THE PIGEONHOLE PRINCIPLE AND ITS APPLICATIONS

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1 International Journal of Recent Innovation in Engineering and Research Scientific Journal Impact Factor by SJIF e- ISSN: THE PIGEONHOLE PRINCIPLE AND ITS APPLICATIONS Gaurav Kumar 1 1 Associate Professor, Department of Mathematics, NAS College, Meerut Abstract-The pigeonhole principle has been one of the most used tools in mathematics. It is a very simple principle and can be applied in different walks of life. The present work contains different statements of pigeonhole principle and includes number of applications of Pigeonhole Principle from different fields. Keywords- Pigeonhole Principle, Applications I. INTRODUCTION The pigeonhole principle is one of the most used tools in combinatorics, and one of the simplest ones. It is applied frequently in graph theory, enumerative combinatorics and combinatorial geometry. The first use of pigeonhole principle is said to be by Dirichlet in 1834 and therefore it is also known as the Dirichlet Drawer Principle or Dirichlet Box Principle. There are different theorems for this principle which are as follows: 1.1 Theorem 1 If n+1 objects are arranged in n places then there must be at least two objects in the same place. Proof: Suppose no more than one pigeon were in each hole then there would be no more than n pigeons altogether which contradicts the assumption that we have n+1 pigeons. This proves the Pigeon Hole Principle. 1.2 Theorem 2 If n objects are arranged in k places then there are at least [n/k] objects in the same place, where [n/k] represents the smallest integer not less than n/k. This is generalized form of pigeonhole principle. 1.3 Theorem 3 Given an infinite set of objects, if they are arranged in a finite number of places then there is at least one place with an infinite number of objects. This is infinite pigeonhole principle. 1.4 Theorem 4 If the sum of n or more numbers is equal to S then among these there must be one or more numbers not greater than S/n and also one or more numbers not less than S/n. 1.5 Theorem 5 (Mathematical Form): Let X and Y be finite sets and let be a function. If X has more elements then Y then f is not one-one. If X and Y have same number of elements and f is onto then f is one-one. If X and Y have same number of elements and f is one-one then f is onto 1.6 Theorem 6 (Infinite Pigeonhole Principle) Given an infinite set of objects, if they are arranged in a finite number of places, there is at least one place with an infinite number of objects Proof: Suppose if in every place there is a finite number of objects, then in total there would be a finite number of objects, which is not true which proves the infinite Pigeon Hole rights Reserved Page 7

2 1.7 Other Principles related to Pigeonhole Principle If n objects are put into n boxes and no box is empty, then each box contains exactly one object. If n objects are put into n boxes and no box gets more than one object, then each box has an object II. APPLICATIONS OF PIGEONHOLE PRINCIPLE The Pigeonhole Principle seems simple but is very powerful. The difficulty comes in where and how to apply this principle. Some of the applications of pigeonhole principle are given as below: 2.1 Example 1 Problem Statement: Consider a bag which contains balls of two colors: black and white. Then to find the smallest number of balls which must be drawn from the bag, without looking, so that among these balls there are two of the same color. Solution: If we consider balls as pigeons and colors as pigeonholes then using pigeonhole principle we can say that we need to draw at least three balls so that two balls are of same color. 2.2 Example 2 Problem Statement: Consider a random list of 8 numbers viz 2, 4, 6, 8, 11, 15, 23 and 34. Is it possible to choose two of them such that their difference is divisible by 7? Solution: There are 7 possible remainders when a number is divisible by 7 viz 0, 1, 2, 3, 4, 5, 6. But we have 8 numbers. If we take numbers as pigeons and remainders as pigeonholes then by pigeonhole principle we can say that there are at least two pigeons sharing two holes i.e. two numbers with the same remainder. The difference of these numbers is therefore divisible by Example 3 Problem Statement: Let there be 25 crates of cold drink bottles to be delivered to a shop. The cold drink bottles are of three different brands and all the bottles in each crate are of the same brand. Then to find the least number of crates containing the same brand of bottles. Solution: Let us consider crates as pigeons and brands as pigeons. So we have 25 pigeons to be put into 3 pigeonholes. Since 25 = 3*8 + 1, we can use generalized form of pigeonhole principle with n = 25 and k = 3. Then [25/8] = 9. Thus some of the pigeonholes must contain at least 9 crates of same brand of bottles. 2.4 Example 4 Problem Statement: In any group of five people, there are at least two people who have an identical number of friends within the group. Solution: There are five possible numbers of friends for any person viz 0, 1, 2, 3 or 4. Consider number of people as pigeons and number of friends as pigeonholes. It would seem that each could have a different number of friends but if any person had 4 friends then no person may have zero friend and hence by pigeonhole principle, two people must have same number of friends. 2.5 Example 5 Problem Statement: Several cricket teams participate in a tournament in which each team plays every other team exactly once. Then at any moment during the tournament, there will be two teams which have played, up to that moment, an identical number of games. Solution: Let there be k teams. Then the number of games played by each team varies from 0 to k-1. If any team has played k-1 games then it has played every other team and no team has played 0 games. So we are fitting k teams into k-1 pigeonholes which are either the numbers from 0 through k-2 or the numbers 1 through k-1. Hence there must be two teams which have played an identical number of games. Available Online at : Page 8

