Job Advert for a Post-Doctoral Research Associate: Mapping cortex evolution through mathematical modelling

Closing Date: Thursday, September 1, 2016 - 12:00

Figure 1. Illustrating the different cortex shapes produced by different animal species. Taken from J. De Felipe. 2011. Front. Neuro. 5:29.

Applications are invited for a 24-month fixed-term post of Research Assistant in St John’s College Research Centre at St John’s College, University of Oxford. The post involves working on a research project entitled ‘Mapping cortex evolution through mathematical modelling’ funded by the St John’s College Research Centre and led by Professor Zoltán Molnár, Professor Philip Maini, and Dr Thomas Woolley. The appointee will take up the post on November 1st or as soon as possible thereafter. The post is full-time and is for 24 months. The appointment will be on the University’s Grade 7 for Academic and Academic-related staff, currently ranging from £30,738 - £41,255 per annum.

Our project proposes to develop and understand a mathematical model for the differentiation of neurons from earlier pluripotent progenitor cell populations. This framework will allow us to map all possible evolutionary pathways of the cortex enabling us to quantitatively and qualitatively highlight multiple possible divergent evolutionary trajectories. Further, by comparing these theories with data we will construct a novel categorisation of different species, and understand the high diversity of cortex development, thus generating an impact in the field of neurobiology. Finally, through applying our framework to pathological cases, we will predict mechanistic links between neuron production failure and resulting phenotypes such as microcephaly, polymicrogyria syndromes and lissencephaly syndromes.

There is no application form. Candidates should email a covering letter, a curriculum vitae with details of qualifications and experience, and a statement of current research interests and publications to academic.vacancies@sjc.ox.ac.uk. Applications should be in the form of a single PDF file. Candidates must also provide the names of two academic referees who should be asked to email their references to the same address. Both applications and references should reach the College no later than noon on 1st September. Late applications will not be accepted. Interviews will be held in Oxford on 16th September or if appropriate via Skype at the same time.

The appointee will hold or be close to completing a PhD in mathematics or in a relevant field of quantitative biology. Research expertise in the field of inference is desirable.

The appointee is expected to take the lead in delivering the programme of research, namely:

1. Construct a mathematical model of differentiation from progenitor cells to neurons.
2. Qualitatively map the possible outcomes of the neuron developmental model in terms of the population distributions and time to full differentiation.
3. Categorise species data using the map.
4. Quantitatively parameterise the model using Bayesian inference techniques.
5. Highlight current gaps in data.

The candidate is expected to work closely with neurobiologists through extended visits to Professor Zoltán Molnár’s laboratory, where they will learn about the various cell lineage tracing methods in detail. Equally, they will attend pertinent biological group meetings.

Finally, they are expected to lead the creation of an interdisciplinary network of St John’s neurobiologists, psychologists and collaborative sciences.

The candidate will contribute to preparing findings for publication and dissemination to academic and non-academic audiences; therefore, excellent communication skills are essential, and an excellent record of academic publication commensurate with stage of career is expected.

St John’s College is an Equal Opportunity Employer.

More information about St John's College can be found here.

Full job advert can be found here.

Further particular can be found here.

 

A Mathematician's Holiday - Problem 7: Fix the Hotel Rooms.

Requirements:

Pen and paper is all that is required. We also use blow up rubber ring to demonstrate the solution.

Description:
Having got to our hotel rooms Dan, Will and I found that our water, electricity and gas supplies had all been damaged in the fire. Being the helpful souls that we are can we connect each of the utility supplies to each room without crossing the utilities?

Figure 1. Problem set up.

This is a classic mathematical problem known as "The Three Utilities Problem". Critically, it is known to be unsolvable on the flat sheet of the paper. However, the hotel we would be staying in would be three-dimensional and, thus, we are asking the audience to extend their ideas beyond the flat surface.

Note that we really do not go into why the solution is impossible on the flat paper. The reason is because the proof of impossibility depends on some deep ideas from topology. However, the interested reader can find the proof here.

Extension
If we add in another utility, can we still solve the problem on a torus, or doughnut?

A Mathematician's Holiday - Problem 6: The Tiny Lift.

Requirements:

We use two beards to go along with the story that we have concocted. However, you can use two hats, two name badges, etc. Anything that will denote two objects as being similar.

