Convergence

EDITION I, 2014

CONVERGENCE A MATHLETES PUBLICATION EDIITION I

Table of Contents The Monty Hall Problem 7 A Bit on Chaos Theory 10 The Human Element in 12 Mathematics Mathematics in Origami 19 Conjecture. Collatz Conjecture 21 Hard for Computers 15 Proof by Contradiction. Or not 23 Fermat’s Last Theorem 24 False for matrices Paradoxes & Fallacies 28 This year in Math 30 Puzzle Up 32

Message from THE HoD The publication of the Mathematics’ department’s annual magazine, ‘CONVERGENCE’ during the joyous occasion of our first department day celebrations is indeed a very happy moment for me, and for the entire department of mathematics. Mathematics education, by its very nature, carries ideals and concepts. It may be vacuous and rather futile to rank such mathematical creativity as it is immeasurable. This beautiful magazine is a translation of such creativity into a tangible artifact, an artifact which depicts our very being. The first ever attempt made in this direction is a testament to the confidence, innovation and progress shown by our faculty and students, and it is thrilling to go through each and every page of the magazine. We have a glowing history of success, and an even a more promising future lies ahead. This year, our department also had the opportunity to celebrate the ‘Department day’, on the eve of the Golden Jubilee celebrations of BITS, Pilani. I anticipate many more such innovative firsts in the immediate future. A magazine must create a bridge between the past and the present as well as between the present and the future. It must chronicle the intellectual status and growth of a department over the years and beyond. I sincerely appreciate the efforts of the members of the editorial board in bringing out this impressive edition in such a short span of time. The steps taken by them in the above directions are laudable and we are keen to assist them in all their future endeavors with the philosophy that in the race for achievement and growth, there is no finish line. With best wishes, Dr. Prasanna Kumar Head of Department, Department of Mathematics

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Editorial Every person blessed with the ability to think rationally values the said ability. There is thrill in the correctness of logic, pleasure in the process of making sense. I speak for a majority of my peers when I say that the happiness that stems from correctly solving a particularly challenging Math question is worth the hours of mind-taxing one undergoes to reach the result, the accuracy of puzzles is almost beautiful, and the joy of writing ‘Hence, proved’ at the end of a problem is incomparable. It is with this assumption that the Mathematical Association of BITS-Pilani, K. K. Birla Goa Campus has worked laboriously to bring out the first edition of its annual magazine, Convergence. The word Convergence, besides bringing a sigh of relief to every mathematician busy struggling with metrics, symbolizes clarity and understanding, and most importantly, the clear-cut definition of a concept. Such ideals are the mandate when one decides to release an issue that is purely Math, because regardless of the supposedly abstract concepts that this subject is famed to delve into, Mathematics makes sense. Yes, there is a degree of groping in the dark and reaching for the stars, but these are ambitions that keep the wonder in the subject alive. There is a little bit of Math in every one of us. This statement is applicable, because the ‘us’ in question, the readership of this magazine, the faculty and the students of our college, primarily hailing from scientific backgrounds, have all at some point in our lives appreciated the beauty that this subject embodies, and its applicability in all realms of life. Convergence strives to bring forth that love for Math in the minds of its readers as they read about famous unsolved problems and paradoxes, or jab furiously at our compilation of puzzles, or marvel at the humor and fallacies that this field is so replete with. It hopes to be a pioneer and an aid in starting a culture of appreciation of our subjects and a love for learning - a culture that our campus is capable of achieving, and a culture that we will truly cherish. Janvi Palan Editor-in-Chief

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The Monty Hall Problem

Sanjana Ramchandran

Consider three doors. Behind one door is everything you have ever wanted: it could be the love of your life, or new shoes, or self-respect, that most evasive version of all known forms of respect. It wouldn’t be remiss of us to assume that this ‘everything’ can find embodiment in the material world in the form of a car, and that this car would fit behind the beautiful, ethereal door. Behind the other two doors are goats. Standing next to the doors is a man who knows exactly what is behind each door. He knows where the car is, and where the goats are. Oh, and you’re also on a game show, where this man says to you, “Pick the door that you think has the car, and if you guess right, it’s yours.” So you pick a door, and are pretty pleased with yourself until the man says, “Okay, so you picked door B. Now I’m going to confuse you!” (Obviously, the doors are A, B, and C, and as any sane person would, you picked B). The man then opens door A, which is revealed to have behind it a goat. He then says, “You picked B, and I later opened A, which you may think has nothing to do with anything, but I’m now going to offer you the chance to revise your choice. Do you want to stick to B, or would you like to switch to C?” How could this move by him have any bearing on whether the car is behind B? Does the knowledge that there was a goat behind A help your chances of getting the car? Should you, in fact, switch to C, and find that the car is behind it? And is there a mathematical foundation to verify any of these outcomes? Anybody who has done an introductory course on Probability is equipped with the technique of finding an answer to all these questions. It simply involves the use of Bayes’ Theorem, whose word form states that the probability that an event A occurred as a result of event B occurring is numerically equal to the product of the probability that B happened (with the knowledge that A is a result of said event) and the total probability of A itself happening, divided by the total probability of B alone happening. Mathematically, it looks like this:

Now the probability that you won the car by switching, by Bayes’ Theorem, can be computed by this: P(you switched | you won the car)=(P(winning the car by switching))/(P(winning the car))

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THE MONTY HALL PROBLEM To win the car by switching, you have to pick wrong the first time, so that equals: P(winning the car by switching)=P(picking the goat and then switching) = 2/3 As can be seen, there are two ways of getting the car: by picking the right door the first time and not switching, or by picking the goat the first time, and switching. Either of these two ways will give you the car, so the probability of winning the car through these methods is 1. P(winning the car)=P(picking it right the first time and not switching) + P(picking the goat and then switching) = 1/3+2/3 Therefore, P(you switched | you won the car)=(2/3)/(2/3+1/3) = 2/3 So it would seem, mathematically at least, that your chances of winning the car by switching are 2 times out of 3. You might be tempted to ask such questions as, ‘The probability of picking the car is always going to be 1/3, whether or not you are shown one door with a goat. How can switching increase that probability to 2/3?’ They call such puzzles veridical in nature, meaning, a paradox that has a result that appears absurd, but is true nevertheless. It was probably in answer to such queries, that Monty Hall, the eponymous host of the game show “Let’s Make a Deal”, where a close form of the puzzle was first posed to players, said, “After one door is seen to have a goat, a player’s chances of getting the car aren’t really 50-50, or any number but what they were in the first place: one out of three. It just seems to the contestant that one box having been eliminated, he stands a better chance. Not so.” While this statement is true, a more intuitive explanation to why your chances of winning double if you switch from your initial choice is this: the chances that you picked a goat the first time are double that of your picking the car the first time, which means that you should switch to another door, simply because you probably picked a goat the first time. This seemingly counter-intuitive solution to the Monty Hall problem initially garnered a rather acrimonious response from the general public and prolific mathematicians alike, most of whose initial answer to whether or not they would switch was either “No” or “There’s always a 1/3 chance of winning the car, so why change?” The important consideration lies in the role that the host plays. For instance, if the formulation of the problem were changed to specify that the host does not actually know where the car, but had instead simply pointed at a door at random and offered you the chance to switch, without opening it, then changing doors would not be beneficial. The probability of finding the car would still be 1/3 and the host would only be doing it to confuse you. Marilyn vos Savant, the American lecturer, listed in the Guiness Book of World Records under ‘Highest IQ’ and to whom the puzzle was initially posed, was widely ostracized for her answer that switching is never to the player’s disadvantage, if we focus on the crucial clause that states that the host always opens a different door than the one chosen by the player, and also always reveals a goat by this action, because he knows where the car is hidden. “Anything else is a different question,” she asserted. Several versions of the Monty Hall problem exist, some considering the results that would ensue should the behaviour of the host vary, or if the number of doors were increased. It is believed that the earliest related version of the Monty Hall problem was the Bertrand Box Paradox, posed by Joseph Bertrand in 1889, and the Three Prisoners’ problem, which is exactly equivalent to the Monty Hall problem, involving three inmates and with ‘freedom’ and ‘execution’ dangerously raising the stakes by replacing the car and the goats respectively. Alhough this problem gets its name from the game show host, Hall’s policy on the show varied slightly from the current known version, in that Hall would open a wrong door just to generate excitement, but offer an assured sum of $100 cash instead of the option to switch doors. He once wrote to Steve Selvin, the man who asked vos Savant the question in her column, “If you ever get on my show, the rules hold fast for you – no trading doors after the selection.”

