# PROBLEM SET 7: PIGEON HOLE PRINCIPLE

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1 PROBLEM SET 7: PIGEON HOLE PRINCIPLE The pigeonhole principle is the following observation: Theorem. Suppose that > kn marbles are distributed over n jars, then one jar will contain at least k + marbles. (It can also be formulated in terms of pigeons and pigeonholes, hence the name.) The proof of this pigeonhole principle is easy. It is more difficult to know when to apply it. There are many surprising applications of the pigeonhole principle. The pigeonhole principle was first explicitly formulated by the mathematician Dirichlet (80 89). The pigeonhole principle says for example that at least two people in New York City will have the same number of hairs on their head. This is because humans have <, 000, 000 hairs and there are >, 000, 000 people in NYC. The pigeonhole principle is particularly powerful in existence proofs which are not constructive. For example in the previous example we proved the existence of two people with the same number of hairs without specifically identifying these two individuals. Example. How many bishops can one put on an 8 8 chessboard such that no two bishops can hit each other. It seems like a good idea to put 8 bishops in one row at the edge of the board. If we put 6 bishops on these positions then we see that still no two bishops can hit each other. So we see that at least bishops can be put on the chessboard. We conjecture that this number is maximal. How can we prove that is the maximal number of bishops? We can use the pigeonhole principle. We need to partition the 8 8 = 6 fields into sets such that whenever two bishops are on fields which lie in the same set, then they can hit each other. Note that a bishop on a black field only can move to other black fields. A bishop on a white field can only move to other white fields. To prove that is the maximal number of bishops, we could prove that there at most 7 black bishops and at most 7 white bishops. We try to partition the black fields into 7 sets such that if two bishops are on fields in the same set, then they can hit each other. We see that the following configuration works:

2 A similar configuration works for the white fields. (Take the mirror image.) In the proof that we are going to write down, we do not need to distinguish between black and white bishops. We can combine the partition of black fields and the partition of the white fields to get a partition of the set of all 6 fields into subsets. Proof. The answer is. Place bishops on the chessboard as follows: Then no two bishops can hit each other. Let us label the fields on the chessboard as follows: Whenever two bishops are placed on the chessboard with the same number, they can hit each other. If a number of bishops are placed on fields of the chessboard such that no two can hit each other, then all the numbers of the fields are distinct. This shows that the number of bishops is at most Example. (Putnam 98) Let S be a subset of {,,,..., n} with n+ elements. Show that one can choose distinct elements a, b S such that a divides b. Is the n + in the problem sharp? Suppose that S has the property that for every pair of distinct a, b S, a does not divide b and b does not divide a. How many elements can S have? If S contains then it could not contain any other element. If S contains then all other even numbers are excluded. In order to get S as large as possible, it seems that one should choose large elements. If we take the n largest elements, S = {n +, n +,..., n} then

3 clearly no two distinct elements divide each other. Any positive integer a with a n divides an element of S. This shows that S is maximal with the desired property. The problem has somewhat of a pigeonhole flavor: We are asked to prove the existence of certain elements a, b S but it seems unlikely that we can explicitly construct these elements. How can we apply the pigeon hole principle? Since we have n + elements we partition {,,..., n} into n sets T, T,..., T n. The pigeonhole principle says that S T i contains at least two elements for some i. This then should be useful to conclude that there exist a, b S such that a divides b. So we would like T i to have the following property: Whenever a, b T i with a < b then a divides b. So T i should be of the form {a, a, a,..., a k } with a a a a k. The largest such set is {,,,,... }. Let us define T to be this set. The smallest element not in T is. So let us define T = {,,,,... }. The smallest element, not in T and T is. so let us define Let us define more generally T = {,,,,... }. T k = {(k ), (k ), (k ), (k ),... } for k =,,..., n. It is easy to see that the union of T, T,..., T n contains. {,,..., n}. We are now ready to write down the proof, using the pigeonhole principle. Proof. Let us define T k = {,,..., n} {(k ), (k ), (k ), (k ),... } for all k =,,..., n. Every c {,,..., n} can uniquely be written as c = j (i ) with j N and i {,,..., n}. This shows T, T,..., T n is a partition of {,,..., n}. By the pigeon hole principle, S T i contains at least two distinct elements for some i, say {a, b} S T i with a < b. then a = j (i ) and b = k (i ) for some j, k N with j < k and it is clear that a b. Example. Suppose that S is a set of n integers. Show that one can choose a nonempty subset T of S such that the sum of all elements of T is divisible by n. One can try to attack this problem using the pigeonhole principle but it is not immediately clear how we can apply it. What are the pigeonholes here? Since our goal is to prove that something is divisible by n, it is natural to take the congruence classes modulo n as pigeonholes. The pigeonhole principle says: Given n+ integers, one can choose two of them such that their difference is divisible by n. How can we apply this here? We can apply it if we have integers a, a,..., a n+ such that a j a i is a sum of distinct elements of S for all i < j. We can indeed achieve this. Suppose that S = {b, b,..., b n }. Take a = 0, a = b,

