How elusive was 42 as the sum of three cubes?

The big mathematical news September 9th 2019 was the final resolution of the decade old Diophantine equation.

As is often the case for super tough questions, the question itself seems completely trivial.
Can all the numbers from 1 to 100 be written as the sum of the cubes of three integers,
i^3 + j^3 + k^3?

Of course the key part of the question is integers, otherwise 0^3 + 0^3 + (cubic root of x)^3 will return x for any x we want!

The other subtlety is that integers can be negative, so each cube can be either positive or negative. So 92 can be written as 9^3 + (-8)^3 + (-5)^3.



This puzzle has been around for at least decades in this current form, but similar equations with integer solutions have been around at least since the third century!

Four our problem, solutions for almost all numbers 1 through 100 had been found, but two remained elusive 33 and 42.

And then 33 was found earlier this year:
   (8866128975287528)^3
+ (−8778405442862239)^3
+ (−2736111468807040)^3 = 33

Which left 42, but advances on the theory and on the computing side have cracked the last remaining elusive number:
   (80435758145817515) ^ 3
+ (-80538738812075974) ^ 3
+  (12602123297335631) ^ 3 = 42

Now I will describe man algorithm to generate all numbers 1 through 1000....
Just kidding.

I was actually curious of the elusiveness of these numbers 33 and 42. Mathematicians had to go really far to get the right numbers. Is that because numbers just weren't very likely to fall in the 1 to 100 range (very likely given how quickly cubes get!) or are certain numbers much more likely to get all the attention. in other words, as all combinations of three integers were tested, were all numbers 1 through 100 obtained in roughly equal proportions or did frequency concentrate around certain ranges?

I wrote a small script which explored this, simply generating all possible combinations of three integers (up to -1000 to 1000 in range, didn't feel like going all the way up to 80538738812075974...) and look at the distribution of i^3 + j^3 + k^3 in the 1 to 100 range.

With i,j,k in -10 to 10

With integers just in that small range, already 61 out of the 100 numbers are solved!
The distribution is far from being randomly uniform though! Two things jump out:

• the spikes: These are perfectly normal! If a number is a perfect cube (i.e. 27), then there is an infinite number of solutions setting the two other values to opposite values: 27 = 3^3 = 3^3 + 0^3 + 0^3 = 3^3 + (1761)^3 + (-1761)^3... And you can add in all the permutations!
• the clusters: it seems the solved numbers go in groups. Again, quite reasonable if we think about it. If a number can be generated as the sum of two cubes, then we can easily generate the one immediately before and after by adding either 1^3 or (-1)^3. We have an extra degree of freedom around the perfect cubes. Being able to generate consecutive solved numbers creates the clusters. There is a further discussion at the end of the post on how unexpected the number of clusters is.

Based on the spike comment, I will truncate at frequencies of 10 to focus on coverage:

With i,j,k in -100 to 100

Expanding the range ten-fold leads to 68 out of the 100 being solved:

With i,j,k in -1000 to 1000

Increasing the range ten-fold once again leads to a single additional number being solved: 51

We now have a better sense of the elusiveness of the solutions. If they aren't found with the range of the smaller integers, then it will get increasingly difficult to identify later, no wonder this has been such an open problem for so many years!

Just for fun, I looked at a new problem switching the power from 3 to 5, only 13 (compared to 69 for cubes) numbers were solved in the -1000 to 1000 range:

Again the clustering is pretty remarkable! While 42 is still unsolved, I'm glad to report a solution for 33: 0^5 + 1^5 + 2^5 !

And to close out this post, I wanted to return to the topic of the clusters. We saw that when exploring the -1000 to 1000 ranges for the integers we were able to cover 69 out of the 100 target numbers. But it didn't appear as though the 69 numbers were randomly distributed in 1-100 but grouped into clusters, more specifically into 13 clusters. We had a high-level explanation for this as consecutive numbers can easily be generated via k^3 with k = -1, 0 and 1. But the question remains, how non-random is this grouping into 13 clusters?

One approach is to sample without replacement 69 out of the 100 first integers, and see how many clusters are obtained. In order to count clusters, I used the rle() function which stands for run length encoding and basically describes a sequence. E.g. for 1,2,2,2,1,0,0 it will return that there is one 1, three 2s, one 1 and two 0s. From there it is easy to get the number of clusters.

