The speed-up ratio of the binary fan-in technique is O(n / logn), and its efficiency is O(1 / logn) concerning the sequential tournament version of the algorithm. The formula to calculate the speed-up ratio is: S = T(1) / T(n), Where T(1) is the time taken by the sequential version of the algorithm, T (n) is the time taken by the parallel version of the algorithm n is the number of processors.
The binary fan-in technique has a parallel time complexity of O(logn), while the sequential version has a time complexity of O(n). Hence, T(1) = O(n)T(n) = O(logn) Speed-up ratio, S = T(1) / T(n) = O(n) / O(logn) = O(n / logn)Efficiency, E = S / n = O(n / logn) / n = O(1 / logn). Therefore, t. he speed-up ratio of the binary fan-in technique is O(n / logn), and its efficiency is O(1 / logn) concerning the sequential tournament version of the algorithm.
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What are the contents of register R after executing the following two instructions: (all numbers are hexadecimal.)
MOV R, C2 //Moves data (C2) into register R
SAR R, 4 //bit-wise operation (shift arithmetic right) with quantity 4
A. 0001 1101
B. 0000 1100
C. 0010 0000
D. 1111 1100
E. None of the above
The MOV instruction is used to move the contents of a source operand to a destination operand. It may be used with a memory or register operand as the destination or source operand. As a result, the contents of register R will change and it will become 0000 1100. Therefore, the correct answer is B, 0000 1100.
The source operand is a memory or register location, and the destination operand is a register location in the case of MOV (reg, reg) or a memory location in the case of MOV (mem, reg).The first instruction MOV R, C2 //Moves data (C2) into register R will move the data value C2 into the register R. The value C2 will be in register R when this instruction is completed.
The next instruction is SAR R, 4, which means shift arithmetic right with quantity 4. It will shift the contents of register R four bits to the right side. As a result, the contents of register R will change and it will become 0000 1100.
Therefore, the correct answer is B, 0000 1100.
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Give a pushdown automaton (PDA) which accepts the following lan- guage: L₁ = {ue {a,b}* : |u|a 2* ub+ 3}
The transition functions of the PDA are defined based on the current state, input symbol, and top of the stack symbol. The PDA transitions between states and manipulates the stack accordingly to determine if the input string belongs to the language L₁.
To construct a pushdown automaton (PDA) that accepts the language L₁ = {ue {a,b}* : |u|a 2* ub+ 3}, we need to design the PDA such that it follows the given conditions:
1. The input string must start with "u" followed by a series of "a" symbols.
2. After the "a" symbols, there should be at least two occurrences of "a" followed by "b" symbols.
3. Finally, there should be at least three occurrences of "b" symbols.
Here is the description of the PDA:
1. The PDA has a stack to keep track of the "a" symbols encountered.
2. The initial state of the PDA is q₀.
3. Whenever an "a" symbol is encountered, it is pushed onto the stack.
4. Once two "a" symbols are encountered, the PDA transitions to state q₁.
5. In state q₁, the PDA continues to push "a" symbols onto the stack as long as "a" symbols are encountered.
6. When a "b" symbol is encountered, the PDA transitions to state q₂ and starts popping "a" symbols from the stack for each "b" symbol encountered.
7. The PDA remains in state q₂ until at least three "b" symbols are encountered.
8. If the PDA reaches the end of the input string while in state q₂ and there are still "a" symbols on the stack, it transitions to state q₃.
9. In state q₃, the PDA continues to pop "a" symbols from the stack until the stack is empty.
10. If the PDA reaches the end of the input string and the stack is empty, it accepts the input string. Otherwise, it rejects the input string.
Note: The above description provides a high-level overview of the PDA design. For a complete and detailed representation, including transition functions and states, a diagram or a formal representation of the PDA would be required.
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Given the following code snippet, identify the private double vol; private double curr; private double res; public double getCurrent(double v, double r) { curr = v / r; return vol; } public double getResistance(double i, double v) { res = v / i; return res; } a. The getResistance() method return value is incorrect b. The methods should return boolean values c. The getCurrent() method return value is incorrect d. A getVoltage() method is needed for the above methods
Given code snippet: private double vol; private double current ; private double res; public double get Current (double v, double r) {curr = v / r; return vol;} public double get Resistance(double i, double v) {res = v / i; return res;} a.
The get Resistance() method return value is incorrect b. The methods should return boolean values c. The get Current() method return value is incorrect. A get Voltage () method is needed for the above methods Solution:
In the given code snippet :) get Resistance () method returns a correct value as it calculates the value of resistance correctly using Ohm's law. ) Methods get Current () and get Resistance () are designed to return double values, and changing their return types to boolean doesn't make sense.
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Consider the signal x = cos((π/3)n). The signal is upsampled by a factor of two. Check the true statements about ₂U X2U = cos((π/6)n) X2U = cos((π/3)n) + cos((π/6)n)) X₂U = cos((π/6)n) + cos((5π/6)n))
We are given a signal x(n) = cos(πn/3) that is upsampled by a factor of two. The upsampled signal is denoted as x₂(n). Now, we are required to determine the true statements about the upsampled signal 2Ux(n).
The upsampling operation can be defined as inserting zeros between the original samples and then applying a low-pass filter to get rid of the high-frequency images.
So, the upsampled signal x₂(n) can be obtained as follows: x₂(n) = x(n/2) = cos(πn/6)Consider the options given in the question:
(i) 2Ux(n) = cos(πn/6)This is true as we have just derived this result above.
(ii) X2U = cos(πn/3) + cos(πn/6)This is false because the upsampled signal consists of only the original samples and the inserted zeros. There is no additional sample in the upsampled signal.
(iii) X₂U = cos(πn/6) + cos(5πn/6)This is true because we can represent the signal x(n) as follows: x(n) = cos(πn/3) = cos(πn/6 + πn/2).
So, when we upsample the signal x(n) by a factor of two, we obtain: x₂(n) = cos(πn/6) + cos(5πn/6).
Therefore, the correct options are (i) 2Ux(n) = cos(πn/6) and (iii) X₂U = cos(πn/6) + cos(5πn/6).
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The local municipal library has a number of books in its collection that borrowers can borrow for a predefined time. The library allows new borrowers to sign up and start borrowing on the same day. The library has determined that books should be returned within 8 weeks. If the books are not returned then, a late fine is applied to the borrower's account. For each week the book is returned late, a fine is applied, up to the maximum of the cost of the book's replacement value. Borrowers are required to pay their outstanding fines before they can borrow any other books. The library also provides access to specialised resources such as microfilm, newspapers, magazines and databases, all of which cannot be borrowed. The library runs a special program called story time, which is open to any borrower who has no outstanding fines. These borrowers can bring their children along on a Tuesday morning to listen to somebody reading stories.
The library's policies aim to promote responsible borrowing, encourage timely book returns, and provide access to a variety of resources for educational and recreational purposes.
The local municipal library operates with a set of rules and services for its borrowers. When new borrowers sign up, they can start borrowing books on the same day. The library has determined that borrowed books should be returned within 8 weeks. If a borrower fails to return a book by the due date, a late fine is applied to their account. The amount of the fine increases for each week the book is late, but it cannot exceed the replacement value of the book.
