CS0441 Discrete Structures Recitation 8 Xiang Xiao
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CS0441 Discrete Structures Recitation 8
Xiang Xiao
Transitive Closure
Transitive Closure of R:
The transitive closure of R is the smallest transitive relation that contains R. It is a subset of every transitive relation containing R.
Finding the transitive closure of R:
Algorithm 1 (P. 603): “The transitive closure of a relation R equals the connectivity relation R*”
* 2 3
If R is a relation on set A with n elements
nR R R R R
*
[2] [3] [ ]n
R R R RRM M M M M
P607 Q26
Use Algorithm 1 to find the transitive closure of these relations on {a, b, c, d, e}.
{(a, c), (b, d), (c, a), (d, b), (e, d)}
Goal: 𝑀𝑅∗ = 𝑀𝑅 ∨ 𝑀𝑅2∨ 𝑀𝑅
3∨⋅⋅⋅∨ 𝑀𝑅
𝑛
0 0 1 0 0
0 0 0 1 0
1 0 0 0 0
0 1 0 0 0
0 0 0 1 0
RM
Solution
𝑀𝑅 =
0 0 1 0 00 0 0 1 01 0 0 0 00 1 0 0 00 0 0 1 0
[2]
0 0 1 0 0 0 0 1 0 0 1 0 0 0 0
0 0 0 1 0 0 0 0 1 0 0 1 0
1 0 0 0 0 1 0 0 0 0
0 1 0 0 0 0 1 0 0 0
0 0 0 1 0 0 0 0 1 0
R R RM M M
0 0
0 0 1 0 0
0 0 0 1 0
0 1 0 0 0
[3] [2]
1 0 0 0 0 0 0 1 0 0 0 0 1 0 0
0 1 0 0 0 0 0 0 1 0 0 0 0
0 0 1 0 0 1 0 0 0 0
0 0 0 1 0 0 1 0 0 0
0 1 0 0 0 0 0 0 1 0
R R RM M M
1 0
1 0 0 0 0
0 1 0 0 0
0 0 0 1 0
Solution
[4] [3]
0 0 1 0 0 0 0 1 0 0 1 0 0 0 0
0 0 0 1 0 0 0 0 1 0 0 1 0
1 0 0 0 0 1 0 0 0 0
0 1 0 0 0 0 1 0 0 0
0 0 0 1 0 0 0 0 1 0
R R RM M M
0 0
0 0 1 0 0
0 0 0 1 0
0 1 0 0 0
[5] [4]
1 0 0 0 0 0 0 1 0 0 0 0 1 0 0
0 1 0 0 0 0 0 0 1 0 0 0 0
0 0 1 0 0 1 0 0 0 0
0 0 0 1 0 0 1 0 0 0
0 1 0 0 0 0 0 0 1 0
R R RM M M
1 0
1 0 0 0 0
0 1 0 0 0
0 0 0 1 0
* [1] [2] [3] [4] [5]
1 0 1 0 0
0 1 0 1 0
1 0 1 0 0
0 1 0 1 0
0 1 0 1 0
R R R R R R RM M M M M M M
Transitive Closure
Transitive Closure of R:
The transitive closure of R is the smallest transitive relation that contains R. It is a subset of every transitive relation containing R.
Finding the transitive closure of R:
Algorithm 1 (P. 603):
Warshall’s algorithm
*
[2] [3] [ ]n
R R R RRM M M M M
[ ][ ]
is the matrix of the transitive closure
k
k ij
n
W w
W
P607 Q28
Use Warshall’s algorithm to find the transitive closure of these relations on {a, b, c, d, e}.
{(a, c), (b, d), (c, a), (d, b), (e, d)}
Let v1 = a, v2 = b, v3 = c, v4 = d, v5 = e.