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THE MONTY HALL PROBLEM The likes of erudite mathematician, Paul Erdos, have been unconvinced by this reasoning and the subsequent result, until verifications of these predictions were shown to him by computer simulations. If, like me, you are convinced that switching the door is indeed ‘where it’s at’, you might share the author’s wonder at why this problem, when given a strict premise, seems arcane to the world at large, so much so that it invoked a psychologist, Piattelli-Palmarini, to remark on it, “No other statistical puzzle comes so close to fooling all the people all the time.”

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A bit on Chaos Theory

Siddhartha Govilkar

“Chaos is when the present determines the future, but the approximate present does not approximately determine the future.”

It’s uncanny how intriguing the phenomenon of chaos is! What do we know about chaos? The fact that it’s fair? Or that it isn’t a pit but is a ladder? Apart from being the subject of various pop culture lines, it is also a field of study in mathematics which deals with the behavior of dynamical systems which are highly sensitive to initial conditions, so much so that a little rounding off or approximation of initial conditions leads to widely diverging outcomes. While most traditional science deals with supposedly predictable phenomena like gravity, electricity, or chemical reactions, Chaos Theory deals with nonlinear things that are effectively impossible to predict or control, for instance, the weather, the stock markets, etc. These phenomena are often described by fractal mathematics, a field which captures the infinite complexity of nature. For a dynamical system to be chaotic, it must have the following three properties:1. It must be sensitive to initial conditions (Butterfly effect) 2. It must show topological transitivity 3. Its periodic orbits must be dense Non-linear mathematics or chaos theory was first applied in the prediction of complex systems such as the weather. Chaotic systems exhibit not only apparently random unpredictability, but also a degree of determinism. The unpredictability arises because minuscule changes in the starting conditions of a chaotic system. Very small changes in initial conditions can produce large changes in the weather, even in the short to medium term, thereby making long term forecasting extremely difficult. American mathematician and meteorologist Edward Lorenz tried to solve this problem during the World War II so that the weather could be more accurately predicted for air-force sorties. Working with a simple computer, he realized that repeated equations in which several decimal places were rounded up (e.g. 3.678658 becomes 3.67866) gave different results, leading to the conclusion that small changes in the initial conditions can lead to highly diverse outcomes. Lorenz predicted that if they had supercomputers (to handle all the decimal places) and means of obtaining accurate information, such as satellites, then we would be able to forecast the day-today weather much more accurately. The advent of supercomputers and satellites has enabled this prediction to come true, at least for short-term to medium-term forecasting. Lorenz himself devised a computer and a program to perform some of these calculations, which forms the basis of what is loosely known as chaos theory. It is important to appreciate that a lot of ‘chaos’ can result from simple re-iteration of simple equations. The Butterfly Effect is one of the most important properties of chaotic system. In chaos theory, the butterfly effect is the sensitive dependency on initial conditions in which a small change at one place in a deterministic nonlinear system can result in large differences in a later state. The name of the effect, coined by Edward Lorenz, is derived from the theoretical example of a hurricane’s formation being dependant on whether or not a distant butterfly had flapped its wings several weeks earlier.

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A BIT ON CHAOS THEORY In 1963, Lorenz developed a simplified mathematical model for atmospheric convection. The model is a system of three ordinary differential equations now known as the Lorenz equations:

Lorenz used the values, = 10, = 8/3 and =28 for which the system exhibit chaotic behaviour and when the solutions were plotted for this particular system, it resembled a butterfly.

Apart from meteorology, chaos theory has various applications in determining the planetary orbits in the solar system, the time evolution of the magnetic field of celestial bodies, the population growth in ecology, the dynamics of the action potentials in neurons, and the molecular vibrations. There is also some controversy over the existence of chaotic dynamics in economics and plate tectonics. Currently, Chaos theory is being applied to medical studies of epilepsy in the prediction of seemingly random seizures by observing initial conditions. Thus, going forward Chaos theory might just be the answer to new developments in modern science. Indeed, chaos isn’t a pit. It is one great ladder.

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The Human Element in mathematics

Sumedh Kaulgud & Suprabhat Bhagavathula

A student of arithmetic who has mastered the first four rules of his art, and successfully striven with money sums and fractions, finds himself confronted with an unbroken expanse of questions, or problems. These are short, adventurous stories with the endings omitted, and though bearing a strong family resemblance, are not devoid of a certain element of romance. The characters in the plot of a problem are three people called A, B, and C. The form of the question is generally the following: A, B, and C do a certain piece of work. A can do as much work in one hour as B in two, or C in four. Find how long they take to complete it. Or, A, B, and C are employed to dig a ditch. A can dig as much in one hour as B can dig in two, and B can dig twice as fast as C. Find how long they take to dig it. Or, A lays a wager that he can walk faster than B or C. A can walk half as fast again as B, and C is only an indifferent walker. Find how far they travel in x hours. The occupations of A, B, and C are many and varied. In old arithmetic, they would simply be doing “a certain piece of work.” It has become the fashion, over the years, to define the job more clearly and to set them at walking matches, ditch-digging, regattas, and piling cord wood. They revel in motion. When they get tired of walking-matches, A rides on horseback, or borrows a bicycle and competes with his weaker-minded associates who are on foot. Or, they race on locomotives or row or become historical and use stagecoaches. If their occupation is actual work, they prefer to pump water into cisterns, two of which leak through holes at the bottom and one of which is water-tight. To a fervent follower of these men, one who has watched them in their leisure hours dallying with cord wood, and seen their panting sides heave in the full frenzy of filling a cistern with a leak in it, they are more than mere symbols. To him, they are creatures, of flesh and blood, each with their own personalities and temperaments. A is a full-blooded blustering fellow, energetic, hot-headed and strong-willed. He proposes everything, challenges B to work, makes the bets, and bends the others as per his will. He is a man of great physical strength and phenomenal endurance, having been known to walk forty-eight hours at a stretch and pump ninety-six. But his life is arduous and perilous, with even the slightest mistake in the working of a sum keeping him digging for a fortnight. An error in a decimal place may even kill him. Nonetheless, he always takes the best cistern, or digs the fastest. A always wins. B is a quiet, easy-going fellow, afraid of A and bullied by him, but very gentle and brotherly to little C, the weakling. Having lost all his money in bets, he lies at the mercy of powerful A. Poor C is an undersized, frail man. Constant walking, digging, and pumping has broken his health and ruined his nervous system. His joyless life has driven him to drink and smoke more than it’s good for him, and

