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Relations and Their Properties

Relations and Their Properties. Section 8.1. Review: Cartesian Products. Definition : An ordered n-tuple ( a 1 , a 2 , …, a n ) is the ordered collection that has a 1 as its first element, a 2 as its second element…, and a n as its nth element. 2-tuples are called ordered pairs .

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Relations and Their Properties

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  1. Relations and Their Properties Section 8.1

  2. Review: Cartesian Products • Definition:An ordered n-tuple (a1, a2, …, an) is the ordered collection that has a1 as its first element, a2 as its second element…, and an as its nth element. • 2-tuples are called ordered pairs. • (a,b) = (c,d)if and only if a=c and b=d • (a,b)≠(b,a)unless a=b

  3. Cartesian Product … • Definition:The Cartesian product of the sets A1, A2,…, An, denoted A1×A2×… ×An, is the set of ordered n-tuples (a1, a2, …, an), where ai belongs to set Aifor i =1, 2, …, n. A1×A2× …×An={(a1, a2, …, an)| aiAifor i=1, 2, …, n}

  4. Binary Relations • Definition:Let A and B be sets. A binary relationR fromAtoB is a subset of the Cartesian product AB. • Notation: xRy means (x,y)R, and x is said to be related toy under R. xRy denotes (x,y)R.

  5. Example • Let A be the set of all cities, and let B be the set of all states in the U.S. Define the relation Rfrom A to B by: (a,b)Rif city a is in state b. • Some ordered pairs belonging to relation R: R= {(East Lansing, Michigan), (Boulder, Colorado), (Chicago, Illinois), (Columbus, Ohio), …} (East Lansing)R(Michigan), (Boulder)R(Colorado), …

  6. a 1 b c 2 d 3 Another Example A= {a, b, c, d } B = {1, 2, 3} R = {(a,2), (b,1), (b, 2), (c,3), (d,2)}. Is this relation a function?

  7. Functions As Relations • Recall that: Let f: A→ B be a function. The graph of fis the set of ordered pairs Gf = {(x, f (x))| x  A}. • The graph of a function from A to B is a relation from A to B. • Conversely, if R is a relation from A to B such that every element in A is the first element of exactly one ordered pair of R, then a function can be defined with R as its graph.

  8. Relations on a Set • Definition: A relation ona setA is a relation from A to A, that is, a subset of AA. • Example: Let A= {1, 2, 3, 4}, and R = {(a,b) | a evenly divides b}. • What are the elements of R? R = {(1,1), (1,2), (1,3), (1,4), (2,2), (2,4), (3,3), (4,4)}

  9. 1 1 2 2 3 3 4 4 Example • R can be displayed graphically and in tabular form/zero-one matrix form: R 1 2 3 4 1 x x x x 1 1 1 1 2 x x 0 1 0 1 3 x 0 0 1 0 4 x 0 0 0 1

  10. Relations on a Set… • Let A be a set with n elements. How many distinct relations are there on A? • Answer: The number of relations is the same as the number of subsets of AA. There are n2 elements in AA. Thus, P(AA), the power set of AA, has 2n2elements. Therefore, there are 2n2relations on a set with n elements.

  11. Types of Relations • Definition:A relation Ron a setAis called reflexive if (a,a)R for every element aA, that is, • Definition:A relation Ron a setAis called irreflexive if (a,a)  R for every element aA, that is, • Note that a relation can be both not reflexive and not irreflexive.

  12. Example • Consider the following relations on {1,2,3,4}. Which are reflexive, irreflexive, neither? • R1 = {(1,1), (1,2), (2,1), (2,2), (3,3), (4,4)} • Reflexive • R2 = {(2,4), (4,2)} • Irreflexive • R3 = {(1,2), (2,3), (3,4)} • Irreflexive • R4 = {(1,1), (2,2), (3,3), (4,4)} • Reflexive • R5 = {(1,3), (1,4), (2,3), (2,4), (3,1), (3,4)} • Irreflexive • R6 = {(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)} • Neither

  13. Symmetric Relations • Definition:A relation Ron a setAis called symmetric if (b,a)R whenever (a,b)R for a, bA, that is, • Definition A relation Ron a setAis called antisymmetric if whenever (a,b)R and (b,a)R, then a = b, for a, b in A, that is, • Note that these definitions are not complementary. • Give a relation that is both symmetric and antisymmetric. • Give a relation that is neither symmetrick not antisymmetric.