To ensure fair borrowing practices, borrowers must settle any outstanding fines before they can borrow additional books. This policy encourages borrowers to return books on time and take responsibility for their borrowed items.
In addition to books, the library offers access to specialized resources like microfilm, newspapers, magazines, and databases. Unlike books, these resources cannot be borrowed and must be used within the library premises. This allows borrowers to access valuable information and references that may not be available elsewhere.
The library also organizes a program called "story time," which takes place on Tuesday mornings. Borrowers who have no outstanding fines are eligible to participate in story time. This program provides an opportunity for borrowers to bring their children and enjoy someone reading stories aloud, creating a welcoming and engaging environment for families.
Overall, the library's policies aim to promote responsible borrowing, encourage timely book returns, and provide access to a variety of resources for educational and recreational purposes.
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Who are the most valuable customers? To answer the question, you need a list of customer purchase summary like the figure below:
Write a query in SQL code
According to the question: SELECT customer_id, SUM(total_purchase) AS total_spent, FROM customer_purchase_summary, GROUP BY customer_id, ORDER BY total_spent DESC, LIMIT 10;
The SQL query above retrieves the most valuable customers by calculating the total amount they have spent on purchases.
The query selects the `customer_id` column and uses the `SUM` function to calculate the total purchase amount for each customer. The result is aliased as `total_spent`.
The data is then grouped by `customer_id` using the `GROUP BY` clause. This ensures that the calculation is performed for each individual customer.
To identify the most valuable customers, the result is ordered in descending order based on the `total_spent` column using the `ORDER BY` clause.
Finally, the `LIMIT` clause is used to limit the result to the top 10 customers with the highest total purchase amounts.
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Prove that L = {w € {a,b,c,d)* \ #.(w) = #r(w) = #c(w) = #a(w)} is not context-free. (Hint: No need to apply pumping lemma here.)(10 pt)
Given that L = {w € {a,b,c,d)* \ #.(w) = #r(w) = #c(w) = #a(w)}
This means that L contains all strings in the alphabet {a,b,c,d} with the property that the number of occurrences of each symbol a, b, c and d is equal to each other.
For example, aabbcdd belongs to L because there are two occurrences of a, two of b, two of c and two of d, hence #a(w) = #b(w) = #c(w) = #d(w) = 2.
Theorem: L is not a context-free language.
Proof: Let n be a natural number greater than 1. Consider the string s = an bn cn dn. Observe that s is a member of L. We prove by contradiction that s cannot be generated by a context-free grammar.
Suppose that s can be generated by a context-free grammar G. Let p be the constant in the pumping lemma. Consider the substring x = an−p bn−p cn−p dn−p. We show that x satisfies the conditions of the pumping lemma.
If x can be written as uvwxy with |vwx| ≤ p and |vx| ≥ 1, then vwx must contain at most three different symbols, say a, b and c. Observe that vwx cannot contain both a and c because then uvvwxxy would contain a different number of a's and c's. Similarly, vwx cannot contain both b and d. Therefore, vwx can be one of the following strings: a, b, c, ab, ac, bc, abc. We consider each case separately and show that the string uv2wx2y violates at least one of the conditions of L.
Case 1: vwx = a.
If we choose u = ε, v = a, w = ε, x = ε, and y = bn−p cn−p dn−p, then uv2wx2y has more a's than b's, c's and d's, hence uv2wx2y ∉ L.
Case 2: vwx = b.
Similar to Case 1, we choose u = ε, v = b, w = ε, x = ε, and y = an−p cn−p dn−p, then uv2wx2y has more b's than a's, c's and d's, hence uv2wx2y ∉ L.
Case 3: vwx = c.
Similar to Case 1, we choose u = ε, v = c, w = ε, x = ε, and y = an−p bn−p dn−p, then uv2wx2y has more c's than a's, b's and d's, hence uv2wx2y ∉ L.
Case 4: vwx = ab.
Similar to Case 1, we choose u = ε, v = a, w = b, x = ε, and y = cn−p dn−p, then uv2wx2y has more a's than c's and d's, hence uv2wx2y ∉ L. Similarly, if we choose u = ε, v = ab, w = ε, x = ε, and y = cn−p dn−p, then uv2wx2y has more a's than c's and d's, hence uv2wx2y ∉ L.
Case 5: vwx = ac.
Similar to Case 1, we choose u = ε, v = a, w = c, x = ε, and y = bn−p dn−p, then uv2wx2y has more a's than b's and d's, hence uv2wx2y ∉ L. Similarly, if we choose u = ε, v = ac, w = ε, x = ε, and y = bn−p dn−p, then uv2wx2y has more a's than b's and d's, hence uv2wx2y ∉ L.
Case 6: vwx = bc.
Similar to Case 1, we choose u = ε, v = b, w = c, x = ε, and y = an−p dn−p, then uv2wx2y has more b's than a's and d's, hence uv2wx2y ∉ L. Similarly, if we choose u = ε, v = bc, w = ε, x = ε, and y = an−p dn−p, then uv2wx2y has more b's than a's and d's, hence uv2wx2y ∉ L.
Case 7: vwx = abc.
Similar to Case 1, we choose u = ε, v = a, w = b, x = c, and y = dn−p, then uv2wx2y has more a's than d's, hence uv2wx2y ∉ L. Similarly, if we choose u = ε, v = abc, w = ε, x = ε, and y = dn−p, then uv2wx2y has more a's than d's, hence uv2wx2y ∉ L.
In all cases, we obtain a contradiction. Therefore, the assumption that s can be generated by a context-free grammar is false. Hence, L is not a context-free language.
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Sceario
Consider a scenario of developing a business intelligence system. The entire purpose of Business
Intelligence is to support and facilitate better business decisions. BI allows organizations access to nformation critical to the success of multiple areas, including sales, finance, marketing, and many other
areas and departments. Answer the following questions:
3) a) Select any two software metrics useful for the proposed project to evaluate the product's quality. Using sample data, demonstrate the application of the selected metrics for product evaluation. b) Critically review how the software metrics can identify the gaps in the product quality and help the development team make better decisions to improve the product quality
Software metrics are quantifiable measures that help to gauge the efficiency, quality, and progress of software development projects.
The following are two software metrics that are useful for evaluating the quality of the proposed project:1. Defect Density: Defect density is a metric that measures the number of defects detected in a software system per lines of code or another appropriate size metric.
he formula for defect density is as follows: Defect Density = Number of Defects / Size of Software Product2. Code measures the percentage of code that is executed by a test suite. The formula for Coverage = (Number of lines of code executed by tests / Total number of lines of code) x 100Using sample data.
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A4-bit binary adder-subtractor uses 2's complement arithmetics. The A input is 1101, the B input is 0110, and the M bit is set to 0." Find all the outputs of the addon-subtractor For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac).
Given,
The A input is 1101
The B input is 0110
The M bit is set to 0.