3 2.6 Example 6 Problem Statement: Let five people receive as wages Rs. 1500/- altogether. Each of them wants to buy an article costing Rs. 320/-. Then at least one of them must wait for next paycheck to make his purchase. Solution: The sum S of their earnings is Rs 1500/-. So at least one worker earned no more than 1500/5 = 300. Such a worker must wait for his purchase. 2.7 Example 7 Problem Statement: There are 16 different time periods during which classes at a university can be scheduled. If there are 377 different classes, then to find the number of different rooms to be needed. Solution: Here n = 377 and k =16. Therefore the number of rooms is given by [377/16] = Example 8 Problem Statement: A computer network consists of 8 computers. Each computer is directly connected to zero or more of the other computers. Then there are at least two computers that are directly connected to the same number of other computers. Solution: Each computer can be directly connected to 0, 1, 2, 3, 4, 5, 6, 7 computers. But there are really only 7 choices, not 8, since if one computer is connected to zero other computers, then no computer can be connected to 7 others. So we have 8 computers and 7 choices. Pigeonhole principle says that at least two must have the same number of direct connections. 2.9 Example 9 Problem Statement: Let there be 50 people in a room then to find the minimum number of people that are born on the same month. Solution: Here the people are pigeons and months are pigeonholes. Therefore we have n = 50 and k = 12. Thus [50/12] = 5. Hence there are at least 5 people who are born on the same month Example 10 Problem Statement: To find the minimum number of students required to be sure that at least 6 will receive the same grades A, B, C, D, E, F. Solution: Consider the grades as pigeonholes. Then there are 5 pigeonholes. One pigeonhole must have at least 6 students. Therefore we have: [number of students/5] = 6 Hence number of students = Example 11 Problem Statement: Let d be a positive integer greater than zero. Then to show that amongst d+1 (not necessarily consecutive) integers there are at least two with the same remainder when divided by d Solution: Here the pigeonholes are the remainders. There are at most d remainders 0,1,2,..d-1. There are d+1 numbers. We consider theses as pigeons. Since we have d+1 numbers and d remainders, therefore by pigeonhole principle there will be at least two numbers with same remainder Example 12 Problem Statement: In a group of 8 people, to find the least number of people born on the same day of the week. Solution: Here days of the week are pigeonholes and numbers of people are pigeons. Since we have 8 pigeons and 7 pigeonholes therefore by pigeonhole principle, at one pigeonhole must contain at least two pigeons. Therefore, at least two people must be born on the same day of the week Example 13 Problem Statement: In a group of 27 English words, to find the least number of words that must start with the same letter. Available Online at : Page 9

