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Relations and Their Properties

Relations and Their Properties. Section 9.1 , 9.3 Tuesday, June 17. 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.

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Relations and Their Properties

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  1. Relations and Their Properties Section 9.1, 9.3 Tuesday, June 17 RELATIONS

  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. • ordered 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 RELATIONS

  3. Review: General relation • Definition:The Cartesian product of the sets A1, A2, …, An: A1×A2× …×An={(a1, a2, …, an)| aiAifor i=1, 2, …, n} • Definition: A relation R over the setsA1, A2, …, An is a subset of A1×A2× …×An : R ⊆ A1×A2× …×An • The number n is called the arity of the relation; equivalently, R is called an n-ary relation. RELATIONS

  4. Binary Relations • Definition:Let A and B be sets. A binary relation Rfrom A to B is a subset of the Cartesian product AB. • Notation: xRy means (x,y)  R if xRy, we say x is related to y under R. xRy is short for (x,y)  R. RELATIONS

  5. Example • Let A be the set of all cities in the U.S., and let B be the set of states in the U.S. • Define the relation S from A to B by: aSb if citya is in state b. • Some ordered pairs belonging to relation S: • (East Lansing, MI)  S • (Chicago, IL)  S • (Boulder, CO)  S • … (East Lansing) S (MI) (Chicago) S (IL) (Boulder) S (CO) RELATIONS

  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? What about the relation S that relates a city to the state that the city is in? What about the relation represented by a Python namespace? RELATIONS

  7. Functions and Relations • Recall: If f: A→ B is a function, the graph of f is depicted by plotting the set of ordered pairs Gf = {(x, f (x))| x  A}. • More generally, we can graph a binary relation. • Example: Define xRy iff x2 + y2 = 5. • Is this relation a function? How can youtell by looking at the graph? 5 -5 5 -5 RELATIONS

  8. Binary Relations on a Set • Definition: A binary relation R on a set A is a relation from A to A, i.e., R AA. • Example: Let A = {1, 2, 3, 4}, and define D = {(a,b) | ∃m (m ∈A∧b = am)}. • Exercise: What pairs does D contain? D = {(1,1), (1,2), (1,3), (1,4), (2,2), (2,4), (3,3), (4,4)} Thus: 1D1, 1D2, 1D3, 1D4, 2D2, 2D4, 3D3, 4D4 RELATIONS

  9. 1 1 2 2 3 3 4 4 Different ways to show a binary relation • D can be displayed graphically, in tabular form, as a zero-one matrix, or as a directed graph: 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 RELATIONS

  10. Directed Graph Representation of a Binary Relation on a Finite Set • Directed Graphs • Vertices, one for each element in the set • Directed edges, one for each ordered-pair in the relation • Example: Let A = {1, 2, 3, 4}, and define aDb ↔ ∃m (m ∈A∧b = am). 1 2 4 3 RELATIONS

  11. Exercise: 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. RELATIONS

  12. Types of Relations • Definition:A relation R on a set A is reflexive if (a,a)  R for every element aA, that is, ∀a ( aRa ) • Definition:A relation R on a set A is irreflexive if (a,a)  R for every element aA, that is, ∀a ( aRa ) RELATIONS

  13. Exercise Reflexive: ∀a ( aRa ) • Consider the following relations on {1,2,3,4}. Classify each as reflexive, irreflexive, both, or neither. • R1 = {(1,1), (1,2), (2,1), (2,2), (3,3), (4,4)} • R2 = {(2,4), (4,2)} • R3 = {(1,2), (2,3), (3,4)} • R4 = {(1,1), (2,2), (3,3), (4,4)} • R5 = {(1,3), (1,4), (2,3), (2,4), (3,1), (3,4)} • R6 = {(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)} • Can a relation on a set A be both reflexive and irreflexive? • Is every relation on A either reflexive or irreflexive? Irreflexive:∀a ( aRa ) RELATIONS

  14. Symmetric Relations • Definition:A relation R on a set A is symmetric if (b,a)  R whenever (a,b)  R, for a, bA; that is, ∀a,b ( aRb → bRa ) • Definition A relationR on a set A is antisymmetric if a = b whenever both (a,b)  R and (b, a)  R, for a, b in A; that is, ∀a,b ( aRb ∧ bRa → a = b ) RELATIONS

  15. Exercise Symmetric: ∀a,b ( aRb → bRa ) • Classify each relation below as “symmetric,” “antisymmetric,” “both”, or “neither.” • R1 = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (4, 4)} on {1, 2, 3, 4} • R2 = {(2, 4), (4, 2)} on {1, 2, 3, 4} • R3 = {(1, 2), (2, 3), (3, 4)} on {1, 2, 3, 4} • R4 = {(1, 1), (2, 2), (3, 3), (4, 4)} on {1, 2, 3, 4} • R5 = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 1), (3, 4)} on {1, 2, 3, 4} • The“less than or equal to” relation (i.e., “≤” ) on the positive integers • The “less than” relation (i.e., “<“) on the positive integers • These definitions are not complementary. • Give a relation that is both symmetric and antisymmetric. • Give a relation that is neither symmetric not antisymmetric. Antisymmetric: ∀a,b ( aRb ∧ bRa → a = b ) RELATIONS

  16. Transitive Relations • Definition:A relationR on a set A is transitiveif whenever (a, b)  R and (b, c)  R, then (a, c)  R, for a, b, c in A, that is, ∀ a, b, c ( aRb ∧ bRc → aRc ) RELATIONS