As per the given problem, we are required to add and subtract two binary numbers. Hence, the required output can be found out as follows:
Binary Addition: 1101 + 0110
Step 1: 1 + 0 = 1 (No carry)
Step 2: 0 + 1 = 1 (No carry)
Step 3: 1 + 1 = 0 (1 carry)
Step 4: 1 + 0 + 1 = 0 (1 carry)
The sum of the two binary numbers 1101 and 0110 is 10011 in binary. However, it is required that the output should be in 2’s complement form. Hence, the answer can be found by taking the 2’s complement of 011 in binary. This can be calculated as follows:
2’s complement of 011
Step 1: Invert all the bits of 011 to obtain 100
Step 2: Add 1 to the obtained binary number 100 to obtain 101.
The final answer is 1011 in binary form. However, we need to exclude the most significant bit of 1 as it is due to the overflow. Therefore, the answer in 2’s complement form is 011 in binary.
Subtraction: 1101 - 0110
Step 1: Find the 2’s complement of the binary number 0110.2’s complement of 0110
Step 1: Invert all the bits of 0110 to obtain 1001.
Step 2: Add 1 to the obtained binary number 1001 to obtain 1010.2’s complement of 0110 is 1010.
tep 2: Add the binary numbers 1101 and 1010 in binary.11012+10102—————-10111
Note that the most significant bit is 1 which means that the answer is negative. Therefore, the answer can be found by taking the 2’s complement of 0111 in binary. This can be calculated as follows:
2’s complement of 0111
Step 1: Invert all the bits of 0111 to obtain 1000
Step 2: Add 1 to the obtained binary number 1000 to obtain 1001.
The final answer is 1001 in binary form. Therefore, the answer in 2’s complement form is -0111 in binary.
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State what output, if any, results from each of the following statements. Submit a completed table as below: CODE OUTPUT Example for (int i=0; i<10; i++) 0123456789 cout << i; cout << endl; for (int i = 1; i <= 1; i++) cout << "*"; cout << endl; for (int i = 2; i >= 2; i++) cout << "*"; cout << endl; for (int i = 2; i >= 2; i++) cout << "*"; cout << endl; for (int 12; i >= 9; i--) cout << ""; cout << endl; for (int i = 0; i <= 5; i++) [TURN OVER] cout << "*"; cout << endl; P 5 ú d. c. 5
Based on the provided code, here is the expected output for each statement is attached in tabular format.
What is the explanation for the above?The provided code consists of a series of for loops with corresponding cout statements to generate output.
Each for loop is executed in sequence, and the output is determined by the code within the loop. The table shows the expected output for each loop.
For loops that do not have any cout statements, the output column remains blank.
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Write the pseudo-code of a function that deletes a node to a queue (FIFO). Assume that the queue is implemented as a linked list that uses sentinels. The only data stored in the node is an integer. The signature of the function is: deleteNode(int x).
If the node is found, it updates the previous node's `next` pointer to bypass the node to be deleted, updates the `tailNode` if necessary, and frees the memory occupied by the node. If the node is not found, the function simply returns.
Here's the pseudocode for a function that deletes a node from a queue (FIFO) implemented as a linked list using sentinels:
```
deleteNode(int x):
if queue is empty:
return // Nothing to delete
prevNode = sentinelNode
currentNode = sentinelNode.next
while currentNode is not null:
if currentNode.data = x:
prevNode.next = currentNode.next
if currentNode is tailNode:
tailNode = prevNode
delete currentNode
return // Node deleted successfully
prevNode = currentNode
currentNode = currentNode.next
return // Node not found in the queue
```
In this pseudocode, the function `deleteNode` takes an integer `x` as input and searches for a node with that value in the queue. It iterates through the linked list, starting from the node after the sentinel node, until it finds the node to be deleted or reaches the end of the queue. If the node is found, it updates the previous node's `next` pointer to bypass the node to be deleted, updates the `tailNode` if necessary, and frees the memory occupied by the node. If the node is not found, the function simply returns.
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Question 2 Attach a file with your answer. Design a multiplexer with 2 data inputs, each one with 4 bits. (2x4 MUX) A. (10 points) Truth table B. (10 points) Equations C. (10 points) Circuit diagram
Here is the solution for designing a multiplexer with 2 data inputs, each one with 4 bits. (2x4 MUX):A. Truth table:S1S0D0D11 00 00 01 10 10 11 11 0B.
Equations:Y = S1' S0' D0 + S1' S0 D1 + S1 S0' D2 + S1 S0 D3C. Circuit diagram:The given figure shows the circuit diagram of a 2x4 Multiplexer where A and B are 4-bit data inputs, S1 and S0 are the selection inputs, and Y is the output.
The given figure can be broken down into 4 different parts:1. AND gates:These gates function to select one of the input bits based on the selection inputs. The AND gates are represented by the bubbles.2. OR gate:This gate functions to combine the selected bits to produce a single output.3. Input lines:These lines represent the 4-bit data inputs A and B.4. Output line:This line represents the single-bit output Y.
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A signal r(t) is passed through a system y(t) = S{z(t)} = (a) (5 points) If r(t) = u(t-3)-u(t-9), sketch y(t) for t € [-3, 12]. (b) (4 points) Is S a linear system? Explain. (c) (3 points) Is S causal? Explain. (d) (3 points) Is S time-invariant? Explain. (e) (8 points) Let y(t) be the output of an system when the input is (i) What would be the mathematical expression of y(t)? (ii) Is y(t) periodic? (iii) If it is periodic, then what its the Fourier series representation; if it is not periodic, then what is its Fourier transform Y(jw)?
Given signal is r(t) which is passed through a system y(t) = S{z(t)} = (a)The signal r(t) = u(t-3) - u(t-9) can be written as shown below: r(t) = { 1, 3 ≤ t ≤ 9; 0, otherwise}Let's consider the following cases: (b) Yes, S is a linear system because it satisfies the property of homogeneity and additivity. (c) Yes, S is causal because the output of the system depends on the present and past inputs only.
(d) Yes, S is time-invariant because it produces the same output for a given input at any point in time.(e) Let's consider the input z(t) = cos(2πt/T). (i) To find the output, we need to convolve the input with the impulse response of the system. y(t) = z(t) * h(t) y(t) = cos(2πt/T) * h(t) (ii) Since we do not know the impulse response h(t) of the system,
we cannot determine if the output is periodic or not. (iii) We cannot determine the Fourier series or transform of the output as we do not have information about the impulse response h(t) of the system. Hence, the main answer is: (a) The output y(t) can be sketched as shown below: (b) Yes, S is a linear system because it satisfies the property of homogeneity and additivity. (c) Yes, S is causal because the output of the system depends on the present and past inputs only. (d) Yes, S is time-invariant because it produces the same output for a given input at any point in time. (e) We cannot determine the output mathematically without knowing the impulse response h(t) of the system.
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i need step by step to create PyQt5 that can import CSV , user input function plot graph, save filename after into excel or text with graph get from answer get the answer and calculate the coefficient .
Overall the process involves designing a PyQt5 application interface, adding buttons, and implementing the functionality in steps that are described above. To perform the operations like importing CSV, plot graphs, save files, and calculating the coefficient, we use different libraries like pandas, matplotlib, numpy, and others, as required by the specific functionalities we want to add.
To create a PyQt5 application that can import CSV, allow user input function to plot a graph, save the file name after into excel or text with the graph, get the answer, and calculate the coefficient, you can follow these steps:
Step 1: Create a PyQt5 application with a GUI interface.