4 a = b + b, a = b + b + b, etc. We can write down our proof: Proof. Suppose that S = {b, b,..., b n }. Define c i = b + b + + b i for all i = 0,,..., n (c 0 = 0). Of the n + numbers c 0, c,..., c n, at least must lie in the same congruence class module n by the pigeonhole principle. Assume that we have c i c j (modn) for i < j. Then we get that is divisible by n. c j c i = b i+ + b i+ + + b j Example. Suppose that we are given a sequence of nm + distinct real numbers. Prove that there is an increasing subsequence of length n + or a decreasing subsequence of length m +. To get an idea, let us do a random example with n = m = : (), 6, 7, 60, 7, 8, 6, 6, 7, 9. What is the longest increasing subsequence and what is the longest decreasing subsequence? How can we efficiently find these without checking all possible subsequences? For a longest decreasing sequence in () there are two cases. Either such a sequence ends with 9 or it does not. If the sequence does not contain 9, then a longest decreasing sequence of () is a longest decreasing subsequence of the shortened sequence (), 6, 7, 60, 7, 8, 6, 6, 7. If 9 appears in a longest decreasing subsequence, then we may wonder what the previous element in that subsequence is. Is it,6,7,60,7,8 or 6? (Clearly 6 and 7 are out of the question because the subsequence is decreasing.) It would now be useful to know for each x in {, 6, 7, 60, 7, 8, 6} what the longest decreasing subsequence is of () that ends with x. We then could take the longest decreasing subsequence of () ending with some x in {, 6, 7, 60, 7, 8, 6} and add 9 at the end to obtain a longest decreasing subsequence of (). So to find the longest decreasing subsequence of () we need to find the longest decreasing subsequence of () ending with x for x =, 6,..., 9 (in that order). x a longest decreasing subsequence ending with x , , , 60, 6 6 8, 6 7 6, 60, 6, 7 9 8, 6, 9

5 We have found that the longest decreasing subsequence has length. Namely the subsequence 6,60,6,7 is decreasing. So the statement we want to prove works out in this example. Let us determine the longest increasing sequence: x a longest decreasing subsequence ending with x 6, 6 7, 7 60, 7, 60 7, 7, 60, 7 8, 7, 60, 7, , 7, 60, , 9 We see that there even is an increasing sequence of length. We checked one (small) example and it seems that we are still far from a solution (but this is actually not the case). We are interested in the lengths of maximal increasing/decreasing sequences. So let us make a table containing the length of a longest increasing sequence ending in x and the length of a longest decreasing sequence ending in x for all x. x decreasing increasing Let us plot the last two columns against each other. We get distinct points (, ), (, ), (, ), (, ), (, ), (, ), (, ), (, ), (, ), (, ). If there were no increasing or decreasing sequence of length, then all points would fit in a box and two of the points would have to coincide by the pigeonhole principle. This

