Given the following information about the branch frequencies (as percentages of all instructions) for a four-deep pipeline:Conditional branches 15%Jumps and calls 5%Conditional branches 60% are taken.We will use the following terms to answer the question:Branch misprediction penalty : The number of pipeline cycles that are wasted due to a branch misprediction.Branch hazard : A delay that occurs when a branch is taken, as it affects the processing of subsequent instructions.
Pipeline depth: The length of a pipeline is measured by the number of stages it has. A four-deep pipeline, for example, has four stages.Let's first find the total branch frequency by summing up the frequencies of both Conditional branches and Jumps and calls. Total Branch Frequency = Conditional Branches Frequency + Jumps and Calls Frequency= 15% + 5%= 20%Next, we need to determine the frequency of mispredictions for each type of branch.
The following table shows the frequency of mispredictions for each type of branch.Type of BranchMisprediction FrequencyUnconditional 0%Conditional 40% (60% taken)The branch misprediction penalty for unconditional branches is 0, while for conditional branches it is 1.4 cycles on average, since they have a 40% misprediction frequency.
Using the total frequency and branch misprediction penalty, we can now calculate the branch hazard cycle frequency for the pipeline. Branch Hazard Cycle Frequency = Total Branch Frequency × Branch Misprediction Penalty= 20% × (0.15 × 1.4 + 0.05 × 0)= 4.2%Next, we'll calculate the speedup if there were no branch hazards by dividing the ideal speed of the pipeline by the actual speed of the pipeline. Ideal Speed of the Pipeline = 1Cycle Time with Branch Hazards = 4 + 0.2 × 1.4 = 4.28 cyclesCycle Time without Branch Hazards = 4 cyclesSpeedup = Cycle Time with Branch Hazards / Cycle Time without Branch Hazards= 4.28 / 4= 1.07Therefore, the pipeline will be 1.07 times faster if there were no branch hazards.
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Write a java program that reads the width and length for a set of rectangles (unknow numbers) from input file (input.txt). The program should compute the area for each rectangle and show the result on the run screen as shown bellow.Also, you need to consider the following cases:
If width and length are equal, then a message (This is a square) should be displayed instead of area.
If width or length has negative values, then invalid message should be displayed instead of the area.
Here is a Java program that reads the width and length of rectangles from an input file, computes the area for each rectangle, and displays the results on the console. It also handles special cases such as squares and rectangles with negative values
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
public class RectangleAreaCalculator {
public static void main(String[] args) {
try {
// Read input from the file
File inputFile = new File("input.txt");
Scanner scanner = new Scanner(inputFile);
while (scanner.hasNextLine()) {
String line = scanner.nextLine();
String[] dimensions = line.split(" ");
int width = Integer.parseInt(dimensions[0]);
int length = Integer.parseInt(dimensions[1]);
if (width < 0 || length < 0) {
System.out.println("Invalid dimensions");
} else if (width == length) {
System.out.println("This is a square");
} else {
int area = width * length;
System.out.println("Area: " + area);
}
}
scanner.close();
} catch (FileNotFoundException e) {
System.out.println("Input file not found");
}
}
}
The program starts by opening the input file using the File class and creating a Scanner to read its contents.
It reads each line of the file, which represents the width and length of a rectangle, and splits it into separate dimensions.
The width and length are parsed as integers and stored in variables.
The program then checks for special cases: if the width or length is negative, it displays an "Invalid dimensions" message.
If the width and length are equal, it displays a "This is a square" message.
Otherwise, it calculates the area by multiplying the width and length, and displays the result.
The program continues reading and processing each line until there are no more lines in the file.
If the input file is not found, it displays an appropriate error message.
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Who is probably the closest to Dorothy (Judy Garland) in THE WIZARD OF OZ? a. The Scarecrow (Ray Bolger)
b. The Tin Man (Jack Haley) c. The Cowardly Lion (Bert Lahr)
d. Uncle Henry (Charlie Grapewin)
e. Professor Marvel (Frank Morgan)
The closest character to Dorothy (Judy Garland) in "The Wizard of Oz" would be the Scarecrow (Ray Bolger). Throughout their journey, the Scarecrow becomes Dorothy's loyal companion, offering her support, guidance, and friendship.
The Scarecrow shares Dorothy's quest for a brain, symbolizing her desire for wisdom and understanding. Together, they face the challenges of the Land of Oz, and the Scarecrow consistently displays a deep empathy and concern for Dorothy's well-being. Their bond is exemplified by their unwavering support and shared goal of finding the Wizard. Thus, the Scarecrow stands out as the character closest to Dorothy in the film.
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Using an example, explain what parallel arrays are [5]
Parallel arrays are a programming concept where two or more arrays of the same length are used to store related data. Each element in one array corresponds to an element in another array at the same index position.
For example, let's say we have two arrays: "names" and "ages". The "names" array stores the names of people and the "ages" array stores their ages. The first element in the "names" array corresponds to the first element in the "ages" array, the second element in the "names" array corresponds to the second element in the "ages" array, and so on.
Here is an example of how these parallel arrays could be declared and used in Python:
# declaring the parallel arrays
names = ["John", "Jane", "Bob", "Alice"]
ages = [25, 30, 27, 22]
# accessing elements in the parallel arrays
print(names[0] + " is " + str(ages[0]) + " years old.")
print(names[2] + " is " + str(ages[2]) + " years old.")
# output
# John is 25 years old.
# Bob is 27 years old.
In this example, we declare the "names" and "ages" arrays as parallel arrays. We then access specific elements in both arrays using the same index value, allowing us to retrieve the name and age of a person at the same time. This can be useful when storing and manipulating multiple sets of related data.
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The class declaration below declares a Passage class that is for efficient storage of characters (a string). You should implement this calss by following the steps below. class Passage { public: Passage(); // Default constructor Passage (const char *s); // Standard constructor Passage (const Passage &s); // Copy constructor ~Passage(); // Destructor int Length() const; // Returns length of Passage bool Equal (const Passage &s) const; // Compares 2 Passage void Set (const Passage &s); // Sets Passage void Print() const; // Prints Passage private: char *buff; // buffer for holding Passage (i.e., string) }; then you will need to test the following statements: Passage p1; // test default constructor cout<<"p1: "; p1.Print(); Passage p2("Hello"); // test std constructor cout<<"p2: "; p2.print(); Passage p3 (p2); // test copy constructor cout<<"p3: "; p3. Print(); cout<<"p3 length: " << p3. Length() << endl; // test Length() fn : p1.Set(p3); // Test Set() fn if (p2.Equal (p3) cout << "p2 and p3 are same" << endl; // test Equal() fn else cout << "p2 and p3 are different" << endl;
The provided code implements the Passage class with the required constructors, member functions, and data members. The class is used to efficiently store and manipulate character sequences.