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THE HUMAN ELEMENT IN MATHEMATICS his hand often shakes as he digs ditches. Hamlin Smith has said, “A can do more work in one hour than C in four.” I was first introduced to these three one evening after a regatta. They had all been rowing in it, and it had transpired that A could row as much in one hour as B could in two, or C in four. B and C had come in dead tired, and C was coughing badly. “Never mind, old fellow,” I had heard B say, “I’ll fix you up on the sofa and get you some hot tea.” Just then, A had come blustering in and shouted, “I say, you fellows, Hamlin Smith has shown me three cisterns in his garden and he says we can pump them until tomorrow night. I bet I can beat you both. Come on. You can pump in your rowing things, you know. Your cistern leaks a little, I think, C.” B had growled, saying that it was a dirty shame and that C was used up now, but they had gone, and I could tell from the sound of the water that A was pumping five times as fast as C. For years after that, I used to see them about town, always busy. I had never heard of any of them eating or sleeping. Then, owing to a long absence from home, I lost sight of them. On my return, I was surprised to no longer find A, B, and C toiling away. They had been replaced by N, M, and O, and some people had even employed foreigners called Alpha, Beta, Gamma, and Delta for the job. One day, I stumbled upon old D, in the little garden in front of his cottage, hoeing in the sun. D is an aged laboring man who used to be occasionally called in to help A, B, and C. “Did I know ‘em, sir?” he answered, “Why, I knowed ‘em ever since they were little fellows in brackets. Master A, he were a fine lad, sir, though I always said, give me Master B for kind-heartedness-like. Many’s the job as we’ve been on together, sir, though I never did no racing nor aught of that, but just the plain labor, as you might say. I’m getting a bit too old and stiff for it nowadays, sir--just scratch about in the garden here and grow a bit of a logarithm, or raise a common denominator or two. But Mr. Euclid he use me still for them propositions, he do.” From the garrulous old man, I learned about the melancholy end of my former acquaintances. Soon after I left town, he told me, C had been taken ill. It seems that A and B had been rowing on the river for a wager, and C had been running on the bank and then had sat in a draught. Of course the bank had refused the draught and C was taken ill. A and B had come home to find C lying helpless in bed. A shook him roughly and said, “Get up C, we’re going to pile wood.” C looked so worn and pitiful that B said, “Look here, A, I won’t stand this. He isn’t fit to pile wood tonight.” C smiled feebly and said, “Perhaps I might pile a little if I sat up in bed.” Then B, thoroughly alarmed, said, “Look here, A, I’m going to fetch a doctor. He’s dying.” A flared up and answered, “You’ve no money to fetch a doctor.” “I’ll reduce him to his lowest terms,” B said firmly, “that’ll fetch him.” C’s life might even then have been saved had they not made a mistake about the medicine. It stood at the head of the bed on a bracket, and the nurse accidentally removed it from the bracket without changing the sign. After the fatal blunder, C sunk rapidly. I think that even A was affected, as he stood with bowed head, aimlessly offering to bet with the doctor on C’s labored breathing. “A,” whispered C, “I think I’m going fast.” “How fast do you think you’ll go, old man?” murmured A. “I don’t know,” said C, “but I’m going, at any rate.” The end came soon after that. C rallied for a moment and asked for a certain piece of work that he had left downstairs. A put it in his arms and he expired. As his soul sped heavenward, A watched its flight with melancholic admiration. B burst into a passionate flood of tears and sobbed, “Put away his little cistern and the rowing clothes he used to wear, I don’t think I can ever dig again.” The funeral was plain and unostentatious. It differed in nothing from the ordinary, except that out of deference to sporting men and mathematicians, A engaged two hearses. Both vehicles started at the same time, with B driving the one which bore the sable parallelopiped containing the last remains of his ill-fated friend.

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THE HUMAN ELEMENT IN MATHEMATICS A, on the box of the empty hearse, generously consented to a handicap of a hundred yards, but arrived first at the cemetery by driving four times as fast as B. As the sarcophagus was lowered, the grave was surrounded by the broken figures of the first book of Euclid. A and B then set about to calculate the distance to the cemetery. After the death of C, it was said that A became a changed man. He lost interest in racing with B, and even dug languidly. He finally gave up his work and settled down to live on the interest of his bets. B never recovered from the shock of C’s death - his grief preyed upon his intellect and he became deranged. He grew moody and spoke only in monosyllables. His disease became rapidly aggravated, and he presently spoke only in words whose spelling was regular and which presented no difficulty to the beginner. Realizing his precarious condition, he voluntarily submitted to be incarcerated in an asylum, where he abjured mathematics and devoted himself to writing the ‘History of the Swiss Family Robinson’ in words of one syllable.

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Hard for computers

Harman Singh

If you’ve ever heard a computer science student claim that a given problem is “NP-complete” and seen fit to ask said student what the term means, you’ve probably received an answer along the lines of “just really really difficult.” That student may have been me. But I’ve read up a bit in preparation for writing this article, and I can now tell you that that isn’t strictly correct. To be precise, what computer scientists and mathematicians term ‘hard’ isn’t actually hard in the sense that the layman understands it. It refers in fact, not to the predicament of a human tasked with solving the problem, but to that of a computer. Computers are simple-minded creatures - an obvious fact given that they don’t really have minds to speak of. In reality, computers are capable of performing exactly the same tasks that human are, just unimaginably faster. So for a computer, ‘hardness’ of a problem isn’t measured by how elusive the solution is, say, in terms of creativity. Difficulty is instead measured by how long it takes to solve the problem. Some terminology before we go further. Complexity of any mathematics is traditionally represented using something called the Big-O notation. For instance, one might say that a certain problem has O(n) complexity (read as ‘order of n’), where n refers to the size of the input to a problem. Let’s get through a formal definition that I promise we’ll never look at again once we’re past it. Let f(x)and g(x) be two functions defined for some subset of real numbers. Then we can say that f(x) = O(g(x)) as x -> ∞ if and only if there exists some positive real number c and some real number x0 such that |f(x)|≤c|g(x)| for all x>x0. In plain English, f(x) is of the order of g(x) if beyond a certain point, g(x) starts to grow faster than f(x). Quick example: Say you have an array of length n, and I ask you to find the largest element in the array. Oh, and you’re a computer. The way for a computer to do this is to check each element of the array, one by one, and keep track of the largest one who it’s encountered so far. So every time it sees a new element, the computer compares it to the largest element so far, and if it’s bigger, then replace it as the new largest element. Clearly, the computer here performs a constant set of operations for each element. Let’s say that set of operations takes some time c to complete. Doing these operations for each element of the array would take a total time equal to c x n, where c is a constant, and n is a variable, specifically the size of the input. This is an example of a problem with complexity O(n), which we now understand to mean that the time taken to solve the problem is a linear function of the size of the input. As problems go, this isn’t so bad for a computer. Quicker example: Your new task is to sort the elements of an array. You’re going to use the most straightforward way of doing this, which is something called selection sort. A simple recursive way of explaining how to sort an array using selection sort is as follows: (i) find the smallest element of the array, (ii) place it at the start of the array, and (iii) perform selection sort on the remaining n-1 elements of the array. Keep doing this till there’s only one element left, which you can then leave where it is. Thus you’re essentially performing an O(n) complexity operation on each suffix subarray of the array. There are n such subarray, so the total time taken increases by a factor of n. The complexity for the whole sorting procedure is therefore O(n2).

