Topic 7 - Relation Student should be able to know how
Transcription
Topic 7 - Relation Student should be able to know how
6/4/2015 Learning Outcomes Student should be able to know how relation is being builds. Students should be able to apply relation and combine the knowledge with other topics in this subject. Students should be able to use relation in daily lives. Topic 7 - Relation 2 5. Equivalence relation 1. Basics of relation Domain and Range (Clear / Not Clear) Complement and Inverse 2. Relation Representation Set representation Matrix representation Digraph representation (Clear / Not Clear) 6. Relations Manipulation Intersection and union (Clear / Not Clear) (Clear / Not Clear) (Clear / Not Clear) (Clear / Not Clear) (Clear / Not Clear) 7. Transitive Closure Matrix product (Clear / Not Clear) (Clear / Not Clear) Warshall Algorithm 3. Concept of degree, path and cycle (Clear / Not Clear) 4. Properties of relations and how to identify it Reflexive and Irreflexive (Clear / Not Clear) Symmetric, Asymmetric and Antisymmetric (Clear / Not Clear) 3 An ordered pair (a, b) is a listing of object a and b in which a and b will appear in dedicated order; a appearing first, followed by b. If two sets A and B are nonempty set, its product set (A x B) or named Cartesian product is a set for ALL (a, b) pairs in which a A and b B. 4 Example 1: Let A = {1, 2, 3} and B = {r, s} So, A x B = {(1, r), (1, s), (2, r), (2, s), (3, r), (3, s)} Alternatively, we can arrange the element of A and B in a tabular array as follows: Written as A x B = {(a, b) | a A and b B} B A 1 r s (1,r) (1,s) 2 (2,r) (2,s) 3 (3,r) (3,s) 1 6/4/2015 Example 2: Let A = {1, 2, 3} and B = {r, s} So if A x B = {(1, r), (1, s), (2, r), (2, s), (3, r), (3, s)} and B x A = {(r, 1), (r, 2), (r, 3), (s, 1), (s, 2), (s, 3)} Example 3: A marketing research firm classifies a person according to the following two criteria: Gender: male (m); female (f) Highest level of education completed: elementary school (e); high school (h); college (c); graduate school (g) Therefore, A x B B x A and |A x B| = |A| x |B| Find all possible categories that classify a person based on the gender and highest level of education completed. Solution to Example 3: Example 4: Let S = {m, f} and L = {e, h, c, g}. The product set S x L A manufacturer offers the following options for its refrigerators: contains all the categories into which the population classified. There are eight categories in this classification scheme.--> {(m, e), (m, h), (m, c), (m, g), (f, e), (f, h), (f, c), (f, g)}. Thus the classification (f, g) represents a female who completed graduate school. Doors: side-by-side (s), over-under (o), three (t) Icemaker: freezer (f), door (d) Finish: standard (r), metallic (m), custom (c) Calculate the number of categories that describe refrigerator options. Solution: |Doors| x |Icemaker| x |Finish| = 3 x 2 x 3 = 18 categories. A relation between two sets where its value was determined by a condition. All elements in that relation must fulfill the condition. Let A and B be nonempty set. A relation R from A to B is a subset of A x B. If R A x B and (a, b) R, we say that a relates to b through R and we write as a R b. Example 6: Let A = {1, 2, 3, 4, 5}. Define the following relation R (less than) on A. Solution: R is a relation ‘less than’ on set A. Therefore, a R b if and only if a < b So R = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)} Example 5: Let A = {1, 2, 3} and B = {r, s}. Then R = {(1, r), (2, s), (3, r)} is a relation from A to B. Note: We only list down the subsets of A x A that satisfy relation ‘less than’ 2 6/4/2015 We can also tabulate the elements of A to identify the subsets Example 7: An airline services the five cities c1, c2, c3, c4 and that satisfy the relation R ‘less than’. c5. Table below gives the cost (in RM)of going from ci to cj. R = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, To 5), (4, 5)} From c1 C1 1 1 c2 c3 c4 c5 140 100 150 200 2 3 4 5 C2 190 (1,2) (1,3) (1,4) (1,5) C3 110 180 (2,4) (2,5) C4 190 200 120 (3,4) (3,5) C5 200 100 200 2 (2,3) 3 4 (4,5) 200 160 220 190 250 150 150 We now define the following relation R on the set of cities A = {c1, c2, c3, c4, c5}: ci R cj