Download Pigeonhole: the box principle

Pigeonhole: the box principle
Salvatore Raucci
May 21, 2014
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Abstract
The Pigeonhole principle is a simple statement which has many applications in combinatorics. In this article we provide examples taken from some
mathematical competitions which can be solved using Pigeonhole, with some
considerations about their deeper mathematical background, in particular
about Ramsey numbers.
If there exist n pigeonholes containing n + 1 pigeons, then one of the pigeonholes
must contain at least two pigeons.
It can be generalized saying that, if there are n boxes and k objects in the boxes,
there exists at least one box containing d nk e objects, where dxe with x ∈ R is the
ceiling function, i.e. the smallest integer greater than or equal to x.
This combinatorical principle was first explicitly used by Dirichlet in number
theory and was generalized by F.P.Ramsey.
How do we recognize a problem solvable by Pigeonhole?
This principle gives us only information about the existence of something, without
saying in which box there is more than one object. This means that many existence
problems about finite or infinite sets are usually solved by Pigeonhole. The main
difficulty is to find out which are the boxes and which are the objects.
And now, let us solve some problems...
E1. There are n students in a room. Prove that among them there are two
students who have the same number of acquaintances in the room.
Proof.
We have n boxes which are the sets of students having 0, 1, 2, ...n−1 acquaintances
in the room. However, box #0 and box #n − 1 cannot both be occupied. Thus
there are n students in n − 1 boxes.
E2. Let a1 , a2 , ..., an be n not necessarily distinct integers. Then there always
exists a subset of these numbers with sum of the elements divisible by n.
Proof.
We consider the n integers
s 1 = a1
, s2 = a1 + a2
, s3 = a1 + a2 + a3
, ... , sn =
n
X
ai
i=1
If one of these integers si is divisible by n, then the proof is done.
Otherwise, we consider their remainders modulo n. They are n − 1, but the numbers si are n.
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Hence there exist sp and sq with WLOG p > q equal modulo n, which means that
sp − sq = aq+1 + aq+2 + ... + ap is divisible by n.
E3. Among n + 1 integers from {1, 2, ..., 2n}
a. there are two coprime integers
b. there is one integer divisible by another
Proof.
a. Among the n + 1 integers there are two successive integers, which are coprime:
(a, a + 1) = (a, 1) = 1
b. We consider the n + 1 integers a1 , a2 , ...an+1 and write them as ai = 2k bi with
bi odd. Now we consider b1 , b2 , ..., bn+1 . They are n + 1 odd numbers from the
interval [1, 2n], while in this interval there are only n distinct odd integers. Hence
there exist p, q such that bp = bq , which means that one of ap and aq is divisible
by the other.
E4. In any convex 2n−gon, there exist a diagonal not parallel to any side
Proof.
2n(2n − 3)
= n(2n − 3) diagonals.
A 2n−gon has
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Now we consider one side of the 2n−gon, say AB. If we consider the parallel lines
to AB through every vertex of the 2n−agon, we note that for every vertex passes
at most one diagonal parallel to AB.
Let P be the vertex with maximum distance to the line AB.
If the parallel to AB passing through P passes through another vertex Q, then
P Q is a side of the 2n−gon, and the maximum number of diagonals parallel to
= n − 2.
AB is 2n−2−2
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Otherwise, there remain 2n − 3 vertices between AB and the parallel thruogh P ,
c = n − 2.
and the maximum number of diagonals parallel to AB is b 2n−3
2
Hence the total number of diagonals parallel to some sides is at most 2n(n − 2) <
n(2n − 3) the total number of diagonals, which means that there exists one diagonal not parallel to any side.
E5. Two of 70 distinct positive integers ≤ 200 have differences of 4, 5 or 9.
Proof.
Let a1 , a2 , ...a70 be these numbers. We consider the following 210 numbers:
a1 , a2 , ..., a70 , a1 + 4, a2 + 4, ..., a70 + 4, a1 + 9, a2 + 9, ..., a70 + 9
None of them is > 209, hence there exist two of them which are equal ai + x and
aj + y with x 6= y because the ai are all distinct, and with x, y ∈ {0, 4, 9}.
Now we have ai + x = aj + y ⇒ ai − aj = y − x and, since x and y can assume
only the three values 0, 4, 9 then |y − x| ∈ {4, 5, 9}
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Now, before solving other problems, we introduce some notation.
Gp is a complete graph with p vertices and p2 edges.
An n−colouring of Gp means that we colour each edge of Gp with one of n colours.
If Gp contains a triangle with all sides of the same colour, then we have a monochromatic triangle.
E6. Among six students there are always three who know each other or three
who are completely strangers.
Proof.
We colour the edges of a G6 , whose vertices are the six students, red if the students
know each other, blue if not.