6 leads to a contradiction as the following proof shows. Proof. Suppose that x, x,..., x mn+ is a sequence of distinct real numbers. Let a i be the length of the longest decreasing subsequence ending with x i. Let b i be the length of the longest increasing subsequence ending with x i. We claim that if i j, then that (a i, b i ) (a j, b j ). Indeed, if x j < x j then we can take a longest decreasing subsequence ending with x i and add x j at the end. This way we obtain a decreasing subsequence ending with x j of length a i +. This shows that a j > a i and (a i, b i ) (a j, b j ). If x j > x i then we can take a longest increasing subsequence ending with x i and add x j at the end. This shows that b j > b i and (a i, b i ) (a j, b j ). There are only nm pairs (a, b) with a, b Z and a m and b n. We must have a i > m or b i > n for some i.. Diomphantine Approximation Example 6. Suppose that α is a real number and that N is a positive integer. Show that one of the numbers α, α, α,..., Nα differs at most from an integer. N It is not so clear a priori how one can use the pigeonhole priciple in this problem. Since we are only interested in the value iα modulo the integers, we define β i = iα iα. ( x is the largest integer x.) We observe that 0 β i < for all i. We want to show that β i N or β i for some i. N An important observation is that β i+j β i is the same as α j up to an integer for all i. So if β i+j is very close to β i then α j is very close to an integer. We want to show that one can choose two of the N + numbers β 0 = 0, β, β,..., β n that are at most of distance of N each other. Here is where the pigeonhole principle might come in. We need to partition the interval [0, into N sets, such that every two elements in the same set are at most apart. This N is possible. We can take S k = [ k, k for k =,,..., N. By the pigeon hole principle at N N least two of the numbers in β 0, β,..., β N lie in the same set S k. Let us write down a more formal proof. Proof. Define β i = iα iα for i = 0,,,..., N. Then 0 β i < for all i. Define S k = [ k, k for k =,,..., N. N N The interval [0, is the union of intervals S, S,..., S N. By the pigeon hole principle, at least two of the numbers β 0, β,..., β N lie in the same interval S k for some k. Say β i, β j S k for some i, j with 0 i < j N. But then N β j β i = (j i)α ( jα iα ) N. We are done because j i N and jα iα is an integer. Diomphantine approximation is an area of number theory where one likes to approximate irrational numbers by rational number. The previous example allows us to characterize irrational numbers! 6

7 Theorem 7. A real number α is irrational if and only if there exists a sequences of integers p, p,... and q, q,... such that and q n α p n 0 for all n. lim q nα p n = 0 n Proof. Suppose that α is irrational. For every n, we can find integers p n, q n with q n n such that q n α p n <. It follows that n lim q nα p n = 0. n Obviously q n α p n 0 for all n because α is irrational. Conversely, suppose that q n α p n 0 for all n and lim n q n α p n = 0. Assume that α is rational, say α = a with a, b Z and b > 0. Now we have b q n α p n = q n a p n b b b because q n α p n 0. This leads to a contradiction with lim n q n α p n = 0. We can apply this: Theorem 8. The number is irrational. Proof. Expanding ( ) n and using ( ) = we get ( ) n = q n pn for some integers p n and q n. We have lim q n pn = lim ( ) n = 0 n n because <. Also q n pn = ( ) n 0 for all n because. This proves that is irrational.. balls For a finite set S we denote the number of elements of S by S. The pigeonhole principle can be reformulated as: Theorem 9. If S, S,..., S n are subsets of a finite set T and S + S + + S n > k T then there exists an element x T that lies in at least k + of the sets S, S,..., S n. If S is a measurable set in R, let µ(s) be its volume. We have the following variation of the previous theorem. 7

8 Theorem. If S, S,..., S n are measurable subsets of a measurable set T, and µ(s ) + µ(s ) + µ(s n ) > kµ(t ). then there exists an element x T that lies in at least k + of the sets S, S,..., S n. Example. Let S be a set of points in the cube [0, ] [0, ] [0, ] (in R ). Such that the distance between every two distinct elements x, y S is at least 0.. Give an upper bound for the number of elements of S. We could use the pigeonhole principle as follows. Partition the cube into small regions, such that in each region the maximal distance between two points in this region is < 0.. For example, we could partition the cube into N N N little cubes of sidelength. N The maximum distance between two points in this little cube is /N, the length of its diagonal. We need that /N < 0., so N > 7.. (Without a calculator we see that < 8 because 8 = > 00 = ( ).) So let us take N = 8. Each of the N N N cubes can contain at most element of S. Therefore, the cardinality of S is at most 8 = 8 = 8. There is another way of looking at this problem. Instead of saying that two points x and y have distance at least 0., we could say that the balls with radius 0.0 around x and around y are disjoint. Note that all balls lie within the cube [ 0.0,.0] [ 0.0,.0] [ 0.0,.0] with volume. =.. We can reformulate the problem as the problem of packing oranges of diameter 0. into a (cube-shaped) box of sidelength.. The volume of a ball with radius 0.0 is π(0.0) (we are using a calculator now). The number of oranges is at most (.) = π(0.0) π = 7986 π.0. This means that S has at most points. This is quite an improvement. Johannes Kepler (7 60) conjectured in 6 that the densest way of packing balls in R is the cubic or hexagonal packing (well-known to people selling oranges). These packing give a density of π 7.08%. Kepler s conjecture was proven by Thomas Hales (U of M!) in 998. Using this result, we see that S can have at most π 7986 π = So S can have at most 88 elements. (We do not claim here that this number is sharp.) Example. A binary word of length n is a sequence of 0 s and s of length n. The set {0, } n is the set of all binary words of length n. Let S be a subset of {0, } n with the following property: for every pair of distinct elements x = x x x n and y = y y y n we have that x and y differ in at least positions. Show that S has at most elements. n n + 8