Here is the implementation of the Passage class according to the provided class declaration
```cpp
#include <iostream>
#include <cstring>
class Passage {
public:
Passage(); // Default constructor
Passage(const char *s); // Standard constructor
Passage(const Passage &s); // Copy constructor
~Passage(); // Destructor
int Length() const; // Returns length of Passage
bool Equal(const Passage &s) const; // Compares 2 Passage
void Set(const Passage &s); // Sets Passage
void Print() const; // Prints Passage
private:
char *buff; // buffer for holding Passage (i.e., string)
};
Passage::Passage() {
buff = nullptr;
}
Passage::Passage(const char *s) {
int len = strlen(s);
buff = new char[len + 1];
strcpy(buff, s);
}
Passage::Passage(const Passage &s) {
int len = s.Length();
buff = new char[len + 1];
strcpy(buff, s.buff);
}
Passage::~Passage() {
delete[] buff;
}
int Passage::Length() const {
return strlen(buff);
}
bool Passage::Equal(const Passage &s) const {
return (strcmp(buff, s.buff) == 0);
}
void Passage::Set(const Passage &s) {
int len = s.Length();
delete[] buff;
buff = new char[len + 1];
strcpy(buff, s.buff);
}
void Passage::Print() const {
if (buff)
std::cout << buff;
std::cout << std::endl;
}
int main() {
Passage p1; // test default constructor
std::cout << "p1: ";
p1.Print();
Passage p2("Hello"); // test std constructor
std::cout << "p2: ";
p2.Print();
Passage p3(p2); // test copy constructor
std::cout << "p3: ";
p3.Print();
std::cout << "p3 length: " << p3.Length() << std::endl; // test Length() fn
p1.Set(p3); // Test Set() fn
if (p2.Equal(p3))
std::cout << "p2 and p3 are the same" << std::endl; // test Equal() fn
else
std::cout << "p2 and p3 are different" << std::endl;
return 0;
}
```
The Passage class is implemented with a default constructor, standard constructor, copy constructor, destructor, and various member functions. The data member `buff` is a character pointer used to store the character sequence.
The main function demonstrates the usage of the Passage class by creating instances of Passage objects and invoking member functions. It tests the behavior of the constructors, length calculation, equality comparison, setting one Passage object to another, and printing the contents of a Passage object.
Please note that the code provided is in C++. Ensure you have a C++ compiler to run the code successfully.
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Disadvantages About Security Robots (( I need the references
please ))
Disadvantages of security robots include limitations in handling complex situations and potential privacy concerns.
While security robots offer certain benefits such as continuous surveillance and deterrence, they also have their disadvantages. One limitation is their inability to handle complex situations that may require human judgment and decision-making. Security robots often rely on pre-programmed responses and algorithms, which may not be suitable for unpredictable or nuanced scenarios. Moreover, there are concerns about privacy as security robots record and monitor activities in public or private spaces. The use of surveillance technology raises questions about the collection, storage, and potential misuse of sensitive data. Additionally, security robots can be vulnerable to hacking or tampering, posing a risk to both the robot itself and the security infrastructure it is meant to protect. It is important to carefully consider these drawbacks when implementing security robot systems.
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Write sql statement to print the product id, product name, average price of all product and difference between average price and price of a product. Now develop PL/SQL procedure to get the product name, product id, product price , average price of all products and difference between product price and the average price.
Now based on the price difference between product price and average price , you
will update the price of the products based on following criteria:
If the difference is more than $100 increase the price of product by $10
If the difference is more than $50 increase the price of the product by $5
If the difference is less than then reduce the price by 0.99 cents.
SQL is a domain-specific language used in programming and made for relational databases that allow manipulation and querying data. It is used to communicate with a database. PL/SQL is a procedural language that Oracle designed to extend SQL by introducing constructs like variables, loops, and conditional statements.
To print the product id, product name, average price of all product and difference between average price and price of a product, we can use the below SQL statement:
SELECT product_id,product_name,AVG(price) OVER(),(AVG(price) OVER() - price) price_differenceFROM products
Now, to develop PL/SQL procedure to get the product name, product id, product price, average price of all products, and difference between product price and the average price, we can use the below code:
CREATE OR REPLACE PROCEDURE update_product_priceISavg_price NUMBER(6,2);
BEGINSELECT AVG(price) INTO avg_price FROM products;
FOR prod IN (SELECT * FROM products) LOOPUPDATE productsSET price = CASEWHEN (avg_price - prod.price) > 100 THEN price + 10WHEN (avg_price - prod.price) > 50 THEN price + 5WHEN (avg_price - prod.price) < 0 THEN price - 0.99ENDWHERE product_id = prod.product_id;
END LOOP;
END update_product_price;
We have created the "update_product_price" procedure that will update the price of the products based on the price difference between product price and average price. We have used the "CASE" statement to check the difference and update the price accordingly. The price will be increased by $10 if the difference is more than $100, and it will be increased by $5 if the difference is more than $50. If the difference is less than $0, the price will be reduced by 0.99 cents. SQL is used to communicate with a database and to manipulate and query data. PL/SQL is a procedural language designed by Oracle to extend SQL by introducing constructs like variables, loops, and conditional statements. We have used the above SQL statement to print the product id, product name, average price of all products, and difference between average price and price of a product. We have also created a PL/SQL procedure that will update the price of the products based on the price difference between product price and average price.
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1. Write a program that returns the number of days between date_1 and date_2. Take into account leap years and correct number of days in each month (e.g., 28 or 29 days in Feb). The accepted input must be in a string format (MM-DD-YYYY). The correct output would also be in a string format (# days). (e.g., input: 06-20-2022 and 06-24-2022 output: 4 days) 2. Assume a user that is active p% of the time with a transfer speed of k Mbps. Write a Python program that computes the network usage of the users between date_1 and date 2 (use your program for Question 1 here). The network usage should be reported in Bytes with the appropriate multiple of binary metric (Kilo, Mega, Giga, ...).. (e.g., input 14-05-2022, 15-05-2022, p = 0.5, k=8 output: 42.1875 GB)
This Python program calculates the number of days between two dates and computes network usage based on user activity and transfer speed.
The program uses the 'datetime' module to parse the input dates and calculate the number of days between them. For Question 2, it utilizes the 'calculate_days' function from Question 1 to obtain the number of days. It then calculates the number of active seconds by multiplying the days with the seconds per day and the user activity percentage.
The total number of bits transferred is computed by multiplying the active seconds with the transfer speed in Mbps. The 'convert_bytes' function converts the total bits into the appropriate binary metric (e.g., KB, MB, GB) and returns the formatted result.
The example usage demonstrates how to input the dates, user activity percentage ('p'), and transfer speed ('k'), and displays the number of days and network usage in the desired format.
import datetime
# Question 1: Calculate the number of days between two dates
def calculate_days(date_1, date_2):
date_format = "%m-%d-%Y"
d1 = datetime.datetime.strptime(date_1, date_format)
d2 = datetime.datetime.strptime(date_2, date_format)
delta = d2 - d1
return str(delta.days) + " days"
# Question 2: Calculate network usage based on user activity and transfer speed
def calculate_network_usage(date_1, date_2, p, k):
days = int(calculate_days(date_1, date_2).split()[0])
seconds_per_day = 24 * 60 * 60
active_seconds = days * seconds_per_day * p
total_bits = active_seconds * k * 1000000
return convert_bytes(total_bits)
def convert_bytes(size):
power = 2**10
n = 0
labels = {0: 'Bytes', 1: 'KB', 2: 'MB', 3: 'GB', 4: 'TB'}
while size > power:
size /= power
n += 1
return "{:.4f} {}".format(size, labels[n])
# Example usage
date_1 = "06-20-2022"
date_2 = "06-24-2022"
p = 0.5
k = 8
print("Number of days:", calculate_days(date_1, date_2))
print("Network usage:", calculate_network_usage(date_1, date_2, p, k))
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A work unit with 20 employees lines up for a building evacuation. The order in which the employees line up is random with each ordering being equally likely. There are two employees in the unit named Karl and Kareem. What is the probability that Kareem will be first in line?
To calculate the probability that Kareem will be first in line, we need to consider the total number of possible orderings and the number of orderings in which Kareem is first.
Given that there are 20 employees in total, the number of possible orderings is equal to the factorial of 20 (20!). This represents all the possible permutations of the employees in line.
To calculate the number of orderings in which Kareem is first, we can fix Kareem's position as the first in line and then consider the remaining 19 employees. The remaining 19 employees can be arranged in any order, which is equal to the factorial of 19 (19!).
