1. CSS 342 Data Structures, Algorithms, and Discrete Mathematics I
Lecture
Carrano Chapt. 15 (partial); CUSACK CHAPT 4.

2. Agenda Queues HW5: Questions / Discussion Peer Design Review
How will we grade this?
Signatures of BST
Peer Design Review
Trees: BST
If time, prop logic

3. Queues

4. A Pointer-Based Implementation2 4 1 7
NULL
front
back

5. class MyQueue {
public:
MyQueue();
~MyQueue();
bool isEmpty() const;
int getCount() const;
int Dequeue() throw (exception);
void Enqueue(const int &val);
private:
struct Node
int val;
Node *next = NULL; };
Node *front;
Node *back;
int count;

6. bool MyQueue::isEmpty() const{
if (front == NULL)
return true; }
else
return false;
int MyQueue::getCount() const
return count;
MyQueue::MyQueue() {
front = NULL;
back = NULL;
count = 0; }
MyQueue::~MyQueue() {
while (getCount() > 0)
Dequeue(); }

7. void MyQueue::Enqueue(const int &value){
Node *insNode;
insNode = new Node;
insNode->val = value;
if (count == 0)
back = insNode;
front = insNode; }
else
back->next = insNode;
count++;
int MyQueue::Dequeue() throw (exception) {
if (count == 0)
throw exception("Queue is empty"); }
Node *temp = front;
int retVal = temp->val;
front = front->next;
delete temp;
if (count == 1)
back == NULL;
count--;
return retVal;

8. Are we done? Do we need to overload =, copy constructor?
Why or why not?
Any other overloads?

9. Queue usage in OS, Network, Services
Message queues
Scheduling queue
TCP Flow control
Payment processing

10. Unix Message Queues 1 2 Message queue (id = msgid)
struct mymesg {
long mytype;
char mtext[512];
} message_body;
int main( void ) {
int msgid = msgget( 100, IPC_CREAT );
strcpy( message_body.mtext, “hello world\n” );
msgsnd( msgid, &message_body, 512, 0 ); }
struct mymesg {
long mytype;
char mtext[512];
} message_body;
int main( void ) {
int msgid = msgget( 100, IPC_CREAT );
msgrcv( msgid, &message_body, 512, 0, 0 );
cout << message_body.mtext << endl; }
Message queue
(id = msgid) 1 2
Some other process can
enqueue and dequeue a message

11. Multilevel Queue Scheduling
Each queue has its own scheduling algorithm,
foreground (interactive) – RR, 80% CPU time
background (batch) – FCFS, 20% CPU time

12. Multilevel Feedback-Queue Scheduling
A new job enters queue Q0 which is served FCFS. When it gains CPU, job receives 8 milliseconds. If it does not finish in 8 milliseconds, job is moved to queue Q1.
At Q1 job is again served FCFS and receives 16 additional milliseconds. If it still does not complete, it is preempted and moved to queue Q2.

13. Flow Control X X X X Sending application Receiving application packet
Send Socket Buffer
Receive Socket Buffer
send
ack
read blocked
packet X
packet
packet retransmitted
packet dropped
Application is slow to read. X
packet
packet retransmitted
packet dropped X X
write blocked

14. Lab 5
PEER DESIGN REVIEW

15. Grading Lab5 Copy all *.cpp and *.h files into a directory
Copy BankTransIn.txt into directory
Open up a command line window for compilation
cl *.cpp
name.exe > BankTrans.out 2> BankTrans.err

16. Trees

17. FIGURE 15-1 (a) A tree; (b) a subtree of the tree in part a
Terminology
Root
Parent
Child
Ancestor
Descendent
Height
Subtree
N-ary tree
Binary Tree
FIGURE 15-1 (a) A tree; (b) a subtree of the tree in part a

18. Binary Tree: Algebraic Expressions.

19. Binary Search Tree Well suited for searching
For each node n, a binary search tree satisfies the following three properties:
n ’s value is greater than all values in its left subtree T L .
n ’s value is less than all values in its right subtree T R .
Both T L and T R are binary search trees.
Well suited for searching
Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

20. Binary Search Tree: Example

21. Binary Search Tree: Example
FIGURE Binary search trees
with the same data as in Figure 15-13

22. Binary Search Tree: Example
FIGURE Binary search trees
with the same data as in Figure 15-13
Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

23. Binary Search Tree: Example
FIGURE Binary search trees
with the same data as in Figure 15-13
Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

