Condition as a compromise

They explained to me: “You have an orange, right? Now you cut this orange into a finite number of pieces, put them back into an orange, and it becomes as big as the sun. True or false? ”
- Between the pieces there is no space? - Not.
- Impossible! This simply can not be.
- Ha! Gotcha! Go all here! This is the theorem of Something about an immense measure!
And when it seems to them that they caught me, I remind them: “But you said an orange! And the orange peel can not be cut into pieces thinner than atoms. "
- But we have a condition of continuity. We can cut endlessly!
- No, you said an orange, so I accepted that you mean a real orange.
So I always won. If I guessed it - great. If he didn’t guess, he could always find something in their simplification that they had missed.

Richard Feynman. “You are joking, of course, Mr. Feynman!”

Prologue

It so happened that since childhood I have been fascinated by entertaining tasks. I solved them, as a rule, well and quickly, although I couldn’t do without funny things. For example, at the Mathematics Olympiad for the seventh grade, where I got, being in the sixth, there was a task: to find such and such a corner in a triangle with such and such properties. My knowledge in the field of geometry at that time was very fragmentary, but for some reason it was still enough. Without thinking, I built this triangle in a notebook with a compass and a ruler, and then measured the desired angle with a protractor. It was almost like in that joke about “find X”, when the student pointed his finger at the letter “x” with a joyful cry “here he is!”.

Another interesting case happened to me in the second year when, having missed several classes in algebra in a row, I finally decided to return to them. Having carefully studied the textbooks and notes, I came for a couple with the firm conviction that I know no less than a lecturer, or even more. And here the lecturer writes down the formulation of the theorem on the blackboard, and I immediately notice that the theorem is, in general, incorrect.

If you are familiar with general algebra, you should remember about such a thing as rings of characteristic 2. For those who do not know: these are numerical systems in which any number folded with itself will give zero. Because of this, they do not hold a bunch of properties to which we are accustomed to native integers / rational / real numbers. For example, knowing the squares of two numbers and the square of their sums, we can no longer find their product by the formula ab = ((a + b) ² - a ² - b ² ) / 2. This is due to the fact that (a + b) ² = a ² + (ab + ab) + b ² = a ² + b ² , and 2 = 1 + 1 = 0.
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So, I see that the theorem written on the blackboard is obviously wrong when characterizing a ring equal to two, and I hasten to share my observation with the lecturer. However, instead of being embarrassed and correcting the wording, the lecturer looks at me condescendingly, as at a person recovering from a hard meningitis and not yet fully recovering his mental tone. It turns out that a few lessons ago, he and the group agreed that all the theorems from this section are proved for rings with a characteristic different from two, and the corresponding reservation in their formulations is simply omitted for brevity.

What conclusion can be drawn from these episodes? The condition of the problem is often (probably even always) implied more than what is written. The task is not formulated “from scratch”, but in a particular context that imposes its own limitations. This is especially pronounced in those cases when the wording uses objects and concepts from the world we are used to, under which, however, pure mathematical abstractions are hidden.

In construction tasks, you can only use a compass and a ruler, and not ordinary ones, but mathematical ones. You can build a circle of arbitrarily large radius, which the real compass, of course, does not allow. But the ruler has only one edge, and therefore it is impossible to build a parallel straight line by drawing a pencil around the other. In transfusion tasks, you cannot drive to the store behind the beaker or use a pre-hidden flask whose capacity is known. On the other hand, the water in them does not evaporate, does not spill, does not change its volume depending on temperature - in general, it does not do a lot of cunning things that complicate accurate measurements in the real world.

The condition of the problem is a fragile compromise between what the author wanted to say and what the reader is prepared to take without serious consequences. Like all fragile, it can easily be broken. The creation of this habropost was inspired by the “solvers” of the prisoner problem, a chessboard and coins , which began to offer offtopic solutions, such as turning a coin, putting a coin on the edge, etc. Let us set aside the question of how witty it is, and think about this: what would happen if the wording of the tasks were perfectly strict?

Pulp

As an example, consider one simple problem with an extremely concise formulation. I chose her precisely because in due time she also got from the “solvers”. So, here it is:

Is it possible to wrap a 1x12 cube with edge 1 in exactly two layers?

• possible, if you break the tape into squares;
• as an option: unwind the tape into threads;
• it is impossible, because the tape has a thickness, and on the second layer it will need more;
• it is impossible, because no matter how wrapped, there will be cracks;
• it is impossible, on a bend of a tape there will be not two layers, but one / one and a half / pi in half;
• and these are just the options that I remember.

Well, let's do the formalization.