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HARD FOR COMPUTERS A sharp reader might observe that we’re not actually performing n2 operations, since we’re only dealing with n elements in the first iteration. So the term should really be more along the lines of n + (n-1) + (n-2) +...+2+1 = (n2 + n)/2. But that’s where the definition of Big-O comes in. Because we’re only concerned about the relation between f(x) and g(x) when x tends to infinity, we can afford to ignore the smaller powered terms. So O(n2+n) is actually exactly the same as O(n2). It is for the same reason that we can let go of the constant term completely, because the constant in the definition of Big-O makes up for it. Both the algorithms mentioned here are examples of algorithms having polynomial complexity. They’re of the order O(nk) where k is some real number (O(1) is simply k=0). You can also have algorithms with logarithmic complexity, such as that of finding the location of an element in a sorted array. Here, we can keep dividing the sorted array into two halves and only searching the half that contains our element. This would take a maximum of log2n divisions, and thus the binary search algorithm has O(log n) complexity. Polynomial complexity isn’t much of a challenge for the machine, since an arbitrarily large input can still be dealt with in what is considered ‘reasonable’ time. The formal term for these problems, which can be handled by a computer, irrespective of the size of the input, is tractable. So what is intractable? Well, O(n!) and O(cn) are. Algorithms that have these complexities are those that have a solving time which grows so ridiculously fast with an increase in the size of the input, that even moderately sized versions of these problems can easily take billions and trillions of years to solve, using any computing power known to man today. Yes, trillions. We’ll get back to that. A simple example of an O(cn) algorithm is your ordinary brute force method of hacking a password-protected account. If there are c possible characters one can use in a password, and the password is a word of length n, then a maximum of cn attempts at guessing the password will get you in. That’s O(cn). A famous example of an O(n!) algorithm is the straightforward way of solving the infamous Travelling Salesman problem (TSP). The problem is: given a list of cities and their pair-wise distances, find the shortest possible route that visits each city exactly once and then returns to the origin city. The naive way of solving this would involve listing all n! permutations of the cities, finding out the total distance travelled in each of them, and returning the minimum. Let’s say we have a list of 30 cities. Just 30. And I have a Sony Vaio. The laptop is capable of evaluating a single city-cycle in one CPU cycle, and happens to be able to do 2.5 billion cycles every second. However, 30 cities means 30! = 265*1030 cycles. So my computer will still take 106*1021 seconds to do it. That’s 33*1014 years, or 3300 trillion years. See? Trillions. And you thought your Python script was slow. And that isn’t because my Vaio is lame. The world’s fastest supercomputer right now is China’s Tianhe-2, which can do 33.86 quadrillion cycles per second. That’s a pretty major improvement, and the Tianhe can therefore solve the 30 city problem in just under 25 million years.

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HARD FOR COMPUTERS

We’ve now got a pretty good idea of what is and isn’t good for the computer. The class of problems that can be solved in polynomial time, and are therefore a cakewalk for any computer, is termed simply P. The other significant class of problems we need to know about in NP, or non-deterministic polynomial. These are problems that may not be solvable in polynomial time themselves, but if a solution were to be proposed for one, then it could be verified in polynomial time. This is a fairly important point. Consider Sudoku. I need to fill every one of 81 boxes with a number from 1-9 such that no column, row, or 3x3 square has a repeated number. This is equivalent to a TSP involving 81 vertices. Now I can’t ask my computer to solve an 81-city TSP, because we’ve seen what tends to happen to my computer if I do that. But if someone were to propose a solution to a Sudoku puzzle, then verifying whether the solution is correct is simple enough. It would involve checking 9 rows, 9 columns, and 9 boxes for a repeated number. All of these are O(n) operations, and therefore easily doable. So a Sudoku puzzle can be verified quickly, but can’t be solved easily, and is therefore, an NP problem. NP problems have the property of being easily verifiable, but they may or may not be solvable in polynomial time. This means all problems in P are also in NP. Many a times, a problem A can be solved by first reducing it to a problem B, and then solving B instead. NP-complete is a special class of NP problems that have a handy property: every problem in NP can be easily (meaning in polynomial time) transformed into any NP-complete problem. Informally, this is saying that any NP-complete problem is an NP problem that is at least as tough as any other problem in NP. So if you actually solve a single NP-complete problem, you’ve effectively solved every other problem in NP as well. Neat. NP-complete is the class of problems that computers struggle to solve, and these include real world problems like vertex cover and graph colouring, and also problems like Sudoku and Minesweeper. Fun fact: A man named G. Aloupis took the time to mathematically prove that Super Mario Bros. is NP-complete. Finally, there is the class NP-hard. This is a class of problems that is, again, at least as hard as the problems in NP, which is to say that any problem in NP can be reduced to any problem in NP-hard. So how is this different from NP-complete? Well, all NP-complete problems are necessarily NP problems too, and are therefore at least verifiable in polynomial time. NP-hard problems have no such obligation, and may therefore be neither solvable nor verifiable in polynomial time. I’m tempted now to place an Euler diagram here that quickly summarizes everything you’ve just read, but I can’t, because of the Question. The Question is the P vs NP problem, and is one of the major unsolved problems in mathematics and computer science. It is considered by many to be the most significant problem in the field, and is one of the seven Millennium Prize Problems selected by the Clay Mathematics Institute to carry a million dollar prize for the first correct solution. The P vs NP problem simply asks this: Is P=NP? That is to say, is every problem that is verifiable in polynomial time (NP), also solvable in polynomial time (P)? The proof of this would revolutionize computer science because people would be able to construct faster algorithms for a lot of important problems, most of which have extremely significant real life applications, not least of which is TSP. All the world’s scientists are yet to put forward a proof either for or against the statement, but the solution itself is maddeningly close. The reason for this is because of what I’ve just explained. Because every problem in NP is reducible to any problem in NP-complete, if someone were to present a polynomial time solution to even a single problem in NP-complete, this would be equivalent to presenting a polynomial time solution to every single problem in NP!