if and only if the cost of going from ci to cj is defined and less than or equal to RM180. Find R. 5 13 Solution: The relation R is the subset of A x A consisting of all cities (ci, cj), where the cost going from ci to cj is less than or equal to RM180. Hence: R = {(c1, c2), (c1, c3), (c1, c4), (c2, c4), (c3, c1), (c3, c2), (c4, c3), (c4, c5), (c5, c2), (c5, c4)} To From C1 C1 C2 190 C3 110 C2 C3 C4 C5 100 150 200 200 160 220 190 250 C4 190 200 120 C5 200 100 200 domain and range for the relation. Domain for R is elements in set A. That is, Dom(R) is a subset for A, is the set of all first element in the pairs that make up R 140 180 When there is a relation from A to B, we can find the Range for R is elements in set B. That is, Ran(R) is a subset for B, is the set of element in B that are second elements of pairs in R. 150 150 Example 8 A relation can be represented in three ways: Consider Example 6. Let A = {1, 2, 3, 4, 5}. Define the 1. following relation R (less than) on A. So R = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)} Solution Dom(R) is {1, 2, 3, 4} Ran(R) is {2, 3, 4, 5} Set Example: R = {(1, r), (2, s), (3, r)} 2. Matrix Matrix representation is only for two finite sets. First element in the pair 3. Digraph A geometrical/pictorial representation of relations Second element in the pair 3 6/4/2015 Example 9: Let R be the relation defined in example 5. Example 10: Consider the matrix MR = 1 0 0 1 Let A = {1, 2, 3} and B = {r, s} and R = {(1, r), (2, s), (3, r)} is a relation from A to B. 0 1 1 0 Then the matrix of R is MR = 1 0 1 0 1 0 1 0 1 if (ai, bj) R mij = And since M is a 3 x 4 matrix, let: 0 if (ai, bj) R 0 1 B r A 1 A = {a1, a2, a3} and B = {b1, b2, b3, b4}. Question: Find R. s Solution: (1,r) 2 (2,s) 3 Based on MR: (3,r) R = {(a1, b1), (a1, b4), (a2, b2), (a2, b3). (a3, b1), (a3, b3)} In example 10, how do we get Consider a relation R from a finite set A to a finite set B where A = {1, 2, 3, 4} and B = {x, y, z} and R = {(1,y), (1,z), (3,y), (4,x), (4,z)} R = {(a1, b1), (a1, b4), (a2, b2), (a2, b3). (a3, b1), (a3, b3)}? Build the array tabulation of A and B. Then write value 1 to Construct MR the cell which satisfy MR. MR = 1 0 0 1 0 1 1 0 1 0 1 0 A = {a1, a2, a3} The number of ‘1’ in the matrix is equal to the number of ordered pair in R B A b1 a1 1 b3 1 1 b4 1 a2 a3 b2 1 1 B = {b1, b2, b3, b4} R = {(a1, b1), (a1, b4), (a2, b2), (a2, b3). (a3, b1), (a3, b3)} 22 3. Digraph It is normally called a directed graph or digraph of R. Each How to draw the digraph in example 11? 1. Draw a small circle for each element of A and label the circle with the corresponding element of A. These circles are called vertices. 2. Draw an arrow, called an edge, from ai to aj, if and only if ai R aj. Example 11: A = {a, b, c, d} Given A = {a, b, c, d} and R = {(a, a), (a, b), (b, a), (b, b) (b, c), (b, d), (c, d), (d, a)} So, the diagraph of R is shown below: R = {(a, a), (a, b), (b, a), (b, b) (b, c), (b, d), (c, d), (d, a)} Digraph is a geometrical/pictorial representation of relations. Works on finite set, and is used only for relation on one set. element is called vertex and the path is called edge. The diagraph of R: b b a a c d c The arrow represents ‘relation’ d 23 24 4 6/4/2015 This is an important concept in relation inspired by the visual form of digraph. For diagraph in example 11, its degrees are: If R is a relation on set A and a A, then in-degree for a is b the number of b A where (b, a) R. Out-degree for a is the number of b A where (a, b) R. a c Note: d The in-degree of a vertex is the number of edges terminating (going in) the vertex. The out-degree of a vertex is the number of edges leaving (going out) the vertex. visualise To read in-degree and out-degree from matrix. Example 12: Let A = {a, b, c, d} and R is given below: MR = 1 0 0 0 Vertex a b c In-degree 3 2 1 2 Out-degree 2 4 1 1 Get the in degree and out degree for diagraph below. Then, write the relation into set and matrix. Read the in degree and out degree from matrix. a b c d a 1 0 0 0 0 1 0 0 b 0 1 0 0 1 1 1 0 c 1 1 1 0 0 1 0 1 d 0 1 0 1 2 1 3 Note: In-degree is in the column, while out-degree is in the 4 row. The answer: d Element a b c d In-degree 2 3 1 1 Out-degree 1 1 3 2 Let say R is a relation on set A. A path with length n in R from a to b is a finite sequence, = a, x1, x2,.... xn-1 , b, which start at a and ends at b in which a R x1, x1 R x2 , …., xn-1 R b Example 13: Consider this digraph: 1 2 3 5 4 1: 1, 2, 5, 4, 3 is a path length 4 from vertex 1 to vertex 3. 