Applications Of The Pigeonhole Principle Mathematics Essay

Applications Of The pl sail through Principle Mathematics EssayWe begin our discussion with a common everyday embarrassing moment. look that in aces dresser drawer, he has socks of iii different colours (all placed in messy order). Having to get up early in the morning wee-arm it is lighten dark, how does he ensure that he gets a matching pair of same coloured socks in the most convenient way without disturbing his retainer? While, the adjudicate is simple He just has to take 4 socks from the drawer The answer behind this is of course, the Pigeonhole Principle which we leave al whiz be exploring in this math Project.What is the Pigeonhole Principle then? Let me give you an example to illustrate this principle. For instance, in that respect atomic number 18 3 pigeonholes around. A pigeon is delivering 4 mails and has to place all its mails into the operable pigeonholes. With nevertheless 3 pigeonholes around, on that point bound to be 1 pigeonhole with at least 2 mail s.Thus, the general rule states when there be k pigeonholes and there be k+1 mails, then they forget be 1 pigeonhole with at least 2 mails. A more go on version of the principle will be the chase If mn + 1 pigeons are placed in n pigeonholes, then there will be at least one pigeonhole with m + 1 or more pigeons in it.The Pigeonhole Principle sounds trivial but its uses are deceiving awe-inspiring Thus, in our project, we aim to learn and explore more about the Pigeonhole Principle and illustrate its numerous interesting lotions in our daily life.We begin with the future(a) simple example2. Pigeonhole Principle and the Birthday problemWe exact always heard of people saying that in a large concourse of people, it is not difficult to find devil persons with their birthday on the same calendar month. For instance, 13 people are involved in a survey to understand the month of their birthday. As we all know, there are 12 months in a year, thus, even if the basic 12 people hav e their birthday from the month of January to the month of December, the 13th person has to have his birthday in any of the month of January to December as well. Thus, we are right to say that there are at least 2 people who have their birthday falling in the same month.In fact, we squirt view the problem as there are 12 pigeonholes (months of the year) with 13 pigeons (the 13 persons). Of course, by the Pigeonhole Principle, there will be at least one pigeonhole with 2 or more pigeonsHeres another example of the application of Pigeonhole Principle with peoples relationship3. Pigeonhole Principle and problems on relationsAssume that the relation to be present with is symmetric if Peter is present with Paul, then Paul is acquainted with Peter. envisage that there are 50 people in the room. Some of them are acquainted with to each one other, while some not. Then we can show that there are deuce persons in the room who have equal total of acquaintances.Lets relieve that there is one person in the room that has no acquaintance at all, then the others in the room will have either 1, 2, 3, 4, , 48 acquaintance, or do not have acquaintance at all. Therefore we have 49 pigeonholes numbered 0, 1, 2, 3, , 48 and we have to distribute between them 50 pigeons. So, there are at least both persons that have the same number of acquaintance with the others.Next, if everyone in the room has at least one acquaintance, we will still have 49 pigeonholes numbered 1, 2, 3, , 48, 49 and we have to be distribute between them 50 pigeonsAlso, we can apply the Pigeonhole Principle in the proving of numerical properties. The followers are two of such examples4. Pigeonhole Principle and divisibilityConsider the following random list of 12 numbers say, 2, 4, 6, 8, 11, 15, 23, 34, 55, 67, 78 and 83. Is it possible to choose two of them such that their difference is divisible by 11? Can we provide an answer to the problem by app hypocrisy the Pigeonhole Principle?There are 11 possibl e remainders when a number is divided by 110, 1, 2, 3, .., 9, 10.But we have 12 numbers. If we take the remainders for pigeonholes and the numbers for pigeons, then by the Pigeon-Hole Principle, there are at least two pigeons sharing the same hole, ie two numbers with the same remainder. The difference of these two numbers is thus divisible by 11In fact, in our example, there are several answers as the two numbers whose difference is divisible by 11 could be 4 15 34 67 or 6 83.5. Pigeonhole