We shall define a square lamina to be a square outline with a square “hole” so that the shape possesses vertical and horizontal symmetry.

Given eight tiles it is possible to form a lamina in only one way: 3 × 3 square with a 1 × 1 hole in the middle. However, using thirty-two tiles it is possible to form two distinct laminae. If t represents the number of tiles used, we shall say that t = 8 is type L(1) and t = 32 is type L(2).

Let N(n) be the number of t <= 1000000 such that t is type L(n); for example, N(15) = 832.

What is sum of N(n) for n=1 to 10?

This question turned out to be fairly straight forward, and is solvable using a brute force method.

Basically, we want to arrange tiles in laminae that are symmetrical, and don’t use more than 1,000,000 tiles. The largest lamina we can arrange would then be 250,000 tiles per side, with a wall thickness of 1.

The way I set this up was to iterate through all lamina side lengths (L) from 3 to our limit, 250,000. Then, for each, iterate for different hole sizes (H) in the middle from L–2 down to 1. (The evenness of L and H have to match.)

Calculating the number of tiles is an easy prospect; t = L² – H². A dictionary then keeps track of how many tiles are used for each combination. Once we’ve brute forced all the combinations, look through the dictionary to count up all the tile values that have 10 solutions or less!

let history = {};
let limit = 1000000;

for (let L=3; L<250000; L++){
    let L2 = L * L;
    // Hole (H) size must be at least 1, smaller than or equal to l-2
    for (let H=0; H<=L-2; H++){
        // Calculate number of tiles (t) in lamina
        let t = L2 - (H * H);

        // L and H evenness must match
        if (L%2 != H%2) continue;
        
        // Ignore answers over limit
        if (t > limit) continue;

        // Store result in history
        if (history[t] == undefined)
            history[t] = [];

        history[t].push(L);
    }
}

let count = 0
for (let i in history)
    if (history[i].length <= 10)
        count++;

console.log(count);

The first time I ran this, it was much slower than anticipated. I let it run overnight, and was happy to see the correct result when I checked on it the next day. I wondered what I needed to optimize to make it run better. It turns out that I had typed the upper limit, L, as 2,500,000! By decreasing that to the proper value, that reduced the compute time to 17 seconds.

I was able to find one more low-hanging optimization opportunity. By iterating through the H values from largest to smallest, the number of tiles, t, would increase. Once that value got over the 1,000,000 limit, I can break out of the loop, and go to the next value of L. This saves a lot of time computing out of bounds answers!

for (let L=3; L<250000; L++){
    let L2 = L * L;
    // Hole (H) size must be at least 1, smaller than or equal to l-2
    for (let H=L-2; H>1; H-=2){
        // Calculate number of tiles (t) in lamina
        let t = L2 - (H * H);

        // Break once we get over limit
        if (t > limit) break;

        // Store result in history
        if (history[t] == undefined)
            history[t] = [];

        history[t].push(L); 
    } 
}

This optimization pushed the compute time down to .278s! That’s a very reasonable time for Project Euler.

It’s been a while since I’ve taken the time to solve some Project Euler problems. I thought today would be a good day to pick it up and try solving one I had not been able to solve the last time I tried.

I’m going to speak generically about the problem because I don’t want to spoil it for anyone who may be trying to solve it. In brief, I had written a brute-force method for solving the problem, but as Project Euler so often does, they made the scale so huge that brute-force methods often fail. I turned to the internet to try to find a clue to point me in the right direction.

Sadly, one of the links I clicked didn’t have a clue, but just said “This is a Fibonacci sequence problem.”

I couldn’t unsee it, and it was clearly the way the problem was intended to be solved at scale. It felt like it was ruined, and I regretted trying to find assistance.

I decided to go do another problem without getting any aids or accidentally cheating. The problem I chose was another problem that scales quickly. The problem is stated as follows:

A row measuring seven units in length has red blocks with a minimum length of three units placed on it, such that any two red blocks (which are allowed to be different lengths) are separated by at least one grey square. There are exactly seventeen ways of doing this.

How many ways can a row measuring fifty units in length be filled?

I decided to start this problem with a brute force approach, even though I knew it wouldn’t scale. I wrote a recursive function to create rows filled with tiles. I started with the length of 7 they provided in the example, and I too got the correct 17.

function recurse(array){
    // Find the length of the array
    let s = tile_sum(array); 

    // If the array it too long, abort
    if (s > limit) 
        return;

    // If the array is correct size,
    // check for uniqueness and record, abort
    if (s == limit){ 
        record(array);
        return;
    }

    // Add tiles sized 1-limit and recurse
    for (let t=1; t<=limit; t++){
        if (t == 2) continue; // Tiles length 2 don't exist

        // Add a single tile spacer in if we are adding a red tile
        if (array[array.length-1] != 1 && t > 1)
            array.push(1);

        // Add tile to the array and recurse
        recurse(array.slice().concat([t]));
    }
}

I slowly increased the length by one to make sure nothing would break in my algorithm. I also noted the numbers as it computed them.

f(7) = 17
f(8) = 27
f(9) = 42
f(10) = 72

I thought that having learned my lesson from the previous problem I might have some luck by identifying this series of numbers. I went to the OEIS (Online Encyclopedia of Integer Sequences) and entered in these integers.

Spoilers below…

The OEIS search provided a result! It was A004695, the Fibonacci sequence… (but halved). Sure enough, computing the correct number in the Fibonacci sequence, and halving it yielded the correct answer for this problem.

I can’t believe two problems in a row ended up having the same shortcut. I wonder if the next one will as well…