# Balancing the centrifuge – a Riddler puzzle

From the riddler:

Quoc’s lab has a microcentrifuge, a piece of equipment that can separate components of a liquid by spinning around very rapidly. Liquid samples are pipetted into small tubes, which are then placed in one of the microcentrifuge’s 12 slots evenly spaced in a circle.

For the microcentrifuge to work properly, each tube must hold the same amount of liquid. Also, importantly, the center of mass of the samples must be at the very center of the circle — otherwise, the microcentrifuge will not be balanced and may break.

Quoc notices that there is no way to place exactly one tube in the microcentrifuge so that it will be balanced, but he can place two tubes (e.g., in slots 1 and 7).

Now Quoc needs to spin exactly seven samples. In which slots (numbered 1 through 12, as in the diagram above) should he place them so that the centrifuge will be balanced? Extra credit: Assuming the 12 slots are distinct, how many different balanced arrangements of seven samples are there?

https://fivethirtyeight.com/features/can-you-break-a-very-expensive-centrifuge/

I’ve seen this one before, but I’ll take it on as a programming challenge.

I’ll say that hole k is at coordinates $(\sin 2\pi k/12, \cos 2\pi k/12)$. This is nonstandard, but the result looks like a clock, so it’s easier to visualize:

Now, there are ${12 \choose 7} = 792$ ways that we could pick 7 out of the 12 holes. To generate a list of these I can use the following R function:

combinations = function(n, k){
if (k == 0 & n == 0){return(list())} else {
if (k == 0){return(list(c()))} else {
if (n == k){return(list(1:n))} else {
without_n = combinations(n-1, k)
with_n = lapply(combinations(n-1, k-1), function(x){c(x, n)})
return(c(without_n, with_n))
}
}
}
}

To generate the subsets of [n] of size k, I just take:

• the subsets that don’t contain n (which are therefore subsets of [n-1] of size k), and
• the subsets that do contain n (which are therefore subsets of [n-1] of size k-1, with n adjoined)
(I learned this a good two decades ago, from section 3.6 of Herb Wilf’s notes East Side, West Side.)

Then iterate over the subsets to find their centers of mass:

our_comb = combinations(12, 7)
com_x = rep(NA, length(our_comb))
com_y = rep(NA, length(our_comb))
for (i in 1:length(our_comb)){
com_x[i] = mean(x[our_comb[[i]]])
com_y[i] = mean(y[our_comb[[i]]])
}


So our_comb runs through the combinations, and com_x and com_y are the x- and y-coordinates of the center of mass. Plotting the centers of mass gives we get an attractive symmetric pattern. Here I’ve plotted transparent points, so the darker points are points that arise from more distinct combinations. The outermost points represent the most imbalanced centrifuges – for example the one closest to the top represents loading slots 9, 10, 11, 12, 1, 2, and 3. The main question is: how many points are on top of each other at the very center.

I can extract those subsets from the matrix our_comb and build them into a matrix for looking at. There’s a small problem in that we used floating-point arithmetic, so the coordinates don’t come out exactly at zero, but it’s enough to ake the ones that are “close enough”:

epsilon = 10^(-6)
balanced_indices = which(abs(com_x) < epsilon & abs(com_y) < epsilon)
balanced_combs =  t(matrix(unlist(our_comb[balanced_indices]), nrow = k))

and the matrix balanced_combs, which has one row for each combination that balances the centrifuge, is

      [,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,]    1    2    3    6    7    9   10
[2,]    1    2    4    5    8    9   10
[3,]    1    2    5    6    7   10   11
[4,]    2    3    4    7    8   10   11
[5,]    2    3    5    6    9   10   11
[6,]    1    2    5    6    8    9   12
[7,]    1    3    4    7    8    9   12
[8,]    1    4    5    6    9   10   12
[9,]    1    4    5    7    8   11   12
[10,]    2    3    6    7    8   11   12
[11,]    3    4    5    8    9   11   12
[12,]    3    4    6    7   10   11   12

(The order here is lexicographic order, but you have to read the rows backwards.)

If you can make sense of this pile of numbers without making some pictures, good for you! But I am human, though a mathematician, so let’s plot. Here big black dots represent the loaded spots, and small blue dots represent the non-loaded spots.

(And yes, it’s in base R graphics. It’s been a while since I’ve used those – at my day job it’s all ggplot all the time.)


par(mfrow = c(4, 3), mar = c(1, 1, 1, 1))
for (i in 1:nrow(balanced_combs)){
full = balanced_combs[i,]
empty = setdiff(1:n, balanced_combs[i,])
plot(x[full], y[full], pch = 19,cex = 2,
asp = 1, axes = FALSE, xlab = '', ylab = '',
xlim = c(-1.25, 1.25), ylim = c(-1.25, 1.25))
points(x[empty], y[empty], pch = 19, cex = 1, col = 'lightblue')
}

So there are 12 ways to load the centrifuge… but we can clearly see that all of them are really just rotations of a single pattern. To me the visually most convenient representative of that pattern is the third one in the third row, which has loadings at 12, 1, 4, 5, 7, 8, and 11.

We can play a little game of connect-the-dots to see why this pattern is balanced. Connect 12, 4, and 8 to form an equilateral triangle. Connect 1 and 7; connect 5 and 11. Like this:

Each of those three subsets has its center of mass at the center of the circle, so the whole arrangement does too.

That decomposition turns out to be the key to the general problem of which centrifuge sizes can be loaded with which number of tubes and remain balanced. Stay tuned.