Feynman Checkerboard an Intro to Algorithmic Quantum Field Theory

Feynman checkerboard An intro to algorithmic quantum field theory

E. Akhmedova, M. Skopenkov, A. Ustinov, R. Valieva, A. Voropaev Summary. It turns out that rainbow patterns on soapbubbles and unbelievable laws of motion of electrons can be explained by means of one simple-minded model. It is a game, in which a checker moves on a checkerboard by certain simple rules, and we count the turnings. This ‘Feynman checkerboard’ can explain all phenomena in the world (with a serious proviso) except atomic nuclei and gravitation. We are going 0.007 to solve mathematical problems related to the game and discuss their 0.006 0.005 physical meaning; no knowledge of physics is assumed. 0.004 0.003 Main results. The main results are explicit formulae for 0.002 0.001 the percentage of light of given color reflected by a glass plate of ∙ Basic model from §1 given width (Problem 24);

∙ the probability to find an electron at a given point, if it wasemit- ted from the origin (Problem 15; see Figure 1). Although the results are stated in physical terms, they are mathematical theorems, because we provide mathematical models of the physical phenomena in question, with all the objects defined rigorously. More precisely, a sequence of models with increasing precision. 0.00010 0.00008 Plan. We start with a basic model and upgrade it step by step 0.00006 0.00004 in each subsequent section. Before each upgrade, we summarize which 0.00002 physical question does it address, which simplifying assumptions does it resolve or impose additionally, and which experimental results does Model with mass from §3 it explain. Our aim is what is called 2-dimensional quantum electro- Figure 1: The probability to dynamics but the last steps on this way (sketched in Sections 8–10) find an electron at a given point still have not been done. (A 4-dimensional one can already explain all (white depicts large oscillations) phenomena — with proviso and exceptions — but we do not discuss it.) The scheme of upgrades dependence might help to choose your way:

1. Basic model 6. External field / 8***. Interaction O O

) * 4. Source o 2. Spin / 7. Identical particles & antiparticles 10***. QED 4

5. Medium o 3. Mass / 9***. Creation & annihilation Conventions. If a problem is a statement of an assertion, then it is requested to prove the assertion. A puzzle is a problem, in which both a precise statement and a proof are requested. Hard problems are marked with stars; you get first rank in Feynman checkerboard for solving one, and masters in Feynman checkerboard for solving three from three different sections. Solutions are accepted in writing but youmay spend an earned star for 10 attempts to submit a solution in oral form (successful or not). If you cannot solve a problem, proceed to the next ones: they may provide hints. Even if you do not reach the top (the proofs of main results), you learn much. You are encouraged to state and try to prove also your own observations and conjectures; you become a grand-master in Feynman checkerboard for discovering a new nontrivial one (and maybe even write your own scientific paper). 1 1 Basic model

Question: what is the probability to find an electron at the point (푥, 푦), if it was emitted from the point (0, 0)? Assumptions: no self-interaction, no creation of electron-positron pairs, unit mass and lattice step, point source; no nuclear forces, no gravitation, electron moves uniformly along the 푦-axis and does not move along the 푧-axis. Results: double-slit experiment, charge conservation. On an infinite checkerboard, a checker moves to the diagonal- neighboring squares, either upwards-right or upwards-left. To each path 푠 of the checker, assign a vector ⃗푎(푠) as follows. Start with a vector of length 1 directed upwards. While the checker moves straightly, the vector is not changed, but each time when the checker changes the direction, the vector is rotated through 90∘ clockwise (independently of the direction the checker turns). In addition, at the very end the vector is divided by 2(푦−1)/2, where 푦 is the total number of moves. The final position of the vector is what we denote by ⃗푎(푠). For instance, for the path in Figure 2 to the top, the vector ⃗푎(푠) = (1/8, 0) is directed to the right and has length 1/8. Denote ∑︀ , where the sum is over all the paths of the ⃗푎(푥, 푦) := 푠 ⃗푎(푠) checker from the square (0, 0) to the square (푥, 푦), starting with the upwards- right move. Set ⃗푎(푥, 푦) := ⃗0, if there are no such paths. For instance, ⃗푎(1, 3) = (0, −1/2)+(1/2, 0) = (1/2, −1/2). The length square of the vector ⃗푎(푥, 푦) is called the probability1 to find an electron in the square (푥, 푦), if it was emitted from the square (0, 0). Notation: 푃 (푥, 푦) := |⃗푎(푥, 푦)|2. Figure 2: Checker paths In Figure 1 to the top, the color of a point (푥, 푦) with even 푥 + 푦 depicts the value 푃 (푥, 푦). The sides of the apparent angle are not the lines 푦 = ±푥 (and nobody knows why!). In what follows squares (푥, 푦) with even and odd 푥 + 푦 are called black and white respectively. 1. Observations for small 푦. Answer the following questions for each 푦 = 1, 2, 3, 4 (and state your own questions and conjectures for arbitrary ): Find the vector and the probability for each . 푦 ⃗푎(푥, 푦) 푃 (0.05푥, 푦) 푥 When 푃 (푥, 푦) = 0? What is ∑︀ 푃 (푥, 푦) for fixed 푦? What are the directions of ⃗푎(1, 푦) and ⃗푎(0, 푦)? 푥∈Z 2 ′ ′ The probability to find an electron in the square (푥, 푦) subject to absorption-1000 in-500 the square 500(푥 , 푦 )1000is defined analogously to 푃 (푥, 푦), only the summation is over those paths 푠 that do not pass through (푥′, 푦′). The probability is denoted by 푃 (푥, 푦 bypass 푥′, 푦′). -0.05 2. Double-slit experiment. Is it true that bypass 푃 (푥, 푦) = 푃 (푥, 푦 0, 2)+ Figure 3: 푎2(푥, 1000) 푃 (푥, 푦 bypass 2, 2)? Is it true that 푃 (푥, 푦) ≥ 푃 (푥, 푦 bypass 푥′, 푦′)? 3. Find 푃 (0, 12). How to table the values ⃗푎(푥, 푦) quickly without exhaustion of all paths? (The first solution grants first rank in Feynman checkers.)

Denote by 푎1(푥, 푦) and 푎2(푥, 푦) the coordinates of ⃗푎(푥, 푦); see Figure 3.

4. Dirac’s equation. Express 푎1(푥, 푦) and 푎2(푥, 푦) through 푎1(푥 ± 1, 푦 − 1) and 푎2(푥 ± 1, 푦 − 1). 5. Probability/charge conservation. For each positive integer 푦 we have ∑︀ 푃 (푥, 푦) = 1. 푥∈Z

6. Symmetry. How the values 푎1(푥, 100) for 푥 < 0 and for 푥 ≥ 0 are related with each other? The same for the values 푎1(푥, 100) + 푎2(푥, 100). 7. Huygens’ principle. What is a fast way to find ⃗푎(푥, 199), if we know ⃗푎(푥, 100) for all integers 푥?

8. Using a computer, plot the graphs of the functions 푓푦(푥) = 푃 (푥, 푦) for various 푦, joining each pair of points (푥, 푓푦(푥)) and (푥 + 2, 푓푦(푥 + 2)) by a segment; cf. Figure 3. The same for the function 푎1(푥, 푦). 9. * Find an explicit formula for the vector ⃗푎(푥, 푦) and the probability 푃 (푥, 푦) (it is allowed to use a sum with at most 푦 summands in the answer). 10.** (Skipable puzzle) Guess a simple “approximate formula” for ⃗푎(푥, 푦) and 푃 (푥, 푦), accurate for |푥| ≪ 푦. 1One should think of the value 푦 as fixed, and the squares (−푦, 푦), (−푦 + 2, 푦),..., (푦, 푦) as all the possible outcomes of an experiment. For instance, the 푦-th horizontal might be a photoplate detecting the electron. Familiarity with probability theory is not required for solving the presented problems. Beware that our rule for the probability computation is valid only for the basic model in question; we are going to change the rule slightly in the upgrades. We make similar remarks each time we break some fundamental principles for simplicity. 2Thus an additional outcome of the experiment is that the electron has been absorbed and has not reached the photoplate. 2 2 Spin

Question: what is the probability to find a right electron at (푥, 푦), if a right electron was emitted from (0, 0)? Assumptions: the same. Results: spin reversal. The trick in the solution of the previous problems has a physical meaning: it is convenient to consider an electron as being in one of the two states: right-moving or left-moving. We write just ‘right’ or ‘left’ for brevity3. This is not just a convenience but reflects an inalienable electron’s property called spin4. Denote ∑︀ , where the sum is over only those paths from to , which both ⃗푎(푥, 푦, +) := 푠 ⃗푎(푠) (0, 0) (푥, 푦) start and finish with an upwards-right move. Define ⃗푎(푥, 푦, −) to be an analogous sum over paths which start with an upwards-right move but finish with an upwards-left move. The length square of the vector ⃗푎(푥, 푦, +) (respectively, ⃗푎(푥, 푦, −)) is called the probability5 to find a right (respectively, left) electron in the square (푥, 푦), if a right electron was emitted from the square (0, 0). Denote by 푃 (푥, 푦, +) := |⃗푎(푥, 푦, +)|2 and 푃 (푥, 푦, −) := |⃗푎(푥, 푦, −)|2 these probabilities.

11. Express ⃗푎(푥, 푦, +) and ⃗푎(푥, 푦, −) through 푎1(푥, 푦) and 푎2(푥, 푦); 푃 (푥, 푦) through 푃 (푥, 푦, +) and 푃 (푥, 푦, −). 12. * Spin reversal. What is the probability 푃 (푦 , −) := ∑︀ 푃 (푥, 푦 , −) to find a left electron on the 0 푥∈Z 0 line 푦 = 푦0 (it is allowed to use a sum with at most 푦0 summands in the answer)? Find the maximal element and the limit of the sequence 푃 (1, −), 푃 (2, −), 푃 (3, −),... . A crash-course in calculus. For this and other problems with stars, the following formuli might be useful. Recall that ∑︀푛 1 1 . Clearly, as increases, the sum becomes closer and closer to . We would like 푘=0 2푘 = 2 − 2푛 푛 2 to write ∑︀∞ 1 ; let us give a definition of such an infinite sum. A sequence has a limit , if 푘=0 2푘 = 2 푎1, 푎2, 푎3,... 푎 for each real 휀 > 0 there is 푁 such that for each integer 푛 > 푁 we have |푎푛 − 푎| < 휀. Notation: 푎 = lim푛→∞ 푎푛. For instance, (︀ 1 )︀ . By definition, put ∑︀∞ ∑︀푛 . Then indeed ∑︀∞ 1 . lim푛→∞ 2 − 2푛 = 2 푘=0 푎푘 = lim푛→∞ 푘=0 푎푘 푘=0 2푘 = 2 The following generalization is called Newton’s binomial theorem (allowed to use without proof):

∞ ∑︁ 푟(푟 − 1) ··· (푟 − 푘 + 1) (1 + 푥)푟 = 푥푘 푘(푘 − 1) ··· 1 푘=0 for each complex with and each real , or for and . In particular, for and 1 we get 푥 |푥| < 1 푟 푥 = 1 푟 > −1 푟 = −1 − 2

∞ ∞ 1 ∑︁ 1 ∑︁ 2푘(2푘 − 1) ··· (푘 + 1) 푥푘 = 푥푘 and √ = . 1 − 푥 1 − 푥 푘(푘 − 1) ··· 1 4푘 푘=0 푘=0 The following Stirling formula allows to estimate the summands in Newton’s binomial theorem: √ 2휋 푘푘+1/2푒−푘 ≤ 푘(푘 − 1) ··· 1 ≤ 푒 푘푘+1/2푒−푘.

푛 Here 푒 denotes lim푛→∞(1 + 1/푛) . It is an irrational number between 2.71 and 2.72.

3 Mass

Question: what is the probability to find a right electron of mass 푚 at (푥, 푦), if it was emitted from (0, 0)? Assumptions: the mass and the lattice step are now arbitrary. Results: a formula for the probability for small lattice step. To check our model against experiment we need the following generalization. Fix 휀, 푚 > 0 called lattice step and particle mass respectively. To each path 푠 of the checker, assign a vector ⃗푎(푠, 푚휀) as follows. Start with the vector (0, 1). While the checker moves straightly, the vector is not changed, but each time when the checker changes the direction, the vector is rotated through 90∘

3Beware that in 3 or more dimensions ‘right’ and ‘left’ mean something very different from the movement direction. Although often visualized as the direction of the electron rotation, these states cannot be explained in nonquantum terms. 4And chirality; beware that the term spin usually refers to a property, not related to the movement direction at all. 5Thus an experiment outcome is a pair (final 푥-coordinate, last-move direction), whereas the final 푦-coordinate is fixed. These are the fundamental probabilities, whereas 푃 (푥, 푦) should in general be defined by the formula from the solution of Problem 11 rather than the above formula 푃 (푥, 푦) = |⃗푎(푥, 푦)|2 (being a coincidence). 3 clockwise and multiplied by 푚휀. In addition, at the very end the vector is divided by (1 + 푚2휀2)(푦−1)/2, where 푦 is the total number of moves. The final position of the vector is what we denote by ⃗푎(푠, 푚휀). The vectors ⃗푎(푥, 푦, 푚휀, ±) and the numbers 푃 (푥, 푦, 푚휀, ±) are defined analogously to ⃗푎(푥, 푦, ±) and 푃 (푥, 푦, ±), only ⃗푎(푠) is replaced by ⃗푎(푠, 푚휀). For instance, 푃 (푥, 푦, 1, +) = 푃 (푥, 푦, +). In Figure 1 to the bottom, the color of a point (푥, 푦) with even 푥 + 푦 depicts the value 푃 (푥, 푦, 0.02, +) + 푃 (푥, 푦, 0.02, −). 13. (Puzzle) Massless and heavy particles. Find 푃 (푥, 푦, 0, +) and define 푃 (푥, 푦, ∞, +) for each 푥, 푦. 14. Solve analogues of Problems 4, 5, and 9 for 푚휀 ̸= 1.

In an experiment, we measure the probabilities to find the electron in intervals 푥0 ≤ 푥 ≤ 푥0 + ∆푥, 푦 = 푦0 rather than at particular points. Here 푥0, 푦0, ∆푥 are not integers but actual lengths measured in meters. If all squares have small size 1 1 , then the interval is approximated by the collection of black 푛 × 푛 squares (︀ ⌊︀ 푛푥 ⌋︀ ⌊︀ 푛푦 ⌋︀)︀ with satisfying the above (in)equalities. This leads to the following problem.6 2 2 , 2 2 푥, 푦 15.* (First main problem) Continuum limit. For each find (︀ ⌊︀ 푛푥 ⌋︀ ⌊︀ 푛푦 ⌋︀ 푚 )︀ and 푥, 푦, 푚 lim푛→∞ 푛⃗푎 2 2 , 2 2 , 푛 , − (︀ ⌊︀ 푛푥 ⌋︀ ⌊︀ 푛푦 ⌋︀ 푚 )︀ In the answer, it is allowed to use the following expressions7 lim푛→∞ 푛⃗푎 2 2 , 2 2 , 푛 , + ∞ ∞ ∑︁ (푧/2)2푘 ∑︁ (푧/2)2푘+1 퐽 (푧) := (−1)푘 and 퐽 (푧) := (−1)푘 . 0 (푘!)2 1 푘!(푘 + 1)! 푘=0 푘=0 4 Source

Question: what is the probability to find a right electron at (푥, 푦), if it was emitted by a source of wave length 휆? Assumptions: the source is now realistic. Results: wave propagation. A realistic source does not produce electrons localized at 푥 = 0 (as in our game) but a rather wide wave impulse instead. For our game, this means that the checker can start from an arbitrary black square on the horizontal line 푦 = 0 (not too far from the origin), but the initial direction of the vector ⃗푎(푠) is rotated through an angle proportional to the distance from the starting square to the origin; see Figure 4. Formally, fix real 휀, 휆 > 0 and odd ∆ called lattice step, wave length, and impulse width respectively. Denote by 푅훼 ⃗푎 the rotation of a vector ⃗푎 through the angle |훼|, which is counterclockwise for 훼 ≥ 0 and clockwise Figure 4: Checker paths may for 훼 < 0. Define the vector now start at distinct squares Δ−1 1 ∑︁ ∑︁ ⃗푎(푥, 푦, 휆/휀, ∆, +) := √ 푅2휋푥0휀/휆 ⃗푎(푠), ∆ 푥0=1−Δ 푠 푥0 even where the second sum is over all checker paths 푠 from the square (푥0, 0) to the square (푥, 푦), starting and ending with an upwards-right move. The length square of the vector is the probability to find a right electron at (푥, 푦), emitted by a source of wave length 휆 and impulse width ∆. It is denoted by 푃 (푥, 푦, 휆/휀, ∆, +). Define ⃗푎(푥, 푦, 휆/휀, ∆, −) and 푃 (푥, 푦, 휆/휀, ∆, −) analogously. For instance, 푃 (푥, 푦, 휆/휀, 1, +) = 푃 (푥, 푦, +) for each 휆, 휀, and ⃗푎(푥 + 1, 1, 휆/휀, ∆, +) = √1 (︀− sin 2휋푥휀 , cos 2휋푥휀 )︀ for even |푥| < ∆. Δ 휆 휆 16. Let ∆ = 3, 휆/휀 = 4. Find the vector ⃗푎(푥, 푦, 4, 3, +) and the probability 푃 (푥, 푦, 4, 3, +) for 푦 = 1, 2, 3 and each 푥. What is ∑︀ (푃 (푥, 푦, 4, 3, +)+푃 (푥, 푦, 4, 3, −)) for fixed 푦 = 1, 2, 3? When 푃 (푥, 3, 4, 3, +) = 0? 푥∈Z 17. Probability/charge conservation. Solve analogues of Problems 4 and 5 for ⃗푎(푥, 푦, 휆/휀, ∆, −), ⃗푎(푥, 푦, 휆/휀, ∆, +), and 푃 (푥, 푦, 휆/휀, ∆, +) + 푃 (푥, 푦, 휆/휀, ∆, −) instead of 푎1(푥, 푦), 푎2(푥, 푦), and 푃 (푥, 푦). √ 18. Causality. Both ∆⃗푎(푥, 푦, 휆/휀, ∆, +) and ∆푃 (푥, 푦, 휆/휀, ∆, +) do not depend on ∆ for ∆ > 푦 + |푥|. √ Denote8 ⃗푎(푥, 푦, 휆/휀, ±)= ∆⃗푎(푥, 푦, 휆/휀, ∆, ±) and 푃 (푥, 푦, 휆/휀, ±)=∆푃 (푥, 푦, 휆/휀, ∆, ±) for any ∆>푦+|푥|. 19. Wave. How to find ⃗푎(푥, 100, 휆/휀, +) for all even 푥, if we know it for just one even 푥? 20.* Wave propagation. For each 푥, 푦, 휆, 휀 find ⃗푎(푥, 푦, 휆/휀,−), 푃 (푥, 푦, 휆/휀,−), 푃 (푥, 푦, 휆/휀,−)+푃 (푥, 푦, 휆/휀,+). 6In the limits, the normalization factor of 푛 is a bit harder to explain; we not discuss it. 7Called Bessel functions, which are almost as well-studied as sine and cosine; but familiarity with them is not required. 8The notation ⃗푎(푥, 푦, 휆/휀, ±) should not be confused with ⃗푎(푥, 푦, 푚휀, ±). 4 5 Medium Question: which part of light of given color reflects from a glass plate of given width? Assumptions: right angle of incidence, no polarization of light; the mass now depends on 푥 but not on the color. Results: thin-film reflection. Our model can be applied also to describe propagation of light in transparent media such as glass9. Light propagates as if it had some nonzero mass inside the media, while the mass remains zero outside10. The wavelength determines the color of light. 21. (Puzzle) Define an analogue of ⃗푎(푠, 푚휀) in the case when the mass 푚 = 푚(푥) depends on 푥 so that analogues of Problems 4 and 5 remain true. Given a mass 푚 = 푚(푥), define ⃗푎(푥, 푦, 푚(푥)휀, 휆/휀, ±) and 푃 (푥, 푦, 푚(푥)휀, 휆/휀, ±) analogously to ⃗푎(푥, 푦, 휆/휀, ±) and 푃 (푥, 푦, 휆/휀, ±), only replace ⃗푎(푠) by ⃗푎(푠, 푚휀) in the definition. First we take 휀 = 1. 22. One-surface reflection. Find 푃 (푥, 푦, 푚(푥), 휆, +) and 푃 (푥, 푦, 푚(푥), 휆, −) for {︃ 0.2, for 푥 = 0; 푚(푥) = 푚0(푥) ≡ 0 and 푚(푥) = 푚1(푥) = 0, for 푥 ̸= 0.

