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Lecture 4.4: Equivalence Classes and Partially Ordered Sets

Lecture 4.4: Equivalence Classes and Partially Ordered Sets. CS 250, Discrete Structures, Fall 2011 Nitesh Saxena * Adopted from previous lectures by Cinda Heeren. Course Admin. Mid-Term 2 Exam Graded Solution has been posted Any questions, please contact me directly

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Lecture 4.4: Equivalence Classes and Partially Ordered Sets

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  1. Lecture 4.4: Equivalence Classes and Partially Ordered Sets CS 250, Discrete Structures, Fall 2011 NiteshSaxena *Adopted from previous lectures by CindaHeeren

  2. Course Admin • Mid-Term 2 Exam Graded • Solution has been posted • Any questions, please contact me directly • Please pick up, if haven’t done so already • HW3 Graded • Solution has been posted • Please contact the TA for questions • Please pick up, if haven’t done so already Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

  3. Final Exam • Thursday, December 8,  10:45am- 1:15pm, lecture room • Heads up! • Please mark the date/time/place • Our last lecture will be on December 6 • We plan to do a final exam review then Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

  4. HW4 • HW4 to be posted by this weekend • Covers the chapter on Relations • Will be due in 10 days from the day of posting Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

  5. Outline • Equivalence Classes • Partially Ordered Sets (POSets) • Hasse Diagrams Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

  6. Defn of [a]R symmetry transitivity symmetry Defn of [b]R Equivalence Classes Lemma: Let R be an equivalence relation on S. Then • If aRb, then [a]R = [b]R • If not aRb, then [a]R [b]R =  Proof: • Suppose aRb, and consider x  S. x  [a]R aRx  xRa  xRb  bRx  x  [b]R Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

  7. Thus, [a]Rand [b]R are either identical or disjoint. Equivalence Classes Lemma: Let R be an equivalence relation on S. Then • If aRb, then [a]R = [b]R • If not aRb, then [a]R [b]R =  Proof: 2. Suppose to the contrary that  x  [a]R [b]R. x  [a]R x  [b]R aRx and bRx  aRx and xRb  aRb, contradicting “not aRb” Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

  8. Each Aiis called a block of the partition. Equivalence Classes So S is the union of disjoint equivalence classes of R. A partition of a set S is a (perhaps infinite) collection of sets {Ai} with • Each Ai non-empty • Each Ai  S • For all i, j, Ai  Aj =  • S = Ai Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

  9. Equivalence Classes Give a partition of the reals into 2 blocks (numbers <= 0) and (numbers > 0) Give a partition of the integers into 4 blocks numbers modulo 4 Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

  10. Partition of, equivalence relation on Subset of, equivalence class of Relation on, partition of Equivalence relation on, partition of I have no clue. Equivalence Classes Theorem: if R is a _____ S, then {[a]R : a  S} is a _____ S. Theorem: if R is an equivalence relation on S, then {[a]R : a  S} is a partition of S. Proof: we need to show that an equivalence relation R satisfies the definition of a partition. Follows from previous arguments and definition of equivalence classes. Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

  11. Equivalence Classes Theorem: if {Ai} is any partition of S, then there exists an equivalence relation R, whose equivalence classes are exactly the blocks Ai. Proof: If {Ai} partitions S then define relation R on S to be R = {(a,b) :  i, a  Ai and b  Ai} Next show that R is an equivalence relation. Reflexive and symmetric. Transitive? Suppose aRb and bRc. Then a and b are in Ai, and b and c are in Aj. But b  Ai  Aj, so Ai = Aj. So, a, b, c  Ai, thus aRc. Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

  12. Example: Partition  Equivalence Relation Give an equivalence relation for the partition: {1,2}, {3, 4, 5}, {6} for the set {1,2,3,4,5,6} Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

  13. a  a for any real If a  b, b  c then a  c If a  b, b  a then a = b Partially Ordered Sets (POSets) Let R be a relation then R is a Partially Ordered Set (POSet) if it is • Reflexive - aRa, a • Transitive - aRb  bRc  aRc, a,b,c • Antisymmetric - aRb  bRa  a=b, a,b Ex. (R,), the relation “” on the real numbers, is a partial order. Reflexive? How do you check? Transitive? Antisymmetric? Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

  14. Partially Ordered Sets (POSets) Let (S, R ) be a PO. If a R b, or b R a, then a and b are comparable. Otherwise, they are incomparable. A total orderis a partial order where every pair of elements is comparable. Ex. (R,), is a total order, because for every pair (a,b) in RxR, either a  b, or b  a. Ex. (people in a queue, “behind or same place”) is a total order Ex. (employees, “supervisor”) is not a total order Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

  15. Yes, or No? A total order? Yes, x|x since x=1x (k=1) a|b means b=ak, b|c means c=bj. Does c=am for some m? c = bj = akj (m=kj) a|b means b=ak, b|a means a=bj. But b = bjk (subst) only if jk=1. jk=1 means j=k=1, and we have b=a1, or b=a Partially Ordered Sets (POSets) Ex. (Z+, | ), the relation “divides” on positive integers. Reflexive? Transitive? Antisymmetric? Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

  16. 3|-3, and -3|3, but 3  -3. Yes, or No? A total order? Yes, x|x since x=1x (k=1) a|b means b=ak, b|c means c=bj. Does c=am for some m? c = bj = akj (m=kj) Not a poset. Partially Ordered Sets (POSets) Ex. (Z, | ), the relation “divides” on integers. Reflexive? Transitive? Antisymmetric? Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

  17. A  B, B  A  A=B Now take an x, and suppose it’s in A. Must it also be in C? A total order? Yes, A  A, A 2S A  B, B  C. Does that mean A  C? A  B means x A  x B B  C means x B  x C A poset. Partially Ordered Sets (POSets) Ex. (2S,  ), the relation “subset” on set of all subsets of S. Reflexive? Transitive? Antisymmetric? Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

  18. Partially Ordered Sets (POSets) When we don’t have a special relation definition in mind, we use the symbol “” to denote a partial order on a set. When we know we’re talking about a partial order, we’ll write “a  b” instead of “aRb” when discussing members of the relation. We will also write “a < b” if a  b and a  b. Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

  19. A total order? 0110  1000 0110  0000 0110  1110 0110  1011 Partially Ordered Sets (POSets) Ex. A common partial order on bit strings of length n, {0,1}n, is defined as: a1a2…an  b1b2…bn If and only if ai  bi,  i. 0110 and 1000 are “incomparable” … We can’t tell which is “bigger.” Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

  20. 4 3 2 Draw edge (a,b) if a b Don’t draw up arrows Don’t draw self loops Don’t draw transitive edges 1 Hasse Diagrams Hasse diagrams are a special kind of graphs used to describe posets. Ex. In poset ({1,2,3,4}, ), we can draw the following picture to describe the relation.

  21. Today’s Reading • Rosen 9.5 and 9.6 Lecture 4.4 -- Equivalence Classes and Partially Ordered Sets

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