�
6.042/18.062J
Mathematics
for
Computer
Science
October
5,
2010
Tom
Leighton
and
Marten
van
Dijk
Notes
for
Recitation
8
1
Buildup
error
Recall
a
graph
is
connected
iff
there
is
a
path
between
every
pair
of
its
vertices.
False
Claim.
If
every
vertex
in
a
graph
has
positive
degree,
then
the
graph
is
connected.
1.
Prove
that
this
Claim
is
indeed
false
by
providing
a
counterexample.
Solution.
There
are
many
counterexamples;
here
is
one:
2.
Since
the
Claim
is
false,
there
must
be
a
logical
mistake
in
the
following
bogus
proof.
Pinpoint
the
first
logical
mistake
(unjustified
step)
in
the
proof.
Proof.
We
prove
the
Claim
above
by
induction.
Let
P
(
n
)
be
the
proposition
that
if
every
vertex
in
an
n
vertex
graph
has
positive
degree,
then
the
graph
is
connected.
Base
cases
: (
n
≤
2).
In
a
graph
with
1
vertex,
that
vertex
cannot
have
positive
degree,
so
P
(1)
holds
vacuously.
P
(2)
holds
because
there
is
only
one
graph
with
two
vertices
of
positive
degree,
namely,
the
graph
with
an
edge
between
the
vertices,
and
this
graph
is
connected.
Inductive
step
:
We
must
show
that
P
(
n
)
implies
P
(
n
+
1)
for
all
n
≥
2.
Consider
an
n
vertex
graph
in
which
every
vertex
has
positive
degree.
By
the
assumption
P
(
n
),
this
graph
is
connected;
that
is,
there
is
a
path
between
every
pair
of
vertices.
Now
we
add
one
more
vertex
x
to
obtain
an
(
n
+
1)vertex
graph:
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2
Recitation
8
z
y
X
n  vertex graph
All
that
remains
is
to
check
that
there
is
a
path
from
x
to
every
other
vertex
z
.
Since
x
has
positive
degree,
there
is
an
edge
from
x
to
some
other
vertex,
y
.
Thus,
we
can
obtain
a
path
from
x
to
z
by
going
from
x
to
y
and
then
following
the
path
from
y
to
z
.
This
proves
P
(
n
+
1).
By
the
principle
of
induction,
P
(
n
)
is
true
for
all
n
≥
0,
which
proves
the
Claim.
Solution.
This
one
is
tricky:
the
proof
is
actually
a
good
proof
of
something
else.
The
first
error
in
the
proof
is
only
in
the
final
statement
of
the
inductive
step:
“This
proves
P
(
n
+
1)”.
The
issue
is
that
to
prove
P
(
n
+
1),
every
(
n
+
1)vertex
positivedegree
graph
must
be
shown
to
be
connected.
But
the
proof
doesn’t
show
this.
Instead,
it
shows
that
every
(
n
+
1)vertex
positivedegree
graph
that
can
be
built
up
by
adding
a
vertex
of
positive
degree
to
an
n
vertex
connected
graph
,
is
connected.
The
problem
is
that
not
every
(
n
+
1)vertex
positivedegree
graph
can
be
built
up
in
this
way.
The
counterexample
above
illustrates
this:
there
is
no
way
to
build
that
4vertex
positivedegree
graph
from
a
3vertex
positivedegree
graph.
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 Fall '10
 TomLeighton,Dr.MartenvanDijk
 Computer Science, Graph Theory, Mathematical logic, Planar graph, Glossary of graph theory

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