Now fix odd 퐿 > 1 called the width of a glass plate. First assume for simplicity that light is reflected only by the two surfaces of the plate and thus take11 ⎧ −0.2, for 푥 = 1; ⎨⎪ 푚2(푥)= +0.2, for 푥 = 퐿; ⎩⎪0, otherwise. The reflection/transmission probabilities12 of light of wavelength 휆 by glass plate of width 퐿 are respectively 푃 (휆, 퐿, −) = lim 푃 (0, 푦, 푚2(푥), 휆, −); 푦→+∞ 푦 even

푃 (휆, 퐿, +) = lim 푃 (퐿 + 1, 푦, 푚2(푥), 휆, +). 푦→+∞ 푦 even 23. Plot the graph of the function 푓(퐿) = 푃 (0, 2퐿, 푚2(푥), 16, −). 24.* (Second main problem) Two-surface reflection. Find 푃 (휆, 퐿, −), 푃 (휆, 퐿, +), 푃 (휆, 퐿, −)+푃 (휆, 퐿, +), and max퐿 푃 (휆, 퐿, −). In fact the light is reflected inside the plate; this should be taken into account for a more accurate computation of the reflection probability (there is no hope for an exact solution for complicated mattersuch as glass). For this purpose we need to modify the model essentially. Take arbitrary 휀 > 0. Fix 푚 > 0 and let 푚3(푥) = 푚 for 0 < 푥 ≤ 퐿/휀 and 푚3(푥) = 0 otherwise. Now for each move starting in a square inside the glass, our vector is additionally rotated through the angle arctan 푚휀 clockwise (independently on if the 13 −푘 arctan 푚휀 checker does or does not turn in the square) . In other words, set ⃗푎m(푠, 푚(푥)휀) := 푅 ⃗푎(푠, 푚(푥)휀), where is the number of moves starting in the strip in the path . Define 휆 푘 0 < 푥 ≤ 퐿/휀 푠 푃m(푥, 푦, 푚(푥)휀, 휀 , −) analogously to 휆 , only replace by in the definition. 푃 (푥, 푦, 푚(푥)휀, 휀 , −) ⃗푎(푠, 푚(푥)휀) ⃗푎m(푠, 푚(푥)휀) ** 휆 25. Thin-film reflection. Find lim lim 푃m(0,푦,푚3(푥)휀, ,−). For which 푚 maximum of the expres- 휀→0+ 푦→+∞ 휀 푦 even 2휋푦휀/휆 휆 sion over 퐿 equals max퐿 푃 (휆, 퐿, −)? (Use existence of lim 푅 ⃗푎(푥,푦,푚3(푥)휀, ,−) without proof.) 푦→+∞ 휀 푦+푥 even 9Beware: in general Feynman checkerboard is inappropriate to describe light; partial reflection is a remarkable exception. √ √ 10The mass is proportional to (푛 − 1)/2 푛, where 푛 is the refractive index; for glass 푛 ≈ 1.5 and (푛 − 1)/2 푛 ≈ 0.2. 11This simplifying assumption requires negative mass for the left surface; the origin of that becomes clear after solving 25. 12It is more conceptual to define ∑︀ 휆 but we do not. 푃 (휆, 퐿, −) = lim휀→0 limΔ→+∞ lim푦→+∞ 푥∈ 푃 (푥, 푦, 푚3(푥)휀, 휀 , Δ, −) 13This additional rotation is explained as follows. The light can be scattered inZ each square inside the glass several times. Each individual scattering gives a factor of −푖푚휀 to our vector (viewed as a complex number) and may or may not change the movement direction. Assume that 푚휀 < 1. Thus a move without changing the direction contributes a factor of 1 1 (no scattering) − 푖푚휀 (1 scattering) + (−푖푚휀)2 (2 scatterings) + ··· = . 1 + 푖푚휀 Without a scattering, the checker moves straightly. Thus a turn in a particular square inside the glass contributes a factor −푖푚휀 −푖푚휀 (1 scattering) + (−푖푚휀)2 (2 scatterings) + (−푖푚휀)3 (3 scatterings) + ··· = . 1 + 푖푚휀 These are the same factors as in the model from §3 but additionally rotated through the angle arctan 푚휀 clockwise. 5 6 External field

Question: what is the probability to find a right electron at (푥, 푦), if it moves in a given magnetic field 푢? Assumptions: the magnetic field vanishes outside the 푥푦-plane, it is not affected by the electron. Results: deflection of electron and spin ‘precession’ in a magnetic field, charge conservation. An external magnetic field changes the motion as follows14. A common point of 4 squares of the checkerboard is called a vertex.A magnetic field 15 is a fixed assignment 푢 of numbers +1 and −1 to all the vertices. For instance, in Figure 5, the magnetic field is −1 at the top-right vertex of each square (푥, 푦) with both 푥 and 푦 even. Modify the definition of the vector ⃗푎(푠) by reversing the direction each time when the checker passes through a vertex with the magnetic field −1. Denote by ⃗푎(푠, 푢) the resulting vector. Formally, Figure 5: Paths in a field put ⃗푎(푠, 푢) = ⃗푎(푠)푢(퐶1)푢(퐶2) . . . 푢(퐶푦), where 퐶1, 퐶2, . . . , 퐶푦 are all the vertices passed by 푠. Define ⃗푎(푥, 푦, 푢, ±) and 푃 (푥, 푦, 푢, ±) analogously to ⃗푎(푥, 푦, ±) and 푃 (푥, 푦, ±) replacing ⃗푎(푠) by ⃗푎(푠, 푢) in the definition. For instance, if 푢(퐶) = +1 identically, then 푃 (푥, 푦, 푢) = 푃 (푥, 푦). 26. Homogeneous field. Let 푢(퐶) = −1, if 퐶 is the top-right vertex of a square (푥, 푦) with both 푥 and 푦 even, and 푢(퐶) = +1 otherwise. Find the vector ⃗푎(푥, 푦, 푢, +) and the probability 푃 (푥, 푦, 푢, +) for 푦 = 1, 2, 3, 4 and each integer 푥. What is ∑︀ (푃 (푥, 푦, 푢, +) + 푃 (푥, 푦, 푢, −)) for fixed 푦 = 1, 2, 3, or 4? 푥∈Z 27. Spin ‘precession’ in a magnetic field. Plot the graph of the function 푓(푦) = ∑︀ 푃 (푥, 푦, 푢, +) 푥∈Z for the field 푢 from the previous problem using a computer. For a given magnetic field 푢, a white square is negative, if 푢 equals −1 at 1 or 3 vertices of the square. 28. Gauge transformations. Changing the signs of the values of 푢 at the 4 vertices of one black square simultaneously does not change 푃 (푥, 푦, 푢, +). 29. Curvature. One can make 푢 to be identically +1 in a rectangle formed by checkerboard squares using the transformations from Problem 28, if and only if there are no negative white squares in the rectangle. 30. Homology. The magnetic field 푢 equals +1 on the boundary of a rectangle 푚 × 푛 formed by checkerboard squares. Which can be the number of negative white squares in the rectangle? 31. Probability/charge conservation. Solve analogues of Problems 4, 5 for 푢 not being identically +1.

7 Identical particles and antiparticles

Question: what is the probability to find electrons (or electron+positron) at 퐹 and 퐹 ′, emitted from 퐴 and 퐴′? Assumptions: the same as in the basic model; 푦-coordinate is interpreted as time. Results: exclusion principle. The motion of several electrons is described by a similar model as follows. To each pair of paths 푠, 푠′ of the checker, consisting of 푦 moves each, assign a vector ⃗푎(푠, 푠′) as follows. Start with the vector (0, 1). Move the checker consecutively along both paths, and rotate the vector according to the same rule as in §1: each time when the checker changes the direction, the vector is rotated through 90∘ clockwise. (Thus the vector is rotated totally 푡(푠) + 푡(푠′) times, where 푡(푆) is the number of turns in a path 푆.) In addition, at the very end the vector is divided by 2푦−1. The final position ′ of the vector is denoted by ⃗푎(푠, 푠 ). For instance, in Figure 2 we have ⃗푎(푠, 푠0) = (−1/4, 0). ′ ′ ′ Fix squares 퐴 = (0, 0), 퐴 = (푥0, 0), 퐹 = (푥, 푦), 퐹 = (푥 , 푦) and their diagonal neighbors 퐵 = (1, 1), ′ ′ ′ 16 퐵 = (푥0 + 1, 1), 퐸 = (푥 − 1, 푦 − 1), 퐸 = (푥 − 1, 푦 − 1), where 푥0 ̸= 0. Denote ∑︁ ∑︁ ⃗푎(퐴퐵, 퐴′퐵′ → 퐸퐹, 퐸′퐹 ′) := ⃗푎(푠, 푠′) − ⃗푎(푠, 푠′), 푠:퐴퐵→퐸퐹 푠:퐴퐵→퐸′퐹 ′ 푠′:퐴′퐵′→퐸′퐹 ′ 푠′:퐴′퐵′→퐸퐹 14Beware that this method of adding the magnetic field, although well-known, is very different from the one from [Feynman]. 15Or electromagnetic vector-potential, to be precise. The field is interpreted as magnetic or electric depending on if the 푦-coordinate is interpreted as position or time. 16Here it is essential that 푠 and 푠′ are paths of particles of the same sort, e.g., two electrons. Otherwise the 2nd sum is omitted. The sign before the 2nd sum is changed to plus for some other sorts of particles, e.g., photons (particles of light). 6 where the first sum is over all pairs consisting of a checker path 푠 starting with the move 퐴퐵 and ending with the move 퐸퐹 , and a path 푠′ starting with the move 퐴′퐵′ and ending with the move 퐸′퐹 ′, whereas in the second sum the final moves are interchanged. The length square 푃 (퐴퐵, 퐴′퐵′ → 퐸퐹, 퐸′퐹 ′) = |⃗푎(퐴퐵, 퐴′퐵′ → 퐸퐹, 퐸′퐹 ′)|2 is called the probability17 to find right electrons at 퐹 and 퐹 ′, if they are emitted from 퐴 and 퐴′. In particular, 푃 (퐴퐵, 퐴′퐵′ → 퐸퐹, 퐸퐹 ) = 0, i.e., two right electrons cannot be found at the same point; this is called exclusion principle.

′ ′ ′ ′ ′ 32. Independence. For 푥0 ≥ 2푦 and 푥 > 푥 express 푃 (퐴퐵, 퐴 퐵 → 퐸퐹, 퐸 퐹 ) through 푃 (푥, 푦, +) and ′ 푃 (푥 − 푥0, 푦, +). 33. Exclusion principle (for intermediate states). Prove that ⃗푎(퐴퐵, 퐴′퐵′ → 퐸퐹, 퐸′퐹 ′) is not changed, if the sums in the definition are over only those pairs of paths 푠, 푠′ which have no common moves. Define 푃 (퐴퐵, 퐴′퐵′ → 퐸퐹, 퐸′퐹 ′) analogously also for 퐸 = (푥 ± 1, 푦 − 1), 퐸′ = (푥′ ± 1, 푦 − 1).

34. Probability/charge conservation. We have ∑︀ ′ ′ ′ ′ , where the 퐸,퐸′,퐹,퐹 ′ 푃 (퐴퐵, 퐴 퐵 → 퐸퐹, 퐸 퐹 ) = 1 sum is over all quadruples 퐹 = (푥, 푦), 퐹 ′ = (푥′, 푦), 퐸 = (푥 ± 1, 푦 − 1), 퐸′ = (푥′ ± 1, 푦 − 1) with fixed 푦 ≥ 1. An electron has an antiparticle called positron. One can think of an antiparticle as a particle moving backwards in time. The motion is described by a similar model as follows. Add one more checker to the checkerboard, which now moves either downwards-right or downwards-left to the diagonal neighbors. The added checker is called black, while the one studied before is called white. For each pair of paths 푠, 푠′ of the white and black checker respectively, define a vector ⃗푎(푠, 푠′) analogously to the above, only for each turn of the black checker rotate the vector counterclockwise rather than clockwise (independently of the direction the checker turns). For instance, ⃗푎(푠, 푠′) = (0, 21−푦), if 푠 and 푠′ is the same path with opposite directions. Denote18 by ∑︁ ⃗푎(퐴퐵, 퐵′퐴′ → 퐸퐹, 퐹 ′퐸′) := ⃗푎(푠, 푠′) 푠,푠′ the sum is over all the pairs of a white-checker path 푠 starting with the move 퐴퐵 and ending with the move 퐸퐹 , and a black-checker path 푠′ starting with the move 퐹 ′퐸′ and ending with the move 퐵′퐴′. The length square 푃 (퐴퐵, 퐵′퐴′ → 퐸퐹, 퐹 ′퐸′) = |⃗푎(퐴퐵, 퐵′퐴′ → 퐸퐹, 퐹 ′퐸′)|2 is called the probability to find an electron at 퐹 and a positron at 퐹 ′, if they are emitted from 퐴 and 퐴′. 35. Independence. Solve analogue of Problem 32 for 푃 (퐴퐵, 퐵′퐴′ → 퐸퐹, 퐹 ′퐸′).

8 Interaction***

Question: the same as in the basic model but the 푦-coordinate is now interpreted as time. Assumptions: the electron now generates a magnetic field affecting the motion; encircling by waveproof walls. Results: interaction with the walls changes the motion of an electron. We have reached an unexplored area: a physically correct rigorous definition of the next upgrade is not yet known. In this section we give a WRONG definition leading to paradoxical results:

∙ the upgrades measures the interaction of the electron with the walls rather than self-interaction;

∙ the interaction (with the walls) propagates at infinite speed rather than the speed of light. But we hope that the introduced ideas still might be of interest. The moving electron itself generates an magnetic field, which in turn affects the motion. The generated field is random, with the probability resulting from summation over all possible intermediate fields. Fix 푔 ≥ 0 called the interaction constant 19 (it is related to the electron charge). Fix a black square (푥, 푦) called the final position. Fix the rectangle 푅 formed by all the squares (푥′, 푦′) such that 1−푦 < 푥′ < 푦

17One should think of the value 푦 as fixed, and the quadruples (퐹, 퐹 ′, 퐸, 퐸′) as the possible outcomes of an experiment. 18 This definition makes sense only for 푥0 ≥ 2푦; otherwise annihilation of particles cannot be ignored, and the model becomes inappropruate. 19If the lattice step 휖 is not fixed, then 푔 might depend on it. This is called renormalization; we do not discuss it. 7 and 0 < 푦′ < 푦. This waveproof box 푅 encircles any possible checker path 푠 from (0, 0) to (푥, 푦) excluding the first and the last move; it is required to make the sum over all intermediate fields belowfinite. Take any such path 푠 and any assignment 푢 of the numbers ±1 to all the vertices in the rectangle 푅. Let 푛 be the number of negative white squares in 푅 for 푢 (they play the role of turnings of the checker). Denote by 푔푛 ⃗푎(푠, 푢, 푔) = 푅−휋푛/2⃗푎(푠, 푢) 2(푦−2)2 (1 + 푔2)(푦−1)2/2 the vector20 ⃗푎(푠, 푢), rotated clockwise through 푛·90∘, multiplied by 푔푛 and divided by 2(푦−2)2 (1+푔2)(푦−1)2/2. Let 푢fin be any assignment of the numbers ±1 to all the vertices on the top side of 푅 such that 푢fin equals +1 at the endpoints of the side. It is called the final magnetic field. Denote by ∑︁ ⃗푎(푥, 푦, 푢fin, 푔, +) := ⃗푎(푠, 푢, 푔) 푠,푢 the sum over all the paths 푠 from (0, 0) to (푥, 푦) starting and ending with an upwards-right move and over all the assignments 푢 of the numbers ±1 to all the vertices in the rectangle 푅 such that {︃ 푢 on the top side of 푅; 푢 = fin +1 on all the other sides of 푅.

2 The length square 푃 (푥, 푦, 푢fin, 푔, +) = |⃗푎(푥, 푦, 푢fin, 푔, +)| is the probability that the final magnetic field equals 푢fin and a right electron is found at (푥, 푦). Define 푃 (푥, 푦, 푢fin, 푔, −) analogously.

36. Find the probabilities 푃 (푥, 푦, 푢fin, 1, +) and 푃 (푥, 푦, 푢fin, 1, −) for 푔=1, 푦=2, 3 and all possible 푥 and 푢fin. The sum21 ∑︁ 푃 (푥, 푦, 푔, +) = 푃 (푥, 푦, 푢fin, 푔, +)

푢fin over all assignments 푢fin of the numbers ±1 to the vertices on the top side of 푅 is the probability to find a right electron at (푥, 푦). Define 푃 (푥, 푦, 푔, −) analogously. For instance, 푃 (푥, 푦, 0, +) = 푃 (푥, 푦, +) (why?).

37. * For a path from to , consider the sum ∑︀ over all assignments of to all 푠 (0, 0) (푥, 푦) 푢 ⃗푎(푠, 푢, 푔) 푢 ±1 the vertices in the rectangle 푅 such that 푢 = +1 on the boundary of 푅. Express the sum through ⃗푎(푠) and the number of white squares in each of the two parts, into which 푅 is divided by the path 푠.

* 38. Let 푢± be equal to ±1 respectively at the top-right corner of the square (2 − 푦, 푦 − 1), and equal to +1 at all the other vertices on the top side of 푅. Solve analogues of Problems 4 and 5 for ⃗푎(푥, 푦, 푢±, 푔, ±) and 푃 (푥, 푦, 푔, +) + 푃 (푥, 푦, 푔, −) instead of 푎1(푥, 푦), 푎2(푥, 푦), and 푃 (푥, 푦).

9 Creation and annihilation*** Question: the same as in the model with identical particles and antiparticles. Assumptions: electron-positron pairs now created and annihilated, no interaction, encircling by reflecting walls. Results: no; this is only an ingredient for more realistic models with interaction.

Finally we have reached an unexplored area: we state an almost 40-years-old open problem. Start with mentioning what is not done in this section:

∙ we do not give a definition of the new upgrade (it is unknown so far);

∙ the upgrade (even if defined) would not explain any new experimental results. But

∙ we do give a precise statement of the problem: which exactly properties of the upgrade are requested;

20It is well-defined because the path 푠 is contained in 푅 except the first and the last move. Weset 푔푛 = 1 for 푔 = 푛 = 0. 21Here we sum probabilities rather than vectors. The notation 푃 (푥, 푦, 푔, +) should not be confused with 푃 (푥, 푦, 푚휀, +). 8 ∙ the upgrade is an important ingredient of further ones fantastically agreeing with experiment. Informally, our plan is as follows. Checker paths turning downwards or upwards or forming cycles mean creation and annihilation of electron-positron pairs. Even if we start with just one electron, we might end up with many electrons and positrons. To each possible configuration of the resulting particles, we want to assign a complex number so that its length square is the probability of the configuration in a sense. The number itself is the sum over all possible transitions from the initial configuration to the final one, that is, all possible paths configurations joining them. To make the sum finite, we put reflecting walls around.

Fix a rectangle 푅 formed by all the squares (푥, 푦) such that 푥min ≤ 푥 ≤ 푥max and 0 ≤ 푦 ≤ 푦max (the lines 푥 = 푥min and 푥 = 푥max are called reflecting walls). Fix 푚, 휀 > 0 called mass and lattice step. An initial configuration is any assignment22 of one of the signs “+” or “−” to some vertices inside 푅 lying between the lines 푦 = 0 and 푦 = 1. Physically the signs mean the initial positions of positrons and electrons respectively (and points without a sign are vacant). For our game, this means that the checkers pass the vertices with the “−” sign upwards-left or -right, and the vertices with the “+” sign — downwards- left or -right (and do not pass through vertices without a sign). Analogously, a final configuration is an assignment to vertices between the lines 푦 = 푦max and 푦 = 푦max − 1. An intermediate configuration is any assignment of one of the signs “+” or “−” to some vertices inside 푅 such that the difference between the number of “+” and “−” signs on the 2 top vertices of each black square in the strip 1 ≤ 푦 ≤ 푦max −1 equals the difference on the 2 bottom vertices. For our game, this means that the checkers start and finish motion in the lines 푦 = 0 or 푦 = 푦max only. In other words, the signs at the vertices of each black square are in one of the 19 positions23 shown in Figure 6 to the left. These 19 positions are called basic configurations.

Figure 6: Basic configurations

Suppose that one has assigned a complex number (depending on 푚휀) to each of the 19 basic configurations. A “right” choice of the numbers is unknown; think of them as fixed parameters of our model. Then take any intermediate configuration. In each black square inside 푅 not having common points with the boundary, write the complex number assigned to the basic configuration in the black square. Assign the product of all the written numbers to the intermediate configuration. Now to any pair of initial and final configurations, sum the complex numbers assigned to all possible intermediate configurations between them. Assign the resulting complex number to the pair.

39. (Puzzle) Restrict to configurations without “+” signs. Let 퐴, 퐴′, 퐵, 퐵′, 퐸, 퐸′, 퐹, 퐹 ′ be the black squares defined in §7. Take any 푥min < −푦 and 푥max > 푥0 + 푦. Fix the initial and final configurations with exactly two “−” signs, located at the top-right vertices of the squares 퐴, 퐴′, and 퐸, 퐸′ respectively. Assign complex numbers to the 6 basic configurations without “+” signs so that the sum of the numbers assigned to all intermediate configurations without “+” signs equals to the vector ⃗푎(퐴퐵, 퐴′퐵′ → 퐸퐹, 퐸′퐹 ′) from §7.

A case of particular interest is when the initial configuration consists of just one “−” sign at the top- right vertex of the square (0, 0) and the final configuration consists of just one“−” sign at the bottom-right 22The assignment has nothing to do with the magnetic field from §6. 23We consider positions of signs, but not possible checker paths in a particular black square like in Figure 6 to the right.

