Tag Archives: mathematics with games

Numbers #12

Subitizing: The ability to recognize(or guess) the number of a small group of objects without counting.

The name subitizing comes from the Latin word “subitus” which means “sudden”.

They all represent 6

Subitizing can be seen in many every-day activities. One of them is a six-pack soda. No matter how they are lined up, we know that the number of soda bottles is 6. We inherit this knowledge without counting the bottles. And if we decide to drink one of them, we automatically know(without having to count them) that the number of soda bottles left is 5.

You don’t have to count the dots on the surfaces. You just know that it is 5.

Another example of subitizing can be given from the game backgammon. Assume that two dices are rolled and you identify them as 2 and 5. The process of identifying the dices can be measured in milliseconds. This can be even shortened as you spend more time playing the game. In short; subitizing is a skill that can be developed if one spends time and work on it.

Research studies showed that 6-month olds can differentiate, visually (a top bounces 3 times) and from sounds (clapping hands 3 times), between 1, 2, and even 3. In other words; humans start developing the number concept when they are just infants.

Kebab Truck & Subitizing
Subitizing is hidden behind the number of customer groups in the game of Kebab Truck. As the game is played, scores become higher and higher. The reason behind this is that players’ subitizing skills are improving.

Let’s check this scene from Kebab Truck:

In the beginning, you will be making certain moves during the game. Nevertheless, in time, your moves will differ substantially. The biggest reason behind this is that your subitizing skills were improved while you were playing the game.

Kebab Truck also helps the players to develop their basic arithmetic skills. These improvements are not limited to adding and subtracting the number of customers. Once you understand how the scoring system formulated, you will realize that (to maximize your scoring) multiplication is an important part of this game as well.

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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:

Red undergoes the green.

The inverse of this rope is:

Green undergoes the red.

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 – Graphs #7

Serkan’s System

Serkan the math teacher, hands out a specific number of problems to his students. Kids who can solve 1 or more of those problems would get a certain prize. At the beginning of each semester, Serkan and his students sit down and agree on what kind of prize is going to be distributed. For the current semester, oreo is chosen as the prize:

If Serkan the math teacher hands out 10 problems:

  • 10 Oreos for the kids who solved 10, 9 or 8 of those problems,
  • 5 Oreos for the kids who solved 7, 6 or 5 of those problems,
  • 2 Oreos for the kids who solved 4, 3 or 2 of those problems,
  • 1 oreo for the kids who solved 1 problem,
  • Absolutely nothing for the students who solved… well… none of those problems.

If you take a careful look at the numbers, you can see that Serkan the math teacher selected those numbers with a kind of logic: 10, 5, 2 and 1.

These are the natural numbers that can divide the number of the problems (that is 10) without any remainder.

Prize Distribution Machine (P.D.M.)

One month later…

Serkan the math teacher had faced some problems 4 weeks into the semester. He realized that it took hours to distribute the prizes since he has 10 classes in total.

Serkan the math teacher had to use almost all his free time in school to distribute the Oreos. This led him to think about a machine that would help him with the distribution:

  • P.D.M. will have 4 different compartments. (Because of 10, 5, 2 and 1.)
  • The volumes of those compartments will be measured with Oreos. They will be 10, 5, 2 and 1 Oreo-sized.
  • Oreos will enter the machine from the 10-Oreo-sized compartment. From there, Oreos will move to the other compartments using the connections that will be established.
  • Golden Rule: To establish a connection between any two compartments, the size of those compartments must be factors of one another.

Connections of the compartments for 10 problems:

  • For 10-Oreo-sized: 5, 2 and 1.
  • For 5-Oreo-sized: 10 and 1.
  • For 2-Oreo-sized: 10 and 1.
  • For 1-Oreo-sized: 10, 5, and 2.

Then, the sketch of the P.D.M. would look like the following:

Is this another graph?!

If you are familiar with graph theory (or if you read the graph section of the blog) you can recognize that the sketch of Serkan the math teacher’s machine is a planar graph:

You should connect the numbers (dots) using lines (connections) according to the golden rule.

One wonders…

What if Serkan the math teacher asks 12 problems?

For 12 problems, the numbers of prizes are going to be: 12, 6, 4, 3, 2 and 1.

In such a situation, can Serkan build his machine? In other words; is it possible to connect the dots for 12-sized P.D.M.?

Hint: First, you should consider where the lines should be. Also, you can arrange the dots in any order you’d like.

M. Serkan Kalaycıoğlu

Real Mathematics – Algorithm #6

The year 2600…

Finally, we discovered a planet where we can live besides Earth. Scientists named this planet as T-489. The living conditions in T-489 seems very much like the Earth’s. Satellite views show that there is water on this planet. And scientists discovered that the atmosphere on this planet is almost the same as Earth’s atmosphere.

T-489 and its system.

Space agencies of the world’s biggest countries gathered a team to discover T-489. According to the plan, the first team that lands on T-489 will secure the landing point and build a headquarter. This point is set beforehand with the use of the satellite photos.

