SO TO WHOM, that is: TRY WHERE YOU CAN - part 2

In the previous episode, we dealt with Sudoku, an arithmetic game in which numbers are basically arranged in various diagrams according to certain rules. The most common variant is a 9×9 chessboard, additionally divided into nine 3×3 cells. The numbers from 1 to 9 must be set on it so that they do not repeat either in a vertical row (mathematicians say: in a column) or in a horizontal row (mathematicians say: in a row) - and, moreover, so that they do not repeat. repeat within any smaller square.

Na fig. 1 we see this puzzle in a simpler version, which is a 6 × 6 square divided into 2 × 3 rectangles. We insert the numbers 1, 2, 3, 4, 5, 6 into it - so that they do not repeat vertically, neither horizontally, nor in each of the selected hexagons.

Let's try shown in the top square. Can you fill it in with numbers from 1 to 6 according to the rules set for this game? It is possible - but ambiguous. Let's see - draw a square on the left or a square on the right.

We can say that this is not the basis for the puzzle. We usually assume that a puzzle has one solution. The task of finding different bases for the "big" Sudoku, 9x9, is a difficult task and there is no chance of completely solving it.

Another important connection is the contradictory system. The bottom middle square (the one with the number 2 in the bottom right corner) cannot be completed. Why?

Fun and Retreats

We play on. Let's use children's intuition. They believe that entertainment is an introduction to learning. Let's go into space. switched on fig. 2 everyone sees the grid tetrahedronfrom balls, for example, ping-pong balls? Recall school geometry lessons. The colors on the left side of the picture explain what it is glued to when assembling the block. In particular, three corner (red) balls will be glued into one. Therefore, they must be the same number. Maybe 9. Why? And why not?

Oh I didn't phrase it problem. It sounds something like this: is it possible to inscribe the numbers from 0 to 9 in the visible grid so that each face contains all the numbers? The task is not difficult, but how much you need to imagine! I will not spoil the pleasure of readers and will not give a solution.

This is a very beautiful and underestimated shape. regular octahedron, built from two pyramids (=pyramids) with a square base. As the name suggests, the octahedron has eight faces.

There are six vertices in an octahedron. It contradicts cubewhich has six faces and eight vertices. The edges of both lumps are the same - twelve each. This double solids - this means that by connecting the centers of the faces of the cube we get an octahedron, and the centers of the faces of the octahedron will give us a cube. Both of these bumps perform ("because they have to") Euler formula: The sum of the number of vertices and the number of faces is 2 more than the number of edges.

3. A regular octahedron in parallel projection and an octahedron lattice composed of spheres in such a way that each edge has four spheres.

Task 1. First, write down the last sentence of the previous paragraph using a mathematical formula. On the fig. 3 you see an octahedral grid, also made up of spheres. Each edge has four balls. Each face is a triangle of ten spheres. The problem is set independently: is it possible to put numbers from 0 to 9 in the circles of the grid so that after gluing a solid body, each wall contains all the numbers (it follows that without repetition). As before, the greatest difficulty in this task is how the mesh is transformed into a solid body. I cannot explain it in writing, so I am not giving the solution here either.

4. Two icosahedrons from ping-pong balls. Notice the different color scheme.

already Plato (and he lived in the XNUMXth-XNUMXth centuries BC) knew all the regular polyhedra: tetrahedron, cube, octahedron, dodecahedron i icosahedron. It's amazing how he got there - no pencil, no paper, no pen, no books, no smartphone, no internet! I won't talk about the dodecahedron here. But the icosahedral sudoku is interesting. We see this lump on illustration 4and its network fig. 5.

5. Regular mesh of the icosahedron.

As before, this is not a grid in the sense in which we remember (?!) from school, but a way of gluing triangles from balls (balls).

Task 2. How many balls does it take to build such an icosahedron? Does the following reasoning remain correct: since each face is a triangle, if there are to be 20 faces, then as many as 60 spheres are needed?

6. Grid of an icosahedron from spheres. Each circle is, for example, a ping-pong ball, but the construction of circles on circles marked with the same color merges into one. So we have twelve spheres (= twelve vertices: red, blue, purple, blue and eight yellow).