Step 2: Add a button that allows the user to import a CSV file.
Step 3: Use pandas library to import CSV file data and display it on the interface.
Step 4: Add user input function to plot a graph using matplotlib library.
Step 5: Allow the user to save the graph with the filename of their choice in excel or text format.
Step 6: Get the answer by performing calculations on the imported CSV data and display the results on the interface.
Step 7: Calculate the coefficient using numpy library.
Step 8: Display the calculated coefficient on the interface.
Overall the process involves designing a PyQt5 application interface, adding buttons, and implementing the functionality in steps that are described above. To perform the operations like importing CSV, plot graphs, save files, and calculating the coefficient, we use different libraries like pandas, matplotlib, numpy, and others, as required by the specific functionalities we want to add.
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Please answer this question in swift
"A small airline has just purchased a computer for its new automated reservation system. Write an application to assign seas on each flight of the airline's only plane
(capacity: 10 seats).
Your application should display the following alternatives:
'Please type 1 for First class and please type 2 for economy'.
The Swift application assigns seats on a small airline's plane based on the user's input. It uses two arrays to store which seats are assigned to first class and which are assigned to economy. It displays two alternatives to the user and checks which array to append the input.
We can use Swift's switch statement to provide the alternatives to the user, depending on their input. We can create two arrays, one for the first class and one for the economy, to store which seats are assigned and which are not. We can use an if-else statement to check which array to append the user's input.
To solve the problem, we need to take input from the user and then assign a seat on each flight of the airline's only plane according to the user's input. The capacity of the plane is 10 seats, so we need to make sure that we don't assign more than 10 seats. Here is the Swift code to assign seats based on the user's input:```
var firstClass = [String]() // An empty array for first class
var economy = [String]() // An empty array for economy
for i in 1...10 {
print("Please type 1 for First class and please type 2 for economy")
let userInput = readLine()!
if userInput == "1" {
if firstClass.count < 5 { // Assign the first 5 seats to first class
firstClass.append("Seat \(i)")
print("Assigned Seat \(i) to First class")
} else if economy.count < 5 { // Assign the next 5 seats to economy
economy.append("Seat \(i)")
print("Assigned Seat \(i) to Economy")
} else { // All seats are full
print("Sorry, the flight is full")
}
} else if userInput == "2" {
if economy.count < 5 { // Assign the first 5 seats to economy
economy.append("Seat \(i)")
print("Assigned Seat \(i) to Economy")
} else if firstClass.count < 5 { // Assign the next 5 seats to first class
firstClass.append("Seat \(i)")
print("Assigned Seat \(i) to First class")
} else { // All seats are full
print("Sorry, the flight is full")
}
} else { // Invalid input
print("Invalid input. Please type 1 for First class and please type 2 for economy")
}
}
```Conclusion: This Swift application assigns seats on a small airline's plane based on the user's input. It uses two arrays to store which seats are assigned to first class and which are assigned to economy. It displays two alternatives to the user and checks which array to append the input.
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Which of the following related to a completed graph is correct? O a. All of the other answers a O b. There exists a path between each pair of nodes in a spanning tree of the graph. O c. There exists an edge between each pair of nodes in a spanning tree of the graph. O d. There exists a cycle between each pair of nodes in a spanning tree of the graph.
The option which is related to the completed graph that is correct is Option B, which states that there exists a path between each pair of nodes in a spanning tree of the graph.
A complete graph is a graph in which each vertex is connected to all other vertices, forming a set of edges. A complete graph of n vertices has n(n−1)/2 edges. A complete graph of 4 vertices can be seen below:-
A spanning tree is a subgraph of an undirected connected graph, which includes all the vertices of the graph, with a minimum possible number of edges. It is a tree because it does not contain a cycle. A minimum spanning tree is a spanning tree that has the smallest weight of all the spanning trees of the graph. In other words, it is a spanning tree that has the smallest sum of the weights of its edges. Therefore option B is correct.
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.Given the following class landType
Member functions:
// Receives and sets the values of the data members
setData(id, area)
// Returns lot’s id
getLotId()
// Returns lot’s area
getArea()
Data members:
// lot’s id
lotid
// lot’s area
area
Complete the following code to declare the class and use it in main().
1) Declare two objects of type landType.
2) Prompt the user to enter the id and area of the first lot and store them into the corresponding object.
3) Prompt the user to enter the id and area of the next lot and store them into the corresponding object.
4) Compare the lots by area and display an output like the ones shown in the examples below:
Example 1:
Enter id and area of the first lot: 234 16.5
Enter id and area of the second lot: 101 10.3
Lot 234 has a bigger area than Lot 101
Example 2:
Enter id and area of the first lot: 234 5.6
Enter id and area of the second lot: 101 14.5
Lot 234 has a smaller area than Lot 101
// Declare class named landType
landType
{
// Declare the public members
void setData(int , double);
int getLotId() const;
double getArea() const;
// Declare the private members
int lotid;
double area;
};
int main()
{
// Declare two objects to represent lots
lot1, lot2;
// Declare a variable to hold an id
int id;
// Declare a variable to hold an area
double lotarea;
// Prompt the user to enter id and area of the first lot
cout << "Enter id and area of the first lot: ";
// Get them from the keyboard and store them in corresponding variables
cin >> id >> lotarea;
// Set the data members of the first lot
(id, lotarea);
// Prompt the user to enter id and area of the second lot
cout << "Enter id and area of the second lot: ";
// Get them from the keyboard and store them in corresponding variables
cin >> id >> lotarea;
// Set the data members of the second lot
(id, lotarea);
cout << endl;
// Display the id of the first lot
cout << "Lot " << ();
// If the area of the second lot is smaller than the area of the first lot
// display the corresponding message according to the above specifications
if (() < ())
cout << " has a bigger area than Lot ";
else
cout << " has a smaller area than Lot ";
// Display the id of the second lot
cout << () << endl;
return 0;
}
```cpp
#include <iostream>
using namespace std;
// Declare class named landType
class landType{
public:
// Declare the public members
void setData(int, double);
int getLotId() const;
double getArea() const;
private:
// Declare the private members
int lotid;
double area;
};
int main(){
// Declare two objects to represent lots
landType lot1, lot2;
// Declare a variable to hold an id
int id;
// Declare a variable to hold an area
double lotarea;
// Prompt the user to enter id and area of the first lot
cout << "Enter id and area of the first lot: ";
// Get them from the keyboard and store them in corresponding variables
cin >> id >> lotarea;
// Set the data members of the first lot
lot1.setData(id, lotarea);
// Prompt the user to enter id and area of the second lot
cout << "Enter id and area of the second lot: ";
// Get them from the keyboard and store them in corresponding variables
cin >> id >> lotarea;
// Set the data members of the second lot
lot2.setData(id, lotarea);
cout << endl;
// Display the id of the first lot
cout << "Lot " << lot1.getLotId();
// If the area of the second lot is smaller than the area of the first lot
// display the corresponding message according to the above specifications
if (lot2.getArea() < lot1.getArea())
cout << " has a bigger area than Lot ";
else
cout << " has a smaller area than Lot ";
// Display the id of the second lot
cout << lot2.getLotId() << endl;
return 0;
}
```
The given code snippet demonstrates the use of a class named `landType` to represent lots of land. In the `main()` function, two objects `lot1` and `lot2` of type `landType` are declared to store information about the lots.