9 or bottom-left vertex of a black square (푥, 푦max). The complex numbers assigned to the pairs in question are denoted by ⃗푎(푥, 푦max, 푚휀, 푥min, 푥max, −) and ⃗푎(푥, 푦max, 푚휀, 푥min, 푥max, +) respectively. The desired continuum limit of these complex numbers involves the following modified Bessel functions and Hankel functions: ∫︁ ∞ −푥푡 ∫︁ ∞ 푖푥푡 푒 (1) 2 푒 퐾0(푥) = √ 푑푡 퐻 (푥) = √ 푑푡 2 0 2 1 푡 − 1 푖휋 1 푡 − 1 ∫︁ ∞ √ ∫︁ ∞ √ −푥푡 2 (1) 2푥 푖푥푡 2 퐾1(푥) = 푥 푒 푡 − 1 푑푡 퐻1 (푥) = − 푒 푡 − 1 푑푡 1 푖휋 1 40.*** Continuum limit. Assign complex numbers to the 19 basic configurations so that for each |푦| < |푥| (︁ ⌊︁ ⌋︁ ⌊︁ ⌋︁ )︁ 푛푥 푛푦 푚 √︀ 2 2 lim lim 푛⃗푎 2 , 2 , , −푥max, 푥max, − = 푚 퐾0(푚 푥 − 푦 ); 푛→∞ 푥max→∞ 2 2 푛 (︁ ⌊︁ ⌋︁ ⌊︁ ⌋︁ )︁ 푛푥 푛푦 푚 푥 + 푦 √︀ 2 2 lim lim 푛⃗푎 2 , 2 , , −푥max, 푥max, + = −푖푚√︀ 퐾1(푚 푥 − 푦 ); 푛→∞ 푥max→∞ 2 2 푛 푥2 − 푦2 and for each |푦| > |푥| we have (︁ ⌊︁ ⌋︁ ⌊︁ ⌋︁ )︁ 푛푥 푛푦 푚 휋 (1) √︀ 2 2 lim lim 푛⃗푎 2 , 2 , , −푥max, 푥max, − = 푖푚 퐻0 (푚 푦 − 푥 ); 푛→∞ 푥max→∞ 2 2 푛 2 (︁ ⌊︁ ⌋︁ ⌊︁ ⌋︁ )︁ 푛푥 푛푦 푚 휋 푥 + 푦 (1) √︀ 2 2 lim lim 푛⃗푎 2 , 2 , , −푥max, 푥max, + = 푚 · √︀ 퐻1 (푚 푦 − 푥 ). 푛→∞ 푥max→∞ 2 2 푛 2 푦2 − 푥2 Let us discuss the physical meaning of the upgrade. For 푦 ≫ |푥| the right-hand sides of the latter two equations are very close to the right-hand sides of the answer to Problem 10. Thus one may wish to interpret the upgrade as a more accurate approximation for the probability to find the electron in a square (푥, 푦). But this faces serious objections. First, it is in principle impossible to measure the coordinates of an electron exactly 24. Such a priori uncertainty has the same order of magnitude as the correction introduced by the upgrade. Thus the upgrade does not actually add anything to description of the electron motion. Second, for fixed initial configuration, the squares of the absolute values of the numbers assignedto all the possible final configurations do not sum up to 1 (even in the continuum limit). The reason is that distinct configurations are not mutually exclusive: for a field in a given configuration, there isapositive probability to find it in a different configuration. We do not know any clear explanation ofthat. To summarize, the upgrade lacks a direct physical interpretation, and should be considered as an ingredient for further upgrades.

10 (1 + 1)-dimensional quantum electrodynamics***

Question: what is the probability to find electrons (or electron+positron) with momenta 푞 and 푞′ in the far future, if they were emitted with momenta 푝 and 푝′ in the far past? Assumptions: interaction now switched on; all simplifying assumptions removed except the default ones: no nuclear forces, no gravitation, electron moves only along the 푥-axis (and 푦-coordinate is interpreted as time). Results: quantum corrections. Unifying the (so far unknown) upgrades discussed in the previous two sections would give an elementary definition of (1 + 1)-dimensional QED.

Future research An algorithmic quantum field theory is a one which for each experimentally observable quantity anda positive number 휀 provides a precise statement of an algorithm giving the predicted value of the quantity within accuracy 휀. (Surely, the predicted value does not have to agree with the experiment for 휀 less than accuracy of theory itself.) This is an extension of constructive quantum field theory, the latter currently being far away from algorithmic one.

24This should not be confused with uncertainty principle, which does not allow simultaneous measurement of the coordinates and momentum. 10 Epilogue (underwater rocks) We hope that at least some of our readers have become interested in elementary particles and want to learn more about them. As an epilogue, let us give a few warnings to such readers. In popular science, theory of elementary particles is usually oversimplified. This sequence of problems is not an exception. The toy models introduced here are very rough and should be considered with a grain of salt. Simplicity is their only advantage; if taken too seriously, the models could even give a wrong physical intuition. Real understanding of particles theory requires excellent knowledge of both physics and mathematics. We should also remark that nowadays there are almost no mathematical results in lattice quantum field theory; what we have is usually just a numeric simulation. Finally, there are“theories of New Physics” which are developed without any objective truth criterion: such theories are supported by neither experimental nor mathematical proofs (and some of them have experimental disproofs).

11 Hints, solutions, answers

2 For any vector ⃗푎 ∈ R denote by 푎1, 푎2 the coordinates of ⃗푎. E.g., ⃗푎(푥, 푦, 푚휀, −) = (푎1(푥, 푦, 푚휀, −), 푎2(푥, 푦, 푚휀, −)). Sometimes the vector ⃗푎 is considered as a complex number 푎1 + 푖푎2 (although complex numbers are not required for the solution of most problems). In what follows assume that 푥 and 푦 have the same parity unless the opposite is indicated. 1. Answer. In the table, the vector in the cell (푥, 푦) is ⃗푎(푥, 푦), and an empty cell (푥, 푦) means that ⃗푎(푥, 푦) = (0, 0).

(︁ )︁ (︁ )︁ (︁ )︁ (︁ )︁ 4 √1 , 0 0, − √1 √1 , − √1 0, √1 2 2 2 2 2 2 2 2 2 3 (︀ 1 , 0)︀ (︀ 1 , − 1 )︀ (︀0, 1 )︀ 2 (︁ )︁ 2 2 (︁ )︁ 2 2 √1 , 0 0, √1 2 2 1 (0, 1) y x −2 −1 0 1 2 3 4

In the following table, the number in the cell (푥, 푦) is 푃 (푥, 푦), and an empty cell (푥, 푦) means that 푃 (푥, 푦) = 0.

4 1/8 1/8 5/8 1/8 3 1/4 1/2 1/4 2 1/2 1/2 1 1 y x −2 −1 0 1 2 3 4

(The vectors ⃗푎(푥, 푦) can be easily computed consecutively using Problem 4.) See also the following figure.

For any positive 푦 we have ∑︀ 푃 (푥, 푦) = 1. (This is Problem 5.) 푥∈Z For −4 ≤ −푦 < 푥 ≤ 푦 ≤ 4 the probability 푃 (푥, 푦) equals 0 if and only if one of the numbers 푥, 푦 is odd and the other one is even. For any 푥 > 푦 and 푥 < −푦 + 1 we have 푃 (푥, 푦) = 0. (In general it is not known, if 푃 (푥, 푦) = 0 can vanish for 푥 and 푦 of the same parity satisfying −푦 < 푥 ≤ 푦.) For each odd 푦 we have ⃗푎(0, 푦) = 0. For each nonnegative integer 푛 we have

the vector ⃗푎(0, 8푛 + 2) is directed to the right, the vector ⃗푎(0, 8푛 + 4) is directed downwards, the vector ⃗푎(0, 8푛 + 6) is directed to the left, the vector ⃗푎(0, 8푛 + 8) is directed upwards.

Analogously, for each even 푦 we have ⃗푎(1, 푦) = 0. For each nonnegative integer 푛 we have

the vector ⃗푎(1, 8푛 + 3) is directed downwards-right, the vector ⃗푎(1, 8푛 + 5) is directed downwards-left, the vector ⃗푎(1, 8푛 + 7) is directed upwards-left, the vector ⃗푎(1, 8푛 + 9) is directed upwards-right. (See the remark after the answer to Problem 9.) 12 Remark. In the table, 2(푛+푘−2)/2⃗푎(푘 − 푛 + 1, 푛 + 푘 − 1) stands at the intersection of 푘-th column and 푛-th row:

(0, 1) (0, 1) (0, 1) (0, 1) (0, 1) (0, 1) (0, 1) (0, 1) (1, 0) (1, −1) (1, −2) (1, −3) (1, −4) (1, −5) (1, −6) (1, −7) (1, 0) (0, −1) (−1, −1) (−2, 0) (−3, 2) (−4, 5) (−5, 9) (−6, 14) (1, 0) (−1, −1) (−2, 0) (−2, 2) (−1, 4) (1, 5) (4, 4) (8, 0) (1, 0) (−2, −1) (−2, 1) (0, 3) (3, 3) (6, 0) (8, −6) (8, −14) (1, 0) (−3, −1) (−1, 2) (3, 3) (6, 0) (6, −6) (2, −12) (−6, −14) (1, 0) (−4, −1) (1, 3) (6, 2) (6, −4) (0, −10) (−10, −10) (−20, 0) (1, 0) (−5, −1) (4, 4) (8, 0) (2, −8) (−10, −10) (−20, 0) (−20, 20)

2. Answer: no for both questions. For example, 푃 (2, 4) = 5/8 ̸= 1/8+1/4 = 푃 (2, 4 bypass 2, 2)+푃 (2, 4 bypass 0, 2) and 푃 (0, 4) = 1/8 < 1/4 = 푃 (0, 4 bypass 2, 2). 3. Answer: 푃 (0, 12) = 25/512. Hint: the answer is obtained immediately by means of Remark after the answer to Problem 9 or it can be quickly computed recursively by means of the answer to Problem 4. {︃ 푎 (푥, 푦) = √1 푎 (푥 + 1, 푦 − 1) + √1 푎 (푥 + 1, 푦 − 1); 4. Answer: 1 2 2 2 1 푎 (푥, 푦) = √1 푎 (푥 − 1, 푦 − 1) − √1 푎 (푥 − 1, 푦 − 1). 2 2 2 2 1 Solution. Let us derive the formula for 푎2(푥, 푦); the derivation for 푎1(푥, 푦) is analogous. Consider any path 푠 from (0, 0) to (푥, 푦). Denote by 푡(푠) the number of turnings in 푠. Denote ⃗푎(푠) = (푎1(푠), 푎2(푠)). Notice that the component 푎2(푠) ̸= 0 if and only if 푡(푠) is even and 푎1(푠) ̸= 0 if and only if 푡(푠) is odd. Therefore, ∑︀ ∑︀ 푎2(푥, 푦) = 푎2(푠) and 푎1(푥, 푦) = 푎1(푠). 푠:푡(푠) even 푠:푡(푠) odd The last move in the path 푠 is made either from (푥 − 1, 푦) or from (푥 + 1, 푦). It is obvious that if 푡(푠) is even, then the last move is directed upwards-right, else it is directed upwards-left. Since we are interested in 푎2(푥, 푦), assume that last move is directed upwards-right. Denote by 푠′ the path 푠 without the last move. If the directions of the last moves in 푠 and 푠′ coincide, then ⃗푎(푠) = √1 ⃗푎(푠′), otherwise ⃗푎(푠) = √1 (푎 (푠′), −푎 (푠′)). Therefore, 2 2 2 1 ⎛ ⎞ ∑︁ 1 ∑︁ ′ ∑︁ ′ 1 푎2(푥, 푦) = 푎2(푠) = √ ⎝ 푎2(푠 ) − 푎1(푠 )⎠ = √ (푎2(푥 − 1, 푦 − 1) − 푎1(푥 − 1, 푦 − 1)). 2 2 푠:푡(푠) even 푠′:푡(푠′) even 푠′:푡(푠′) odd

5. Let us prove by induction over 푦 that ∑︀ 푃 (푥, 푦) = 1 for all 푦 ≥ 1: Obviously, ∑︀ 푃 (푥, 1) = 1. The step of 푥∈ 푥∈ induction follows immediately from the followingZ computation: Z

∑︁ ∑︁ [︀ 2 2]︀ ∑︁ 2 ∑︁ 2 푃 (푥, 푦 + 1) = 푎1(푥, 푦 + 1) + 푎2(푥, 푦 + 1) = 푎1(푥, 푦 + 1) + 푎2(푥, 푦 + 1) = 푥∈Z 푥∈Z 푥∈Z 푥∈Z 1 ∑︁ 1 ∑︁ = (푎 (푥 + 1, 푦) + 푎 (푥 + 1, 푦))2 + (푎 (푥 − 1, 푦) − 푎 (푥 − 1, 푦))2 = 2 1 2 2 2 1 푥∈Z 푥∈Z 1 ∑︁ 1 ∑︁ ∑︁ ∑︁ = (푎 (푥, 푦) + 푎 (푥, 푦))2 + (푎 (푥, 푦) − 푎 (푥, 푦))2 = [︀푎 (푥, 푦)2 + 푎 (푥, 푦)2]︀ = 푃 (푥, 푦). 2 1 2 2 2 1 1 2 푥∈Z 푥∈Z 푥∈Z 푥∈Z

A generalization of conservation law by Gleb Minaev and Ivan Russkikh, participants of Summer conference of Tournament of towns. Perform the change coordinates 푥+푦 푦−푥 , i.e., rotate the coordinate (푥, 푦) ↦→ (푡, 푢) = ( 2 − 1, 2 ) system through 45 degrees clockwise and shift it by the vector (−1, 0). Denote ⃗푏(푡, 푢) := ⃗푎(푡 − 푢 + 1, 푡 + 푢 + 1), 푄(푡, 푢) := 푃 (푡 − 푢 + 1, 푡 + 푢 + 1), 퐵⃗ (푡, 푢) := (1 + 푚2휀2)(푡+푢)/2 ⃗푏(푡, 푢). The coordinates of the vectors ⃗푏(푡, 푢) and 퐵⃗ (푡, 푢) denote by 푏1(푡, 푢), 푏2(푡, 푢), 퐵1(푡, 푢), 퐵2(푡, 푢). Remark. In the new coordinate system, a checker moves between neighbouring points of the integer lattice rather than along the diagonals between black squares. Also we suppose that we start at (0, 0) and move to any 2 point of (N ∪ {0}) , and the additional “pre-move” from (−1, 0) to (0, 0) is taken into account only to compute the number of turns. For a subset 2 denote ⃗ bypass ∑︀ ⃗ , where we sum is over all paths from 푀 ⊂ (N ∪ {0}) 푏(푡, 푢 푀) := 푠 푏(푠) 푠 (0, 0) to (푡, 푢), which bypass the points of the set 푀. Analogously, define 푄(푡, 푢 bypass 푀), ⃗푏(푡, 푢 bypass 푀, ±), 푄(푡, 푢 bypass 푀, ±). Also for a set 푀 denote ∑︁ 푄(푀) := 푄(푝 bypass 푀 ∖ {푝}). 푝∈푀 13 Remark. Note that for an infinite set 푀 the sum becomes infinite as well. The order of a summation is irrelevant because all the summands are positive.

2 Theorem 1. For each finite set 푀 ⊂ (N ∪ {0}) such that there are no infinite paths from (0; 0) bypassing the points of 푀 we have 푄(푀) = 1.

Proof. Prove the theorem by induction over 푚 := max(푡,푢)∈푀 (푡+푢), i.e., the maximal number of a downwards-right diagonal containing at least one point of the set 푀. The diagonal is called maximal. Base: 푚 = 0. In this case 푀 = (0, 0) and 푄(푀) = 푄(0, 0) = 1. Step. Let us prove a lemma.

2 2 Lemma 1. If a set 퐴 ⊂ (N ∩ {0}) , which does not contain the points (푡, 푢), (푡, 푢 + 1), (푡 + 1, 푢) ∈ (N ∩ {0}) , then

푄(푡, 푢 bypass 퐴) = 푄(푡, 푢 + 1 bypass 퐴, −) + 푄(푡 + 1, 푢 bypass 퐴, +).

Proof. This is straightforward:

2 2 푄(푡, 푢 + 1 bypass 퐴, −) + 푄(푡 + 1, 푢 bypass 퐴, +) = 푏1(푡, 푢 + 1 bypass 퐴) + 푏2(푡 + 1, 푢 bypass 퐴) = (푏 (푡, 푢 bypass 퐴) + 푏 (푡, 푢 bypass 퐴))2 + (푏 (푡, 푢 bypass 퐴) − 푏 (푡, 푢 bypass 퐴))2 = 1 2 2 1 = 2 2 2 = 푏1(푡, 푢 bypass 퐴) + 푏2(푡, 푢 bypass 퐴) = 푄(푡, 푢 bypass 퐴). Suppose that the point (푡, 푢) ∈ 푀 is such that 푡 + 푢 is maximal. Suppose that there is a checker path starting at the point (0, 0) and ending at (푡, 푢) with the last move, say, in the upwards direction bypassing all the points of the set 푀 ∖ {(푡, 푢)}. Then there exists a path to the point (푡, 푢 − 1), bypassing all the points of the set 푀 ∖ {(푡, 푢 − 1)}. Notice that in this case (푡, 푢 − 1) ∈/ 푀, because the move to (푡; 푢) is in the upwards direction. Hence there exists a path to the point (푡 + 1, 푢 − 1), bypassing all the points of the set 푀 ∖ {(푡 + 1, 푢 − 1)}. If (푡 + 1, 푢 − 1) ∈/ 푀 then there exists a path going to infinity passing through (푡 + 1, 푢 − 1) and bypassing all the points of the set 푀. For example, the path turning right at the point (푡, 푢) and only going right from there passes through (푡, 푢) and bypasses all the points in 푀 because the diagonal is maximal. Therefore (푡 + 1, 푢 − 1) ∈ 푀. 2 Notice that 푄(푎, 푏 bypass 퐴) = 푄(푎, 푏 bypass 퐴, +) + 푄(푎, 푏 bypass 퐴, −) for any set 퐴 ⊂ (N ∩ {0}) . Denote 퐾 := 푀 ∖ {(푡, 푢); (푡 + 1, 푢 − 1)}. Then we have the following chain of equalities (where we use 1 for the set 퐴 = 퐾):

푄(푀) = 푄(퐾) + 푄(푡, 푢 bypass 푀 ∖ {(푡, 푢)}) + 푄(푡 + 1, 푢 − 1 bypass 푀 ∖ {(푡 + 1, 푢 − 1)}) = = 푄(퐾) + 푄(푡, 푢 bypass 푀 ∖ {(푡, 푢)}, +) + 푄(푡, 푢 bypass 푀 ∖ {(푡, 푢)}, −)+ + 푄(푡 + 1, 푢 − 1 bypass 푀 ∖ {(푡 + 1, 푢 − 1)}, +) + 푄(푡 + 1, 푢 − 1 bypass 푀 ∖ {(푡 + 1, 푢 − 1)}, −) = = 푄(퐾) + 푄(푡, 푢 bypass 푀 ∖ {(푡, 푢)}, +)+ + 푄(푡, 푢 − 1 bypass 푀 ∖ {(푡, 푢 − 1)}) + 푄(푡 + 1, 푢 − 1 bypass 푀 ∖ {(푡 + 1, 푢 − 1)}, −) = 푄(푀 ∪ {(푡, 푢 − 1)}).

Thus if the point (푡, 푢 − 1) is added to the set 푀 then the probability 푄(푀) is not changed. This way we put new points onto the diagonal 푡 + 푢 = 푚 − 1, therefore, the maximal diagonal is not changed. Thus if we add a few points to the set 푀, then the paths bypassing other points of the set 푀 bypass the points of 푀 in the maximal diagonal as well. Therefore, if we remove all the points from the maximal diagonal, then 푄(푀) is not changed and no infinite path bypassing the points of the set 푀 appears. This way we change 푀 keeping 푄(푀) fixed but decreasing the number of a maximal diagonal. By the inductive hypothesis thenew 푄(푀) equals 1, hence the old one also equals 1.

6. Answer. For each 푥 we have 푎1(푥, 푦) = 푎1(−푥, 푦) and 푎2(푥, 푦) + 푎1(푥, 푦) = 푎2(2 − 푥, 푦) + 푎1(2 − 푥, 푦). Solution. For each path 푠 denote by 푓(푠) the reflection of 푠 with respect to the 푦 axis. Then if 푠 is a path to (푥, 푦), then 푓(푠) is a path to (−푥, 푦). For each path 푠 denote by 푔(푠) the path consisting of the same moves as 푠, but in the opposite order. Notice that reordering of moves does not affect the endpoint. Now consider a path 푠 to (푥, 푦) such that 푡(푠) is odd. Then the last move in 푠 is upwards-left. Therefore, the last move in 푓(푠) is upwards-right, hence the first move in 푔(푓(푠)) is upwards-right. Thus 푓 ∘ 푔 is a bijection between paths to (푥, 푦) with odd number of tunings beginning with an upwards-right move and paths to (−푥, 푦) with odd number of tunings beginning with an upwards-right move. To prove the second equation use the result of Problem 4: √ √ 푎1(푥, 푦) + 푎2(푥, 푦) = 2 · 푎1(푥 − 1, 푦 + 1) = 2 · 푎1(1 − 푥, 푦 + 1) = 푎1(2 − 푥, 푦) + 푎2(2 − 푥, 푦).

14 Remark. These identities can also be proved simultaneously by induction over 푦 using Problem 4. Also Problem 9 implies another identity (푦 − 푥) 푎2(푥, 푦) = (푦 + 푥 − 2) 푎2(2 − 푥, 푦). 7. Answer: for each 0 < 푦′ < 푦 we have

∑︁ [︀ ′ ′ ′ ′ ′ ′ ′ ′ ]︀ 푎1(푥, 푦) = 푎2(푥 , 푦 )푎1(푥 − 푥 + 1, 푦 − 푦 + 1) + 푎1(푥 , 푦 )푎2(푥 − 푥 + 1, 푦 − 푦 + 1) , 푥′ ∑︁ [︀ ′ ′ ′ ′ ′ ′ ′ ′ ]︀ 푎2(푥, 푦) = 푎2(푥 , 푦 )푎2(푥 − 푥 + 1, 푦 − 푦 + 1) − 푎1(푥 , 푦 )푎1(푥 − 푥 + 1, 푦 − 푦 + 1) . 푥′

The required expression for ⃗푎(푥, 199) is obtained by taking 푦 = 199 and 푦′ = 100. Solution. Fix any positive integer 푦′ < 푦. Consider a path 푠 from (0, 0) to (푥, 푦). Denote by (푥′, 푦′) the square ′ ′ ′ at which 푠 intersects the line 푦 = 푦 . Denote by 푠1 the part of 푠 that joins (0, 0) with (푥 , 푦 ) and by 푠2 the part ′ starting at the intersection square of 푠 with the line 푦 = 푦 − 1 and ending at (푥, 푦). Translate the path 푠2 so that it starts at (0, 0).