Let's start with what a cube is. First, it is the set of points in some three-dimensional Euclidean space E ³ . The interior of the cube in the context of the problem is of little interest to us, so we will deal with its surface. The surface consists of six faces, this is a non-trivial statement we will need in the future. We denote them as A ₁ ... A ₆ and clarify that we consider closed faces — that is, include corresponding edges. We cite the entire surface of the cube S. In this case, S = A ₁ ∪ A ₂ ∪ A ₃ ∪ A ₄ A ₅ A ₆ .

The tape, in turn, is a subset of some two-dimensional Euclidean space E ² . In order not to produce extra space, you can put it in the above space E ³ . It is convenient to define subsets of Euclidean spaces using equations and inequalities connecting the coordinates of their points. We introduce in our space a Cartesian coordinate system, and agree to designate the coordinates of points by triples (x, y, z). In this case, the tape can be set with the following system of equations and inequalities:

0 ≤ x ≤ 1
0 ≤ y ≤ 12
z = 0

Tape obzovim letter L. And what about the cube? Now that we have determined the tape so accurately, we would be nice to define the cube and its faces with the same accuracy. Well, let it be a single cube whose faces are described by the following systems:

A ₁ = {P (x, y, z) ∈ E ³ | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, z = 0}
A ₂ = {P (x, y, z) ∈ E ³ | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, z = 1}
A ₃ = {P (x, y, z) ∈ E ³ | 0 ≤ x ≤ 1, 0 ≤ z ≤ 1, y = 0}
A ₄ = {P (x, y, z) ∈ E ³ | 0 ≤ x ≤ 1, 0 ≤ z ≤ 1, y = 1}
A ₅ = {P (x, y, z) ∈ E ³ | 0 ≤ y ≤ 1, 0 ≤ z ≤ 1, x = 0}
A ₆ = {P (x, y, z) ∈ E ³ | 0 ≤ y ≤ 1, 0 ≤ z ≤ 1, x = 1}

Good. It remains to define such concepts as “wrapping” and “in two layers”. What is a wrapping? First is the mapping. Each point L we associate some point S. However, if we restrict ourselves to this definition. everything will turn out too easy: divide the tape into squares and display two squares on each face. To prohibit this hack, we could require that the map be homeomorphic, i.e. continuous. In this case, roughly speaking, the infinitely close points of the tape will go to the infinitely close points of the cube (I don’t want to tire the reader with the epsilon-delta language). However, this requirement will also be too weak: it describes an infinitely stretchable rubber band, and with such a thing in hand, you can wrap anything in any number of layers. We need a requirement that reflects the "non-rubber" tape.

In differential geometry there are such dialectically interrelated concepts as “bending” and “internal geometry”. In short, bending is a surface deformation in which the lengths of all the curves lying on it are preserved, and the internal geometry is surface properties that do not change during bending. It would be possible to use them, if not three circumstances:

1. Differential geometry deals with smooth manifolds, and we are going to stretch the tape on a nonsmooth cube.
2. The task is intended for students of the seventh-eighth grade, and if about these Euclidean spaces and the theory of sets these mischievous people still could overhear somewhere, then the words “differential geometry” will definitely be perceived by them as obscene.
3. Since we are talking about wrapping in two layers, obviously, the mapping will not be injective, and two different tape points may (and will) be displayed at the same cube point, which will obviously have a destructive effect on the internal geometry.

Fortunately, we have a way to get by with simpler concepts. Take the tape, carefully examine it and remember the distances between all pairs of its points (I hope you have a good memory). In this case, we will measure distances not within the framework of internal geometry, but in the way they are usually measured in E ³ . For the tape in its direct, initial state, these two methods of measurement will give the same results. Now we will doubt the tape and ask ourselves: for which pairs of points did the distances between them change?

As is known from the course of differential geometry (do not worry, we will need it only at the stage of drawing up the condition), the Eulerian curvature is preserved during bending. We, however, will not be engaged in bending the tape, but by crushing it - in our country, we can (and will) form straight folds in which the surface of the tape ceases to be smooth, and the word “differential” ceases to make sense. However, smooth pieces of tape between these folds due to the preservation of the Euler curvature will have a completely predictable shape: flat or cylindrical. The latter option disappears, since the cube, as a rule, has no rounded parts.

Perhaps the statement of the problem does not imply this explicitly, but at least out of charity we need to accept an additional restriction: the number of folds must be finite. In this case, the tape will be divided by these folds into a number of polygons. The distance between the pairs of points inside the same polygon will not change when the ribbon is crumpled. At the same time, points on the folds are in a particularly convenient position: for them, distances to points from all adjacent polygons are preserved. However, I would not like to give some points an unfair advantage over the others, so we will try to derive the property common to the points of both types.