I generated 10,000 samples of "solved" numbers between 1-100 and in each case computed the number of clusters obtained. Here is the associated histogram to which I've added the 13 we obtained in our case as a red vertical line:

While we'd typically expect to get about 20-25 clusters, only in extreme cases do we get a 15 or even a 14. Our 13 was not obtained a single time in all ten thousand simulations. This should settle the question of whether the low number of clusters comes as a surprise or not!

Solving Mathematical Challenges with R: #3 Sum of all grids

Welcome to our third challenge!

You can find more details on where these challenges come from and how/why we will be approaching them in the original posting.

Challenge #3

The original link to the challenge (in French) can be found here. The idea is to consider an empty three-by-three grid, and then place two 1s in any two of the nine cells:

The grid is the completed as follows:

• pick any empty cell
• fill it as the sum of all its neighbors (diagonals as well)

I've here detailed an example, the new cell is indicated at each step in red:

The challenge is now to find a way to complete the grid in such a manner as to obtain the greatest value in a cell. In the previous example, the maximum obtained was 30, can we beat it?

Power grid

I strongly encourage you to try out a few grids before continuing to read. Any guesses as to what the greatest obtainable value is?

As you try it out, you might realize that the middle cell is rather crucial, as it can contribute its value to every other cell, thus offering the following dilemma:

• fill it too soon and it will be filled with a small value
• fill it too late and it will not be able to distribute its value to a sufficient number of other cells

Gotta try them all!
In the spirit of our previous posts, let's try all possibilities and perform an exhaustive search.
The first thing we need to ask ourselves is the number of possible ways of filling up the square, and then generating all those combinations in R.
Ignoring how the cells are filled up, there are 9 possibilities to fill up the first cell (with a 1), 8 possibilities for the second (also a 1), 7 for the third (with the rule)... and 1 for the ninth final cell. Which amounts to 9*8*7*...*1 = 9! = 362880, piece of cake for R to handle.

Generating permutations
So we want to generate all these possibilities which are actually permutations of numbers 1 through 9, where each number corresponds to a cell. {7,1,5,2,4,9,3,8,6} would mean filling up cells 7 and 1 first with a 1, then filling up cell #5, then #2 and do on.
A naive method would be to use expand.grid() once again for all digits one through 9, which would also give unacceptable orders such as {7,1,5,2,1,3,6,8,9} because cell #1 is filled twice at different times! We could easily filter those out by checking if we have 9 unique values. But this would amount to generating all 9*9*9..*9=9^9>=387 million combinations and narrow down to just the 362K of interest. That seems like a lot of wasted resources and time!

Instead, I approached the problem from a recursive perspective (R does have a package combinat that does this type of work, but here and in general in all the posts I will try to only use basic R):

• If I only have one value a, there is only a single permutation:{a}
• If I only have two values a and b, I fix one of the values and look at all permutations I have of the left-over value, then append the value that was fixed:
  - fix a, all permutations of b are {b}, then append the a: {b,a}
  - fix b, all permutations of a are {a}, then append the b: {a,b}
  - the result is {a,b} and {b,a}
• with three values, repeat the process now that we know all permutations of 2 values:
  - fix a, all permutations of b,c are {b,c} and {c,b}, then append the a: {b,c,a}, {c,b,a}
  - fix b, all permutations of a,c are {a,c} and {c,a}, then append the b: {a,c,b}, {c,a,b}
  - fix c, all permutations of a,b are {a,b} and {b,a}, then append the c: {a,b,c}, {b,a,c}

In R:
SmartPermutation <- function(v) {
  if (length(v) == 1) {
    return(as.data.table(v))
  } else {
    out <- rbindlist(lapply(v, function(e) {
      tmp <- SmartPermutation(setdiff(v, e))
      tmp[[ncol(tmp) + 1]] <- e
      return(tmp)
    }))
    return(out)
  }
}

Computing the score

The next and final step is to fill out the grid for each possible permutation.

This is done by defining a static mapping between cell ID and its neighbors:

We then go through each permutation, fill out a virtual grid (in our case it's not a 3-by-3 but a vector, which doesn't change anything as long as the relationships between neighbors is maintained) and track maximum value.