After the completion of the headquarters, three separate teams will be sent to T-489 to discover certain points on the planet that was set beforehand. These teams will be searching for the possible existence of life forms as well as testing the life conditions on each location.

The map: Headquarters is at the yellow point. Other points must be discovered by the teams. In an emergency, teams can use the roads and return to the headquarters.

The headquarter of T-489 should develop a map for those three teams which will be scouting the planet. This map is crucial since it will show how those teams should travel on the planet, and it will also explain how a team can return to the headquarters in case of an emergency.

In certain situations such as not knowing where a team is at a given time, headquarters must add an algorithm to the map. This way, in an emergency, any team can use this algorithm and return to the headquarters safely.

Road Coloring Problem

In 1970, Roy Adler suggested the road coloring problem. After almost 40 years, an Israeli mathematician called Trahtman solved this problem.

Yellow point the headquarters. Select a point and use the roads that in this order three times: blue-red-red. You will always end up at the yellow point.

Trahtman thought of the map as a graph shown above; each road had a direction and a distinct color. Trahtman created these directions and colors such that, following a certain algorithm, your travel would always end up on a specific point. In this example, the algorithm is following blue-red-red roads three times and always ending up at the yellow point. You can try at any point and see by yourself.

Chaos on T-489

You must develop a map for the teams that will be scouting the planet. Headquarters and the locations to visit are shown as follows:

M is the headquarters. A, B and C are the points to be discovered by the teams.

You should prepare for the worst possible scenario. If the connection breaks between the headquarters and the scouting teams, and if those teams can’t access to the map, your algorithm might save their lives.

The idea is: Place a sign on the entry of each point. These signs will have only two pieces of information: The color and the direction of the roads. (Why don’t you just hang the map on these signs? Because you are not sure if there is an alien life form on the planet. And in case they exist, such maps might put the headquarters and the whole mission at danger.)

Let’s say you created a map as follows:

Such a map would have an algorithm like this: Follow red-blue twice, and you will always return to the headquarters.

One wonders…

Create a map and an algorithm that will let you arrive at the point M in each time.

Let’s say that there is one more point added to the discovery. Create a map and an algorithm such that, whenever you use the algorithm you will end up at the headquarters.

(Try to use the least number of roads on your map.)

M. Serkan Kalaycıoğlu

Real Mathematics – What are the chances?! #8

Hooligans

This year’s cup final will be between two arc-rivals of the country; it is both a final and a derby game. Supporters of both sides are waiting anxiously for the game. Though, Steve has no interest in this game. His team was eliminated in the semi-finals.

Two of Steve’s close friends, Jack and Patrick, have been arguing over the final for the past two weeks. Jack thinks his team FC Ravens will be the winner as Patrick thinks his team AC Bolognese will be victorious. Meanwhile, Steve is thinking about what to eat at supper.

  • Jack and Patrick

Jack

“We haven’t lost to them in years. The cup is 90% ours. But, this is football; anything is possible. Well, that makes 10% anyway. Yeah, I am pretty sure of my team. If you want, let’s bet on it!”

Patrick

“We are the best team of the season. Plus, our striker scored 69 goals in 27 games. The cup is 75% ours. But, we have been unlucky against them. Which is why I am giving them 25%. Steve, I can bet on it right this second!”

Smart Steve

Steve is aware of the fact that either of these teams will be victorious as this is a cup final and there is no chance for a tie.

He would like to take advantage of his friends’ blindness. Steve would like to set such conditions for the bets such that he would profit whatever the result is.

What would you do if you were in Steve’s situation?

M. Serkan Kalaycıoğlu

Real Mathematics – Game #13

Crossing the bridge

Years and years ago, there was a village known as Togan. It was located in Mesopotamia’s fertile lands, next to rivers and beautiful waterfalls. There were only a few hundred people who lived in Togan. All of the Togan were farmers, except one: Berkut.

Berkut

Berkut was the oldest man in the village. He had a long white beard. His story had become a kind of a legend within time. According to the people of Togan, whoever entered his property would never be seen again. This is why Berkut’s house was the only house that stayed on the other side of the river of Togan.

Berkut’s home.

For kids, Berkut was a mystery. Whenever Berkut was out on his garden, kids would gather and watch him from the other side of the river.

People of Togan were hardworking farmers. They would be working on their farms starting from their childhood. Like the rest of the Togan, Ali starting helping his family at an early age. Ali would work from sunrise to the sunset.

For Ali and his friends, Berkut’s situation was one of the hot topics. One day, these four friends decided that they would skip working and go across the river to investigate Berkut’s house. This legend had to be questioned!

The next day, while he was out on his garden, Berkut realized that four kids were about to cross the river, using the old bridge. He watched them crossing the bridge in pairs, as the old bridge was not strong enough to carry more than 2 of them at the same time.

He, then, went inside his house as Ali and his friends were approaching the house. When kids were inside the garden, they saw that a jar of cookies and a steamy teapot was waiting for them. While they were checking the table, Berkut went outside and greeted them. Kids, dumbfounded, started screaming as they all dispersed out of the garden in different ways.

