Constructing a rectangle of size nx2 with dominos and L-shaped trominos

Question

The Question is from DP tiling a 2xN tile with L shaped tiles and 2x1 tiles? I want an explanation about this question or theory

Answer 1

• Let $f[m]$ be the number of ways to cover the shape shown below, an $m$ by $2$ rectangle. Our ultimate goal is $f[n]$.

     ┌───────────┐
   2 │           │ 
     └───────────┘
          m

• Let $g[m]$ be the number of ways to cover the first shape shown below, an $m$ by $2$ rectangle with an extra 1x1 square at the top-right corner. By symmetry, $g[m]$ is also the number of ways to cover the second shape shown below.

           m+1                     m
     ┌─────────────┐        ┌───────────┐
   2 │           ┌─┘      2 │           └─┐
     └───────────┘          └─────────────┘
          m                        m+1

To find the recurrence relation, try covering the space at the rightmost boundary of the above shapes in all possible ways.

Consider $f[m]$. We have the following 4 ways to cover the rightmost space.

              ┌─────────┬─┐     ┌──────┬────┐     ┌───────┬───┐     ┌─────────┬─┐
              │         │ │     │      ├────┤     │       └─┐ │     │       ┌─┘ │
              └─────────┴─┘     └──────┴────┘     └─────────┴─┘     └───────┴───┘
What is left:   (m-1)x2         (m-2)x2            (m-2)x2+1         (m-2)x2+1

So, we have $\quad\quad f[m] = f[m - 1] + f[m - 2] + g[m - 2] \cdot 2 $ for $m\ge2$.

Consider $g[m]$. We have the following 2 ways to cover the rightmost space of the first shape.

                     m+1                 m+1
               ┌─────────┬───┐     ┌─────────┬───┐
               │         │ ┌─┘     │         └─┬─┘
               └─────────┴─┘       └───────────┘
 What is left:   (m-1)x2              (m-1)x2+1

So we have $\quad\quad g[m] = f[m - 1] + g[m-1]$ for $m\ge1$.

Applying the above two recurrence equations, we can compute all $f[m]$ and $g[m]$, in order of increasing $m$, starting from $m=2$, given the initial conditions, $f[0]=1$, $f[1]=1$, $g[0]=0$ and $g[1]=1$.

# Python program to compute the first n+1 values of f 
def show_number_of_ways(n):
    f = [0] * (n+1)
    g = [0] * (n+1)
    f[0] = f[1] = g[1] = 1
    g[0] = 0
    for m in range(2, n+1):
        f[m] = f[m - 1] + f[m - 2] + 2 * g[m - 2]
        g[m] = f[m - 1] + g[m - 1]
    print(f)

show_number_of_ways(10)
# [1, 1, 2, 5, 11, 24, 53, 117, 258, 569, 1255] 

---

Here is a way to derive a simpler recurrence relation that involves $f$ only.

Glorfindel's answer explains how to compute the number of patterns by "cutting the rightmost elementary block." To recap, there are one elementary block of size $1\times2$, one elementary block of $2\times2$ and two elementary blocks of $n\times2$ for $n\ge3$. Let $f(n)$ be the number of patterns for $2\times n$. we have the following base cases and recurrence relation, $$f(0)=1,\ f(1)=1,\ f(2)=2,$$ $$f(n)=f(n-1)+f(n-2)+2f(n-3)+2f(n-4)+\cdots+2f(0),\text{ for }n\ge3 $$

The above formulas leads to an algorithm that computes $f(n)$ with $O(n^2)$ time-complexity and $O(n)$ space-complexity.

We can do better. Replacing $n$ with $n-1$, we have $$f(n-1)=f(n-2)+f(n-3)+2f(n-4)+2f(n-5)+\cdots+2f(0),\text{ for }n\ge4 $$

Subtracting the above two equations, we get $$f(n)-f(n-1)=f(n-1)+f(n-3)$$ So we have for $n\ge4$, $$f(n)=2f(n-1)+f(n-3)\tag{simple}$$ Since $f(3)=5=2f(2)+f(0)$, the above recurrence relation holds for all $n\ge3$. This leads to an algorithm that computes $f(n)$ with $O(n)$ time-complexity and $O(1)$ space-complexity.

# Python program to compute the first n+1 values of f
def show_number_of_ways(n):
    f = [0] * (n+1)
    f[0] = f[1] = 1
    f[2] = 2
    for i in range(3, n+1):
        f[i] = 2 * f[i - 1] + f[i - 3]
    print(f)

show_number_of_ways(10)
# [1, 1, 2, 5, 11, 24, 53, 117, 258, 569, 1255] 

---

We can also derive the simple recurrence relation directly from the first two mutual recursive relations between $f$ and $g$.

The equality $g[m] = f[m - 1] + g[m-1]$ tells us $f[m-1] = g[m]-g[m-1]$, and, hence, $f[m] = g[m+1]-g[m]$ and $f[m-2] = g[m-1]-g[m-2]$. Applying them to eliminate $f$ away in $f[m] = f[m - 1] + f[m - 2] + g[m - 2] \cdot 2$, we get $g[m+1]=2g[m]+g[m-2]$.

Since $f[m]$ is a linear combination of $g[m+1]$ and $g[m]$, $f$ satisfies the same kind of recurrence relation, i.e., $f[m]=2f[m-1]+f[m-3]$. Checking the valid range of the index $m$, we see that it is value for all $m\ge3$.

Answer 2

One way to think of it would be elementary blocks, i.e. blocks that cannot be split by cutting them vertically.

• There's one elementary block of size 1x2 (one vertical B).
• There's one elementary block of size 2x2 (two horizontal Bs).
• There are two elementary blocks of size 3x2 (the last two shown in your figure, Example I).
• There are two elementary blocks of size 4x2 (they both appear in the final pattern of Example II).
• In general, there are two elementary blocks of size $n \times 2$, $n \ge 3$.

Now, an $n \times 2$ rectangle can be divided into two smaller blocks by cutting the rightmost elementary block. (This includes the case where the elementary block is the entire rectangle).

• A 0x2 rectangle can be tiled in 1 way.
• From a 1x2 rectangle, we can cut the rightmost elementary block (we have 1 of them) and we're left with a 0x2 rectangle, so it can be tiled in 1 way.
• From a 2x2 rectangle, we can either cut a 2x2 block (1 way) or cut a 1x2 block and be left with a 1x2 rectangle (1 * 1 = 1 way), for a total of 2 ways.
• From a 3x2 rectangle:
  • cut a 3x2 block (2 ways)
  • cut a 2x2 block (1 way) and be left with a 1x2 rectangle (1 way)
  • cut a 1x2 block (1 way) and be left with a 2x2 rectangle (2 ways)
  • Total: 2 + 1 * 1 + 1 * 2 = 5 ways.

This way, you can work all the way to 10x2.

After computing a few terms, I found the sequence is A052980 in The On-Line Encyclopedia of Integer Sequences:

a(n) is the number of possible tilings of a 2 X n board, using dominos and L-shaped trominos. - Michael Tulskikh, Aug 21 2019

so the final answer is 1255 for $n=10$.