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HARD FOR COMPUTERS

The effects of proving that P=NP are breathtaking. It would mean that there is simply no problem that a computer cannot efficiently solve, and would significantly change life as we know it. It would mean that a computer would be capable of finding a formal proof for any theorem where the proof is of reasonable length, since such a proof is easily verifiable. This may well include all other six open problems listed by the Clay Institute. Essentially, a man who solved this particular problem would be entitled to not one, but seven million dollars. However, most scientists expect that P ≠ NP. A proof for this would be significantly less exciting, but important nonetheless. It would allow one to show, formally, that some problems cannot be solved in a reasonable amount of time, so that the attention of researchers can be focused on partial solutions, or on other problems. Due to widespread belief in P ≠ NP, much of this focusing of research has already taken place. On a more philosophical front, MIT professor Scott Aaronson best explains why the world believes the P ≠ NP: “If P = NP, then the world would be a profoundly different place than we usually assume it to be. There would be no special value in ‘creative leaps,’ no fundamental gap between solving a problem and recognizing the solution once it’s found. Everyone who could appreciate a symphony would be Mozart; everyone who could follow a step-by-step argument would be Gauss...” And here’s that Euler Diagram I promised you.

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ARTICLE

mathematics in origami

Rutuja Surve

Origami is the Japanese art of paper folding. In traditional origami, constructions are done using a single sheet of coloured paper. In modular origami, a number of individual “units,” each folded from a single sheet of paper, are combined to form a compound structure. Origami is an extremely rich art form, and constructions for thousands of objects, from dragons to buildings to vegetables have been devised. Many mathematical shapes can also be constructed, especially using modular origami. So how exactly are math and origami related to each other? The connection with geometry is clear and yet multifaceted. A folded model is both a piece of art and a geometric figure. Just unfold it and take a look! You will see a complex geometric pattern, even if the model you folded was a simple one. A beginner student of geometry might want to figure out the types of triangles on the paper. What are the angles that can be seen? What are the shapes that are notable? How did those angles and shapes get there? An advanced geometry student or teacher might want to investigate more in depth the relationships between math and origami. For instance, the traditional crane unfolded provides a crease pattern from which we can learn a lot. Pick a point (vertex) on the crease pattern. How many creases originate at this vertex? Is it possible for a flat origami model to have an odd number of creases coming out of a vertex on it’s crease pattern? How about the relationship between mountain and valley folds? Can you have a vertex with only valley folds or only mountain folds? How about the angles around this point? The answers to these questions lie in the following theorem formulated by Japanese mathematician, Toshikazu Kawasaki, and is called the Kawasaki’s Theorem The Theorem This theorem gives the criterion for an origami construction to be flat. It states that a given crease pattern can be folded to a flat origami if, and only if all the sequences of angles , ..., surrounding each (interior) vertex fulfil the following condition For example, around the central vertex in the figure below, 1 + 3 + 5 = 2 + 4 + 6 = 180. (Note that the number of angles is always even; each of them corresponds to a layer of the folded sheet)

The rule evidently applies to the case of a rectangular sheet of paper folded twice, where the crease pattern is formed by the bisectors. But there are many more interesting examples where the above property can be checked . The mathematician Humiaki Huzita developed six axioms (and later a seventh) based on origami construction. What’s especially great about all this is that these axioms are not just theoretical - they have been put

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ARTICLE

MATHEMATICS IN ORIGAMI into real use. Origami and Topology The study of origami and mathematics can be classified as topology, although some feel that it is more closely aligned with graph theory. Here’s an origami theorem which can be seen from both points of view. The theorem states: “Every flat-foldable pattern is 2-colourable”. In other words, suppose you have folded an origami model which lies flat. If you completely unfold the model, the crease pattern that you will see has a special property. You require only two colours to colour the whole region such that no two bordering crease pattern regions have the same colour. In Graph Theory The above statement may remind you of the famous map-maker’s problem - what is the least number of colours you need to colour countries on a map so that two neighbouring countries aren’t the same colour? The answer is four, and this is known as the Four Colour Theorem. But back to our theorem. Can you see that you need only two colours to colour a crease pattern? Try it yourself. You will see that anything you fold (as long as it lies flat) will need only two colours to colour in the regions on its crease pattern. Here’s an easy way to see it: fold something that lies flat. Now colour all of the regions facing towards you red and the ones facing the table blue (remember to only colour one side of the paper). When you unfold, you will see that you have a proper 2-colouring. A more rigorous proof goes as follows: First, show that each vertex in the crease pattern has an even degree. Then you know the crease pattern is an Eulerian graph, that is, a graph containing a path which starts and ends at the same point and travels along every edge. Finally, it is well known that Eulerian graphs are 2-colorable. This clearly establishes the result as a part of Graph Theory. In Topology Before we look into the result from a topological point of view, let us understand the basics of topology. Topology developed as a field of study out of geometry and set theory, and it deals with properties like connectedness and continuity in space that are preserved under continuous deformations including stretching and bending, but not tearing or gluing. It is sometimes referred to as “rubber sheet geometry”, meaning that in topology, changing the shape of an object will not affect it (as long as you do not create any holes or patch up any holes). In fact, to a topologist, a coffee cup and a doughnut are the same, while a geometer sees them as completely different. This is because a sufficiently pliable doughnut can be reshaped into a coffee cup by creating a dimple and progressively enlarging it, while shrinking the hole into a handle. How is the above result a topological one? Well, the 2-colouring gives us an easy way of determining the orientation of each region we colour in. All regions coloured blue will be facing up (or down) while all regions coloured red will be facing the opposite way. Try it. This time, fold a model, unfold it, and colour the crease pattern regions with red and blue. Now, refold the model and see for yourself. (For those of you who are inquisitive about math in origami, watch this documentary: ‘ Independent Lens: Between the Folds’)

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ARTICLE

Conjecture Collatz conjecture

Janvi Palan

“Stating the obvious” is a phrase used by most rational thinking students to express their annoyance at a remark, the logic and reason behind which is so distastefully obvious that its mere stating merits an expression of annoyance. Quite like the preceding sentence, Mathematics necessitates the establishment of explanations for statements that the average layman takes for granted. Quite unlike the first statement, however, fields in Mathematics, abreast with paradoxes and infinities, often venture into undefined territories, and hence reiteration and a concrete establishment of EVERY blatantly obvious statement on EVERY domain becomes a must, or else it is doomed for-(almost)-ever to serve as a Conjecture. Conjectures are statements which are believed (and not assumed) to be true, but lack a definite proof that spans all applicable domains. Although all conjectures remain proof-less as long as they remain conjectures, they are usually based on educated guesses and approximations based on known information. Mathematicians tinker (they call it Recreational Mathematics), and their tinkering raises conundrums, one example of which is the Collatz Conjecture. Various other conjectures have been posed as open questions related to the fields of Number theory and Graph theory, such as the abc conjecture, the Goldbach conjecture and the lonely runner conjecture. Their unsolved nature, after all these years, have led to the coinage of the term “mathematical diseases”, for their level of notoriety have led to many mathematicians spending years on them only to realize that the solution is beyond the scope of the subject as of their day and age. The Collatz Conjecture Consider any arbitrary positive integer n. • If n is even, divide it by two. • If n is odd, triple it and add one. Or,

Form a sequence of the values of this function for any initial positive integer, taking the value of f as the input for the next iteration. The Conjecture states that this process will eventually reach the number 1, regardless of which positive integer is chosen initially. It was posed by German Mathematician Lothar Collatz in 1937. Take, for instance n = 5. This gives rise to the sequence: 5 16 8 4 2 1 First Glance The longest progression for any initial starting number less than 100 million is 63,728,127, which has 949 steps. For starting numbers less than 1 billion it is 670,617,279, with 986 steps, and for numbers less than 10 billion it is 9,780,657,630, with 1132 steps. It should be noted that the moment we encounter a number that is a power of 2, it converges to one rapidly as it is halved n times. If the conjecture is false, it can only be because there is some starting number which gives rise to a sequence which does not contain 1. Such a sequence might enter a repeating cycle that excludes 1, or increase without bound. No such sequence has been found.