Neighbors <- function(i) {
  neighbor.list <-
    list(c(2, 4, 5), c(1, 3, 4, 5, 6), c(2, 5, 6),
         c(1, 2, 5, 7, 8), c(1, 2, 3, 4, 6, 7, 8, 9), c(2, 3, 5, 8, 9),
         c(4, 5, 8), c(4, 5, 6, 7, 9), c(5, 6, 8))
  return(neighbor.list[[i]])
}

CompleteGrid <- function(row) {
  grid <- rep(0, 9)
  grid[row[1 : 2]] <- 1
  for (i in 3 : 9) {
    grid[row[i]] <- sum(grid[Neighbors(row[i])])
  }
  return(max(grid))
}

Final answer
Based on our exhaustive search, we were able to identify the following solution yielding a max value of...57!

Little bonus #1: Distribution
The natural question is around the distribution of values? Were you also stuck around a max value in the forties? 57 seems really far off, but the distribution graph indicates that reaching values in the forties is already pretty good!

Little bonus #2: Middle cell
Do you remember our original trade-off dilemma? How early should the middle cell be filled out? In the optimal solution we see it was filled out exactly halfway through the grid.
To explore further, I also kept track for each permutation when the middle cell was filled out and what its filled value was. The optimal solution is shown in green, the middle cell being filled out in 5th position with a seven.

And a similar plot of overall max value against position in which the middle cell was filled:

It clearly illustrates the dilemma of filling neither too late nor too early: some very high scores can be obtained when the middle cell is filled as 4th or 6th cell, but absolute max of 57 will only be achieved when it is filled in 5th. Filling it in the first or last three positions will only guarantee a max of 40.

Solving Mathematical Challenges with R: #2 Slicing Cubes

Welcome to the second challenge!
You can find more details on where these challenges come from and how/why we will be approaching them in the original posting.

Challenge #2

The original link to the challenge (in French) can be found here. It's easy to understand but non-trivial to wrap one's head around...
Consider a nice perfect cube. So far so good. Consider slicing this cube with a Samurai sword in one swift straight cut. Now take a look at the 2D shape created by the cut on the cube. What types of shape can it be? We are primarily interested in whether or not it can be a regular polygon (equilateral triangle, square, regular pentagon, regular hexagon...).

Starting simple: Square

Let's look at some of the simpler shapes. Ironically, the simplest solution is not the equilateral triangle but the square. Can the slice lead to a square? After a few virtual cuts in your mind, you will have noticed that if the cut is parallel to any side, the shape of the slice will be a square!

Equilateral triangle

A little trickier, but a few additional virtual slicing and dicing should reveal the answer. Select a corner of the cube (technically, a vertex). Then move away from that corner along the three edges that make up the corner by an equal amount. The dimensional plane going through those three points (there is one and only one such plane: in 2D there is one and only one line going through two distinct points, in 3D there is one and only one plane going through three distinct points) will lead to an equilateral triangle:

The proof is straightforward: because we moved away from the corner by a constant value in all three directions, the three sides of the triangle are all equal to each other (square root of twice the square of the constant value, thanks Pythagoras!).

3-D Plotting in R

Before continuing, one question you might have is how the previous visuals were generated. Let's remind ourselves these posts are not as much about solving the challenge but also about expanding our R skills!
There are multiple functions and packages, but I was aiming for simplicity. I used the lattice package which has the benefit of containing lots of other functions so probably a good idea to have it installed once and for all, not only for the purpose of this challenge. Within that package, wireframe() is the function of interest.
The simplest way to use it is to provide a matrix where the value of the i-th row and j-th column is the altitude at the i-th index on the x-axis and j-th index on the y-axis. This assumes that all points on the x-axis and y-axis are equidistant. Alternatively, you can create a data.frame (say df) with columns for the x, y and z values. Then simply call wireframe(z ~ x + y, data = df) and you're good! There are multiple parameters for the shading, viewpoint angle and so forth...
Let's just run the basic example from the help file using the pre-loaded volcano dataset which is a 87 * 61 matrix:
wireframe(volcano, shade = TRUE, aspect = c(61/87, 0.4), light.source = c(10, 0, 10))

The z-axis label was slightly truncated, but aside from that it's a pretty nice plot for one-line of code!

Back to our cube
We solved the 3 and 4 sided regular polygons, but what about, 5, 6, 7...
We could continue playing around, but visualizing these slices is pretty tricky you'll have to admit!
So how can R help? Well how about generating a whole bunch of planes and looking at the type of slices we obtain?

