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CSE15 Discrete Mathematics 05/03/17

CSE15 Discrete Mathematics 05/03/17. Ming-Hsuan Yang UC Merced. 9.4 Equivalence relation. In traditional C, only the first 8 characters of a variable are checked by the complier

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CSE15 Discrete Mathematics 05/03/17

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  1. CSE15 Discrete Mathematics05/03/17 Ming-Hsuan Yang UC Merced

  2. 9.4 Equivalence relation • In traditional C, only the first 8 characters of a variable are checked by the complier • Let R be relation on the set of strings of characters s.t. sRt where s and t are two strings, if s and t are at least 8 characters long and the first 8 characters of s and t agree, or s=t • Easy to see R is reflexive, symmetric, and transitive

  3. Example • The integers a and b are related by the “congruence modulo 4” relation when 4 divides a-b • We will see this relation is reflexive, symmetric and transitive • The above-mentioned two relations are examples of equivalence relations, namely, relations that are reflexive, symmetric, and transitive

  4. Equivalence relation • A relation on a set A is called an equivalence relation if it is reflexive, symmetric, and transitive • Important property in mathematics and computer science • Two elements a and b are related by an equivalence relation are called equivalent, denoted by a ∼ b, a and b are equivalent elements w.r.t. a particular equivalence relation

  5. Example • Let R be the relation on the set of integers s.t. aRb iff a=b or a=-b • We previously showed that R is reflexive, symmetric, and transitive • R is reflexive, aRa iff a=a or a=-a • Ris symmetric, aRb if a=b or a=-b, then b=a or b=-a and so bRa (also for the only if part) • R is transitive, if aRb and bRc, then (a=b or a=-b) and (b=c or b=-c). So a=c or a=-c. Thus aRc (also for the only if part) • If follows that R is an equivalence relation

  6. Example • Let R be the relation on the set of real numbers s.t. aRb iff a-b is an integer. Is R an equivalence relation? • Since a-a=0 is an integer for all real number a, aRa for all a. Hence R is reflexive • Now suppose aRb, then a-b is an integer, so b-a is an integer. Hence bRa. Thus R is symmetric • If aRb and bRc, then a-b and b-c are integers. So, a-c=(a-b)+(b-c) is an integer. Hence aRc. Thus R is transitive • Consequently, R is an equivalence relation

  7. Example • Let m be a positive integer with m>1. Show that the relation R={(a,b)|a ≡b (mod m)} is an equivalence relation on the set of integers • Recall a≡b (mod m) iff m divides a-b. Note that a-a=0 is divided by m. Hence a≡a (mod m). So congruence modulo m is reflexive • Suppose a≡b(mod m), then a-b is divisible by m, so a-b=km, where k is an integer. It follows b-a=(-k)m, so b≡a(mod m). Hence, congruence modulo m is symmetric

  8. Example • Suppose a≡b(mod m) and b≡c(mod m). Then m divides both a-b and b-c. Thus, there are integers k and l with a-b=km and b-c=lm • Put them together a-c=(a-b)+(b-c) = km+lm = (k+l)m. Thus a≡c(mod m). So, congruence modulo m is transitive • It follows that congruence modulo m is an equivalence relation

  9. Example • Let R be the relation on the set of real numbers s.t. xRy iff x and y are real numbers that differ by less than 1, that is |x-y|<1. Show that R is not an equivalence relation • R is reflexive as |x-x|=0<1 where x ∊ R • R is symmetric for if xRy, then |x-y|<1, which tells us |y-x|<1. So yRx • R is not transitive. Take x=2.8, y=1.9, z=1.1, so that |x-y|=0.9<1, |y-z|=0.8<1, but |x-z|=1.7>1 • So R is not an equivalence relation

  10. 9.6 Partial orderings • Often use relations to order some or all of the elements of sets • Example: order words, schedule projects • A relation R on a set S is called partial ordering or partial order if it is reflexive, antisymmetric, and transitive • A set S together with a partial ordering R is called partially ordered set, or poset, and is denoted by (S,R) • Members of S are called elements of the poset

  11. Example • Show that the “greater than or equal” relation ( ≥) is a partial ordering on the set of integers • ≥ is reflexive as a ≥ a • ≥ is antisymmetric as if a ≥ b and b ≥ a then a=b • ≥ is transitive as if if a≥b and b≥c then a ≥ c • (Z, ≥) is a poset