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CONJECTURE. COLLATZ CONJECTURE.

Why do we believe it’s true? The conjecture is believed to be true mainly due to the probabilistic heuristic justification that can be put forward. Given the above defined function, if n is odd, then

Replace

by

: N --> N

if n is even if n is odd Now, we see that if one chooses n “at random”, it is odd with probability 1/2 and even with probability 1/2, and so increases n by a factor of roughly 3/2 half the time, and decreases it by a factor of 3/2 in the other half. Furthermore, if n is uniformly distributed modulo 4, one easily verifies that is uniformly distributed modulo 2, and so should be roughly 3/2 times as large as half the time, and roughly 1/2 times as large as the other half of the time. Continuing this, we expect generically that half the time, and the other half of the time. The logarithm of this orbit can then be modeled with steps log(3/2) and log(1/2) occurring with equal probability (=1/2). The expectation is given by:

E is negative, and so we expect the orbit to decrease over a very long period of time, and ultimately result in 1. This is why the Collatz Conjecture ‘makes sense’ heuristically. Current Status The conjecture has even been checked by computers for all starting values up to 5 × 260 ≈ 5.764×1018, and has proven to be correct for all tested values. However, this sequential examination of all numbers is not a finite process, and hence, such an approach can never demonstrate that the conjecture is true, merely rule out the possibly of any known counter-example. The evasiveness of the proof of this statement, despite its simplicity in understanding, led to the prolific mathematician, Paul Erdős (of Erdős number fame) offering a prize of $500 to anyone who could come up with a solution for it. He is alleged to have said, “Mathematics is not yet ready for such problems.” Despite various attempts at arriving at a solution, mathematicians have still not been able to solve this elusive problem. The concrete importance of this conjecture remains an enigma, but it is believed that for the solving of such problems genuinely new mathematical ideas will have to be created, and it is these ideas that may help in the solution of other domains where truly important problems are at stake.

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ProOF by contradiction or not

Bhavul Gauri

The journey probably started when we first heard of irrational numbers. Proving the irrationality of √2 seemed so elegant after employing the Reductio ad Impossibilem technique, also fondly known as ‘Proof by Contradiction’. A proof by contradiction, as the name suggests, is a form of proof that establishes the validity of a proposition by showing that the case where the proposition is false would lead to a contradiction. Personally, I’m not too fond of proving by contradiction, because to me, it seems like an easy way out - if you can’t prove it directly, just prove that the alternative just isn’t feasible. Curious about these types of proofs and their necessity, I tried to figure out if every proof by contradiction could also be shown without contradiction. To do this, let us first formalize the notion of a ‘proof’ using symbolic logic. Let represent a proposition that is never true. Let A be any statement. Then the negation of A is equivalent to A . We use the latter as the definition of ‘negation of A’, so that we have it in terms of . There are two principles that make up any proof by contradiction: 1. Principle of explosion : For any statement A, we can take “ implies A” as an axiom. 2. Law of excluded middle : For any statement A, we can take “A or ¬A” as an axiom. Now, there are three logic systems in proof theory: 1. Minimal Logic : It contains neither of the principles above, but only basic proof rules to manipulate logical connectives (other than negation) and quantifiers ( and ). This system avoids negation and hence comes more close to “direct proofs”. 2. Intuitionistic logic : It is a superset of Minimal logic, additionally containing the Principle of Explosion. 3. Classical logic : This contains Intuitionistic logic, as well as the law of Excluded Middle. Now, if we try to prove a formula of the form x¬A(x) B(x), where B(x) doesn’t have as a subformula, we cannot prove it using just minimal logic. However, it can be proved with the help of Principle of Explosion. Hence, there exist statements that are provable in intuitionistic logic but not in minimal logic. In Intuitionistic logic, if a statement implies a contradiction then the negation of the statement is true, but in classical logic we also have that if the negation of a statement implies a contradiction, then the original statement is true, and the latter is not provable in intuitionistic logic, and in particular is not provable directly. So, it is known that there are statements that are provable in classical logic but not in intuitionistic logic. In this sense, the Principle of explosion allows us to prove things that are not provable without it, and the Law of excluded middle allows us to prove things that we could not even prove with the principle of explosion. So, there are statements that are provable by contradiction but cannot be proved directly. Hence, if we were to assume that every proof by contradiction can also be shown without contradiction, we would certainly arrive at some contradiction.

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ARTICLE

Fermat's last theorem false for matrices

Dr. Jajati K Sahoo

1. Introduction Fermat’s last theorem is a well-known theorem in number theory. It states that the Dio- phantine equation xn + yn = zn (1) has no non trivial solution if n ≥ 3, i.e. it is impossible to find three non-zero integers such that (1) holds. This famous statement was finally proved by Andrew Wiles in 1995. Most doubt whether Fermat really had a proof, but it is an intriguing part of the history of this famous problem. If Fermat had a solution, he certainly did not have the brilliant one which Wiles found. In this article, I would like to shed some light about some results that generalize Fermat’s Little Theorem to matrices in different sense. They do not seem to be well known, but are quite pretty. Perhaps they may have applications to some complexity theory problems. Without discussing any more on number theory, we will straight forward go to the theory of matrices with an aim to discuss the parallel statement of Fermat’s last theorem. The parallel statement of Fermat’s last theorem says that there exist three non-zero matrices A, B, C having integer entries such that An + Bn = Cn (2) For all n ≥ 3. In fact, this will hold for all n ≥ 1. This means Fermat’s last theorem for matrices is false. The structure of this article is as follows. In Section 2, we will discuss the above problem in two different ways, see [1]. One way is to begin with an idempotent matrix A, that is, the matrix A satisfies A2 = A. Another way is to begin with an n × n pseudo magic square A, i.e. a matrix whose row sums and column sums are all same. In Section 3, we will discuss some examples corresponding to the theorem. We will also discuss another construction of matrices, which will easily verify that there are infinite number of matrices for which (2) holds. 2. Main Theorems In this section, we will straight forward go to the construction of matrices satisfying (2) and describe in two different theorems. Theorem 2.1: There exist matrices A and B such that An + Bn = (A + B)n For all integer n ≥ 1. Proof 2.2: Let A be an idempotent matrix, i.e. A2 = A. Then clearly An = A. Take B = I − A. Thus B2 = (I − A)(I − A) = I − 2A + A2 = I − A = B and so Bn = I − A = B. Therefore we get An + Bn = A + I − A = I = In = (A + B)n This shows that there exist matrices A, B, C which satisfies (2) with C = A + B.