To do that, a tiny little bit of algebra first. The general equation of a 2D plane in 3D is:
z = a + b * x + c * y
[well the purists might disagree and they would be correct, how about the vertical plane satisfying x = 0? It's not covered by the previous equation]
However the planes not covered are all the x = constants and y = constants which either don't slice the cube, or slice parallel to certain sides of the cube which we showed leads to squares!

So now, all we need to do is generate a whole bunch of a, b and c values, and see where the resulting plane intersects the cube.

Step 1: Edge intercepts
Again a little algebra, but there are 12 edges the plane could potentially intercept:

• the four vertical edges, parallel to the z axis, corresponding to {x=0 and y=0}, {x=0 and y=1}, {x=1 and y =0} and {x=1 and y=1}
• the four edges parallel to the x-axis
• the four edges parallel to the y-axis

In each case we can fit two of three values of x, y and z into the equation of the plane and get the third value.
Example: for x=0 and y=1, the equation becomes z = a + c. So the plane intercepts that edge at {x=0, y=1,z=a+c}.

So we end up with 12 points.

Step 2: Valid edge intercepts
For an edge to be valid, the third computed coordinate should be between 0 and 1. In our previous example, {x=0, y=1,z=a+c} is part of the cube if and only if 0 <= a + c <= 1.
We then filter to only keep the valid intercepts.

Step 3: Resulting figure?
Once we know how the cube was intercepted, we need to determine if we luckily stumbled across a special polygon!
To do that, I did the following:

• compute all distances between intercepts (only once so as not to compute everything twice, needless to have both A --> B and B --> A)
• keep the smallest value(s)
• count the number of occurrences of that smallest value
• if the number of occurrences is equal to the number of intercepts, we have ourselves a winner!

Illustration

Consider the square ABCD of side 1. I would compute all 6 possible distances:

• A --> B: 1
• A --> C: sqrt(2)
• A --> D: 1
• B --> C: 1
• B --> D: sqrt(2)
• C --> D: 1

I keep the minimal values:

• A --> B: 1
• A --> D: 1
• B --> C: 1
• C --> D: 1

I have four matches of the minimal value, same as number of points, I have a square!

The reason this works is that a regular polygon is convex, so the sides are the smallest distances between points (convex polygons on the left, concave on the right):

Running through a whole bunch of values for a, b and c (remember expand.grid() from the first challenge?) I got the following results for number of identified regular polygons based on number of intercepts:

#Intercepts   Not Regular   Regular
          0       2965368         0
          1         39049         0
          2           564         0
          3       1680834     54885
          4       1761091      3884
          5       1084221         0
          6        530505       196
          7             4         0


So a whole bunch of equilateral triangles which makes sense given the technique we developed to generate them ourselves earlier.
Quite a few squares, including some non-trivial ones (non-parrelel to a side):

We did find some regular hexagons:

However, not a single pentagon nor heptagon in sight.
Doesn't mean they don't exist, but again, we used R here to give us some leads, not fully resolve geometrical proofs!

Hope you enjoyed!

Solving Mathematical Challenges with R: #1 Palindromes

This post is the first challenge from what will hopefully be a long series!

You can find more details on where these challenges come from and how/why we will be approaching them in the original posting.

Challenge #1

Let's jump right in with the first challenge! You can find the original presentation of the challenge (in French!) here.

It goes as follows: a number is said to be a palindrome if the digits read left to right are the same as if read right to left. An example would be 151, or 70355307. The first question is simply to determine the number of palindrome numbers that have 351 digits. As a bonus question: what is the minimal difference between two palindrome numbers with 351 digits?

Naturally the first step is to get a better understanding of what we are dealing with, independently of how we want to tackle the problem.

Definitely seems like a small pattern is emerging! If we were to continue with the mathematical approach we would look at N = 4, possibly N = 5 to better understand the dynamics, and start separating the case of N even with N odd. Let's take the R route for now.

How can we generate/identify all palindromes with N digits?