Description:
Once we got into the hotel we found that the lift to our floor was very small. This meant that only  one of us could go up to the first floor at once. This leads to a problem: you see Will and I have bonded over our beards, whilst Dan does not have any facial hair! Thus, neither Will nor I want to be left alone with Dan because we would have nothing to discuss. How do you get us all from the ground floor to the first floor, without ever leaving an hirsute person with a beardless person?

Figure 1. Problem set up.

This problem is simply the fox, chicken, grain problem, a classic brain teaser that has been around for centuries. The audience should have no problem producing an adhoc solution in no time. However, there are two key factors that should be extracted. Firstly, having them demonstrate the problem is a very fun, visual and hilarious thing to do, if you are using beards and, so, it rejuvenates a flagging audience. The second (mathematical) point of the solution to extract is that the audience probably solved the problem by trial and error, however, the solution can be solved neatly by using a graph, just like the "airport security" problem.

Extension 1

Can you solve these two related, but different transport problems:

Vampires and maidens

Three maidens and three vampires must cross a river using a boat which can carry at most two people, under the constraint that, for both banks, if there are maidens present on the bank, they cannot be outnumbered by vampires (otherwise the vampires would bite the maiden).
Jealous husbands
Three married couples want to cross the river in a boat that can only hold two people. Unfortunately, no woman can be in the presence of another man unless her husband is also present.

Extension 2
How is the vampire and maidens puzzle related to the Jealous husbands puzzle?

A Mathematician's Holiday - Problem 5: Hotel Fire!

Requirements:

Some pieces of string, or rope, depending on the size of your demonstration and a play to draw the diagram below in Figure 1. Note that it does not have to be too accurate.

Description:
Having got to the hotel we unfortunately find that it is on fire. Thankfully, there is a river very close to our location, where we can load up buckets with water and, thus, help put the fire out. What is the quickest route from our location, to the river and to the hotel?

Figure 1. Problem set up.

This problem is very similar to "Late for the Plane" and works well if they are done in combination with one another. Again, the question relies on using the ropes to measure distance, in order to measure time.

A critical point for the audience to understand is that the helpers all run at the same speed, even when they are carrying water.

Extension:
Suppose we run slower once we are carrying water. How does this extra facet influence the solution?  Note that you can solve this problem as well, however, but it is far more involved and involves calculus.

A Mathematician's Holiday - Problem 4: Bags Mix-up.

Requirements:

Three bags with different named labels. A t-shirt or jumper. Of course, you can dress this problem up in many different ways so if you have got a t-shirt or jumper to hand you could have three pencil cases with blue and yellow pencils, for example.

Description:

En route to China our luggage got mixed up. Each of our bags look the same, but the name tags have all been mixed up such that we know that no bag has the correct label (a key point to reiterate). We are allowed to put our hand in a bag of our choice and pull out an object at random (we can look in the bag). Can we identify which bag belongs to each person, from this single piece of information?

You should ensure that the audience understands that we are looking for a definite strategy. We do not need to consider probabilities at all.

Extension:

To be honest, I cannot think of one. This question is pretty much self contained. However, the idea behind this question, i.e. logical deduction, is one of the crucial weapons in the mathematical arsenal. If you have any suggestions for extensions please write them in the comments below.

A Mathematician's Holiday - Problem 3: Late for the Plane.

Requirements:

Some pieces of string, or rope, depending on the size of your demonstration and a play to draw the diagram below in Figure 1. Note that it does not have to be too accurate.

Description:
The critical point to ensure for this question is that you're audience is comfortable with the idea of a straight line between two points being the shortest distance between those two points and, thus, the route of shortest distance.

We are late for our plane and so we want to take the quickest route from the terminal our aeroplane. However, there are two active runways between us and our destination. Since these are dangerous places to be we want to ensure that we minimise our time over the runways and, so, we always run directly across, at right angles to the runway direction. What path minimises our total distance?

Figure 1. The problem set up.

Unless they've seen the problem before it is highly unlikely that your audience will generate the solution shown in the video. However, that does not matter. The fundamental concept we are trying to introduce here is that we are looking for a method, rather than an exact solution. Thus, it should be made clear to the audience that the problem set up does not need to be exact.

Extension
In what case is the shortest route from the terminal to the plane an actual straight line?

A Mathematician's Holiday - Problem 2: Airport security.