Therefore, the probability that Kareem will be first in line is given by:
Probability = Number of orderings with Kareem first / Total number of possible orderings
Probability = (19! / 20!)
To simplify this expression, we can cancel out common terms:
Probability = 1 / 20
Hence, the probability that Kareem will be first in line is 1/20 or 0.05.
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Ask user for an Integer input called "limit" * write a while loop to print first limit Even numbers
In this program, the user is prompted to enter the value of "limit." The while loop will continue until the counter variable "count" reaches the specified limit.
Inside the loop, each even number is printed starting from 2, and then the number and count variables are updated accordingly.
Here's an example of a program in Python that asks the user for an integer input called "limit" and then uses a while loop to print the first "limit" even numbers:
python
Copy code
limit = int(input("Enter the limit: ")) # Ask user for the limit
count = 0 # Initialize a counter variable
number = 2 # Start with the first even number
while count < limit:
print(number) # Print the current even number
number += 2 # Increment by 2 to get the next even number
count += 1 # Increase the counter
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Write SQL command to find the average temperature for a specific
location.
In this command, replace 'specific_location' with the actual location for which you want to calculate the average temperature. The command will calculate the average temperature for that specific location and return it as "average_temperature".
To find the average temperature for a specific location in SQL, you would need a table that stores temperature data with columns such as "location" and "temperature". Assuming you have a table named "temperatures" with these columns, you can use the following SQL command:
sql
Copy code
SELECT AVG(temperature) AS average_temperature
FROM temperatures
WHERE location = 'specific_location';
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(Algo) The following data have been recorded... The following data have been recorded for recently completed Job 450 en its job cost sheet. Direct materials cost was $2.059 A total of 4t diect labor-heurs and 200 machine-hours were worked on the job. The direct labor wage rate is $21 per iabor-hour. The Corporation applies marufocturing overhead on the basis of machine-hours. The predetermined overhed eate is $29 per machine hour The total cost for the job on its job cost sheet would be: Mukipie Chaice- seobs 35.76 \$10.065 18.720
The total cost for Job 450 on its job cost sheet can be calculated by considering the direct materials cost, direct labor cost, and manufacturing overhead cost.
1. Direct materials cost: The question states that the direct materials cost was $2.059. So, this cost is already given.
2. Direct labor cost: The question mentions that 4 direct labor-hours were worked on the job and the direct labor wage rate is $21 per labor-hour. To calculate the direct labor cost, multiply the number of labor-hours (4) by the labor wage rate ($21): 4 labor-hours x $21/labor-hour = $84.
3. Manufacturing overhead cost: The question states that the manufacturing overhead is applied based on machine-hours. It also provides the predetermined overhead rate of $29 per machine hour. The total machine-hours worked on the job is given as 200. To calculate the manufacturing overhead cost, multiply the number of machine-hours (200) by the predetermined overhead rate ($29): 200 machine-hours x $29/machine-hour = $5,800.
4. Total cost: To find the total cost for the job, add the direct materials cost, direct labor cost, and manufacturing overhead cost: $2.059 + $84 + $5,800 = $6,943.059.
Therefore, the total cost for Job 450 on its job cost sheet would be $6,943.059.
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Using: C Language & tinkercad.com & arduino uno r3
Implement and test a function called get_elapsed_time which computes the elapsed time from power-up in a ATMEGA328P microcontroller. The program will use a designated 16-bit timer in normal mode, with overflow interrupt handling. Time calculation will be accurate to the nearest timer update "tick"
Your task is to adapt the sample program provided in "Lecture 9: Implementing Timer Overflow ISR" to implement a new library function called get_elapsed_time () which is capable of tracking elapsed time for a reasonably long period.
Use Timer 1, and set it up in normal operational mode so that it overflows approximately once every 0.25 seconds. Create a global 32-bit unsigned integer variable called counter. Implement an interrupt service routine which increments counter by 1 every time the timer overflows. Implement a function called get_elapsed_time() which returns the elapsed time since program start, accurate to the nearest timer stick", as a double-precision floating point value. To implement the function, follow the detailed specification laid out in comments in the program skeleton below.
Notes • Use this test driver to implement and test your function in TinkerCad Circuits prior to submission. #include
#include
#include
#include
#include
#include
#include
#include
void uart_setup(void);
void uart_put_byte(unsigned char byte_val);
void uart_printf(const char * fmt, ...);
void setup(void) {
// (a) Initialise Timer 1 in normal mode so that it overflows
// with a period of approximately 0.25 seconds.
// Hint: use the table you completed in a previous exercise.
// (b) Enable timer overflow for Timer 1.
// (c) Turn on interrupts.
// (d) Send a debugging message to the serial port using
// the uart_printf function defined below. The message should consist of
// your student number, "n10507621", followed immediately by a comma,
// followed by the pre-scale factor that corresponds to a timer overflow
// period of approximately 0.25 seconds. Terminate the
// debugging message with a carriage-return-linefeed pair, "\r\n".
}
// (e) Create a volatile global variable called counter.
// The variable should be a 32-bit unsigned integer of type uint32_t.
// Initialise the variable to 0.
// INSERT GLOBAL VARIABLE HERE
// (f) Define an interrupt service routine to process timer overflow
// interrupts for Timer 1. Every time the interrupt service
// routine is called, counter should increment by 1.
// INSERT ISR HERE
// (g) Define a function called get_elapsed_time which has
// no parameters, but returns a value of type double which contains
// the total elapsed time measured up to the time at which it is called.
// Use the method demonstrated in the Topic 9 lecture to compute the
// elapsed time, taking into account the fact that the timer counter has
// 16 bits rather than 8 bits.
// INSERT FUNCTION HERE
// -------------------------------------------------
// Helper functions.
// -------------------------------------------------
// Make sure this is not too big!
char buffer[100];
void uart_setup(void) {
#define BAUD (9600)
#define UBRR (F_CPU / 16 / BAUD - 1)
UBRR0H = UBRR >> 8;
UBRR0L = UBRR & 0b11111111;
UCSR0B = (1 << RXEN0) | (1 << TXEN0);
UCSR0C = (3 << UCSZ00);
}
void uart_printf(const char * fmt, ...) {
va_list args;
va_start(args, fmt);
vsnprintf(buffer, sizeof(buffer), fmt, args);
for (int i = 0; buffer[i]; i++) {
uart_put_byte(buffer[i]);
}
}
#ifndef __AMS__
void uart_put_byte(unsigned char data) {
while (!(UCSR0A & (1 << UDRE0))) { /* Wait */ }
UDR0 = data;
}
#endif
int main() {
uart_setup();
setup();
for (;;) {
double time_now = get_elapsed_time();
uart_printf("Elapsed time = %d.%03d\r\n", (int)time_now, (int)((time_now - (int)time_now) * 1000));
_delay_ms(1000);
}
return 0;
}
• Do not use the static qualifier for global variables. This causes variables declared at file scope to be made private, and will prevent AMS from marking your submission.
For implementing the code, you can run it on your Arduino Uno board or simulate it using Tinkercad to test the functionality and verify the elapsed time calculations.
The implementation and testing of the function called get_elapsed_time that computes the elapsed time from power-up in an ATMEGA328P microcontroller is a crucial part of microcontroller programming. The program would use a designated 16-bit timer in normal mode, with overflow interrupt handling.
Time calculation would be accurate to the nearest timer update "tick."Here is a sample program that you can use for your implementation and testing of the function in TinkerCad Circuits, which is provided in "Lecture 9: Implementing Timer Overflow ISR." Use Timer 1 and set it up in normal operational mode so that it overflows about once every 0.25 seconds. Create a global 32-bit unsigned integer variable called counter.
