# The Second Arrangement

To validate our analyses, I’ve been using randomisation to show that the results we see would not arise due to chance. For example, the location of pixels in an image can be randomised and the analysis rerun to see if – for example – there is still colocalisation. A recent task meant randomising live cell movies in the time dimension, where two channels were being correlated with one another. In exploring how to do this automatically, I learned a few new things about permutations.

Here is the problem: If we have two channels (fluorophores), we can test for colocalisation or cross-correlation and get a result. Now, how likely is it that this was due to chance? So we want to re-arrange the frames of one channel relative to the other such that frame i of channel 1 is never paired with frame i of channel 2. This is because we want all pairs to be different to the original pairing. It was straightforward to program this, but I became interested in the maths behind it.

The maths: Rearranging n objects is known as permutation, but the problem described above is known as Derangement. The number of permutations of n frames is n!, but we need to exclude cases where the ith member stays in the ith position. It turns out that to do this, you need to use the principle of inclusion and exclusion. If you are interested, the solution boils down to

$$n!\sum_{k=0}^{n}\frac{(-1)^k}{k!}$$

Which basically means: for n frames, there are n! number of permutations, but you need to subtract and add diminishing numbers of different permutations to get to the result. Full description is given in the wikipedia link. Details of inclusion and exclusion are here.

I had got as far as figuring out that the ratio of permutations to derangements converges to e. However,  you can tell that I am not a mathematician as I used brute force calculation to get there rather than write out the solution. Anyway, what this means in a computing sense, is that if you do one permutation, you might get a unique combination, with two you’re very likely to get it, and by three you’ll certainly have it.

Back to the problem at hand. It occurred to me that not only do we not want frame i of channel 1 paired with frame i of channel 2 but actually it would be preferable to exclude frames i ± 2, let’s say. Because if two vesicles are in the same location at frame i they may also be colocalised at frame i-1 for example. This is more complex to write down because for frames 1 and 2 and frames n and n-1, there are fewer possibilities for exclusion than for all other frames. For all other frames there are n-5 legal positions. This obviously sets a lower limit for the number of frames capable of being permuted.

The answer to this problem is solved by rook polynomials. You can think of the original positions of frames as columns on a n x n chess board. The rows are the frames that need rearranging, excluded positions are coloured in. Now the permutations can be thought of as Rooks in a chess game (they can move horizontally or vertically but not diagonally). We need to work out how many arrangements of Rooks are possible such that there is one rook per row and such that no Rook can take another. If we have an 7 frame movie, we have a 7 x 7 board looking like this (left). The “illegal” squares are coloured in. Frame 1 must go in position D,E,F or G, but then frame 2 can only go in E, F or G. If a rook is at E1, then we cannot have a rook at E2. And so on.

To calculate the derangements:

$$1 + 29 x + 310 x^2 + 1544 x^3 + 3732 x^4 + 4136 x^5 + 1756 x^6 + 172 x^7$$

This is a polynomial expansion of this expression:

$$R_{m,n}(x) = n!x^nL_n^{m-n}(-x^{-1})$$

where $$L_n^\alpha(x)$$ is an associated Laguerre polynomial. The solution in this case is 8 possibilities. From 7! = 5040 permutations. Of course our movies have many more frames and so the randomisation is not so limited. In this example, frame 4 can only either go in position A or G.

Why is this important? The way that the randomisation is done is: the frames get randomised and then checked to see if any “illegal” positions have been detected. If so, do it again. When no illegal positions are detected, shuffle the movie accordingly. In the first case, the computation time per frame is constant, whereas in the second case it could take much longer (because there will be more rejections). In the case of 7 frames, with the restriction of no frames at i ±2, then the failure rate is 5032/5040 = 99.8%. Depending on how the code is written, this can cause some (potentially lengthy) wait time. Luckily, the failure rate comes down with more frames.

What about it practice? The numbers involved in directly calculating the permutations and exclusions quickly becomes too big using non-optimised code on a simple desktop setup (a 12 x 12 board exceeds 20 GB). The numbers and rates don’t mean much, what I wanted to know was whether this slows down my code in a real test. To look at this I ran 100 repetitions of permutations of movies with 10-1000 frames. Whereas with the simple derangement problem permutations needed to be run once or twice, with greater restrictions, this means eight or nine times before a “correct” solution is found. The code can be written in a way that means that this calculation is done on a placeholder wave rather than the real data and then applied to the data afterwards. This reduces computation time. For movies of around 300 frames, the total run time of my code (which does quite a few things besides this) is around 3 minutes, and I can live with that.

So, applying this more stringent exclusion will work for long movies and the wait times are not too bad. I learned something about combinatorics along the way. Thanks for reading!

Further notes

The first derangement issue I mentioned is also referred to as the hat-check problem. Which refers to people (numbered 1,2,3 … n) with corresponding hats (labelled 1,2,3 … n). How many ways can they be given the hats at random such that they do not get their own hat?

Adding i+1 as an illegal position is known as problème des ménages. This is a problem of how to seat married couples so that they sit in a man-woman arrangement without being seated next to their partner. Perhaps i ±2 should be known as the vesicle problem?

The post title comes from “The Second Arrangement” by Steely Dan. An unreleased track recorded for the Gaucho sessions.

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