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The Towers of Hanoi
In the great temple of Brahma in Benares, India, on a brass
plate under the dome that marks the center of the world, there
are 64 disks of pure gold that the priests carry one at a time be-
tween three diamond needles according to Brahma’s immutable
law: No disk may be placed on a smaller disk. In the beginning
of the world all 64 disks formed the Tower of Brahma on one
needle. Now, however, the process of transfer of the tower from
one needle to another is in mid course. When the last disk is
finally in place, once again forming the Tower of Brahma but
on a different needle, then will come the end of the world and
all will turn to dust.
Question? Suppose n disks are to be moved. How many
moves are necessary?
1
A bit of logic
To prove a proposition like “If P , then Q”, abbreviated “P ⇒Q” there are two options: prove the proposition directly, or
prove the logically equivalent contrapositive of the proposi-
tion ¬Q⇒ ¬P .
In the second method, called proof by contradiction, one
assumes that Q is false and shows that that assumption leads
to a contradiction of P .
Proposition. The square root of 2 is irrational.
The Pigeonhole Principle: If k + 1 or more pigeons are
placed into k pigeonholes, then there must be one pigeonhole
containing two or more pigeons.
Finally, we will need a bit of number theory:
Proposition. If two numbers have the same remainder when
divided by c, then their difference is divisible by c.
Problem. A group of 30 students wrote a dictation. John
Bull made 13 errors, and each of the rest made fewer than 13
errors. Prove that at least 3 students made the same number
of errors.
2
Problem. Prove that, given any 12 distinct natural numbers,
we can choose two of them such that their difference is divisible
by 11.
Problem. Given a set S of 10 distinct numbers between 1
and 100, inclusive, there exist two distinct, disjoint subsets A
and B of S whose elements sum to the same number.
3
Modular arithmetic
Modular arithmetic was developed by
Carl Friedrich Gauss and published
in his Disquisitiones Arithmeticae in
1801 when he was 21. It offers a con-
venient approach to questions of divis-
ibility.
Carl Friedrich Gauss
(1777 – 1855)
Definition: Let m be a fixed integer. For integers a and b we
say that a is congruent to b modulo m and write
a ≡ b (modm)
whenever m|(a− b). If m 6 |(a− b), we write a 6≡ b (modm).
Example. Clocks and modular arithmetic.
4
Proposition. Let m be a fixed integer. Let a, b, and c be
integers. Then
(i) a ≡ a (modm).
(ii) If a ≡ b (modm), then b ≡ a (modm).
(iii) If a ≡ b (modm) and b ≡ c (modm), then a ≡ c (modm).
The properties enumerated in the Proposition are called, re-
flexive, symmetric, and transitive, respectively.
Proposition. If a ≡ a′, (modm) and b ≡ b′ (modm), then
(i) a + b ≡ a′ + b′ (modm)
(ii) a− b ≡ a′ − b′ (modm)
(iii) a · b ≡ a′ · b′ (modm)
Question? So addition, subtraction, and multiplication be-
have nicely! How about division? The following example illus-
trates the difficulty with division.
5
Example. Note that 28 ≡ 10 (mod 6). Also, 2 divides both
28 and 10 so we might expect to divide by 2 and have 14 be
equivalent to 5 modulo 6.
But 14 6≡ 5 (mod 6).
What happened? Why can’t we “cancel” the 2 in this case?
The answer requires a couple of definitions. . .
The greatest common divisor of two integers a and b
is the largest integer d such that d|a and d|b. The greatest
common divisor of two integers is denoted by gcd(a, b).
Integers a and b are said to be relatively prime if gcd(a, b) =
1.
This allows an answer to when we can “cancel” in modular
arithmetic. . .
Proposition. If ac ≡ bc (modm) and gcd(c,m) = 1, then
a ≡ b (modm).
So we should not have expected 14 to be equivalent to 5 modulo
6 since gcd(2, 6) = 2.
Example. 20 · 2 ≡ 9 · 2 (mod 11) and therefore 20 ≡ 9
(mod 11) since gcd(11, 2) = 1.
So we can cancel a constant in a modular equivalence only when
the constant and the modulus are relatively prime.
6
Proposition. a ≡ b (modm) if and only if a and b have
the same remainder when divided by m.
Proposition. An integer n is divisible by 9 if and only if the
sum of its digits is divisible by 9.
Example. Determine all solutions to 8x + 12y = b where x
and y are positive integers and 75 < b < 80.
Example. Determine the remainder when 237 is divided by7.
7
Example. Divisibility by 7
To test an integer for divisibility by 7, subtract twice the last
digit from the other digits repeatedly until a single digit number
is obtained. If the number obtained is 0 or 7 then the integer is
divisible by 7.
Proposition. The test for divisibility by 7 described above is
correct.
Proof. Let m = an10n + an−110n−1 + · · · + a110 + a0 be the
number to test. We need to show m ≡ 0 (mod 7) and
an10n−1 + an−110n−2 + · · · + a1 − 2a0 ≡ 0 (mod 7)
are equivalent. Multiplying the second equation by 10 yields
an10n + an−110n−1 + · · · + a110− 20a0 ≡ 0 (mod 7),
and adding 21a0, which is zero modulo 7, yields
an10n + an−110n−1 + · · · + a110 + 21a0 − 20a0 ≡ 0 (mod 7).
But the left side is just m so
m = an10n + an−110n−1 + · · · + a110 + a0 ≡ 0 (mod 7)
The equivalence established here for one step in the processis maintained throughout the steps needed and thus the testworks. �
Proof suggested by Nathan Vallapureddy, PGSS 2016.
