Imagine you and your friend Alice are two thieves and you stole a necklace together. The necklace has beads of two colors, red and blue. Now, you would like to split it evenly, or in other words, each of you would like half of the red beads and half of the blue beads. To do this, you open the necklace at the clasp and get an interval of beads. We assume the number of red beads and blue beads are both even, so the desired splitting is possible. But you also don't want to cut the necklace in too many pieces. How many cuts are enough to guarantee a desired division? Do you have any guesses?

The above is the simplest setting of Necklace Splitting. Last year I took a course on Algorithmic Game Theory and got interested in how algorithms and complexity appear in economic applications. During my personal investigation into the field, this problem interested me a lot because of its roots in combinatorics. I believe this is an interesting topic because it’s easy to describe but the solution to the problem is non-trivial at all. Therefore, I would like to share it with you. Are you ready? Let’s go dive into the world of necklace splitting!

As said in the beginning, you and your friend Alice stole a necklace which consists of beads of two colors, red and blue. And you would like to split it evenly. For example, the interval of beads can look like this:

*Figure 1: An example of a necklace with beads of two colors.*

You would like to cut the necklace into several pieces and distribute the beads evenly, in this example, each of you get 4 blue beads and 3 red beads, so 7 beads in total. Moreover you would like to use only a small number of cuts. Think about it a bit: in this example, could you use only one cut to split the necklace and take two intervals containing exactly 4 blue and 3 red beads?

If each one wants half of the beads and use a single cut, then the necklace should obviously be cut in the middle. However, then you may find the left interval has only one red bead, which is less than required. Conversely, the number of red beads in the right interval is more than required. So, it is not possible with a single cut.

But then you could shift the left interval to the right, one bead per step. We represent this below using the green box, you start moving from left to right until you get an interval with 4 blue beads. Now you need to use two cuts instead, which divide the necklace into three parts. As you can see the middle interval (in the green box) achieves the goal of 4 blue and 3 red beads. Similarly, combining the other two intervals also achieves the goal.

*Figure 2: Moving the green box from the left to the right you can find the two cuts.*

But are 2 cuts always enough for splitting an arbitrary necklace with an even number of blue and red beads? The answer is yes! Image a long necklace with an even number of blue and red beads, and you cut it in the middle. Then similarly, if the left has too few red beads, the right must have too many red beads. Observe that if you shift the left interval to the right by one bead, the number of red beads in it can only change by at most 1. Since at the beginning you get too few red beads and you get too much in the end, and in each step the number of red beads can only change by 1, then there must exists a stage that you get the right number of red beads. Because you always get half beads in this way, then the number of blue beads you get is also correct.

This argument shows that 2 cuts are always sufficient! In the proof above, we used a theorem in mathematics, the discrete version of the *Intermediate Value Theorem*. The Intermediate Value Theorem is one of those results we learn at school when studying functions but we don't appreciate its importance at that moment.

**Intermediate Value Theorem**

The Intermediate Value Theorem says any continuous function over a closed interval achieves all values between the function values at the endpoints of the closed interval.

Furthermore, it is not possible, in general, to achieve such a splitting with only one cut. To see this, consider, next to the example in Figure 1, also the following example: it is necessary to use 2 cuts to achieve the desired splitting, can you see why?

*Figure 3: Putting beads of the same color next to eachother shows that 2 cuts are sometimes necessary.*

What if the necklace is more complicated in the sense that the beads have more colors? If the necklace has beads of 3 colors for example, how many cuts do you need to split it evenly among two thieves? How about 4 or in general colors? The problem is defined formally as follows:

**General form of necklace-splitting problem **

Given a -types and -beads necklace consisting of beads of each type ( is even), and the beads are fixed on the string, how many cuts are needed to guarantee a fair splitting such that each thief gets beads of each type ?

In fact, it turns out that for a necklace with -colors cuts are always enough! Furthermore, cuts are sometimes necessary by considering a -colored necklace using a similar construction as above, by putting all beads of the same color together. This result was obtained by mathematicians Professor Noga Alon and Professor Douglas West in the 80s using the Borsuk-Ulam theorem from Algebraic Topology. In fact, the name of the problem was also due to them.

Let us see together what the Borsuk-Ulam theorem states and how it solves this problem.

**The Borsuk-Ulam theorem**

The Borsuk-Ulam theorem says that if is continuous, then there exists a point such that .

Don't feel intimidated by the formal statement. When we write we mean the set of -dimensional vectors of real numbers, i.e.

With we mean a sphere as you know it, but in dimensions, is just the circle, is the "ordinary" sphere. The formal definition is

In other words, every continuous function from an -dimensional sphere to the real vectors always maps some pair of antipodal points to the same number. Antipodal points are in exactly opposite directions form the sphere’s center. **Couple of fun facts**

Do you know that you can apply the Borsuk-Ulam theorem to the earth to get a classical conclusion: On earth, there are always two antipodal points with same temperature and same pressure.

A second interesting fact is that in 1978 László Lovász (you can read an interview with Lovász here) proved the Knesser conjecture using the Borsuk-Ulam theorem. This breakthrough gave rise to the field of topological combinatorics.

To apply the Borsuk-Ulam theorem, we need to translate the necklace splitting problem to a so-called *topological problem*, that is to write as a problem concerning a function from the sphere to the real numbers.

We can use a function to keep track of where the necklace is split and the amount of each type of beads secured by each thief. This is achieved by representing the splitting points on the necklace using the unit -sphere (we denote the unknown needed number of cuts as ), where each point corresponds to a specific way the necklace is split. Each point on the sphere encodes information about the lengths of the necklace pieces, with positive and negative roots allowing differentiation between the two thieves. We’ll illustrate this using the following example

The green area in the middle represents a splitting in two pieces. This specific splitting can be represented as the vector

Try to verify this using the definition of above.

