The blue dots below are draggable.
What is this about?
If five nodes of a plane lattice are chosen at random, any two of them
define a line that contains other lattice points, and more specifically,
the midpoint of the segment between the two points is necessarily a lattice
point. Four points may not possess such a pair. The applet above
illustrates this statement in case of the square grid.
The validity of the statement for four points is easily established with
the help of the applet. The case of five points is also simple.
The midpoint of the segment joining points P(x1,
y1) and Q(x2, y2) is
given by M((x1 + x2)/2, (y1 +
y2)/2). M is the lattice point if and only if
x1 and x2 are of the same parity and so are
y1 and y2. With 2 coordinates, there are four
different parity pairs: odd/odd, odd/even, even/odd and even/even. Among
five points, at least two are bound to have the same parity
combination. The line joining any two such points is bound to contain other
lattice nodes.
The proof makes use of the
Pigeonhole
Principle: If n objects are distributed between fewer than n boxes,
at least one box must contain at least two of the objects. Most often the
principle is taken for granted, but sometimes it is offered as an exercise,
e. g., see [Graham], Ex. 3.8 and [Averbach], Ex. 6.32. [Graham]
actually deals with a more general Dirichlet's box
principle using integer arithmetic. The latter is also shown in Chapter
20 of Proofs from THE BOOK. [Averbach] proves the Principle by invoking the
(bipartite) graph theory. The nicest ever and most convincing proof
(without words) was recently published for n =
7.
The Pigeonhole Principle is simple and natural. Given its simplicity,
the number and variety of its applications is utterly surprising. Its place
in the tool chest of any professional problem solver is indispensible. As
E. Barbeau, M. Klamkin and W. Moser put it ([Barbeau],
p. 211), with an inspired choice of objects and boxes, several problems
are rendered completely innocuous by this principle. It is equally true
that those whose inspiration is begging may, in most cases, have difficulty
fingering both objects and boxes. Chapter 20 of Proofs
from THE BOOK collects several examples, to which applicability of
the Pigeonhole is not at all obvious.
In the recent proceedings of a gathering in honor of
Martin Gardner, we find two relevant and entertaining applications.
Place (by dragging) the ten words at the bottom of the applet into the
ten cells that form a chain such that any two words adjacent in the chain,
share one common letter.
The puzzle comes in two variants [Rodgers] that differ
from one another ever so slightly. The words SON, HUT in one are replaced
with SUN, HOT in the other. The latter is solvable, the former is not. The
reason is that the two variants are represented by different graphs:

Obviously, a solution to the puzzle constitutes a Hamiltonian circuit on
its graph. (A circuit in a graph is Hamiltonian if it passes through every
node of the graph exactly once.) The first of the two is the famous
Petersen graph that is known not to house any Hamiltonian circuits. In the
second, Hamiltonian circuits are easily found even by trial and
error. (M. Gardner's The Last Recreations
contains a chapter on the Pigeonhole Principle and a chapter on Snarks that
can't live without the Petersen graph.) A generalized version of the
Pigeonhole is naturally
used to show that the
Petersen graph does not have Hamiltonian circuits.
The inventor, Tom Rodgers, suggests using the puzzle to amuse oneself at
the expense of befuddled friends. Have 12 word scrub tiles ready and use 10
of them to arrange publicly in a chain. Then drop the tiles on the
table. "Be careful to hold onto the HOT and SUN scrub with your thumbs,
letting HUT and SON fall in their place. Your poor, befuddled victim will
have no chance at finding a loop."
In the same volume, B. Epstein offers a no less
bewildering party trick which is also based on the Pigeonhole
Principle. But before I describe it, let's play a game. The task is to
recognize a permutations of the numbers 1 through 10. Some numbers are
hidden, others (at most 6 of them) are clearly shown. Sometimes the numbers
appear in blue, sometimes in red. With such preliminaries, can you now name
the hidden numbers? (For a training session we can use 5 numbers, of which
at most 2 are shown.)
Believe me you can, for I shall presently reveal the secret of my
selection. It follows from Dirichlet's
box principle, that in any permutation of 10 distinct numbers there exists
an increasing subsequence of at least 4 numbers or a decreasing subsequence
of at least 4 numbers. (A general statement is due to P. Erdös and
G. Szekeres, see [Aigner], p. 124.) These are the numbers
that remain hidden. You'll know that the sequence is increasing if the
numbers are red. It's decreasing if they are blue.
For a party trick, two persons should coordinate their efforts. The
mathematical origins of the trick could be easily concealed by picking ten
playing cards whose order is agreed upon in advance. Whenever a decreasing
sequence turns up, deal cards in the reverse order, thus avoiding the
dependence on color clues.

Dirichlet's box principle asserts that if n objects are put
into m boxes, some box must contain at least ceil(n/m) objects, some box
must contain at most floor(n/m). As an application, the task of writing
this column consisted of five pieces (m = 5): programming the
three applets (3), writing up the text in-between (1), and coming up with
this paragraph (1) to end the column with. It took me n = 6
hours to complete the task. It follows from Dirichlet's box principle that
one of the subtasks was bound to take at least ceil(6/5) = 2 hours. As a
matter of fact, the last one did.
References
- M. Aigner, G. Ziegler, Proofs from THE BOOK, Springer, 2000
- B. Averbach, O. Chein, Problem Solving Through Recreatinoal Mathematics, Dover, 2000
- E. Barbeau, M. Klamkin, W. Moser, Five Hundred Mathematical Challenges, MAA, 1995
- B. Epstein, All You Need Is Cards, in Puzzlers' Tribute, D. Wolfe, T. Rodgers (eds), A K Peters, 2002
- M. Gardner, The Last Recreation, Copernicus, 1997
- R. Graham, D. Knuth, O. Patashnik, Concrete Mathematics, 2nd edition, Addison-Wesley, 1994.
- R. Libeskind-Hadas, PWW: The Pigeonhole Principle, Math Magazine 75 (1) 2002, p. 39
- T. Rodgers, A Scrub Tile Puzzle, in Puzzlers' Tribute, D. Wolfe, T. Rodgers (eds), A K Peters, 2002
- D. Wolfe, T. Rodgers, Puzzlers' Tribute, A K Peters, 2002

Alex Bogomolny has started and still maintains a popular Web site Interactive Mathematics Miscellany and Puzzles
to which he brought more than 10 years of college instruction and, at least as much, programming experience. In March 2002 the site has welcomed its 4,000,000th visitor. Alex holds M.S. degree in Mathematics from the Moscow State University and Ph.D. in Applied Mathematics from the Hebrew University of Jerusalem. He can be reached at alexb@cut-the-knot.com
Copyright © 1996-2002 Alexander Bogomolny
|