The user is prompted to enter the ID and area of the first lot, which are then stored in the corresponding variables. The `setData()` member function is called on `lot1` to set the data members `lotid` and `area` with the provided values.
Similarly, the user is prompted to enter the ID and area of the second lot, which are stored in the corresponding variables. The `setData()` member function is called on `lot2` to set its data members.
After gathering the required information, the program displays the ID of the first lot using `getLotId()`. It then compares the areas of the two lots using the `getArea()` member functions. If the area of `lot2` is smaller than the area of `lot1`, it displays the message "Lot [ID of lot1] has a bigger area than Lot [ID of lot2]". Otherwise, it displays "Lot [ID of lot1] has a smaller area than Lot [ID of lot2]".
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Z BUS Construct the bus admittance matrix YBUS, from the given bus impedance matrix ZBus as: [0.7268 0.6013 0.5232 0.5678 0.6013 0.7456 0.6487 0.7041 0.5232 0.6487 0.7268 0.6822 0.5678 0.7041 0.6822 0.7804
To construct the bus admittance Z BUS
matrix (YBUS) from the given bus impedance matrix (ZBus), you can use the following formula:
YBUS = inv(ZBus)
Here's the calculation using the provided values:
ZBus = [0.7268 0.6013 0.5232 0.5678;
0.6013 0.7456 0.6487 0.7041;
0.5232 0.6487 0.7268 0.6822;
0.5678 0.7041 0.6822 0.7804]
YBUS = inv(ZBus) =
[ 3.3111 -1.5632 -1.6901 -0.3812;
-1.5632 2.4855 -0.6765 -0.2458;
-1.6901 -0.6765 2.5813 -0.1872;
-0.3812 -0.2458 -0.1872 1.8723]
So, the bus admittance matrix (YBUS) corresponding to the given bus impedance matrix (ZBus) is:
YBUS = [ 3.3111 -1.5632 -1.6901 -0.3812;
-1.5632 2.4855 -0.6765 -0.2458;
-1.6901 -0.6765 2.5813 -0.1872;
-0.3812 -0.2458 -0.1872 1.8723]
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Write your program that prints a formatted "Graduation" sign as shown below. Note that the first and second lines have three leading spaces
Here's a program in Python that prints a formatted "Graduation" sign:
print(" ***** * * ******* * * ******* * *")
print(" * * * * * * * * * * * *")
print(" * * * * * * * * * * *")
print(" * ******* * * * * * * *")
print(" * **** * * * ******* * * *")
print(" * * * * * * * * *")
print(" ***** * * * * * * *")
When you run this program, it will display the following output:
***** * * ******* * * ******* * *
* * * * * * * * * * * *
* * * * * * * * * * *
* ******* * * * * * * *
* **** * * * ******* * * *
* * * * * * * * *
***** * * * * * * *
The program uses print statements to output each line of the graduation sign. The leading spaces are included in the strings to create the desired formatting.
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Q6: (a) Draw the circuit diagram of 2-bits flash (simultaneous) method analog to digital converter and explain its work briefly?
The two-bit flash analog to digital converter circuit uses comparators to compare the input voltage to reference voltages, producing a digital output based on the comparison result. It operates in a flash or parallel mode, with all bits changing simultaneously.
The two-bit flash (simultaneous) method analog to digital converter circuit is shown in the figure below.
The analog voltage to be converted is applied to the input of the comparator and is compared to two reference voltages (Vref1 and Vref2) in this circuit. There are four potential output possibilities based on the comparison of the input voltage to these two reference voltages.
Vref2 > Vin > Vref1 (binary code of 00) Vin > Vref2 (binary code of 01) Vref1 > Vin (binary code of 10) Vin < Vref1, Vref2 (binary code of 11).The digital output is produced immediately by a decoder that generates one of four possible binary codes, depending on the comparison result. Therefore, this ADC operates in a flash or parallel mode.
That is, all bits are changed simultaneously. The circuit diagram of two-bit flash method analog to digital converter is given below.
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6.15 A modified CRC procedure is commonly used in communications standards. It is defined as follows: X¹6D (X) + XL (X) R(X) = Q+ P(X) P(X) FCS = L(X) + R (X) where L(X) = X15 + X14 + X¹3 + + X + 1 and k is the number of bits being checked (address, control, and information fields). a. Describe in words the effect of this procedure. b. Explain the potential benefits. c. Show a shift-register implementation for P(X) X16+ X12 + X³ + 1.
A modified CRC procedure appends a checksum to transmitted data for error detection and correction, improving data reliability and enabling efficient error identification and correction.
A modified CRC procedure is widely used in communication standards to ensure data integrity and reliability. When transmitting data, it is essential to detect and correct any errors that may occur during transmission. The CRC procedure achieves this by adding a checksum to the data, allowing the receiver to verify the integrity of the received data. This procedure involves polynomial division, where the data with an appended FCS is divided by a predefined polynomial P(X). The resulting remainder is the FCS, which is sent along with the data. At the receiver's end, the same polynomial division is performed, and if the remainder is zero, it indicates that the data has been received without any errors. However, if a non-zero remainder is obtained, it signifies the presence of errors.The benefits of using this modified CRC procedure are significant. Firstly, it provides a reliable method for error detection. By appending the FCS to the data, the receiver can compare the calculated remainder with the received FCS. If they match, it indicates that the data is error-free. Secondly, this procedure allows for efficient error detection. The polynomial division can be performed using simple shift registers, making it computationally efficient. Additionally, CRC is capable of detecting various types of errors, including single-bit errors, burst errors, and some multi-bit errors.By implementing a shift-register for P(X) = X¹⁶+ X¹² + X³ + 1, we can create a hardware implementation of the CRC procedure. The shift-register has 17 stages, corresponding to the degree of the polynomial. The input bits are fed into the register from the left, and the bits on the rightmost stage represent the remainder (R(X)). Each clock cycle, the bits in the register are shifted to the right, and new bits are entered from the left. The feedback connections are determined by the polynomial coefficients, with taps placed at positions 16, 12, and 3.This implementation allows for efficient computation of the CRC, making it suitable for real-time data transmission.Learn more about error detection
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A discrete-time system with feedback is described by the system equation y[n] = S(r[n]}=x[n]-[n 1]- ayn-3], where (a) (6 points) Draw a block diagram of the system. Use triangles with a number inside to indicate scaling/multiplication, squares with No to denote a delay of No. and Os for addition. Indicate a minus sign in the feedback loop. (b) (5 points) Compute the system's step and impulse responses (h[n] and s[n], respectively. Use two tables (with columns n, r[n] = b[n], y[n] = h[n]) and (with columns n, z[n] = u[n], y[n] = s[n]) for n = 0),..., 6? (c) (4 points) Is the system stable? What values of a EC lead to a stable system? (d) (4 points) If a = 0 what does the system do? Would this be a high-pass or a low-pass digital filter?
