(Header image generated from here).

**ECDSA: Elliptic Curve Signatures**

This is the second post on this topic – following on from the first post here. Read that first for more of the maths behind this! In this post I’ll look at this from a computational angle – and make a simple Python code to create and verify Elliptic Curve Signatures.

**Why Elliptical Curve Signatures?**

Say I create 100 MATHSCOINS which I sell. This MATHSCOIN only has value if it can be digitally verified to be an original issued by me. To do this I share some data publicly – this then allows anyone who wants to check via its digital signature that this is a genuine MATHSCOIN. Once you understand this idea you can (in theory!) create your own digital currency or NFT – complete with a digital signature that allows anyone to check that it has been issued by you.

**Python code**

This code will revolve around solutions mod M to the following elliptical curve:

We can run a quick Python code to find these solutions for a defined M:

This Python code then needs to use the algorithms for repeated addition of the base pair. It then needs to store all the coordinate pairs in a list (one list for the x coordinates and one for the y coordinates). These can then follow the algorithm for creating the digital signature. Note that we need to define the mod of the curve (M), the starting base pair (a,b), the order of the base pair (n), our data to digitally sign (z1), our private key (k1) and a public key (k2).

**The full code for digital signatures**

**Running this code**

I have put this code online at Replit.com here – so you can see how it works. It should look something like this:

**Checking a digital signature is genuine**

We might also want to work backwards to check if a digital signature is correct. The following code will tell us this – as long as we specify all the required variables below. Note we need the digital signature (s1, s2) as well as (r1,r2) – which is worked out by the previous code.

**Running this code**

You can run this code here – again on Replit.com. You should see something like this:

**Try it yourself!**

To create your own digital signatures you need to find a mod M and a base pair with order n, such that both M and n are prime. Remember you can use this site to find some starting base pairs mod M. Here are some to start off with

(1)

M = 907. Base pair = (670,30). n = 967

(2)

M = 79. Base pair = (60, 10). n = 67

(3)

M = 97. Base pair = (85, 92). n = 79

(4)

M = 13. Base pair = (8,8). n = 7

Can you run the code to create a digital signature, and then run the verification code to check that it is indeed genuine?

]]>**The mathematics behind blockchain, bitcoin and NFTs.**

If you’ve ever wondered about the maths underpinning cryptocurrencies and NFTs, then here I’m going to try and work through the basic idea behind the Elliptic Curve Digital Signature Algorithm (ECDSA). Once you understand this idea you can (in theory!) create your own digital currency or NFT – complete with a digital signature that allows anyone to check that it has been issued by you.

Say I create 100 MATHSCOINS which I sell. This MATHSCOIN only has value if it can be digitally verified to be an original issued by me. To do this I share some data publicly – this then allows anyone who wants to check via its digital signature that this is a genuine MATHSCOIN. So let’s get into the maths! (Header image generated from here).

**Elliptical curves**

I will start with an elliptical curve and a chosen prime mod (here we work in mod arithmetic which is the remainder when dividing by a given mod). For this example I will be in mod 13 and my curve will be:

First I will work out all the integer solutions to this equation. For example (7,5) is a solution because:

The full set of integer solutions is given by:

Now we define addition of 2 non equal points (p_1, p_2) and (q_1, q_2) on the curve mod M by the following algorithm:

And we define the addition of 2 equal points (p_1, p_2) on the curve mod M by the following algorithm:

So in the case of (8,8) if we want to calculate (8,8) + (8,8) this gives:

This is a little tedious to do, so we can use an online generator here to calculate the full addition table of all points on the curve:

This shows that (say) (7,5) + (8,5) = (11,8) etc.

I can then chose a base point to find the order of this point (how many times it can be added to itself until it reaches the point at infinity). For example with the base point (8,8):

We can also see that the order of our starting point A(8,8) is 7 because there are 7 coordinate points (including the point at infinity) in the group when we calculate A, 2A, 3A…

**ECDSA: Elliptic Curve Signatures**

So I have chosen my curve mod M (say):

And I choose a base point on that curve (p_1, p_2) (say):

And I know the order of this base point is 7 (n=7). (n has to be prime). This gives the following:

I now chose a private key k_1:

Let’s say:

This is super-secret key. I never share this! I use this key to generate the following point on the curve:

I can see that 5(8,8) = (11,5) from my table when repeatedly adding (8,8) together.