Denote ⃗푎(푠) = (푎1(푠), 푎2(푠)). Then {︃ 푎 (푠 )푎 (푠 ) if the move to (푥′, 푦′) is upwards-left, 푎 (푠) = 1 1 2 2 1 ′ ′ 푎2(푠1)푎1(푠2) if the move to (푥 , 푦 ) is upwards-right, {︃ −푎 (푠 )푎 (푠 ), if the move to (푥′, 푦′) is upwards-left, 푎 (푠) = 1 1 1 2 2 ′ ′ 푎2(푠1)푎2(푠2), if the move to (푥 , 푦 ) is upwards-right. Therefore, ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ∑︁ ∑︁ ∑︁ ∑︁ ⎢ ∑︁ ∑︁ ⎥ 푎1(푥, 푦) = 푎1(푠) = 푎1(푠) = ⎢ 푎2(푠1)푎1(푠2) + 푎1(푠1)푎2(푠2)⎥ = ′ ′ ′ ′ ⎢ ′ ′ ′ ′ ⎥ 푠 푥 푠 through (푥 ,푦 ) 푥 ⎣푠 through (푥 ,푦 ), 푠 through(푥 ,푦 ), ⎦ move to (푥′,푦′) move to (푥′,푦′) is upwards-right is upwards-left ∑︁ [︀ ′ ′ ′ ′ ′ ′ ′ ′ ]︀ = 푎2(푥 , 푦 )푎1(푥 − 푥 + 1, 푦 − 푦 + 1) + 푎1(푥 , 푦 )푎2(푥 − 푥 + 1, 푦 − 푦 + 1) . 푥′

The formula for 푎2(푥, 푦) is proven analogously. Remark. Formulae illustrating the Huygens principle have the form of a convolution. In combinatorics the convolution of two given sequences {푎푘} and {푏푙} (푘, 푙 ≥ 0) is the sequence {푐푛} (푛 ≥ 0), defined by the equality ∑︀ A convolution of two sequences corresponds to the product of their generating functions: if 푐푛 = 푘+푙=푛 푎푘푏푙.

∞ ∞ ∞ ∑︁ 푘 ∑︁ 푙 ∑︁ 푛 퐴(푡) = 푎푘푡 , 퐵(푡) = 푏푙푡 , 퐶(푡) = 푐푛푡 , 푘=0 푙=0 푛=0 then 퐶(푡) = 퐴(푡)퐵(푡). Therefore, choosing the generating functions in a right way, we can write down the Huygens principle in a more compact form. Define two series of Laurent polynomials (i.e., polynomials in variables 푡 and 푡−1):

푛 푛 ∑︁ 푚 ∑︁ 푚 푃푛 = 푃푛(푡) = 푎1(푚, 푛)푡 , 푄푛 = 푄푛(푡) = 푎2(푚, 푛)푡 . 푚=−푛+2 푚=−푛+2

15 These polynomials can be considered as generating fuctions of the sequences 푎1,2(푚, 푛) with fixed 푛. From the answer to the Problems 6 and 7 we have

∑︁ [︀ ′ ′ ′ ′ ′ ′ ′ ′ ]︀ 푎2(푥, 푦) = 푎2(푥 , 푦 )푎2(푥 − 푥 + 1, 푦 − 푦 + 1) − 푎1(푥 , 푦 )푎1(푥 − 푥 + 1, 푦 − 푦 + 1) = 푥′ ∑︁ [︀ ′ ′ ′ ′ ′ ′ ′ ′ ]︀ = 푎2(푥 , 푦 )푎2(푥 − 푥 + 1, 푦 − 푦 + 1) − 푎1(푥 , 푦 )푎1(푥 − 푥 − 1, 푦 − 푦 + 1) . 푥′ This formula is equivalent to the following addition theorem for 푄–polynomials: 1 푄 = 푄 푄 − 푡푃 푃 . 푛+푚 푡 푛 푚+1 푛 푚+1 Also from the Problems 6 and 7 we have

∑︁ [︀ ′ ′ ′ ′ ′ ′ ′ ′ ]︀ 푎1(푥, 푦) = 푎2(푥 , 푦 )푎1(푥 − 푥 + 1, 푦 − 푦 + 1) + 푎1(푥 , 푦 )푎2(푥 − 푥 + 1, 푦 − 푦 + 1) = 푥′ ∑︁ ′ ′ ′ ′ = [푎2(푥 , 푦 )푎1(푥 − 푥 + 1, 푦 − 푦 + 1)+ 푥′ ′ ′ ′ ′ ′ ′ ′ ′ + 푎1(푥 , 푦 )(푎2(푥 − 푥 + 1, 푦 − 푦 + 1) + 푎1(푥 − 푥 + 1, 푦 − 푦 + 1) − 푎1(푥 − 푥 − 1, 푦 − 푦 + 1))]. This formula respectively is equivalent to the following addition theorem for 푃 –polynomials:

1 (︂1 )︂ 푃 = (푄 푃 + 푃 푄 ) + − 푡 푃 푃 . 푛+푚 푡 푛 푚+1 푛 푚+1 푡 푛 푚+1 In particular substitution of 푚 = 1 gives Dirac’s equation written in terms of 푃 – and 푄–polynomials: {︃ 푃 = 1 푄 푃 + 푃 · ( 1 푄 + 1 푃 − 푡푃 ) = √1 (푄 + 푃 ), 푛+1 푡 푛 2 푛 푡 2 푡 2 2 푡 2 푛 푛 푄 = 1 푄 푄 − 푡푃 푃 = √푡 (푄 − 푃 ). 푛+1 푡 푛 2 푛 2 2 푛 푛 It follows immediately that √ √ √ {︃ 2 {︃ 2 푄푛+1 {︃ 1 1 푡 2푃 − 푄 = 푃 = 푄 − 푄 푃 = 푄 − 푃푛+1 = √ (푡 + )푃푛 − 푃푛−1 √ 푛+1 푛 푛 푛 √푡 푛+1 ⇒ 푛+1 푡 푛 푡√2 ⇒ 2 푡 푡 2푃 − 푃 = 푄 = 푃 + 2 푄 푄 = 푡2푃 − 푡 2푃 푄 = √1 (푡 + 1 )푄 − 푄 . 푛+1 푛 푛 푛 푡 푛+1 푛+1 푛+1 푛 푛+1 2 푡 푛 푛−1

The latter formulae give us yet another recursive definitions for 푃푛 and 푄푛. 8. Answer: the graphs of 푃 (푥, 1000), 푎1(푥, 1000), and 푎2(푥, 1000) respectively as functions in 푥: 0.0030

0.06 0.0025 0.05 0.04 0.0020 0.02 0.0015

-1000 -500 500 1000 -1000 -500 500 1000 0.0010 -0.02

0.0005 -0.04 -0.05 -0.06 -1000 -500 500 1000 In the above graph of 푃 (푥, 1000), large values are cut out, to make the oscillations more visible. Compare with the following graphs of 푃 (푥, 1000) and 푃 (푥, 푦) without cutting (kindly provided by A. Daniyarkhodzhaev and F. Kuyanov, participants of the Summer School in Contemporary Mathematics in Dubna, 2019):

16 9. Answer 1: for each 푦 > |푥| we have

푦−|푥|−2 sign (︂ )︂(︂ )︂ (1−푦)/2 ∑︁ 푦−푥−2+2푤· (푥) |푥| 2 푎 (푥, 푦) = 2 (−1) 4 , 1 푤 푦−푥−2−2푤 푦−푥−2 4 푤≡2 2 푦−|푥|−2 sign (︂ )︂(︂ )︂ (1−푦)/2 ∑︁ 푦−푥+2푤· (푥) |푥| 2 푎 (푥, 푦) = 2 (−1) 4 ; 2 푤 푦−푥−2푤 푦−푥 4 푤≡2 2 for each 푦 = 푥 > 0 we have

푎1(푦, 푦) = 0, (1−푦)/2 푎2(푦, 푦) = 2 ; for each 0 < 푦 < |푥| or 푦 = −푥 > 0 we have

푎1(푥, 푦) = 0,

푎2(푥, 푦) = 0.

Answer 2: for each 푦 > |푥| we have

⌊푦/2⌋ ∑︁ (︂(푥 + 푦 − 2)/2)︂(︂(푦 − 푥 − 2)/2)︂ 푎 (푥, 푦) = 2(1−푦)/2 (−1)푟 , 1 푟 푟 푟=0 ⌊푦/2⌋ ∑︁ (︂(푥 + 푦 − 2)/2)︂(︂(푦 − 푥 − 2)/2)︂ 푎 (푥, 푦) = 2(1−푦)/2 (−1)푟 ; 2 푟 푟 − 1 푟=1 For other 푦 the answer is the same as Answer 1. Answer 3. For even 푝 + 푞 and 2 − 푞 ≤ 푝 ≤ 푞 we have25

푞−2 푞 (푞−3)/2 ∑︁ −푘 푘! 푎1(푝, 푞) = (−1) 2 · (−2) · (푞 − 2 − 푘)!( 푝−푞 + 1 + 푘)!(푘 − 푝+푞 + 1)! 푞+|푝|−2 2 2 푘= 2 (︂ )︂ 푞−2 (1−푞)/2 푞 − 1 푞 (푞−3)/2 ∑︁ −푘 푘!(푘 + 푝 + 1) 푎2(푝, 푞) = 2 · − (−1) 2 (−2) · 푞−푝 (푞 − 1 − 푘)!( 푝−푞 + 1 + 푘)!(푘 − 푝+푞 + 1)! 2 푞+|푝|−2 2 2 푘= 2

푦−푘−1 푦−푘 Remark. Answer 2 shows that 푎1(−푦 + 2푘, 푦) and 푎2(−푦 + 2푘, 푦) are the coefficients before 푡 and 푡 in the expansion of 2(1−푦)/2(1 + 푡)푦−푘−1(1 − 푡)푘−1 for 1 ≤ 푘 ≤ 푦 − 1. In particular,

푛 (︂ )︂ (−1) 2푛 푎1(0, 4푛) = 0, 푎1(0, 4푛 + 2) = , 2(4푛+1)/2 푛 and (−1)푛 (︂2푛 − 1)︂ 푎2(0, 4푛) = . 푎2(0, 4푛 + 2) = 0, 2(4푛−1)/2 푛

The sums in the expressions for 푎1(푥, 푦) and 푎2(푥, 푦) for 푦 > |푥| are particular cases of well-known Gaussian hypergeometric functions (but familiarity with them is not required for what follows):

(︂ 푥 + 푦 푥 − 푦 )︂ 푎 (푥, 푦) = 2(1−푦)/2 퐹 1 − , 1 + , 1; −1 , 1 2 1 2 2 (︂ 푥 + 푦 )︂ (︂ 푥 + 푦 푥 − 푦 )︂ 푎 (푥, 푦) = 2(1−푦)/2 1 − 퐹 2 − , 1 + , 2; −1 . 2 2 2 1 2 2

Hint. Let us find 푎1(푥, 푦) for 푦 > |푥|. Consider a path from the square (0, 0) to the square (푥, 푦) with an odd number of turns (the others do not contribute to 푎1(푥, 푦)). Denote by 푅 and 퐿 the number of upwards-right 25Fractions in those equalities are respectively (︂ )︂ (︂ )︂ 푘 and 푘 푘 + 푝 + 1 푝−푞 푝+푞 푝−푞 푝+푞 · . 푞 − 2 − 푘, 2 + 1 + 푘, 푘 − 2 + 1 푞 − 2 − 푘, 2 + 1 + 푘, 푘 − 2 + 1 푞 − 1 − 푘

17 and upwards-left moves respectively. These numbers are uniquely determined by the conditions 퐿 + 푅 = 푦 and 푅 − 퐿 = 푥. Denote by 2푟 + 1 the number of turns in the path. Let 푥1, 푥2, . . . , 푥푟+1 be the number of consecutive upwards-right moves before the first, the third, . . . , the last turn respectively. Let 푦1, 푦2, . . . , 푦푟+1 be the number of consecutive upwards-left moves after the first, the third, . . . , the last turn respectively. Then 푥푘, 푦푘 ≥ 1 and

푅 = 푥1 + ··· + 푥푟+1;

퐿 = 푦1 + ··· + 푦푟+1.

Conversely, each collection (푥1, 푥2, . . . , 푥푟+1, 푦1, 푦2, . . . , 푦푟+1) satisfying the resulting equations determines a path from (0, 0) to (푥, 푦) with an odd number of turns. The problem now reduces to a combinatorial one: find the number of positive integer solutions of the resulting equations. For the first equation, this number equals to the number of ways toput 푟 sticks between 푅 coins in a row, that is, (︀푅−1)︀. Thus 푟 ⌊푦/2⌋ ∑︁ (︂푅 − 1)︂(︂퐿 − 1)︂ 푎 (푥, 푦) = 2(1−푦)/2 (−1)푟 . 1 푟 푟 푟=0 Since 퐿 + 푅 = 푦 and 푅 − 퐿 = 푥, the required formula from Answer 2 follows. The formula for 푎1(푥, 푦) from Answer 1 can be derived using the polynomial from the remark after the answer. Depending on the sign of 푥 we get

{︃ (1−푦)/2 2 푦−푥−2 푥 2 (1 − 푡 ) 2 (1 − 푡) , for 푥 ≥ 0; 2(1−푦)/2(1 + 푡)푦−푘−1(1 − 푡)푘−1 = (1−푦)/2 2 푦+푥−2 −푥 2 (1 − 푡 ) 2 (1 + 푡) , for 푥 < 0.

푥−푦−2 푥−푦 For each |푥| < 푦 the numbers 푎1(푥, 푦) and 푎2(푥, 푦) are equal to the coefficients before 푡 2 and 푡 2 in the expansion of the polynomials. Considering the cases 푥 ≥ 0 and 푥 < 0 separately we get the required formula. The proof of the formulae for 푎2(푥, 푦) for 푦 > |푥| is analogous. The case 푦 ≤ |푥| is obvious.

Solution by Alexander Kudryavtsev (participant of Summer conference of Tournament of Towns). Denote 푞/2 푞/2 푐푝,푞 = 푎1(2푝 − 푞 + 1, 푞 + 1) · 2 , 푑푝,푞 = 푎2(2푝 − 푞 + 1, 푞 + 1) · 2 (they are well-defined for 0 ≤ 푝 ≤ 푞 − 1). Let 퐶(푥, 푦), 퐷(푥, 푦) be generating functions of 푐푝,푞 and 푑푝,푞 respectively. From the answer to Problem 4 we obtain the following equations: ⎧퐶(푥, 푦) − 퐶(푥, 0) ⎪ = 퐷(푥, 푦) + 퐶(푥, 푦) ⎨ 푦 퐷(푥, 푦) − 퐷(푥, 0) ⎩⎪ = 푥 · (퐷(푥, 푦) − 퐶(푥, 푦)) 푦 Since 퐶(푥, 0) = 0 and 퐷(푥, 0) = 1, we get {︃ 퐶(푥, 푦) = 푦 · (퐷(푥, 푦) + 퐶(푥, 푦)) 퐷(푥, 푦) − 1 = 푥푦 · (퐷(푥, 푦) − 퐶(푥, 푦))

푦 1 − 푦 The solution of this system is 퐶(푥, 푦) = and 퐷(푥, 푦) = . Denote 1 − 푦 − 푥푦 + 2푥푦2 1 − 푦 − 푥푦 + 2푥푦2

퐶(푥, 푦) 퐷(푥, 푦) 퐸(푥, 푦) := = = 1 + 푦(1 + 푥 − 2푥푦) + 푦2(1 + 푥 − 2푥푦)2 + ... 푦 1 − 푦

The coefficient of 푥푛 · 푦푚 in the 퐸(푥, 푦) is equal to

푚 ∑︁ 푘! (−2)푚−푘 · , (푚 − 푘)!(푛 − 푚 + 푘)!(푘 − 푛)! 푘=max(푛,푚−푛) since from every summand of the form 푦푘(1 + 푥 − 2푥푦)푘 we must take exactly one combination of factors:

∙ for the power of 푦 to be equal to 푚, the number of factors −2푥푦 must be exactly 푚 − 푘;

∙ for the power of 푥 to be equal to 푛, the number of factors 푥 must be exactly 푛 − (푚 − 푘);

∙ for the total number of factors to be 푘, the number of factors 1 must be 푘 − (푛 − (푚 − 푘)) − (푚 − 푘) = 푘 − 푛.

18 And the number of ways to choose 푚 − 푘, 푘 − 푛 and 푛 − (푚 − 푘) factors (1 + 푥 − 2푥푦) equals to the fraction in the formula for 퐸(푥, 푦). Now Answer 3 can be easily obtained from the equations 퐶(푥, 푦) = 푦 · 퐸(푥, 푦) and 퐷(푥, 푦) = 퐸(푥, 푦) − 퐶(푥, 푦). 10. Answer:

√ (︁ (︁ 휋푦 2 )︁ (︁ 휋푦 푥(2−푥) )︁)︁ 푥 푥/푦 2 2 sin 4 − 2푦 , 푒 cos 4 + 2푦 (︂log 푦 )︂ √ ⃗푎(푥, 푦) = √ + O for |푥| = O ( 푦) . 휋푦 푦

Remark. Compare the graphs of 푃 (푥, 400), 푎1(푥, 400), and 푎2(푥, 400) (blue) with the graphs of their approxi- mations in the right-hand side for 푦 = 400 (orange): 0.007 0.10

0.006 0.10

0.005 0.05 0.05 0.004

0.003 -400 -200 200 400 -400 -200 200 400

0.002 -0.05

0.001 -0.05 -0.10

-400 -200 200 400 Hint. Note that by Problem 9

−푛/2 푘−1 푘 푛−푘−1 푘 푎2(−푛 + 2푘 + 1, 푛 + 1) = 2 [푥 ](−1) (1 + 푥) (1 − 푥) ; −푛/2 푘 푘 푛−푘−1 푘 푎1(−푛 + 2푘 + 1, 푛 + 1) = 2 [푥 ](−1) (1 + 푥) (1 − 푥) . Thus ∫︁ 1 −푛/2 푘 2휋푖푡 푛−푘−1 2휋푖푡 푘 −2휋푖(푘−1)푡 푎2(−푛 + 2푘 + 1, 푛 + 1) = 2 (−1) (1 + 푒 ) (1 − 푒 ) 푒 푑푡. 0 For an odd denote 푛−1 . We have 푛 푙 = 푘 − 2 푘 ∫︁ 1 푖 푛−1 푙 −2휋푖(푙−1)푡 푎2(−푛 + 2푘 + 1, 푛 + 1) = √ (sin(2휋푡)) 2 (tan(휋푡)) 푒 푑푡. 2 0 Since we can calculate the integral in the areas , with good accuracy. 푙 << 푛, √ (1/4 − 훿, 1/4 + 훿) (3/4 − 훿, 3/4 + 훿) Denote 푈 ≪ 푛 log 푛, |푙| ≤ 푈. In (1/4 − 훿, 1/4 + 훿) we have 푡 = 1/4 + 휏, |휏| ≤ 훿,

2 2 sin(2휋푡) = cos(2휋휏) = 1 − 2휋2휏 2 + 푂(휏 4) = 푒−2휋 휏 + 푂(휏 4); tan(휋푡) = 1 + 2휋휏 + 푂(휏 2) = 푒2휋휏 + 푂(휏 2); 푒−2휋푖(푙−1)푡 = 푖1−푙푒−2휋푖(푙−1)휏 and we get 푘+1−푙 ∫︁ 훿 푖 −2휋2휏 2 4 푛−1 2휋휏 2 푙 −2휋푖(푙−1)휏 √ (푒 + 푂(휏 )) 2 (푒 + 푂(휏 )) 푒 푑휏. (*) 2 −훿 As 푛휏 4 ≪ 1 and 푙휏 2 ≪ 1, i.e. 푛훿4 ≪ 1 and 푈훿2 ≪ 1,

푛+1 훿 푖 2 ∫︁ 2 2 (*) = √ (푒−(푛−1)휋 휏 + 푂(휏 4))(푒2휋휏푙 + 푂(휏 2))푒−2휋푖(푙−1)휏 푑휏. 2 −훿

If we open the brackets we get the summand 푂(푛휏 4푒2휋푙휏 ) = 푂(푛훿4푒2휋푈훿) = 푂(푛−Δ), since we ask 푈훿 ≪ log 푛. Then 푛+1 훿 푖 2 ∫︁ 2 2 (*) = √ (푒−(푛−1)휋 휏 +2휋휏푙−2휋푖(푙−1)휏 + 푂(푛−Δ) + 푂(푈훿2))푑휏. 2 −훿 Now estimate the error

∫︁ ∞ ∫︁ ∞ 2 ∫︁ ∞ 2 2 −(푛−1)휋2휏 2+2휋휏푙 −(푛−1)휋2(휏− 푙 )2+ 푙 −(푛−1)휋2 휏 + 푙 1 −(푛−1)훿2 푒 푑휏 = 푒 휋(푛−1) 푛−1 푑휏 ≪ 푒 4 푛−1 푑휏 ≤ √ 푒 , 훿 훿 훿 푛 − 1 since 휏 푙 , i.e. 2 ≥ 휋(푛−1) 훿 ≥ 푈/(푛 − 1). 19 So the necessary condition is 푛훿2 ≫ log 푛. Then we can calculate

푛+1 ∞ 푛+1 ∞ 푖 2 ∫︁ 2 2 푖 2 ∫︁ 2 2 (*) = √ 푒−(푛−1)휋 휏 +2휋휏푙−2휋푖(푙−1)휏 푑휏+푂(푛−Δ)+푂(푈훿3) = √ 푒−(푛−1)휋 휏 +2휋휏푙−2휋푖(푙−1)휏 푑휏+푂(log2 푛/푛), 2 −∞ 2 −∞ since 푈훿3 ≪ log 푛 · 훿2 ≪ log2 푛/푛. Finally, from the formula ∫︀ ∞ −푎푡2+푏푡 √︀ 푏2/4푎 we get −∞ 푒 푑푡 = 휋/푎푒

푛+1 (2푙−1)−2푖(푙−1) 푖 2 2 (*) = 푒 푛−1 + 푂(log 푛/푛). √︀2휋(푛 − 1)

푛+1 (2푙−1)+2푖(푙−1) 3 2 In the area of 3/4 the integral approximately equals 푖 √ 푒 푛−1 and, since 3 푛+1 ≡ − 푛+1 mod 4, 2 2 2 √︃ 2 (︂휋(푛 + 1) 2푙2 2푙 )︂ 푎 (−푛 + 2푘 + 1, 푛 + 1) = 푒2푙−1 cos − + − + 푂(log2 푛/푛) = 2 휋(푛 − 1) 4 푛 − 1 푛 − 1 √︃ 2 (︂휋(푛 − 1) 2푙2 2푙 )︂ = −푒2푙−1 sin − + + 푂(log2 푛/푛). 휋(푛 − 1) 4 푛 − 1 푛 − 1 Analogously, √︃ 2 (︂휋(푛 − 1) 2푙2 )︂ 푎 (−푛 + 2푘 + 1, 푛 + 1) = cos − + 푂(log2 푛/푛). 1 휋(푛 − 1) 4 푛 − 1 For an even 푛 the answer is √︃ 2 (︂휋(푛 + 1) 2(푙2 − 1/4))︂ 푎 (−푛 + 2푘 + 1, 푛 + 1) = 푒2푙 cos − + 푂(log2 푛/푛); 2 휋(푛 − 2) 4 푛 − 2 √︃ 2 (︂휋(푛 + 1) 2(푙 + 1/2)2 )︂ 푎 (−푛 + 2푘 + 1, 푛 + 1) = sin − + 푂(log2 푛/푛), 1 휋(푛 − 2) 4 푛 − 2 where 푙 = 푘 − 푛/2.

