Topology

Real Mathematics – Strange Worlds #18

Every year in December, each city changes drastically. Suddenly we find ourselves surrounded by decorations that remind us of the upcoming new year.

Steve the teacher starts to decorate his classrooms for the new year like he does every year. Though, Steve the teacher set his mind on using new year decorations for his mathematics lessons.

New Year Decorations Game (N.Y.D.G.)

Steve’s creation N.Y.D.G. is a multiplayer game. This is why the game is played in knockout stages/rounds. The winner of the game wins the new year decorations and gets to decorate the classroom as he/she wishes.

Content of N.Y.D.G.

In each knockout round, students are given 4 decorations as follows:

Players wind the decorations one another.
The winding procedure should be done secretly from the opponent.
Each player has at most four moves for winding.

Let’s use an example to explain what a “move” means during the winding procedure.

Assume that the first move is made with the red decoration as follows:

This counts as one move. The red one undergoes the blue and green decorations in this move. Let the next two moves are as follows:

In the second move, the yellow decoration undergoes the green and red ones, while the blue one passes over the green and yellow decorations. The illustration (up-right) shows us how the winding looks after these 3 moves.

In the end, winding gives us a braid.

The Goal of The Game

In any round, to knock your opponent out, you should solve the braid of your opponent faster than your opponent solves yours. (Solving a braid means, bringing the decorations to their first state. For instance, in the example given up, the first state is yellow-green-blue-red in order.)

Braids

Braids have a very important part in daily life. We encounter them not just in new year decorations, but also in a piece of cheese, a hairstyle, a basket or even in a bracelet:

In case you wish to understand what braids mean in mathematics; one can take a look at Austrian mathematician Emil Artin’s works from the 1920s.

Let’s call the following an identity braid from now on:

In Steve the teacher’s game, the ambition is to go back to the identity braid from a complex braid in the shortest amount of time. To do that, we can use Artin’s work on braids.

Example One: Solving two ropes.

Assume that we have two ropes tangles with each other as follows:

The inverse of this rope is:

If we combine these two ropes, when each rope to be stretched, the result will give us the identity braid:

Example Two: Solving three ropes.

Take three ropes and make a braid as follows:

There are three intersections in this braid:

1: Green over the blue.

2: Red over the green.

3: Blue over the red.

Now, you should repeat these steps, but from last to the first this time. Then, you should do these moves:

Move #1: Blue over the red.

Move #2: Red over the green.

Move #3: Green over the blue.

Finally, the combination will give you the identity braid. Try and see yourself.

Paper and Braids

Take an A4 paper and cut the paper using a knife like the following:

Then, hold the paper from its sides and rotate it 90 degrees to the left. You will end up with some kind of a braid:

One wonders…

How can you use Emil Artin’s work in the game of Steve the teacher?
In “example two”, rotate the ropes 90 degrees to the left. Start investigating the intersections from left to right. What do you notice?
Play Steve the teacher’s game with an A4 paper. (It is more than enough to use 3 or 4 cuts on the paper.)

M. Serkan Kalaycıoğlu

Real Mathematics – Strange Worlds #17

Topology On Your Head

Often you see me writing about changing our perspective. For example, when you encounter a baby first thing you do is to make baby sounds and try to make the baby laugh. Whereas if you’d looked carefully at baby’s hair, you could have seen a very valuable mathematical knowledge hidden on the baby’s head:

As shown above, there is a point on each baby’s head. You can see that the hair besides that point is growing in different directions. Can you tell me which direction the hair grows at that exact point?

Hairy ball theorem can give us the answer.

Hairy Ball Theorem

Hairy ball theorem asks you to comb a hairy ball towards a specific direction. The theorem states that there is always at least one point (or one hair) that doesn’t move into that direction.

You can try yourself and see it: Each time at least one hair stands high. This hair (or point) is a sort of singularity. That hair is too stubborn to bend.

Baby’s hair is some kind of a hairy ball example. (I use the expression “some kind of”because the hairy ball has hair all over its surface. Though the baby’s head is not covered with hair completely.) This is why the point on the baby’s head is a singularity. It is the hair that gives a cowlick no matter how hard you comb the baby’s hair.

Torus

Hairy ball theorem doesn’t work on a torus that is covered with hair. In other words, it is possible to comb the hair on a torus towards a single direction.

No Wind

Hairy ball theorem can be used in meteorology. The theorem states that there is a point on earth where there is no wind whatsoever.