Next I have some data z_1 which I want to give a digital signature to – this signature will show anyone who examines it that the data is authentic, has been issued by me and has not been tampered with. Let’s say:

I choose another integer k_2 such that:

Let’s say:

I am now ready to create my digital signature (s_1, s_2) by using the following algorithm:

Note, dividing by 2 is the same as multiplying by 4 in mod 7 (as this is the multiplicative inverse).

I can then release this digital signature alongside my MATHSCOIN (represented by the data z_1 = 100). Anyone can now check with me that this MATHSCOIN was really issued by me.

**Testing a digital signature**

So someone has bought a MATHSCOIN direct from me – and later on wants to sell to another buyer. Clearly this new buyer needs to check whether this is a genuine MATHSCOIN. So they have check the digital signature on the data. To allow them to do this I can share all the following data (but crucially not my private key):

This gives:

To verify someone then needs to do the following:

To verify that the data z_1 has a valid digital signature we need:

So with the shared data we have:

This verifies that the data had a valid digital signature – and that the MATHSCOIN is genuine! This is basically the foundation of all digital assets which require some stamp of authenticity.

In real life the numbers chosen are extremely large – private keys will be 256 digits long and primes very large. This makes it computationally impossible (at current speeds) to work out a private key based on public information, but still relatively easy to check a digital signature for validity.

I have also made some simple Python code which will provide a digital signature as well as check that one is valid. You can play around with these codes here:

(1) Digital signature code, (2) Checking a digital signature for validity

So time to create your own digital currency!

]]>**Finding planes with radar**

PlusMaths recently did a nice post about the link between ellipses and radar (here), which inspired me to do my own mini investigation on this topic. We will work in 2D (with planes on the ground) for ease of calculations! A transmitter will send out signals – and if any of these hit an object (such as a plane) they will be reflected and received by a receiver. This locates the object as somewhere on the ellipse formed with the receiver and transmitter as the 2 foci. When we add a second receiver as shown above then if both receivers receive a signal, then we can narrow down the location of the object as the intersection of the 2 ellipses.

So, for this mini exploration I wanted to find the equations of 2 ellipses with a shared focus so that I could plot them on Desmos. I then would be able to find the intersection of the ellipses in simple cases when both ellipses’ major axis lies on the x axis.

**Defining ellipses**

For an ellipse centred at the origin shown above, with foci at c and -c we have:

where c is linked to a and b by the equation:

**Rotating an ellipse**

Next we can imagine a new ellipse in a coordinate system (u,v)

This coordinate system is created by rotating the x and y axis by an angle of theta radians anticlockwise about the origin. The following matrix transformation achieves this rotation:

This therefore gives:

and we can substitute this into our new coordinate system to give:

When we plot this we can therefore rotate our original ellipse by any given theta value:

We can use basic Pythagoras to see that the focus point c will become the point c1 shown above with coordinates:

By the same method we can see that the point c2 will have coordinates:

**Transformation**

Next we want to translate this new ellipse so that it shares a focus point with our original green ellipse. To do this we need to translate the point c2 to the point c. This is given by the translation:

So we can therefore translate our ellipse:

Which becomes:

When we plot this we get:

This then gives the 2nd ellipse in blue which does indeed share a focus point at c:

**Finding points of intersection**

The coordinates of when the 2 ellipses intersect is given by the solution to:

This looks a bit difficult! So let’s solve an easier problem – the points of intersection when the theta value is 0 (i.e when the ellipses both lie on the x axis). This simplifies things to give:

and we can find the y coordinates by substituting this into the original ellipse equation.

So the coordinates of intersection are given by:

So – in the above case we would be able to narrow down the location of the plane to 2 locations. With a 3rd ellipse we could pinpoint the location exactly.

]]>**Proving Pythagoras Like Einstein?**

There are many ways to prove Pythagoras’ theorem – Einstein reputedly used the sketch above to prove this using similar triangles. To keep in the spirit of discovery I also just took this diagram as a starting point and tried to prove this myself, (though Einstein’s version turns out to be a bit more elegant)!