Principle and numerical propertyWe can also apply the Pigeonhole Principle in determining useful numerical properties. Consider a sequence of any 7 distinct real numbers. Is it possible to select two of them say x and y, which satisfy the inequality that 0 The problem sounds difficult as we whitethorn need to consider more advanced calculus and trigonometrical methods in the determination of the result. Well, to answer the above problem, one will be surprised to know that we just need a simple trigonometrical identity and apply the Pigeonhole Principle in the lead breaking to answer the problem, we first note that given any real number x, we can always find a real number a where n1 = tan a1, n2 = tan a2, .., n7 = tan a7Now, if we were to divide the interval (-p, p) into 6 equal intervals, we experience the following sub-intervals( -p, -p ), -p, -p ), -p, 0 ) , 0, p ), p,p ) and p, p ).For the 7 distinct numbers a1, a2, .a7, by the Pigeonhole Principle, there should be two determine say, ai and aj such that aI aj and ai aj are in the same interval For these two values ai and aj, we should have 0 We may recall an important trigonometrical identity B ) = .Thus, if ni = ai and nj = aj , then == tan ( ai aj )As 0 0 and so, 0 which is the result we are seekingWe may also apply the Pigeonhole Principle in the proving of useful daily geometrical results.. The following examples illustrate such usages6. Pigeonhole Principle and Geometrya. Dartboard applicationsAnother c ommon type of problem requiring the pigeonhole principle to solve are those which involve the dartboard. In such questions, a given number of darts are thrown onto a dartboard, the general shape and size of which are known. mathematical maximum distance between two certain darts is then to be determined. As with most questions involving the pigeonhole principle, the hardest part is to identify the pigeons and pigeonholes.Example 1Seven darts are thrown onto a circular dartboard of radius 10 units. Can we show that there will always be two darts which are at most 10 units apart?To grow that the final statement is always true, we first divide the circle into sesteter equal sectors as shownAllowing each sector to be a pigeonhole and each dart to be a pigeon, we have seven pigeons to go into sestet pigeonholes. By pigeonhole principle, there is at least one sector containing a minimum of two darts. Since the greatest distance between two points lying in a sector is 10 units, the st atement is proven to be true in any case.In fact, it is also possible to prove the scenario with only six darts. In such a case, the circle is this time divided into five sectors and all else follows. However, take note that this is not always true any longer with only five darts or less.Example 2Nineteen darts are thrown onto a dartboard which is shaped as a regular hexagon with side length of 1 unit. Can we prove that there are two darts within units of each other ?Again, we identify our pigeonholes by dividing the hexagon into six equilateral trilaterals as shown below.With the six triangles as our pigeonholes and the 19 darts as pigeons, we find that there must be at least one triangle with a minimum of 4 darts in it.Now, considering the best case scenario, we will have to try an equilateral triangle of side 1 unit with 4 points inside.If we try to put the points as far apart from each other as possible, we will end up assigning each of the first three points to the vertices o f the triangle. The last point will then be at the exact centre of the triangle. As we know that the distance from the centre of the triangle to each vertex is two-third of the altitude of this triangle, that is, units, we can see that it is definitely possible to find two darts which are units apart within the equilateral triangleb. Encompassing problemsConsider the following problem51 points are placed, in a random way, into a square of side 1 unit. Can we prove that 3 of these points can be c overed by a circle of radius units ?To prove the result, we may divide the square into 25 equal littler squares of side units each. Then by the Pigeonhole Principle, at least one of these small squares (so call pigeonholes) should contain at least 3 points (ie the pigeons). Otherwise, each of the small squares will contain 2 or less points which will then mean that the total number of points will be less than 50 , which is a contradiction to the fact that we have 51 points in the first case Now the circle circumvented around