EASY VERSION
Requirements:

Two containers in the ratio 3:5, for example 75ml and 125ml. We used the bottom of drinks bottles and scaled up the quantities in order to make them more visible for a large audience.

Description:

Airports only allow 100ml of fluid through onto a plane. Can we measure out exactly 100ml of fluid using containers that are of the sizes given above and none of the containers have graduation marks? Beyond the trial and error that the audience may try you can actual solve this problem very easily using a graph.

Figure 1 shows a graph illustrating the volume on liquid contained in each. Filling and emptying each container corresponds to horizontal movements, all the way to the left and all the way to the right. Transferring the liquid from one container to the other corresponds with travelling diagonally as far as you can go. By using these two rules you can easily bounce around the graph and successfully reach 100ml.

Figure 1. Solving the fluid problem graphically. The volume of water in the 75ml bottle in along the vertical axis and the volume of water in the 125ml bottle in along the horizontal axis.

Extensions:
In Figure 1 we filled the 125ml bottle first. Can you solve the problem by filling the 75ml bottle first?
What values of fluid can you not possibly make using this method of pouring between the containers?

HARD VERSION
Requirements:

Three containers in the ratio 3:5:8, for example 75ml, 125ml and 200ml.

Description:
The task is the same as above. However, this time we do not have a reservoir of water to fill from, and empty to. We only have a container of 200ml of water, which can be transferred amongst the different sized bottles.

Critically, the solution to this problem depends on ternary coordinates, which are plotted in a triangle form, as seen in Figure 2, see the video for more details. Once the students have seen this form of graph they are able to solve the problem in exactly the same way as the previous question.

Figure 2. Ternary coordinate plot. Each side of the equilateral triangle represents the volume in one of the bottles. Each of the stars represents one of the points on the left. See the video for more details.

Extensions:
Again, what values of fluid can you not possibly make using this method of pouring between the containers?
Instead of starting with 200ml, suppose we started with 125ml only. What fluid volumes can now be made by transferring the fluid between the bottles?

A Mathematician's Holiday - Problem 1: Planning the tour.

Requirements:
Each participant should have a pencil, paper and rubber to draw and modify the diagram shown in Figure 2.

Description: 
In this activity we set the scene of the whole presentation. In particular, we talk about how all the activities were motivated by problems that we faced on our tour. The first problem is then, of course, organising the tour.

As mentioned previously we visited a number of locations around China and South Korea. These locations are shown in Figure 1.

Figure 1. All of the locations we visited on our tour.

Being huge tourists, we didn't want to have to travel between these locations in the same way twice, because each transportation route would allow us to see something different. In Figure 2 we represent each method of transportation by a black line between each of the cities. For example, the O and L mean Oxford and London and the two black lines represent the two different transportations of train and bus.

Figure 2. Representing the city connections.

The challenge is then to find a way around all the cities using all possible forms of transport. We found that we were not able to solve this problem. Can you? Of course the answer is contained in the video at the top.

Extension
From watching the video you should be able to see that identifying whether a set of paths are completely traversable with no repeating is pretty easy. However, suppose we specified the time each path took, how difficult would it be to find the quickest path around all cities?

A Mathematician's Holiday - Prologue.

One of the first activities that got me into outreach was joining Marcus' Marvellous Mathemagicians (M3), which was started in 2008 by Marcus du Sautoy. The idea was that hundreds of schools contact Marcus every year to give presentations to their kids, but unfortunately he simply doesn't have the time to answer all the requests. Thus, myself and a number of other undergraduates and graduates go out to schools and take a number of fun mathematical workshops

Multiple Oxford scientists coming out in force at the Newbury Science Festival

As part of M3, I've been up and down the country giving maths workshops, over to Ireland and Wales, and presented at numerous science festivals across the country.

However, perhaps the best experience I've ever had was when myself, Will Binzi and Dan Martin were invited to tour around the Dulwich Colleges of China and South Korea and demonstrate mathematics to all the kids out there.

From left to right: Dan, Will and myself enjoying the view from the Great Wall.

During our two weeks out there we travelled to Beijing, Suzhou, Shanghai, Zhuhai and Seoul. By the end of the two weeks we were of course exhausted but extremely happy because our workshops had gone down so well.

We presented in classrooms and theatres. The Dulwich college kids were great audiences.