**Step 1: Finding some links between triangles**

We can see that our large right angled triangle has sides *a,b,c* with angles alpha and beta. Hopefully it should also be clear that the two smaller right angled triangles will also have angles alpha and beta. Therefore our triangles will all be similar. It should also be clear that the area of the 2 small triangles will be the same as the area of the large triangle.

**Step 2: Drawing a sketch to make things clearer:**

It always helps to clarify the situation with some diagrams. So, let’s do that first.

**Step 3: Making some equations**

As the area of the 2 small triangles will be the same as the area of the large triangle this gives the following equation:

We also can make the following equation by considering that triangles 2 and 3 are similar

We can now substitute our previous result for x into this new equation (remember our goal is to have an equation just in terms of* a,b,c* so we want to eliminate *x* and *y* from our equations).

We can also make the following equation by considering that triangles 1 and 2 are similar:

And as before, our goal is to remove everything except a,b,c from these equations, so let’s make the substitution for y using our previous result:

And if by magic, Pythagoras’ theorem appears! Remember that the original *a,b,c *related to any right angled triangle with hypotenuse *c, *so we have proved that this equation must always be true for right angled triangles.

You can explore some other ways of proving Pythagoras here. Which is the most elegant?

]]>These now have some great free resources for students to help them with the IB maths course – including full course notes, formula books, Paper 3s, an Exploration guides and a great mind-map. Make sure to check these all out to get some excellent support for the IB maths course.

These now have over 25 worksheets, investigations, paper 3s, treasure hunts and more resources – both with question pdfs and markscheme pdfs. I’ve added a lot of enriching activities that would support explorations and paper 3 style problems and also put a selection of some excellent other resources from IB teachers too.

So be sure to check these both out!

]]>**Finding the average distance in a polygon**

Over the previous couple of posts I’ve looked at the average distance in squares, rectangles and equilateral triangles. The logical extension to this is to consider a regular polygon with sides 1. Above is pictured a regular pentagon with sides 1 enclosed in a 2 by 2 square. The points N and O represent 2 randomly chosen points which we find the distance between. On average what is the distance between these randomly chosen points N and O?

**Starting with a hexagon**

It’s a little easier to start with a hexagon as we get some nicer coordinate points. So, our first challenge is to find the coordinates of a regular hexagon with sides 1. Luckily we can use the complex roots of unity to do this. We start by finding the 6th roots of unity and then converting these to coordinates in an Argand diagram:

This then allows us to plot the following:

We can then work out the inequalities which define the inside of the hexagon when we generate points within the 2×2 square centred at (0,0). This gives:

We can then run the following code to find the average distance:

This gives the following result:

We can check this result as the exact value is:

which is 0.8262589495. So we can see we are accurate here to 3 sf.

**Pentagon**

For the pentagon we can find the coordinates by finding the 5th roots of unity:

We then need to scale all coordinate points by a factor, because in a pentagon the distance from the centre to the points is not 1 (as is the case in roots of unity). We can find the distance from the centre to the edge of a pentagon by the following trigonometry:

So, when we scale all coordinate points by this factor we get:

And we can then do the same method as before and run the following Python code:

This gives:

**n-sided polygon**

We can now consider an n-sided polygon with sides 1. Let’s start with the values we’ve found for an equilateral triangle (0.364), a square (0.522), a pentagon (0.697) and a hexagon (0.826.

When we plot these they appear to follow a linear relationship:

average distance = 0.14n

We can check that this is correct by considering the fact that an n sided polygon will approximate a circle when n gets large. So an n sided polygon with sides length 1 can be approximated by a circle with circumference n. This allows us to work out the radius.

We can then substitute this into the equation for the average distance of 2 points in a circle.

So we would expect the average distance between 2 points in a regular polygon of sides 1 to approach the equation (as n gets large):

average distance = 0.144101239n

And we’ve finished! Everything cross-checks and works nicely. We’ve been able to use a mixture of complex numbers, geometry, coding and trigonometry to achieve this result.

]]>**Finding the average distance in an equilateral triangle**

In the previous post I looked at the average distance between 2 points in a rectangle. In this post I will investigate the average distance between 2 randomly chosen points in an equilateral triangle.

**Drawing a sketch.**

The first step is to start with an equilateral triangle with sides 1. This is shown above. I sketched this using Geogebra – and used some basic Pythagoras to work out the coordinates of point C.