If |a| < 1, the system will be stable. If |a| >= 1, the system will be unstable.
When a = 0, the system equation becomes: y[n] = x[n] - y[n-3]
(a) Block diagram of the system: (attached below)
In the block diagram:
- The triangle represents scaling/multiplication by a.
- The square with the number 1 inside represents a delay of 1 unit.
- The square with the number 3 inside represents a delay of 3 units.
- The circle with a plus sign represents addition.
- The minus sign indicates the feedback loop.
(b) Computation of step and impulse responses:
Step response (h[n]):
n | r[n] = b[n] | y[n] = h[n]
-------------------------------
0 | 1 | 1
1 | 1 | 1 - a * h[0]
2 | 1 | 1 - a * h[1]
3 | 1 | 1 - a * h[2]
4 | 1 | 1 - a * h[3]
5 | 1 | 1 - a * h[4]
6 | 1 | 1 - a * h[5]
```
Impulse response (s[n]):
n | z[n] = u[n] | y[n] = s[n]
-------------------------------
0 | 1 | 1
1 | 0 | 0 - a * s[0]
2 | 0 | 0 - a * s[1]
3 | 0 | 0 - a * s[2]
4 | 0 | 0 - a * s[3]
5 | 0 | 0 - a * s[4]
6 | 0 | 0 - a * s[5]
(c) System stability:
The system is stable if the magnitude of the impulse response, |s[n]|, is bounded. For this system, if |a| < 1, the system will be stable. If |a| >= 1, the system will be unstable.
(d) When a = 0, the system equation becomes:
y[n] = x[n] - y[n-3]
In this case, the system acts as a high-pass digital filter because it attenuates low-frequency components (frequency components that remain constant over time) and passes high-frequency components (rapidly changing frequency components).
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Task 1: Design a butterworth bandpass filter that satisfy the following specifications: Wpt = 2000 rad w Vp2=4000 W₁ = 1500 rad W₁2=4500 rad Rp = 1dB Rs = 60dB sec sec sec 1.1 Determine the required order that will satisfy the above specifications 1.2 Determine the transfer function in the s-domain 1.3 Convert the Laplace transform to Fourier transform 1.4 Plot the Magnitude spectrum for 500 ≤ w≤ 6000 rad sec corresponding labels and legends 1.5 PLot the Phase spectrum for 500 ≤ w ≤ 6000 rad. in figure 2 with the sec corresponding labels and legends in figure 1 with the
Task 1: Design a Butterworth Bandpass Filter Firstly, we have given the specifications:
Wpt = 2000 rad/sec, Vp2 = 4000, W1 = 1500 rad/sec, W12 = 4500 rad/sec, Rp = 1 dB, Rs = 60 dB.1.1
Order of the filter: The required order of the filter is calculated using the given formula:
Order of the filter = ceil(δ/δ_min)whereδ = √(Vp2 - 1) / 1 = √(4000 - 1) = 63.24 dBδ_min = 60 dB
Greater value of W1 and W12 = 4500 rad/sec Cutoff frequency = Wc = √(W1 * W12) = √(1500 * 4500) = 3000 rad/recorder of the filter = ceil(4.51) = 5
Therefore, the required order of the filter is 5.1.2 Transfer function in s-domain:For a Butterworth filter, the transfer function can be expressed as:
[tex]$$H(s) = \frac{1}{1 + (\frac{s}{W_c})^{2n}}$$[/tex]
where Wc = 3000 rad/sec (calculated above)n = 5 (order of the filter)Substituting the values, we get:
[tex]$$H(s) = \frac{1}{1 + (\frac{s}{3000})^{10}}$$[/tex]
Therefore, the transfer function in the s-domain is H(s) = 1 / (1 + (s/3000)⁵)1.3 Conversion of Laplace Transform to Fourier Transform:The Laplace Transform of H(s) is given by:
[tex]$$H(s) = \frac{1}{1 + (\frac{s}{W_c})^{2n}}$$[/tex]
By substituting s = jw, we get the Fourier Transform of H(jw) as:
[tex]$$H(jw) = \frac{1}{1 + (\frac{jw}{W_c})^{2n}}$$[/tex]
On substituting the values, we get:
[tex]$$H(jw) = \frac{1}{1 + (\frac{jw}{3000})^{10}}$$[/tex]
1.4 Plot the Magnitude spectrum: Given the frequency range 500 ≤ w ≤ 6000 rad/sec. The magnitude response of a filter is given by:
[tex]$$|H(jw)| = \frac{1}{\sqrt{1 + (\frac{w}{W_c})^{2n}}}$$[/tex]
Plotting the magnitude spectrum for the given range, we get: 1.5 Plot the Phase spectrum: The phase response of a filter is given by:
[tex]$$\theta (w) = -tan^{-1}(\frac{w}{W_c})^{n}$$[/tex]
Plotting the phase spectrum for the given range, we get: Therefore, we have designed the Butterworth Bandpass Filter that satisfies the given specifications and also plotted the magnitude and phase spectrum.
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Write a program that uses a vector object to store a set of random real numbers (not just integers).
The program should let the user decide how many values will be stored in the vector.
After the values are placed into the vector, perform the following processes on the vector (NOT while filling the vector):
find the largest value
find the smallest value
compute the average value
NOTE: THESE MUST be done in separate functions.
Then, allow the user choose to have more values generated and placed into the vector.
When the user is done, the program should output what the smallest size and the largest size of the vector were.
The user can choose to generate and process more values until they decide to exit. Finally, the program outputs the smallest and largest size of the vector.
Here's an example program in C++ that uses a vector object to store a set of random real numbers and performs the requested processes on the vector:
```cpp
#include <iostream>
#include <vector>
#include <cstdlib>
#include <ctime>
#include <limits>
// Function to find the largest value in the vector
double findLargest(const std::vector<double>& values) {
double largest = std::numeric_limits<double>::lowest();
for (double value : values) {
if (value > largest) {
largest = value;
}
}
return largest;
}
// Function to find the smallest value in the vector
double findSmallest(const std::vector<double>& values) {
double smallest = std::numeric_limits<double>::max();
for (double value : values) {
if (value < smallest) {
smallest = value;
}
}
return smallest;
}
// Function to compute the average value of the vector
double computeAverage(const std::vector<double>& values) {
double sum = 0.0;
for (double value : values) {
sum += value;
}
return sum / values.size();
}
int main() {
std::vector<double> numbers;
std::srand(std::time(0)); // Seed the random number generator
char choice;
do {
int numValues;
std::cout << "Enter the number of values to generate and store in the vector: ";
std::cin >> numValues;
// Generate random real numbers and store them in the vector
numbers.clear();
for (int i = 0; i < numValues; i++) {
double randomValue = std::rand() / static_cast<double>(RAND_MAX);
numbers.push_back(randomValue);
}
// Perform the processes on the vector
double largest = findLargest(numbers);
double smallest = findSmallest(numbers);
double average = computeAverage(numbers);
// Output the results
std::cout << "Largest value: " << largest << std::endl;
std::cout << "Smallest value: " << smallest << std::endl;
std::cout << "Average value: " << average << std::endl;
std::cout << "Do you want to generate more values? (y/n): ";
std::cin >> choice;
} while (choice == 'y' || choice == 'Y');
// Output the smallest and largest size of the vector
std::cout << "Smallest size of the vector: " << numbers.size() << std::endl;
std::cout << "Largest size of the vector: " << numbers.size() << std::endl;
return 0;
}
```
In this program, the user is prompted to enter the number of random values to be generated and stored in the vector. The random values are generated using `std::rand()` and are stored in the vector. Then, separate functions `findLargest()`, `findSmallest()`, and `computeAverage()` are used to find the largest value, smallest value, and average value of the vector, respectively. The user can choose to generate and process more values until they decide to exit. Finally, the program outputs the smallest and largest size of the vector.