Implement an interrupt service routine that increments counter by 1 every time the timer overflows. Implement a function called get_elapsed_time() that returns the elapsed time since program start, accurate to the nearest timer stick", as a double-precision floating-point value. Follow the detailed specification laid out in comments in the program skeleton below.
The code implementation for the function called get_elapsed_time that computes the elapsed time from power-up in a ATMEGA328P microcontroller is as follows:
#include
#include
#include
#include
#include
#include
#include
#include
void uart_setup(void);
void uart_put_byte(unsigned char byte_val);
void uart_printf(const char * fmt, ...);
void setup(void) {
// (a) Initialise Timer 1 in normal mode so that it overflows
// with a period of approximately 0.25 seconds.
TCCR1B |= (1 << WGM12) | (1 << CS12) | (1 << CS10);
OCR1A = 62499;
TCCR1A = 0x00;
TIMSK1 = (1 << TOIE1);
sei();
// (d) Send a debugging message to the serial port using
// the uart_printf function defined below. The message should consist of
// your student number, "n10507621", followed immediately by a comma,
// followed by the pre-scale factor that corresponds to a timer overflow
// period of approximately 0.25 seconds. Terminate the
// debugging message with a carriage-return-linefeed pair, "\r\n".
uart_printf("n10507621,256\r\n");
}
volatile uint32_t counter = 0;
ISR(TIMER1_OVF_vect)
{
counter++;
}
double get_elapsed_time()
{
double tick = 1.0 / 16000000.0; // clock tick time
double elapsed = (double)counter * 0.25;
return elapsed;
}
// -------------------------------------------------
// Helper functions.
// -------------------------------------------------
// Make sure this is not too big!
char buffer[100];
void uart_setup(void) {
#define BAUD (9600)
#define UBRR (F_CPU / 16 / BAUD - 1)
UBRR0H = UBRR >> 8;
UBRR0L = UBRR & 0b11111111;
UCSR0B = (1 << RXEN0) | (1 << TXEN0);
UCSR0C = (3 << UCSZ00);
}
void uart_printf(const char * fmt, ...) {
va_list args;
va_start(args, fmt);
vsnprintf(buffer, sizeof(buffer), fmt, args);
for (int i = 0; buffer[i]; i++) {
uart_put_byte(buffer[i]);
}
}
#ifndef __AMS__
void uart_put_byte(unsigned char data) {
while (!(UCSR0A & (1 << UDRE0))) { /* Wait */ }
UDR0 = data;
}
#endif
int main() {
uart_setup();
setup();
for (;;) {
double time_now = get_elapsed_time();
uart_printf("Elapsed time = %d.%03d\r\n", (int)time_now, (int)((time_now - (int)time_now) * 1000));
_delay_ms(1000);
}
return 0;
}
Notes: Ensure you do not use the static qualifier for global variables, as this causes variables declared at file scope to be made private and will prevent AMS from marking your submission.
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Answer ALL questions on this question paper. Circle the correct answer.
Which of the following is a valid functional requirement?
The software shall be written in C++.
The software shall respond to all requests within 5 seconds.
The software shall be composed of the following twenty-one modules.
The software shall use the following fifteen menu screens whenever it is communicating with the user.
Requirements can only be drawn up in collaboration with stakeholders. Which of the following statements concerning communication between people is INCORRECT?
Communication occurs simultaneously at several levels.
Communication can lead to misunderstandings between the sender and the receiver.
Different rules apply to communication in different cultures.
For the receiver, only the factual content is relevant.
Which of the following is NOT a reason why requirements are important?
The resulting software may not satisfy user’s real needs.
The later in the development life cycle that a software error is detected, the more expensive it will be to repair.
Both time and money may be wasted building the wrong system.
Stakeholder’s inability to communicate proper system objective.
For which of the following practices does requirements engineering provide appropriate mechanisms and tools?
Analysing need
Unambiguous specification of the solution
Validating the specification
All of the above.
Which of the following elicitation methods is most suitable for acquiring requirements from existing documents?
Field Observation
System Archaeology
Apprenticing
CRC Cards
Which one of the listed problems leads to the greatest difficulties in requirements engineering?
Too little experience with new technology.
A too complex model.
Changing company objectives.
Communication problems between project team and stakeholder.
Which of the following elicitation methods is most suitable for acquiring basic requirements categorization according to Kano Model?
Interview
Prototyping
Brainstorming
Work observation
Requirements can be documented in THREE (3) perspectives where each one has its own suitable diagram. The following perspectives – diagrams are correct EXCEPT ________.
Behavioural – state model
Data – entity relationship model
Functional – activity diagram
Temporal – data flow diagram
Which of the following is the standard used in writing requirement documents?
DI-MCCR-80025A
SMAP-DID-P200_SW
IEEE Standard 830 - 1984
IEEE Standard 830 - 1998
A large, regional railway company with 70 stations is purchasing a new communications device for station employees. Which of the elicitation techniques would you implement for determination of requirements?
Use-case specification.
Use of self-recording with all stakeholders.
Assessment using products obtainable on the market.
Observation of work of selected stakeholders at selected stations.
The questions revolve around requirements engineering, including functional requirements, communication in requirements engineering, the importance of requirements, practices in requirements engineering, elicitation methods, and requirement documentation standards. The questions require selecting the correct answer from multiple choices.
1. A valid functional requirement is a statement that describes what the software system should do. Among the options given, "The software shall respond to all requests within 5 seconds" is a valid functional requirement as it specifies a desired behavior or functionality of the software.
2. The incorrect statement regarding communication between people is "For the receiver, only the factual content is relevant." In communication, the factual content is important, but other aspects like tone, context, and emotions also play a role in understanding the message. Misunderstandings can arise from non-factual elements in communication.
3. The reason "Stakeholder’s inability to communicate proper system objective" is NOT a reason why requirements are important. Requirements are necessary to ensure that the resulting software meets the user's needs and avoids wasting time and money building the wrong system, and they help identify errors early in the development life cycle.
4. Requirements engineering provides appropriate mechanisms and tools for "Analysing need, Unambiguous specification of the solution, and Validating the specification." These practices involve understanding user requirements, creating clear and precise specifications, and verifying that the specifications meet the desired objectives.
5. The most suitable elicitation method for acquiring requirements from existing documents is "System Archaeology." System Archaeology involves studying existing documentation, code, and other artifacts to extract requirements and gain insight into the system's design and functionality.
6. Communication problems between the project team and stakeholders lead to the greatest difficulties in requirements engineering. Effective communication is crucial for understanding and capturing stakeholders' needs, managing expectations, and ensuring alignment between the project team and stakeholders.
7. The perspective-diagram combination that is NOT correct is "Temporal – data flow diagram." Temporal perspective typically focuses on the sequence of events and time-related aspects, while data flow diagrams represent the flow of data between processes, making them more suitable for the functional perspective.
8. The standard used in writing requirement documents is "IEEE Standard 830 - 1998." IEEE Standard 830 provides guidelines for writing software requirements specifications, ensuring clarity, completeness, and consistency in documenting requirements.
9. For determining requirements in a large, regional railway company purchasing a communications device, the elicitation technique "Observation of work of selected stakeholders at selected stations" would be appropriate. By observing stakeholders' work at various stations, their needs and requirements can be identified and incorporated into the new communications device.
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Exercise 2. Mini Logo
Consider the simplified version of Mini Logo (without macros), defined by the following abstract syntax.
type alias Point = (Int,Int)
type Mode = Up | Down
type Cmd = Pen Mode
| MoveTo Point
| Seq Cmd Cmd
ThesemanticsofaMiniLogoprogramisasetofdrawnlines. However,forthedefinitionofthesemanticsa"drawing
state" must be maintained that keeps track of the current position of the pen and the pen’s status (Up or Down). This
state should be represented by values of the following type.
type alias State = (Mode,Point)
The semantic domain representing a set of drawn lines is represented by the type Lines.
type alias Line = (Point,Point)
type alias Lines = List Line
Define the semantics of Mini Logo via two Elm functions. First, define a function semS that has the following type.
semCmd : Cmd -> State -> (State,Lines)
This function defines for each Cmd how it modifies the current drawing state and what lines it produces. After that
define the function lines with the following type.
lines : Cmd -> Lines
The function lines should call semCmd. The initial state is defined to have the pen up and the current drawing
position at (0,0).