To prove that, you can use a hairy ball. Let’s assume that there is wind all over the earth from east to west. If you comb the ball like that, you will realize that north and south poles will have no wind at all.

On Maps

The hairy ball theorem is a kind of a fixed point theorem. Actually, it is also proven by L. E. J. Brouwer in 1912.

One of the real-life examples of the fixed point theorem uses maps. For example, print the map of the country you live in, and place it on the ground:

There is a point on the printed map that is exactly the same as the map’s geographical location.

“You are here” maps in malls or bus stops can be seen as an example of this fact.

One wonders…

Assume that all the objects below are covered with hair. Which one(s) can be combed towards the same direction at all its points? Why is that?

M. Serkan Kalaycıoğlu

Real Mathematics – Strange Worlds #16

The Walk

Select two points in the classroom.
Draw a line between them.
Send a student to one of those points.
Once the student starts his/her walk, he/she should arrive at the other point exactly 10 seconds later.
Everybody in the classroom would count to 10 to help the walker.
Ask the student to do the same walk twice while recording the walk using a camera.

The goal of the experiment

After the experiment is done, the following question is asked to the classroom:
“Is there a moment during both walks when the student stands at the exact point?”
In other words, the student walks the same distance in the same amount of time at different speeds. The goal is to find if there is a moment in both walks when the student passes the exact point on the line.
First of all, we should give time to the students for them to think and brainstorm on the problem. Then, using the video shots, the answer is given.
The most important question comes at last: Why so?

Weeding out the stone

In my childhood, one of my duties involved weeding out the stones inside a pile of rice. To be honest, I loved weeding out. Because I was having fun with the rice as I was making different shapes with it.

Years later when I was an undergrad mathematics student I heard of a theorem that made me think of my weed out days. This theorem stated that after I finish the weed out, there should be at least one rice particle that sits in the exact point where it was before the weed out started. (Assuming that the rice particles are covering the surface completely.) In other words; no matter how hard to stir the rice particles, there should be at least one rice particle that has the exact spot where it was before stirring.

This astonishing situation was explained by a Dutch mathematician named L.E.J. Brouwer. Brouwer’s fixed point theorem is a topology subject and it is known as one of the most important theorems in mathematics.

The answer to the walking problem,

The walking problem is an example of Brouwer’s fixed point theorem. This is why the answer to the question is “yes”: There is a moment in both walks when the student stands at the exact point on the line.

I will be talking about Brouwer’s fixed point in the next article.

One wonders…

A man leaves his home at 08:00 and arrives at another city at 14:00. Next morning at 08:00 he leaves that city and arrives at his home at 14:00, using the exact roads.

Conditions

Starting and finishing points are the same, as well as the time intervals of both trips.
The first condition means that the man could travel in his choice of speed as long as he sticks to the first condition.

Is there a point on these trips where the man passes at the exact time during both trips?

Hint: You could assume that the distance is 600 km and the man must finish that in 6 hours. For instance, he could have been traveling 100 km/h the first day, and the next day 80 km/h in the first 2 hours; 100 km/h in the next 2 hours, and 120 km/h in the last 2 hours of the trip.

M. Serkan Kalaycıoğlu

Real MATHEMATICS – Strange Worlds #15

Human Knot Game

Inside a classroom divide students into groups such that each group has at least 5 students. Groups should stand up and form a circle before following the upcoming instructions:

1. If there is an even number of student in a group:

Have each student extend his/her right hand and take the right hand of another student in the group who is not adjacent to him/her.
Repeat the same thing for the left hand.

2. If there is an odd number of student in a group:

Have each student (except one of them) extend his/her right hand and take the right hand of another student in the group who is not adjacent to him/her.
Then take the extra student and have him/her extend his/her right hand so that the extra student can hold the left hand of another student who is not adjacent to him/her.
Finally, repeat the process for the students whose left hands are free.

In the end, students will be knotted.

Now, each group should find a way to untangle themselves without letting their hands go. To do that, one can use Reidemeister moves.

Reidemeister Moves

Back in 1926 Kurt Reidemeister discovered something very useful in knot theory. According to him, in the knot theory one can use three moves which we call after his name. Using these three moves we can show if two (or more) knots are the same or not.
For example, using Reidemeister moves, we can see if a knot is an unknot (in other words, if it can be untangled or not).

What are these moves?

