n − 1) board, fn, is just fn−1 + fn−2. Hence, the number of tilings do indeed follow the Fibonacci numbers. I will use this fact twice: in problem 4 and problem 5.
منابع مشابه
Fall 2017, MATH 4707 (UMN) Homework 1 solutions
f1 = f2 = 1. Now, we need to show the recurrence relation. Suppose that fn−1 and fn−2 count the tilings on a 2 × (n − 2) and 2 × (n − 3) board respectively. Now, consider a 2 × (n − 1) board. For each tiling, we can either have a vertical domino in the first column, or a horizontal domino that covers the first and second columns. In the first case, we just have a 2 × (n − 2) board remaining. In...
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تاریخ انتشار 2017