Specifically, because this was a very special tour we decided to write a brand new presentation for the Dulwich students. The presentation was called A Mathematician's Holiday and collected eight different problems that you might come across whilst planning and presenting on such an international tour.

When we came back we decided to film the presentation so it can be used by anybody. The videos can either be found on YouTube, iTunes U or through the University of Oxford's Podcasts page. Over the next 8 weeks I will be presenting each video and discussing any particular techniques that we found aided us in our presenting the workshop.

Crucially, because we were taking this tour on the road, we had to make sure all the props were light, transportable, easily fixable, reproducible, or cheap to buy. In the end we simply used:

• rope;
• plastic bottles;
• bears;
• a t-shirt;
• a jumper;
• three hold alls; and
• a blow up ring.

As a final note on the videos, they're pretty low energy. Normally we would have been giving this presentation in front of 30-50 excitable kids. Here we had 6 sixth-form students and a doctoral researcher, not exactly our prime demographic, but it had to do!

We begin next time by organising the tour!

Diffusion of the dead - Epilogue. Part 2, The results.

Last time, I presented a question that I had been given by PrettyGreen, a PR company that asked me to get involved with their Halloween publicity campaign for the mobile network operator GiffGaff. We also saw how I modelled the problem and this week I present the results and also the Matlab codes that I used, should you want to recreate, or play around, with the solutions.

The advert presenting the characters involved can be found below.

________________________________________________________________

________________________________________

Results

The survival time of each character in ascending order can be found in the table below. This is further complemented by Figure 1, in which we illustrate the health of each combatant over a 60 minute time span.

The first two characters to die are the Dead Girl and Skin Face. Although these two characters are relatively strong their low initial healths mean that the other combatants are able to dispatch them with relative ease. Thus, neither of them last longer than a minute demonstrating that neither of them are able to use their aggression to its fullest. Similarly, Baby Man is next to die, because although he is able to defend himself better, lasting just over a minute, the combatants once again go after the weakest.

Name	Survival time
Dead Girl	3 secs
Skin Face	13 secs
Baby Man	1 mins 44 secs
Pumpkin Head	7 mins 9 secs
Tall Guy	10 mins 7 secs
Hero Girl	55 mins 37 secs
Evil Clown	Survivor

The next two to die are quite interesting, these are Pumpkin Head and Tall Guy, lasting approximately 7 and 10 minutes respectively. Up until this point the group has simply been killing the current weakest. At the five minute marker this would be Tall Guy, however, as seen in Figure 1, he is able to outlast Pumpkin Head because Tall guys higher agility allows him to defend better.

Eventually, only Evil Clown and Hero Girl are the surviving combatants. This is interesting as it is often the plot of a horror film to have the heroine survive up until the last few moments, when the final kill is unleashed during extended scenes of heightening tensions. This is exactly what is seen in Figure 1. Initially, things are looking good for Hero Girl as she begins the final battle with much more health than Evil Clown. Furthermore, the agility of the Hero Girl and Evil Clown are equal and the highest out of the group. This has kept them safe up until this point because they were able to run away from danger. Similarly, for the 30 minutes after Tall Guy’s death the two characters are able to capably defend themselves and escape each others attacks.

These tropes are used in many horror films: although the heroine is no physical match for the villain, she is able to make up for the lack of strength in terms of speed. It is also a standard horror trend that it should look like the heroine will survive, but then she is subjected to escalating bouts of terror. Although she is able to put up a valiant fight Hero Girl’s life is slowly drained by Evil Clown’s higher strength, ultimately leading to his victory after 55 minutes and with 45 health remaining.

Who'd have thought that the Evil Clown was so dangerous?! And who could have foreseen where such a simple piece of mathematics, based on the modelling of zombies, would take us?

This brings us to the end of this current set of posts. Next time: something completely different.

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function Battle_Royale
%Clear computer memory and plots.
clear
close all
clc

options = odeset('RelTol',1e-4,'AbsTol',1e-5); %Options set for the ODE solver.

S=[9  7  6  8  3  8  9]; %Vector of strength values.
A=[7  8  4  3  8  6  3]; %Vector of agility values.
H=[45 65 60 40 90 10 20];%Vector of initial health.
C=S.*A; %The interaction function is scaled by the strength multiplied by the agility.