How did we do this translation? Observe that this splitting breaks the necklace in three pieces with lengths equal to and of the total length. We represent these numbers as a vector on the unit sphere , observe that , and . Moreover

Thief 1 receives the first and third pieces (because they are with + sign) and thief 2 receives the second (because it is with - sign). This step shows how to translate the points at which the necklase is cut to points on the sphere, cutting the necklase at points yields a point on . The next step is to construct a continuous function from the sphere to the real vectors.

This function will record how many beads of each type of bead one of the two thieves receives. Say that the necklace consists of beads of different colors, then this function will send a cut of the necklace to a vector of numbers in . In the example above it will be

This means that the first thief receives 4 blue beads and 3 red ones. By considering the necklace as a continuous map from the -sphere to , the problem is rigorously formulated. Recall the Borsuk-Ulan theorem, every continuous function from an -dimensional sphere to always maps some pair of antipodal points to the same number. To apply the theorem the dimensions of the sphere and the reals should be the same.

Let's choose the number of cuts to be equal to the number of the types of beads , and suppose there is an even number of each type of bead. Then according to the Borsuk-Ulam theorem there are two andipodal points such that . But remember what represents, and what does it mean to flip the sign in . If an element in , say is positive, then this means that thief one takes the beads in that cut. Hence the andipodal point gives essentially the same information as since only the signs are flipped. The equality means thus that in the cut both thieves receive the same number of beads for each color!

The theorem states that there must be some way to split the necklace evenly between the thieves. This way we have shown the sufficiency of cuts to guarantee a fair splitting between two thieves using the Borsuk-Ulam theorem. The last question that remains open is whether cuts are necessary. If you have a necklace with beads of colors, could you in general split it with less than cuts? In general no. Consider for example a necklace where beads of the same color are placed next to eachother.

I hope you were excited about learning how to use a topological method to solve an “easy” problem seemingly having nothing about topology at the first place, which is necklace splitting. However, even though we have shown that there always exists a solution with at most cuts, this proof is non constructive, which means that we have no clue from it how could we efficiently find a -cut does the job.

In summary, we discussed a constructive proof of necklace splitting problem between two thieves when we restrict the type of beads be at most two. Moreover, a general, non-constructive proof via the Borsuk-Ulam theorem was given for the general cases. Moreover, we learned that sometimes a math problem can (or even must) be solved using tools from other fields in math, showing the deep connections between different subjects inside the world of mathematics!

The featured image was taken from this amazing video from Numberphile. You can also watch this very nice video from 3blue1brown on the necklace splitting problem and the Borsuk-Ulam theorem.

]]>Imagine you’re in a foreign country and you want to visit the capital city. There’s a river that divides the capital in different areas. To get from one area to the other, they build beautiful bridges which you all would like to see. But they’re still just bridges, so after crossing each bridge exactly once, you’ve seen enough. At the moment you’re at a really nice café, so you want to end your walk at the same spot as you are now, you want it to be a cycle.

Believe it or not, this problem was the foundation of an entire field of mathematics.

An old map of Königsberg with the seven bridges. Image By Bogdan Giuşcă - Public domain (PD),based on the image, CC BY-SA 3.0.

*The roads connecting the bridges of Königsberg represented as a graph.*

Mathematician Leonhard Euler wrote an article called “the 7 bridges problem” in which he tried to find a walk through the bridges of Königsberg. He drew a dot/node for each area, and a line connecting two points for each bridge connecting two areas. This collection of dots and lines is what we call a ** graph**. If we draw arrows in the direction of the walk instead of lines, we call this a

Euler found out that such a cycle, which he called an ** Eulerian cycle**, that crosses each bridge exactly once, exists if and only if the graph is

A connected graph is simply a graph that consists of one piece. Take a pencil and paper and draw two triangles, this is not a connected graph. If we now draw a line from a point of the first triangle to a point of the second one, then we do have a connected graph. Unfortunately, Euler proved that there is no Eulerian cycle possible in Königsberg. Try and figure out why it is not possible to find an Eulerian cycle in Figure 1a below. The fun in math is doing it yourself!

*Figure 1a. There is no Eulerian cycle in Königsberg.*

*Figure 1b. If we remove three bridges, there is an Eulerian cycle*.

**Euler's theorem**

An Eulerian cycle exists in a directed and connected graph.

An equal amount of arrows are going in each vertex as there are going out.

Although graph theory was born back in 1736, it keeps revealing wonderful results until today, and amazingly enough many of its magical aspects still remain hidden. Researchers are passionately trying to reveal them, making it a very vibrant area of modern mathematics. We will take you back to 1946, two-hundred years after Euler's discovery, and in the office of the Dutch mathematician Nicolaas Govert (Dick) de Bruijn, who in one of his mathematical discoveries managed to reveal some magic!

Try to imagine the following situation. You are in a classroom and you have eight cards. You let the cards go round and everyone is allowed to perform a cut of the deck. Next, three people pick the first three cards. Then you ask them to raise their hand if they have a red card. Would you believe it if I told you there is a way to always be able to guess the three cards, one for one?

This is a pretty cool trick on itself but using the magic of mathematics I will show you later on that this trick doesn’t only work for three people, but for any number of people you want. There are a few components to this magic trick:

- the ordering of the cards such that cutting the deck doesn’t affect it,
- a unique sequence of red and black cards,
- and a "cheat sheet" to help you in the end.

First, let’s have a look at what happens when someone performs a cut of the deck. Suppose the ordering of the eight cards is as follows (denote the red cards by 1 and the black cards by 0). Now suppose that we perform a cut of the deck with the two first cards.

*Figure 1a. Before cutting the deck.*

*Figure 1b. After cutting the deck.*

This is exactly the same as shifting the starting position of our sequence by two. If we write the sequence on a toll then this would be the same as turning the toll two steps counterclockwise. As you can see, the order of the sequence hasn’t changed by shuffling the cards, the only thing that changed was the starting position on our toll.

