CS 173: Discrete Structures, Spring 2010
Homework 11 Solutions
This homework contains 3 problems worth a total of 42 regular points and 4 bonus points. It
is due on Friday, April 30th at 4pm. Put your homework in the appropriate dropbox in the
Siebel basement.
1.
Partial Orders [16 points]
(a) Suppose that
A
is the set of finite nonzero length
sequences of integers
. E.g. 5
,
2
is an element of
A
. So is 3
,
4
,
3
,
3, and so is 3
,
1
,

37
,
67
,
0
,
3. A generic element of
A
would be
x
1
, x
2
, . . . , x
n
where
n
≥
1 (since each sequence contains at least one
integer.) Let
S
be the “subsequence of” relation on
A
defined as follows:
x
1
, . . . , x
m
S y
1
, . . . , y
n
if and only if
m
≤
n
and there is an integer
k
such that
x
1
=
y
k
, x
2
=
y
k
+1
,
. . . , x
m
=
y
k
+
m

1
.
Prove that
S
is a partial order.
Solution:
To prove that
S
is a partial order, we show that it is reflexive, antisymmetric and tran
sitive:
i.
Reflexive:
x
1
, . . . , x
m
S x
1
, . . . , x
m
holds if we set
k
= 1
since
m
≤
m
and
x
1
=
x
k
=
x
1
, x
2
=
x
k
+1
=
x
2
, ... x
m
=
x
k
+
m

1
=
x
m
.
ii.
Antisymmetric:
Suppose
x
1
, . . . , x
m
S y
1
, . . . , y
n
and
y
1
, . . . , y
n
S x
1
, . . . , x
m
. If
we take the first relation,
m
≤
n
(by the definition of
S
) and there must be an
integer
k
1
such that:
x
1
=
y
k
1
, x
2
=
y
k
1
+1
,
. . . , x
m
=
y
k
1
+
m

1
.
For the second relation,
n
≤
m
and there must be an integer
k
2
such that:
y
1
=
x
k
2
, y
2
=
x
k
2
+1
,
. . . y
n
=
x
k
2
+
n

1
.
By the antisymmetric property of the
≤
operator,
m
≤
n
and
n
≤
m
must imply
that
n
=
m
. If we substitute
m
for
n
in the second relation, we get:
y
1
=
x
k
2
, y
2
=
x
k
2
+1
,
. . . y
n
=
x
k
2
+
m

1
.
Therefore,
k
1
=
k
2
= 1
, which implies:
x
1
=
y
1
, x
2
=
y
2
,
. . . , x
m
=
y
m
.
iii.
Transitive:
Suppose
x
1
, . . . , x
m
S y
1
, . . . , y
n
and
y
1
, . . . , y
n
S z
1
, . . . , z
p
such that
m
≤
n
≤
p
. We want to prove that
x
1
, . . . , x
m
S z
1
, . . . , z
p
.
Note that
m
≤
n
≤
p
implies that
m
≤
p
because
≤
is a transtive operator.
Let
k
1
and
k
2
be two integers such that:
x
1
=
y
k
1
, x
2
=
y
k
1
+1
, . . . , x
m
=
y
k
1
+
m

1
and
y
1
=
z
k
2
, y
2
=
z
k
2
+1
, . . . , y
n
=
z
k
2
+
n

1
We can deduce the
k
1
th element in the
y
sequence (by substituting
k
1
for
n
) is
y
k
1
=
z
k
2
+
k
1

1
. Thus,
x
1
=
z
k
2
+
k
1

1
, x
2
=
z
k
2
+
k
1
, . . . , x
m
=
z
k
2
+
k
1
+
m

2
Therefore,
x
1
, x
2
, . . . x
m
S z
1
, z
2
, . . . z
p
because
m
≤
p
, which implies that
S
is transitive.
1
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(b) Suppose you are given two sets,
X
and
Y
with partial orders
X
and
Y
. We define
a new relation
X
×
Y
on
X
×
Y
(the Cartesian product of sets X and Y) such that:
(
a, b
)
X
×
Y
(
c, d
) if and only if the following is true:
i.
a
6
=
c
and
a
X
c
or
ii.
a
=
c
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 Spring '08
 Erickson
 Equivalence relation, Transitive relation, relation, partial order

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