DISCRETE COMPUTATIONAL STRUCTURES

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DISCRETE COMPUTATIONAL STRUCTURES

• CSE 2353
• Fall 2005

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CSE 2353 OUTLINE

1. Sets
2. Logic
3. Proof Techniques
4. Integers and Induction
5. Relations and Posets
6. Functions
7. Counting Principles
8. Boolean Algebra

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Learning Objectives

• Learn about relations and their basic properties
• Explore equivalence relations
• Become aware of closures
• Learn about posets
• Explore how relations are used in the design of
  relational databases

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Relations

• Relations are a natural way to associate objects
  of various sets

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Relations

• R can be described in
• Roster form
• Set-builder form

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Relations

• Arrow Diagram
• Write the elements of A in one column
• Write the elements B in another column
• Draw an arrow from an element, a, of A to an
  element, b, of B, if (a ,b) ? R
• Here, A 2,3,5 and B 7,10,12,30 and R
  from A into B is defined as follows For all a ?
  A and b ? B, a R b if and only if a divides b
• The symbol ? (called an arrow) represents the
  relation R

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Relations
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Relations

• Directed Graph
• Let R be a relation on a finite set A
• Describe R pictorially as follows
• For each element of A , draw a small or big dot
  and label the dot by the corresponding element of
  A
• Draw an arrow from a dot labeled a , to another
  dot labeled, b , if a R b .
• Resulting pictorial representation of R is called
  the directed graph representation of the relation
  R

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Relations
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Relations

• Directed graph (Digraph) representation of R
• Each dot is called a vertex
• If a vertex is labeled, a, then it is also called
  vertex a
• An arc from a vertex labeled a, to another
  vertex, b is called a directed edge, or directed
  arc from a to b
• The ordered pair (A , R) a directed graph, or
  digraph, of the relation R, where each element of
  A is a called a vertex of the digraph

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Relations

• Directed graph (Digraph) representation of R
  (Continued)
• For vertices a and b , if a R b, a is adjacent to
  b and b is adjacent from a
• Because (a, a) ? R, an arc from a to a is drawn
  because (a, b) ? R, an arc is drawn from a to b.
  Similarly, arcs are drawn from b to b, b to c , b
  to a, b to d, and c to d
• For an element a ? A such that (a, a) ? R, a
  directed edge is drawn from a to a. Such a
  directed edge is called a loop at vertex a

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Relations

• Directed graph (Digraph) representation of R
  (Continued)
• Position of each vertex is not important
• In the digraph of a relation R, there is a
  directed edge or arc from a vertex a to a vertex
  b if and only if a R b
• Let A a ,b ,c ,d and let R be the relation
  defined by the following set
• R (a ,a ), (a ,b ), (b ,b ), (b ,c ), (b
  ,a ), (b ,d ), (c ,d )

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Relations

• Domain and Range of the Relation
• Let R be a relation from a set A into a set B.
  Then R ? A x B. The elements of the relation R
  tell which element of A is R-related to which
  element of B

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Relations
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Relations
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Relations
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Relations

• Let A 1, 2, 3, 4 and B p, q, r. Let R
  (1, q), (2, r ), (3, q), (4, p). Then R-1
  (q, 1), (r , 2), (q, 3), (p, 4)
• To find R-1, just reverse the directions of the
  arrows
• D(R) 1, 2, 3, 4 Im(R-1), Im(R) p, q, r
  D(R-1)

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Relations
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Relations
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Relations

• Constructing New Relations from Existing
  Relations

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Relations
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Relations

• Example
• Consider the relations R and S as given in Figure
  3.7.
• The composition S ? R is given by Figure 3.8.