*Figure 2a. Before cutting the deck.*

*Figure 2b. After cutting the deck.*

So cutting the deck does not "really" change the ordering of the cards. This can also be seen in the animation below. If you know the sequence, then you cut the deck a number of times, and you know the first card, then you know the whole sequence.

*Figure 3: What changes is the first card, not their order.*

Additional to not changing the red/black sequence while shuffling, we also wish to be able to directly figure out which specific cards the three people are holding. For this we need to take all the possible ** combinations **of red/black in consideration. These are the different possible sequences of red/black of the three cards the people are holding. The first person can have a red or black card, which gives 2 possibilities. The second and the third person also have 2 different possibilities. Therefore, the total number of combinations would be , which was exactly the amount of cards that were passed through the class.

If we want to be able to match cards to a 0/1 sequence, 0 for a black card and 1 for a red card, we want the cards to be ordered in such a way that each possible combination of length 3 occurs only once in the ordering. Hence, we want a sequence of 0/1 that contains all different combinations, and contains them exactly once. An example of such a sequence is **11100010**. Do you recognize this sequence from before?

Figure 4: In the sequence 11100010 all possible combinations of 0/1 of length 3 appear. On the contrary, if you take the sequence 10101010 this does not happen.

Lets go back to the cards in Figure 1 and the magic trick. You have given these cards to eight people, after letting them cut the deck. Suppose that I know that the first person has a black card, the second person a red card and the third person a black card. Then this sequence is 010 and using the ordering above we know that the cards are 9 spades, 10 hearts and 10 spades. And by knowing this sequence you also know what everybody else has!

Figure 5: If you know the cards the last three people have, then you know what each one has!

So how would you perform this trick? First, order all the cards in the specific order shown above. Let the audience shuffle the deck by cutting it, let three people pick the cards and remember who took first, who second, and who third. Afterwards, let them rise their hand if they have a red card. Finally look at the cheat sheet to find out which cards they’re holding!

*Figure 6: To perform the magic trick it is good to use a "cheat sheet" with the sequences.*

You now understand the mathematics behind the trick, but mathematics has one more trick up its sleeve. This trick doesn’t only work for 3 people and 8 cards: it could be done with 5 people and 32 cards as well. Actually, if we extend the deck of cards in a particular way, the trick could even be performed with 10 people and 1024 cards. Just like Euler did when he went from a statement about bridges in real life to a general statement about Eulerian cycles in graphs, this trick can be generalized to an even more impressive mathematical statement: this trick is possible with cards and people for each number !

As discussed before, the main ingredient of this trick was the existence of a sequence of 0s and 1s of length , in which each 0/1 combination of length three appeared exactly one time. Such sequences are called ** de Bruijn sequences**. We found a de Bruijn sequence for , namely 11100010, or to put differently: for the situation where 3 people are involved. Thus, showing that this trick is possible is equivalent to showing that for each number of people, for each , there exists a de Bruijn sequence of length .

The trick is possible with cards and people.

There exists a de Bruijn sequence for .

To prove that a de Bruijn sequence exists, we use a special graph which is, not so surprisingly, called a ** de Bruijn graph**. Its nodes for represent the different combinations of sequences of length 2. Draw an arrow from to if there exists a 0/1 sequence of length 3 with on its left and on its right. Take for example and , then 001 is a sequence with on its left and on its right, hence we draw an arrow from 00 to 01.

De Bruijn graphs have three beautiful properties.

- They exists for all numbers .
- They are always connected.
- In each node of such a graph there is an equal amount of arrows going in as there are going out.

Hopefully this rings a bell: these are the conditions in Euler's theorem! Hence there exists an Eulerian cycle in each de Bruin graph! But what has an Eulerian cycle in a graph that, supposedly, was pulled out of thin air to do with our magic trick? Let’s look at the cards and find out.

The first three cards had sequence 111. This is the same as the lowest arrow in the picture on the right. The next cards had sequence 110, this is the same as the arrow from 11 in de graph to 10. The next sequence, 100, is the same as the arrow from 10 to 00. If we keep doing this, we see that each arrow corresponds to a combination and once we finish we see that the last sequence is 011. Hence we are exactly back at our starting point. The Eulerian cycle in our de Bruin graph, corresponds exactly to a de Bruijn cycle!

*Figure 7: A de Bruijn cycle yields the order of the cards!*

We now are at the final part of our proof: there exists a de Bruijn graph for each , these are connected graphs, and there is an equal amount of arrows going in each node as going out of it. Therefore we know that there must be an Eulerian cycle in this graph. As Eulerian cycles in de Bruijn graphs correspond to de Bruijn cycles, we now know that de Bruijn cycles exist for each . Hence we must conclude that the magic trick is possible for each amount of people. Isn’t that just wonderful? Enchanting? Magical? I’d like to welcome you in the fascinating world of mathematics.

This article was inspired by the really nice book "Magical Mathematics: the mathematical ideas that animate great magic tricks" by Persi Diaconis, Ron Graham, Martin Gardner. The magic trick is also described in this book.

]]>*Both pictures are taken from the site of the Abel Prize.*

Every year the Abel Prize is awarded in March, this is always a nice opportunity to take a short break from everything else we are doing and have a glimpse into an area of modern mathematics that many researchers are enthusiastic about, and passionately try to unravel its mysteries. Moreover, it is often unknown to the public what research in modern mathematics concerns. For example, can you name ten great mathematicians of our time? Or can you name some recent major discoveries in mathematics?

When we think of great results from mathematics we often come to results of great mathematicians from the past, like Euler, Gauss, Hilbert, Noether, Wiles. But what is modern mathematics about? Undoubtedly mathematics has a rich history, and modern discoveries rely heavily on the legacy left by the previous generations. Mathematics has a continuous history; discoveries made 2000 years ago by Euclid and Archimedes remain valid today. But at the same time, the mathematical community must also communicate the beautiful discoveries in modern mathematics. These results may be more abstract and advanced, which shows how important it is to keep developing new insights and techniques to communicate mathematics.