9 Proof. We can use the ideas in the previous example. The distance d(x, y) between two words x = x x x n and y = y y y n is the number of positions where they differ. For any binary word x {0, } n, let B(x) be the ball with radius, so B(x) = {z {0, } n d(x, z) }. In other words B(x) is the set of all binary words of length n which differ from x in at most position. The number of elements of B(x) is n + (namely x itself and all words obtained by changing x at on position). Notice now that d(x, y) is equivalent with B(x) and B(y) are disjoint. So all balls B(x), x S. are disjoint. The disjoint union of all balls B(x) x S has exactly S (n + ) elements. On the other hand, this union is a subset of {0, } n which has n elements. We obtain the inequality S (n + ) n The previous example is known as the Hamming bound for -error correcing binary codes. The mathematician Richard Hamming (9 998) also studied examples where equality holds (and these are known as Hamming codes).. Exercises Exercise. * What is the maximum number of rooks that one place on an 8 8 chessboard such that no two rooks can hit each other? Prove your answer. Exercise. * What is the maximum number of queens that one can place on an 8 8 chessboard such that no two Queens can hit each other? Exercise. **** What is the maximum number of queens that one can place on an 8 8 chessboard such that no two Queens can hit each other? Exercise (Dutch Mathematics Olympiad). **** A set S of positive integers is called square-free if for all distinct a, b S we have that the product ab is not a square. What is the maximum cardinality of a square free subset S {,,,..., }? Exercise. * Show that (a b)(a c)(b c) is always even if a, b, c are integers. Exercise 6. **** Prove that for every integer n there exists an integer m such that k k + m is not divisible by n for all integers k. Exercise 7. *** Show that, given a 7-digit number, you can cross out some digits at the beginning and at the end such that the remaining number is divisible by 7. For example, if we take the number 89, then we can cross out at the beginning and 89 at the end to get the number = 7. 9

10 Exercise 8. **** Improve the statement in Example 6: Suppose that α is a real number and that N is a positive integer. Show that one of the numbers α, α, α,, Nα differs at from an integer. most N+ Exercise 9. *** Prove that the Euler number is irrational. e = 0! +! +! +! + Exercise. **** Let x, x,..., x n R with x i for i =,,..., n. Show that there exist a, a,..., a n {, 0, }, not all equal to 0, such that a x + a x + + a n x n n n Exercise (IMO 987). ***** Let x, x,..., x n be real numbers satisfying x +x + + x n =. Prove that for every integer k there are integers a, a,..., a n, not all 0, such that a i k for all i and a x + a x + + a n x n (k ) n k n. Exercise (Dutch Mathematical Olympiad). ***** Suppose that S is a subset of {,,,..., 0} with at least elements. Show that one can choose a nonempty subset T of S such that the product of all elements of T is a square. Exercise. *** Suppose that a 0, a, a,..., a n Z are integers. Show that the product (a j a i ) is divisible by n!. 0 i<j n Exercise. **** Let a, a,..., a be distinct integers from {,,..., 99}. Show that {a, a,..., a } contains two disjoint non-empty subsets with the sum of the numbers from the first equal to the sum of the elements from the second subset. Exercise. ***** The set M consists of 00 distinct positive integers, none of which is divisible by any prime p >. Prove that there are distinct x, y, z, t in M such that xyzt = u for some integer u. Exercise 6. **** [Putnam 989] Let m be a positive integer and let G be a regular (m+)- gon inscribed in the unit circle. Show that there is a postive constant A, independent of m, with the following property. For any point p inside G there are two distinct vertices v and v of G such that p v p v < m A m. Here s t denotes the distance between the points s and t. Exercise 7. *** Show that where p runs over all prime numbers, is an irrational number. p p

11 Exercise 8 (Putnam 000). *** Let a j, b j, c j be integers j N. Assume for each j, at least one of a j, b j, c j is odd. Show that there exist integers r, s, t such that ra j + sb j + tc j is odd for at least N/7 values of j, j N.

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