They found each other only after it was dark. Now, kids, who had only one lamp with themselves, had to cross the bridge as fast as they can. But they couldn’t risk crossing the old bridge at once. Each time, only two of them could cross the bridge.

Crossing Times:
Jane: 1 minute
Ali: 2 minutes
Tom: 6 minutes
Jenny: 10 minutes

Since they had to cross the bridge in pairs while sharing a lamp, their speed would be at the rate of the slowest of the pair.

Now, you must solve this problem:

What is the fastest route to the other side of the river?

M. Serkan Kalaycıoğlu

Real MATHEMATICS – Geometry #20

Escape From Alcatraz

Imagine a classroom that has 5 meters between its walls in length. Tie a 6-meter long rope between these walls. Let the rope be 2 cm high off the ground. Since the rope strained to its limits, its 1-meter long part hangs from either side of the rope.

The ultimate goal is to escape from the classroom from under this rope, without touching the rope.

Rules

  • Escape should be from the middle point of the rope.
  • One should use the extra part of the rope to extend it.
  • One of the students will help you during the escape. He/she will strain the rope for you so that you can avoid touching the rope.
  • Each student has exactly one try for his/her escape.

Winning Condition: Using the least amount of rope for your escape.

Football Field

Legal-size for a football field is between 90 and 120 meters in length. Assume that we strain a rope on a football field that is 100 meters long. We fixed this rope right in the middle of both goals while the rope is touching the pitch.

The middle of the rope sits right on the starting point of the field. This is also known as the kick-off point.

Let us add 1 meter to the existing rope. Now, the rope sits flexed, not strained, on the field.

Question: If we try to pick the rope up at the kick-off point, how high will the rope go?

Solution

We can express the question also as follows:

“Two ropes which have length 100m and 101m are tied between two points sitting 100m apart from each other. One picks the 101m-long rope up from its middle point. How high the rope can go?”

If we examine the situation carefully, we can realize that there are two equal right-angled triangles in the drawing:

Using Pythagorean Theorem, we can find the length h:

(50,5)2 = 502 + h2

h ≈ 7,089 meters.

Conclusion

Adding only 1 meter to a 100-meter long rope helps the rope to go as high as 7 meters in its middle point. This means that a 1-meter addition could let an 18-wheeler truck pass under the rope with ease.

M. Serkan Kalaycıoğlu

Real MATHEMATICS – What are the chances?! #7

How Close?

Game: In a group, everyone is asked to pick a number between 0 and 100. Even though it is possible for more than one person to pick the same number, it is forbidden for participants to communicate with each other.

Winner: Winner is the person who is closest to the two-thirds of the average of the picked numbers.

Question: Is there any way for you to optimize your chance to win the game?

At first glance, one might think that it is not important which number you pick between 0 and 100. Because the winning number depends on the choices of others’. Although, if there is a player who has probability knowledge, he/she could maximize his/her chance for winning the game.

Step #1
Assume that we have a group of 12 people, and every individual selects 100. Then the average becomes:

12*100/12 = 100.

The winning number is the one that is closest to the two-thirds of the average. That means 100*2/3 = 66,666…

66,666… is the highest winning number for this game. If you are aware of this fact; then you would select a number that is between 0 and 66.

0-66

Obviously, there is a chance for you to win the game even though you select a number higher than 66. Then again; why would you select such a number if you know that the winning number is between 0 and 66?!

What if everyone realizes…?!

Let’s assume that you are aware of this fact. Then while others will pick a number between 0 and 100, you will be picking a number between 0 and 66. This is a huge advantage. But, suddenly you realized something else: What if everyone came to the same conclusion?

Step #2

If everyone knows that the winning number can’t exceed 66,666…, then no one will choose a number higher than 66. Hence, everyone will choose a number between 0 and 66.

In this situation, the highest average can be 66:

66*12/12 = 66 average.

66*2/3 = 44 is the highest winning number.

0-44

This means that if everyone selects between 0 and 66; the winning number can’t exceed 44. Then, why would you choose a number that is higher than 44?!

Step #3

If everyone comes to the same conclusion, then no one within the group will select a number that is higher than 44. This causes a new calculation. Since everyone knows that the winner will be between 0 and 44, the winning number can at most be:

44*12/12 = 44 (average)

44*2/3 = 29,333…

This means that the winning number is at most 29. Then no one will choose a number that is higher than 29.

0-29

If one follows the same logic, at the end of the 11th step he/she will find 0 (zero) as a result. This is why picking zero for everyone is the most logical move for the whole group. Using your probability knowledge, one will eventually conclude that zero is the most reasonable choice for each individual.

Conclusion

Mathematics can help a group find a solution that benefits everyone within the group, even though there is no communication inside that group of people.

One wonders…

You know a person inside the group who isn’t good at mathematics. In this situation, would you change your logic? Give your answers using probabilistic calculations.

Ps. You can click here and create yourself an example set.

M. Serkan Kalaycıoğlu