We prove that there always exists a monochromatic triangle.
Take one of the six vertices, say R: at least three of the five edges which start at
R are of the same colour, say red.
These red edges end at A, B and C.
If any side of the triangle ABC is red, we have a red triangle. If not, ABC is a
blue triangle.
Hence, in any case, we have a monochromatic triangle.
What if the students were 5? In this case we can colour the graph so that there is
no monochromatic triangle.
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E7. An international society has members from six different countries. The
list of members contains 1978 names, numbered 1, 2, ..., 1978. Prove that there is
at least one member whose number is the sum of the numbers of two members
from his own country or twice as large as the number of one member from his own
country.(IMO 1978)
Proof.
We prove this by contradiction.
We note that the difference between two numbers of two members is still a number
of another member.
Let A, B, C, D, E and F the sets representing the six countries. In one of these,
e = 330 elements, i.e. members
WLOG A there are at least d 1978
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a1 < a2 < a3 < ... < a330
The 329 differences (which are numbers belonging to members) a330 − ai with
i = 1, 2, 3, ..., 329 do not lie in A, hence they must lie in B to F .
e = 66 of these differences
One of these, say B, has at least d 329
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b1 < b2 < ... < b66
The 65 differences b66 − bi with i = 1, 2, ..., 65 do not lie in A or B.
In fact, if they lay in B, then in B there would be three numbers such that one is
the sum of the others.
If they lay in A, then b66 − bi ∈ A for some i, and there would exist ah and ak such
that b66 = a330 −ah and bi = a330 −ak , which would mean that b66 −bi = ak −ah ∈ A,
which is not possible.
Hence the 65 differences lie in the other sets.
e = 17 of these differences
Then the set C has at least d 65
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c1 < c2 < ... < c17
The 16 differences c17 − ci with i = 1, 2, ..., 16 do not lie in A to C for the same
above-mentioned reasons, so they lie in D to F .
Then the set D has at least d 16
e = 6 of these differences
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d1 < d2 < ... < d6
The 5 differences d6 − di with i = 1, 2, ..., 5 hence lie in E or F .
Then E has at least d 25 e = 3 of these differences
e1 < e2 < e3
The two differences f1 = e3 − e2 and f2 = e3 − e1 lie in F .
However, f1 − f2 = e1 − e2 cannot lie in F , in E, or in any other set for the same
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above-mentioned reasons.
Hence, having all the members numbers between 1 and 1978, the difference between the numbers of two of them is still a number of a member, but we have
shown that f1 − f2 does not lie in any set, which is a contradiction.
Were all the 1978 members necessary? No, 1957 were enough
Why?
Ramsey’s theorem If q1 , ..., qn ≥ 2 are integers, there is a minimal number
R(q1 , ..., qn ) so that, for p ≥ R(q1 , ..., qn ) for at least one i = 1, ..., n, Gp contains
at least one monochromatic Gqi
The numbers R(q1 , ..., qn ) are called Ramsey numbers
Some trivial Ramsey numbers are R(s, 1) = 1 and R(s, 2) = s.
However, there are few known non-trivial Ramsey numbers: R(3, 3) = 6,
R(3, 3, 3) = 17, R(4, 3) = 9, R(5, 3) = 14, R(5, 4) = 25, R(4, 4) = 18 and others.
The main problems in finding other Ramsey numbers are their upper and lower
bounds, which are not generally known with any accuracy. They can be improved
only with some deeper mathematical understanding.
The bounds we have, which are continually improved[3], are not enough: for example, if 43 ≤ R(5, 5) ≤ 49, to prove that R(5, 5) 6= 43 by drawing every possible
43
2−colouring of G43 we should draw 2( 2 ) = 2903 graphs.
As we take larger Ramsey numbers, we have to draw more graphs, which is impossible without better computers[4].
The following example is a simple upper bound for R(3, 3, ..., 3)
| {z }
n
E8. In space there are pn = ben!c + 1 points. Each pair of points is connected by a line, and each line is coloured with one of n colours. Prove that there
is at least one triangle with sides of the same colour.
Hint.
We find pn+1 − 1 as a function of pn − 1, and then we use the truncated series for
e ...
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References
[1] Arthur Engel Problem solving strategies Springer-Verlag, New York, 1998
[2] Gregory E. W. Taylor Ramsey Theory http://web.mat.bham.ac.uk/D.Kuehn/RamseyGreg.pdf
[3] Kathleen Johst and Yury Person On the multicolor Ramsey number of a graph
with m edges arXiv.org: 1311.5471
[4] Zhengbing Bian, Fabian Chudak, William G. Macready, Lane Clark, Frank
Gaitan Experimental determination of Ramsey numbers arXiv.org: 1201.1842
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