It is easy to see that three numbers in the icosahedron are not enough. More precisely: it is impossible to enumerate vertices with numbers 1, 2, 3 so that each (triangular) face has these three numbers and there are no repetitions. Is it possible with four numbers? Yes it is possible! Let's look at Rice. 6 and 7.

7. Here is how to number the spheres that make up the icosahedron so that each face contains numbers other than 1, 2, 3, 4. Which of the bodies in fig. 4 is colored like this?

Task 3. Three of the four numbers can be chosen in four ways: 123, 124, 134, 234. Find five such triangles in the icosahedron in fig. 7 (as well as from illustrations 4).

Task 4 (requires very good spatial imagination). The icosahedron has twelve vertices, which means that it can be glued together from twelve balls (fig. 7). Note that there are three vertices (=balls) labeled with 1, three with 2, and so on. Thus, balls of the same color form a triangle. What is this triangle? Maybe equilateral? Look again illustrations 4.

The next task for the grandfather / grandmother and grandson / granddaughter. Parents can finally try their hand too, but they need patience and time.

Task 5. Buy twelve (preferably 24) ping-pong balls, some four colors of paint, a brush and the right glue - I don't recommend fast ones like Superglue or Droplet because they dry too quickly and are dangerous for children. Glue on the icosahedron. Dress your granddaughter in a t-shirt that will be washed (or thrown away) immediately afterwards. Cover the table with foil (preferably with newspapers). Carefully color the icosahedron with four colors 1, 2, 3, 4, as shown in fig. fig. 7. You can change the order - first color the balloons and then glue them. At the same time, tiny circles must be left unpainted so that the paint does not stick to the paint.

Now the most difficult task (more precisely, their entire sequence).

Task 6 (More specifically, the general theme). Plot the icosahedron as a tetrahedron and an octahedron on Rice. 2 and 3 This means that there should be four balls on each edge. In this variant, the task is both time-consuming and even costly. Let's start by finding out how many balls you need. Each face has ten spheres, so the icosahedron needs two hundred? No! We must remember that many balls are shared. How many edges does an icosahedron have? It can be painstakingly calculated, but what is the Euler formula for?

w–k+s=2

where w, k, s are the number of vertices, edges, and faces, respectively. We remember that w = 12, s = 20, which means k = 30. We have 30 edges of the icosahedron. You can do it differently, because if there are 20 triangles, then they have only 60 edges, but two of them are common.

Let's calculate how many balls you need. In each triangle there is only one internal ball - neither at the top of our body, nor on the edge. Thus, we have a total of 20 such balls. There are 12 peaks. Each edge has two non-vertex balls (they are inside the edge, but not inside the face). Since there are 30 edges, there are 60 marbles, but two of them are shared, which means you only need 30 marbles, so you need a total of 20 + 12 + 30 = 62 marbles. Balls can be bought for at least 50 pennies (usually more expensive). If you add the cost of glue, it will come out ... a lot. Good bonding requires several hours of painstaking work. Together they are suitable for a relaxing pastime - I recommend them instead of, for example, watching TV.

Digression 1. In Andrzej Wajda's film series Years, Days, two men play chess "because they have to somehow pass the time until dinner." It takes place in Galician Krakow. Indeed: newspapers have already been read (then they had 4 pages), TV and telephone have not yet been invented, there are no football matches. Boredom in the puddles. In such a situation, people came up with entertainment for themselves. Today we have them after pressing the remote control ...

Digression 2. At the 2019 meeting of the Association of Teachers of Mathematics, a Spanish professor demonstrated a computer program that can paint solid walls in any color. It was a little creepy, because they only drew the hands, almost cut off the body. I thought to myself: how much fun can you get from such a "shading"? Everything takes two minutes, and by the fourth we don’t remember anything. Meanwhile, old-fashioned “needlework” calms and educates. Who does not believe, let him try.

Let's go back to the XNUMXth century and to our realities. If we do not want relaxation in the form of laborious gluing of balls, then we will draw at least a grid of an icosahedron, the edges of which have four balls. How to do it? Chop it right fig. 6. The attentive reader already guesses the problem:

Task 7. Is it possible to enumerate the balls with numbers from 0 to 9 so that all these numbers appear on each face of such an icosahedron?

