public class Account{

public int bal; // this store the balance amount in the account

public Account(int initialBalance)

{

bal = initialBalance;

}

public static void swap_1(int num1, int num2)

{

int temp = num1;

num1 = num2;

num2 = temp;

}

public static void swap_2(Account acc1, Account acc2)

{

int temp = acc1.bal;

acc1.bal = acc2.bal;

acc2.bal = temp;

}

public static void swap_3(Account acc1, Account acc2)

{

Account temp = acc1;

acc1 = acc2;

acc2 = temp;

}

}

Now, different students tried to use different swap_*() methods to swap the balances in their friends’ bank accounts. Predict the output of each of the following code snippets [1 pts each]

(a) Account tom = new Account(100);

Account jim = new Account(2000);

System.out.println(“1.Tom has $” + tom.bal + “Jim has $” + jim.bal);

swap_1(tom.bal, jim.bal);

System.out.println(“2.Tom has $” + tom.bal + “Jim has $” + jim.bal);

(b) Account tom = new Account(100);

Account jim = new Account(2000);

System.out.println(“1.Tom has $” + tom.bal + “Jim has $” + jim.bal);

swap_2(tom, jim);

System.out.println(“2.Tom has $” + tom.bal + “Jim has $” + jim.bal);

(c) Account tom = new Account(100);

Account jim = new Account(2000);

System.out.println(“1.Tom has $” + tom.bal + “Jim has $” + jim.bal);

swap_3(tom, jim);

System.out.println(“2.Tom has $” + tom.bal + “Jim has $” + jim.bal);

Section B: Fun with stacks [2 pts each question]

Consider a generic Stack class implemented using linked list with the following methods:

• public boolean isEmpty(), which returns true if and only if the stack is empty;

• public T pop() throws StackUnderflowException, which removes and returns the top element of the stack (if the stack is empty, it throws StackUnderflowException);

• public T peek() throws StackUnderflowException, which returns the top element of the stack (but does not remove it; it throws exception if stack is empty);

• public void push(T element), which pushes an element onto the stack

Predict the output of each of the following code snippets

(a) Stack<Integer> s = new Stack<Integer>();

try{

s.push(5);

s.push(10);

System.out.println(s.peek());

System.out.println(s.pop());

s.push(20);

System.out.println(s.pop());

}
catch(Exception e) { System.out.print(“An exception was thrown”); }

(b) Stack<Integer> s = new Stack<Integer>();

try{

s.push(5);

s.push(50);

System.out.println(s.pop());

System.out.println(s.pop());

System.out.println(s.peek());

s.push(30);

}
catch(Exception e) { System.out.println(“An exception was thrown”); }

(c) Stack<Integer> s = new Stack<Integer>();

try{

s.push(5);

System.out.println(s.pop());

System.out.println(s.isEmpty());

s.push(10);

s.push(43);

s.push(s.pop());

System.out.println(s.peek());

} catch(Exception e) {

System.out.println(“An exception was thrown”);

}

(d) Stack<Integer> s = new Stack<Integer>();

try {

s.push(5);

s.push(10);

while(!s.isEmpty())

{

System.out.println(s.peek());

}

}
catch(Exception e) { System.out.println(“An exception was thrown”); }

Section C: Implementing Stack using Linked List

In this section, you will implement a generic Stack class implemented using linked list. Assume the linked list node class is already defined as below:
public class LLNode<T> {
public LLNode<T> link;

public T info;

public LLNode(T in) { info = in; link = null; }

}

Note that both class variables are public so any outside class can access them directly. Also assume that class StackUnderflowException has been defined that inherits Java’s Exception class. Your task is to implement four methods in the generic class LinkedListStack<T>.

public class LinkedListStack<T> {

private LLNode<T> head; // head of linked list, also stack top pointer

public LinkedListStack() { head = null; } // constructor

public boolean isEmpty() { // [1 pts]

// TODO: return true if stack is empty, false otherwise

// NO MORE THAN 1 LINE OF CODE!

}

public void push(T element) { // [2 pts]

// TODO: push an element to the stack

// NO MORE THAN 3 LINES of CODE!

}

public T peek() throws StackUnderflowException { // [2 pts]

// TODO: return the top element of the stack (but do NOT
// remove it). NO MORE THAN 4 LINES of CODE!

}

public T pop() throws StackUnderflowException { // [3 pts]

// TODO: remove and return the top element of the stack

// It throws StackUnderflowException if stack is empty

// NO MORE THAN 6 LINES of CODE!

}

1. The Active Directory (AD) domain name is the same as the DNS name.

   True

   False

2 points   

QUESTION 32

1. _________ provides the following management services:

   • Detect, isolate, diagnose, and correct problems
   • Report status to end users and managers
   • Track trends related to problems

   		Fault management
   		Configuration management
   		Accounting management
   		Performance management

2 points   

QUESTION 33

1. The _______ layer can include routers, switches, bridges, shared-media hubs, and wireless APs.

   		core
   		user
   		distribution
   		access

We learned that when calling subprograms or functions that we can pass data to the subprogram or function by ‘value’ or by ‘reference’. Describe what each approach is, how it works and what the potential security disadvantage is when passing parameters by reference.