The two great awards in mathematics, the Abel Prize and the Fields medal, offer very good oportunities to have a glimpse into modern mathematics. On the website of the Abel Prize you can find very nice short articles about the work of Michel Talagrand.

We cite below some parts of a short article written by Matt Parker about Talagrand's work.

*Michel Talagrand is an expert at understanding and taming complicated random processes. Randomness can arise in a wide range of ways, and Talagrand has explored many different types. One of the most common, and arguably most important, types of randomness arises from “Gaussian processes”. The Gaussian distribution has been a constant feature of Talagrand’s career so it’s worth considering it as a tangible example when exploring his work.*

*A Gaussian distribution (sometimes called the “normal distribution” or “bell curve”) occurs with surprising frequency in the world around us. The mass of babies at birth, the test results students get at school and the ages athletes retire at are all seemingly random things which neatly follow the Gaussian distribution. These are characterised by* *having an average in the middle which most values are close to, and then decreasing numbers of cases as values move further above and below the average. When observing a random process there are some things which it could be nice to know. For example, if you take the average of the values being produced, how close is that likely to be to the true average value of the underlying Gaussian process? How big or small are the possible values in the future likely to be? Talagrand produced rigorous and tight thresholds with specific uncertainties so when these kind of random, stochastic processes are observed we know exactly how confident we can reasonably be about what the process will do going forward.*

And a little bit further

*The ubiquity of Gaussian processes means there are many possible applications of Talagrand’s work. One specific one is the condensed matter physics problem of a “spin glass”. This is an arrangement of matter which is not a ‘glass’ like in a window, but rather a random structure of magnetic moments.*

*Spin glasses sit between the highly organised magnetic properties of ferromagnet materials and the random arrangement of paramagnetic materials. If a ferromagnetic substance is heated and an external magnetic field used to align all the internal magnetic moment, this neat alignment will persist even after the substance is cooled. A paramagnetic substance will exponentially lose its internal magnetic field once the external field is removed. A spin glass however, will also lose its own magnetic arrangement but at a rate and in a way which currently defies our understanding.*

*The internal magnetic structures within a spin glass are arranged randomly, but with some order which causes their complex behaviours.*

On the website of the Abel Prize you can also find much more information on Talagrand's work, we recommend the popular science articles by Arne B. Sletsjøe on the work of the laureate. A broad article about his work, one about stochastic processes, and one about concentration of measure.

]]>

The Anton de Kom university is the only university in Suriname and is located in the west of Paramaribo. It currently has around 4000 students spread over four faculties. They offer a three year bachelor degree in mathematics, which is offered every two years to a fixed number of twenty students. The teaching staff consists of about ten teachers, offering a curriculum that looks similar to European mathematics curricula apart from the fact that they offer fewer optional courses in year two and three.

*The campus of the Anton de Kom University.*

This visit was inspired from various other previous collaborations between the two programs, and the two countries in general. Professor Shanti Venetiaan is one of the driving forces of these collaborations. She studied mathematics at the University of Leiden and then obtained a PhD from the University of Amsterdam in 1994 (you can read an interview with her here). Currently the Aïda Paalman-de Miranda foundation, named in honor of Ietje Paalman-de Miranda, a former professor from the UvA of Surinamese origin, offers two students from the Anton de Kom University the possibility to do an exchange semester at the UvA. Building on Diletta's experience with establishing collaborations with the Global South, we decided to organize a visit to the Anton de Kom University to find ways to expand this connection!

*Some images from Paramaribo.*

During our stay we held several meetings with the teaching staff and students of the mathematics program at Anton de Kom University. These discussions centered around crucial topics such as structuring bachelor projects, offering joint advanced courses, and fostering research collaborations. It was a very inspiring experience to be in a country on the other side of the globe with a very different culture, and realize that we have so many ideas in common regarding mathematics education.

Overall, the atmosphere during the visit was very positive, and offered a good stepping stone to enhance the collaboration between the Anton de Kom University and the Mathematics Institute at the UvA. We listened with mutual respect to each other's ideas and the discussions converged rather quickly to small and concrete joint projects which could support them in achieving some of their goals. This trip was only the beginning of a new journey, with mathematics and education as its building blocks!

]]>Education and work shifted online. One positive change brought about by the lockdown that we enjoy to this day is the rise in the online delivery sector. Even though the world has opened up again, we have embraced online delivery, and at the moment, it looks like it’s made to stay for the years to come. Go down the memory lane, and compare for yourself - how frequently did you order groceries/food online before and after the Covid-19 lockdown?

Suppose you are returning home from a long day at work, and looking forward to the weekend. If you are an introvert, then on a Friday night, one might argue that there are only a handful of things more fun than eating a pizza while playing video games or binge-watching your favourite TV series. There are several pizza shops in your city, two of your usual go-to ones being ‘Papa Tom’s’ and ‘Pizza Shed’, with a preference for the former. You therefore login to your ’Papa Tom’s account and build a custom pizza with your favourite toppings (pineapple is not on the menu!). When checking out, you are provided an estimated waiting time of 55 minutes before delivery. No way! Why does it have to take so long? You are starving, and need the pizza asap, and you cannot wait even a second longer than your patience threshold of 30 minutes. Since 55 minutes is way too long to bear, you logout from ‘Papa Tom’s’ and login to the Pizza Shed website, and re-build your customer order. This time, you are notified an expected waiting time of 25 minutes, which you gladly accept.

Why was the expected waiting time at Papa Tom’s (55 minutes) so high? Probably because there were not enough chefs and/or delivery agents to expedite the service. As a result, they lost a potential customer in you. But what is important is that they never realized that you were a potential customer. You had to look for other alternatives, due to their unreasonable waiting time. Let us think of another situation where we encounter similar experiences - you go to a hair salon and before you even enter, you see the barbers tending to one person each, along with few other customers reading magazines on the couch. You estimate that it will be your turn after a long time, so you decide to go to another hair salon. It is highly probable that not just you, but several other people decided to go to another hair salon after observing the congestion in the shop at different moments during the day. And once again, the owner of the hair salon never realized how many potential customers they were losing. In all such instances, you represent someone who brings revenue to the service provider. But due to large waiting times, you abandon, without them even realizing that you have abandoned. We call this a *loss of opportunity*. This leads to customer dissatisfaction; but more importantly, is it possible that service providers do not even realize a loss of opportunity?