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ARTICLE

FERMAT’S LAST THEOREM FALSE FOR MATRICES Before entering to next theorem, we recall the definition of pseudo magic square and required properties. A pseudo magic square is a square matrix whose row sums and column sums are all identical. Readers should not confuse with magic square as no requirement is made in our discussion about the diagonals. It is not difficult to prove the following property. Lemma 2.3: A matrix A is a pseudo magic square with common row and column sums equal to d if and only if AJ = J A = dJ , where J is the matrix whose all entries are 1. Theorem 2.4: There exists n × n matrices A, B and C such that An = Bn + Cn For any integer n ≥ 1. Proof 2.5: Let A be a n × n pseudo magic square having common row and column sums equal to d. Define

and

Now we see that BC = By Lemma 2.3, we have J A = dJ . Since J2 = nJ , we obtain BC = A similar computation yields that C B = 0. We see that A = B + C and hence A2 = (B + C )2 = B2 + BC + C B + C2 = B2 + C2 . Also we have A3 = (B2 + C2)(B + C ) = B3 + C (C B) + B(BC ) + C3 = B3 + C3. In a similar fashion, we see that An = Bn + Cn For all n ≥ 1. Hence proved.

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FERMAT’S LAST THEOREM FALSE FOR MATRICES Examples: Here, we give one example (without detail) for each theorem mentioned in previous section. Example 3.1: One can easily verify Theorem 2.1 with the matrix

Example 3.2: One can also easily verify Theorem 2.4 for the matrix

We will end our discussion with additional examples of n x n matrices (n ≥ 3) satisfying (2). Let us first look at the 3 x 3 matrices:

A simple computation gives,



which shows A3 + B3 = C3.

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ARTICLE

FERMAT’S LAST THEOREM FALSE FOR MATRICES

In general, for any integer a, given an n x n matrix such as



A=

We see that An would produce a resulting scalar identity matrix of the form (we skip the Proof):

An =

In a similar way we conclude that for any n × n matrices A, B and C of the form (4), An, Bn and Cn will produce scalar identity matrices. Finally, these ideas of the construction of matrices show that there are infinite number of solutions to the formula (2). This proves that Fermat’s last theorem is false for the matrix of the above form as well.

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MATHEMATICIAN’S VIEW

Paradoxes and Fallacies 1. The fallacy of the isosceles triangle purports to show that every triangle is isosceles, meaning that two sides of the triangle are congruent. This fallacy has been attributed to Lewis Carroll. Given a triangle ABC, prove that AB = AC: i. Draw a line bisecting A ii. Call the midpoint of line segment BC, D iii. Draw the perpendicular bisector of segment BC, which contains D iv. If these two lines are parallel, AB = AC; otherwise they intersect at point O v. Draw line OR perpendicular to AB, line OQ perpendicular to AC vi. Draw lines OB and OC vii. By ASA Test of Congruence, RAO QAO (AO = AO; OAQ OAR since AO bisects A; ARO AQO are both right angles) viii. By SAS Test of Congruence, ODB ODC ( ODB, ODC are right angles; OD = OD; BD = CD because OD bisects BC) ix. By HL, ROB QOC (RO = QO since RAO QAO; BO = CO since ODB ODC; ORB and OQC are right angles) x. Thus, AR AQ, RB QC, and AB = AR + RB = AQ + QC = AC Catch The error in the proof is the assumption in the diagram that the point O is inside the triangle. In fact, whenever AB ≠ AC, O lies outside the triangle. Furthermore, it can be shown that, if AB is longer than AC, then R will lie within AB, while Q will lie outside of AC (and vice versa). (Any diagram drawn with sufficiently accurate instruments will verify the above two facts.) Because of this, AB is still AR + RB, but AC is actually AQ − QC; and thus the lengths are not necessarily the same.

2. The sum of series: 1 + 2 + 3 + 4+ … = -1/12 We know that the sum of all natural numbers 1 + 2 + 3 + 4 + • • • is a divergent series. The nth partial sum of the series is the triangular number

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PARADOXES AND FALLACIES Which increases without bound as n goes to infinity. Because the sequence of partial sums fails to converge to a finite limit, the series is divergent, and it does not have a sum in the usual sense of the word. Srinivasa Ramanujan presented two derivations of “1 + 2 + 3 + 4 + … = −1/12” in chapter 8 of his first notebook. He transformed the series 1 + 2 + 3 + 4 + … into 1 − 2 + 3 − 4 + …, one can subtract 4 from the second term, 8 from the fourth term, 12 from the sixth term, and so on. The total amount to be subtracted is 4 + 8 + 12 + 16 + · · ·, which is 4 times the original series. Let c = 1 + 2 + 3 + 4 + … 4c = 4+ 8+ 12 + … -3c = 1 - 2 + 3 - 4 + 5 … Consider f(x) = 1/(1+x) 2 = 1 - 2x + 3x2 - 4x3 … f(x = -1) = 1 - 2 + 3 - 4 + 5 … = 1/(1+1) 2 = 1/4 Therefore, -3c = 1/4 Which implies c = -1/12. Catch The catch lies in the step 4c = 0 + 4 + 0 + 8 + 0 + … Addition of such arbitrary zeroes in diverging sums is not stable. In the following step, when we subtract 4c from c, we treat infinite sums as finite sums. Doing the same leads to highly inconsistent results. The level of instability that is attained when we add arbitrary zeroes to diverging sums and then perform arithmetic operations on them can be illustrated as below: Let 1 + 2 + 3 + ...=x then adding 0 to both sides gives => 0 +1 + 2 + ... = 0 + x = x Subtracting gives => 1 + 1 + 1 + ... = x – x = 0 Adding 0 to both sides again gives => 0 + 1 + 1 + 1 +... = 0 Subtracting the last two series gives => 1 + 0 + 0 + ... = 0 This is a contradiction. 3. In the paradox of Achilles and the Tortoise, Achilles is in a footrace with the tortoise. Achilles allows the tortoise a head start of 100 metres, for example. If we suppose that each racer starts running at some constant speed (one very fast and one very slow), then after some finite time, Achilles will have run 100 metres, bringing him to the tortoise’s starting point. During this time, the tortoise has run a much shorter distance, say, 10 metres. It will then take Achilles some further time to run that distance, by which time the tortoise will have advanced farther; and then more time still to reach this third point, while the tortoise moves ahead. Thus, whenever Achilles reaches somewhere the tortoise has been, he still has farther to go. Therefore, because there are an infinite number of points Achilles must reach where the tortoise has already been, thus one cannot exactly determine the time when he overtakes the tortoise. 4. 1 = √1 = √(-1 x -1) = √-1 x √-1 = i x i = -1 Thus, 1 = -1. Catch √(axb) = √(a)x √(b) only if a,b>0