  14. Example • Consider the following relations on {1,2,3,4}. Which are symmetric, antisymmetric, both, or neither? • R1 = {(1,1), (1,2), (2,1), (2,2), (3,3), (4,4)} • Symmetric • R2 = {(2,4), (4,2)} • Symmetric • R3 = {(1,2), (2,3), (3,4)} • Antisymmetric • R4 = {(1,1), (2,2), (3,3), (4,4)} • Both • R5 = {(1,3), (1,4), (2,3), (2,4), (3,1), (3,4)} • Neither • The“divides” relation on the set of positive integers • Antisymmetric • R6 = {(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)} • Neither

  15. Transitive Relations • Definition:A relation Ron a setAis called transitiveif whenever (a,b)R and (b,c)R, then (a,c)R, for a, b, c in A, that is, Is the “football team relationship” transitive? A defeats B and B defeats C implies A defeats C ? Is the “subclass” relation on classes in an OO program transitive?

  16. Example • Consider the following relations on {1,2,3,4}. Which are transitive? • R1 = {(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)} • Transitive • R2 = {(1,1), (1,2), (2,1), (2,2), (3,3), (4,4)} • Transitive • R3 = {(2,4), (4,2)} • Not Transitive • R4 = {(1,2), (2,3), (3,4)} • Not Transitive • R5 = {(1,1), (2,2), (3,3), (4,4)} • Transitive • R6 = {(1,3), (1,4), (2,3), (2,4), (3,1), (3,4)} • Not Transitive • The“divides” relation on the set of positive integers • Transitive

  17. Example Summary • Consider the following relations on {1,2,3,4}. Name its properties? • R1 = {(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)} • T/A • R2 = {(1,1), (1,2), (2,1), (2,2), (3,3), (4,4)} • R/S/T • R3 = {(2,4), (4,2)} • S • R4 = {(1,2), (2,3), (3,4)} • A • R5 = {(1,1), (2,2), (3,3), (4,4)} • R/S/A/T • R6 = {(1,3), (1,4), (2,3), (2,4), (3,1), (3,4)} • None • The“divides” relation on the set of positive integers • R/A/T • Keys:R = Reflexive, S = Symmetric, A = Antisymmetric, T=Transitive

  18. 2 1 3 4 Pictorial Representation of Binary Relations on a Finite Set • Directed Graphs • Vertices, one for each element in the set • Arcs, one for each ordered-pair in the relation • Example: Set S = {1,2,3,4} and relation R on S R= {(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)}

  19. Boolean matrix representation of binary relations on a finite set • Recall: If R is a binary relation over set S • Order the elements of S = { x1, x2, …, xn } • Represent by matrix with a 1 in position (i, j) iff xiRxj • Example: Set S = {1,2,3,4} and relation R on S R= {(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)}

  20. Equivalence Relations • A relation Ron a set A is called an equivalence relation if it is reflexive, symmetric and transitive. • In such a relation, for each element aA, the set of all elements related to a under R is called the equivalence class of a, and is denoted by [a]. [a] = { (a,b) | a  A  b  A aRb }

  21. Example • Determine the properties of each of the following relations defined on the set of all real numbers R: • R = {(x,y) | x + y = 0} • R = {(x,y) | x = y x = – y} • R = {(x,y) | x – y is a rational number} • R = {(x,y) | x = 2y} • R = {(x,y) | xy≥ 0} • R = {(x,y) | xy = 0} • R = {(x,y) | x = 1} • R = {(x,y) | x = 1 or y = 1}

  22. R = {(x,y) | x + y = 0} • Solution: • It is not reflexivesince, for example, (1,1)  R. • It is not irreflexivesince, for example, (0,0) R. • Since x + y = y + x, it follows that if x + y = 0 then y + x = 0, so the relation is symmetric. • It is not antisymmetric since, for example, (–1,1) and (1, –1) are both in R, but 1 ≠ –1 • The relation is not transitive since, for example, (1,–1) R and (–1,1) R, but (1,1)  R.