[T,Y] = ode45(@(t,y) ODES(t,y,C),[0 7000],H,options); % Solve the ODE.

cc=lines(7); % Set the legend colours.
for i=1:7
    %Find the last moment alive.
    YY=Y(:,i);
    YY(YY<0.1)=0;
    index=find(YY,1,'last');
    YY(YY==0)=nan;
    YY(index)=0;

    plot(T/60,YY,'linewidth',3,'color',cc(i,:)); %Plot results.
    t(i)=T(index)/60;
    hold on
end
t %Print out the times of death.
axis([0 60 0 90]) %Set axes.

%Make the plot look nice.
legend('Tall guy','Evil Clown','Pumpkin Head','Baby Man','Hero Girl','Dead Girl','Skin Face','location','northoutside','orientation','horizontal')
xlabel('Time in minutes')
ylabel('Combatant Health')

function dy = ODES(t,y,C)
Terms=y.*C'; %The value of the each combatant's attack.
dy=zeros(7,1); %Preallocation of the variables.
for i=1:7
    indices=[1:i-1,i+1:7]; %Indices to be summed over.
    dy(i)=-heaviside(y(i))/((1+y(i)^2)*2*C(i))*sum(Terms(indices)); %Differential equation.
end

Diffusion of the dead - Epilogue. Part 1, Horror Battle Royale.

Having written and published the article on zombie invasion I thought the story would end there. Of course the media picked it up and I had my 15 minutes of fame on the radio and TV. However, it appears I was very wrong. A PR company, PrettyGreen, contacted me about getting involved with their Halloween publicity campaign for GiffGaff, a mobile network operator. 

Their question was: suppose seven stereotypical characters from horror movies are released in an arena leading to a battle royale, who would survive? The characters are:

• Tall Guy – A very tall (6,9 ft) escaped convict;
• Evil Clown – An evil, twisted clown;
• Pumpkin Head – A man with a pumpkin as a head;
• Baby Man – A somewhat mental man who has spent is life trapped in a basement;
• Hero Girl – A normal young girl;
• Dead Girl – A dead bride;
• Skin Face – A man with excess skin stapled over his face.

My first thought was how was I going to do this? As mentioned when we first started this set of posts there are many different ways mathematics could answer this question. I decided to use differential equations to model the health of each character. As the characters interact their health would reduce depending on certain factors. Thus, I asked PrettyGreen to provide me with a strength, agility and initial health for each combatant and I used these to simulate who will be the last person standing. They were even kind enough to send me a picture to use with this post.

Figure 1. Hero Girl, Evil Clown, Dead Girl, Tall Guy and Pumpkin Head all having a nice jog in London. Image courtesy of PrettyGreen.

This post is a little more mathematical then normal as I present all the gory equation details. However, do not let this scare you. Hopefully, I provide all the intuitive details you should need in the text. If you just want the brief that appeared in the news, then please look here.

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________________________________________

Method

The battle simulation is based on the reduction of each combatant’s initial health at a rate proportional to their meeting with an opponent. Once a combatant’s health has dropped to zero, they are immediately removed from the conflict, whilst the survivors fight on.

The interaction outcome is based on a scaled form of the empirical rule known as the “Law of Mass Action”. Explicitly, the rate of reduction in health is proportional to the attacker’s health, scaled by an inverse square of the defender’s health. Intuitively, this simply means that a reduction in health can only occur when the attacker and defender meet. Moreover as the attacker gets weaker, their attacks do less damage. Equally, as the defender becomes weaker their defence is less effective. The precise form of the interaction term is,
\begin{equation}
\frac{H_a}{1+H_d^2},
\end{equation}

where $H_a$ is the health of the attacker and $H_d$ is the health of the defender. It should be understood that this interaction term is known as a "constitutive equation". This means that it is not based on any fundamental law, but it is postulated because it produces the right kind of dynamics that we would expect. There are many other functions I could have used in its place, however, in using this equation, I am hoping that it is simple enough to capture the general outcomes of the system and, thus, the results will be robust to any small changes that might occur.