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Relations
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Relations
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Relations
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Relations
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Relations
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Relations
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Relations
Example 3.1.26 continued
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Relations
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Relations

• Example 3.1.27 continued

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Relations
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Relations

• Example 3.1.31 continued

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Relations
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Relations

• Example 3.1.32 continued

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Relations
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Relations
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Relations

• Consider the relation R (1, 1), (2, 2), (3,
  3), (4, 4), (5, 5), (1, 4), (4, 1), (2, 3), (3,
  2) on the set S 1, 2, 3, 4, 5.
• The digraph is divided into three distinct
  blocks. From these blocks the subsets 1, 4, 2,
  3, and 5 are formed. These subsets are
  pairwise disjoint, and their union is S.
• A partition of a nonempty set S is a division of
  S into nonintersecting nonempty subsets.

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Relations
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Relations

• Example Let A denote the set of the lowercase
  English alphabet. Let B be the set of lowercase
  consonants and C be the set of lowercase vowels.
  Then B and C are nonempty, B n C ?, and A B ?
  C. Thus, B, C is a partition of A.
• Let A be a set and let A1, A2, A3, A4, A5 be a
  partition of A. Corresponding to this partition,
  a Venn diagram, can be drawn, Figure 3.13

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Relations
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Relations
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Relations
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Relations
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Relations
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Relations

• Let A a, b, c , d, e , f , g , h, i, j . Let
  R be a relation on A such that the digraph of R
  is as shown in Figure 3.14.
• Then a, b, c , d, e , f , c , g is a directed
  walk in R as a R b,b R c,c R d,d R e, e R f , f R
  c, c R g. Similarly, a, b, c , g is also a
  directed walk in R. In the walk a, b, c , d, e ,
  f , c , g , the internal vertices are b, c , d, e
  , f , and c , which are not distinct as c
  repeats.
• this walk is not a path. In the walk a, b, c , g
  , the internal vertices are b and c , which are
  distinct. Therefore, the walk a, b, c, g is a
  path.

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Relations
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Partially Ordered Sets
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Partially Ordered Sets
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Partially Ordered Sets
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Partially Ordered Sets
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Partially Ordered Sets
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Partially Ordered Sets
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Partially Ordered Sets
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Partially Ordered Sets

• Digraphs of Posets
• Because any partial order is also a relation, a
  digraph representation of partial order may be
  given.
• Example On the set S a, b, c, consider the
  relation R (a, a),
  (b, b), (c , c ), (a, b).
• From the directed graph it follows that the given
  relation is reflexive and transitive.
• This relation is also antisymmetric because there
  is a directed edge from a to b, but there is no
  directed edge from b to a. Again, in the graph
  there are two distinct vertices a and c such that
  there are no directed edges from a to c and from
  c to a.

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Partially Ordered Sets

• Digraphs of Posets
• Let S 1, 2, 3, 4, 6, 12. Consider the
  divisibility relation on S, which is a partial
  order
• A digraph of this poset is as shown in Figure 3.20

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Partially Ordered Sets
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Partially Ordered Sets

• Closed Path
• On the set S a, b, c consider the relation
  R (a, a), (b, b), (c , c ),
  (a, b), (b, c ), (c , a)
• The digraph of this relation is given in Figure
  3.21
• In this digraph, a, b, c , a form a closed path.
  Hence, the given relation is not a partial order
  relation

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Partially Ordered Sets
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Partially Ordered Sets

• Hasse Diagram
• Let S 1, 2, 3. Then P(S) ?, 1, 2, 3,
  1, 2, 2, 3, 1, 3, S
• Now (P(S),) is a poset, where denotes the set
  inclusion relation. The poset diagram of (P(S),)
  is shown in Figure 3.22

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Partially Ordered Sets
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Partially Ordered Sets

• Hasse Diagram
• Let S 1, 2, 3. Then P(S) ?, 1, 2, 3,
  1, 2, 2, 3, 1, 3, S
• (P(S),) is a poset, where denotes the set
  inclusion relation
• Draw the digraph of this inclusion relation (see
  Figure 3.23). Place the vertex A above vertex B
  if B ? A. Now follow steps (2), (3), and (4)

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Partially Ordered Sets
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Partially Ordered Sets