2: 1, 2, 5, 1 is a path length 3 from vertex 1 to itself. 3: 2, 2 is a path length 1 from vertex 2 to itself. 4: 2, 5, 1, 2 is a path length 3 from vertex 2 to itself. (visualize) Also, path length n will involve n + 1 elements of A, although they are not necessarily distinct. 1 1 2 2 5 4 1 1 2 5 4 3 3 3 5 4 2 4 5 6/4/2015 To calculate the path, calculate the number of edges, not vertex Example 13: Consider this digraph: 1 2 5 4 3 1: 1, 2, 5, 4, 3 is a path length 4 from vertex 1 to vertex 3. 2: 1, 2, 5, 1 is a path length 3 from vertex 1 to itself. 3: 2, 2 is a path length 1 from vertex 2 to itself. 4: 2, 5, 1, 2 is a path length 3 from vertex 2 to itself. A path that start and ends at the same vertex is called a cycle. 2, 3 and 4 are all cycles. Paths in a relation R can be used to defined new relations that are quite useful. If n is a fixed positive integer, we defined relation Rn on A as follows: x Rn y means that there is a path of length n from x to y in R. For example, x R2 y mean there is a path of length 2 from vertex x to vertex y. 3 and 4 are cycle that starts and ends at the same vertex but different length. 3 is cycle length 1, respectively. It is clear that paths of length 1 can be identified with the ordered pairs (x,y) that belong to R. For example (2,2) a Example 14: Consider the following digraph: b We may also defined a relation R on A. c d Compute R2. x R y mean that there is some path (of any length) in R from x to y. e Solution: The length of the path is depend on x and y. List down all path of length 2 (Also remember that path of length n The relation R is sometimes called the connectivity relation involves n + 1 elements of A ): a R2 a a R2 b a R2 c b R2 d b R2 e c R2 e since since since since since since aRa aRa aRb bRc bRc cRd and and and and and and aRa aRb bRc cRd cRe dRe for R. (3 elements: a, a, a) (3 elements: a, a, b) (3 elements: a, b, c) (3 elements: b, c, d) (3 elements: b, c, e) (3 elements: c, d, e) R2 = {(a,a), (a,b), (a,c), (b,d), (b,e), (c,e)} a Example 15: Consider the following digraph: Compute R . c d e Solution: List down all ordered pair of vertices for which there is a path of any length from the first vertex to the second. R = {(a,a), (a,b), (a,c), (a,d), (a,e), (b,c), (b,d), (b,e), (c,d), (c,e), (d,e)} a b The set Rn(x) consists of all vertices that can be reached from x by means of a path in R of length n. Find R2(b): Solution: R2(b) = {a, d, e} b c d e The set R (x) consists of all vertices that can be reached from x by some path in R. Find R (b): Solution: R (b) = {a, c, d, e} We can derive the path of length n using set, matrix or diagraph representation. 6 6/4/2015 Example 16: Let A = {a, b, c, d, e} Example 17: Let A = {a, b, c, d, e} R = {(a, a), (a, b), (b, c), (c, a), (c, d), (c, e), (d, e)} R = {(a, a), (a, b), (b, c), (c, a), (c, d), (c, e), (d, e)} Find all paths with length 2. Find all paths with length 2. Solution: Solution: Using Set: Using Matrix: MR = R2 = {(a, a), (a, b), (a, c), (b, a), (b, d), (b, e), (c, a), (c, b), (c, e)} 1 1 0 0 0 0 0 1 0 0 1 0 0 1 1 a 0 0 0 0 1 b Advisable to draw the digraph of R first before you answer the question. c d 0 0 0 0 0 Find MR2 e We can use Boolean product of MR and MR to get MR2, where MR2 = MR ⊙ MR. MR2 = MR ⊙ MR. MR2 = 1 1 0 0 0 0 0 1 0 0 1 0 0 1 1 1 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 = Reachability is a concept in which there is a relation between x and y in whatever length possible. 1 1 0 0 0 0 0 1 0 0 = 1 1 1 0 0 R2 ⊙ We can use Boolean product of MR and MR to get MR2, where MR2 = MR ⊙ MR. Written as x R y where x and y have relations. Consider Example 13. Identify all x R y. a b c a 1 1 1 b 1 c 1 1 d e 1 1 1 d e = {(a, a), (a, b), (a, c), (b, a), (b, d), (b, e), (c, a), (c, b), (c, e)} List all path length 2 from Exercise 1 using three ways of representation. List all path length 1, 2 and 3 for Example 10 and Example 13. 41 7