8
For example, consider 5,355, 22,968, and 103,224:
535 2,296 10,322
-10 -16 -8
525 2,280 10,314
52 228 1,031
-10 -0 -8
42 228 1,023
4 22 102
-4 -16 -6
0 6 96
9
-12
-3
Since
5, 355 = 32·5·7·17, 22, 968 = 23·32·11·29, and 103, 224 = 2·3·4·11·17·23,
the test is three for three.
9
Introduction to Graph Theory
A graph is a set of vertices and a set of edges such that each
edge is associated with an (unordered) pair of vertices.
A graph is simple if it contains at most one edge between any
pair of vertices and no loops, i.e., no edges that start and end
on the same vertex.
A path is a sequence of vertices such that consecutive vertices
are adjacent via an edge and no edge is used twice.
A graph is connected if there is a path joining each pair of
distinct vertices.
A path is Eulerian if it uses every edge in the graph exactly
once.
A circuit is a path that returns to its starting point.
The degree of a vertex is the number of edges incident on the
vertex.
10
The Konigsberg Bridge Problem
The river Pregel flows through Konigsberg, and in Euler’s day
there were seven bridges connecting the North and South shores
and the two islands in the river. A popular puzzle in the town
was whether or not it was possible to walk in such a way as to
cross each bridge exactly once.
The Bridges of Konigsberg
Leonhard Euler
(1707 – 1783)
Euler’s Theorem. A graph has an Euler path if and only if
. . .
11
A graph is planar if it can be drawn in the plane without its
edges crossing.
A graph is complete if every pair of distinct vertices is con-
nected by an edge. The complete graph with n vertices is
denoted by Kn.
Example. K4 is planar while K5 is not.
Utilities Puzzle
Connect all of the circles with all of the squares without arcs
crossing.
Unsolved problem. The question of the the minimum num-
ber of edge crossings required to draw Kn is unsolved for most
cases where n ≥ 13.
A graph is bipartite if it includes two disjoint sets of vertices
such that all edges connect a vertex in one of the sets with a
vertex in the other.
12
Kn,m denotes the bipartite graph that includes all possible
edges joining a set of n vertices with a set of m vertices.
A graph H is a subgraph of a graph G if every vertex and
every edge of H is also a vertex or an edge of G.
Kuratowski’s Theorem. A graph is planar if and only if it
does not contain K5 or K3,3 as a subgraph.
Two graphs are isomorphic if (ignoring their vertex labels)
one can be re-drawn to look like the other.
Thus, unlike a one-to-one function that establishes an isomor-
phism for sets, the one-to-one correspondence between the ver-
tices of isomorphic graphs must be accompanied by a one-to-one
correspondence between their edges in such a way that the in-
cidence relationships are preserved.
Question? So who cares if a graph is planar, or at least, close
to planar?
13
In addition to the number of edges, e, and the number of ver-
tices, v, for a planar graph there is also the number f of faces.
When a planar graph is drawn without edge intersections, the
plane is divided into contiguous regions called faces. There is
also an unbounded face consisting of the region lying entirely
outside the graph.
Theorem. (Euler’s Formula) In a connected, planar
graph
e− v + 2 = f,
i.e., the number of edges minus the number of ver-
tices plus two equals the number of faces.
Proof. Use induction on the number of edges . . .
Theorem 6.1.7. In any graph G, the sum of the
degrees of the vertices is twice the number of edges.
Corollary 6.1.8. In any graph G, the number of
nodes with odd degree is even.
14
Question? Can there exist a hydrocarbon with five
carbon atoms and three hydrogen atoms?
Two hydrocarbons
•
• • •
•
H
H
H HC
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.
..........................................................................................................
....................................................................................................................................................................................................................
METHANE
CH4
• •
• •
• •
C
H H
H H
C
..........................................................................................................
..........................................................................................................
..........................................................................................................
..........................................................................................................
ETHYLENE
C2H4
..........................
............................................................................................................................................
...............................................................................................................................................
.......................
What is the degree of each hydrogen atom?
What is the degree of each carbon atom?
Question? How many edges are there in Kn? How
about Kn,m?
Proposition. C(n + m, 2) = C(n, 2) + C(m, 2) + nm.
15
Question? (The Party Problem.) Six guests are
invited to a party, and each pair of guests are ei-
ther friends or strangers. Is it true that among the
guests there are always three who are all friends or
three who are all strangers?
Example. Provide an example of a graph for a party
with five guests without a set of three who are all
friends or three who are all strangers.
Ramsey’s Theorem. For any pair of positive inte-
gers (r, s), there exists a least positive integer R(r, s)
such that for any complete graph on R(r, s) vertices,
whose edges are coloured red or blue, there exists
either a complete subgraph on r vertices which is
entirely blue, or a complete subgraph on s vertices
which is entirely red.
Unsolved problems. The Ramsey number R(r, s) is
known for small values of r and s, and there are
estimates for others, for example 40 ≤ R(10, 3) ≤ 42,
or 205 ≤ R(7, 7) ≤ 540. And 43 ≤ R(5, 5) ≤ 49. But the
search for exact values is full of open problems.
16
Just how hard are these problems? Here is a quote
from Paul Erdos:
“Imagine an alien force, vastly more powerful than
us landing on Earth and demanding the value of
R(5, 5) or they will destroy our planet. In that case,
we should marshal all our computers and all our
mathematicians and attempt to find the value. But
suppose, instead, that they asked for R(6, 6), we
should attempt to destroy the aliens.”
17