the particular square with the three points inside should haveradius = = = It will be worthwhile to note the above technique can be useful in analyzing accuracy of weapons in shooting practices and tests.Next, we will like to proceed to a more creative aspect of the application of Pigeonhole Principle by showing how it can be used to physical body interesting games7. Application of pigeonhole principle in mentality gamesWe like to introduce the application of pigeonhole principle in two exciting score government note tricksa. Combinatorial Card Trick Heres the trickA genius asks an unsuspecting observer to randomly choose five card game from a standard ornament of playing separate. The participant does not show these cards to the magician, but does show them to the magicians accomplice. The accomplice looks at the five cards, chooses four of them, and shows these four to the magician in a certain ordered manner. The magician immediately ident ifies the fifth mysterious card.How does the trick work?The following is an explanation of our working strategy(1) First of all, notice that in any hand of five cards there must be two cards of the same suit (an application of Pigeonhole Principle). The first card that the accomplice shows to the magician is one of these two cards. The other card of the same suit is never shown it is the secret card, the card which the magician must discover. Thus, the accomplice can easily communicate the suit of the hidden card the hidden card has same suit as the first card shown to the magician. Specifying the rank of the mystery card (ie its value) is a little trickier but can be accomplished with a little circular counting manner which we will explained belowNumber the cards in a suit circularly from 1(ace) to 11 (jack), 12 (queen) and 13 (king) so that 1 follows 13 i.e. the list is ordered in a dextrorotatory direction.Now, given any two cards A and B, define distance (A,B) as the clockwi se distance from A to B. It is easy to see that for any two cards A and B either distance(A,B) or distance(B,A) must always be less than or equal to 6. Again as an application of the Pigeonhole Principle, we note that if they are both 7 or more, then there will be at least 2 x 7 = 14 cards in a standard suit of cardsExampleCards 3 and Jack (11)distance(Jack, 3) = 5 distance (3, Jack) = 8Cards Ace(1) and 7distance (Ace, 7) = 6 distance (7,Ace) = 7(2) Our working strategy thus proceeds as follows.From those two cards of the same suit, A and B, the accomplice shows the magician card A such that distance(A, B) is 6 or less.For example, given the choice between the three of clubs and the Jack of clubs, the accomplice reveals the Jack (since distance (Jack ,3) = 5 and distance(3, Jack)= 8). The three of clubs remains hidden.If the two same-suit cards are the five of hearts and the six of hearts, the accomplice chooses the five (since distance (5,6) = 1 but distance (6,5) = 12) leaving the six of hearts as the mystery card.(3) Finally, the accomplice arranges the last three cards to encode a number from 1 to 6 the distance from the value of first card to that of the hidden card. A quick calculation allows the magician to discover the value of the mystery card. Notice that although the magician must decode only one of 6 possibilities, it should not present a problem, even to the slowest of magicians.To facilitate the explanation for the last step involved, we may assign each card a number from 1 to 52 for ranking purpose. For example,the ace of spade can be numbered 1 (the highest ranking card),ace of heart numbered 2,ace of club numbered 3,ace of diamond numbered 4,king of spade numbered 5,..,queen of spade numbered 9,.,jack of spade numbered 13,.,10 of spade numbered 17,. ,. ,2 of diamond numbered 52 (the lowest ranking card).We will now proceed to explain the last step using the following exampleExampleSuppose the five cards chosen are the following3 of Hearts (nu mbered 46)5 of Spades (numbered 37)6 of Clubs (numbered 35)7 of Hearts (numbered 30)2 of Diamonds (numbered 52)The accomplice notices that the 3 and the 7 have the same suit hearts. Since the distance( 3 ,7) = 4 and distance(7, 3) = 9, the accomplice chooses the 3 as the first card to show the magician, leaving the 7 of hearts as the hidden card. The magician now knows that the suit of the mystery card is hearts. The accomplices next task is thus to let the magician know that he must add the value 4 to the number 3 to obtain the final value of 7 for the hidden