Discrete Mathematics CS 2610 January 27, 2009 - part 2 Agenda Previously: Set theory Subsets (proper subsets) & set equality StSet cardilitdinality Power sets n-Tuples & Cartesian product Set operations Union, Intersection, Complement, Difference Venn diagrams Now Symmetric difference Proving properties about sets Sets as bit-strings Functions 2 Symmetric Difference The symmetric difference, A ⊕ B, is: A ⊕ B = { x | (x ∈ A ∧ x ∉ B) v (x ∈ B ∧ x ∉ A)} (i.e., x is in one or the other, but not in both) IitIs it commu tti?tative ? 3 Set Identities Identity: A ∪∅= A , A ∩ U = A Domination: A ∪ U = U , A ∩∅= ∅ Idempotent: A ∪ A = A = A ∩ A DblDouble complement : ( A ) = A Commutative: A ∪ B = B ∪ A , A ∩ B = B ∩ A Associative: A ∪ (B ∪ C) = (A ∪ B) ∪ C A ∩ (B ∩ C) = (A ∩ B) ∩ C 4 Set Identities Absorption: A ∪ (A ∩ B) = A A ∩ (A ∪ B) = A Complement: A ∪ A¯ = U A ∩ A¯ = ∅ Distributive: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) 5 De Morgan’s Rules De Morgan’s I (A U B) = A ∩ B DeMorgan’s II (A ∩ B) = A U B 6 Generalized Union n UAi = A1 ∪ A2 ∪... ∪ An i=1 The union of a collection of sets contains those elements that belong to at least one set in the collec tion. 7 Generalized Intersection n ∩Ai = A1 ∩ A2 ∩... ∩ An i=1 The intersection of a collection of sets contains those elements that belong to all the sets in the collec tion. 8 Proving Set Identities How would we prove set identities of the form S1 = S2 Where S1 and S2 are sets? 1. Prove S1 ⊆ S2 and S2 ⊆ S1 separately. Use previously proven set identities. Use logical equivalences to prove equivalent set definitions. 2. Use a membership table. 9 Proof Using Logical Equivalences Prove that (A U B) = A ∩ B Proof: First show (A U B) ⊆ A ∩ B, then the reverse. Let c ∈ (A U B) c ∈ {x | x ∈ A ∨ x ∈ B} (Def. of union) ¬ (c ∈ A ∨ c ∈ B) (Def. of complement) ¬ (c ∈ A) ∧¬(c ∈ B) (De Morgan’s rule) (c ∉ A) ∧ (c ∉ B) (Def. of ∉) (c ∈ A) ∧ (c ∈ B) (Def. of complement) c ∈ {x | x ∈ A ∧ x ∈ B} (Set builder notation) c ∈ A ∩ B (Def. of intersection) Therefore, (A U B) ⊆ A ∩ B. Each step above is reversible, therefore A ∩ B ⊆ (A U B). 10 Proof Using Membership Table Using membership tables (A U B) = A ∩ B 1 : means x is in the Set 0 : means x is not in the Set A B A BAA ∩ B U BA U B 1 1 0 0 0 1 0 1 0 0 1 0 1 0 0 1 1 0 0 1 0 0 0 1 1 1 0 1 The two columns are the same. Therefore, x ∈ (A U B) iff x ∈ A ∩ B – i.e., the equality holds. 11 Sets as Bit-Strings For a finite universal set U = {a1, a2, …,an} 1. Assign an arbitrary order to the elements of U. 2. Represent a subset A of U as a string of n bits, B = b1b2…bn ⎧0 if a i ∉ A bi = ⎨ ⎩1 if a i ∈A Example: U = {a1, a2, …, a5}, A = {a1, a3, a4 } B = 10110 12 Sets as Bit-Strings Set theoretic operations A 1 0 1 0 1 B 0 0 1 1 0 Bit-wise OR A ∪ B 1 0 1 1 1 Bit-wise AND A ∩ B 0 0 1 0 0 Bit-wise XOR A ⊕ B 1 0 0 1 1 13 Functions (Section 2.3) Let A and B be nonempty sets. A function f from A to B is an assignment of exactly one element of B to each element of A. We write f(a ) = b if b is the unique ele me nt of B assig ne d by the function f to the element a in A. If f is a function from A to B, we write f : A → B. Functions are sometimes called mappings. 14 Example A = {Mike, Mario, Kim, Joe, Jill} B = {John Smith, Edward Groth, Jim Farrow} Let f:A → B where f(a) means father of a. f Mike John Smith Mario Edward Gro th Kim Richard Boon Joe Jill A B Can grandmother of a be a function ? 15.