Take a look at Figures 1 and 2. They represent 2 distinct service systems. In particular, we can assume that Figure 1 corresponds to Papa Tom’s, and Figure 2 corresponds to Pizza Shed.

Figure 1: Papa Tom's service capacity

Figure 2: Pizza Shed's service capacity

The service capacity at Pizza Shed is 4 times that of the service capacity at Papa Tom’s, which essentially means that Pizza Shed processes work 4 times faster than Papa Tom’s. The blue lines represent the demand (how many customers want to receive service at any given time) which varies with time because in most practical systems, the demand fluctuates during the day. For instance, food delivery companies often find higher demands during lunch and dinner hours. The red line represents the number of customers who are actually using the service at a particular time.

Now we might wonder - why is the red line in Figure 1 so flat? Let us try to understand this phenomenon. Say it is 19:00 and a lot of people are trying to order a pizza. Since the demand is high, we expect a high waiting time. This discourages additional people after a certain point from joining. A consequence of this is that Papa Tom’s receives more or less the same number of orders at all times of the day, regardless of the demand, which can mislead the owners into believing that the demand for pizza is uniform throughout the day. Pizza Shed on the other hand with a quick service causes smaller waiting times, and therefore is able to cope up to some degree with the demand during peak hours and thereby generate a larger revenue.

We note again that the real danger lies not in the fact that Papa Tom’s is understaffed, but rather that is easily possible to remain ignorant of the actual demand during the day. Both the companies do not know the demand (the blue line). The only information that they possess is the red line, from which it is easy to assume incorrectly in Figure 1 that the demand is uniform during the day. To be able to serve more customers, Pizza Shed had to hire 3 additional servers than Papa Tom’s, and they need to be compensated as well! This means that there is a tradeoff between revenue that a company can generate, and the costs it incurs for the resources it has to keep functional. Sometimes, the resource costs exceed the additional revenue generated. This tradeoff can be studied once we know the true demand and we can then take care of the staffing.

We therefore conclude that getting an understanding of the true demand (the blue line) is crucial! It is easy to be misled if the realized demand is not analyzed well. There are numerous other real world services such as - ride sharing companies, hospitals, call centres, where similar analysis can be performed. A branch in operations research called queueing theory concerns with modelling and analyzing the performance of such systems. In this area of research, under certain mathematical assumptions such as the distribution of the interarrival times between arriving customers and their service times, the performance of service systems can be studied.

]]>The collaboration, reignited after the challenges posed by the pandemic, had its roots in a successful joint meeting in 2018 at the Indian Statistical Institute, Kolkata, where 70 participants engaged in fruitful discussions. Building on this momentum, several one-day workshops were organized in The Netherlands in 2017 and 2018, setting the stage for the latest gathering in Bangalore.

The ICTS campus, established in 2015, provided an ideal backdrop for the workshop. Nestled away from the bustling downtown, the campus exuded a serene atmosphere, boasting beautiful housing facilities and a welcoming canteen. The venue, with its quiet charm and adorned with blackboards, proved to be the perfect setting for discussions among participants.

A total of 45 attendees, including one-third from The Netherlands affiliated with NETWORKS, participated in the workshop. The agenda featured 20 talks and a comprehensive discussion session. The director of ICTS extended a warm welcome on the second day, acquainting the participants with the various facilities available.

This workshop presented an opportunity for European participants to meet their Indian colleagues, and, arguably more importantly, also for young Indian researchers to meet senior European researchers on the topic of the mathematics of networks. Such an exchange would not have been possible without this physical meeting and the vibrant discussions that it has sparked.

The workshop covered a diverse array of topics, delving into the static and dynamic properties of networks, geometric features, optimization and control issues, and the intersections of network science with physics, computer science, social sciences, and medical sciences.

Tuesday saw students presenting posters, followed by two-minute speeches and engaging discussions with the audience. Midway through the week, a strategic afternoon unfolded, focusing on major future questions in network science. Spearheaded by Remco van der Hofstad (TU/e), Rajesh Sundaresan (IISc), and Diego Garlaschelli (IMT, Lucca and Leiden), this two-hour animated discussion session sparked new research ideas and promising directions. How to decide what is a good network model? How can the mathematical networks community collaborate with its more applied counterparts to understand network statistics? How to use clever control to steer a network in a wanted direction? How can statistical physics ideas be fruitfully used in order to understand network modelling and functionality?

Wednesday morning offered a refreshing break as many participants embarked on a scenic walk to a nearby lake, culminating in a rejuvenating sip of coconut water from a local shop. Thursday brought the group together for a well-deserved evening of drinks and dinner at a pub, providing a delightful respite after four intensive days. The culinary journey throughout the workshop was a highlight, with the canteen offering delicious South Indian breakfasts, diverse lunch options, and satisfying dinners. For those seeking refreshment, the beautiful swimming pool provided a perfect escape.

During the workshop ongoing collaborations were strengthened and new collaborations were set up. Talks were followed by extensive discussion sessions, and for those eager to revisit the presentations, the ICTS YouTube channel offered a digital archive. Looking ahead, there are plans to organize a follow-up workshop in January 2026, promising another exciting chapter in the collaborative journey of NETWORKS.

]]>It's well-known how traditional publishing is cut-throat. In general, competition is fierce, margins are tight, and it is costly, specialised, painstaking work to source, polish, market, distribute high-quality material.