Twist

First Reidemeister move is twist. We can twist (or untwist) a part of a knot within the knot theory rules.

Poke

Second one is poke. We can poke a part of a knot as long as we don’t break (or cut) the knot.

Slide

Final one is slide. We can slide a part of a knot according to Reidemeister.

One wonders…

After you participating in a human knot game, ask yourselves which Reidemeister moves did you use during the game?

M. Serkan Kalaycıoğlu

Real MATHEMATICS – Strange Worlds #14

In my youth, I would never step outside without my cassette player. Though, I had two knotting problems with my cassette player: the First one was about the cassette itself. Rewinding cassettes was a big issue as sometimes the tape knotted itself. Whenever I was lucky, sticking a pencil would solve the whole problem.

The second knotting problem was about the headphone. Its cable would get knotted so bad, it would take me 10 minutes to untangle it. Most of the time I would bump into a friend of mine and the whole plan about listening to music would go down the drain.

The funny thing is I would get frustrated and chuck the headphone into my bag which would guarantee another frustration for the following day.

A similar tangling thing happens in our body, inside our cells, almost all the time.

DNA: A self-replicating material that is present in nearly all living organisms as the main constituent of chromosomes. It is the carrier of genetic information.

DNA has a spiral curve shape called “helix”. Inside the cell, the DNA spiral sits at almost 2-meters. Let me give another sight so that you can easily picture this length in your mind: If a cell’s nucleus was at the size of a basketball, the length of the DNA spiral would be up to 200 km!

You know what happens when you chuck one-meter long headphones into your bags. Trying to squeeze a 200 km long spiral into a basketball?! Lord; knots everywhere!

Knot Theory

This chaos itself was the reason why mathematicians got involved with knots. Although, knot-mathematics relationships existed way before DNA researches started. In the 19th century, a Scottish scientist named William Thomson (a.k.a. Lord Kelvin) suggested that all atoms are shaped like knots. Though soon enough Lord Kelvin’s idea was faulted and knot theory was put aside for nearly 100 years. (At the beginning of the 20th century, Kurt Reidemeister’s work was important in knot theory. I will get back to Reidemeister in the next post.)

Is there a difference between knots and mathematical knots?

For instance the knot we do with our shoelaces is not mathematical. Because both ends of the shoelace are open. Nevertheless a knot is mathematical if only ends are connected.

The left one is a knot, but not mathematically. The one on the right is mathematical though.

Unknot and Trefoil

In the knot theory we call different knots different names with using the number of crosses on the knot. A knot that has zero crossing is called “unknot”, and it looks like a circle:

A rubber band is an example of an unknot.

Check out the knots below:

They look different from each other, don’t they? The one on the left has 1 crossing as the one on the right has 2:

But, we can use manipulations on the knot without cutting (with turning aside and such) and turn one of the knots look exactly like the other one!

This means that these two knots are equivalent to each other. If you take a good look you will see that they are both unknots. For example, if you push the left side of the left knot, you will get an unknot:

Is there a knot that has 1 crossing but can’t be turned into an unknot?

The answer is: No! In fact, there are no such knots with 2 crossiongs either.

How about 3 crossings?

We call knots that have 3 crossings and that can’t be turned into unknots a “trefoil”.

Even though in the first glance you would think that a trefoil could be turned into an unknot, it is in fact impossible to do so. Trefoil is a special knot because (if you don’t count unknot) it has the least number of crossings (3). This is why trefoil is the basis knot for the knot theory.

One of the most important things about trefoils is that their mirror images are different knots. In the picture, knot a and knot b are not equivalent to each other: In other words, they can’t be turned into one another.

Möbius Strip and Trefoil

I talked about Möbius strips and its properties in an old post. Just to summarize what it is; take a strip of paper and tape their ends together. You will get a circle. But if you do it with twisting one of the end 180 degrees, you will get a Möbius band.

Let’s twist one end 3 times:

Then cut from the middle of the strip parallel to its length:

We will get a shape like the following:

After fixing the strip, you can see that it is a trefoil knot:

To be continued…

One wonders…

1. In order to make a trefoil knot out of a paper strip, we twist one end 3 times. Is there a difference between twisting inwards and outwards? Try and observe what you end up with.
2. If you twist the end 5 times instead of 3, what would you get? (Answer is in the next post.)