• Hasse Diagram
• Consider the poset (S,), where S 2, 4, 5, 10,
  15, 20 and the partial order is the
  divisibility relation
• In this poset, there is no element b ? S such
  that b ? 5 and b divides 5. (That is, 5 is not
  divisible by any other element of S except 5).
  Hence, 5 is a minimal element. Similarly, 2 is a
  minimal element

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Partially Ordered Sets

• Hasse Diagram
• 10 is not a minimal element because 2 ? S and 2
  divides 10. That is, there exists an element b ?
  S such that b lt 10. Similarly, 4, 15, and 20 are
  not minimal elements
• 2 and 5 are the only minimal elements of this
  poset. Notice that 2 does not divide 5.
  Therefore, it is not true that 2 b, for all b ?
  S, and so 2 is not a least element in (S,).
  Similarly, 5 is not a least element. This poset
  has no least element

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Partially Ordered Sets
Figure 3.24

• Hasse Diagram
• There is no element b ? S such that b ?15, b gt
  15, and 15 divides b. That is, there is no
  element b ? S such that 15 lt b. Thus, 15 is a
  maximal element. Similarly, 20 is a maximal
  element.
• 10 is not a maximal element because 20 ? S and 10
  divides 20. That is, there exists an element b ?
  S such that 10 lt b. Similarly, 4 is not a maximal
  element.

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Partially Ordered Sets
Figure 3.24

• Hasse Diagram
• 20 and 15 are the only maximal elements of this
  poset
• 10 does not divide 15, hence it is not true that
  b 15, for all b ? S, and so 15 is not a
  greatest element in (S,)
• This poset has no greatest element

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Partially Ordered Sets
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Partially Ordered Sets
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Partially Ordered Sets
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Partially Ordered Sets
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Partially Ordered Sets
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Partially Ordered Sets
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Partially Ordered Sets
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Partially Ordered Sets
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Partially Ordered Sets

• Non-distributive Lattice
• Because a ? (b ? c ) a ? 1 a 0 0 ? 0
  (a ? b) ? (a ? c ), this is not a distributive
  lattice

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Partially Ordered Sets
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Partially Ordered Sets

• Complement
• Let D30 denote the set of all positive divisors
  of 30. Then D30 1, 2, 3, 5, 6, 10, 15, 30.
• (D30,) is a poset where a b if and only if a
  divides b ( is the divisibility relation)
• D30,) is a poset where a b if and only if a
  divides b ( is the divisibility relation).
  Because 1 divides all elements of D30, it follows
  that 1 m, for all m ? D30. Therefore, 1 is the
  least element of this poset.

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Partially Ordered Sets

• Complement
• Let a, b ? D30. Let d gcda, b and m lcma,
  b. Now d a and d b. Hence, d a and d b.
• This shows that d is a lower bound of a, b. Let
  c ? D30 and c a, c b.
• This shows that d is a lower bound of a, b. Let
  c ? D30 and c a, c b. Then, c a and c b
  and because d gcda, b, it follows that c d,
  so c d. Thus, d glba, b.

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Partially Ordered Sets

• Complement
• Because all the positive divisors of a, b are
  also divisors of 30, d ? D30, so d a ? b
• It can be shown that m ? D30 and m a ? b
• Hence, (D30,) is a lattice with the least
  element integer 1 and the greatest element 30

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Partially Ordered Sets

• Complement
• For any element a ? D30, (30/a) ? D30
• Using the properties of gcd and lcm, it can be
  shown that a ? (30/a) 1 and a ? (30/a) 30
• 10 ? (30/10) gcd10, 3 1 and 10 ? (30/10)
  lcm10, 3 30
• Hence, 3 is a complement of 10 in this lattice

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Partially Ordered Sets
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Application Relational Database

• A database is a shared and integrated computer
  structure that stores
• End-user data i.e., raw facts that are of
  interest to the end user
• Metadata, i.e., data about data through which
  data are integrated
• A database can be thought of as a well-organized
  electronic file cabinet whose contents are
  managed by software known as a database
  management system that is, a collection of
  programs to manage the data and control the
  accessibility of the data

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Application Relational Database