11. Answer: 푃 (푥, 푦) = 푃 (푥, 푦, +) + 푃 (푥, 푦, −); ⃗푎(푥, 푦, +) = (0, 푎2(푥, 푦)); ⃗푎(푥, 푦, −) = (푎1(푥, 푦), 0). Solution. Note that if a path 푠 from (0, 0) to (푥, 푦) starts and finishes with an upward-left move, then the checker 1−푦 changes the direction an odd number of times. Therefore, ⃗푎(푠) = (±2 2 , 0). Analogously, if a path 푠 finishes with 1−푦 an upward-right move, then ⃗푎(푠) = (0, ±2 2 ). Hence, ⃗푎(푥, 푦, +) = (0, 푎2(푥, 푦)) and ⃗푎(푥, 푦, −) = (푎1(푥, 푦), 0). For the probability, the following formula comprehensively explains the answer:

2 2 2 푃 (푥, 푦) = |⃗푎(푥, 푦, +) + ⃗푎(푥, 푦, −)| = 푎1(푥, 푦) + 푎2(푥, 푦) = 푃 (푥, 푦, +) + 푃 (푥, 푦, −).

12. Answer: for each integer we have 1 ∑︀⌊푦/2⌋−1 1 (︀2푘)︀. We have also 푦 ≥ 2 푃 (푦, −) = 2 푘=0 (−4)푘 푘 max푦 푃 (푦, −) = 푃 (2, −) = 푃 (3, −) = 1/2 and lim 푃 (푦, −) = √1 . 푦→∞ 2 2 Remark. The graph of the sequence kindly provided by Gleb Minaev and Ivan Russkikh, participants of Summer Conference of Tournament of Towns:

Solution. Denote

∑︁ 2 ∑︁ 2 ∑︁ 푆1(푦) = 푎1(푥, 푦); 푆2(푦) = 푎2(푥, 푦); 푆12(푦) = 푎1(푥, 푦)푎2(푥, 푦). 푥 푥 푥 By Problem 5 for each 푦 ≥ 1 we have 푆1(푦) + 푆2(푦) = 1. 20 By the answers to Problem 4, Problem 6 and Problem 7 we have 1 ∑︁ 1 푎1(0, 2푦) = √ 푎1(푥, 푦)(푎2(푥, 푦) − 푎1(푥, 푦)) + 푎2(푥, 푦)(푎2(푥, 푦) + 푎1(푥, 푦)) = √ (푆2(푦) + 2푆12(푦) − 푆1(푦)). 2 푥 2 By definition and the answer to Problem 4 wehave

푆1(푦 + 1) − 푆2(푦 + 1) = 2푆12(푦). Hence, √ 푆1(푦 + 1) − 푆2(푦 + 1) = 푆1(푦) − 푆2(푦) + 푎1(0, 2푦) 2. Since 푆 (푦) + 푆 (푦) = 1, we have the recurrence relation 푆 (푦 + 1) = 푆 (푦) + √1 푎 (0, 2푦). 1 2 1 1 2 1 The remark from the solution of Problem 9 implies by induction that

⌊푦/2⌋−1 1 ∑︁ 1 (︂2푘)︂ 푆 (푦) = . 1 2 (−4)푘 푘 푘=0

It only remains to recall that 푃 (푦, −) = 푆1(푦) by Problem 11. Since the sequence (︀2푗)︀ 1 decreases, it follows that for each , thus 푗 4푗 푃 (푦, −) < 푃 (3, −) 푦 > 3 max푦 푃 (푦, −) = 푃 (2, −) = 푃 (3, −) = 1/2. By the Newton binomial theorem we get ∑︀∞ (︀2푘)︀푥푘 = √ 1 for each 푥 ∈ [︀− 1 , 1 )︀. Setting 푥 = − 1 we 푘=0 푘 1−4푥 4 4 4 푘 obtain lim 푃 (푦, −) = lim 1 ∑︀∞ (︀2푘)︀− (︀ 1 )︀ = √1 . 푦→∞ 푦→∞ 2 푘=0 푘 4 2 2 Remark. Using the Stirling formula (see §2) one can estimate the convergence rate:

1 (︂2⌊푘/2⌋)︂ 푒 1 |푃 (푘, −) − lim 푃 (푦, −)| < < < √ . 푦→∞ 2 · 4⌊푘/2⌋ ⌊푘/2⌋ 2휋√︀2⌊푘/2⌋ 2 푘 13. Answer: {︃ 1, for 푥 = 푦; 푃 (푥, 푦, 0, +) = 0, for 푥 ̸= 푦. {︃ 1, for 푥 = 1, 푦 ≡ 1 mod 2; 푃 (푥, 푦, ∞, +) = 0, otherwise. Solution. By the definition, if the number of turns in apath 푠 is nonzero then ⃗푎(푠, 0) = 0. Therefore, ⃗푎(푥, 푦, 0, +) ̸= 0 if and only if 푥 = 푦. Hence, we get the answer for 푃 (푥, 푦, 0, +). The case 푥 = 푦 is obvious. One cannot prove the answer for 푃 (푥, 푦, ∞, +) because it is a definition; but let us provide a motivation for the definition. If 푚휀 is “very large”, then ⃗푎(푠, 푚휀) is “small” unless the checker turns at each move; this gives the formula for 푃 (푥, 푦, ∞, +). Remark. 푃 (푥, 푦, ∞, +) = lim푚휀→∞ 푃 (푥, 푦, 푚휀, +). Indeed, {︃ (푚휀)푡(푠) 1 1, for 푡(푠) = 푦 − 1; lim |⃗푎(푠, 푚휀)| = lim = lim = 푚휀→∞ 푚휀→∞ (√︀1 + (푚휀)2)푦−1 푚휀→∞ (√︀1 + (푚휀)2)푦−1−푡(푠) 0, otherwise; where 푡(푠) is the number of turns in the path 푠. To compute 푃 (푥, 푦, 푚휀, +), we take only paths with 푡 ≡ 0 mod 2, hence we get the answer. 14. Answer. Recall the notation introduced at the beginning of section “Hints. . . ”. We have 1 푎1(푥, 푦, 푚휀, −) = √ (푎1(푥 + 1, 푦 − 1, 푚휀, −) + 푚휀 푎2(푥 + 1, 푦 − 1, 푚휀, +)), 1 + 푚2휀2 1 푎2(푥, 푦, 푚휀, +) = √ (푎2(푥 − 1, 푦 − 1, 푚휀, +) − 푚휀 푎1(푥 − 1, 푦 − 1, 푚휀, −)), 1 + 푚2휀2 푎2(푥, 푦, 푚휀, −) = 푎1(푥, 푦, 푚휀, +) = 0.

For each 푦 > |푥| we have

⌊푦/2⌋ ∑︁ (︂(푥 + 푦 − 2)/2)︂(︂(푦 − 푥 − 2)/2)︂ 푎 (푥, 푦, 푚휀) = (1 + 푚2휀2)(1−푦)/2 (−1)푟 (푚휀)2푟+1, 1 푟 푟 푟=0 ⌊푦/2⌋ ∑︁ (︂(푥 + 푦 − 2)/2)︂(︂(푦 − 푥 − 2)/2)︂ 푎 (푥, 푦, 푚휀) = (1 + 푚2휀2)(1−푦)/2 (−1)푟 (푚휀)2푟; 2 푟 푟 − 1 푟=1 21 For each 푦 = 푥 > 0 we have

푎1(푦, 푦, 푚휀) = 0, 2 2 (1−푦)/2 푎2(푦, 푦, 푚휀) = (1 + 푚 휀 ) ;

For each 0 < 푦 < |푥| or 푦 = −푥 > 0 we have

푎1(푥, 푦, 푚휀) = 0,

푎2(푥, 푦, 푚휀) = 0.

Hint. First let us solve the analogue of Problem 4 for 푚휀 ̸= 1. Let us derive the formula for 푎2(푥, 푦, 푚휀, +). Consider any path 푠 from (0, 0) to (푥, 푦). The last move in the path 푠 is made either from (푥 − 1, 푦) or from (푥 + 1, 푦). Denote by 푠′ the path 푠 without the last move. If the directions of the last moves in 푠 and 푠′ coincide, then ⃗푎(푠, 푚휀) = √ 1 ⃗푎(푠′, 푚휀), otherwise 1+푚2휀2 ⃗푎(푠, 푚휀) = √ 푚휀 (푎 (푠′, 푚휀), −푎 (푠′, 푚휀)). 1+푚2휀2 2 1 Therefore, 푎 (푥, 푦, 푚휀, +) = √ 1 (푎 (푥 − 1, 푦 − 1, 푚휀, +) − 푚휀푎 (푥 − 1, 푦 − 1, 푚휀, −)). The formula for 2 1+푚2휀2 2 1 푎1(푥, 푦, 푚휀, −) is proved analogously. Now let us solve the analogue of Problem 5: we prove that ∑︀ 푃 (푥, 푦, 푚휀) = 1 for all 푦 ≥ 1 by induction over 푥∈Z 푦 using these results. Obviously, ∑︀ 푃 (푥, 1, 푚휀) = 1. The step of induction follows immediately from the following 푥∈ computation: Z

∑︁ ∑︁ [︀ 2 2]︀ ∑︁ 2 ∑︁ 2 푃 (푥, 푦+1, 푚휀) = 푎1(푥, 푦 + 1, 푚휀, −) + 푎2(푥, 푦 + 1, 푚휀, +) = 푎1(푥, 푦+1, 푚휀, −) + 푎2(푥, 푦+1, 푚휀, +) = 푥∈Z 푥∈Z 푥∈Z 푥∈Z 1 ∑︁ 1 ∑︁ = (푎 (푥+1, 푦, 푚휀, −)+푚휀푎 (푥+1, 푦, 푚휀, +))2+ (푎 (푥−1, 푦, 푚휀, +)−푚휀푎 (푥−1, 푦, 푚휀, −))2 = 1 + 푚2휀2 1 2 1 + 푚2휀2 2 1 푥∈Z 푥∈Z 1 ∑︁ 1 ∑︁ = (푎 (푥, 푦, 푚휀, −) + 푚휀푎 (푥, 푦, 푚휀, +))2 + (푎 (푥, 푦, 푚휀, +) − 푚휀푎 (푥, 푦, 푚휀, +))2 = 1 + 푚2휀2 1 2 1 + 푚2휀2 2 1 푥∈Z 푥∈Z ∑︁ [︀ 2 2]︀ ∑︁ = 푎1(푥, 푦, 푚휀, −) + 푎2(푥, 푦, 푚휀, +) = 푃 (푥, 푦, 푚휀). 푥∈Z 푥∈Z

Let us now solve the analogue of Problem 9 for 푚휀 ̸= 1. Let us find 푎1(푥, 푦) for 푦 > |푥|. We need to consider paths with an odd number of turns only. Denote by 2푟 + 1 the number of turns in a path. Denote by 푅 and 퐿 the number of upwards-right and upwards-left moves respectively. Let 푥1, 푥2, . . . , 푥푟+1 be the number of upwards-right moves before the first, the third, . . . , the last turn respectively. Let 푦1, 푦2, . . . , 푦푟+1 be the number of upwards-left moves after the first, the third, . . . , the last turn respectively. Then 푥푘, 푦푘 ≥ 1 for 1 ≤ 푘 ≤ 푟 + 1 and

푅 = 푥1 + ··· + 푥푟+1;

퐿 = 푦1 + ··· + 푦푟+1. The problem now reduces to a combinatorial one: find the number of positive integer solutions of the resulting equations. For the first equation, this number equals to the number of ways toput 푟 sticks between 푅 coins in a row, that is, (︀푅−1)︀. Thus 푟

⌊푦/2⌋ ∑︁ (︂푅 − 1)︂(︂퐿 − 1)︂ 푎 (푥, 푦, 푚휀) = (1 + 푚2휀2)(1−푦)/2 (−1)푟 (푚휀)2푟+1. 1 푟 푟 푟=0

Since 퐿 + 푅 = 푦 and 푅 − 퐿 = 푥, the required formula follows. The proof of the formula for 푎2(푥, 푦) for 푦 > |푥| is analogous. The case 푦 ≤ |푥| is obvious. 15. Answer: for |푥| < 푦 we have

(︁ ⌊︁푛푥⌋︁ ⌊︁푛푦 ⌋︁ 푚 )︁ √︀ 2 2 lim 푛⃗푎 2 , 2 , , − = (푚퐽0(푚 푦 − 푥 ), 0); 푛→∞ 2 2 푛 (︁ ⌊︁푛푥⌋︁ ⌊︁푛푦 ⌋︁ 푚 )︁ 푥 + 푦 √︀ 2 2 lim 푛⃗푎 2 , 2 , , + = (0, −푚 퐽1(푚 푦 − 푥 )); 푛→∞ 2 2 푛 √︀푦2 − 푥2 for |푥| ≥ 푦 both limits vanish. 22 Remark. The limit coincides with the retarded fundamental solution of Dirac’s equation, describing motion of an electron in the plane.

Solution by Ivan Gaidai-Turlov, Timofey Kovalev and Alexey Lvov, participants of Summer conference of Tournament of towns. Denote 푛푥 , 푛푦 , 푥푛+푦푛 푦푛−푥푛 . 푥푛 := 2⌊ 2 ⌋ 푦푛 := 2⌊ 2 ⌋ 퐴 := 2 , 퐵 := 2 푚 2 푦푛−1 Lemma 2. lim (1 + ( ) ) 2 = 1. 푛→∞ 푛

푦 −1 √︁ 푚 2 푛 푚 2 푛푦 푚 2 푛푦 푛 푚 2 Proof. We have 1 < (1 + ( ) ) 2 < (1 + ( ) ) , therefore if 푛 → ∞ then (1 + ( ) ) = (1 + ( )2)푛 푦 → √ 푛 푛 푛 푛 푛 2 푚 2 푦푛−1 푒푚 푦 → 1. Hence by the squeeze theorem we have lim (1 + ( ) ) 2 = 1. 푛→∞ 푛

푥+푦 푙 푦−푥 푙 2푙+1 푙 푙 푙 푚 2푙+1 푙 ( 2 ) ( 2 ) ·푚 Lemma 3. lim 푛(−1) 퐶 퐶 ( ) = (−1) · 2 . 푛→∞ 퐴−1 퐵−1 푛 (푙!) Proof. Clearly, 푙 푙 푙 푚 2푙+1 푙 (퐴−1)(퐴−2)...(퐴−푙)·(퐵−1)(퐵−2)...(퐵−푙) 푚2푙+1 . 푛 · (−1) 퐶퐴−1퐶퐵−1( 푛 ) = (−1) · (푙!)2 · 푛2푙 ⌊ 푛푥 ⌋+⌊ 푛푦 ⌋ 푛푥 + 푛푦 +표(푛) Thus lim 퐴−푖 = lim 퐴 = lim 2 2 = lim 2 2 = 푥+푦 . Analogously, lim 퐵−푖 = 푦−푥 . 푛→∞ 푛 푛→∞ 푛 푛→∞ 푛 푛→∞ 푛 2 푛→∞ 푛 2 Therefore,

2푙+1 푥+푦 푙 푦−푥 푙 2푙+1 푙 푙 푙 푚 2푙+1 푙 (퐴−1)...(퐴−푙)·(퐵−1)...(퐵−푙) 푚 푙 ( 2 ) ( 2 ) · 푚 lim 푛 · (−1) 퐶 퐶 ( ) = lim (−1) · 2 · = (−1) · . 푛→∞ 퐴−1 퐵−1 푛 푛→∞ (푙!) 푛2푙 (푙!)2

( 푥+푦 )푙( 푦−푥 )푙·푚2푙+1 For each we have 푙 푙 푙 푚 2푙+1 푙 2 2 because 퐴−푖 퐴 푦+푥 and 푛 |(−1) 퐶퐴−1퐶퐵−1( 푛 ) | < |(−1) · (푙!)2 | 푛 < 푛 < 2 퐵−푖 퐵 푦−푥 . 푛 < 푛 < 2 ∞ Lemma 4. The series ∑︀ 푙 푙 푙 푚 2푙+1 converges absolutely for all . 푛 · (−1) 퐶퐴−1퐶퐵−1( 푛 ) 푛 푙=0 Proof. We have ∞ ∞ 푥+푦 푦−푥 ∑︁ 푚 ∑︁ ( )푙( )푙 · 푚2푙+1 푛 · |(−1)푙퐶푙 퐶푙 ( )2푙+1| < |(−1)푙 · 2 2 | = 퐴−1 퐵−1 푛 (푙!)2 푙=0 푙=0

∞ 푥+푦 푙 푦−푥 푙 2푙 ∞ 푥+푦 푙 푦−푥 푙 2푙 2 2 ∑︁ ( 2 ) ( 2 ) · 푚 ∑︁ ( 2 ) ( 2 ) · 푚 푦 −푥 푚2 = 푚 · < 푚 · = 푚 · 푒 4 . (푙!)2 (푙!) 푙=0 푙=0

∞ ∞ ∞ Hence lim ∑︀(−1)푙퐶푙 퐶푙 ( 푚 )2푙+1 = ( lim ∑︀ 퐶푙 퐶푙 ( 푚 )2푙+1) − ( lim ∑︀ 퐶푙 퐶푙 ( 푚 )2푙+1). 푛→∞ 퐴−1 퐵−1 푛 푛→∞ 퐴−1 퐵−1 푛 푛→∞ 퐴−1 퐵−1 푛 푙=0 푙=0;2|푙 푙=0;2-푙

Lemma 5. Suppose ({푎0(푛)}, {푎1(푛)} ... ) is a sequence of nonnegative sequences such that lim 푎푖(푛) = 푏푖 and 푛→∞ ∞ ∞ ∑︀ ∑︀ 푎푖(푛) < 푏푖 for each 푖, 푛. Then lim 푎푖(푛) = 푏푖. 푛→∞ 푖=0 푖=0 ∞ ∞ 푁 ∑︀ ∑︀ ∑︀ Proof. Denote 푏 := 푏푖. Then for each 푛 we have 푎푖(푛) < 푏. Take any 휀 > 0. Take such 푁 that 푏푖 > 푏 − 휀. 푖=0 푖=0 푖=0 For each take such that 휀 . Then for all we 0 ≤ 푖 ≤ 푁 푀푖 ∀푡 ≥ 푀푖 : 푎푖(푡) > 푏푖 − 2푖+1 푛 > max(푀0, 푀1, . . . , 푀푁 ) ∞ ∞ ∑︀ ∑︀ have 푎푖(푛) > 푏 − 휀. Therefore, lim 푎푖(푛) = 푏. 푖=0 푛→∞ 푖=0 From Lemmas 5 and 3 it follows that ∞ ∞ 푥+푦 푦−푥 ∑︁ 푚 ∑︁ ( )푙( )푙 · 푚2푙+1 lim 푛 · 퐶푙 퐶푙 ( )2푙+1 = 2 2 . 푛→∞ 퐴−1 퐵−1 푛 (푙!)2 푙=0;2|푙 푙=0;2|푙 Therefore, by Problem 14 and Lemma 2 we have

∞ 푚 ∑︁ 푙 푙 푙 푚 2푙+1 lim 푛⃗푎(푥푛, 푦푛, , −) = lim 푛 · 푛(−1) 퐶 퐶 ( ) = 푛→∞ 푛 푛→∞ 퐴−1 퐵−1 푛 푙=0 ∞ 푥+푦 푦−푥 ∞ 푥+푦 푦−푥 ∑︁ ( )푙( )푙 · 푚2푙+1 ∑︁ ( )푙( )푙 · 푚2푙+1 √︀ = ( · 2 2 ) − ( · 2 2 ) = 푚 · 퐽 (푚 푦2 − 푥2). (푙!)2 (푙!)2 0 푙=0;2|푙 푙=0;2-푙 23 푚 푥+푦 √︀ 2 2 It can be proven analogously that lim 푛⃗푎(푥푛, 푦푛, , +) = −푖푚√ · 퐽1(푚 푦 − 푥 ). 푛→∞ 푛 푦2−푥2 Solution. Denote modified Bessel functions

∞ ∞ ∑︁ (푧/2)2푘 ∑︁ (푧/2)2푘+1 퐼 (푧) := and 퐼 (푧) := . 0 (푘!)2 1 푘!(푘 + 1)! 푘=0 푘=0 The answer follows from the following stronger theorem.