M. Serkan Kalaycıoğlu

Real MATHEMATICS – Graphs #6

Bequeath Problem

King Serkan I decides to allocate his lands to his children. Obviously he had set up some ground rules for the allocation:

Each child will get at least one land.
Same child can’t have adjacent lands.

Problem: At least how many children should Serkan I has so that allocation can be done without a problem?

Map #1

Let’s start from a simple map:

In this case Serkan I can have two children:

As you can remember from the previous article a map and a graph is irreversible. If we represent lands with dots, and let two dots be connected with a line if they are adjacent, we can show maps as graphs:

Let’s add another land to this map:

Adding a land on a map is the same thing as adding a dot on a graph:

Map #2

Let’s assume there are three lands on a map:

We can convert this map into graph as follows:

As seen above, three children are needed in order to fulfill Serkan I’s rules:

Map #3

Let the third map be the following:

According to Serkan I’s rules, we will need four children for such map:

Map #3 can be shown as a graph like the following:

Map #4

For the final map, let’s assume Serkan I left a map that looks like USA’s map:

Surprisingly four children are enough in order to allocate the lands on the map of USA:

What is going on?

Careful readers already noticed that adding a dot that connects the other dots in the second map’s graph gives the graph of the third map. Same thing is true for the first and second maps:

Hence, adding a new dot to the graph means adding a new child.

Q: Is it possible to create a map that requires at least five children?

In other words: Is it possible to add a fifth dot to the graph so that it has connection to all existing four dots? (Ps: There can be no crossing in a graph as our maps are planar.)

Then, all we have to do is to add that fifth dot… Nevertheless I can’t seem to do it. When I add the fifth dot outside of the following graph:

It is impossible to connect 1 and 5 without crossing another line. No matter what I try, I can’t do it:

Four Color Theorem

About 160 years ago Francis Guthrie was thinking about coloring maps:

“Can the areas on any map be colored with at most four colors such that no pair of neighboring areas get the same color?”

Incredibly the answer is yes.

This simple problem was introduced for the first time by Francis Guthrie in 1852. Not until 1976 there was no proof for Guthrie’s conjecture. Only then with the help of computers the conjecture was proved. This proof is crucial for mathematics world as it is known as the mathematical theorem that was proven with the help of computers.

One wonders…

Add a fourth dot to the graph you see above and connect that dot to the existing dots. (You are free to place the fourth dot wherever you want on the graph.)

Now check your graph: One of those four dots is trapped inside the lines, isn’t it?

Can you fix that?

Explain how you can/can’t do it.

M. Serkan Kalaycıoğlu

Real MATHEMATICS – Graphs #5

Cruel Traffic Light

I have to drive past the same crossroad almost every single day. Obviously I stop at the longest-lasting traffic light of the crossroad. Within time I started loving these moments because it gave me a chance to think my life over. Though, it doesn’t take too long for me to start thinking about mathematics.

One of those days I found myself questioning the traffic lights and their relationship with mathematics. (After all mathematics is everywhere; isn’t it?) Soon after I realized that there was my beloved graph theory behind traffic lights.

Light #1

Let’s assume a one-way street with two traffic lights: One for the vehicles (we’ll call it A), other for the pedestrians (we’ll call it B):

In such situation we have to avoid accidents. This means whenever A has green light, B must have red light and vice versa. We will not take account of the situation when both lights are red. Because even though there won’t be any accident, neither of the sides will be standing still (which is nonsense):

All these can be shown using graph theory: Lights will be represented by dots. Dots in a graph will be connected with lines if they are not in the same color:

Same thing can be shown with maps: If A and B are two neighboring countries, they should be colored in different colors to avoid confusion:

Light #2

This time we will assume a two-way street with three traffic lights: Two for the vehicles from opposite sides (we’ll call them A and B), and one for the pedestrians (we’ll call that C):

In such a situation when C is red, either or both of A and B should be green. When C is green, then both A and B should be red:

We’ll skip the situation where all three of them are red as no one would move in such situation.

We can show these using graph theory and map coloring as follows:

Since we set all the rules graphs make it much easier and clearer to understand the situations.

Light #3

Finally we have a two-way street (A and B), a right turn (C) and two pedestrian lights (D and E) as follows:

This time it is a much complex situation:

Although using graph theory makes it all easier for us to understand:

How Many Colors?

We will color the following graphs using the same rules we just established above: If two dots are connected with a line, then those dots must have different colors.