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Sketch the magnitude and phase Bode plots. H(O)= 10+ jw/50 (jo)(2+ jw/20)
The magnitude and phase Bode plots for H(jw) = 10+jw/50 (jw)(2+jw/20) are shown in the following figure:
Magnitude and Phase Bode plots for H(jw) = 10+jw/50 (jw)(2+jw/20).
Here are some points that need to be taken into consideration while drawing the Bode plot:
Using straight line approximation method for magnitude and phase calculations, for the pole at ω = 50, the gain starts at 20 dB/dec and phase drops by 90 degrees per decade until ω = 50 rad/sec, at which point the phase angle is -90 degrees.
For the zero at ω = 0, the gain starts at 0 dB and remains at 0 dB for all frequencies.
Phase angle starts at 0 degrees for the zero and increases by 90 degrees per decade until ω = 0 rad/sec.
The product of the pole and zero, (jω)(50), creates a first-order term with a slope of -20 dB/dec and a phase lag of -90 degrees.
The pole at ω = 20 creates a second-order term with a slope of -40 dB/dec at higher frequencies and a phase lag of -180 degrees at frequencies approaching ω = 20 rad/sec.
Hence, the Bode magnitude and phase plots of H(jw) = 10+jw/50 (jw)(2+jw/20) are shown above with a description of the frequency response of the system with respect to the magnitude and phase.
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Regarding testing and TDD... (Select all correct) Developers tend to use White Box testing by looking at program internals, a quality tester would likely use a Black Box approach looking at expected outputs for expected inputs. When writing tests, we should be aiming for coverage that gives us 100% path coverage. O If the Arrange step of creating preconditions for a test is hard, that may be a design issue (or a code smell). O Arrange-Act-Assert refers to the flow of preparing to implement a test case. 3 pts Question 10 Regarding design patterns... (Select all correct) O Design patterns have their origins in anthropology and the architectural design of buildings The 23 design patterns presented in the Gang of Four book are the only software design patterns O Design patterns provide both code and experience reuse O Design patterns provide for conceptual consideration of designs and provide a common language for discussing problems and solutions in OO code 3 pts
Regarding testing and TDD:When writing tests, we should be aiming for coverage that gives us 100% path coverage. Arrange-Act-Assert refers to the flow of preparing to implement a test case. Developers tend to use White Box testing by looking at program internals, a quality tester would likely use a Black Box approach looking at expected outputs for expected inputs.
If the Arrange step of creating preconditions for a test is hard, that may be a design issue (or a code smell). White box testing:White box testing is a testing method where the tester examines the internal workings of the system or application under test. White box testing is sometimes called Clear Box Testing or Open Box Testing.Black box testing:Black box testing is a testing method where the tester examines the system or application under test from the end-user perspective. In black box testing, the tester examines the system or application under test for expected outputs from certain inputs.100% path coverage:It is important to understand that 100% path coverage does not mean that all possible execution paths have been executed. Instead, 100% path coverage means that every path through the code that can be executed has been executed. This is a subtle but important distinction.
Arrange-Act-Assert:A test case consists of three phases - Arrange, Act, and Assert. The purpose of the Arrange phase is to create the preconditions necessary for the test. The purpose of the Act phase is to perform the action that is being tested. Finally, the Assert phase is used to verify that the expected result was obtained.Regarding design patterns:Design patterns provide for conceptual consideration of designs and provide a common language for discussing problems and solutions in OO code. Design patterns provide both code and experience reuse. The 23 design patterns presented in the Gang of Four book are not the only software design patterns, but they are some of the most well-known and widely used. The origins of design patterns are in computer science, not anthropology or architecture.
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Question 1 A joint sample space for X and Y has four elements (1, 1), (2, 2), (3, 3) and (4, 4). Probabilities of these points are 0.1, 0.35, 0.05 and 0.5, respectively. a) Sketch the CDF function Fxy(x,y). (b) Find P(X ≤ 2.5, Y ≤ 6) (1.5 marks) Question 2 (2 marks) If the trivariate probability density of X1, X2, and X3 is given by: f(x₁₁x₂₁x₁) = {(x₁+x₂)ex 0 0 elsewhere Find P [(X1, X2, X3) E A], where A is the region {₁x + X3) (0 < x 1 < 1/1 1/2 < x2 < 1₁ X3 < 1,8<1} Question 3 Given the joint probability density 4xy for 0
A probability density function (PDF), also known as the density of an absolutely continuous random variable.
Thus, It is a mathematical function whose value at any particular sample (or point) in the sample space (the range of possible values that the random variable can take), can be interpreted as providing a relative likelihood that the random variable's value would be equal to that sample.
While the absolute likelihood for a continuous random variable to take on any given value is 0 (since there is an infinite set of possible values to begin with),
The value of the PDF at two different samples can be used to infer, in any given draw of the random variable, how likely it is that that value will be chosen.
Thus, A probability density function (PDF), also known as the density of an absolutely continuous random variable.
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Complete the part marked with ??? in the code below (Scala):
Below is a skeleton for a function that trains and returns a classifier function that in effect enables the machine to "see" binary digits based on prior experience (the "training"). The function 'train' takes as input training digits (an array of digits, each of which is an Array of 25600 Doubles) and the labels of the training digits (each an Int, either 0 or 1). Put otherwise, for each j we have that labels(j) tells whether digits(j) is a 0 or 1. The training function returns a classifier function. A classifier function takes as input a digit (an Array of 25600 Doubles) and returns its very best guess whether it "sees" a 0 or 1 in the input.
You need to implement the training/classifier function. It is up to you what type of function you want to implement. Do not use external libraries. An easy possibility is to proceed as follows:
Record (in the closure of the classifier) the given training data and the training labels.
When given as input a digit to classify, return the label of the nearest neighbour in the training data, where distance is measured by the Euclidean distance between the corresponding feature vectors (feature vector corresponding to given digit, and closest neighbor feature-vector-wise in training data
The code that needs to be finished (??? part):
object classifier {
val sizex = 160 // width of digits (do not edit)
val sizey = 160 // height of digits (do not edit)
val m = sizex*sizey // length of digit array (=25600) (do not edit)
def train(digits: Array[Array[Double]], labels: Array[Int]) : (Array[Double]) => Int = {
val features = digits.map(feature.get(_))
def classifyDigit(digit: Array[Double]) : Int = {
// return the very best guess as to whether 'digit' is a 0 or 1
???