In startling contrast, the segment devoted to **scientific** publishing, with a global market capitalisation similar to that of the film industry, is booming. In one of the most inefficient markets in the world, the main players consistently report to their shareholders profit margins around 30%! Even more startlingly, despite all the most costly/valuable work done for journals on an **unpaid** basis by academics and despite decreasing infrastructure costs for publishers due to technological advances, such eye-watering profits are extracted purely out of the (government-funded) budgets of knowledge institutions and funding agencies worldwide.

To offer an idea of the scale, the annual Dutch share of the profit margin sent to just one of the big academic publishing houses (starting with the letter E) is currently in the ballpark of 5 million euro.

Clearly something has gone awry (since decades), but what are the alternatives?

A new journal project, Innovations in Graph Theory, was conceived against this backdrop. After extensive and meticulous preparations, this high-standard journal launched, with a stellar founding editorial board, in August 2023 at an established European conference in combinatorics. It is hosted within a French nonprofit publishing platform, Centre Mersenne, with startup costs supported through a grant of NWO.

Crucially IGT is a **diamond** open access journal. The term **diamond** means costs are incurred neither for contributors nor for readers. The journal is owned by a nonprofit organisation governed by editorial board members.

The diamond model of open access is a key plank in building up sustainable, fair access to scientific publishing. The essential idea for the IGT journal came about by asking the following: can a scientific professional in the field establish and/or maintain their career solely through publication in diamond journals? (One can obviously ask this too for fields other than graph theory.)

Within a chorus of closely-aligned, recent diamond journal launches/flips in discrete mathematics (Discrete Analysis, Advances in Combinatorics, Algebraic Combinatorics, Combinatorial Theory), the IGT initiative is aimed at further fostering the chances that a graph theorist can answer yes to the above question.

]]>If you see six people when you enter a class, it might be difficult to guess their personalities, but there is an amazing thing you can be sure of (mathematically prove): there are three of them who either all know each other, or who have never met. On the contrary, if there are only five people, there is a chance that you might fail to find such a triple: consider five people sitting around a circular table in a way that everybody knows only their two neighbors. Whoever three people you choose there will always be two people who know each other and two people who don't know each other, never all three.

*Figure 1: Five people sitting around a circular table.*

Why does this argument fail when you have six people sitting around a circular table? Think about this for a moment.

This simple observation due to Frank Ramsey in 1930 was the first step in mathematicians building a great theory on it ever since -- nowadays called Ramsey theory. It mainly concerns the structures emerging as the group gets larger. People in a class who may (or may not) have met before can be represented as a network of points that may (or may not) be linked to each other by an edge. There is a whole branch of mathematics that represents mathematical problems about networks in that way, called *graph theory*, and Ramsey theory is a subfield of graph theory.

Let us call a group of people *homogeneous* if either all know each other, or who have never met. Suppose instead of a group of three, you want to find a homogeneous group of size four. You can easily convince yourself that if, for instance, there are 10000 people in the class, this should be an easy task. But, what about the minimum size of the class in order to guarantee existence of such four people? The following example in the figure below shows you might fail if there are only 17 people:

*Figure 2:* *An example of a graph on 17 vertices with no homogeneous set of size four*

Indeed, it is the best example! It can be shown that one can always find a homogeneous group of size four in every group of 18 people.

Ready to be surprised? Mathematicians have no answer (yet) for the next obvious question:

**Question:** What is the minimum number of people to guarantee the existence of a homogeneous group of size five?

The question above is just one example of problems in graph theory which are innocent-looking but enormously difficult! Mathematicians are only able to establish that the answer should be a number between 43 and 48. How can this problem be not fully answered? I can hear you saying a computer would do the job, but it turns out that we need more than computations, which is quite beyond today's computer's abilities.

Let us start formalizing the concepts. In graph theory language, homogeneous groups correspond to substructures in graphs which are either all connected or all disconnected, respectively called *cliques* and *independent sets*. Letting be the minimum integer such that every graph on vertices has a homogeneous substructure of size , the previous discussion is mathematically equivalent to saying that and . Recall that what we could only say about was that , so it seems there is no hope for when , desperate! But, mathematicians never give up.

If something is quite difficult to determine exactly, it is natural to try to understand what it looks like by giving some good upper and lower bounds. The term *asymptotic behavior* for a function informally means the categorization of when grows towards infinity in terms of well-understood functions such as or . For the case of our problem, the behavior of when gets larger, the famous mathematician Paul Erdős, as the founder of the *probabilistic method*, used a probabilistic argument to show that . For the upper bound, Ramsey himself proved that by using a mathematical method called induction, still a huge gap!

The exact asymptotics of -- whether exists or not -- is still one of the biggest open problems in Ramsey theory, but now we are moving to the part in which we discuss another aspect of homogeneous structures. You can and look at proofs of aforementioned upper and lower bounds at the great reference *Proofs from the book, Chapter 45: Probability makes counting (sometimes) easy*. Also you can read Nicos Starreveld's article about the latest breakthrough about the upper bound for and further discussion.

Let us interpret the upper and lower bounds for from another perspective: shows that every graph on vertices has a homogeneous substructure of size . On the other hand, Erdős' result says that in an -vertex graph, you can avoid homogeneous substructures of size . The main message from these two facts can be summarized as follows:

- In an arbitrary graph, logarithmic sized homogeneous substructures are unavoidable.
- There are graphs whose largest homogeneous substructure has logarithmic size.

Erdős and his collaborator Andras Hajnal asked a question in a paper published in 1989 that is still quite open: Take any substructure you want, say . If a graph does not have a part that exactly looks like in terms of edges and non-edges, it is called -free. They conjectured that any -free graph has to have a polynomial size homogeneous substructure -- rather than only logarithmic size you would see in an arbitrary graph (note that for any fixed positive numbers and , becomes much larger than when gets larger.).

**Erdős-Hajnal conjecture**

For every graph , there exists a constant such that every

free graph on vertices has a homogeneous substructure of size .

