Lecture 4.4: Equivalence Classes and Partially Ordered Sets CS 250, Discrete Structures, Fall 2011 Nitesh Saxena *Adopted from previous lectures by Cinda Heeren

Lecture Equivalence Classes and Partially Ordered Sets 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

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 Equivalence Classes and Partially Ordered Sets

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 Equivalence Classes and Partially Ordered Sets

Outline Equivalence Classes Partially Ordered Sets (POSets) Hasse Diagrams

Lecture Equivalence Classes and Partially Ordered Sets Equivalence Classes Lemma: Let R be an equivalence relation on S. Then 1. If aRb, then [a] R = [b] R 2. If not aRb, then [a] R  [b] R =  Proof: 1. Suppose aRb, and consider x  S. x  [a] R  aRx  xRa  xRb  bRx  x  [b] R Defn of [a] R symmetry transitivity symmetry Defn of [b] R

Lecture Equivalence Classes and Partially Ordered Sets Equivalence Classes Lemma: Let R be an equivalence relation on S. Then 1. If aRb, then [a] R = [b] R 2. 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  aRb, contradicting “not aRb”  aRx and xRb Thus, [a] R and [b] R are either identical or disjoint.

Lecture Equivalence Classes and Partially Ordered Sets 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 {A i } with Each A i non-empty Each A i  S For all i, j, A i  A j =  S =  A i Each A i is called a block of the partition.

Lecture Equivalence Classes and Partially Ordered Sets Equivalence Classes Give a partition of the reals into 2 blocks (numbers 0) Give a partition of the integers into 4 blocks numbers modulo 4

Lecture Equivalence Classes and Partially Ordered Sets Equivalence Classes Theorem: if R is a _____ S, then {[a] R : a  S} is a _____ S. A.Partition of, equivalence relation on B.Subset of, equivalence class of C.Relation on, partition of D.Equivalence relation on, partition of E.I have no clue. 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 Equivalence Classes and Partially Ordered Sets 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  A i and b  A i } Next show that R is an equivalence relation. Reflexive and symmetric. Transitive? Suppose aRb and bRc. Then a and b are in A i, and b and c are in A j. But b  A i  A j, so A i = A j. So, a, b, c  A i, thus aRc.

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 Equivalence Classes and Partially Ordered Sets

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. How do you check? a  a for any real Reflexive? Transitive? Antisymmetric? If a  b, b  c then a  c If a  b, b  a then a = b

Lecture Equivalence Classes and Partially Ordered Sets 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 order is 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 Equivalence Classes and Partially Ordered Sets Partially Ordered Sets (POSets) Ex. (Z +, | ), the relation “divides” on positive integers. Yes, x|x since x=1x (k=1) Reflexive? Transitive? Antisymmetric? 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 Yes, or No? A total order?

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

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

Lecture Equivalence Classes and Partially Ordered Sets 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 Equivalence Classes and Partially Ordered Sets Partially Ordered Sets (POSets) Ex. A common partial order on bit strings of length n, {0,1} n, is defined as: a 1 a 2 …a n  b 1 b 2 …b n If and only if a i  b i,  i and 1000 are “incomparable” … We can’t tell which is “bigger.” A.0110  1000 B.0110  0000 C.0110  1110 D.0110  1011 A total order?

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. 1.Draw edge (a,b) if a  b 2.Don’t draw up arrows 3.Don’t draw self loops 4.Don’t draw transitive edges

Lecture Equivalence Classes and Partially Ordered Sets Today’s Reading Rosen 9.5 and 9.6