Theorem 2. Let , √︀ 2 2, 3푧 2 Then 0 < 휀0 ≤ 1/2 min{푦 + 푥, 푦 − 푥} ≥ 휀0, 푧 := 푚 푦 − 푥 푛 ≥ 푛0 = 푛0(푧, 휀0) = 4푒 /휀0. (︁ ⌊︁푛푥⌋︁ ⌊︁푛푦 ⌋︁ 푚 )︁ 푚 (︂ (︂log2 푛 )︂)︂ 푎1 2 , 2 , , − = 퐽0(푧) + 푂 퐼0(푧) , 2 2 푛 푛 휀0푛 (︁ ⌊︁푛푥⌋︁ ⌊︁푛푦 ⌋︁ 푚 )︁ 푚 푦 + 푥 (︂ (︂log2 푛 )︂)︂ 푎2 2 , 2 , , + = − √︀ 퐽1(푧) + 푂 퐼1(푧) . 2 2 푛 푛 푦2 − 푥2 휀0푛 Proof. Denote ⌊︀ 푛푥 ⌋︀, ⌊︀ 푛푦 ⌋︀. By Problem 9 we have 푥푛 = 2 2 푦푛 = 2 2

1−푦 푛 [푦푛/2] (︂ 2 )︂ 2 (︂ )︂(︂ )︂ (︁ 푚 )︁ 푚 푚 ∑︁ (푦푛 + 푥푛 − 2)/2 (푦푛 − 푥푛 − 2)/2 (︁푚)︁2푟 푎 푥 , 푦 , , − = 1 + (−1)푟 , 1 푛 푛 푛 푛2 푛 푟 푟 푛 푟=0 1−푦 푛 [푦푛/2] (︂ 2 )︂ 2 (︂ )︂(︂ )︂ (︁ 푚 )︁ 푚 푚 ∑︁ (푦푛 + 푥푛 − 2)/2 (푦푛 − 푥푛 − 2)/2 (︁푚)︁2푟−1 푎 푥 , 푦 , , + = 1 + (−1)푟 . 2 푛 푛 푛 푛2 푛 푟 푟 − 1 푛 푟=1

1−푦푛 2 2 2 2 (︁ 2 )︁ 2 푚 1−푦푛 1−푦 −푦푛 Since 푚 푚 /푛 it follows that 푚 푛2 ( 2 ) because . Also 푛 . Hence, 1 + 푛2 ≤ 푒 1 + 푛2 ≥ 푒 푦푛 ≥ 1 2 ≥ 2

1−푦푛 (︂ 2 )︂ 2 2 2 푚 푚 ( −푦푛 ) 푚 푦 1 ≥ 1 + ≥ 푒 푛2 2 ≥ 1 − . 푛2 2푛

1−푦푛 (︁ 2 )︁ 2 Thus 푚 2 푚 푦 1 + 푛2 = 1 + 푂( 푛 ). To find the asymptotic form for 푎 and 푎 consider the 푇 -th partial sums with 푇 = ⌊log 푛⌋ + 1 summands. The 1 2 √ inequalities and 4 guarantee that the total number of summands 푦푛 because 푛푦 푦 ≥ 휀0 푛 ≥ 2 ⌊ ⌋ ≥ 푇 ≥ 푛 ≥ log 푛+1 휀0 2 2 for 푛 > 0. For 푟 ≥ 푇 the ratios of consecutive summands in the expressions for 푎1 and 푎2 equal respectively (︁푚)︁2 ((푦 + 푥 − 2)/2 − 푟)((푦 − 푥 − 2)/2 − 푟) (︁푚)︁2 푛(푦 + 푥) 푛(푦 − 푥) 푧2 푛 푛 푛 푛 < · · = , 푛 (푟 + 1)2 푛 2푇 2푇 4푇 2 (︁푚)︁2 ((푦 + 푥 − 2)/2 − 푟)((푦 − 푥 − 2)/2 − 푟 + 1) (︁푚)︁2 푛(푦 + 푥) 푛(푦 − 푥) 푧2 푛 푛 푛 푛 < · · = . 푛 (푟 − 1)푟 푛 2푇 2푇 − 2 4푇 (푇 − 1)

From the condition 3푧+1 it follows that and 푧2 푧2 1 Therefore, in 푛 > 푒 푇 = ⌊log 푛⌋ + 1 ≥ 3푧 + 1 4푇 2 < 4푇 (푇 −1) < 2 . both cases the error term (i.e., the sum over ) is less then the sum of geometric series with ratio 1 Hence, 푟 ≥ 푇 2 . (︁ 푚 )︁ 푚 (︂ (︂푚2푦 )︂)︂ 푎 푛 , 푛 , , − = 1 + 푂 · 1 푥 푦 푛 푛 푛 [︃푇 −1 (︂ )︂(︂ )︂ (︂ 2 2 푇 )︂]︃ ∑︁ (푦푛 + 푥푛 − 2)/2 (푦푛 − 푥푛 − 2)/2 (︁푚)︁2푟 (푦 − 푥 ) (︁푚)︁2푇 · (−1)푟 + 푂 · , 푟 푟 푛 (푇 !)2 2 푟=0 (︁ 푚 )︁ 푚 (︂ (︂푚2푦 )︂)︂ 푎 푛 , 푛 , , + = 1 + 푂 · 2 푥 푦 푛 푛 푛 [︃푇 −1 (︂ )︂(︂ )︂ (︂ 푇 푇 −1 )︂]︃ ∑︁ (푦푛 + 푥푛 − 2)/2 (푦푛 − 푥푛 − 2)/2 (︁푚)︁2푟−1 (푦 + 푥) (푦 − 푥) (︁푚)︁2푇 −1 · (−1)푟 + 푂 · . 푟 푟 − 1 푛 (푇 − 1)!푇 ! 2 푟=1 √ The error term in the latter formulae are estimated as follows. Since 푇 ! ≥ (푇/3)푇 and 푇 ≥ √︀푦2 − 푥2 3푚 푒 = √ 2 3 푒 , it follows that 2 푧 (푦2 − 푥2)푇 (︁푚)︁2푇 (푦2 − 푥2)푇 (︂3푚)︂2푇 1 · ≤ · ≤ 푒−푇 ≤ . (푇 !)2 2 (푇 )2푇 2 푛 24 Now transform binomial coefficients in the following way:

(︂훼푛 − 1)︂ (훼푛 − 1) ··· (훼푛 − 푟) (훼푛)푟 (︂ 1 )︂ (︁ 푟 )︁ = = 1 − ··· 1 − . 푟 푟! 푟! 훼푛 훼푛 √ 푟 푇 푛 1 휀0 1 1 −2푥 −푥 If 훼 ≥ 휀0 and 푟 < 푇 then < < = √ ≤ · = . Hence from the inequalities 푒 ≤ 1 − 푥 < 푒 훼푛 훼푛 푛휀0 푛휀0 2 휀0 2 for 0 ≤ 푥 ≤ 1/2 it follows that

2 2 (︂ 1 )︂ (︁ 푟 )︁ −2 −4 −2푟 −푟(푟+1) −푇 2 푇 푇 1 − ··· 1 − ≥ 푒 훼푛 푒 훼푛 ··· 푒 훼푛 = 푒 훼푛 ≥ 푒 훼푛 ≥ 1 − ≥ 1 − . 훼푛 훼푛 훼푛 휀0푛

푟 (︁ (︁ 2 )︁)︁ Therefore, (︀훼푛−1)︀ = (훼푛) 1 + 푂 푇 . 푟 푟! 휀0푛 2 Denote 푅 = 푇 퐼 (푧). Replacing (︀(푦푛+푥푛−2)/2)︀ and (︀(푦푛−푥푛−2)/2)︀ in the previous expression for 푎 we get 휀0푛 0 푟 푟 1

[︃푇 −1 ]︃ (︁ 푚 )︁ 푚 (︂ (︂푚2푦 )︂)︂ ∑︁ (︁푚)︁2푟 (푦2 − 푥2)푟 푎 푥 , 푦 , , − = 1 + 푂 · (−1)푟 + 푂(푅) = 1 푛 푛 푛 푛 푛 2 (푟!)2 푟=0 [︃ ∞ ]︃ 푚 (︂ (︂푚2푦 )︂)︂ ∑︁ (︁푚)︁2푟 (푦2 − 푥2)푟 푚 (︂ (︂푚2푦 )︂)︂ = 1 + 푂 · (−1)푟 + 푂(푅) = 1 + 푂 · (퐽 (푧) + 푂(푅)). 푛 푛 2 (푟!)2 푛 푛 0 푟=0 Here we can replace the sum with 푇 summands by an infinite sum because the “tail” of alternating series with decreasing absolute value of the summands can be estimated by the first summand:

⃒ ∞ ⃒ ⃒∑︁ (︁푚)︁2푟 (푦2 − 푥2)푟 ⃒ (푧/2)2푇 1 ⃒ (−1)푟 ⃒ ≤ ≤ ⃒ 2 (푟!)2 ⃒ (푇 !)2 푛 ⃒푟=푇 ⃒ (the latter inequality has been proved before). 2 2 2 2 2 2 2 2 We have 푚 푦휀0 ≤ 푚 · max{푦 + 푥, 푦 − 푥}· min{푦 + 푥, 푦 − 푥} = 푚 (푦 − 푥 ) = 푧 ≤ 9푧 ≤ 푇 , thus 푚2푦 푇 2 hence (︀ 푚 )︀ 푚 as required. 푛 퐽0(푧) ≤ 휀 푛 퐼0(푧), 푎1 푥푛, 푦푛, 푛 , − = 푛 (퐽0(푧) + 푂(푅)) Analogously,0

[︃푇 −1 2푟−1 ]︃ (︂ (︂ 2 )︂)︂ 2푟−1 2 2 2 (︁ 푚 )︁ 푚 푚 푦 푦 + 푥 ∑︁ 푟 (︁푚)︁ (푦 − 푥 ) 2 푇 푎2 푥푛, 푦푛, , + = 1 + 푂 · · (−1) + 푂( 퐼1(푧)) = 푛 푛 푛 √︀ 2 2 2 (푟 − 1)!푟! 휀 푛 푦 − 푥 푟=1 0 푚 푦 + 푥 (︂ (︂ 푇 2 )︂)︂ = − · √︀ 퐽1(푧) + 푂 퐼1(푧) . 푛 푦2 − 푥2 휀0푛

16. Answer: see the following table, where the pair (푐, 푑) in the cell (푥, 푦) means that ⃗푎(푥, 푦, 4, 3, +) = √1 (푐, 푑), + 3 and (푐, 푑) means that ⃗푎(푥, 푦, 4, 3, −) = √1 (푐, 푑); an empty cell (푥, 푦) means that ⃗푎(푥, 푦, 4, 3, +) = ⃗푎(푥, 푦, 4, 3, −) = − 3 (0, 0).

3 (0, 0)+ (0, 1)+ (0, −2)+ (0, 2)+ (0, −1)+ (−1, 0)− (0, 0)− (0, 0)− (−1, 0)− (0, 0)− 2 (0, 0)+ (0, −1)+ (0, 1)+ (0, −1)+ (−1, 0)− (1, 0)− (−1, 0)− (0, 0)− 1 (0, −1)+ (0, 1)+ (0, −1)+ (0, 0)− (0, 0)− (0, 0)− y x −3 −2 −1 0 1 2 3 4 5

Hint. Note that we are interested only in even , thus 휋푥0 is a multiple of . Analogously to 푥0 2휋푥0휀/휆 = 2 휋 Problem 1, that means that ⃗푎(푥, 푦, 4, 3, +) is parallel to 푂푦 and ⃗푎(푥, 푦, 4, 3, −) is parallel to 푂푥 (be careful: this is not true for an arbitrary 휀/휆). 17. Let us give a more general formula; the solution of the problem is the particular case 푚휀 = 1. Denote

푎−(푥, 푦) := ⃗푎(푥, 푦, 푚휀, 휆/휀, Δ, −);

푎+(푥, 푦) := ⃗푎(푥, 푦, 푚휀, 휆/휀, Δ, +),

25 where the vectors in the right-hand side are defined analogously to ⃗푎(푥, 푦, 휆/휀, Δ, ±) and ⃗푎(푥, 푦, 푚휀, ±). (Beware that 푎−(푥, 푦) and 푎+(푥, 푦) are vector-valued functions; unlike 푎1(푥, 푦) and 푎2(푥, 푦) they are not coordinates of any vector.) Then we have the Dirac equation (where 푖 := 푅휋/2 is the counterclockwise rotation through 90∘) {︃ 푎 (푥, 푦) = √ 1 푎 (푥 + 1, 푦 − 1) + √ −푖푚휀 푎 (푥 + 1, 푦 − 1); − 1+푚2휀2 − 1+푚2휀2 + 푎 (푥, 푦) = √ −푖푚휀 푎 (푥 − 1, 푦 − 1) + √ 1 푎 (푥 − 1, 푦 − 1). + 1+푚2휀2 − 1+푚2휀2 + From Problem 14 we know that this equation holds for Δ = 1. It can be generalised for any greater Δ because one can change the order of rotations 푅2휋푥0휀/휆 and 푖 = 푅휋/2. Using this equation, analogously to Problem 14 we get ∑︀ 푃 (푥, 푦, 휆/휀, Δ) = 1 for 푦 ≥ 1 by induction over 푦. √ 푥∈Z √ 18. Since Δ푃 (푥, 푦, 휆/휀, Δ, +) = | Δ⃗푎(푥, 푦, 휆/휀, Δ, +)|2, it suffices to prove that Δ⃗푎(푥, 푦, 휆/휀, Δ, +) does not √ 푥0=Δ−1 depend on Δ for Δ > 푦 +|푥|. By definition Δ⃗푎(푥, 푦, 휆/휀, Δ, +) = ∑︀ ∑︀ 푅2휋푥0휀/휆⃗푎(푠). But there are no paths 푥0=1−Δ 푠 푥0 even 푠 starting at (푥0, 0) and ending at (푥, 푦) for |푥 − 푥0| > 푦, thus increasing Δ over 푦 + |푥| does not change the sum.

19. Answer: if 푥1 and 푥2 are both even, then

(푥1−푥2)2휋휀/휆 ⃗푎(푥1, 푦, 휆/휀, +) = 푅 ⃗푎(푥2, 푦, 휆/휀, +),

(푥1−푥2)2휋휀/휆 ⃗푎(푥1, 푦, 휆/휀, −) = 푅 ⃗푎(푥2, 푦, 휆/휀, −).

Solution. From the solution of Problem 18 it follows that ⃗푎(푥, 푦, 휆/휀, +) = ∑︀ ∑︀ 푅2휋푥0휀/휆⃗푎(푠), where the 푥0 even 푠 second sum is over those paths from (푥0, 0) to (푥, 푦), which both start and finish with an upwards-right move. To each path 푠1 from (푥0, 0) to (푥1, 푦) assign the unique path 푠2 from (푥0 + (푥2 − 푥1), 0) to (푥2, 푦), obtained from 푠1 by translation by (푥1 − 푥2, 0) and vice versa. Thus we sum over the same vectors, but for 푥 = 푥1 they are rotated through 2휋푥0휀/휆 while for 푥 = 푥2 they are rotated through 2휋(푥0 + (푥2 − 푥1))휀/휆, and we are done. 20. Let us give a more general formula; the solution of the problem is the particular case 푚휀 = 1. (︁ cos(2휋휀/휆) )︁ Answer. Denote 퐸 = arccos √ . Then for even 푥 + 푦 we have: 1+푚2휀2

(︁푚휀 cos(2휋휀푥/휆) sin((푦 − 1)퐸) 푚휀 sin(2휋휀푥/휆) sin((푦 − 1)퐸))︁ ⃗푎(푥, 푦, 휆/휀, −) = √ , √ ; 1 + 푚2휀2 sin 퐸 1 + 푚2휀2 sin 퐸 (︁cos(2휋휀(푥 − 2)/휆) sin(푦퐸) − cos(2휋휀푥/휆) sin((푦 − 2)퐸) ⃗푎(푥, 푦, 휆/휀, +) = , 2 cos(2휋휀/휆) sin 퐸 − sin(2휋휀(푥 − 2)/휆) sin(푦퐸) + sin(2휋휀푥/휆) sin((푦 − 2)퐸))︁ ; 2 cos(2휋휀/휆) sin 퐸 푚2휀2 sin2((푦 − 1)퐸) 푃 (푥, 푦, 휆/휀, −) = ; 푚2휀2 + sin2(2휋휀/휆) 푚2휀2 cos2((푦 − 1)퐸) + sin2(2휋휀/휆) 푃 (푥, 푦, 휆/휀, +) = ; 푚2휀2 + sin2(2휋휀/휆) 푃 (푥, 푦, 휆/휀, −) + 푃 (푥, 푦, 휆/휀, +) =1.

Remark. In complex form,

푚휀 sin((푦 − 1)퐸) ⃗푎(푥, 푦, 휆/휀, −) = 푟푥 √ ; 1 + 푚2휀2 sin 퐸 (︁sin((푦 − 1)퐸) sin(2휋휀/휆) )︁ ⃗푎(푥, 푦, 휆/휀, +) = 푟푥−1 √ + 푖 cos((푦 − 1)퐸) , 1 + 푚2휀2 sin 퐸 where 푟 = cos(2휋휀/휆) + 푖 sin(2휋휀/휆). This is the solution of the Dirac equation (see the solution of Problem 17) with the (quasi)periodic initial condition and 2휋푥휀 2휋푥휀 . The number has 푎−(푥, 0) = 0 푎+(푥, 0) = (− sin 휆 , cos 휆 ) 퐸 physical meaning of energy. Hint. Formulas for ⃗푎(푥, 푦, 휆/휀, +) and ⃗푎(푥, 푦, 휆/휀, −) can be proven by induction over 푦. The base is obvious; the inductive step follows from Problem 17. Then 푃 (푥, 푦, 휆/휀, +) and 푃 (푥, 푦, 휆/휀, −) are computed directly. Path to solution. Let us view vectors as complex numbers. Then 푅훼⃗푎 = (cos 훼 + 푖 sin 훼)⃗푎. Denote 푟 = cos(2휋휀/휆) + 푖 sin(2휋휀/휆); then 푟−1 = cos(2휋휀/휆) − 푖 sin(2휋휀/휆). Denote 푓(푦) = 푟−푦⃗푎(푦, 푦, 휆/휀, +) and 푔(푦) = 푟−푦⃗푎(푦, 푦, 휆/휀, −). Denote 푀 = √ 푚휀 and 푈 = √ 1 . By Problem 19 for even 푥+푦 we have ⃗푎(푥, 푦, 휆/휀, +) = 1+푚2휀2 1+푚2휀2 푟푥푓(푦) and ⃗푎(푥, 푦, 휆/휀, −) = 푟푥푔(푦). By Problem 17 we have 푟푥푓(푦) = 푈푟푥−1푓(푦 − 1) − 푖푀푟푥−1푔(푦 − 1) and 26 푟푥푔(푦) = 푈푟푥+1푔(푦 − 1) − 푖푀푟푥+1푓(푦 − 1). Since 푟 ̸= 0, we can divide each side by 푟푥 and obtain the following system:

⎧ 푓(푦) = 푈푟−1푓(푦 − 1) − 푖푀푟−1푔(푦 − 1) (1) ⎪ ⎨⎪ 푔(푦) = 푈푟푔(푦 − 1) − 푖푀푟푓(푦 − 1) (2) −1 ⎪ 푓(1) = 푖푟 (3) ⎪ ⎩ 푔(1) = 0 (4) There are multiple ways of solving this system (we have thought about generating functions and about diagonalising the ‘transfer matrix’) but the one which is probably most accessible (to math high-school students in Russia) is transforming this system into a linear recurrence relation of degree . From (1) we have 1 2 푔(푦 − 1) = 푀 푖푟푓(푦) − 푈 . From this and (2) we have 푈 2 1 . From this and (1) we have 푀 푖푓(푦 − 1) 푔(푦) = 푀 푖푟 푓(푦) − 푀 푖푟푓(푦 − 1) 푓(푦) = 푈(푟 + 푟−1)푓(푦 − 1) − 푓(푦 − 2). The solution of this recurrence equation is 푛 푛 , where 푓(푛) = 푐1푑+ + 푐2푑−

푈(푟 + 푟−1) + √︀푈 2(푟 + 푟−1)2 − 4 √︁ 푑 = = 푈 cos(2휋휀/휆) + 푖 푈 2 sin2(2휋휀/휆) + 푀 2 + 2

푈(푟 + 푟−1) − √︀푈 2(푟 + 푟−1)2 − 4 √︁ 푑 = = 푈 cos(2휋휀/휆) − 푖 푈 2 sin2(2휋휀/휆) + 푀 2 − 2 2 −1 are the roots of the equation 푡 − 푈(푟 + 푟 )푡 + 1 = 0, and 푐1 and 푐2 are constants that are found using (3) and (4). Since we are dealing with complex numbers, the notation √︀푈 2(푟 + 푟−1)2 − 4 is ambiguous. But we can choose any of the two roots; √︀푈 2(푟 + 푟−1)2 − 4 denotes 2푖√︀푈 2 sin2(2휋휀/휆) + 푀 2. It follows that

−1 −1 푟 − 푈푑− 푟 − 푈푑+ 푐1 = 푖 and 푐2 = 푖 . 푑+ − 푑− 푑− − 푑+

Note that 퐸 = arccos(푈 cos(2휋휀/휆)). We have 푑+ = cos 퐸 + 푖 sin 퐸 and 푑− = cos 퐸 − 푖 sin 퐸. Thus we have (푟−1−푈푑 )푑푦 −(푟−1−푈푑 )푑푦 ⃗푎(푥, 푦, 휆/휀, +) = 푟푥푓(푦) = 푖푟푥 − + + − = 푖푟푥−1 sin(푦퐸) − 푖푈푟푥 sin((푦−1)퐸) . 푑+−푑− sin 퐸 sin 퐸 Now we compute 푃 (푥, 푦, 휆/휀, +) = ⃗푎(푥, 푦, 휆/휀, +) ·⃗푎(푥, 푦, 휆/휀, +). It follows from formulas above that 푟 = 푟−1, 푑 = 푑 , 푑 = 푑 , 푐 = 푖 푟−푈푑+ and 푐 = 푖 푟−푈푑− . From this (with some simplification) we finally have + − − + 1 푑+−푑− 2 푑−−푑+ 2 2 푦−1 푦−1 2 2 2 (푑+−푑−) −푀 (푑+ −푑− ) 푀 sin ((푦−1)퐸) 푃 (푥, 푦, 휆/휀, +) = 2 = 1 − 2 . (푑+−푑−) sin 퐸 2 2 Analogously we get ⃗푎(푥, 푦, 휆/휀, −) = 푟푥푀 sin((푦−1)퐸) and 푃 (푥, 푦, 휆/휀, −) = 푀 sin ((푦−1)퐸) . sin 퐸 sin2 퐸 21. Solution. Let (0, 0), (1, 1), (푥2, 2), ··· , (푥푦, 푦) be the squares passed by the path 푠. To define the vector ⃗푎(푠, 푚(푥)휀), start with the vector (0, 1). While the checker moves straightly, the vector is not changed, but if the checker changes the direction after the 푛-th move, the vector is rotated through 90∘ clockwise and multiplied by . In addition, at the very end the vector is divided by ∏︀푦−1 √︀ 2 2, where is the total number 푚(푥푛)휀 푛=1 1 + 푚 (푥푛)휀 푦 of moves. The final position of the vector is what we denote by ⃗푎(푠, 푚(푥)휀). The numbers 푎1(푥, 푦, 푚(푥)휀) and 푎2(푥, 푦, 푚(푥)휀) are defined analogously to 푎1(푥, 푦) and 푎2(푥, 푦), only ⃗푎(푠) is replaced by ⃗푎(푠, 푚(푥)휀). The formulas analogous to the ones from Problem 4 are the following:

1 푎1(푥0, 푦, 푚(푥)휀) = (푎1(푥0 + 1, 푦, 푚(푥)휀) + 푚(푥0 + 1)휀 · 푎2(푥0 + 1, 푦, 푚(푥)휀)), √︀ 2 2 1 + 푚 (푥0 + 1)휀 1 푎2(푥0, 푦, 푚(푥)휀) = (푎2(푥0 − 1, 푦, 푚(푥)휀) − 푚(푥0 − 1)휀 · 푎1(푥0 − 1, 푦, 푚(푥)휀)). √︀ 2 2 1 + 푚 (푥0 − 1)휀 22. Answer: For each 푥, 푦 we have 푃 (푥, 푦, 푚0(푥), 휆/휀, +) = 1; 푃 (푥, 푦, 푚0(푥), 휆/휀, −) = 0. {︃ 1 25 for 1.04 = 26 , 0 < 푥 < 푦; 푃 (푥, 푦, 푚1(푥), 휆/휀, +) = 1, otherwise; {︃ 0.04 1 for 1.04 = 26 , − 푦 < 푥 < 0; 푃 (푥, 푦, 푚1(푥), 휆/휀, −) = 0, otherwise; Solution. Analogously to Problem 13 we have 푃 (푥, 푦, 푚0(푥), 휆/휀, +) = 1, 푃 (푥, 푦, 푚0(푥), 휆/휀, −) = 0 for each 푥, 푦.