W0 is the matrix of the relation. Hence
0
0 0 1 0 0
0 0 0 1 0
1 0 0 0 0
0 1 0 0 0
0 0 0 1 0
RW M
Solution
𝑊0 =
0 0 1 0 00 0 0 1 01 0 0 0 00 1 0 0 00 0 0 1 0
𝑊1 =
0 0 1 0 00 0 0 1 01 0 1 0 00 1 0 0 00 0 0 1 0
𝑊2 =
0 0 1 0 00 0 0 1 01 0 1 0 00 1 0 1 00 0 0 1 0
wij = 1 if there is a path from vi to vj that has only v1 = a as an interior vertex.
wij = 1 if there is a path from vi to vj that has v1 = a and/or v2 = b as an interior vertex.
Solution
wij = 1 if there is a path from vi to vj that has v1 = a, v2 = b and/or v3 = c as an interior vertex. 𝑊3 =
1 0 1 0 00 0 0 1 01 0 1 0 00 1 0 1 00 0 0 1 0
wij = 1 if there is a path from vi to vj that has v1 = a, v2 = b, v3 = c and/or v4 = d as an interior vertex.
𝑊4 =
1 0 1 0 00 1 0 1 01 0 1 0 00 1 0 1 00 1 0 1 0
wij = 1 if there is a path from vi to vj that has v1 = a, v2 = b, v3 = c, v4 = d and/or v5 = e as an interior vertex.
𝑊5 =
1 0 1 0 00 1 0 1 01 0 1 0 00 1 0 1 00 1 0 1 0
Equivalence relations
Equivalence relations:
The relation is reflexive, symmetric, and transitive.
Which of these relations are equivalence relations?
a. {(a, b) | a and b have the same parents}
b. {(a, b) | a and b share a common parent}
c. {(a, b) | a and b have met}
Yes
No, not transitive.
No, not transitive.
P615 Q16 Let R be the relation on the set of ordered pairs of positive integers such that ((a, b), (c, d)) ϵ R if and only if ad = bc. Show that R is an equivalence relation.
Proof:
1. Show R is reflexive: For every two positive integers a and b, ab = ba;
Therefore ((a, b), (a, b)) ϵ R. Hence R is reflexive.
2. Show R is symmetric: For every four positive integers a, b, c, d, if ((a, b), (c, d) ϵ R, then ad = bc.
So cb = da. Therefore ((c, d), (a, b) ϵ R. Hence R is symmetric.
3. Show R is transitive: For every six positive integers a, b, c, d, e, f. if ((a, b), (c, d)) ϵ R and
((c, d), (e, f) ϵ R, then it follows that ad = bc and cf = ed.
Therefore, acdf = bcde. remove cd from both sides, we have af = be.
If af = be, then ((a, b), (e, f) ϵ R. Hence R is transitive.
P616 Q40
• What is the equivalence class of (1, 2) with respect to the equivalence relation in the previous question?
Goal: find the set of all pairs that are related to (1,2)
Suppose a pair (a, b) is related to (1, 2), ((1, 2), (a, b)) ϵ R.
Therefore, b = 2a.
Hence, [(1, 2)] = {(1, 2), (2, 4), (3, 6), …… } = {(a, b) | b = 2a, a ϵ Z+}
P616 Q40
• Give an interpretation of the equivalence classes for the equivalence relation R in the previous question. [Hint: look at the ratio a/b corresponding to (a, b).]
Goal: for every element (a, b) ϵ R, a, b ϵ Z+, find all pairs that are related to it.
Suppose a pair (c, d) is related to (a, b), ((a, b), (c, d)) ϵ R.
Therefore, ad = bc. It follows that a/b = c/d.
Therefore, for every element (a, b) ϵ R, its equivalence class can be interpreted as a positive rational number a/b.
Each equivalence class can be interpreted as an rational number.
What is the equivalence classes of the equivalence relation {(0, 0), (1, 1), (1, 2), (2, 1), (2, 2), (3, 3)}
[0] = {0}
[1] = {1, 2}
[2] = {1, 2}
[3] = {3}
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