Math 15B
Final Exam Solutions
8 December 2000
1. (60 pts) In each case, give an example or explain why none exists.
(a) A tree with exactly nine vertices and exactly nine edges.
Ans
: Impossible since a tree always has one less edge than it has vertices.
(b) A permutation
f
on
{
1
,
2
,
3
,
4
}
such that
f
100
6
=
f
.
Notes
:
Remember that
f
6
=
g
for functions with the same domain means there
is at least one
x
such that
f
(
x
)
6
=
g
(
x
).
Also remember that
f
100
(
x
) means
f
(
f
(
· · ·
f
(
x
)
· · ·
)), NOT (
f
(
x
))
100
.
Ans
: The question is equivalent to asking that
f
99
not be the identity function. Such
a permutation must have at least one cycle whose length does not divide 9. Since
we are permuting a 4set, the possible cycle lengths are 1, 2, 3 and 4. Thus
f
must be either a 4cycle (there are six possibilities) or a product of two 2cycles
(there are three possibilities). Any of the nine possible answers is acceptable.
(c) A simple graph with exactly five vertices that has a cycle containing six edges.
Ans
: Impossible, if we had six edges, the cycle would pass through at least one vertex
twice.
(d) A sample space
U
with a probability function
P
and two different elements
s
and
t
of
U
such that
P
(
s
)
>
1
/
2 and also
P
(
t
)
>
1
/
2.
Note
:
P
(
s
) means the same thing as
P
(
{
s
}
).
Ans
: Impossible since
P
(
x
)
≥
0 and 1 =
∑
x
∈
U
P
(
x
)
≥
P
(
s
) +
P
(
t
).
(e) A sample space
U
with a probability function
P
and two different subsets
S
and
T
of
U
such that
P
(
S
)
>
1
/
2 and also
P
(
T
)
>
1
/
2.
Ans
: There
are
many
possibilities.
For
example,
the
uniform
distribution
on
U
=
{
1
,
2
,
3
}
with
S
=
{
1
,
2
}
and
T
=
{
1
,
2
,
3
}
.
(f) Two functions
f
(
n
) and
g
(
n
) such that “
f
(
n
) is
O
(
g
(
n
))” is TRUE and,
at the same time
, “
g
(
n
) is
O
(
f
(
n
))” is FALSE.
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 Fall '01
 Bender
 Math, Probability theory, Democrats

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