• In a relational database system, tables are
  considered as relations
• A table is an n-ary relation, where n is the
  number of columns in the tables
• The headings of the columns of a table are called
  attributes, or fields, and each row is called a
  record
• The domain of a field is the set of all
  (possible) elements in that column

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Application Relational Database

• Each entry in the ID column uniquely identifies
  the row containing that ID
• Such a field is called a primary key
• Sometimes, a primary key may consist of more than
  one field

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Application Relational Database

• Structured Query Language (SQL)
• Information from a database is retrieved via a
  query, which is a request to the database for
  some information
• A relational database management system provides
  a standard language, called structured query
  language (SQL)
• A relational database management system provides
  a standard language, called structured query
  language (SQL)

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Application Relational Database

• Structured Query Language (SQL)
• An SQL contains commands to create tables, insert
  data into tables, update tables, delete tables,
  etc.
• Once the tables are created, commands can be used
  to manipulate data into those tables.
• The most commonly used command for this purpose
  is the select command. The select command allows
  the user to do the following
• Specify what information is to be retrieved and
  from which tables.
• Specify conditions to retrieve the data in a
  specific form.
• Specify how the retrieved data are to be
  displayed.

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CSE 2353 OUTLINE

1. Sets
2. Logic
3. Proof Techniques
4. Integers and Induction
5. Relations and Posets
6. Functions
7. Counting Principles
8. Boolean Algebra

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Learning Objectives

• Learn about functions
• Explore various properties of functions
• Learn about sequences and strings
• Become familiar with the representation of
  strings in computer memory
• Learn about binary operations

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Functions
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Functions
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Functions

• Every function is a relation
• Therefore, functions on finite sets can be
  described by arrow diagrams. In the case of
  functions, the arrow diagram may be drawn
  slightly differently.
• If f A ? B is a function from a finite set A
  into a finite set B, then in the arrow diagram,
  the elements of A are enclosed in ellipses rather
  than individual boxes.

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Functions

• To determine from its arrow diagram whether a
  relation f from a set A into a set B is a
  function, two things are checked
• Check to see if there is an arrow from each
  element of A to an element of B
• This would ensure that the domain of f is the set
  A, i.e., D(f) A
• Check to see that there is only one arrow from
  each element of A to an element of B
• This would ensure that f is well defined

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Functions

• Let A 1,2,3,4 and B a, b, c , d be sets
• The arrow diagram in Figure 5.6 represents the
  relation f from A into B
• Every element of A has some image in B
• An element of A is related to only one element of
  B i.e., for each a ? A there exists a unique
  element b ? B such that f (a) b

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Functions

• Therefore, f is a function from A into B
• The image of f is the set Im(f) a, b, d
• There is an arrow originating from each element
  of A to an element of B
• D(f) A
• There is only one arrow from each element of A to
  an element of B
• f is well defined

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Functions

• The arrow diagram in Figure 5.7 represents the
  relation g from A into B
• Every element of A has some image in B
• D(g ) A
• For each a ? A, there exists a unique element b ?
  B such that g(a) b
• g is a function from A into B

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Functions

• The image of g is Im(g) a, b, c , d B
• There is only one arrow from each element of A to
  an element of B
• g is well defined

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Functions

• Let h be the relation described by the arrow
  diagram in Figure 5.8
• Every element of A has some image in B i.e.,
  there is an arrow originating from each element
  of A to an element of B. Therefore,
• D(h) A
• However,element 1 has two images in B i.e.,
  there are two arrows originating from 1, one
  going to a and another going to b, so h is not
  well defined. Thus, the first condition of
  Definition 5.1.1 is satisfied,but the second one
  is not.
• Therefore, h is not a function

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Functions

• The arrow diagram in Figure 5.9 represents a
  relation from A into B
• Not every element of A has an image in B. For
  example, the element 4 has no image in B. In
  other words, there is no arrow originating from 4
• Therefore, 4 ? D(k), so D(k) ? A
• This implies that k is not a function from A into
  B