Chromatic Numbers: Whenever a graph is being colored, ambition is the use the least number of colors. This number is also known as the chromatic number of a graph.

Here we need 3 colors. Hence chromatic number of the graph is 3. Let’s add one more dot and line to the graph:

This time chromatic number of the graph becomes 2. We’ll add another dot and line:

Now chromatic number becomes 3 again. Let’s add a dot and a line for the last time:

Chromatic number is back to 2.

To be continued…

One Wonders…

What did just happen? What did you notice? Why is it happening?

How can you increase the chromatic number?

M. Serkan Kalaycıoğlu

Real MATHEMATICS – Strange Worlds #12

“More to this than meets the eye…”

I will be using a real life example in order to explain Mandelbrot’s answer to the coastline paradox.

Maps of Norway and USA are as shown below:

It is clear to the naked eye that total coastline of USA is enormous comparing to Norway’s coastline. Nevertheless there is more to this than meets the eyes. Norway’s coastline is a lot longer than USA’s:

USA: 19.924 km

Norway: 25.148 km

There are more than 5000 km between the coastlines which is a really surprising result. But when you zoom into the maps it is easy to see that Norway’s coastline is way more irregular than USA’s. In other words Norway’s coastline has more roughness. Mandelbrot expresses this in his fractal geometry as follows: Norway coastline has a bigger fractal dimension than USA coastline.

But this doesn’t necessarily mean that a bigger fractal dimension has more length. Length and fractal dimension are incomparable.

Measuring Device

In the coastline paradox we learned that one decreases the length of his/her measuring device, then length of the coastline will increase. This information brings an important question with itself: How did they decide the length of the measuring device for Norway-USA comparison?

This is where fractal dimension works perfectly: Finding the appropriate length for the measuring device.

Q: This is all very well how can a coastline length be measured exactly?

Unfortunately it can’t be done. Today, none of the coastline or border lengths are 100% accurate. Although we are certain about one thing: We can make comparisons between coastlines and borders with the help of fractal dimension. In short, today we are able to compare two coastlines or borders even though we are not sure about their exact length.

Box Counting

Finding fractal dimension is easier than you’d think. All you need to do is to count boxes and know how to use a calculator.

Let’s say I want to calculate the fractal dimension of the following shape:

Assume that this shape is inside a unit square. First I divide the square into little squares with side length ¼ units. Then I count the number of boxes which the shape passes through:

This shape passes through exactly 14 squares.

Up next, I divided the unit square into even smaller squares which have side length 1/8 units. And again I count the number of boxes which the shape passes through:

This time the shape passes through 32 squares.

Then I use a calculator. In order to find the fractal dimension of the shape, I must find the logarithms of the number of boxes (32/14) and length of the squares ({1/8}/{1/4}). Then I must divide them multiply the answer with -1.

This random shape I drew on my notebook has around 1,19 fractal dimension.

One wonders…

Calculate the fractal dimension of the following shape:

M. Serkan Kalaycıoğlu

Real Mathematics – Strange Worlds #11

Fractional Dimension

When you try to measure the length of a coastline, your finding will increase as your measuring device decreases. It means that there is a proportion between these magnitudes. This is why it is possible to find different (even infinite) lengths for a random coastline.

Mathematician Mandelbrot named this proportion as “fractal dimension”.

In the Euclidean geometry a dot has 0, a line has 1, a plane has 2 and a cube has 3 dimensions. But, in the nature shapes of objects are not regular as shown in the Euclidean geometry. In the early 20^th century a mathematician named Felix Hausdorff discovered that some shapes have non-integer dimensions. Later on we started calling this non-integer dimension idea as Hausdorff-Besicovitch dimension. This idea was basis for fractal geometry’s development.

In the previous article I showed how one can calculate dimension of a shape in the Euclidean geometry. Same formula can be used in order to calculate objects that don’t have regular shapes. For that, I will be talking about a couple special fractals.

Snowflake

Swedish mathematician Helge von Koch created a geometrical shaped named after him: Koch snowflake.

To create a Koch snowflake, one can start drawing a straight line. Then that line should be divided into thirds as the middle part gets erased:

Draw sides of an equilateral triangle above the removed segment: (In other words, add a peak where there is a gap.)

Continue the same process forever and you will get Koch fractal:

Here are the segments and all of Koch snowflake:

Now let’s use the dimension formula to the Koch snowflake. We only need the number of parts and their lengths in each step of the construction of the Koch snowflake.