27 The condition ⃗푎(푠, 푚1(푥)) ̸= 0 means that the checker, if at all changed the direction, turned in one of the squares with the coordinates [︀ 푦 ]︀ . Thus the number of turns in any path with (0, 2), (0, 4), ··· , (0, 2 · 2 ) 푠 ⃗푎(푠, 푚1(푥)) ̸= 0 is not greater than 1. To compute 푃 (푥, 푦, 푚1(푥), 휆/휀, +) we need to consider only the paths ending with an upwards-right move – i.e. without turns. If the path 푠 does not contain any of the squares with coordinates (0, 1), (0, 2),..., (0, 푦 − 1) (amounting to 푥 ≤ 0 or 푥 ≥ 푦) then 푃 (푥, 푦, 푚1(푥), 휆/휀, +) = 1. Otherwise, the length of the vector ⃗푎(푠, 푚(푥)) equals √ 1 . 1+0.22 To compute 푃 (푥, 푦, 푚1(푥), 휆/휀, −), consider paths having exactly one turn in the square with zero 푥-coordinate. If there is such a path 푠 (i.e. −푦 < 푥 < 0), then the length of the vector ⃗푎(푠, 푚(푥)) equals √ 0.2 . Otherwise, it 1+0.22 equals zero.

23. Solution. If ⃗푎(푠, 푚2(푥)) ̸= 0 then the checker changed the direction only in the squares with the 푥-coordinate equal to 1 or 퐿. To compute 푃 (0, 2퐿, 푚2(푥), 16, −) we need to consider only the paths with an odd number of turns. Since for 퐿 > 1 we have 푦 = 2퐿 < 3(퐿 − 1) (for 퐿 = 1 the function 푓(퐿) is undefined), it follows that the number of turns must be exactly 1. Therefore, either the path starts in the square (0, 0) and changes the direction in the square (퐿, 퐿), or the path starts in the square (2 − 2퐿, 0) and changes the direction in (1, 2퐿 − 1). The vectors for these two paths are related by the rotation through the angle 휋(퐿−1) and division by . Since 휋 − 4 1.04 the vector for the latter path has square 1 and the one for the former one has square 1 by the law of cosines 26 26·1.042 , we get 1 (︂ 1 2 휋(퐿 − 1))︂ (︂ 휋(퐿 − 1))︂ 푃 (0, 2퐿, 푚 (푥), 16, −) = 1 + − cos ≈ 0.074 1 − cos . 2 26 1.042 1.04 4 4 The graph of this function is shown in the following figure (kindly provided by Punnawith Thuwajit, a participant of Summer Conference of Tournament of Towns).

24. Answer: 0.1664 0.1664 푃 (휆, 퐿, −) = ; max 푃 (휆, 퐿, −) = ≈ 0.14; 2 2휋(퐿−1) 퐿 1.1664 1.1664 + cot 휆 1 푃 (휆, 퐿, +) = ; 푃 (휆, 퐿, −) + 푃 (휆, 퐿, +) = 1. 2 2휋(퐿−1) 1 + 0.1664 sin 휆 Hint. Denote and 2휋 . For we have 푚 = 0.2 휑 = 휆 (퐿 − 1) 푦 > 2퐿 − 2

푃 (0, 푦, 푚2(푥), 휆, −) =

⃒ ⎛ [︁ 푦−2 ]︁ ⎞⃒2 ⃒ 2 (︂ 2 )︂ 2(퐿−1) 2(퐿−1) ⃒ ⃒ 2휋푖 (2−푦) 푚 1 2푖휑 푚 2푖휑 2 1 푚 2푖휑 ⃒ = ⃒푒 휆 √ ⎝−1 + 푒 + (푒 ) + ··· + 푒 ⎠⃒ , ⃒ 1 + 푚2 1 + 푚2 (1 + 푚2)2 푚2 1 + 푚2 ⃒ ⃒ ⃒

Then, since cos 2휑 = 1 − 2 sin2 휑, we have

⃒ ∞ 푘⃒2 푚2 ⃒ 1 ∑︁ (︂ 푚2 )︂ ⃒ 푃 (휆, 퐿, −) = · ⃒1 − 푒2푖휑 · 푒2푖휑 ⃒ = 1 + 푚2 ⃒ 1 + 푚2 1 + 푚2 ⃒ ⃒ 푘=0 ⃒ ⃒ ⃒2 푚2 ⃒ 1 1 ⃒ 푚2 |1 − 푒2푖휑|2 = · ⃒1 − 푒2푖휑 ⃒ = · = 1 + 푚2 ⃒ 1 + 푚2 푚2 2푖휑 ⃒ 1 + 푚2 푚2 2푖휑 2 ⃒ 1 − 1+푚2 푒 ⃒ |1 − 1+푚2 푒 | 푚2(1 + 푚2)(2 − 2 cos(2휑)) 4푚2(1 + 푚2) 0.1664 = = = . 2 2 4 2 2 2 2 2 2 2휋 (1 + 푚 ) + 푚 − 2푚 (1 + 푚 ) cos(2휑) (1 + 2푚 ) + cot 휑 1.1664 + cot 휆 (퐿 − 1)

28 Therefore, 0.1664 . Analogously, max퐿 푃 (휆, 퐿, −) = 1.1664 ≈ 0.14

⃒ ∞ 푘⃒2 ⃒ ⃒2 ⃒ 1 ∑︁ (︂ 푚2 )︂ ⃒ ⃒ 1 1 ⃒ 푃 (휆, 퐿, +) = ⃒ · 푒2푖휑 ⃒ = ⃒ · ⃒ = ⃒1 + 푚2 1 + 푚2 ⃒ ⃒1 + 푚2 푚2 2푖휑 ⃒ ⃒ 푘=0 ⃒ ⃒ 1 − 1+푚2 푒 ⃒ 1 1 1 = = = . 2 2 4 2 2 2 2 2 2 2휋 (1 + 푚 ) + 푚 − 2(1 + 푚 )푚 cos(2휑) 1 + 4푚 (1 + 푚 ) sin 휑 1 + 0.1664 sin 휆 (퐿 − 1) Note that 1 0.1664 0.1664 1 − = = . 2 2휋 1 2 2휋 1 + 0.1664 sin 휆 (퐿 − 1) 2 2휋 + 0.1664 1.1664 + cot 휆 (퐿 − 1) sin 휆 (퐿−1) Hence, 푃 (휆, 퐿, −) + 푃 (휆, 퐿, +) = 1. 25. Answer: √ (푛2 − 1)2 (︀0.3328 + 0.864 1.04)︀ · 휋 and 푚 = , 2 2 2 2 2휋퐿푛 휆 (푛 + 1) + 4푛 cot 휆 where 푛 = √︀1 + 푚휆/휋 (the refractive index; notice that the relation between 푚 and 푛 has changed when we passed to reflection inside the glass). For the latter 푚 the limit equals 0.1664 √ . 2 2휋퐿(1.08+0.08 1.04) 1.1664 + cot 휆 Remark. The first formula in the answer is well-known and confirmed by experiment. A remarkable detailed popular science discussion of Problems 23–25 can be found in [Feynman], and a derivation of the equations on a physical level of rigor — in [Landafshitz, Chapter X, §86, Problem 4]. Path to solution. Consider Dirac’s equation from the solution of Problem 17. Search for a solution of the form

−2휋푖휀푦/휆 (푎+(푥, 푦), 푎−(푥, 푦)) = (푎+(푥), 푎−(푥))푒 .

Use Problem 20 to guess 푎±(푥) separately for 푥 ≤ 0, 0 ≤ 푥 ≤ 퐿/휀, and 푥 ≥ 퐿/휀. Assume for simplicity that 1/휀 is an odd integer. Abbreviate 푚3(푥) =: 푚(푥). First consider the half-plane 푥 ≤ 0 (to the left from glass). We need to specify what is meant by a solution of Dirac’s equation in the half-plane. Analogously to Problem 17, by Dirac’s equation in a black square (푥, 푦) we mean the system ⎧ 1 −푖푚(푥)휀 푎−(푥 − 1, 푦 + 1) = √ 푎−(푥, 푦) + √ 푎+(푥, 푦); ⎨ 1+푚(푥)2휀2 1+푚(푥)2휀2 −푖푚(푥)휀 1 푎+(푥 + 1, 푦 + 1) = √ 푎−(푥, 푦) + √ 푎+(푥, 푦). ⎩ 1+푚(푥)2휀2 1+푚(푥)2휀2

A solution of Dirac’s equation in the half-plane 푥 ≤ 0 is a pair (푎+(푥, 푦), 푎−(푥, 푦)) satisfying the equation in each black square in the half-plane. In particular, 푎−(푥, 푦) is defined for 푥 ≤ 0, whereas 푎+(푥, 푦) is defined for 푥 ≤ 1. 2휋푖푥휀/휆 −2휋푖푥휀/휆 Search for a solution of the form (푎+(푥), 푎−(푥)) = (푖 푒 , 푟(휀) 푒 ) for some 푟(휀) ∈ C to be determined later, where the formulae for 푎+(푥) and 푎−(푥) hold for 푥 ≤ 1 and 푥 ≤ 0 respectively. (Physically, 푎+(푥, 푦) is the 2 incident wave, 푎−(푥, 푦) is the reflected wave, and | lim휀→0 푟(휀)| is the reflection probability.) 2휋푖푥휀/휆 For the half-plane 푥 ≥ 퐿/휀 take (푎+(푥), 푎−(푥)) = (푡(휀) 푒 , 0) for some 푡(휀) ∈ C to be determined later, where the formulae for 푎+(푥) and 푎−(푥) hold for 푥 ≥ 퐿/휀 + 1 and 푥 ≥ 퐿/휀 respectively. (Physically, 푎+(푥) is the 2 transmitted wave, and | lim휀→0 푡(휀)| is the transmission probability.) For 0 ≤ 푥 ≤ 퐿/휀 search for a solution of the form

2휋푖푥휀/휆′(휀) −2휋푖푥휀/휆′(휀) 푎+(푥) = 푎(휀) 푒 + 푏(휀) 푒 , 2휋푖푥휀/휆′(휀) −2휋푖푥휀/휆′(휀) 푎−(푥) = 푐(휀) 푒 + 푑(휀) 푒 .

′ for some coefficients 푎, 푏, 푐, 푑 ∈ C and 휆 > 0 depending on 휀 (we have a combination of refracted and reflected waves). In what follows 푎, 푏, 푐, 푑, 휆′, 푟, 푡 denote the limits of the above numbers as 휀 → 0. The limit 휆′ (wavelength of refracted wave) is determined by Dirac’s equation as follows. Substituting the expressions for 푎±(푥) into Dirac’s equation, canceling common factors, and taking the limit 휀 → 0 we get (︂ 2휋 2휋 )︂ 푚푎 = 푚 − + 푐, 휆′ 휆 (︂ 2휋 2휋 )︂ 푚푐 = 푚 + + 푎. 휆′ 휆 29 We also get an analogous equation with 푏 and 푑 instead of 푐 and 푎. This implies that 휆′ = 휆/푛, where 푛 = √︀1 + 푚휆/휋 (refractive index). Sewing together the solutions (namely, 푎+(푥) at 푥 = 1 and 푥 = 퐿/휀 + 1, whereas 푎−(푥) at 푥 = 0 and 푥 = 퐿/휀) gives a system of linear equations in 푎, 푏, 푐, 푑, 푟, 푡:

2휋푖퐿/휆′ −2휋푖퐿/휆′ 푎−(0) = 푟(휀) = 푐(휀) + 푑(휀), 푎−(퐿/휀) = 0 = 푐(휀) 푒 + 푑(휀) 푒 , 2휋푖퐿/휆 2휋푖퐿/휆′ −2휋푖퐿/휆′ 2휋푖(퐿+휀)/휆 2휋푖(퐿+휀)/휆′ −2휋푖(퐿+휀)/휆′ 푎+(1) = 푖푒 = 푎(휀) 푒 + 푏(휀) 푒 , 푎+(퐿/휀 + 1) = 푡(휀) 푒 = 푎(휀) 푒 + 푏(휀) 푒 .

Passing to the limit 휀 → 0 and solving the resulting system of linear equations we find 푎, 푏, 푐, 푑, 푟, 푡 Surely, one still needs to check that even before passing to the limit, the expressions for 푎±(푥) determined by our equations indeed satisfy Dirac’s equation in each black square. We omit that. 2휋푦휀/휆 휆 2 It remains to prove that (푎+(푥), 푎−(푥)) = lim 푅 ⃗푎(푥,푦,푚3(푥)휀, ,−); then the number |푟| is the required 푦→+∞ 휀 푦+푥 even limit. We do not know any elementary proof of that assertion and thus omit the details.

Informally, the idea is to show that the limit does not depend on the initial values 푎+(푥, 0) for 푥 > 0 and 푎−(푥, 0) for 푥 < 퐿/휀; then one can replace these initial values by the solution we have found above and get the result. Independence on 푎+(푥, 0) for 푥 > 퐿/휀 and 푎−(푥, 0) for 푥 < 0 is obvious. To prove independence on 푎±(푥, 0) for 0 ≤ 푥 ≤ 퐿/휀, assume that 푎±(푥, 0) vanishes outside the segment 0 ≤ 푥 ≤ 퐿/휀 and consider the linear transformation

푈 :(푎−(0, 0), 푎−(2, 0), . . . , 푎−(퐿/휀 − 1, 0), 푎+(0, 0), . . . , 푎+(퐿/휀 − 1, 0))

↦→ (푎−(0, 2), 푎−(2, 2), . . . , 푎−(퐿/휀 − 1, 2), 푎+(0, 2), . . . , 푎+(퐿/휀 − 1, 2)).

Let us show that the absolute values of all the eigenvalues of 푈 are less than 1. Consider an eigenvector 푣 with the coordinates 푣1 = 푎−(0, 0), . . . , 푣퐿/휀+1 = 푎+(퐿/휀 − 1, 0). Let 푣푘 be the leftmost nonzero coordinate. If 푘 > 1 then 푈푣 has nonzero (푘 − 1)-th coordinate; but 푈푣 is proportional to 푣, a contradiction. Thus 푘 = 1. Hence 푎−(−2, 2) ̸= 0 by Dirac’s equation. Then by the probability conservation

2 2 2 ‖푣‖ = |푎−(0, 0)| + |푎−(2, 0)| + ··· + |푎+(퐿/휀 − 1, 0)| 2 2 2 = |푎−(−2, 2)| + |푎−(0, 2)| + ··· + |푎+(퐿/휀 − 1, 2)| 2 2 > |푎−(0, 2)| + ··· + |푎+(퐿/휀 − 1, 2)| = ‖푈푣‖.

Thus the absolute value of the eigenvalue 푣 is strictly less than 1. Thus ‖푈 푦‖ → 0 as 푦 → ∞, which proves that the values 푎±(푥, 0) for 0 ≤ 푥 ≤ 퐿 do not affect the limit. 26. Answer: In the table, the vector in the cell (푥, 푦) is ⃗푎(푥, 푦, 푢, +), and an empty cell (푥, 푦) means that ⃗푎(푥, 푦, 푢, +) = (0, 0). (︁ )︁ (︁ )︁ 4 0, √1 (0, 0) 0, √1 2 2 2 2 3 (︀0, − 1 )︀ (︀0, 1 )︀ 2 (︁ )︁ 2 2 0, − √1 2 1 (0, −1) y x −2 −1 0 1 2 3 4

In the following table the number in the cell (푥, 푦) is 푃 (푥, 푦, 푢, +), and an empty cell (푥, 푦) means that 푃 (푥, 푦, 푢, +) = 0. 4 1/8 0 1/8 3 1/4 1/4 2 1/2 1 1 y x −2 −1 0 1 2 3 4

(The vectors ⃗푎(푥, 푦, 푢, +) can be easily computed using Problem 31.) For any positive 푦 we have ∑︀ (푃 (푥, 푦, 푢, +)+푃 (푥, 푦, 푢, −)) = 1. (This is one of the statements of Problem 31.) 푥∈Z 27. Answer (kindly provided by Gleb Minaev and Ivan Russkikh, participants of Summer Conference of Tourna- ment of Towns):

30 28. Solution. Suppose that 푢′ is the result of changing the values of 푢 in all the vertices of a black square (푎, 푏). Notice that if a path 푠 neither starts nor ends in the square (푎, 푏), then it either passes through two vertices with a changed value of 푢 or does not pass through such vertices at all. Therefore, ⃗푎(푠, 푢) = ⃗푎(푠, 푢′). The equation 푃 (푥, 푦, 푢′, +) = 푃 (푥, 푦, 푢, +) for any (푥, 푦) ̸= (푎, 푏) follows immediately. The only case left to prove is (푎, 푏) = (푥, 푦) or (푎, 푏) = (0, 0). For each path 푠 starting or ending in (푎, 푏) we have ⃗푎(푠, 푢′) = −⃗푎(푠, 푢) because 푠 passes through exactly one vertex with a changed value of 푢. Therefore, 푃 (푎, 푏, 푢′, +) = 푃 (푎, 푏, 푢, +). 29. Solution. The top-right vertex of the square (푥, 푦) is called vertex (푥, 푦). Denote by 푢(푥, 푦) the value of 푢 at the vertex (푥, 푦). Reversing a black square (푥, 푦) means changing the values of 푢 at all the four vertices of the square. Notice that reversing a black square does not change the sign of any white squares, because the number of changed vertices in each white square is even. If all values of a field in a rectangle are positive, then all the white squares inside the rectangle are positive. Therefore, it is impossible to make 푢 identically +1 by reversing black squares, if initially there is a negative white square. Suppose that there are no negative white squares in a rectangle 푀 × 푁, where 푁 is the height of the rectangle. Let the vertex (0, 0) be at the bottom-left of the rectangle. Let us prove that 푢 can be made identically +1 through reversal of black squares by induction over 푁. Base: 푁 = 0. Let us describe the algorithm for making 푢 identically +1 in the rectangle 푀 × 0 (i.e in vertices (0, 0), (1, 0),..., (푀, 0)). Note that for all 푘 we can reverse either square (푘 + 1, 0) or square (푘 + 1, 1), thereby we can change the signs of the values of 푢 in vertices (푘, 0) and (푘 + 1, 0). Denote this operation reversing 푘. The algorithm is as follows: Find the minimum 0 ≤ 푘 ≤ 푀 such that 푢(푘, 0) = −1, and reverse 푘. If after that there still exists 0 ≤ 푘 ≤ 푀 such that 푢(푘, 0) = −1, than the minimum such 푘 is at least 1 more than before using the operation. Therefore, after at most 푀 + 1 operations we make 푢 to be identically +1 in the rectangle 푀 × 0. Step: Suppose that the statement is true for 푁 and prove it for 푁 + 1 by giving an explicit algorithm. First make 푢 identically +1 in the lower rectangle 푀 × 푁 by applying the algorithm for 푁. Then for each white square (푘, 푁 + 1) in the rectangle reverse the square (푘, 푁 + 2) if 푢(푘 − 1, 푁 + 1) = 푢(푘, 푁 + 1) = −1, otherwise do nothing. Finally, if the square (0, 푁 + 1) is black and 푢(−1, 푁 + 1) = −1 then reverse the square (−1, 푁 + 2), and if the square (푀, 푁 + 1) is black and 푢(푀, 푁 + 1) = −1 then reverse the square (푀 + 1, 푁 + 2). The result is a field, positive in the whole rectangle (푀 + 1) × 푁. 30. Answer: Zero, if one of the sides of the rectangle is 1, and any even number no greater than the number of white squares in the rectangle, if both sides are greater than 1. Hint. Denote by 푛(푥, 푦, 푢) the number of vertices of the square (푥, 푦) where the value of 푢 is negative. Notice that any vertex that is strictly inside the rectangle (not on the boundary) is a vertex of exactly four squares in that rectangle, two of which are white. Therefore, the sum of 푛(푥, 푦, 푢) over all white squares in the rectangle is even. Hence the number of negative white squares is even. Let us show that any even number no greater than the number of white squares in the rectangle is indeed possible. If both the length and the width of the rectangle are even, then for any field that is positive everywhere in the rectangle except for 푛 vertices with both coordinates even there are exactly 2푛 negative white squares. For odd length or width (greater than 1) see the figure below. The identically +1 field has no negative white squares. The field 푢2푛+2 with 2푛 + 2 negative white squares is obtained from the field 푢2푛 by changing all the values of 푢2푛 to the opposite in any single encircled area where they have not been changed before.