At the first glance, it might not be intuitive why forbidding a substructure can drastically change the size of the largest homogeneous substructure in a graph compared to arbitrary graphs of the same size, but, of course, there are (mathematical) reasons for it. Maybe it is best to think about it first through an example. Let be a graph that contains three vertices and two edges, i.e. a substructure looks like a path of two edges, depicted in Figure 3 below.

*Figure 3: A path graph with two edges.*

Can we describe -free graphs? Let be a -free graph. Now we will try to understand how this affects the structure of . Consider a vertex in which has at least two other neighbors (vertices adjacent to through a link). Call them as , depicted in Figure 4 below.

*Figure 4: *The vertex and its neighbors.

If and are not linked for some , then observe that would be the same graph with (see Figure 5), a contradiction.

*Figure 5: The vertex and are not connected, then contains .*

Therefore, all the neighbors of should be connected to each other, and this is true for any vertex with at least two neighbors. If you think about it for a few more seconds, you can conclude that should be a disjoint union of parts such that each part is a clique -- all vertices should be linked to each other within the substructure (treat a single vertex as a clique as well). Suppose consists of many cliques. Observe that the largest part has at least vertices, so we find a homogeneous structure of size . On the other hand, we can choose one vertex from each part, they form again a homogeneous substructure because they are all disconnected. As a result, we can find a homogeneous substructure of size . Since , either or , so . So, we proved the conjecture when is path graph consisting of two edges! One case is done among infinitely many graphs. You can try to prove one more case yourself.

**Exercise:** Let be the triangle graph, as depicted in Figure 6. Prove that every -free graph on vertices has a homogeneous set of size .

*Figure 6: A triangle graph consisting of three vertices which are all connected.*

**Solution:** Let be a triangle-free graph on vertices. We will prove that has an independent set of size .

- Suppose there exists a vertex of at least neighbors. Then, any two such neighbors …

- Suppose all the vertices have less than neighbors. Consider the largest independent set , then …

You can see that the conjecture is so strong because it is about any graph . Desperately, since Erdős and Hajnal, there has been no significant progress with the exception examining some special cases of the graph , just like we did. However, even the case of being a path of four edges (see Figure 7) is still unknown!

*Figure 7: A path graph with four edges.*

You can say that even if we could solve this case, it would not mean a huge step towards the general solution of the conjecture (because there are infinitely many graphs). You would be right, but as mathematician we do what we can do!

You can look at the papers listed at the end of this article to have an idea about the recent progress towards the conjecture, which makes mathematicians more optimistic about a proof in the near future. But who knows, maybe we still need to wait for a long time...

- The Erdős-Hajnal Conjecture for Bull-free Graphs by Maria Chudnovsky and Shmuel Safra.
- The Erdős-Hajnal Conjecture by Maria Chudnovsky.
- Erdős-Hajnal for graphs with no 5-hole by Maria Chudnovsky, Alex Scott, Paul Seymour, and Sophie Spirkl.
- Towards the Erdős-Hajnal conjecture for $P_5$-free graphs by Pablo Blanco and Matia Bucic.
- Induced subgraph density I: A loglog step towards Erdős-Hajnal by Matia Bucic, Tung Nguyen, Alex Scott, and Paul Seymour.

This is the opening paragraph of the by-laws of the EMS (European Mathematical Society) Young Academy. From July 15th to 19th the European Congress of Mathematics (9ECM) will take place in Sevilla. The programme looks amazing!

On the website of the event we read:

The European Mathematical Young Academy organizes the following activities during the 9ECM:

**EMYA Lightning talks**

An opportunity for PhD students and early career researchers to present their research in a short, concise, and energetic format. Speakers are tasked with presenting the key-ideas and/or results of their research in just 5-minutes and with a maximum of 3 slides. Keep in mind that session chairs will be strict about the 5-minutes time limit.

To be eligible to give a lightning talk you must be a PhD student or early career researcher (from 2nd year PhD up to 3 years post PhD) and not be presenting elsewhere at the ECM.

Submissions for this activity must be sent through the “abstracts submission” form, selecting “EMYA Lightning talks” as the desired thematic session.

**EMYA ice breaking session**

An occasion for young people to get to know each other and connect with peers in an informal environment. The main target group of the event is PhD students and early stage researchers, especially if they have never participated in big conferences. During the session, there will be the opportunity to talk not only about Mathematics, and participants will be encouraged to interact through organised activities like games as well.

**Sustainability panel & group discussion**

The theme of sustainability of research life is growing in importance in the academic debate. Sustainability can be intended in different ways: in terms of mental health of researchers or, for example, focusing on the environmental point of view. This session is intended as an occasion to discuss in small groups about such themes, sharing our own experiences and ideas on how it is like to live and work in academia and what kind of actions can be taken to mitigate climate change.

**Young KWG**

Also in the Netherlands the mathematics association (KWG) is trying to create a community of early career mathematicians, by establishing the young division Young KWG. The main goal of Young KWG is to attract and support young mathematicians in the early stages of their careers. They aim to create a vibrant network in the Netherlands, a place where individuals can connect, interact, and learn from one another.

]]>Are you one of those people who always seems to pack just a bit too much for your car to handle when packing for a vacation? Then you’re probably familiar with the packing-your-car-Tetris game: given a number of items and bags, can you fit all of these in the back of your car? Some will argue that this is a fun game, for other this might be a straight-up nightmare. However: most will agree, this can be a pretty hard game to win. In this article, I will introduce you to the magical world of computational complexity and will let you *feel* what makes packing your car generally hard.

So why is packing your suitcases into the car so complex? Obviously, it is a 3-dimensional packing problem: we want to pack certain 3D items called suitcases into a 3D space called the trunk. I claim that this is even hard in 2 dimensions. For example, take the puzzle of Steward Coffin on the right with only five(!) pieces which need to be placed into a square tray. You can see these pieces as 2-dimensional suitcases that need to fit into a 2D space, namely the square.