Note. You can test your semantics as follows.
(1) If you haven’t done already, initialize Elm in your current directory with the command elm init to ensure the
presence of a proper elm.json file and the subdirectory src that contains your homework Elm files.
(2) Install the Elm SVG package with the following shell command elm install elm/svg.
(3) Download the file with the name HW3_MiniLogo.elm from Canvas into the src subdirectory. It looks as follows.
module HW3_MiniLogo exposing (..)
...
----- BEGIN HW3 solution
semCmd : Cmd -> State -> (State,Lines)
semCmd = ...
lines : Cmd -> Lines
lines = ...
logoResult : Lines
logoResult = lines (Seq (Seq (Seq (Pen Up) ...
(4) Insert your function definitions after the BEGIN HW3 solution comment.
(5) In the current directory, execute the command elm reactor.
(6) Inyourwebbrowser,entertheURLhttp://localhost:8000. ThiswillallowyoutoloadthefileHW3_MiniLogo.elm,
which will then render the Lines value logoResult (currently, two steps) in your browser.
In this exercise, we are tasked with defining the semantics of a simplified version of Mini Logo in Elm. The semantics are defined using two functions: `semCmd` and `lines`.
To define the semantics of Mini Logo in Elm, we can implement the `semCmd` and `lines` functions as follows:
```elm
type alias Point = (Int, Int)
type Mode = Up | Down
type Cmd = Pen Mode
| MoveTo Point
| Seq Cmd Cmd
type alias State = (Mode, Point)
type alias Line = (Point, Point)
type alias Lines = List Line
semCmd : Cmd -> State -> (State, Lines)
semCmd cmd state =
case cmd of
Pen mode ->
(mode, snd state)
MoveTo point ->
let
newState = (fst state, point)
in
(newState, [])
Seq cmd1 cmd2 ->
let
(newState, lines1) = semCmd cmd1 state
(finalState, lines2) = semCmd cmd2 newState
in
(finalState, lines1 ++ lines2)
lines : Cmd -> Lines
lines cmd =
let
initialState = (Up, (0, 0))
(_, resultLines) = semCmd cmd initialState
in
resultLines
logoResult : Lines
logoResult =
lines (Seq (Seq (Seq (Pen Up) (MoveTo (0, 1))) (Pen Down)) (MoveTo (2, 3)))
```
In the above code, we define the `semCmd` function to handle the different command cases. For the `Pen` command, it updates the pen mode in the state. For the `MoveTo` command, it updates the current drawing position in the state. For the `Seq` command, it recursively calls `semCmd` on both sub-commands and combines their resulting lines.
The `lines` function uses `semCmd` to process the given command and extract the lines from the resulting state. It starts with the initial state and returns the lines produced.
Finally, we define the `logoResult` value as an example usage of the `lines` function, representing a sequence of Mini Logo commands.
To test the semantics, follow the provided instructions to set up the Elm environment, install the necessary packages, and run the program. The rendered result will display the lines produced by the Mini Logo commands defined in `logoResult`.
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1. = (a) (6%) Let A[1..n) and B[1..m] be two arrays, each represents a set of numbers. Give an algorithm that returns an array C[] such that C contains the intersection of the two sets of numbers represented by A and B. Give the time complexity of your algorithm in Big-O. As an example, if A = [6, 9, 2, 1, 0, 7] and B = [9, 7, 11, 4, 8,5,6,0], then C should , contain (9,7,6,0] (the ordering of the numbers in array C does not matter). (b) (6%) Let A[1..n] be an array of n numbers. Each number could appear multiple times in array A. A mode of array A is a number that appears the most frequently in A. Give an algorithm that returns a mode of A. (In case there are more than one mode in A, your algorithm only needs to return one of them.) Give the time complexity of your algorithm in Big-O. As an example, if A = [9, 2, 7, 7, 1, 3, 2,9,7,0,8, 1], then mode of A is 7. - 2 2 > 2
Find the intersection of two arrays by using a hash set. Time complexity: O(n + m)and find the mode of an array using a hash map. Time complexity: O(n).
(a) To find the intersection of two arrays A and B, we can use a hash set to efficiently store the elements of one array and then iterate over the other array to check for common elements. The algorithm involves three main steps: 1. Create an empty hash set.
2. Iterate through array A and insert each element into the hash set.
3. Iterate through array B and check if each element is present in the hash set. If it is, add it to the result array C.
The time complexity of this algorithm is O(n + m), where n and m are the lengths of arrays A and B, respectively.
(b) To find the mode of array A, we can use a hash map to store the frequency count of each element in A. The algorithm involves two main steps:1. Create an empty hash map.
2. Iterate through array A, and for each element, update its frequency count in the hash map.
3. Iterate through the hash map and find the element(s) with the highest frequency count. Return one of the modes.
The time complexity of this algorithm is O(n), where n is the length of array A.
Find the intersection of two arrays by using a hash set. Time complexity: O(n + m) and find the mode of an array using a hash map. Time complexity: O(n)
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What is the output of the following code that is part of a complete C++ Program? a. Int a = 5, b = 8, c = 12; b. cout << b + c/2 + c << " c. cout << a* (2*3/2) << endl; d. cout << a % b<<"
e. cout << b/c<<< endl;
The output of the following code is : 16 , 15 , 1 , 2. The code first declares three variables: a, b, and c. Then, it performs four operations on these variables and prints the results.
The first operation is b + c/2 + c. This operation first divides c by 2, then adds the result to b and c. The result of this operation is 16.
The second operation is a * (2*3/2). This operation first multiplies 2 by 3, then divides the result by 2. The result of this operation is 15.
The third operation is a % b. This operation calculates the modulus of a and b, which is the remainder when a is divided by b. The result of this operation is 1.
The fourth operation is b/c. This operation calculates the quotient of b and c, which is the number of times c fits into b. The result of this operation is 2.
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1. List down the similarities and differences between structures and classes
Structures and classes are both used in programming languages to define custom data types and encapsulate related data and behavior. They share some similarities, such as the ability to define member variables and methods. However, they also have notable differences. Structures are typically used in procedural programming languages and provide a lightweight way to group data, while classes are a fundamental concept in object-oriented programming and offer more advanced features like inheritance and polymorphism.
Structures and classes are similar in that they allow programmers to define custom data types and organize related data together. Both structures and classes can have member variables to store data and member methods to define behavior associated with the data.
However, there are several key differences between structures and classes. One major difference is their usage and context within programming languages. Structures are commonly used in procedural programming languages as a way to group related data together. They provide a simple way to define a composite data type without the complexity of inheritance or other advanced features.
Classes, on the other hand, are a fundamental concept in object-oriented programming (OOP). They not only encapsulate data but also define the behavior associated with the data. Classes support inheritance, allowing for the creation of hierarchical relationships between classes and enabling code reuse. They also facilitate polymorphism, which allows objects of different classes to be treated interchangeably based on their common interfaces.
In summary, structures and classes share similarities in their ability to define data types and encapsulate data and behavior. However, structures are typically used in procedural programming languages for lightweight data grouping, while classes are a fundamental concept in OOP with more advanced features like inheritance and polymorphism.
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Can u solve this questions in C++ please?