  17. Exercise Transitive: ∀ a, b, c ( aRb ∧ bRc → aRc ) • Classify each relation below as “transitive” or “not transitive” • R1 = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (4, 4)} on {1, 2, 3, 4} • R2 = {(2, 4), (4, 2)} on {1, 2, 3, 4} • R3 = {(1, 2), (2, 3), (3, 4)} on {1, 2, 3, 4} • R4 = {(1, 1), (2, 2), (3, 3), (4, 4)} on {1, 2, 3, 4} • The “less than” relation (i.e., “<“) on the positive integers • The “successor” relation S on the integers defined by: nSmiffm = n + 1 • The “subtype” (“subclass”) relation on the set of types (classes) in an OO program. • The “defeated” relation on the set of football teams in the Big 10 RELATIONS

  18. Exercise • Determine which properties hold for the relations defined below 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} RELATIONS

  19. R = {(x, y) | x + y = 0} • Solution: • R is not reflexivesince, for example, 1R1. • R is not irreflexivesince, for example, 0R0. • If x + y = 0 then y + x = 0, so xRy implies yRx, which means R is symmetric. • R is not antisymmetric since, for example, (–1)R1 and 1R(–1) are both in R, but 1 ≠ –1 • R is not transitive since, for example, 1R(–1) and (–1)R1, but 1R1. RELATIONS

  20. R = {(x,y) | x = y x = – y} • Solution: • Since xRx, for each x, relation R is reflexive. • Since R is reflexive, it is not irreflexive. • Since x = ±y if and only if y = ±x, xRy if and only if xRy, for all x and y; thus, R is symmetric. • R is not antisymmetric since, for example, (–1)R1 and 1R(–1) are both in R, but 1 ≠ –1 . • R is also transitive because essentially the product of ±1 and ±1 is ±1. RELATIONS

  21. R = {(x,y) | x – y is a rational number} • Solution: • R is reflexive since, for all x, x – x = 0 is a rational number. • Since R is reflexive, it is not irreflexive. • R is symmetric because, if x – y isrational, then so is – (x – y) = y – x . • R is not antisymmetric because, for example, (–1)R1 and 1R(–1) are both in R, but 1 ≠ –1 . • R is transitive because if x – y is a rational and y – z is a rational, so is x – y + y – z = x – z. RELATIONS

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

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

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

  25. R = {(x, y) | x = 1} • Solution: • R is not reflexive since (2, 2) R. • R is not irreflexive since (1, 1) R. • R is not symmetricsince, for example, (1, 2) R but (2, 1) R. • R is antisymmetric because (x, y) R and (y, x) R only ifx = y = 1. • R istransitive since (x, y) R and (y, z) R only if x = 1, which implies (x, z) R . RELATIONS

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

  27. Summary RELATIONS

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

  29. Combining Relations – Examples • Since relations are sets, all of the operators we used for combining sets can be used to combine relations. • Example: Let R1 and R2 on A = { 1, 2, 3, 4} be defined as follows: • R1 = {(1,1), (2,2), (3,3)} • R2 = {(1,1), (1,2), (1,3), (1,4)} • Then the following are also relations on A • R1R2 = {(1,1), (2,2), (3,3), (1,2), (1,3), (1,4)} • R1R2 = {(1,1)} • R1−R2 = {(2,2), (3,3)} RELATIONS

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

  31. Exercise • 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 S a relation from B to C: • S= {(x,0), (y,0), (w,1), (w,2), (z,1)} • What isS oR? • S oR= {(1,0), (1,1), (2,1), (2,2), (3,0), (3,1)} RELATIONS

  32. Powers of Relations • Definition:Let R be a relation on the set A. The powers of R, denoted Rn, for n = 1, 2, 3,…, are defined recursively by: • R1 = R, • Rn+1=RnoR, for n  1. • So: • R2 = R o R, • R3 = R2 oR = (R oR) o R • R4 = R3 oR = ((R oR) o R) o R • … RELATIONS

  33. Exercise • Consider the relation R= {(1, 2), (1, 3), (3, 4)} on{1, 2, 3, 4}. • Is R transitive? • Calculate R2 • Consider R={(2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4)} on {1, 2, 3, 4} • Is R transitive? • Calculate R2 • Calculate R3 • What will you get if you calculate R153 ? RELATIONS

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

  35. 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, ...(*) We prove (*) inductively. • Basis Step: The basis holds trivially since R1 R, for all relations R. RELATIONS

  36. Characterizing Transitive Relations • Induction Step: Assume k ≥ 1 and that if R is transitive then R kR (IH). We need to show: if R is transitive, then Rk+1R (**). Assume that R is transitive, and let (a,b) Rk+1. By the defn of Rk+1, there is somexA such that (a,x) R and (x,b) Rk. Since R is transitive, the IH implies RkR, which implies (x,b) R. Since R is transitive and (a,x) R, this latter result implies (a,b) R. Thus, Rk+1R RELATIONS

  37. Stopped here (Spring 2012) RELATIONS

  38. Applications in Relational Databases Section 8.2 RELATIONS

  39. Records as n-tuples (with fields) • A record is an n-tuple whose entries, calledfields, associate a value from an attribute domain to each of a fixed set of attributes, where n is the cardinality of the attributed set. • e.g., astudent record might be 4-tuple giving values for the attributes: (name, ID#, Major, GPA) • Relational Data Base: a set of records • e.g. a 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) } • Primary key: attribute whose value uniquely determines a record. RELATIONS

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

  41. 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) • , , –, … RELATIONS

  42. Projection example: 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” RELATIONS

  43. Projection of Students: P1,4 (Students) P1,4 RELATIONS

  44. P1, 2 (Enrollments) P1,2 RELATIONS

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

  46. SMajor = “CSE” (Enrollments) SMajor = “CSE” RELATIONS

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

  48. Join: Jp (R, S) • Join 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 RELATIONS

  49. Table 5 Teaching_Assignments RELATIONS

  50. Table 6 Class_schedule RELATIONS

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