For each point inside the polygon, you can choose a neighborhood that lies entirely inside this polygon. For each point on the border, you can choose a neighborhood that lies entirely inside and on the borders of adjacent polygons. In this way,

for a given kneading of the tape L for each point P, one can choose a neighborhood Ω such that for any point Q from this neighborhood the distance between P and Q when kneading remains unchanged

We take this property as a definition of crushing. However, we do not need to artificially increase the volume and complexity of the formulation, because mathematicians have already come up with a name for this class of mappings: they are called locally isometric . In contrast to simply isometric mappings that preserve distances for each pair of points, locally isometric maps preserve distances only on “microscales”, in some neighborhoods of points.

So, let's try to make an exact statement of the problem. We have a cube S and a ribbon L. Is there a locally isometric map L in S such that ... How to formulate a “two-layer wrapping”? The obvious answer is when each point S has exactly two pre-images in L. However, not everything is so simple, no wonder one of the "solvers" mentioned the edges, and the other about bends. These are problem areas for our task, because of which, precisely in such a formulation, it will not have a solution. Strict proof of this fact, I leave to the meticulous reader, but here I will mention that the bends of the tape can somehow be managed, but the edges create an insurmountable obstacle - special points through which the edge of the tape will pass too many times. You can, of course, instead of a closed tape, take an open one - in the system of inequalities defining the set L, we replace all "≤" by "<". But in this case, at the specific points of the tape layers will be too small. You can include some boundary points in L, and some - do not include. But it will be, sorry, some garbage.

Let's do it easier: let's give ourselves the right to ignore a small number of points on the surface of the cube. Actually, this approach is implied in most cutting tasks. When someone says “cut a square into four parts from which an equilateral triangle can be folded,” the last thing he wants is to think about what happens to a miserable handful of boundary points. But why this handful is pathetic? A square has a non-zero area (or, to put it more generally, a Jordan measure), and the set of boundary points of any measurable figure has a measure (area) equal to zero. Mathematicians sometimes consider themselves entitled to neglect such “small” sets - in such cases they usually add the magic phrase “with the exception of the set of measure zero” at the end of the wording.

So that we could say such a phrase, we need to define a measure on S. Of course, Jordan’s measure is already defined in E ³ , but bad luck - from its point of view, the whole set S has a zero measure entirely, because in three-dimensional space, a measure is a volume, not square. So, we need to take a flat measure and somehow extend it to the surface of the cube. Actually, this is not difficult to do: for each face of the cube, you need to take a flat measure of that part of the set that is located on this face, and then add them all.

Total

Now we are ready to give a rigorous statement of the condition.

Let the space E ^{3 be a} Cartesian coordinate system. Consider the following sets in this space:

A ₁ = {P (x, y, z) ∈ E ³ | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, z = 0}
A ₂ = {P (x, y, z) ∈ E ³ | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, z = 1}
A ₃ = {P (x, y, z) ∈ E ³ | 0 ≤ x ≤ 1, 0 ≤ z ≤ 1, y = 0}
A ₄ = {P (x, y, z) ∈ E ³ | 0 ≤ x ≤ 1, 0 ≤ z ≤ 1, y = 1}
A ₅ = {P (x, y, z) ∈ E ³ | 0 ≤ y ≤ 1, 0 ≤ z ≤ 1, x = 0}
A ₆ = {P (x, y, z) ∈ E ³ | 0 ≤ y ≤ 1, 0 ≤ z ≤ 1, x = 1}

S = A ₁ ∪ A ₂ ∪ A ₃ ∪ A ₄ ∪ A ₅ A ₆

L = {P (x, y, z) ∈ E ³ | 0 ≤ x ≤ 1, 0 ≤ y ≤ 12, z = 0}

On the set S, define the measure m _S as follows:

∀T ⊂ S m _S (T) = ∑ _{i = 1 ... 6} m ₂ (T ⋂ A _i ), where m ₂ is a flat Jordan measure.

Does there exist a locally isometric map F: L → S such that m _S ({P∈S | (| {Q ∈L | F (Q) = P} | ≠ 2)}) = 0?

The last formula, of course, is easier to express in words (the set of points on the surface of a cube, which do not have two types, has a measure of zero), but I just could not deny myself the pleasure of writing it in this creepy form.

It is time for the final words. As the old Kant used to say: "Act so that the maxim of your will could be a universal law." Frankly speaking, I did not fully understand what he wanted to say by this, but for myself I reformulated it in the following form: use such an action algorithm so that if all people on Earth begin to act according to the same algorithm, the world will not plunged into the chaos and mist of hell. So, comrades: if one day everyone starts to “solve” problems as I described at the beginning of the article, then one not quite beautiful day all the wording of the tasks will become the same as at the end of the article. Do not do it this way.

Peace to all, good, love and cookies.