I can then draw a square of sides 1 around this triangle as shown above. I’m then going to run a Python program to randomly select points and then work out the distance between them – but I need to make sure that the 2 points chosen are both inside this triangle. For this I need to work out the equation of the line AC and CB.

Using basic coordinate geometry we can see that the line AC has equation y = √3x. We want the inequality y < √3x so that we are on the correct side of this line.

The line BC has equation y = -√3x + √3. Therefore the triangle must also satisfy the inequality y < -√3x + √3.

I can then run the following code on Python, with finds the average distance between points (a,c) and (b,d) both within the unit square but also subject to the 2 inequality constraints above.

When this is run it performs 999,999 trials and then finds the average distance. This returns the following value:

So we can see that the average distance is just over a third of a unit.

**Finding the average distance of an equilateral triangle of length n.**

We can then draw the sketch above to find the equation of lines AC and CB for an equilateral triangle with lengths n. This leads to the following inequalities:

y < √3x

y < -√3x + √3n

So we can then modify the code as follows:

This then returns the average distances for equilateral triangles of sizes 1 to 10.

And when we plot this on Desmos we can see that there is a linear relationship:

The regression line has gradient 0.36 (2sf) so we can hypothesise that for an equilateral triangle of size *n*, the average distance between 2 points is approximately 0.36*n*.

**Checking the maths**

I then checked the actual equation for the average distance between 2 points in an equilateral triangle of sides n:

This gives us:

So we can see that we were accurate to 2 significant figures. So this is a nice mixture of geometry, graphing and computational power to provide a result which would be otherwise extremely difficult to calculate.

]]>**What is the average distance between 2 points in a rectangle?**

Say we have a rectangle, and choose any 2 random points within it. We then could calculate the distance between the 2 points. If we do this a large number of times, what would the average distance between the 2 points be?

**Monte Carlo method**

The Monte Carlo method is perfect for this – we can run the following code on Python:

This code will find the average distance between 2 points in a 10 by 10 square. It does this by generating 2 random coordinates, finding the distance between them and then repeating this process 999,999 times. It then works out the average value. If we do this it returns:

This means that on average, the distance between 2 random points in a 10 by 10 square is about 5.21.

**Generalising to rectangles**

I can now see what happens when I fix one side of the rectangle and vary the other side. The code below fixes one side of the rectangle at 1 unit, and then varies the other side in integer increments. For each rectangle it then calculates the average distance.

This then returns the first few values as:

This shows that for a 1 by 1 square the average distance between two points is around 0.52 and for a 1 by 10 rectangle the average distance is around 3.36.

**Plotting some Desmos graphs**

Because I have included the comma in the Python code I can now copy and paste this straight into Desmos. The dotted green points show how the average distance of a 1 by x rectangle changes as x increases. I then ran the same code to work out the average distance of a 10 by x rectangle (red), 20 by x rectangle (black), 30 by x rectangle (purple) and 100 by x rectangle (yellow).

We can see if we continue these points further that they all appear to approach the line y = 1/3 x (dotted green). This is a little surprising – i.e when x gets large, then for any n by x rectangle (with n fixed), an increase in x by one will tend towards an increase in the average distance by 1/3.

**Heavy duty maths!**

There is actually an equation that fits these curves – and will give the average distance, a(X) between any 2 points in a rectangle with sides a and b (a≥b). Here it is:

I added this equation into Desmos, by changing the a to x, and then adding a slider for b. So, when I set b=1 this generated the case when the side of a rectangle is fixed as 1 and the other side is increased:

Plotting these equations on Desmos then gives the following:

Pleasingly we can see that the points created by our Monte Carlo method fit pretty much exactly on the lines generated by these equations. By looking at these lines at a larger scale we can see that they do all indeed appear to be approaching the line y = 1/3 x.

**General equation for a square**

We can now consider the special case when a=b (i.e a square). This gives:

Which we can simplify to give:

We can see therefore that a square of side 1 (a=1) will have an average distance of 0.52 (2dp) and a square of side 10 (a=10) will have an average distance of 5.21 – which both agree with our earlier results.

]]>**Plotting Pi and Searching for Mona Lisa**

This is a very nice video from Numberphile – where they use a string of numbers (pi) to write a quick Python Turtle code to create some nice graphical representations of pi. I thought I’d quickly go through the steps required for people to do this by themselves.