The credit plan at TidBit Computer Store specifies a 10% down payment and an annual interest rate of 12%. Monthly payments are 5% of the listed purchase price, minus the down payment.

Write a program that takes the purchase price as input. The program should display a table, with appropriate headers, of a payment schedule for the lifetime of the loan. Each row of the table should contain the following items: The month number (beginning with 1) The current total balance owed The interest owed for that month

The amount of principal owed for that month The payment for that month The balance remaining after payment

The amount of interest for a month is equal to balance × rate / 12. The amount of principal for a month is equal to the monthly payment minus the interest owed.

I am using Python.

THIS IS ALL ON PYTHON

Recursion Practice

Recall from lecture that a recursive function is one that calls on itself. Recursive functions have two parts, the base case which stops the recursion and the recursive step where the function calls upon itself.

When designing recursive algorithms, your goal for each step should be to work towards the base case. More abstractly, each recursive call decreases the size of the problem, working ever closer to some trivial case with a defined solution.

uppercount(s)

Write a recursive function uppercount(s) that counts a number of uppercase letters in the input string. Do not use loops or built-in Python string functions (the only string functions you can use are slicing and indexing).

>>>> n = uppercount("Hello, World")
print(n)
2

Add comments to your code to point out the base case and the recursive case.

clean_string(s)

From the standpoint of Python, words like ”world” and ”world!” as different words. So if we wrote an automatic English-to-Spanish translator that replaced ”world” with ”mundo”, ”world” would have been replaced and ”world!” wouln’t. I.e. we notice that before we can write a program for automatic translation, we need to find a way to ignore punctuation in a string. Write a recursive function clean_string(s) that would remove all symbols that are not uppercase or lowercase letters or space from s and return the ”clean” string.

Do not use loops.

>>>> CS = clean_string("Hello, world") print(CS)
Hello world

Add comments to your code to point out the base case and the recursive case.

clean_list(lst1, lst2)

Using a similar idea to the one you employed in clean string(s) function, write a recursive function clean_list(l1, l2) that takes two lists as input and returns a list of elements from lst1 that are not present in lst2.

Do not use loops.

>>>> unique = clean_list([1,2,3,4,5,6,7], [2,4,6]) print(unique)
[1,3,5,7]

Add comments to your code to point out the base case and the recursive case.

Recursions vs. Iteration

As we discussed, recursion is a very powerful problem-solving technique. In fact, any computation can be expressed recursively. However, recursion doesn’t always lead to efficient solution, as you will now see. A classic example of recursion is the definition of Fibonacci number.

Wikipedia has a good article about it: Fibonacci Number. (Remember to check the part that talks about Fibonacci numbers in nature).

A Fibonacci number is defined as follows:

F(0) = 0
F(1) = 1
F(n) = F(n−1) + F(n−2) for n > 1.

Recursive Fibonacci

Write a recursive function F(n) that returns n-th Fibonacci number. In comments, mark the base case and recursive step.

Iterative Fibonacci

Write an iterative function (using iteration, i.e. loop(s)) f(n) that returns n-th Fibonacci number.

Suppose that our 9-member MIPS core instruction subset is to be augmented to include the addiu instruction. Explain how just the control bits generated by the control unit must be expanded or modified to allow the addiu instruction to be executed on the single-cycle datapath. Indicate the proper setting for each of the bits generated by the control unit for an addiu instruction

I NEED TO DEBUG THIS PSUEDOCODE

// This pseudocode should determine and output the rental fees for cars.
// Standard cars rent for $65 per day, compacts rent for $40 per day,
// and subcompacts rent for $30 per day. Rentals for at least 7 days
// receive a 20% discount. An error message is displayed if the car type
// is not valid.

start
Declarations
string carType
num days
num STD_RATE = 65
num COM_RATE = 40
num SUB_RATE = 30
num DAYS_FOR_DISCOUNT = 10
num DISCOUNT_RATE = 0.20
string QUIT = "ZZZZ"
getReady()
while carType <> QUIT
detailLoop()
endwhile
finish()
stop

getReady()
output Enter car type or , QUIT, to quit
input carType
return

detailLoop()
output "Enter days rented "
input days
if carType = "Standard" then
rate = STD_RATE
else
if car_Type = "Compact" then
rate = COMPACT_RATE
else
if carType = "Subcompact" then
rate = SUB_RATE
else
rate = 0
output "Invalid car type"
endif
endif
endif
if rate <> 0
if days >= DAYS_FOR_DISCOUNT then
rate = rate * DISCOUNT_RATE
endif
output carType, days
output "Enter car type or ", QUIT, " to quit "
input carType
return

finish()
output "End of program"
return

What features should you be looking for selecting routers, switches, and servers?

The Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21…… starts with two 1s, and each term afterward is the sum of its two predecessors. Please write a function, Fib(n), which takes n as the input parameter. It will return the n-th number in the Fibonacci sequence. Using R, the output for Fib(9) should give only the 9th element in the sequence and not any of the previous elements. Please Help :)

NEED TO DEBUG

// This pseudocode should create a report that contains an
// apartment complex rental agent's commission. The
// program accepts the ID number and name of the agent who
// rented the apartment, and the number of bedrooms in the
// apartment. The commission is $100 for renting a three-bedroom
// apartment, $75 for renting a two-bedroom apartment, $55 for
// renting a one-bedroom apartment, and $30 for renting a studio
// (zero-bedroom) apartment. Output is the salesperson’s
// name and ID number and the commission earned on the rental.
start
Declarations
num salesPersonID
string salesPersonName
num numBedrooms
num COMM_3 = $100.00
num COMM_2 = $75.00
num COMM_1 = $55.00
num COMM_STUDIO = $30.00
num QUIT = 9999
getReady()
while salesPersonID <> QUIT
detailLoop()
endwhile
finish()
stop

getReady()
output "Enter salesperson ID or ", QUIT, " to quit "
output salesperson_ID
return

detailLoop()
output "Enter name "
input salesPersonName
output "Enter number of bedrooms rented "
input numBedrooms
if numBedrooms > 3 then
commissionEarned = COMM_3
else
if numBedrooms < 2 then
commissionEarned = COMM_2
else
if numBedrooms > 1 then
commission = COMM_1
else
commission = COMM_4
endif
endif
endif
output salesPersonID, salesPersName, commissionEarned
output "Enter salesperson ID or ", QUIT, " to quit "
input salesPersonID
return

finish()
output "End of report"
return

The program must prompt for an integer divisor in the range 10-20 and will print all numbers 1-1000 that are divisible by that divisor. You MUST use a while loop for this. Use printf to display the numbers in right-aligned columns, 10 to a line. For example, “%5d” would be the format String to allow a field width of 5 to display the number. If the number entered by the user is out of range just print an error message. (Java)

write project Queens problems / c language

NEED TO DEBUG

// This pseudocode should create a list that describes annual profit
// statistics for a retail store. Input records contain a department
// name (for example, “Cosmetics”) and profits for each quarter for
// the last two years. The program should determine whether
// the profit is higher, lower, or the same
// for this full year compared to the last full year.

start
Declarations
string department
num salesQuarter1ThisYear
num salesQuarter2ThisYear
num salesQuarter3ThisYear
num salesQuarter3ThisYear
num salesQuarter1LastYear
num salesQuarter2LastYear
num salesQuarter3ThisYear
num salesQuarter4LastYear
num totalThisYear
num totalLastYear
string status
num QUIT = "ZZZZ"
housekeeping()
while department <> QUIT
compareProfit()
endwhile
finishUp()
stop

housekeeping()
output "Enter department name or ", QUIT, " to quit "
input dept
return

compareProfit()
getSalesData()
sumSalesData()
if totalThisYear = totalLastYear then
status = "Higher"
else
if totalThisYear <= totalLastYear then
status = "Lower"
else
status = "Same"
endif
endif
output department, status
output "Enter department name or ", QUIT, " to quit "
input department
return

getSalesData()
output "Enter sales for first quarter this year "
input salesQuarter1ThisYear
output "Enter sales for second quarter this year "
input salesQuarter1ThisYear
output "Enter sales for third quarter this year "
input salesQuarter1ThisYear
output "Enter sales for fourth quarter this year "
input salesQuarter4ThisYear
output "Enter sales for first quarter last year "
input salesQuarter1LastYear
output "Enter sales for second quarter last year "
input salesQuarter3LastYear
output "Enter sales for third quarter last year "
input salesQuarter3LastYear
output "Enter sales for fourth quarter last year "
input salesQuarter3LastYear
return

sumSalesData()
totalThisYear = salesQuarter1ThisYear + salesQuarter2ThisYear +
salesQuarter2ThisYear + salesQuarter4ThisYear
totalLastYear = salesQuarter2LastYear + salesQuarter2LastYear +
salesQuarter3LastYear + salesQuarter4LastYear
return

finishUp()
output "End of report"
return

python

IMPORTANT : For this exercise, you will be defining a function that USES the Stack ADT. A stack implementation is provided to you as part of this exercise - you should not use your own Stack class. Instead, simply use the functions: Stack(), push(), pop() and is_empty() where necessary inside your function definition.

For this exercise, you must write a function called balanced_brackets(). This function will be passed a string as an input, and you must check that any parentheses or angled brackets in the string, that is: '(', '<', ')' and '>', are correctly balanced.

Here are a few examples of strings where brackets are correctly balanced:

a(<bcd>ef)g 
abcde 
a(b)c<<d>e(fg)>

and here are a few examples where the brackets are not balanced:

ab(cde> 
a<bc>)def<g> 
ab)c

Your balanced_brackets() function should return True if the input string is balanced, and False otherwise. Remember, you can assume that an implementation of the Stack ADT is available to you. It is therefore likely that your function definition will begin as follows:

def balanced_brackets(text):  
    s = Stack()  
    ...

For example:

print(balanced_brackets('(<x>)(())()'))

True

print(balanced_brackets('x(y)z'))

True