From these assumptions we can construct a coupled set of ordinary differential equations that will evolve the battle and predict who will win. The exact form of the equation for combatant $i=1,2,…,7$ is

\begin{equation}
\tau\frac{dH_i}{dt}=\frac{-1}{1+H_i^2}\frac{1}{S_iA_i}\sum_{j\neq i}S_jA_jH_j.
\end{equation}
Note that the equation is only active while $H_i>0$. Also notice that we have scaled the terms with the strength parameter, $S_i$, and agility parameter, $A_i$. Thus, speed is, potentially, just as important as strength. Parameters can be found in the Table 1, below. Finally, we define the initial condition as:
\begin{equation}
H_i{0}=H_{i0}.
\end{equation}

Name and variable	Strength, $S_i$	Agility, $A_i$	Initial health, $H_{i0}$
Tall Guy, $H_1$	9	7	45
Evil Clown, $H_2$	7	8	65
Pumpkin Head, $H_3$	6	4	60
Baby Man, $H_4$	8	3	40
Hero Girl, $H_5$	3	8	90
Dead Girl, $H_6$	8	6	10
Skin Face, $H_7$	9	3	20

The parameters were chosen such that strength and agility were on a scale from 1 to 10, such that higher numbers represent higher strengths and speeds, respectively. The health parameter is on a scale of 0 to 100, with the relative size determining how healthy each character is. It should be noted that in the original data Dead Girl’s health was 0. Understandably, this would make her “dead”, however, it would also suggest that she was unkillable, resulting in her inevitable win. To ensure such a case does not occur I changed her initial health to 10.

Next time I will present the results of the above simulations and demonstrate that although this is only a simple model it produces an outcome that you would expect to see in any good (or bad) horror film. In the mean time, have a go at simulating the system yourself and perhaps vary the interaction rule to see how the results are influenced by this equation.

Diffusion of the dead - The maths of zombie invasions. Part 9, The complete strategy.

Over the past 8 posts we have investigated the mathematical implications of a zombie invasion. If you really can't be bothered to read all of the details, please see part 1, where you'll find some links to presentations that I have done and I'll give you all the details inside of one hour!

For the more patient reader, we began in parts 2, 3 and 4 where we discussed the philosophy of modelling, the assumptions that would make up the zombie invasion model and some basic mathematical simulation techniques. Although important for the recreational mathematician if you're really in a pinch you'll want to skip right to the results.

In particular, in parts 5 and 6 we showed that running away from zombies increases the initial interaction time far more than trying to slow the zombies down. Thus, fleeing for your life should be the first action of any human wishing to survive. However, we cannot run forever; interaction with zombies is inevitable.

In part 8 we showed that the best long-term strategy is to create a fortified society that can sustain the population, whilst allowing us to remove the zombies as they approach. This, in effect, reduces the risk to zero and we will survive.

In the event of the apocalypse, it is unlikely that we would be able to support such a commune without raiding parties scavenging for medicine, food and fuel. Thus, in this case, we fall back on the maxim of being more deadly than the zombies which allows us to survive as shown in part 7.

Over all, it is difficult for us to survive simply because zombies do not have a natural death rate, and by biting us, they are able to increase their own ranks whilst reducing ours. Thus reinforcing our greatest fear that human allies could quickly become our biggest nightmares.

Our conclusion is grim, not because we want it to be so but because it is so. It had always been the authors' intention to try and save the human race. So to you, the reader, who may be the last survivor of the human race, we say: run. Run as far away as you can get; an island would be a great choice. Only take the chance to fight if you are sure you can win and seek out survivors who will help you stay alive.

Good luck.
You are going to need it.

Diffusion of the dead - The maths of zombie invasions. Part 8, Surfing the infection wave.

Last time we considered the interaction rules

1. humans kill zombies;
2. zombies kill humans; and
3. zombies can transform humans into zombie,

and produced the following system of equations,

\begin{align}\frac{\partial H}{\partial t}&=D_H\frac{\partial^2 H}{\partial x^2}-\alpha HZ\label{Human_PDE}\\\frac{\partial Z}{\partial t}&=D_Z\frac{\partial^2 H}{\partial x^2}+\beta HZ\label{Zombie_PDE},\end{align}

which predicts the evolution of the human population, $H$, and the zombie population, $Z$.

To see how quickly the infection moves through the human population, we look for a particular type of solution, known as a "Fisher wave". This type of wave travels at a specific speed and does not change its shape as its travels. Such a solution can be seen below.

 Movie: The black line represents the human population. The red dashed line represents the zombie population. Initially, there are only a small number of zombies, but over time the infected population spreads out and transforms the susceptible population.