Firstly you can run the Turtle code on trinket.io. If you type the above code this will take the decimal digits of pi one at a time and for each one move forward 10 steps and then turn by 36 degrees times by that digit. So for example the 1 will lead to a right turn of 36 degrees and the 4 will lead to a right turn of 36 x 4 = 144 degrees.

Next it would be nice to have more digits of pi to paste in rather than type. So we can go to the onlinenumbertools website and generate as many digits of pi as we want. Select them to be comma separated and also to not include the first digit 3. You can then copy and paste this string in place of the 1,4,1 in the code above.

**1000 digits of pi**

If we run this program after pasting the first 1000 digits of pi we get (after waiting a while!) the above picture. There are a number of questions that they then raise in the video – if this program was ran infinitely long would the whole screen eventually be black? Would this create every possible image that can be created by 36 degree turns? Would it be possible to send this picture (say to an alien civilization) and for the recipient to be able to reverse engineer the digits of pi?

**2000 digits of pi**

If you increase the digits of pi to around 2000 you get the above picture. The graph spends a large time in the central region before finally “escaping” to the left. It then left my screen at the top.

**3000 digits of pi**

We can see that the turtle “returned” from off the top of the screen and then joined back up with the central region. This starts to look like a coastline – maybe the south of the UK!

**Different bases: Base 3**

We can consider the digits of pi in base three – which means that they are all equivalent to 0,1,2. This means that we can use these to specify either 0 degree, 120 degree or 240 degree turns. We can change the code as shown above to achieve this. Note the i%3 gives i mod 3. For example if the digit is 8, then 8 mod 3 is 2 (the remainder when 8 is divided by 3) and so this would turn 120 x 2 = 240 degrees.

This then creates a pattern which looks similar to the Sierpinski triangle fractal design:

**Base 4**

Using a similar method, we can create the following using a base 4 design:

This creates what looks like a map layout.

**Base 5:**

In base 5 the turtle quickly departed from my screen! With turns of 72 we don’t see the tessellating shapes that we do with base 3 and 4.

**Base 6:**

With a 60 degree turn we can now see a collection of equilateral triangles and hexagons.

You can explore with different numbers and different bases to see what patterns you can create!

]]>**Witness Numbers: Finding Primes**

The Numberphile video above is an excellent introduction to primality tests – where we conduct a test to determine if a number is prime or not. Finding and understanding about prime numbers is an integral part of number theory. I’m going to go through some examples when we take the number 2 as our witness number. We have a couple of tests that we conduct with 2 – and for all numbers less than 2047 if a number passes either test then we can guarantee that it is a prime number.

**Miller-Rabin test using 2 as a witness number:**

We choose an odd number, n >2. First we need to write it in the form:

Then we have to conduct a maximum of 2 different tests:

If either of the above are true then we have a prime number.

**Testing whether n = 23 is prime.**

First we need to write 23 in the following form:

Next we need to check if the following is true:

Remember that mod 23 simply means we look at the remainder when we divide by 23. We can do this using Wolfram Alpha – but in this case let’s see how we could do this without a calculator:

Therefore this passes the test – and we can say that it is prime.

**Testing whether 1997 is prime**

For 1997 we have:

So we need to first test if the following is true:

However using Wolfram Alpha we get:

So this fails the first part of the test.

Trying the second part of the test, we need:

We have already tested the case when r=0 (this gives the earlier result), so just need to look at what happens when r=1. Again we use Wolfram Alpha to get:

This passes the 2nd part of the test and so confirms that 1997 is prime.

**What happens with 2047?**

2047 is not prime as we can write it as 2 x 3 x 11 x 31. However it is the first number for which the witness 2 gives a false positive (i.e we get a positive result even though it is not prime). We write 2047 as:

But we do indeed get:

So we can call 2047 a pseudoprime – it passes this prime number test but is not actually prime.

**Larger primes**

For numbers larger than 2047 you can combine witnesses – for example if you use both 2 and 3 as your witness numbers (and regard a positive result as at least one of them returning a positive result) then this will find all primes for n < 1,373,653.

More interestingly for extremely large numbers you can use this test to provide a probability estimate for the likelihood that a number is prime. Lots to explore here!

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