  23. R = {(x,y) | x = y x = – y} • Solution: • Since for each x, xRx then it is reflexive. • Since it is reflexive, it is not irreflexive. • Since x = ±y if and only if y = ±x, then it is symmetric. • It is not antisymmetric since, for example, (1,–1) and (–1,1) are both in R but 1 ≠ –1 • It is also transitive because essentially the product of ±1 and ±1 is ±1.

  24. R = {(x,y) | x – y is a rational number} • Solution: • It is reflexive since, for all x, x – x = 0 is a rational number. • Since it is reflexive, it is not irreflexive. • It is symmetric because, if x – y isrational, then – (x – y) = y – x is also. • It is not antisymmetric because, for example, (1, –1) and (– 1,1) are both in R but 1 ≠ –1. • It is transitive because if x – y is a rational and y – z is a rational, so is x – y + y – z = x – z.

  25. R = {(x,y) | x = 2y} • Solution: • It is not reflexive since, for example, (1,1) R. • It is not irreflexive since, for example, (0,0) R. • It is not symmetric since, for example, (2,1) Rbut (1,2) R. • It is antisymmetric because x = y =0 is the only time that (x,y) and (y,x) are both in R. • It is not transitivesince, for example, (4,2) R and (2,1) R but (4,1) R.

  26. R = {(x,y) | x y ≥ 0} • Solution: • It is reflexive since x·x≥ 0. • Since it is reflexive, it is not irreflexive. • It is symmetric as the role of x and y are interchangeable. • It is not antisymmetric since, for example, (2,3) and (3,2) are both in R, but2 ≠ 3. • It is not transitive because, for example, (1,0) and (0, – 2) are both in R but (1, – 2) is not.

  27. R = {(x,y) | xy = 0} • Solution: • It is not reflexive since (1,1) R. • It is not irreflexive since (0,0) R. • It is symmetric as the role of x and y are interchangeable. • It is not antisymmetric since, for example, (2,0) and (0,2) are both in R but2 ≠ 0. • It is not transitive because, for example, (1,0) and (0, – 2) are both in R but (1, – 2) is not.

  28. R = {(x,y) | x = 1} • Solution: • It is not reflexive since (2,2) R. • It is not irreflexive since (1,1) R. • It is not symmetricsince, for example, (1,2) R but (2,1) R. • It is antisymmetric because if (x,y) R and (y,x) R it means x = y = 1. • It istransitive since if (1,y) R and (y,z) R, so is (1,z) R .

  29. R = {(x,y) | x = 1 or y = 1} • Solution: • It is not reflexive since (2,2) R. • It is not irreflexive since (1,1) R. • It is symmetricas the roles of x and y are interchangeable. • It is not antisymmetric since, for example, (2,1) and (1,2) are both in R but2 ≠ 1. • It isnot transitive since, for example, (3,1) and (1,7) are both in R but (3,7) R.

  30. Summary

  31. Combining Relations • Since relations are special kind of sets, all of the operators we used for combining sets could be used to combine relations.

  32. Combining Relations – Examples • Let A={1, 2, 3} and B={1, 2, 3, 4} • Let R1 and R2 be relations from A to B • R1 = {(1,1), (2,2), (3,3)} • R2 = {(1,1), (1,2), (1,3), (1,4)} • R1R2 = {(1,1), (2,2), (3,3), (1,2), (1,3), (1,4)} • R1R2 = {(1,1)} • R1R2 = {(2,2), (3,3)}

  33. S oR R S B c A Composition of Relations • Definition: Let R be a relation from a setA to a set B, and S be a relation from Bto a set C. The composition of R and S, denoted by S o R, is the relation consisting of ordered pairs (a,c), where: • aA,cC, and • bB | (a,b)R and (b,c)S.