4 Solution: Here numbers of letters are pigeonholes and numbers of English words are pigeons. Since there are 26 letters (pigeonholes) and 27 English words (pigeons) therefore by pigeonhole principle, one pigeonhole must contain at least two pigeons. Therefore, at least two English words must start with the same letter Example 14 Problem Statement: There are 50 baskets of apple. Each basket contains no more than 24 apples. To find the least number of baskets containing the same number of apples. Solution: Here we consider baskets as the pigeons and we place each of them in one of the 24 pigeonholes. Then we have n = 50 and k =24. Now [n/k] = [50/24] = 3. Therefore by general pigeonhole principle, there are at least 3 baskets containing same number of apples Example 15 Problem Statement: To show that among any 4 numbers one can find 2 numbers so that their difference is divisible by 3. Solution: There are 3 possible remainders when a number is divided by 3 viz 0, 1, 2. Thus by pigeonhole principle, since we have 4 numbers, some of them must have the same remainder when divided by 3. So we can write these two numbers as: n 1 = 3k 1 + r and n 2 = 3k 2 + r where r is the remainder when divided by 3. Then the difference of two numbers is: n 1 n 2 = (3k 1 + r) (3k 2 + r) = 3k 1 3k 2 = 3(k 1 k 2 ) which is divisible by Example 16 Problem Statement: To show that for any natural number n, there is a number composed of digits 5 and 0 only and divisible by n. Solution: We know that given any set of n + 1 numbers, some two of them have a difference that is divisible by n. So we should try to find a set of n + 1 numbers with the property that for any two of them, the difference is a number composed of digits 5 and 0 only. One possibility is the sequence of numbers 5; 55; 555; 5555; 55555; : : :, since the difference of any two of these will be some number of 5's followed by some number of 0's. So we can take the first n+1 numbers whose only digits are 5, and there must be some pair whose difference is composed of only 5's and 0's, and divisible by n Example 17 Problem Statement: Given 12 different 2-digit numbers, to show that one can choose two of them so that their difference is a two-digit number with identical first and second digit. Solution: Here again we shall use the result that given any n+1 numbers, one can find 2 numbers so that their difference is divisible by n. Since we have 12 numbers, therefore there must be two numbers whose difference is divisible by 11. But this difference can't have more than two digits, and since it's divisible by 11, it can't have fewer than two digits, so it must have exactly two digits. And any two-digit number divisible by 11 has identical first and second digit Example 18 Problem Statement: Given a triangle in the plane, prove that there is no line that does not go through any of its vertices but intersects all three sides. Solution: We know that any line divides the plane into two parts. By the pigeonhole principle, since there are three vertices, there must be at least two on the same side. The triangle side formed by those two vertices does not intersect the line. Available Online at : Page 10

5 2.19 Example 19 Problem Statement: (Ramsey Theory) In a group of 6 people, in which each pair consists of 2 friends or 2 enemies, there must be 3 mutual friends or 3 mutual enemies in the group assuming that anyone who is not a friend is an enemy. Solution: This can be proved using generalized pigeonhole principle. Let A be one of the persons and of the other 5, 3 or more are either friends or enemies of A because by the generalized pigeonhole principle, when 5 objects are divide into 2 sets, one set has at least [5/2] = 3 elements. Suppose that B, C, D are friends of A then if any of 2 of these 3 are friends then that pair and along with A make 3 mutual friends. Else if they are not friends then they are mutual enemies. Hence the theorem Example 20 Problem Statement: Let there be 30 buses to carry 2000 students from NAS College to CCS University, Meerut. If each bus has 80 seats then to show that one of the buses will have 14 empty seats and one of the buses will carry at least 67 students. Solution: Here total number of seats is = 80x30 = Therefore total number of empty seats are = 400. By the pigeonhole principle, one bus must have [400/30] = 14 empty seats. Again we have 2000 students in 30 buses and therefore one bus must have [2000/30] = 67 students. III. CONCLUSION Although, the pigeonhole principle seems simple and trivial, it is extremely useful in helping one to formulate and facilitate calculation and proving steps for number of mathematical problems in different walks of life. In this work, substantial numbers of examples are included so as to show that a simple mathematical concept like the pigeonhole principle does have numerous interesting and beneficial applications in our daily life. REFERENCES [1] A.I. Orlov, Printcip Dirikhle, Kwant, 3, 1971 (Russian) [2] Baclace, P.E., Competitive agents for information filtering, Communication ACM, 35-50, 1992 [3] Brualdi, Richard A., Introductory Combinatorics, Prentice Hall, 2010 [4] G. Polya, Mathematical discovery: on understanding, learning and teaching problem solving, New York, Wiley, 1981 [5] J. Krajicek, On the weak pigeonhole principle, Manuscript, August [6] R.B.J.T. Allenby and A. Slomson, How to count: An introduction to Combinatorics, Chapman and Hall/CRC, Boca Raton, Florida, 2010 [7] S. R. Buss and T. Pitassi, Resolution and the weak pigeonhole principle, In Computer Science Logic, 1997, pp [8] S. R. Buss and G. Turin, Resolution proofs of generalized pigeonhole principles, Theoret. Comput.Sei., 62, 1988, pp [9] Tague, N.R., The Quality Toolbox, Second Edition, Milwaukee:ASQ Quality Press, 2005 [10] V. K. Balakrishnan, Theory and Problems of Combinatorics, Schaum s Outline Series, Tata McGraw-Hill, 1995 Available Online at : Page 11