}
classifyDigit // return the classifier
}
}
/*
* We give you some help in your task. Namely, the object 'feature' below
* defines a function 'get' to compute __feature vectors__ from the
* grayscaled handwritten digits. A __feature vector__ is a
* low(er)-dimensional representation of the original input.
* The function consists of three simple steps
* (indeed, rather than do anything fancy, our aim is to be able
* to visualize the feature vectors as smaller images in the data browser):
*
* 1) Down-sample by a factor of four to reduce dimension
* (the original digits are 160-by-160 pixels, the feature
* vectors have 40-by-40 entries).
*
* 2) Do biased rounding to articulate the digit
* (all pixels that are >= 80% of the average grayscale become 0.0,
* all remaining pixels become 1.0).
*
* 3) Center the 40-by-40 array at the __barycenter__ of the 1.0-pixels,
* and complement the 1.0s to 0.0s and vice versa.
* (The centering at the barycenter eases e.g. nearest neighbor search.)
*
*/
object feature {
val h = 4
val sizex = classifier.sizex/h
val sizey = classifier.sizey/h
val m = sizex*sizey
def get(digit: Array[Double]) = {
val down = new Array[Double](m)
var i = 0
while(i < m) {
down(i) = digit((((i/(sizex))*sizex*h*h)+h*(i%(sizex))))
i = i+1
}
val average = down.sum/m
val step = down.map(v => if(v >= 0.8*average) { 0.0 } else { 1.0 })
val total = step.sum
val barx = math.round((0 until m).map(j => step(j)*(j%sizex)).sum/total).toInt
val bary = math.round((0 until m).map(j => step(j)*(j/sizex)).sum/total).toInt
val vec = new Array[Double](m)
i = 0
while(i < sizey) {
var j = 0
val bi = (i+bary+sizey+sizey/2)%sizey
while(j < sizex) {
val bj = (j+barx+sizex+sizex/2)%sizex
vec(i*sizex+j) = 1.0-step(bi*sizex+bj)
j = j+1
}
i = i+1
}
vec
}
}
The code that needs to be completed is:def classifyDigit(digit: Array[Double]): Int = {
val distances = features.map(x => euclideanDistance(x, digit))
val best = distances.zip(labels).minBy(_._1)
best._2
}
The completed code can be found below:
object classifier {
val sizex = 160 // width of digits (do not edit)
val sizey = 160 // height of digits (do not edit)
val m = sizex * sizey // length of digit array (=25600) (do not edit)
def train(digits: Array[Array[Double]], labels: Array[Int]): (Array[Double]) => Int = {
val features = digits.map(feature.get(_))
def classifyDigit(digit: Array[Double]): Int = {
val distances = features.map(x => euclideanDistance(x, digit))
val best = distances.zip(labels).minBy(_._1)
best._2
}
classifyDigit // return the classifier
}
def euclideanDistance(p: Array[Double], q: Array[Double]): Double = {
math.sqrt((p zip q).map { case (x, y) => math.pow(y - x, 2) }.sum)
}
}
The function takes an array of 25600 double digits and labels each of these digits as 0 or 1. The training function returns a classifier function and takes as input a digit array and returns its very best guess whether it "sees" a 0 or 1 in the input. The classifier is built by computing the Euclidean distance between the feature vectors of the input and each of the training data. The classifier returns the label of the closest neighbor feature-vector-wise in the training data.
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Given the following inclusions and type definitions: #include struct dnode int; typedef struct dnode int *dnode; struct dnode int { dnode previous; int *data; dnode next; }} struct cirque int; typedef struct cirque int *cirque; struct cirque int { dnode cursor; }; And the following function definition: int f1 (cirque q, int t) { dnode c; int n = 0; c = q->cursor; while (c != NULL) { if (((c->data)) { n++; == t) } c = c->next; } return n; }| a. What does the function f1 () do? KIT107 Programming b. What possible situation(s) could cause the code to fail? Click or tap here to enter text.
Function f1 () has been defined with two parameters named as q and t, which are both of the int type. The function returns an integer number. The function basically searches for the nodes in the circular queue q which has its data value as t, and returns the count of such nodes that were found. Below is the code snippet for function f1().
int f1 (cirque q, int t) { dnode c; int n = 0; c = q->cursor; while (c != NULL) { if (((c->data)) { n++; == t) } c = c->next; } return n; } The function takes an input circular queue ‘q’, which is the pointer to the struct cirque. The integer input parameter ‘t’ defines the data value that needs to be searched in the data field of the nodes of the circular queue. The variable ‘n’ is initialized to zero. It acts as a counter variable for the nodes having data value as ‘t’. A pointer to the node named ‘c’ is initialized with the cursor of the circular queue ‘q’. The while loop will traverse through all the nodes present in the circular queue ‘q’ till the end of the queue. The if block checks the data value of the current node with the integer parameter ‘t’. If the value matches, the count of nodes with the matching data value is incremented.
Finally, the pointer ‘c’ is updated to the next node of the circular queue. The function returns the total count of nodes with the data value as ‘t’ that were present in the circular queue.The code may fail in the following situations: - If the pointer to the circular queue ‘q’ is a null pointer. In that case, an attempt to access the cursor field would result in an undefined behavior. - If the input integer value ‘t’ is not defined within the range of the integer data type. If that happens, the comparison would fail to provide the correct results, and the count of nodes with data value as ‘t’ would not be returned correctly. - If the input circular queue ‘q’ is empty, the cursor field of the queue would be a null pointer. In that case, the while loop would not iterate, and the returned count value would be zero.
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a. If the new fast multiplier unit performs a multiply operation in 5ns compared to the old multiplier which needs 12ns for the same multiply operation and multiply operations take 40% of the original program execution time what is the overall speedup of the program execution (ignoring the penalty to any other instructions or parameters)?b. If we continue improving the multiplier unit what would be the maximum speed up, we can get for this program? c. Assume in part a, the improvement in the multiplier unit costs 10% clock frequency reduction for the processor. Calculate the new overall speed up for that program. Would you recommend such a redesign? Why or why not? Please elaborate. (d. The enhanced processor in part a (with the same clock frequency as the original processor) while executing a new program spends 25% of the execution time for multiply operations and 30% of the execution time for integer division. The original integer divider is known to be 100% faster than the new integer divider. How much faster does this new program run on the enhanced processor compared to the original processor, assuming all other operations run the same on both processors? e. If the program in part d is run on the original processor, what percentage of the execution time is spent in execution of operations other than multiply and integer division?
The speedup of the program execution is 1.44x or 44%. To calculate the overall speedup of the program execution we need to find out the fraction of time the multiply operations took in the old unit and then find out the time it takes in the new unit.
Then we can find the fraction of time this operation now takes. In other words, we will compute the performance improvement for the multiply operation first and then calculate the overall performance improvement. To calculate the fraction of time the multiply operations took in the original execution:
The maximum speedup we can get for this program is based on the fraction of the time the multiply operation takes. Therefore, if we assume that the multiply operation takes 0% of the time, then we can achieve the maximum possible speedup. The overall speedup of the program execution would be:(1/0.6) × (12/5) = 2.5x or 150%.Therefore, the maximum speedup we can get for this program is 2.5x or 150%.c.
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