  12. Example • The divisbility relation | is a partial ordering on the set of positive integers, as it is reflexive, antisymmetric, and transitive • We see that (Z+, |) is a poset

  13. Example • Show that inclusion ⊆ is a partial ordering (the relation of one set being a subset of another is called inclusion) on the power set of a set S • Example: power set of {0,1,2} is P({0,1,2})= {∅, {0}, {1}, {2}, {0,1}, {0,2}, {1,2}, {0,1,2}} • A ⊆ A whenever A is a subset of S, so ⊆ is reflexive • It is antiysmmetric as A ⊆ B and B ⊆ A imply that A=B • It is transitive as A ⊆ B and B ⊆ C imply that A ⊆ C. Hence ⊆ is a partial ordering on P(S), and (P(S),⊆) is a poset

  14. Example • Let R be the relation on the set of people s.t. xRy if x and y are people and x is older than y. Show that R is not a partial ordering • R is antisymmetric if a person x is order than a person y, then y is not order than x • R is transitive • R is not reflexive as no person is older than himself/herself

  15. Relation in any poset • In different posets different symbols such as ≤, ⊆ and | are used for a partial ordering • Need a symbol we can use when we discuss the ordering in an arbitrary poset • The notation a ≼ b is used to denote (a,b)∊R in an arbitrary poset (S,R) • ≼ is used as the “less than or equal to” relation on the set of real numbers is the most familiar example of a partial ordering, and similar to ≤ symbol • ≼ is used for any poset, not just “less than or equals” relation , i.e., (S, ≼)

  16. Lexicographic order Definition: Given two posets (A1,≼1) and (A2,≼2), the lexicographic ordering on A1 ⨉A2 is defined by specifying that (a1, a2) is less than (b1,b2), that is, (a1, a2) ≺ (b1,b2), either if a1 ≺1 b1 or if a1 =b1 and a2 ≺2 b2. • This definition can be easily extended to a lexicographic ordering on strings (see text). Example: Consider strings of lowercase English letters. A lexicographic ordering can be defined using the ordering of the letters in the alphabet. This is the same ordering as that used in dictionaries. • discreet≺discrete, because these strings differ in the seventh position and e≺t. • discreet≺discreetness, because the first eight letters agree, but the second string is longer.

  17. Hasse diagrams Definition: A Hasse diagram is a visual representation of a partial ordering that leaves out edges that must be present because of the reflexive and transitive properties. A partial ordering is shown in (a) of the figure above. The loops due to the reflexive property are deleted in (b). The edges that must be present due to the transitive property are deleted in (c). The Hasse diagram for the partial ordering (a), is depicted in (c).

  18. Comparable • The elements a and b of a poset (S, ≼) are called comparable if either a≼b or b≼a. When a and b are elements of S s.t. neither a≼b nor b≼a, a and b are incomparable • In the poset (Z+, |), are the integers 3 and 9 comparable? Are 5 and 7 comparable? • The integers 3 and 9 are comparable as 3|9. the integers 5 and 7 are incomparable as 5 does not divide 7 and 7 does not divide 5

  19. Total ordering • Pairs of elements may be incomparable and thus we have “partial” ordering • When every 2 elements in the set are comparable, the relation is called total ordering • If (S, ≼) is a poset and every two elements of S are comparable, S is called a totally ordered or linearly ordered set, and ≼ is called a total order or a linear order • A totally ordered set is also called a chain

  20. Example • The poset (Z, ≤) is totally ordered as a ≤ b or b ≤ a whenever a and b are integers • The poset (Z+, |) is not tally ordered as it contains elements that are incomparable, such as 5 and 7

  21. Well-ordered set • (S, ≼) is a well-ordered set if it is a poset s.t. ≼ is a total ordering and every nonempty subset of S has a least element • The set of ordered pairs of positive integers, Z+×Z+ with (a1, a2) (b1, b2) if a1 < a2 or if a1=b1 and a2≤b2 (the lexicographic ordering), is a well-ordered set • The set Z, with the usual ordering, is not well-ordered as the set of negative integers has no least element

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