COT 3100 Spring 2001
Homework #3 Solution
Assigned: Feb. 20/21
Due: March 7,8 in recitation
Problem #1
With
A
= {1, 2, 3, 4}, let
R
= {(1, 1), (1, 2), (2, 3), (3, 3), (3, 4), (4, 4)} be a
relation on
A
. Find two relations
S
and
T
on
A
where
S
≠
T
but
R
S
=
R
T
= {(1, 1), (1,
2), (1, 4)}.
Ans. One possible solution:
S
= {(1, 1), (1, 2), (2, 4)},
T
= {(2, 1), (2, 2), (1, 4)}.
We can get (1, 1) in
R
S
in the following ways: (1, 1)+(1, 1), (1, 2)+(2, 1)
We can get (1, 2) in
R
S
in the following ways: (1, 1)+(1, 2), (1, 2)+(2, 2)
We can get (1, 4) in
R
S
in the following ways: (1, 1)+(1, 4), (1, 2)+(2, 4)
So, any combination will do the job.
Problem #2.
If
A
={
w
,
x
,
y
,
z
}, determine the number of relations on
A
that are
(a) reflexive;
The total number of pairs we can either include or not is 4
2
-4=16. Any reflexive
relation is a subset of this set of 12 elements; we know there exist 2
12
such subsets.
(b) symmetric;
The number of decisions we can make for any symmetric relation is 4+(16-
4)/2=4+6=10.
The number od possible symmetric relations is 2
10.
(c) reflexive and symmetric;
Now we are free to make (16-4)/2=6 choices about couples of pairs from distinct
elements. The number of possible relations is 2
6
.
(d) reflexive and contain (
x
,
y
);
We must put 4 self-loops and (
x
,
y
), i. e. 5 out of total 16 pairs. We can make choices
about remaining 11 pairs. This allows to make 2
11
reflexive relations that include (
x
,
y
).
(e) symmetric and contain (
x
,
y
);
For a symmetric relation we have 4 choices for self-loops and 5 for coupled pairs of
distinct elements, i. e. 2
9
relations.
(f) anti-symmetric;
We can have 2
4
⋅
3
6
anti-symmetric relations. The reason is that we have 4 self-loops
that we can either place in the relation or not. The choices of these are independent
from choosing the other items in the relation. For the other items, you pair up each
ordered pair of the form (a,b) with (b,a). There are 6 such pairs. For each pair, you
may choose to NOT have either in the relation, OR to have only one of them in the
relation. This is three distinct choices for each pair. Invoke the multiplication
principle to get the final answer.