31 31. Answer. Recall that 푢(푥, 푦) is the value of 푢 in the top-right vertex of the square (푥, 푦). Then 1 푎1(푥, 푦, 푢) = √ 푢(푥, 푦 − 1)(푎1(푥 + 1, 푦 − 1, 푢) + 푎2(푥 + 1, 푦 − 1, 푢)), 2 1 푎2(푥, 푦, 푢) = √ 푢(푥 − 1, 푦 − 1)(−푎1(푥 − 1, 푦 − 1, 푢) + 푎2(푥 − 1, 푦 − 1, 푢)). 2 Hint. The solution of the analogue of Problem 4 is very similar to the original one, only the factors 푢(푥, 푦 − 1) and 푢(푥 − 1, 푦 − 1) are added due to the last step of the path 푠 passing through the top-right vertex of either (푥, 푦 − 1) or (푥 − 1, 푦 − 1). Let us prove the analogue of Problem 5 by induction over 푦. The step of induction follows immediately from the following computation:

∑︁ ∑︁ [︀ 2 2]︀ ∑︁ 2 ∑︁ 2 푃 (푥, 푦 + 1, 푢) = 푎1(푥, 푦 + 1, 푢) + 푎2(푥, 푦 + 1, 푢) = 푎1(푥, 푦 + 1, 푢) + 푎2(푥, 푦 + 1, 푢) = 푥∈Z 푥∈Z 푥∈Z 푥∈Z 1 ∑︁ 1 ∑︁ = 푢(푥, 푦)2(푎 (푥 + 1, 푦, 푢) + 푎 (푥 + 1, 푦, 푢))2 + 푢(푥 − 1, 푦)2(푎 (푥 − 1, 푦, 푢) − 푎 (푥 − 1, 푦, 푢))2 = 2 1 2 2 2 1 푥∈Z 푥∈Z 2 2 ∑︁ (푎1(푥, 푦, 푢) + 푎2(푥, 푦, 푢)) ∑︁ (푎2(푥, 푦, 푢) − 푎1(푥, 푦, 푢)) ∑︁ ∑︁ = + = [︀푎 (푥, 푦, 푢)2 + 푎 (푥, 푦, 푢)2]︀ = 푃 (푥, 푦, 푢). 2 2 1 2 푥∈Z 푥∈Z 푥∈Z 푥∈Z

⎧ ′ ′ ′ ⎪푃 (푥, 푦, −)푃 (푥 − 푥0, 푦, −), if 퐸 = (푥 + 1, 푦), 퐸 = (푥 + 1, 푦), ⎪ ⎨⎪푃 (푥, 푦, −)푃 (푥′ − 푥 , 푦, +), if 퐸 = (푥 + 1, 푦), 퐸′ = (푥′ − 1, 푦), 32. Answer: 푃 (퐴퐵, 퐴′퐵′ → 퐸퐹, 퐸′퐹 ′) = 0 푃 (푥, 푦, +)푃 (푥′ − 푥 , 푦, −), if 퐸 = (푥 − 1, 푦), 퐸′ = (푥′ + 1, 푦), ⎪ 0 ⎪ ′ ′ ′ ⎩푃 (푥, 푦, +)푃 (푥 − 푥0, 푦, +), if 퐸 = (푥 − 1, 푦), 퐸 = (푥 − 1, 푦). Solution. Notice that ⃗푎(푠, 푠′) = −푖⃗푎(푠)⃗푎(푠′), where ⃗푎⃗푏 denotes the product of complex numbers ⃗푎 and ⃗푏. ′ Without loss of generality suppose that 푥 > 푥. Due to the condition 푥0 ≥ 2푦 there are no paths starting at 퐴 and ending in 퐹 ′ and no paths starting at 퐴′ and ending in 퐹 . Therefore ∑︀ ⃗푎(푠, 푠′) = 0 and ⃗푎(퐴퐵, 퐴′퐵′ → 퐸퐹, 퐸′퐹 ′) = ∑︀ ⃗푎(푠, 푠′). 푠:퐴퐵→퐸′퐹 ′ 푠:퐴퐵→퐸퐹 푠′:퐴′퐵′→퐸퐹 푠′:퐴′퐵′→퐸′퐹 ′ If 퐸 = (푥 − 1, 푦) and 퐸′ = (푥′ − 1, 푦) then ⃗푎(퐴퐵, 퐴′퐵′ → 퐸퐹, 퐸′퐹 ′) = −푖 ∑︀ ⃗푎(푠)⃗푎(푠′), where the sum is over 푠,푠′ ′ ′ ′ ′ ′ ′ 푠 and 푠 ending with an upwards-right move. Therefore, ⃗푎(퐴퐵, 퐴 퐵 → 퐸퐹, 퐸 퐹 ) = −푖⃗푎(푥, 푦, +)⃗푎(푥 − 푥0, 푦, +). ′ ′ ′ ′ ′ Hence 푃 (퐴퐵, 퐴 퐵 → 퐸퐹, 퐸 퐹 ) = 푃 (푥, 푦, +)푃 (푥 − 푥0, 푦, +). If 퐸 = (푥 + 1, 푦) and 퐸′ = (푥′ − 1, 푦) then ⃗푎(퐴퐵, 퐴′퐵′ → 퐸퐹, 퐸′퐹 ′) = −푖 ∑︀ ⃗푎(푠)⃗푎(푠′), where the sum is over 푠 푠,푠′ ending with an upwards-left move and 푠′ ending with an upwards right move. Hence 푃 (퐴퐵, 퐴′퐵′ → 퐸퐹, 퐸′퐹 ′) = ′ 푃 (푥, 푦, −)푃 (푥 − 푥0, 푦, +). If 퐸 = (푥 − 1, 푦) and 퐸′ = (푥′ + 1, 푦) then ⃗푎(퐴퐵, 퐴′퐵′ → 퐸퐹, 퐸′퐹 ′) = −푖 ∑︀ ⃗푎(푠)⃗푎(푠′), where the sum is over 푠 푠,푠′ ending with an upwards-right move and 푠′ ending with an upwards left move. Hence 푃 (퐴퐵, 퐴′퐵′ → 퐸퐹, 퐸′퐹 ′) = ′ 푃 (푥, 푦, +)푃 (푥 − 푥0, 푦, −). 32 If 퐸 = (푥 + 1, 푦) and 퐸′ = (푥′ + 1, 푦) then ⃗푎(퐴퐵, 퐴′퐵′ → 퐸퐹, 퐸′퐹 ′) = −푖 ∑︀ ⃗푎(푠)⃗푎(푠′), where the sum is over 푠,푠′ ′ ′ ′ ′ ′ ′ 푠 and 푠 ending with an upwards-left move. Hence 푃 (퐴퐵, 퐴 퐵 → 퐸퐹, 퐸 퐹 ) = 푃 (푥, 푦, −)푃 (푥 − 푥0, 푦, −). 33. Hint. Suppose that paths 푠 : 퐴퐵 → 퐸퐹 and 푠′ : 퐴′퐵′ → 퐸′퐹 ′ have a common move. Suppose that 퐶퐷 is the first common move of 푠 and 푠′. Denote by 푠˜ the path that coincides with 푠 up to the square 퐶 and with 푠′ after 퐶. and 푠˜′ the path that coincides with 푠′ before 퐶 and with 푠 after. The paths 푠,˜ 푠˜′ start with the moves 퐴퐵, 퐴′퐵′ and end with moves 퐸′퐹 ′ and 퐸퐹 respectively. The total number of turns in the paths 푠 and 푠′ equals the total number of turns in 푠˜ and 푠˜′ , because 퐶퐷 is a common move for all of them. Therefore, ⃗푎(푠, 푠′) = ⃗푎(˜푠, 푠˜′). Notice that 푓 :(푠, 푠′) ↦→ (˜푠, 푠˜′) is a bijection, because 퐶퐷 is exactly the first common move. One of (푠, 푠′) and (˜푠, 푠˜′) contributes to ⃗푎(퐸퐹, 퐸′퐹 ′) with the coefficient (+1) and the other one with the the coefficient (−1). Therefore, these pairs cancel each other. 34. Hint. Let us prove this equality by induction over 푦. The base is obvious. Note that by analogy with Problem 11, ⃗푎(퐴퐵, 퐴′퐵′ → 퐸퐹, 퐸′퐹 ′) is parallel to either the line 푥 = 0 or the line ′ ′ ′ ′ ′ ′ ′ ′ 푦 = 0. Denote by 푎1(퐴퐵, 퐴 퐵 → 퐸퐹, 퐸 퐹 ) and 푎2(퐴퐵, 퐴 퐵 → 퐸퐹, 퐸 퐹 ) its coordinates. ′ ′ ′ ′ ′ ′ ′ Fix the starting moves 퐴퐵 and 퐴 퐵 of the paths. Denote by 푎(퐸, 퐸 , ↑, +, −) = 푎1(퐴퐵, 퐴 퐵 → 퐸퐹, 퐸 퐹 ), ′ ′ ′ ′ ′ ′ where 퐸퐹 is an upwards-right move, 퐸 퐹 is an upwards-left move, and 푎1(퐴퐵, 퐴 퐵 → 퐸퐹, 퐸 퐹 ) is the (possibly) non-zero coordinate of the corresponding vector. 푎(퐸, 퐸′, ↑, +, +), 푎(퐸, 퐸′, ↑, −, +) and 푎(퐸, 퐸′, ↑, −, −) are defined ′ ′ ′ ′ ′ ′ ′ analogously. Denote 푎(퐸, 퐸 , ↓, +, −) = 푎1(퐴퐵, 퐴 퐵 → 퐷퐸, 퐷 퐸 ), where 퐷퐸 is an upwards-right move and 퐷 퐸 is an upwards-left move. 푎(퐸, 퐸′, ↓, +, +), 푎(퐸, 퐸′, ↓, −, +) and 푎(퐸, 퐸′, ↓, −, −) are defined analogously. Note that the induction step is a proof of the equality ∑︁ ((푎(퐸, 퐸′, ↑, +, +))2 + (푎(퐸, 퐸′, ↑, +, −))2 + (푎(퐸, 퐸′, ↑, −, +))2 + (푎(퐸, 퐸′, ↑, −, −))2) = 퐸,퐸′ ∑︁ = ((푎(퐸, 퐸′, ↓, +, +))2 + (푎(퐸, 퐸′, ↓, +, −))2 + (푎(퐸, 퐸′, ↓, −, +))2 + (푎(퐸, 퐸′, ↓, −, −))2). 퐸,퐸′

We will prove this equation by proving that for every pair (퐸, 퐸′), where 퐸 and 퐸′ have the same 푦-coordinate,

(푎(퐸, 퐸′, ↑, +, +))2 + (푎(퐸, 퐸′, ↑, +, −))2 + (푎(퐸, 퐸′, ↑, −, +))2 + (푎(퐸, 퐸′, ↑, −, −))2 = =(푎(퐸, 퐸′, ↓, +, +))2 + (푎(퐸, 퐸′, ↓, +, −))2 + (푎(퐸, 퐸′, ↓, −, +))2 + (푎(퐸, 퐸′, ↓, −, −))2. Analogously to Problem 4 we obtain 1 푎(퐸, 퐸′, ↑, +, +) = (푎(퐸, 퐸′, ↓, +, +) − 푎(퐸, 퐸′, ↓, +, −) − 푎(퐸, 퐸′, ↓, −, +) − 푎(퐸, 퐸′, ↓, −, −)), 2 1 푎(퐸, 퐸′, ↑, +, −) = (푎(퐸, 퐸′, ↓, +, +) − 푎(퐸, 퐸′, ↓, +, −) + 푎(퐸, 퐸′, ↓, −, +) + 푎(퐸, 퐸′, ↓, −, −)), 2 1 푎(퐸, 퐸′, ↑, −, +) = (푎(퐸, 퐸′, ↓, +, +) + 푎(퐸, 퐸′, ↓, +, −) − 푎(퐸, 퐸′, ↓, −, +) + 푎(퐸, 퐸′, ↓, −, −)), 2 1 푎(퐸, 퐸′, ↑, −, −) = (−푎(퐸, 퐸′, ↓, +, +) − 푎(퐸, 퐸′, ↓, +, −) − 푎(퐸, 퐸′, ↓, −, +) + 푎(퐸, 퐸′, ↓, −, −)). 2 With this, we can prove our equation by substitution and expansion. ⎧ ′ ′ ′ ⎪푃 (푥, 푦, −)푃 (푥0 − 푥 , 푦, −), if 퐸 = (푥 + 1, 푦), 퐸 = (푥 + 1, 푦), ⎪ ⎨⎪푃 (푥, 푦, −)푃 (푥 − 푥′, 푦, +), if 퐸 = (푥 + 1, 푦), 퐸′ = (푥′ − 1, 푦), 35. Answer: 푃 (퐴퐵, 퐵′퐴′ → 퐸퐹, 퐹 ′퐸′) = 0 푃 (푥, 푦, +)푃 (푥 − 푥′, 푦, −), if 퐸 = (푥 − 1, 푦), 퐸′ = (푥′ + 1, 푦), ⎪ 0 ⎪ ′ ′ ′ ⎩푃 (푥, 푦, +)푃 (푥0 − 푥 , 푦, +), if 퐸 = (푥 − 1, 푦), 퐸 = (푥 − 1, 푦). Hint. Notice that ⃗푎(푠, 푠′) = 푖⃗푎(푠)⃗푎*(푠′), where ⃗푎⃗푏 denotes the product of complex numbers ⃗푎 and ⃗푏, and ⃗푎* denotes conjugation. For the remaining problems, we only give preliminary draft solutions.

36. Answer: for each 푢fin we have

푃 (2, 2, 푢fin, 1, +) = 1/4, 푃 (0, 2, 푢fin, 1, −) = 1/4,

푃 (3, 3, 푢fin, 1, +) = 1/32, 푃 (1, 3, 푢fin, 1, −) = 1/32,

푃 (1, 3, 푢fin, 1, +) = 1/32, 푃 (−1, 3, 푢fin, 1, −) = 1/32; all the other probabilities vanish for 푦 = 2, 3. 3푦−4 2 푦−1 37. Answer: ∑︀⃗푎(푔, 푠, 푢) = 2 (1 + 푔 ) 2 cos((푟 − 푙) arctan(푔))⃗푎(푠). 푢 33 Hint. For brevity, denote 1 . 푍 = (푦−1)2 2 2(푦−2) (1+푔2) 2 Suppose that 푟 and 푙 are the numbers of white squares in 푅 to the right and to the left from 푠 respectively. Notice that 푟 and 푙 are odd if and only if the number of “−” signs on the path 푠 is odd (that is, ⃗푎(푠, 푢) = −⃗푎(푠)). Easy-to-prove lemma: if 푏 is the number of black squares in 푅 not touching the boundary, then the number of fields 푢 with exactly 2푘 negative white squares equals 2푏 times the number of ways to assign “−” to 2푘 white squares in 푅. The factor 2푏 is the number of gauge transformations from the Problem 28. By the lemma we have

⎡ ⎤ ∑︁ ∑︁ ∑︁ (︂푟)︂(︂ 푙 )︂ ∑︁ (︂푟)︂(︂ 푙 )︂ ⃗푎(푔, 푠, 푢) = 푍 (−푖푔)푛⃗푎(푠, 푢) = 2푏푍 (−푖푔)푘+푚 − (−푖푔)푘+푚 ⃗푎(푠) = ⎣ 푘 푚 푘 푚 ⎦ 푢 푢 푘,푚 푒푣푒푛 푘,푚 표푑푑 [︁ (︁ )︁ (︁ )︁]︁ = 2푏−2푍 ((1 + 푖푔)푟 + (1 − 푖푔)푟) (1 + 푖푔)푙 + (1 − 푖푔)푙 − ((1 + 푖푔)푟 − (1 − 푖푔)푟) (1 + 푖푔)푙 − (1 − 푖푔)푙 ⃗푎(푠) =

푏 (︁ 푟 푙)︁ 푏 2 푟+푙 = 2 푍푅푒 (1 + 푖푔) (1 − 푖푔) ⃗푎(푠) = 2 (1 + 푔 ) 2 푍 cos((푟 − 푙) arctan(푔))⃗푎(푠).

2 3푦−4 2 푦−1 Since 푏 = 푟 + 푙 = 푦 − 푦, it follows that ∑︀⃗푎(푔, 푠, 푢) = 2 (1 + 푔 ) 2 cos((푟 − 푙) arctan(푔))⃗푎(푠). 푢 38. Answer. Denote

푎−−(푥, 푦) = 푎2(푥, 푦, 푢−, 푔, −), 푎−+(푥, 푦) = 푎1(푥, 푦, 푢−, 푔, +),

푎+−(푥, 푦) = 푎1(푥, 푦, 푢+, 푔, −), 푎++(푥, 푦) = 푎2(푥, 푦, 푢+, 푔, +).

Denote by 푅(푦) = 푅 the rectangle formed by all the squares (푥′, 푦′) such that 1 − 푦 < 푥′ < 푦 and 0 < 푦′ < 푦. Denote 훼 = 훼(푥, 푦) = (Δ푙 − Δ푟) arctan 푔, where Δ푙 and Δ푟 are the numbers of white squares in 푅(푦) outside 푅(푦 − 1) lying to the left and to the right from the square (푥 + 1, 푦 − 1). Denote 훽 = 훼(푥 − 2, 푦). Then ⎧ 푎 (푥, 푦) = cos√ 훽 (푎 (푥 − 1, 푦 − 1) − 푎 (푥 − 1, 푦 − 1)) − sin√ 훽 (푎 (푥 − 1, 푦 − 1) − 푎 (푥 − 1, 푦 − 1)) ; ⎪ ++ 2 ++ +− 2 −+ −− ⎪ ⎪푎 (푥, 푦) = cos√ 훼 (푎 (푥 + 1, 푦 − 1) + 푎 (푥 + 1, 푦 − 1)) − sin√ 훼 (푎 (푥 + 1, 푦 − 1) + 푎 (푥 + 1, 푦 − 1)) ; ⎨ +− 2 ++ +− 2 −+ −− 푎 (푥, 푦) = sin√ 훽 (푎 (푥 − 1, 푦 − 1) − 푎 (푥 − 1, 푦 − 1)) + cos√ 훽 (푎 (푥 − 1, 푦 − 1) − 푎 (푥 − 1, 푦 − 1)) ; ⎪ −+ 2 ++ +− 2 −+ −− ⎪ ⎪푎 (푥, 푦) = sin√ 훼 (푎 (푥 + 1, 푦 − 1) + 푎 (푥 + 1, 푦 − 1)) + cos√ 훼 (푎 (푥 + 1, 푦 − 1) + 푎 (푥 + 1, 푦 − 1)) . ⎩ −− 2 ++ +− 2 −+ −−

Hint. The latter formula defines an orthogonal transformation, which implies that ∑︀ (푃 (푥, 푦, 푔, +) + 푥∈Z 푃 (푥, 푦, 푔, −)) = 1 by induction over 푦. 39. Answer. The number assigned to a basis configuration is shown below it:

1 √ −푖푚휀 √ 1 √ 1 √ −푖푚휀 − 1. 1+푚2휀2 1+푚2휀2 1+푚2휀2 1+푚2휀2 Hint. Let us prove that the sum over all pairs of paths passing through vertices with “−” signs only equals the product of all the numbers assigned to the black squares in the rectangle. Notice that the squares of type (1) do not contribute to the result at all. Squares of types (2) and (5) can only be passed through by a path with a turn in that square, which means contributing a factor of √ −푖푚휀 to 1+푚2휀2 ⃗푎(푠, 푠′). Squares of types (3) and (4) can only be passed through by a path with a turn in that square which means contributing a factor of √ 1 to ⃗푎(푠, 푠′). Squares of type (6) can be passed through in two different ways: by 1+푚2휀2 (︁ )︁2 two turning or two not turning paths. When the paths do turn, the square contributes √ −푖푚휀 and when they 1+푚2휀2 do not, it contributes −1 , which sums up to . 1+푚2휀2 −1 Acknowledgements The authors are grateful to V. Skopenkova for some of the figures, G. Chelnokov, I. Ivanov for useful discus- sions, and all participants of Summer Conference of Tournament of Towns in Arandelovac and Summer School in Contemporary Mathematics in Dubna for their contribution.

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