Counting the ways to build a wall with two tile sizes

Question

You are given a set of blocks to build a panel using 3”×1” and 4.5”×1" blocks.

For structural integrity, the spaces between the blocks must not line up in adjacent rows.

There are 2 ways in which to build a 7.5”×1” panel, 2 ways to build a 7.5”×2” panel, 4 ways to build a 12”×3” panel, and 7958 ways to build a 27”×5” panel. How many different ways are there to build a 48”×10” panel?

This is what I understand so far:

with the blocks 3 x 1 and 4.5 x 1

I've used combination formula to find all possible combinations that the 2 blocks can be arranged in a panel of this size

C = choose --> C(n, k) = n!/r!(n-r)! combination of group n at r at a time

Panel: 7.5 x 1 = 2 ways -->

1 (3 x 1 block) and 1 (4.5 x 1 block) --> Only 2 blocks are used--> 2 C 1 = 2 ways

Panel: 7.5 x 2 = 2 ways

I used combination here as well

1(3 x 1 block) and 1 (4.5 x 1 block) --> 2 C 1 = 2 ways

Panel: 12 x 3 panel = 2 ways -->

2(4.5 x 1 block) and 1(3 x 1 block) --> 3 C 1 = 3 ways

0(4.5 x 1 block) and 4(3 x 1 block) --> 4 C 0 = 1 way

3 ways + 1 way = 4 ways

(This is where I get confused)

Panel 27 x 5 panel = 7958 ways

6(4.5 x 1 block) and 0(3 x 1) --> 6 C 0 = 1 way

4(4.5 x 1 block) and 3(3 x 1 block) --> 7 C 3 = 35 ways

2(4.5 x 1 block) and 6(3 x 1 block) --> 8 C 2 = 28 ways

0(4.5 x 1 block) and 9(3 x 1 block) --> 9 C 0 = 1 way

1 way + 35 ways + 28 ways + 1 way = 65 ways

As you can see here the number of ways is nowhere near 7958. What am I doing wrong here?

Also how would I find how many ways there are to construct a 48 x 10 panel? Because it's a little difficult to do it by hand especially when trying to find 7958 ways.

How would write a program to calculate an answer for the number of ways for a 7958 panel? Would it be easier to construct a program to calculate the result? Any help would be greatly appreciated.

Answer 1

I don't think the "choose" function is directly applicable, given your "the spaces between the blocks must not line up in adjacent rows" requirement. I also think this is where your analysis starts breaking down:

Panel: 12 x 3 panel = 2 ways -->

2(4.5 x 1 block) and 1(3 x 1 block) --> 3 C 1 = 3 ways

0(4.5 x 1 block) and 4(3 x 1 block) --> 4 C 0 = 1 way

3 ways + 1 way = 4 ways

...let's build some panels (1 | = 1 row, 2 -'s = 1 column):

+---------------------------+
|      |      |      |      |
|      |      |      |      |
|      |      |      |      |
+---------------------------+

+---------------------------+
|          |         |      |      
|          |         |      |      
|          |         |      |      
+---------------------------+

+---------------------------+
|      |          |         |            
|      |          |         |      
|      |          |         |      
+---------------------------+

+---------------------------+
|          |      |         |                  
|          |      |         |      
|          |      |         |      
+---------------------------+

Here we see that there are 4 different basic row types, but none of these are valid panels (they all violate the "blocks must not line up" rule). But we can use these row types to create several panels:

+---------------------------+
|      |      |      |      |
|      |      |      |      |
|          |         |      |      
+---------------------------+

+---------------------------+
|      |      |      |      |
|      |      |      |      |
|      |          |         |      
+---------------------------+

+---------------------------+
|      |      |      |      |
|      |      |      |      |
|          |      |         |     
+---------------------------+

+---------------------------+
|          |         |      |      
|          |         |      |
|      |      |      |      |
+---------------------------+

+---------------------------+
|          |         |      |      
|          |         |      |
|      |          |         |
+---------------------------+

+---------------------------+
|          |         |      |      
|          |         |      |
|          |      |         |
+---------------------------+

...

But again, none of these are valid. The valid 12x3 panels are:

+---------------------------+
|      |      |      |      | 
|          |      |         |
|      |      |      |      |
+---------------------------+

+---------------------------+
|          |      |         |
|      |      |      |      |
|          |      |         |
+---------------------------+

+---------------------------+
|          |         |      |
|      |          |         |
|          |         |      |
+---------------------------+

+---------------------------+
|      |          |         |
|          |         |      |
|      |          |         |
+---------------------------+

So there are in fact 4 of them, but in this case it's just a coincidence that it matches up with what you got using the "choose" function. In terms of total panel configurations, there are quite more than 4.

Answer 2

1. Find all ways to form a single row of the given width. I call this a "row type". Example 12x3: There are 4 row types of width 12: (3 3 3 3), (4.5 4.5 3), (4.5 3 4.5), (3 4.5 4.5). I would represent these as a list of the gaps. Example: (3 6 9), (4.5 9), (4.5 7.5), (3 7.5).

2. For each of these row types, find which other row types could fit on top of it.

   Example:

   a. On (3 6 9) fits (4.5 7.5).

   b. On (4.5 9) fits (3 7.5).

   c: On (4.5 7.5) fits (3 6 9).

   d: On (3 7.5) fits (4.5 9).