24. Computer Scientist of the Week
J. Alex Halderman
Professor at University of Michigan
Expert on Computer security and privacy
Focus on issues which impact public policy and society
Coursera: Securing Digital Democracy
2016: Advised for a recount in WI, MI
Explanation
Popular science top-10 brightest mind:
Popular Science

25. class BinarySearchTree{
public:
BinarySearchTree();
~BinarySearchTree();
void Insert(Object *);
bool Remove(const Object &);
bool Retrieve(const Object &, Object * &);
void Flush();
bool Contains(const Object &);
void Display() const;
bool isEmpty() const;
int Height() const;
int getCount() const;
private:
struct Node
Object *pObj;
Node *right;
Node *left; };
Node *root;

26. Searching a Binary Search Tree
Data Structures and Poblem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

27. Searching a Binary Search Tree: pseudo-code
Search(Obj *root) {
if (root == NULL) return NULL;
else if (target == *(root->pObj))
return (root->pItem); }
else if (target < *(root->pObj))
return Search(root->left);
else
return Search(root->right)

28. Creating a Binary Search Tree
FIGURE Empty subtree where the search algorithm terminates when looking for Frank
Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

29. Efficiency of BST Operations
Retrieve:
Insert:
Removal:
Traversal:

30. Class Bell

31. Propositional Logic

32. Text Book
An Active Introduction to Discrete Mathematics and Algorithms (version 2.6), Charles Cusack, David Santos, GNU Free Software, 2016:
athematics%20and%20Algorithms/ActiveIntroToDiscreteMathAndAlgorithms.2.6.pdf
Chapter 4: Logic.

33. Propositions
Proposition: a statement that is either true or false, but not both
Examples of propositions
Proposition p: All mathematicians wear sandals.
Proposition r: Discrete Math is fun.
Proposition v: This is not a proposition.
Proposition s: The sum of the first n odd integer is equal to n2 for n >1.
These are not propositions:
What is your name?
Go to the store and get me a steak?
Propositions have a truth value: True or False.

34. Propositions Math History Programming languages
The only positive integers that divide 7 are 1 and 7 itself: (true)
For every positive integer n, there is a prime number larger than n: (true)
History
Alejandro Inarritu won an Academy Award in 1999 for directing “Amores perros”: (false, he won in 2015 for Birdman)
Seattle held the World’s Fair, Expo 62: (true )
Programming languages
Boolean expressions in if-else, while, and for statements
for ( index = 0; index < 100; index++ ) {
…….; }
A proposition
Not a proposition

35. Compound propositions
Conjunction (AND) of p and q
notations: p ^ q, p && q
True only if both p and q are true
Truth table
Disjunction (OR) of p or q
Notations: p v q, p || q
True if either p or q or both are true
truth table p q
p ^ q F T p q
p v q F T

36. Compound Propositions
The negation (NOT) of p
Notations: not p, ¬p !p
Examples
P: = 3 (false)
!p: !(1 + 1 = 3) ≡ ≠ 3 (true)
Exclusive OR (XOR)
Notations: p xor q, p !p F T p q
p q F T

37. Binary Expressions in C++
How do you examine the behavior of if-else?
if ( a > = 1 && a <= 100 )
true if 1 ≤ a ≤ 100
if ( a >= 1 || a <= 100 )
always true
a < 1
1 <= a <= 100
100 < a
a >= 1
a <= 100
false
true
condition
proposition
a < 1
1 <= a <= 100
100 < a
a >= 1
a <= 100
false
true
condition
proposition

38. Theorems for Boolean algebra
Commutative Law
p v q = q v p
p ^ q = q ^ p
Associative Law
p v (q v r) = (p v q) v r
p ^ (q ^ r) = (p ^ q) ^ r
Distributive Law
p ^ (q v r) = (p ^ q) v (p ^ r)
p v (q ^ r) = (p v q) ^ (p v r)
Identity
p v F = p
P ^ T = p
Domination
p v T = T
p ^ F = F

39. Theorems for Boolean algebra
Complement Law (tautologies)
p ^ ¬p = F
p v ¬p = T
Square Law
p ^ p = p
p v p = p
Double Negation
¬(¬p) = p
Absorption
p ^ (p v q) = p
p v (p ^ q) = p

40. Precedence of Operators
NOT: ¬
AND: ^
XOR:
OR: v
Conditional: →
iff: ↔

41. Prove the following Boolean equations
¬p ^ q v p ^ ¬q = (¬p v ¬q) ^ (p v q)
p ^ q v p ^ q ^ r v p ^ ¬ q = p
(p ^ q v r ) ^ q = p ^ q ^ ¬r v ¬p ^ q ^ r v p ^ q ^ r