Define a template of a function finding the maximum of three values
Define a class MyStack supporting the stack data structure storing integers, with methods: push, pop, size, print
Convert the class into a template capable of generating stacks of any data types
Check how this template works
The code provides a template function to find the maximum of three values and a class MyStack supporting stack operations for integers. The class MyStack can be converted into a template to generate stacks of any data types by specifying the template argument when instantiating the class.
Here's the implementation of the requested functions in C++:
1. Template function to find the maximum of three values:
#include <iostream>
template <typename T>
T maximum(T a, T b, T c) {
T maxVal = a;
if (b > maxVal)
maxVal = b;
if (c > maxVal)
maxVal = c;
return maxVal;
}
int main() {
int a = 5, b = 10, c = 7;
int maxInt = maximum(a, b, c);
std::cout << "Maximum integer value: " << maxInt << std::endl;
double x = 3.14, y = 2.71, z = 2.99;
double maxDouble = maximum(x, y, z);
std::cout << "Maximum double value: " << maxDouble << std::endl;
return 0;
}
2. Class MyStack implementation:
#include <iostream>
#include <vector>
class MyStack {
private:
std::vector<int> stack;
public:
void push(int value) {
stack.push_back(value);
}
void pop() {
if (!stack.empty())
stack.pop_back();
}
int size() {
return stack.size();
}
void print() {
for (int value : stack) {
std::cout << value << " ";
}
std::cout << std::endl;
}
};
int main() {
MyStack stack;
stack.push(5);
stack.push(10);
stack.push(7);
stack.print(); // Output: 5 10 7
stack.pop();
stack.print(); // Output: 5 10
return 0;
}
To convert the class into a template, you can modify the class definition as follows:
template <typename T>
class MyStack {
// ...
};
You can then create stacks of any data type by specifying the template argument when instantiating the class, for example:
MyStack<double> doubleStack;
doubleStack.push(3.14);
doubleStack.push(2.71);
You can similarly test the template version of the MyStack class with different data types.
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React Js questions
Q7 Observe the custom hook used to get/set the address of the user. Anu called the useToManageAddress custom hook from Profile page and set the address. After that she is calling the same custom hook from Delivery page to get the address which was set from Profile page. Assume that all the required import and export statements are written. useToManageAddress.js const useToManageAddress = (addObj) => { const [address, setAddress] = useState({}); const setUserAddress = () => { setAddress(addObj);
} return { address, setUserAddress }; } Profile.js const Profile = () => { const { setUserAddress } = useToManageAddress(); const setAddress = () => { setUser Address({ name: "Arjun", district: "Mangalore", state: "Karnataka" }); return Set Address 6/14 } Delivery.js const Delivery = () => { const { address } = useToManageAddress(); return {address.name} ) } Predict the output of Delivery page. a) Prints the value ‘Arjun b) Nothing will be displayed address.name will be undefined c) Error d) None of the above is the right answer
The output of the Delivery page will be 'Arjun' (option a). The useToManageAddress custom hook maintains the address state.
It is set with the values { name: "Arjun", district: "Mangalore", state: "Karnataka" } in the Profile page using the setAddress function. When the Delivery page calls the useToManageAddress custom hook, it retrieves the address from the state, and since it has been previously set in the Profile page, the value of address.name will be 'Arjun'.
The useToManageAddress custom hook returns an object with two properties: address and setUserAddress. In the Profile page, the setAddress function is called, which internally calls setUserAddress to set the address object with the values { name: "Arjun", district: "Mangalore", state: "Karnataka" }. This sets the address state in the custom hook.
In the Delivery page, the useToManageAddress custom hook is called again, and it retrieves the address object from the state. Since the address was previously set in the Profile page, the value of address.name will be 'Arjun'. Therefore, the output of the Delivery page will be 'Arjun'.
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11. We can review the values in the TVM registers by simply pressing the key of the value we want to review. (T or F) 12. Values can be entered in the TVM registers in any order. (T or F ) 13. When entering dollar amounts in the PV, PMT, and FV registers, we should enter amounts paid as positive numbers, and amounts received as negative numbers. ( T or F ) 14. Suppose you are entering a negative $300 in the PMT register. Keystrokes are: [−]300 [PMT]. (T or F) 15. If you make a total of ten $50 payments, you should enter $500 in the PMT register. (T or F)
We can review the values in the TVM registers by simply pressing the key .False. Values cannot be entered in the TVM registers in any order. Hence, the answer is as follows:True. False. True. True. True.
When entering dollar amounts in the PV, PMT, and FV registers, we should enter amounts paid as positive numbers, and amounts received as negative numbers.True. Suppose you are entering a negative $300 in the PMT register. Keystrokes are: [−]300 [PMT].15. True. If you make a total of ten $50 payments, you should enter $500 in the PMT register.True. We can review the values in the TVM registers by simply pressing the key of the value we want to review.
False. Values cannot be entered in the TVM registers in any order. TVM refers to time value of money which is a financial concept.13. True. When entering dollar amounts in the PV, PMT, and FV registers, we should enter amounts paid as positive numbers, and amounts received as negative numbers.True. Suppose you are entering a negative $300 in the PMT register. Keystrokes are: [−]300 [PMT]. True. If you make a total of ten $50 payments, you should enter $500 in the PMT register.In total, there are five statements given, each of which is either true or false.
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Arif a photography enthusiast, was looking for a new digital camera. He was going on a holiday to Melaka after 5 day (October 5), so he needed the camera to arrive by then. He went to "Easybuy" website, and he quickly found the camera he wanted to buy. He checked the delivery time and upon seeing "Free delivery by October 3 (Three days later)", added it to the cart, and without incident, confirmed the order and select COD as the payment option. Quick and easy - he was pleased and excited to receive the camera. He was also received an e-mail of the tracking no. from the courier partner when the item was shipped. After two days, he wanted to check the delivery status, so he went to the "Easybuy" website, but he was frustrated to find that could not track the package. He had to go to a third-party website to track it. The courier website was badly designed, and he was not able to figure out how to get the details. Then he called up customer support of "Easybuy", where he talked with the customer support executive and came to know that his order was delayed a bit due to logistics issues at the courier's side. He was unhappy about the whole process and asked to cancel the order as he needed the camera urgently. But the customer support executive told him that COD order can only be cancelled after delivery and not during while the item was in transit. Arif explained to him that no one would be there to receive the package when it arrived. He was frustrated with the whole situation and finally had to buy the camera offline at higher price. After the "Easybuy" package arrived, the courier partner tried to deliver the package for three days before they send it back to the seller. Everyday, a new delivery boy kept calling Arif about the house was locked and where should he deliver the package and whom should he deliver to? Arif was frustrated with the whole experience and decided that he will never buy from "Easybuy" again and instead use some other website. QUESTION 1 [10 marks]: A. Illustrate a user journey map for Arif from the scenario A above (see Figure 1 for guide). [10 marks]
User Journey Map for Arif from the scenario: The user journey map for Arif from the given scenario is as follows: Step 1: Need Recognition:Arif was going on a holiday to Melaka after five days and needed a new digital camera for the trip.Step 2: Research:He visited the Easybuy website and found the camera he wanted to buy.
He checked the delivery time and found that it would be delivered for free by October 3.Step 3: Purchase:Arif confirmed the order and selected COD as the payment option.Step 4: Delivery:After two days, he wanted to check the delivery status, so he went to the "Easybuy" website, but he was frustrated to find that could not track the package. He went to a third-party website to track it, but the courier website was badly designed and he was not able to get the details.
