discrete mathematics cs 2610 january 27, 2009 - part 2
DESCRIPTION
3 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) Is it commutative ?TRANSCRIPT
Discrete Mathematics CS 2610
January 27, 2009 - part 2
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AgendaPreviously: Set theory Subsets (proper subsets) & set equality Set cardinality 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
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Symmetric DifferenceThe 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)
Is it commutative ?
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Set IdentitiesIdentity:
A = A , A U = A
Domination: A U = U , A =
Idempotent: A A = A = A A
Double complement:
Commutative: A B = B A , A B = B A
Associative: A (B C) = (A B) C A (B C) = (A B) C
AA =)(
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Set IdentitiesAbsorption:
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)
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De Morgan’s Rules De Morgan’s I
DeMorgan’s II
(A U B) = A B
(A B) = A U B
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Generalized Union
The union of a collection of sets contains those elements that belong to at least one set in the collection.
Ai
i1
n
U A1 A2 An
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Generalized Intersection
The intersection of a collection of sets contains those elements that belong to all the sets in the collection.
Ai
i1
n
A1 A2 An
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Proving Set IdentitiesHow 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.
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Proof Using Logical Equivalences
Prove thatProof: 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).
(A U B) = A B
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Proof Using Membership Table
Using membership tables
1 : means x is in the Set0 : means x is not in the Set
(A U B) = A B
1000
1000
0111
A U BA B
1010B
100110001011
A U BABA
The two columns are the same. Therefore, x (A U B) iff x A B – i.e., the equality holds.
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Sets as Bit-StringsFor 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
Example: U = {a1, a2, …, a5}, A = {a1, a3, a4 } B = 10110
A a if1A a if0
bi
ii
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Sets as Bit-StringsSet theoretic operations
A BA BA B
BA
00100111010110010101
Bit-wise ORBit-wise AND
11001Bit-wise XOR
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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 element of B assigned 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.
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ExampleA = {Mike, Mario, Kim, Joe, Jill}B = {John Smith, Edward Groth, Jim Farrow}
Let f:A B where f(a) means father of a.
Can grandmother of a be a function ?
Mike Mario
Kim Joe Jill
John SmithEdward GrothRichard Boon
f
A B