# Competitive Programming: POJ 2663 – Tri Tiling

### Description

In how many ways can you tile a 3xn rectangle with 2×1 dominoes?

Here is a sample tiling of a 3×12 rectangle.

### Input

Input consists of several test cases followed by a line containing -1. Each test case is a line containing an integer 0 <= n <= 30.

### Output

For each test case, output one integer number giving the number of possible tilings.

2
8
12
-1

### Sample Output

3
153
2131

Solution below . . .

### Steps for Solving Dynamic Programming Problems

#### 1. Define subproblems

Let $f(n)$ be the number of tilings of a 3xN rectangle (which is what we’re looking for) and let $g(n)$ denote the number of tilings of a 3xN rectangle with one of its corner squares removed.

********
********
********
f(n)

********
********
*******
g(n)

#### 2. Write down the recurrence that relates subproblems

The recurrence functions are $f(n) = f(n-2) + 2g(n-1)$ and $g(n) = f(n-1) + g(n-2)$. The functions can be illustrated as follows:

********   AA*******   AA******   A*******
******** = BB******* + B******* + A*******
********   CC*******   B*******   BB******
f(n)   =  f(n-2)   +  g(n-1)  +  g(n-1)

********   A********   AA*******
******** = A******** + BB*******
*******    ********    CC******
g(n)   =   f(n-1)  +   g(n-2)

If you’re wondering why patterns like

AA*******
BC*******
BC*******

or

AB*******
AB*******
CC******

are not represented, notice that they are subsets of existing patterns and thus already accounted for.

#### 3. Recognize and solve the base cases

The base cases are: $f(0) = 1, f(1) = 0, g(0) = 0, g(1) = 1$. (Re $g(0) = 0$, there’s no way to remove the corner from a 3×0 rectangle.)

import java.util.Scanner;

public class Main {
public static void main(String[] args) {

Scanner sc = new Scanner(System.in);

int n = sc.nextInt();

while (n != -1) {

// Optimization: If n is odd, the number of tilings = 0.
if (n % 2 == 1) {
System.out.println(0);
}
else {
System.out.println(f(n));
}

n = sc.nextInt();
}
sc.close();
}

private static int f(int n) {
if (n == 0) return 1;
if (n == 1) return 0;

return f(n-2) + 2 * g(n-1);
}

private static int g(int n) {
if (n < 2) return n;

return f(n-1) + g(n-2);
}
}