1.
[5 marks]
(a) Determine
[
x
6
]
x
2
(1 + 3
x
2
)

2
.
Solution:
[
x
6
]
x
2
(1 + 3
x
2
)

2
= [
x
4
](1 + 3
x
2
)

2
= [
x
4
]
∞
X
j
=0
±
2 +
j

1
2

1
¶
3
j
x
2
j
= [
x
4
]
∞
X
j
=0
(
j
+ 1)3
j
x
2
j
= 27
.
(b) Prove that
X
even
j
±
n
j
¶
2
j
=
X
odd
j
±
n
j
¶
2
j
+ (

1)
n
where the summations are over even and odd integers
j
satisfying 0
≤
j
≤
n
.
(Hint: Consider (1

2
x
)
n
. )
Solution:
By the binomial theorem, we know that
(1

2
x
)
n
=
n
X
j
=0
(

2)
j
x
j
.
Setting
x
= 1 gives
(

1)
n
=
n
X
j
=0
(

2)
j
and moving the odd
j
values to the left hand side gives
X
odd
j
2
j
+ (

1)
n
=
X
even
j
2
j
as required.
2.
[4 marks]
For each of the following sets, ﬁnd a decomposition that uniquely creates the elements of that set.
(a) The
{
0
,
1
}
strings that have no blocks of 1s with length 3 or more, and no blocks of 0s of length 2.
Solution:
We can start with a block decomposition that creates all
{
0
,
1
}
strings,
{
1
}
*
(
{
0
}{
0
}
*
{
1
}{
1
}
*
)
*
{
0
}
*
and modify it to exclude blocks of 1s with length 3 or more, and blocks of 0s of length 2. This gives us
{
²,
1
,
11
}
(
{
0
,
000
,
0000
,...
}{
1
,
11
}
)
*
{
²,
0
,
000
,
0000
,...
}
.
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View Full Document(b) The set of
{
0
,
1
}
strings in which the substring 01110 does not occur.
Solution:
We start with the block decomposition
{
1
}
*
(
{
0
}{
0
}
*
{
1
}{
1
}
*
)
*
{
0
}
*
. In the
(
{
0
}{
0
}
*
{
1
}{
1
}
*
)
*
part, we
can only allow a 111 block if there are no 0’s following it. So we can write
{
1
}
*
(
{
0
}{
0
}
*
{
1
,
11
,
1111
,
11111
,...
}
)
*
(
{
0
}{
0
}
*
{
111
} ∪ {
0
}
*
)
.
Alternatively, we could start with the 0decomposition and in the
(
{
0
}{
1
}
*
)
part omit the string 0111
unless it occurs at the very end of the string. This gives us
{
1
}
*
(
{
0
}{
²,
1
,
11
,
1111
,...
}
)
*
(
{
²
} ∪ {
0
}{
111
}
)
.
3.
[11 marks]
(a) For the following decomposition, prove that the strings are not uniquely created (by ﬁnding a string
that can be created at least two diﬀerent ways).
{
0
}
*
(
{
10
,
1
}{
00
,
01
}
)
*
{
1
}
*
Solution:
e.g. 10 01 = 1 00 1 can be created in two diﬀerent ways.
(b) Let
S
=
{
1
}
*
(
{
0
}{
111
}
*
)
*
. Prove that this decomposition creates the elements of
S
uniquely.
Solution:
A string
σ
with
k
zeroes must come from the
{
1
}
*
(
{
0
}{
111
}
*
)
k
part of
S
. Any 1s preceding any zeroes
(if there are any) must come from the initial
{
1
}
*
part, and any 1s between the
j
th and (
j
+1)th 0 must
come from the
{
111
}
*
in the
j
th repetition of
{
0
}{
111
}
*
in the
(
{
0
}{
111
}
*
)
k
part. Thus any string
σ
∈
S
can be created in only one way.
Solution:
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 Spring '09
 M.PEI
 Combinatorics

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