By manipulating the equations we can show that the wave speed, $v$, has a value of

\begin{equation}
v^2=4D_Z\beta H_0,
\end{equation}

where $D_Z$ is diffusion rate of the zombies, $\beta$ is the net-rate of zombification (a.k.a how quickly the zombie population grows) and $H_0$ is the initial human population. Critically, note that the left-hand side of the equation is positive, because it is a squared value.

In order to slow the infection, we should try to reduce the right-hand side of the above equation.

• Reducing $D_Z$ amounts to slowing the zombies down; thus, an effective fortification should have plenty of obstructions that a human could navigate but a decaying zombie would find challenging.
• Reducing $\beta$ occurs through killing zombies quicker than they can infect humans. If we are really effective in our zombie destroying ways then we can make $\beta$ negative. This would make the right-hand side of the equation negative, but the left-hand side of the equation is positive. Since we get a contradiction no wave can exist and the infection wave is stopped.
• Reducing $H_0$ involves reducing the human population. This could mean geographically isolating the population on an island because if the zombies are unable to get to you then they can't infect you.

The last tactic of reducing the human populations could also lead to a rather controversial tactic. Suppose you don't live near a deserted fortified island. Rather, you work in an office surrounded by people who would not be able to protect themselves from the oncoming hordes. Then you may consider removing the humans around you in a more drastic and permanent way. This is because everyone that surrounds you is simply a potential infection!

I do not recommend this cause of action, as reducing the human population also reduces the number of people able to fight the zombies and humans sacrificing other humans would only speed the extinction of their own species. The human population will have enough trouble trying to survive the hordes of undead, without worrying about an attack from their own kind!

This, pretty much, brings us to the end of the mathematical zombie story. It's been a long journey, thus, to ensure that you haven't missed any of the critical strategies along the way the next post simply focuses on summarising and concluding what we're been discussing since Halloween. See you then.

Diffusion of the dead - The maths of zombie invasions. Part 7, Face to face with a zombie.

So far we have only considered zombie motion. It has been an incredibly simply model, but it has been able to furnish us with a wealth of information. In particular, we have been able to predict how long it will take the zombies to get to us and we have shown that the best strategy is to run away.

Unfortunately, you can only run so far. At some point you are no longer running away from the living dead, but actually running towards a different mob of zombies. So what should you do when you finally end up having to go hand to hand with a zombie?

To model human-zombie interactions, we suppose that a meeting between the

two populations can have three possible outcomes. Either

1. the human kills the zombie;
2. the zombie kills the human; or
3. the zombie infects the human and so the human becomes a zombie.

These outcomes are illustrated in Figure 1.

Allowing $H$ to stand for the human population and $Z$ to stand for the zombie population, these three rules can be written as though they were chemical reactions:

\begin{align}
H+Z&\stackrel{a}{\rightarrow}H \text{ (human kills zombie)}\\
H+Z&\stackrel{b}{\rightarrow}Z \text{ (zombie kills human)}\\
H+Z&\stackrel{c}{\rightarrow}Z+Z \text{ (human becomes zombie).}
\end{align}

The letters above the arrows indicate the rate at which the transformation

happens and are always positive. If one of the rates is much larger than the

other two, then this "reaction" would most likely happen.

Figure 1. The possible outcomes of a human-zombie interaction. Either (a)

humans kill zombies, (b) zombies kill humans, or (c) zombies convert humans.

To transform these reactions into a mathematical equation, we use the "Law of Mass Action". This law states that the rate of reaction is proportional to the product of the active populations. Simply put, this means that the above reactions are more likely to occur if we increase the number of humans and/or zombies. Thus, we can produce the following equations which govern the population dynamics

\begin{align}\frac{\partial H}{\partial t}&=D_H\frac{\partial^2 H}{\partial x^2}-\alpha HZ\\\frac{\partial Z}{\partial t}&=D_Z\frac{\partial^2 H}{\partial x^2}+\beta HZ.\end{align}

where $b+c=\alpha$ is the net death rate of humans and $c-a=\beta$ is the net creation rate of zombies.

If we ignore the reactions for a second, we have seen the first part of the equations before. Explicitly we are assuming that both the zombies and humans randomly diffuse throughout their domain. Now we have previously justified the zombies' diffusive motion as they are mindless monsters. However, humans are not usually known for their random movement. Here, we use the fact that if the dead should start to rise from their graves, then panic would set in and humans would start to run away and spread out randomly from location of high population density. Thus, human movement could also be described by diffusion, although their diffusion rate is likely to be much larger than the zombies'.