cardHow can he achieve this? Basically, he can arrange the other three cards in 3 = 6 ways. Based on the numbering method explained earlier, the 3 be cards can be ranked 1st, 2nd and 3rd . In our example, the 6 of Clubs will be ranked 1, the 5 of Spades will be ranked 2 and the 2 of Diamonds will be ranked 3. The accomplice may agree with the magician earlier that the arrangement of these 3 cards represent specific numbers as shown belowOrder in which 3 remaining cards are shownNumber represented by the arrangement1, 2, 311, 3, 222, 1, 332, 3, 143, 1, 253, 2, 16Thus in our example, the accomplice should display the cards in the following manner firstly, the 5 of Spades, then the 2 of Diamonds and lastly, the 6 of Clubs b. Permutation Card TrickHeres the trickA magician asks an unsuspecting observer to randomly arrange 10 cards which are labelled 1 to 10 in a hidden face down manner. The participant does not show the arrangement of these cards to the magician, but does show them to the magicians accomplice. The accomplice looks at the ten cards and flips over six of the cards in a certain ordered manner to reveal their values to the magician. The magician immediately identifies the values of the four remaining unknown cards.How does the trick work?We first note that by applying the Pigeonhole Principle, we can show that in any permutation of 10 distinct numbers there exists an increasing sequel of at l east 4 numbers or a decreasing subsequence of at least 4 numbers. (refer next section of our discussion). These are the numbers that remain hidden in our trick. The magician will know that the sequence is increasing if the accomplice flips over the other six cards from the left to right and it is decreasing if the other six cards are flipped over from the right to the left.We will now proceed to explain the trick behind the gameThe trick behind the gameGiven any sequence of mn+1 real numbers, some subsequence of (m+1) numbers is increasing or some subsequence of (n+1) numbers is decreasing.We shall prove the result by Contradiction method.Assume that the result is false. For each number x in the sequence, we have the ordered pair (i,j), where i is the length of the longest increasing subsequence beginning with x, and j is the length of the longest decreasing subsequence ending with x. Then, since the result is false, 1 i m and 1 j n. Thus we have mn+1 ordered pairs, of which at most mn are distinct. Hence by the Pigeonhole Principle, two members of the sequence, say a and b, are associated with the same ordered pair (s,t). Without loss of generality, we may assume that a precedes b in the sequence.If aThus, in our trick, we should have an increasing subsequence of at least (3+1) numbers or a decreasing subsequence of at least ( 3+ 1) numbers in a permutation of (33+ 1) distinct numbersHere is an example of how the trick can be performedExampleSuppose the participant arranges the 10 cards in the following manner (value faced down from left to right) 3, 5, 8, 10, 1, 7, 4, 2, 6, 9.Upon careful inspection, the accomplice notices that an increasing subsequence can be 3, 5, 8, 10 while a decreasing subsequence can be 10, 7, 4, 2.If he decides to use the increasing subsequence, he should leave the first four cards untasted and flips the other six cards over in a leftward manner as shown1 7 4 2 6 9 watchfulness of flipThe magician on realising that the four missi ng numbers are 3, 5, 8 and 10 and the leftward direction of flip, will thus proclaim the 4 hidden numbers to be 3, 5, 8, and 10 respectivelyIf the accomplice decides to use the decreasing subsequence, he should leave the cards bearing the numbers 10, 7, 4, 2 untouched and flips the other six cards over in a rightward manner as shown3 5 8 1 6 9Direction of flipThe magician on realising that the four missing numbers are 2, 4, 7 and 10 and the rightward direction of flip, will thus proclaim the 4 hidden numbers (from left to right) to be 10, 7, 4, 2 respectively8. ConclusionAlthough the Pigeonhole Principle seems simple and trivial, it is extremely useful in helping one to phrase and facilitate calculation and proving steps for numerous important Mathematical results and applications. We have included just a substantial amount of its applications in our project discussion. to a greater extent importantly, we will like to show that a simple Mathematical concept like the Pigeonhole Prin ciple does have numerous interesting and beneficial application in our daily life End of Report