The courier partner finally sent an e-mail to Arif with the tracking number. However, the delivery of the package was delayed due to logistics issues at the courier's side. Step 5: Frustration and Cancellation:Arif called up the customer support executive of "Easybuy" and asked to cancel the order as he needed the camera urgently. But the customer support executive told him that COD order can only be cancelled after delivery and not during while the item was in transit. After the package arrived, the courier partner tried to deliver the package for three days before they sent it back to the seller. Arif was frustrated with the whole experience and decided that he would never buy from "Easybuy" again and instead use some other website.
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Challenge two Write a query to list the event IDs and the total sales for each event in descending order.
Assuming you have a table named "events" with columns "event_id" and "sales", the above query will calculate the total sales for each event by summing up the "sales" values for each event ID. The results will be sorted in descending order based on the total sales.
To list the event IDs and the total sales for each event in descending order, you can use the following SQL query:
```sql
SELECT event_id, SUM(sales) AS total_sales
FROM events
GROUP BY event_id
ORDER BY total_sales DESC;
Please note that you may need to adjust the table name and column names according to your specific database schema.
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TSP: Lower Upper Bounds; Minimum Spanning Tree; Optimal
Route.
The Traveling Salesman Problem (TSP) is a well-known combinatorial optimization problem in computer science and operations research.
It involves finding the shortest possible route that visits a set of cities and returns to the starting city, while visiting each city exactly once.
1. Lower Bound: In the TSP, the lower bound refers to an estimate or approximation of the minimum possible cost of the optimal solution. Various lower bound techniques can be used, such as the minimum spanning tree (MST) approach.
2. Upper Bound: The upper bound in the TSP represents an estimate or limit on the maximum possible cost of any feasible solution. It can be used to evaluate the quality of a given solution or as a termination condition for certain algorithms. Methods like the nearest neighbor heuristic or 2-opt optimization can provide upper bounds.
3. Minimum Spanning Tree (MST): The minimum spanning tree is a graph algorithm that finds the tree that connects all vertices of a graph with the minimum total edge weight. In the context of the TSP, the MST can be used as a lower bound estimation. By summing the weights of the edges in the MST and doubling the result, we obtain a lower bound on the TSP's optimal solution.
4. Optimal Route: The optimal route in the TSP refers to the shortest possible route that visits all cities exactly once and returns to the starting city. It is the solution that minimizes the total distance or cost. Finding the optimal route is challenging because the problem is NP-hard, meaning that as the number of cities increases, the computational time required to find the optimal solution grows exponentially.
To solve the TSP optimally for small problem sizes, exact algorithms such as branch and bound, dynamic programming, or integer linear programming can be used. However, for larger instances, these exact methods become infeasible, and heuristic or approximation algorithms are employed to find near-optimal solutions. Popular heuristic approaches include the nearest neighbor algorithm, genetic algorithms, and ant colony optimization.
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Task 3:
The Driver Relationship team wants to create some workshops and increase communication with the active drivers in InstantRide. Therefore, they requested a new database table to store the driver details of the drivers that have had at least one ride in the system. Create a new table, ACTIVE_DRIVERS, from the DRIVERS and TRAVELS tables which contains the following fields:
DRIVER_ID CHAR(5) (Primary key)
DRIVER_FIRST_NAME VARCHAR(20)
DRIVER_LAST_NAME VARCHAR(20)
DRIVER_DRIVING_LICENSE_ID VARCHAR(10)
DRIVER_DRIVING_LICENSE_CHECKED BOOL
DRIVER_RATING FLOAT
Here's the tables already created
Tables_in_InstantRide
CARS
DRIVERS
TRAVELS
USERS
InstantRide
A new table named ACTIVE_DRIVERS is created from the existing DRIVERS and TRAVELS tables to store driver details for drivers with at least one ride. The table includes fields for driver ID, first name, last name, driving license ID, license check status, and driver rating.
To create the new table ACTIVE_DRIVERS from the existing DRIVERS and TRAVELS tables, you can use the following SQL statement:
CREATE TABLE ACTIVE_DRIVERS (
DRIVER_ID (5) PRIMARY KEY,
DRIVER_FIRST_NAME (20),
DRIVER_LAST_NAME (20),
DRIVER_DRIVING_LICENSE_ID (10),
DRIVER_DRIVING_LICENSE_CHECKED BOOL,
DRIVER_RATING FLOAT
);
This SQL statement creates a new table named ACTIVE_DRIVERS with the specified fields: DRIVER_ID (primary key), DRIVER_FIRST_NAME, DRIVER_LAST_NAME, DRIVER_DRIVING_LICENSE_ID, DRIVER_DRIVING_LICENSE_CHECKED, and DRIVER_RATING. The table is created based on the provided data types, such as CHAR, VARCHAR, BOOL, and FLOAT.
By creating this table, you can now store the driver details for drivers who have had at least one ride in the system.
Note: Ensure that you have appropriate access rights and privileges to create tables in the InstantRide database.
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Write a public static method that returns a boolean value of true if the three numbers of type int in its parameter list are all equal to each other. Otherwise, the method should return a boolean value of false.
A public static method is implemented to determine whether three given integer numbers are equal to each other. It returns true if they are all equal, and false otherwise.
In order to check if three integers are equal, we can compare them pairwise. The method takes three integer parameters and uses an if statement to compare them. If the first number equals the second number and the second number equals the third number, then all three numbers are equal, and the method returns true. Otherwise, the method returns false.
The implementation of this method involves using a conditional statement, specifically the equality operator (==), to compare the integers. By comparing each number to its adjacent number, we can determine if all three numbers are equal. This approach ensures that the method accurately identifies whether the numbers are equal or not.
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What is the largest square plate that can cover a rectangular plate of the size 330 150? (try solving this problem with Euclid's algorithm) 30 O 10 0 50 025
Euclid's algorithm is a simple and efficient way to find the greatest common divisor (GCD) of two numbers. It involves dividing the larger number by the smaller number and taking the remainder, then dividing the smaller number by the remainder until the remainder is zero. The last non-zero remainder is the GCD of the two numbers.
In the case of finding the largest square plate that can cover a rectangular plate of size 330 x 150, we use Euclid's algorithm to find the GCD of 330 and 150. We first divide 330 by 150 to get a quotient of 2 and a remainder of 30. Then we divide 150 by 30 to get a quotient of 5 and a remainder of 0. Since the remainder is now zero, we can stop dividing, and the last non-zero remainder (30) is the GCD of 330 and 150.
The significance of this result is that we now know that the largest square plate that can cover the rectangular plate has a side length of 30 units. We can cut out a square plate with sides of this length and place it over the rectangular plate so that it covers the entire area without overlapping or leaving any gaps.
Using Euclid's algorithm can be helpful in many applications such as cryptography, computer science, and engineering. It provides a quick and efficient way to find the GCD of two numbers, which is a fundamental concept in many mathematical and computational problems.
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Create a code that illustrates matlab loop example Loop should has 5 iterations Loop should invoke disp('Hello'); function (in other words you programm will print "Hello" 5 times (command disp('Hello')), please use loops)
The code to print "Hello" five times using a loop in MATLAB is shown below:
for i=1:5 disp('Hello')end
In MATLAB, a loop is a programming construct that repeats a set of instructions until a certain condition is met. Loops are used to iterate over a set of values or to perform an operation a certain number of times.
The for loop runs five iterations, as specified by the range from 1 to 5.
The disp('Hello') command is invoked in each iteration, printing "Hello" to the command window each time. This loop can be modified to perform other operations by replacing the command inside the loop with different code.
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1. Questions on Recurrence Analysis and Master Theorem. (50 marks)
(a) Consider the time-complexity of an algorithm with respect to the problem size being Tሺሻ ൌ 2Tሺ⌊ 2⁄ ⌋ሻ . Formally demonstrate that Tሺሻ ∈ Θሺ ∙ lg ሻ . Full marks for using basic definitions and concepts, such as those found in lecture materials.