If we now include the interaction formulation once again then the equations immediately highlight some important components of this problem. Firstly, because $b$, $c>0$ and $H$, $Z\geq 0$ then the human interaction term, $-\alpha HZ$, is always negative. Thus, the  human population will only ever decrease over time.

We could add a birth term into this equation, which would allow the population to also increase in the absence of zombies but, as we have seen previously, the time scale on which we are working on is extremely short, much shorter than the 9 months it takes for humans to reproduce! Thus we ignore the births since they are not likely to alter the populations a great deal during this period.

Interpreting the zombie equation is not so easy. The term $c-a=\beta$ may either be positive or negative. If $(c-a)>0$ then the creation rate of zombies, $c$, must be greater than the rate which we can destroy them, $a$. In this case the humans will be wiped out as our model predicts that the zombie population will grow and the human population will die out. However, there is a small hope for us. If the rate at which humans can kill zombies is greater than the rate at which zombies can infect humans then $(c-a) < 0$. In this case both populations are decreasing, thus our survival will come down to a race of which species becomes extinct first.

Next week we will delve into the equations more and consider the spread of infection. We will then be able to derive expressions that really tell us how to survive, or at least delay, the zombie uprising.

Diffusion of the dead - The maths of zombie invasions. Part 6, Run, don't fight.

Last time I demonstrated how to approximately find the time of your first interaction with a zombie using the diffusion equation and the bisection method. The Time-Distance-Diffusion graph is illustrated below.

Figure 1. Time in minutes until the density of zombies reaches one for various rates of diffusion and distances.

When the apocalypse does happen, we have to ask ourselves the question: do we want to waste time computing solutions when we could be out scavenging? In order to speed up the computational process we consider the diffusive time scale:

\begin{equation}t=\frac{L^2}{\pi^2 D}.\label{Time_scale}\end{equation}

You may recognise this group of parameter, as we saw it back in Part 4. In particular, in the solution to the diffusion equation, we can see that this is the time it takes for the first term of the infinite sum to fall to $\exp(-1)$ of its original value. The factor of $\exp(-1)$ is used due to its convenience.

Only the first term of the expansion is considered because as $n$ increases, the contribution from the term

\begin{equation}\exp\left(-\left( \frac{n\pi}{L} \right)^2Dt\right)\end{equation}

rapidly decreases. Thus, the first term gives an approximation to the total solution and, so, equation \eqref{Time_scale} gives a rough estimate of how quickly the zombies will reach us.

For example, being 90 metres away and with zombies who have a diffusion rate of 100m$^2$/min, $t\approx 26$ minutes, comparable to the solution in Figure 1. We have had to use no more computing power than you would find on a standard pocket calculator. More importantly this parameter grouping also implies a very important result about delaying the human-zombie interaction.There are two possible ways we could increase the time taken for the zombies to reach us. We could:  

1. run away, thereby increasing $L$; or
2. slow the zombies down, thereby decreasing $D$.

Since the time taken is proportional to the length squared, $L^2$ and inversely proportional to the diffusion speed, $D$. This means that if we were to double the distance between ourselves and the zombies, then the time for the zombies to reach us would approximately quadruple. However, if we were to slow the zombies down by half, then the time taken would only double.

Since we want to delay interaction with the zombies for as long as possible then, from the above reasoning, we see that it is much better to expend energy running away from the zombies than it is to try and slow them down. Note that we are assuming that zombies are hard to kill without some form of weaponry. If they weren't difficult to destroy then we need not worry about running away.

These conclusions are confirmed in Figure 1. Slowing a zombie down from 150 m/min to 100 m/min only gains you a couple of minutes when you are 50 metres away. However, running from 50 m to 90 m increases the time by over 10 minutes, even in the scenario of relatively fast zombies.

It should be noted that the time derived here is a lower bound. In reality, the zombies would be spreading out in two dimensions and would be distracted by obstacles and victims along the way, so the time taken for the zombies to reach us may be longer. The fact that this is a conservative estimate though will keep us safe, since the authors would prefer to be long gone from a potential threat rather than chance a few more minutes of scavenging!

Of course, we can't run forever. Next week we will begin to ask what happens when we finally meet this horrific horde!