  34. Example • Let A={1, 2, 3}, B={x, y, w, z}, C={0, 1, 2}, R a relation from A to B: • R= {(1, x), (1, z), (2, w), (3, x), (3, z)}, and Sa relation from B to C: • S= {(x,0), (y,0), (w,1), (w,2), (z,1)} • What is So R? • So R= {(1,0), (1,1), (2,1), (2,2), (3,0), (3,1)}

  35. Powers of Relations • Definition:Let R be a relation on the set A. The powers Rn, n = 1, 2, 3,… are defined recursively by: • R1 = R, • Rn+1=Rn o R, for n  1. • So: • R2 = RoR, • R3 = R2oR = (RoR) o R • …

  36. Rn+1 Rn+1=Rn o R R Rn A A A In the digraph representation, Rn consists of all the order pairs (x,y) where y is reachable from x using a directed path of length n.

  37. Example • Consider the following relations on {1,2,3,4}. • R= {(1,2), (1,3), (3,4)} • Is R transitive? • No • Find R2=R o R • R2 = {(1,4)} • Consider the following relation: • R= {(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)} • IsR transitive? • Yes • Find R2=R o R • R2 = {(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)}

  38. Characterizing Transitive Relations • Theorem: The relation R on a set A is transitive if and only if Rn  R for n = 1, 2, 3, ... • If Part: If Rn  R for n = 1, 2, 3, ... then R is transitive • See exercise 6.1 • Only If Part: If R is transitive then Rn  R for n = 1,2,3, ... • Proof By induction • Basis: If R is transitive then R1 R (trivially true) • Hypothesis: Assume true for n = 1, 2, 3, ... , k • Inductive: Given thatif R is transitive then Rk R, we want to show that if R is transitive then Rk+1 R • Proof: Assume that R is transitive, and let (a,b) be inRk+1. So xA | (a,x) R and (x,b)  Rk,by definition of Rk+1. Since R is transitive then Rk R by inductive hypothesis. So (x,b) R. Since R is transitive, then (a,b)  R. So Rk+1 R

  39. Applications in Relational Databases Section 8.2

  40. Records as n-tuples (with fields) • Record is an n-tuple of fields • e.g., student record might be 4-tuple of the form: (name, ID#, Major, GPA) • Relational Data Base: collection of records • e.g. student database { (Anderson, 231455, CSE, 3.88), (Adams, 888323, PHY, 3.45), (Chou, 102147, CSE, 3.49), (Gould, 453876, MTH, 3.45), (Rao, 678543, MTH, 3.90), (Stevens, 786576, PSY, 2.99) } • Field associates the attribute (name, ID#, etc.) with a value from a domain (set of student names, set of ID#s, etc.)

  41. Table 1 Student Data:(Student_Name, PID, Major, GPA)

  42. Operations on relations (tables) • Projection : projects a table on one or more attributes (columns) • Selection : selects records (rows) that satisfy a condition • Join : combines two tables into one based on shared fields • Of course, you can also use the “usual” operations on relations (sets) • , , –, …

  43. Projection of Students: P1,4 (Students) • Columns 1 and 4 are taken from Students • Columns 2 and 3 are ignored • Name is paired with GPA (perhaps so a department chair can send congratulations or condemnations to students with special GPAs) • Beware: the key ID# is lost so there may be some ambiguity • Perhaps there were 2 students “Smith”

  44. Projection of Students: P1,4 (Students) P1,4

  45. P1,2 (Enrollments) P1,2

  46. Selection: SC • Select all records (rows) that satisfy the condition C

  47. SMajor = “CSE” (Enrollments) SMajor = “CSE”

  48. SMajor = “CSE”  GPA > 3.5 (Students) SMajor = “CSE”  GPA > 3.5

  49. Join: Jp (R, S) • Joins records (rows) from R and S on the last p fields (columns) of R and the first p fields (columns) of S • To join two records corresponding fields must be identical • The join of (a1, …, am ) and (b1, …, bn ) is (a1, …, am, bp+1 , …, bn ), if am – p + 1 = b1 and am – p + 2 = b2 and . . . and am = bp

  50. Example: Joining two tables (on 2 cols) …

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