(i) Prove via induction that Tሺሻ has a function form of Tሺ2ሻ ൌ 2ሺTሺ1ሻ ሻ. Hint: start with an appropriate variable substitution ൌ2, ∈ ℕଵ , and iterate through ൌ 1,2,3, … to discover the inductive structure of Tሺሻ. Full marks for precise mathematical statements and proofs for both the basis and induction step. [20 marks]
(ii) Prove that Tሺሻ ∈ Θሺ ∙ lg ሻ. You can use the multiplication rule with drop smaller terms directly without its formal construction, as well as apply other results as claimed in lecture materials. For the rest of your answer, justify any assumption you have to make. [16 marks]
(iii) If this algorithm involves a partitioning process, what does Tሺ1ሻ ൌ Θሺ1ሻ mean or suggest? [6 marks]
(b) Given Tሺሻ ൌ 81Tሺ 3⁄ ሻ , 3 27, use the Master Theorem to determine its asymptotic runtime behaviour. [8 marks]
(a) (i) T(n) = 2^(log₂) ∈ Θ(log ) by definition of asymptotic notation.
(ii) , T(n) has a time complexity of Θ(n^(log₂3)).
(iii) possible value (i.e., n=1), the algorithm can solve it in constant time.
b) T(n) = Θ(f(n)) = Θ(n^3).
(a) (i)
We need to prove that the function form of T() is T() = 2^(log₂) ∈ Θ(log ), where log denotes base-2 logarithm.
Basis Step: For n=1, we have T(1) = 2^(log₂1) = 1, which is a constant. Thus, T(1) is in Θ(1) and the basis step is true.
Inductive Hypothesis: Assume that for all k < n, the statement T(k) = 2^(log₂k) ∈ Θ(log k) holds.
Inductive Step: We need to show that T(n) = 2^(log₂n) ∈ Θ(log n).
We can write T(n) as:
T(n) = 2^log₂n + T(⌊n/2⌋)
Using the inductive hypothesis,
T(⌊n/2⌋) = 2^(log₂⌊n/2⌋) ∈ Θ(log ⌊n/2⌋)
Since log is an increasing function, we have log ⌊n/2⌋ ≤ log n - 1. Therefore,
T(⌊n/2⌋) = 2^(log₂⌊n/2⌋) ∈ O(2^(log n))
Substituting this in the original equation, we get:
T(n) ∈ O(2^log n + 2^(log n)) = O(2^(log n))
Similarly, T(n) = 2^log₂n + T(⌊n/2⌋) ∈ Ω(2^log n) since T(⌊n/2⌋) ∈ Ω(2^log ⌊n/2⌋) by the inductive hypothesis.
Thus, T(n) = 2^(log₂) ∈ Θ(log ) by definition of asymptotic notation.
(ii)
Using the multiplication rule and ignoring lower order terms, we have:
T(n) = 2^(log₂n) + T(⌊n/2⌋)
= 2^(log₂n) + 2^(log₂(⌊n/2⌋)^log₂3) + ...
= 2^(log₂n) + (2^(log₂n - 1))^log₂3 + ...
= 2^(log₂n) + n^(log₂3) * (2^(log₂n - log₂2))^log₂3 + ...
= Θ(n^(log₂3))
Therefore, T(n) has a time complexity of Θ(n^(log₂3)).
(iii)
If T(1) = Θ(1), then this suggests that the base case takes constant time to solve. In other words, when the problem size is reduced to its smallest possible value (i.e., n=1), the algorithm can solve it in constant time.
(b)
Using the Master Theorem, we have:
a = 81, b = 3, f(n) = n^(log₃27) = n^3
Case 3 applies because f(n) = Θ(n^3) = Ω(n^(log₃81 + ε)) for ε = 0.5.
Therefore, T(n) = Θ(f(n)) = Θ(n^3).
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Suppose we have a relational database with five tables. table key Attributes S(sid, A) Sid T(tid, B) Tid U(uid, C) Uid R(sid, tid, D) sid, tid Q(tid, uid, E) tid, uid Here R implements a many-to-many relationship between the entities implemented with tables S and T, and Q implements a many-to-many relationship between the entities implemented with tables T and U. A. Write an SQL query that returns all records of the form sid, uid where sid is the key of an S- record and uid is the key of a U-record and these two records are related through the relations R and Q. Use SELECT and not SELECT DISTINCT in your query. B. Write an SQL query that returns records of the form A, C where the A-value is from an S- record and the C-value is from a U-record and these two records are related through the relations R and Q. Use SELECT and not SELECT DISTINCT in your query. C. Could one of your queries from parts (a) and (b) return more records than the other? If so, which one? Justify your answer. D. Suppose you replaced SELECT with SELECT DISTINCT in your queries from parts (a) and Could one of these modified queries return more records than the other? If so, which one? Justify your answer. E. Consider again your query from part (a). If pair sid, uid is returned by this query then there must exist at least one "path" that goes from from table S to table T (via relation R) and then from table T to table U (via relation Q). Note that there can be many such paths for a given pair sid, uid. Write an SQL query that returns records of the form tid, total where tid is a key of a record from table T and total indicates the total number of such paths that "go through" that record.
A. SQL query to return all records of the form sid, uid where sid is the key of an S-record and uid is the key of a U-record related through relations R and Q:
SELECT R.sid, Q.uid
FROM R
JOIN Q ON R.tid = Q.tid
B. SQL query to return records of the form A, C where the A-value is from an S-record and the C-value is from a U-record related through relations R and Q:
SELECT S.A, U.C
FROM S
JOIN R ON S.sid = R.sid
JOIN Q ON R.tid = Q.tid
JOIN U ON Q.uid = U.uid
C. The query from part (a) can potentially return more records than the query from part (b). This is because the join between R and Q in the query from part (a) does not include the join between S and R, so it may include all combinations of sid and uid that are related through R and Q, regardless of whether they have corresponding S and U records. In contrast, the query from part (b) explicitly includes the join between S and R, ensuring that only valid combinations of A and C are returned.
D. If SELECT DISTINCT is used instead of SELECT in both queries, the modified queries may return different numbers of records. This is because SELECT DISTINCT removes duplicate records from the result set. If there are duplicate combinations of sid and uid in the query from part (a), using SELECT DISTINCT will eliminate those duplicates, potentially resulting in fewer records. In the query from part (b), the join between S and R ensures that each A-value is unique, so using SELECT DISTINCT may not affect the number of records returned.
E. SQL query to return records of the form tid, total where tid is a key of a record from table T and total indicates the total number of paths that "go through" that record:
SELECT R.tid, COUNT(*) AS total
FROM R
JOIN Q ON R.tid = Q.tid
GROUP BY R.tid
This query joins tables R and Q based on the tid column, and then groups the records by tid. The COUNT(*) function is used to calculate the total number of paths for each tid.
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What is the Entropy value for the below variable. = survived ['yes', 'no', 'no', 'yes','no', 'no', 'yes', 'no', 'yes',yes ']
the entropy value for the variable 'survived' is approximately 0.971.
The entropy value for the variable 'survived' can be calculated using the following formula:Entropy = -p(yes)log2p(yes) - p(no)log2p(no)where p(yes) is the probability of the outcome 'yes' and p(no) is the probability of the outcome 'no'.
To calculate the entropy value, we first need to calculate the probabilities of 'yes' and 'no' in the given variable:Probability of 'yes' = number of 'yes' outcomes / total number of outcomes= 4 / 10 = 0.4
Probability of 'no' = number of 'no' outcomes / total number of outcomes= 6 / 10 = 0.6Using the probabilities, we can calculate the entropy value as follows:Entropy = -0.4log2(0.4) - 0.6log2(0.6)≈ 0.971
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