In the first step, we had a straight line that was divided into 1/3s:

In the second step we ended up with 4 of those 1/3s:

If we examine each step of the Koch snowflake we will end up with 4 parts that have 1/3 lengths. Therefore fractal dimension of Koch snowflake (which I call d) can be found as follows:

(1/3)^d = 4

d ≈ 1,26.

Koch Curve

Let’s try a variant of the Koch snowflake, which we call Koch curve. This time we will draw sides of a square instead of an equilateral triangle.

So, we will start with a straight line that is divided into thirds. Then we will remove the middle part and draw sides of a square that has no bottom line:

Next few stages of the Koch curve will look like the following:

Here we see that in each step, we end up with 5 parts that have length 1/3:

Apply this to the dimension formula and this fractal’s dimension will be as follows:

(1/3)^d = 5

d ≈ 1,4649.

What does this difference in dimensions mean?

Between the curve and the snowflake, curve has more roughness and it takes up more area than the snowflake. Hence one can conclude that higher dimension means more roughness and more area for Koch fractals:

To be continued…

One wonders…

Another handmade fractal is Sierpinski triangle. This famous fractal was first discovered more than 100 years ago and named after a mathematician named Waclaw Sierpinski.

To construct Sierpinski triangle, one must start with an equilateral triangle:

Then mark middle of each side and connect those points to form a new triangle:

At this point, there are four smaller versions of the original triangle. Cut the middle one out and you will have three equilateral triangles that have half of the side lengths of the original triangle:

Repeat the steps forever and you will get Sierpinski triangle:

Show that Sierpinski triangle is a fractal.
Calculate the dimension of the Sierpinski triangle and compare your result with Koch snowflake.

M. Serkan Kalaycıoğlu

Real Mathematics – Strange Worlds #9

Island S

I own a private island near New Zealand. (In my dreams) Unfortunately I put it on the market due to the economical crisis. If I can’t sell my island I will have to use charter flights to Nice instead of using my private jet.

I created an ad on Ebay. But I choose a different approach when I set a price for my island:

“A slightly used island on sale for 100.000 dollars times the coastline length of the island.

Note: Buyer must calculate the length of the coastline.”

Soon enough I got an offer from a potential buyer. Buyer said he calculated the coastline length as follows:

Buyer used three straight lines in order to measure the length of the coastline. He took each line 8 km long which gave 8*3=24 km. Hence his offer was 24*100.000 = 2.400.000 dollars.

I thought my island worth more than that. Hence I asked the buyer to evaluate his bid again. He came up with a new bid:

This time buyer used seven 5-km-long lines: 5*7=35 km. Thus his second offer was 3.500.000 dollars.

Even though new offer is higher, I thought the buyer can do better. This is why I asked the buyer to measure the length of the coastline one more time:

At last buyer used sixteen 3-km-long lines: 3*16=48 km. Therefore buyer’s final last offer was 4.800.000 dollars.

Q: What is the highest bid I can get from a buyer?

Give yourself a second and think about the answer before continuing the article.

Coastline Paradox

As the buyer decreases the length of the ruler, length of the coastline will get bigger. What is the smallest length for the ruler?

1 cm?

1 mm?

1 mm divided by 1 billion?

There isn’t any answer for the smallest length of a ruler; it can be decreased up to a point where it is infinitely small.

Since there is a disproportion between the length of the ruler and the length of the coastline, coastline can have infinity length.

This is a paradox. Because it is a known fact that there isn’t any land on earth which has infinitely long coastline. Although using buyer’s measurement method, one can’t find an upper limit for the coastline of Island S.

Root of the Problem

British mathematician Lewis Fry Richardson (1881-1953) had done a very interesting research in the first part of the 20^th century. He wanted to know what factors would reduce the frequency of wars between any two country. One of the questions he asked was the following:

“Is there any correlation between the probability of war and the shared border length among two neighbor countries?”

Richardson took Spain and Portugal as an example. Therefore he wanted to know the border length between them. Richardson was really surprised when two countries reported their measurements. Even though they measured the same length, there was a difference of 200 km between two values.

This huge difference led Richardson to pursue the topic and he eventually came up with the coastline paradox.

Is there a sensible explanation for this paradox?

To be continued…

One wonders…

How can my island’s coastline be measured if I want to sell my island for more than 6.000.000 dollars?

M. Serkan Kalaycıoğlu