Even if the pieces are just rectangles, for example in the figure on the left, this is an extremely hard problem. You can __try this for yourself here__!

If it is already hard to do this in two dimensions, I hope you agree three dimensions is even harder.

The Arithmeum museum in Bonn has an exposition on Chip Design, which can be partly viewed on their website. If you go to here and choose **Placement** and then **Game** you can play a 2D packing game with rectangles, which is part of this exposition. I would encourage you to start with only a few items and after each success increase the number of items to pack by two. Hopefully you’ll notice it will get a **lot** harder with each extra item!

When we play this game in only 1 dimension, we call it `bin packing’. In this case, we are given so-called bins with a certain capacity and items of a length that need to be distributed over these bins. This type of problem occurs for example when you are packing several suitcases for flying, where each suitcase can weigh at most 23kg. In business, this problem appears when distributing tasks (the items) over the workday of employees (which can be seen as the bins).

Even though Bin Packing is a problem played in one dimension, it remains hard to solve. Let us first look at a simple example, say that I have 6 items of lengths: 8, 7, 7, 5, 2, and 1, can we fit it into two `bins’ of length 15? Try to solve this in the interactive game below, you can move the items with your cursor.

Probably, you can solve this in a couple of seconds, either by just trying or by remarking that and that , so indeed this would fit.

In the following game you can try it yourself. You can choose the number of bins you want, and you have to put the items in them so that they can fit precisely.

However, what if I give you the following set of lengths of items: {4, 12, 17, 28, 34, 37, 49, 54, 59, 65, 96} and I want to distribute them over two bins of capacity 228? Can you do that? How much harder did the problem get, even though the number of items only doubled?

Well yes—and no. Of course we can use computers for this, however, a computer needs a set of instructions to follow, also called an *algorithm*. In other words, we need a structured way to find a solution. An example of an algorithm for Bin Packing with two bins could be:

* For all combinations of items *:

*compute whether all of**fits into one bin, and**compute whether all items***not**in fit into one bin.

*If both are yes: we found a way to distribute the items.If the answer is no for each combination of items*

In the example of and two bins of capacity 228, the algorithm will find the solution when it chooses as {4, 28, 34, 37, 59, 65}.

So, we can use a computer to solve Bin Packing. How long will the computer take to compute this?

This obviously depends on which computer you use and how fast it is: The computers that are produced today are many times faster than those that run on Windows XP from 2001. However, we **can** say something about the *number of computations* the algorithm (and therefore the computer) takes. The algorithms above will do two computations (namely a. and b.) per combination of items. So how many combinations of items are there?

Well, if is the number of items, then each item can be either a part of, or not a part of any combination. This gives us a total of different combinations to check. Hence, the number of computations scales **exponentially** in the number of items. If you recall exponential growth from corona, then you’re probably aware that this is not a good thing. Because of this combinations we check, the number of combinations **doubles **if we add one item to the problem (see also the table on the right). So, if my computer takes 1 second to do Bin Packing with 10 items, it will take for Bin Packing with 20 items around seconds which is around 17 **minutes**. And if you want to compute this for 50 items, well…. It would take about 35.678.376 **years**. I don’t think we would have time to wait for that. So yes, we are able to compute it using computers, but only up to a small number of items.

1 | 2 |

2 | 4 |

3 | 8 |

4 | 16 |

5 | 32 |

10 | 1.024 |

11 | 2.048 |

12 | 4.096 |

13 | 8.192 |

14 | 16.384 |

15 | 32.768 |

16 | 65.536 |

17 | 131.072 |

18 | 262.144 |

19 | 524.288 |

20 | 1.048.576 |

Well- not really! Or at least: not that we know of. Of course, the fact that we do not know of an efficient algorithm, does not necessarily prove that it does not exist.

For two bins, the best algorithms need about computations. For three bins, the best algorithms need about computations. For any other number of bins, there was a recent breakthrough by my co-authors and me, where we present an algorithm that needs about computations.

However, the computer science community believes that efficient algorithms (those that only need for example around or computations) do not exist for Bin Packing. This belief is based on a hypothesis referred to as “”; If you would be able to design an efficient algorithm for Bin Packing it would show that . Why does that matter?

Well first of all, you would earn a million dollars, as showing is one of the millennium price problems. In theory, which would be nice. However, it would also imply that we can solve **many other** **problems** efficiently. That may sound wonderful in theory, until you realize that our security systems are then suddenly also easy to crack! So, giving an efficient algorithm for Bin Packing would actually have a lot of impact on our society.

No! There are actually many, many so-called -hard problems such as Bin Packing. Giving an efficient algorithm for any of these NP-hard problems would also show and therefore have the same impact. These are often fun games to solve with many applications. Let me give you some examples of such -hard problems.

**Steiner Tree:** In this problem, you’re given points that need to be connected with using as few connecting lines as possible. You can play this game here and then clicking ‘Routing’. The website explains one of the applications of this problem: connecting parts of a chip while using the least amount of connecting fluids.

**Travelling Salesman Problem:** In this problem, you want to find the fastest route visiting a set of cities. You can play this game here. You can encounter this problem for example when a Picnic or PostNL car has to visit a set of customers.

**3-coloring:** In this problem, you’re given a set of points and lines between the points. You need to color the points red, blue or green, such that for each line, the endpoints receive different colors.

There is actually some good news: not all packing problems are hard! Sometimes the solution is actually relatively easy to find, and we humans seem to be remarkably good at finding these types of structures. So: maybe you’re lucky and with a bit of hard Tetris work, you can manage to pack your car. But if you can’t, there is actually a really easy solution:

**Just pack a bit less next time.**

Thanks for reading this article about complexity theory. If you’re searching for a subject for a school project or a 'profielwerkstuk’ related to computer science or mathematics, I can recommend studying any of the problems above, matching problems, or for example puzzles like Sudoku. Understanding (the complexity of) the problem, being able to implement/compare some of the algorithms for it, and looking for applications for the problem can all be part of the project.

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