What are we being paid for?

Today we often ask ourselves the question of the purpose of our activities, and the "gray taxpayer" will ask why he should pay mathematicians to solve such puzzles?

The answer is pretty simple. Such "puzzles", interesting in themselves, are "a fragment of something more serious." After all, military parades are only an external, spectacular part of a difficult service. I will give just one example, but I will start with a strange but internationally recognized mathematical subject. In 1852, an English student asked his professor if it was possible to color a map with four colors so that neighboring countries are always shown in different colors? Let me add that we do not consider "neighbors" those that meet at only one point, such as the states of Wyoming and Utah in the US. The professor didn't know... and the problem had been waiting for a solution for over a hundred years.

8. Icosahedron from RECO blocks. Flash reflectors show what the icosahedron has in common with the triangle and pentagon. Five triangles converge at each vertex.

It happened in an unexpected way. In 1976, a group of American mathematicians wrote a program to solve this problem (and they decided: yes, four colors will always be enough). This was the first proof of a mathematical fact obtained with the help of a "mathematical machine" - as a computer was called half a century ago (and even earlier: "electronic brain").

Here is a specially shown “map of Europe” (fig. 9). Those countries that have a common border are connected. Coloring the map is the same as coloring the circles of this graph (called the graph) so that no connected circles are the same color. A look at Liechtenstein, Belgium, France and Germany shows that three colors are not enough. If you wish, Reader, color it with four colors.

9. Who borders with whom in Europe?

Well, yes, but is it worth the taxpayers' money? So let's look at the same graph a little differently. Forget that there are states and borders. Let the circles symbolize information packets to be sent from one point to another (for example, from P to EST), and the segments represent possible connections, each of which has its own bandwidth. Send as soon as possible?

First, let's look at a very simplified, but also very interesting situation from a mathematical point of view. We have to send something from point S (= as start) to point M (= finish) using a connection network with the same bandwidth, say 1. We see this in fig. 10.

10. Network of connections from Statsyika Zdrój to Megapolis.

Let's imagine that about 89 bits of information need to be sent from S to M. The author of these words likes problems about trains, so he imagines that he is a manager at Stacie Zdrój, from where he has to send 144 wagons. to metropolis station. Why exactly 144? Because, as we will see, this will be used to calculate the throughput of the entire network. The capacity is 1 in each lot, i.e. one car can pass per unit of time (one information bit, possibly also Gigabyte).

Let's make sure that all cars meet at the same time in M. Everyone gets there in 89 units of time. If I have a very important information packet from S to M to send, I break it up into groups of 144 units and push it through as above. The math guarantees that this will be the fastest. How did I know that you need 89? I actually guessed, but if I didn't guess, I would have to figure it out Kirchhoff equations (does anyone remember? - these are equations describing the flow of current). The network bandwidth is 184/89, which is approximately equal to 1,62.

About joy

By the way, I like the number 144. I liked to ride the bus with this number to the Castle Square in Warsaw - when there was no restored Royal Castle next to it. Perhaps young readers know what a dozen are. That's 12 copies, but only older readers remember that a dozen dozen, ie. 122=144, this is the so-called lot. And everyone who knows mathematics a little more than the school curriculum will immediately understand that fig. 10 we have Fibonacci numbers and that the network bandwidth is close to the "golden number"

In the Fibonacci sequence, 144 is the only number that is a perfect square. One hundred forty-four is also a "joyful number." That's how an Indian amateur mathematician Dattatreya Ramachandra Caprecar in 1955, he named numbers that are divisible by the sum of their constituent digits:

If he knew it Adam Mickiewicz, he would certainly have written no in Dzyady: “From a strange mother; his blood is his old heroes / And his name is forty-four, only more elegant: And his name is one hundred and forty-four.

Take entertainment seriously

I hope I've convinced readers that Sudoku puzzles are the fun side of questions that certainly deserve to be taken seriously. I can't develop this topic any further. Oh, full network bandwidth calculation from the diagram provided on fig. 9 writing a system of equations would take two or more hours - perhaps even tens of seconds (!) of computer work.