One first approach is to generate all numbers of N digits, flip them, and keep those where the flip left the number unchanged:

PalindromesSlow <- function(n) {
  numbers <- seq(10 ^ (n-1), 10 ^ n-1)
  flipped <- as.numeric(sapply(
    lapply(strsplit(as.character(numbers), split = ""), rev),
    paste0, collapse = ""))
  palindromes <- numbers[numbers == flipped]
  return(palindromes)
}

Why the “Slow” in the name? While this function performs exactly as expected, it won’t scale well if we start looking at numbers 7-digit long and larger. The main reason is that we are generating every single potential number and only selecting an ever-decreasing fraction of those meeting our palindrome criteria. Why ever-decreasing? Let’s look at the trend for our first values of N:

• N = 1: 9 palindromes, 9 total numbers: 100%
• N = 2: 9 palindromes, 90 total numbers: 10%
• N = 3: 90 palindromes, 900 total numbers: 10%
• N = 4: 90 palindromes, 9000 total numbers: 1%

Because palindromes are determined by the first half of their digits, we shouldn’t try generating all possible numbers with N digits, but only all possible first halves, then mirror image them to generate the second half. For example, if we want to generate all 6-digit palindromes, let us first generate all combinations of 3-digits XYZ, then flip and switch to get XYZZYX. For odd number of digits, it’s only slightly trickier as the middle element acts as the mirror and has no constraints. For 7-digits we would simply generate the first 4 digits, but flip the first 3:
XYZT --> XYZTZYX.

length.firsthalf <- ceiling(n / 2)
digit.list <- rep(list(0 :9), length.firsthalf)
# remove 0 as option for first digit
digit.list[[1]] <- 1 : 9
# expand grid
eg <- expand.grid(digit.list)

A quick word on the expand.grid() function which has saved me on multiple occasions. It essentially generates all possible cross-combinations of arguments provided. Easier to see through a simple example:

expand.grid(c(1, 2, 3), c("a", "b"))

Var1 Var2
   1    a
   2    a
   3    a
   1    b
   2    b
   3    b

For our PalindromeQuick function it is incredibly valuable as we want something generic enough to adjust to any n. So we create a list of length n with all possible options at each position, and stick that into expand.grid and voila! No need for complicated nested for loops, especially as we wouldn't know how many for loops to nest in the first place!

We can then wrap that in a function to determine how many palindromes were found as well as minimal distance between consecutive palindromes. The only trick is carefully handling n being even or odd as discussed previously.

PalindromesQuick <- function(n) {
  length.firsthalf <- ceiling(n / 2)
  digit.list <- rep(list(0 :9), length.firsthalf)
  # remove 0 as option for first digit
  digit.list[[1]] <- 1 : 9
  # expand grid
  eg <- expand.grid(digit.list)
  if (n %% 2 == 0) {
    flipped.ending <- apply(eg, 1, function(row) {
        paste0(rev(row), collapse = "")
      })
  } else {
    flipped.ending <- apply(eg, 1, function(row) {
        paste0(rev(row[- length(row)]), collapse = "")
      })
  }
  eg$ending <- flipped.ending
  palindromes <- sort(as.numeric(
                   apply(eg, 1, paste0, collapse = "")))
  return(palindromes)
}

Having written a slow and quick function, we should pause to think about how much faster quick is, and talk about timing our R functions.

Timing in R

When I started using R, my trick for timing functions was simply to record the absolute time before and after execution and take the difference:

t1 <- Sys.time()
output.slow.7 <- PalindromesSlow(7)
t2 <- Sys.time()
t2 - t1
1.174394 mins

I later discovered it could all be done in one go with the built-in system.time() function:
system.time(output.slow.7 <- PalindromesSlow(7))
   user  system elapsed 
 68.348   0.544  69.016 

They each have small caveats one should be aware of in how they track elapsed time, but for most timing purposes return exactly what you care about. system.time() has the advantage of being a one-liner, but the former solution is easier to use if you have multiple lines you want to time and also makes it easier to comment in/out of code.

From now on we'll solely focus on the Quick version, a 600X improvement is definitely something we'll take!

A quick comment on efficiency and speed: it's always important to weigh the pros and cons. While a function that runs faster always seems preferable on paper, one must also consider the necessity of speeding in the first place, as well as the additional complexity of speeding things up. Here the quick version was just a few lines longer and simple to understand and the increase in speed will allow us to explore numbers that are 14 digits long as quickly as the slow version took to explore 7-digit numbers. However, this is not always the case, we'll see this in challenge #3 where trying to be efficient is not going to be straightforward to implement and scale, and the gain might still be insufficient to explore the higher dimensions.

