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**The Telephone Numbers – Graph Theory**

The telephone numbers are the following sequence:

1, 1, 2, 4, 10, 26, 76, 232, 764, 2620, 9496…

(where we start from n=0).

This pattern describes the total number of ways which a telephone exchange with n telephones can place a connection between pairs of people.

To illustrate this idea, the graph below is for n=4. This is when we have 10 telephones:

Each red line represents a connection. So the first diagram is for when we have no connections (this is counted in our sequence). The next five diagrams all show a single connection between a pair of phones. The last three diagrams show how we could have 2 pairs of telephones connected at the same time. Therefore the 4th telephone number is 10. These numbers get very large, very quickly.

**Finding a recursive formula**

The formula is given by the recursive relationship:

**T(n) = T(n-1) + (n-1)T(n-2)**

This means that to find (say) the 5th telephone number we do the following:

**T(5) = T(5-1) + (5-1)T(5-2)**

**T(5) = T(4) + (4)T(3)**

**T(5) = 10 + (4)4**

**T(5) = 26**

This is a quick way to work out the next term, as long as we have already calculated the previous terms.

**Finding an nth term formula
**

The telephone numbers can be calculated using the nth term formula:

This is going to be pretty hard to derive! I suppose the first step would start by working out the total number of connections possible between n phones – and this will be the the same as the graphs below:

These clearly follow the same pattern as the triangular numbers which is 0.5(n² +n) when we start with n = 1. We can also think of this as n choose 2 – because this gives us all the ways of linking 2 telephones from n possibilities. Therefore n choose 2 also generates the triangular numbers.

But then you would have to work out all the permutations which were allowed – not easy!

Anyway, as an example of how to use the formula to calculate the telephone numbers, say we wanted to find the 5th number:

We have n = 5. The summation will be from k = 0 and k = 2 (as 5/2 is not an integer).

Therefore T(5) = 5!/(2^{0}(5-0)!0!) + 5!/(2^{1}(5-2)!1!) + 5!/(2^{2}(5-4)!2!)

T(5) = 1 + 10 + 15 = 26.

**Finding telephone numbers through calculus**

Interestingly we can also find the telephone numbers by using the function:

y = e^{0.5x2+x}

and the nth telephone number (starting from n = 1) is given by the nth derivative when x = 0.

For example,

So when x = 0, the third derivative is 4. Therefore the 3rd telephone number is 4.

The fifth derivative of the function is:

So, when x =0 the fifth derivative is 26. Therefore the 5th telephone number is 26.

If you liked this post you might also like:

Fermat’s Theorem on the Sum of two Squares – A lesser known theorem from Fermat – but an excellent introduction to the idea of proof.

Unbelievable: 1+2+3+4…. = -1/12 ? A result that at first glance looks ridiculous – and yet can be shown to be correct. How?

**The Chinese Postman Problem**

*There is a fantastic pdf resource from Suffolk Maths which goes into a lot of detail on this topic – and I will base my post on their resource. Visit their site for a more in-depth treatment.*

The Chinese Postman Problem was first posed by a Chinese mathematician in 1962. It involved trying to calculate how a postman could best choose his route so as to mimise his time. This is the problem that Kuan Mei-Ko tried to solve:

How could a postman visit every letter on the graph in the shortest possible time?

Solving this requires using a branch of mathematics called graph theory, created by Leonard Euler. This mathematics looks to reduce problems to network graphs like that shown above. Before we can solve this we need to understand some terminology:

Above we have 3 graphs. A graph which can be drawn without taking the pen off the paper or retracing any steps is called *traversable (*and has a Euler trail*). *Graph 1 is not traversable. Graph 2 is traversable as long as you start at either A or D, and Graph 3 is traversable from any point that you start. It turns out that what dictates whether a graph is traversable or not is the order of their vertices.

Looking at each letter we count the number of lines at the vertex. This is the order. For graph 1 we have 3 lines from A so A has an order of 3. All the vertices on graph 1 have an order of 3. For graph 2 we have the orders (from A, B, C, D, E in turn) 3, 4, 4, 3, 2. For graph 3 we have the orders 4,4,4,4,2,2.

This allows us to arrive at a rule for working out if a graph is traversable.

*If all orders are even then a graph is traversable. If there are 2 odd vertices then we can find a traversable graph by starting at one of the odd vertices and finishing at the other. We need therefore to pair up any odd vertices on the graph.
*

Next we need to understand how to pair the odd vertices. For example if I have 2 odd vertices, they can only be paired in one way. If I have 4 vertices (say A,B,C,D) they can be paired in 3 different ways (either AB and CD or AC and BD or AD and BC) . The general term rule to calculate how many ways n odd vertices can be paired up is (n-1) x (n-3) x (n-5) … x 1.