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Functions

• Numeric Functions
• If the domain and the range of a function are
  numbers, then the function is typically defined
  by means of an algebraic formula
• Such functions are called numeric functions
• Numeric functions can also be defined in such a
  way so that different expressions are used to
  find the image of an element

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Functions
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Functions
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Functions
Example 5.1.16

• Let A 1,2,3,4 and B a, b, c , d. Let f
  A ? B be a function such that the arrow diagram
  of f is as shown in Figure 5.10
• The arrows from a distinct element of A go to a
  distinct element of B. That is, every element of
  B has at most one arrow coming to it.
• If a1, a2 ? A and a1 a2, then f(a1) f(a2).
  Hence, f is one-one.
• Each element of B has an arrow coming to it. That
  is, each element of B has a preimage.
• Im(f) B. Hence, f is onto B. It also follows
  that f is a one-to-one correspondence.

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Functions
Example 5.1.18

• Let A 1,2,3,4 and B a, b,
  c , d, e
• f 1 ? a, 2 ? a, 3 ? a, 4 ? a
• For this function the images of distinct elements
  of the domain are not distinct. For example 1 ?
  2, but f(1) a f(2) .
• Im(f) a ? B. Hence, f is neither one-one nor
  onto B.

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Functions

• Let A 1,2,3,4 and B a, b,
  c , d, e
• f 1 ? a, 2 ? b, 3 ? d, 4 ? e
• For this function, the images of distinct
  elements of the domain are distinct. Thus, f is
  one-one. In this function, for the element c of
  B, the codomain, there is no element x in the
  domain such that f(x) c i.e., c has no
  preimage. Hence, f is not onto B.

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Functions
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Functions

• Let A 1,2,3,4, B a, b, c , d, e,and C
  7,8,9. Consider the functions f A ? B, g B
  ? C as defined by the arrow diagrams in Figure
  5.14.
• The arrow diagram in Figure 5.15 describes the
  function h g ? f A ? C.

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Functions
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Functions
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Special Functions and Cardinality of a Set
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Special Functions and Cardinality of a Set
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Special Functions and Cardinality of a Set
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Special Functions and Cardinality of a Set
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Special Functions and Cardinality of a Set
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Special Functions and Cardinality of a Set
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Special Functions and Cardinality of a Set
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Special Functions and Cardinality of a Set
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Special Functions and Cardinality of a Set
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Special Functions and Cardinality of a Set
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Special Functions and Cardinality of a Set
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Special Functions and Cardinality of a Set
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Special Functions and Cardinality of a Set
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Special Functions and Cardinality of a Set
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Special Functions and Cardinality of a Set
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Sequences and Strings
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• m is the lower limit of the sum, n the upper
  limit of the sum, and ai the general term of
  the sum

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• m is the lower limit of the sum, n the upper
  limit of the sum, and ai the general term of the
  product.

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Sequences and Strings

• Representing Strings into Computer Memory
• A convenient way of storing a string into
  computer memory is to use an array.
• Programming languages such as C and Java
  provide a data type to manipulate strings.
• This data type includes algorithms to implement
  operations such as concatenation, finding the
  length of a string,determining whether a string
  is a substring in another string, and finding a
  substring into another string.

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Binary Operations
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Binary Operations

• Let A be an alphabet. Let A be the set of all
  nonempty strings on the alphabet A. If u, v ?
  A,i.e., u and v are two nonempty strings on
  A,then the concatenation of u and v, uv ? A.
  Thus, the concatenation of strings is a binary
  operation on A. Also, by Theorem 5.3.23(ii), the
  concatenation operation is associative.
• Hence, A becomes a semigroup with respect to the
  concatenation operation.
• Called a free semigroup generated by A.

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Binary Operations

• Let A denote the set of all words including
  empty word ?. By Theorem 5.3.23(i), for all s ?
  A, ?s s s?. Thus, A becomes a
  monoid with identity 1 ?.
• This monoid is called a free monoid generated by
  A.
• If A contains more than one element, then A is a
  noncommutative monoid.