Results

Using our PalindromeQuick function, let's look at the number of palindromes and minimal distance between consecutive palindromes:

N	Number of Palindromes	Minimum Difference
1	9	1
2	9	11
3	90	10
4	90	11
5	900	11
6	900	11
7	9000	11
8	9000	11
9	90000	11
10	90000	11
11	900000	11
12	900000	11
13	9000000	11
14	9000000	11

Although we haven’t provided a formal mathematical proof, I think we at  least have a good idea of the challenge's answers.

If n is even, we have as many palindromes as there are combinations of n / 2 digits with the first one not being a 0, that’s 9 * 10 ^ (n/2 - 1).
If n is odd, we have as many palindromes as there are combinations of (n + 1) / 2 digits with the first one not being a 0, that’s 9 * 10 ((n + 1) / 2 - 1).

Minimal difference is somewhat trickier...

Based on the simulations, minimal value fluctuates a bit at first, but definitely seems to quickly stabilize at 11 for numbers with 4 or more digits.

It’s an interesting problem: if we want to make a small change to an existing palindrome to create the next small palindrome, we would typically want to make small changes to the right hand side of the number (units, tens...). But because of the palindromic nature of our numbers, small changes to the right will lead to the same change on the far left and thus a huge delta between our two numbers. The obvious option is to aim for the middle, but that’s still an absolute change of the order of 10 ^ (n / 2).

If we look at examples with small n, we see we can do better when we have nine(s) in the middle: 393 and 404, 5995 and 6004. This is basically our initial trick of incrementing the “middle” but after 9 we can 0 and need to increment the digits to the immediate right and left: 37973 increments to 38983 (delta of 1010), but 39993 increments to 40004 (delta of 11).
So we've reached the point where we've proven the delta is no bigger than 11 as X99...99X (with X any digit between 1 and 8) and (X+1)00...00(X+1) will have a delta of 11.There will be 8 such instances, no matter the number of digit of the palindrome. Proving that 11 is actually the best we can do is a much tougher question.

Back to the challenge, we can quite confidently answer the first challenge that there are 9e175 351-digit palindromes, and that minimal difference will be 11.

Before we sign off, let’s finish with a small plot of the distribution of differences between consecutive palindromes. Given our previous comments and at a very high-level, we should expect a big spike around 10 ^ (n/2), then a spike about one-tenth of that for 10 ^ (n / 2 - 1) whenever the middle number is 9, then a spike one-hundredth smaller at 10 ^ (n / 2 - 2) whenever the middle number is 999 and so on. And as noted previously,  8 cases with a delta of 11.
Turns out we weren’t too far off !

For N = 13:

For N = 14:

For the graphs, despite the numerical nature of the x-axis, we naturally went for barplots instead of histograms given the very peculiar discrete distribution!

Solving Mathematical Challenges with R

In a new series of posts, I will explore how the free statistical tool R can be used to answer mathematical challenges.

Le Monde is a French daily newspaper with one of the highest circulation numbers. There used to be a small section at the very end which varied by day of the week containing a small intellectual puzzle, either a crossword, a Sudoku, a math challenge...
I haven't lived in France for quite some time, but checking online recently it appears they have slightly changed the format, adding mathematical challenges which are presented by Cedric Villani, French Mathematician and recipient of the 2010 Fields Medal (shameless fun fact, his PhD Advisor Pierre-Louis Lions and also recipient of the Fields Medal in 1994 was my math teacher during my Master's).

While these weekly challenges are mathematical in nature and can be (should be!) resolved with pen and paper, I will take a computer-oriented approach, namely with the R programming language for statistical computing. The idea is not necessarily to fully solve the challenge (computationally or not), but indirectly use these challenges as R tutorials, dig deeper into the mathematical challenges along the way, and with a focus on visualisations. Because of the focus on R, we'll also take slight detours to look into R performance, best practices, try different packages and approaches.

Think of this as a road trip, we kind of know where we want to go, but have no idea if we'll reach our destination, and much less certain of the path we'll take, but whatever the path we'll see some really cool stuff along the way!

Challenge #1: Palindromes

Challenge #2: Slicing Cubes

Challenge #3: Sum of all grids

Challenge #4: Match-ing game (coming up soon!)

Challenge #5: Winning ropes (coming up soon!)