So now we are ready to actually solve the Chinese Postman Problem. Here is the algorithm:

So, we can now apply this to the Chinese Postman Problem below:

Step 1: We can see that the only odd vertices are A and H.

Step 2: We can only pair these one way (AH)

Step 3 and 4: The shortest way to get from A to H is ABFH which is length 160. This is shown below:

Step 5 and 6: The total distance along all the lines in the initial diagram is 840m. We add our figure of 160m to this. Therefore the optimum (minimum) distance it is possible to complete the route is 1000m.

Step 7: We now need to find a route of distance 1000m which includes the loop ABFH (or in reverse) which starts and finishes at one of the odd vertices. One solution provided by Suffolk Maths is ADCGHCABDFBEFHFBA. There are others!

**The Bridges of Konigsburg**

Graph theory was invented as a method to solve the Bridges of Konigsburg problem by Leonard Euler. This was a puzzle from the 17oos – Konigsburg was a Russian city with 7 bridges, and the question was, could anyone walk across all 7 without walking over any bridge twice. By simplifying the problem into one of connected lines, Euler was able to prove that this was in fact impossible.

If you like this post you might also like:

Knight’s Tour – This puzzles dates over 1000 years and concerns the ways in which a knight can cover all squares on a chess board.

Game Theory and Tic Tac Toe – Tic Tac Toe has already been solved using Game Theory – this topic also brings in an introduction to Group Theory.

**The Telephone Numbers – Graph Theory**

The telephone numbers are the following sequence:

1, 1, 2, 4, 10, 26, 76, 232, 764, 2620, 9496…

(where we start from n=0).

This pattern describes the total number of ways which a telephone exchange with n telephones can place a connection between pairs of people.

To illustrate this idea, the graph below is for n=4. This is when we have 10 telephones:

Each red line represents a connection. So the first diagram is for when we have no connections (this is counted in our sequence). The next five diagrams all show a single connection between a pair of phones. The last three diagrams show how we could have 2 pairs of telephones connected at the same time. Therefore the 4th telephone number is 10. These numbers get very large, very quickly.

**Finding a recursive formula**

The formula is given by the recursive relationship:

**T(n) = T(n-1) + (n-1)T(n-2)**

This means that to find (say) the 5th telephone number we do the following:

**T(5) = T(5-1) + (5-1)T(5-2)**

**T(5) = T(4) + (4)T(3)**

**T(5) = 10 + (4)4**

**T(5) = 26**

This is a quick way to work out the next term, as long as we have already calculated the previous terms.

**Finding an nth term formula
**

The telephone numbers can be calculated using the nth term formula:

This is going to be pretty hard to derive! I suppose the first step would start by working out the total number of connections possible between n phones – and this will be the the same as the graphs below:

These clearly follow the same pattern as the triangular numbers which is 0.5(n² +n) when we start with n = 1. We can also think of this as n choose 2 – because this gives us all the ways of linking 2 telephones from n possibilities. Therefore n choose 2 also generates the triangular numbers.

But then you would have to work out all the permutations which were allowed – not easy!

Anyway, as an example of how to use the formula to calculate the telephone numbers, say we wanted to find the 5th number:

We have n = 5. The summation will be from k = 0 and k = 2 (as 5/2 is not an integer).

Therefore T(5) = 5!/(2^{0}(5-0)!0!) + 5!/(2^{1}(5-2)!1!) + 5!/(2^{2}(5-4)!2!)

T(5) = 1 + 10 + 15 = 26.

**Finding telephone numbers through calculus**

Interestingly we can also find the telephone numbers by using the function:

y = e^{0.5x2+x}

and the nth telephone number (starting from n = 1) is given by the nth derivative when x = 0.

For example,

So when x = 0, the third derivative is 4. Therefore the 3rd telephone number is 4.

The fifth derivative of the function is:

So, when x =0 the fifth derivative is 26. Therefore the 5th telephone number is 26.

If you liked this post you might also like:

Fermat’s Theorem on the Sum of two Squares – A lesser known theorem from Fermat – but an excellent introduction to the idea of proof.

Unbelievable: 1+2+3+4…. = -1/12 ? A result that at first glance looks ridiculous – and yet can be shown to be correct. How?

**IB Revision**

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

The Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions. This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.