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JOURNAL

This year in math The Department of Mathematics of BITS-Pilani, K.K.Birla Goa Campus has conducted a wide foray of events in the last year. These include guest lectures, workshops, projects, quizzes and the upcoming MathFest. Lectures and workshops The department conducted various talks by respected Mathematicians during the academic year 2013-14. 1. December 6, 2013: Dr.Goutam Paul, Assistant Professor, R.C. Bose Centre for Cryptology and Security, Indian Statistical Institute, Kolkata delivered a talk on “No Cloning Theorem and Quantum Key Distribution”. 2. October 11, 2013: Prof. A. S. Vasudeva Murthy, TIFR Centre for Applicable Mathematics, Bangalore, delivered talks on “Searching and Sorting Million Needles in Zillion Haystacks” and “On the string equation of Narasimha”. 3. October 10, 2013: Prof. A. S. Vasudeva Murthy, TIFR Centre for Applicable Mathematics, Bangalore, delivered a talk on “Far field boundary conditions and their numerical approximation” 4. October 3, 2013: Prof.Mridul Nandi, Applied Statistics Unit, ISI Kolkata delivered a lecture on “Analysis of Modes of Hash Functions” 5. September 14, 2013: Prof.Sudhir Ghorpade, Professor & Head, Department of Mathematics, Indian Institute of Technology Bombay delivered a talk on “Matrices and Linear Recurrences over Finite Fields” 6. September 14, 2013: Prof. M. K. Kadalbajoo, Professor, Department of Mathematics, Indian Institute of Technology Kanpur delivered a talk on “Computational Mathematics-Some Challenges” An international workshop on Mathematical Foundation of Advanced Finite Element Methods (MFAFEM-2013) was held at BITS, Pilani-K K Birla Goa Campus from 26th December to 3rd January 2014. The coordinators were Dr P. Dhanumjaya and Dr Anil Kumar. Sponsored research projects 1. Project Title: Mathematical modelling and simulation of crack in smart materials Principal Investigator: Dr. Amit Setia 2. Project Title: Portfolio Optimization and its Implementation in Financial Markets Principal Investigator: Dr MayankGoel 3. Project Title: Studies on neighbourhood magic graphs Principal Investigator: Dr. T. Singh 4. Project title: Multidimensional dark energy cosmological model in modified theories of gravitation Principal Investigator: Dr.G.Samanta 5. Project Title: Establishment of Scientific computation Laboratory Principal Coordinator: Dr.Dhanumjaya

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JOURNAL

THIS YEAR IN MATH Events Math Ecstasy - To tickle the mathematical bone we know all of us wannabe-engineers have, the department organized a Mathematics puzzles event called Math Ecstacy. Not only did the event see enthusiastic participation from all the batches, there were some very tempting cash prizes for the sharpest of the lot. Numb3rs - The Math department’s contribution to the Techno-managerial fest of the campus, Quark, is Numb3rs, the notorious math quiz which spans over three days. While the first round was based on comprehensions on basic math concepts, the second round involved solving the ken-ken puzzle, after which teams were to participate in a special relay round, where only one person could solve a question at a time while the other participant waited. The third round involved three long questions which basically tested participant’s level of relating a real-life problem to mathematics. Numb3rs 2014 saw enthusiastic participation from BITS as well as other colleges and was touted to be the ‘most participated event’ of the fest. Department day The Department of Mathematics is conducting its department day on the 23rd of March, as a part of Golden Jubilee Celebrations. On this day, Prof.Govindan Rangarajan will be honoured as the chief guest and main speaker for the day. Following the talk by the chief guest, the Mathematics association, or the ‘Mathletes’ will conduct the first ever ‘Mathfest’. Mathfest - This is a one day event consisting of various types of competitions which test participants’ I.Q, puzzle solving skills, algorithm-forming-and-coding skills. ‘Problem-creation’, a first-of-its-kind event, will see teams of participants designing questions based on the abstract of a mathematical concept given to them. Various other quizzes that test the mathematical grit of any participant will be held, followed by guest lectures and a cultural show. Mathematics Portal An exclusive Mathematics portal was set up by Bhavul Gauri for modular, better and easy discussions and interaction amongst the Mathletes. With an interface that one can easily get familiar with, the portal can be used to discuss notes on various Mathematics classes, have debates on the forums, as well as subscribe to updates about relevant areas. URL : http://www.bits-maths.org/

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PUZZLES

PUZZLE uP Puzzle One Imagine that you are on an island with 5 people. One of them is the truth-teller, and always speaks the truth. The other four toggle between truth and lie states, i.e. if they are initially in truth state, then they will tell the truth for that question and then switch to the other state and lie for the next question. The initial states of these four people are not known. The task is to find out the truth-teller by asking at the most two questions. What should the questions be and who should they be asked to? Puzzle Two Find 10 distinct coordinates of vertices (xi , yi) in R x R such that they satisfy the following: 1. 0 ≤ xi ≤ 3 2. 0 ≤ yi ≤ 3 3. Euclidean distance between any pair of vertices should be greater than the square root of two. Puzzle Three Galaxy Puzzle: In Japan, Galaxies are known as Tentai Show. They are also known as Spiral Galaxies, and Sym-a-pix. In these puzzles, you fill in the horizontal and vertical edges to form a galaxy shaped island around each circle, which represents the galactic center. The galaxy shapes must be rotationally (or 180°) symmetric. Each galaxy puzzle has a unique solution. Try this one.

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PUZZLES

PUZZLE UP

Solutions 1. As we initially don’t know who the toggling people are, it doesn’t matter who we ask the questions to. We will ask the following two questions, and to the same person. Question 1: Are you the truth-teller? If the answer we get is Yes, then there are 2 possibilities: either the person is the truth-teller that we are looking for or he is the toggling chap in lie state. In both cases, he will tell the truth for the next question. Question 2: Who is the truth teller? Obviously, if the answer to this question turns out to be ‘Yes’, we have found the truth-teller! Now, we look at the case where the answer to our first question is No. This definitely means that the person is toggling in the truth state. So, he will lie for the next question. Question 2: Who is not the truth teller? 2. It is impossible to find 10 distinct vertices. Proof : This is a simple application of Pigeonhole Principle. We divide the 3x3 square into 9 distinct 1x1 squares constructed by the lines x = 0 , x = 1 , x = 2 , x = 3 , y = 0 , y = 1 , y = 2 , y = 3. Now since there are 10 vertices and 9 squares, by the Pigeonhole principle, at least 1 square must have more than 2 points in it. Now, what can be the maximum distance between these two points? It is the length of the diagonal of that square i.e. square root of 2. Contradiction. Tada. 3.

Courtesy puzzle 1 & 2 : Krunal Patel

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Editorial Team Janvi Palan Rutuja Surve Siddhartha Govilkar Bhavul Gauri Rutuja Kshirsagar

Designing Bhavul Gauri

Download your own soft copy here http://www.scribd.com/Convergence2014

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Convergence 2014

BITS Pilani K K Birla Goa Campus