3. Enumerate the ways to build stacks of the given height from these rules. Dynamic programming is applicable to this, as at each level, you only need the last row type and the number of ways to get there.

Edit: I just tried this out on my coffee break, and it works. The solution for 48x10 has 15 decimal digits, by the way.

Edit: Here is more detail of the dynamic programming part:

Your rules from step 2 translate to an array of possible neighbours. Each element of the array corresponds to a row type, and holds that row type's possible neighbouring row types' indices.

0: (2)
1: (3)
2: (0)
3: (1)

In the case of 12×3, each row type has only a single possible neighbouring row type, but in general, it can be more.

The dynamic programming starts with a single row, where each row type has exactly one way of appearing:

1 1 1 1

Then, the next row is formed by adding for each row type the number of ways that possible neighbours could have formed on the previous row. In the case of a width of 12, the result is 1 1 1 1 again. At the end, just sum up the last row.

Complexity:

• Finding the row types corresponds to enumerating the leaves of a tree; there are about (/ width 3) levels in this tree, so this takes a time of O(2^w/3) = O(2^w).

• Checking whether two row types fit takes time proportional to their length, O(w/3). Building the cross table is proportional to the square of the number of row types. This makes step 2 O(w/3·2^2w/3) = O(2^w).

• The dynamic programming takes height times the number of row types times the average number of neighbours (which I estimate to be logarithmic to the number of row types), O(h·2^w/3·w/3) = O(2^w).

As you see, this is all dominated by the number of row types, which grow exponentially with the width. Fortunately, the constant factors are rather low, so that 48×10 can be solved in a few seconds.

Answer 3

This looks like the type of problem you could solve recursively. Here's a brief outline of an algorithm you could use, with a recursive method that accepts the previous layer and the number of remaining layers as arguments:

• Start with the initial number of layers (e.g. 27x5 starts with remainingLayers = 5) and an empty previous layer
• Test all possible layouts of the current layer
  • Try adding a 3x1 in the next available slot in the layer we are building. Check that (a) it doesn't go past the target width (e.g. doesn't go past 27 width in a 27x5) and (b) it doesn't violate the spacing condition given the previous layer
  • Keep trying to add 3x1s to the current layer until we have built a valid layer that is exactly (e.g.) 27 units wide
  • If we cannot use a 3x1 in the current slot, remove it and replace with a 4.5x1
  • Once we have a valid layer, decrement remainingLayers and pass it back into our recursive algorithm along with the layer we have just constructed
• Once we reach remainingLayers = 0, we have constructed a valid panel, so increment our counter

The idea is that we build all possible combinations of valid layers. Once we have (in the 27x5 example) 5 valid layers on top of each other, we have constructed a complete valid panel. So the algorithm should find (and thus count) every possible valid panel exactly once.

Answer 4

This is a '2d bin packing' problem. Someone with decent mathematical knowledge will be able to help or you could try a book on computational algorithms. It is known as a "combinatorial NP-hard problem". I don't know what that means but the "hard" part grabs my attention :)

I have had a look at steel cutting prgrams and they mostly use a best guess. In this case though 2 x 4.5" stacked vertically can accommodate 3 x 3" inch stacked horizontally. You could possibly get away with no waste. Gets rather tricky when you have to figure out the best solution --- the one with minimal waste.

Answer 5

Here's a solution in Java, some of the array length checking etc is a little messy but I'm sure you can refine it pretty easily.

In any case, I hope this helps demonstrate how the algorithm works :-)

import java.util.Arrays;

public class Puzzle
{
    // Initial solve call
    public static int solve(int width, int height)
    {
        // Double the widths so we can use integers (6x1 and 9x1)
        int[] prev = {-1};      // Make sure we don't get any collisions on the first layer
        return solve(prev, new int[0], width * 2, height);
    }

    // Build the current layer recursively given the previous layer and the current layer
    private static int solve(int[] prev, int[] current, int width, int remaining)
    {
        // Check whether we have a valid frame
        if(remaining == 0)
            return 1;

        if(current.length > 0)
        {
            // Check for overflows
            if(current[current.length - 1] > width)
                return 0;

            // Check for aligned gaps
            for(int i = 0; i < prev.length; i++)
                if(prev[i] < width)
                    if(current[current.length - 1] == prev[i])
                        return 0;

            // If we have a complete valid layer
            if(current[current.length - 1] == width)
                return solve(current, new int[0], width, remaining - 1);
        }

        // Try adding a 6x1
        int total = 0;
        int[] newCurrent = Arrays.copyOf(current, current.length + 1);
        if(current.length > 0)
            newCurrent[newCurrent.length - 1] = current[current.length - 1] + 6;
        else
            newCurrent[0] = 6;
        total += solve(prev, newCurrent, width, remaining);

        // Try adding a 9x1
        if(current.length > 0)
            newCurrent[newCurrent.length - 1] = current[current.length - 1] + 9;
        else
            newCurrent[0] = 9;
        total += solve(prev, newCurrent, width, remaining);

        return total;
    }

    // Main method
    public static void main(String[] args)
    {
        // e.g. 27x5, outputs 7958
        System.out.println(Puzzle.solve(27, 5));
    }
}