i. The router will create 5 fragments.
The size of the TCP segment is 4800 bytes, including a header of 20 bytes. The maximum transmission unit (MTU) of the outgoing line is 1120 bytes, including a header of 20 bytes. To determine the number of fragments, we divide the size of the TCP segment (4800 bytes) by the MTU (1120 bytes), resulting in 4 fragments. However, since there is a remaining size of 400 bytes, it will require an additional fragment, making a total of 5 fragments.
ii. For each fragment:
Fragment 1: Length = 1120 bytes, ID = 333, Offset = 0, Flag = More Fragments (MF)
Fragment 2: Length = 1120 bytes, ID = 333, Offset = 112, Flag = MF
Fragment 3: Length = 1120 bytes, ID = 333, Offset = 224, Flag = MF
Fragment 4: Length = 1120 bytes, ID = 333, Offset = 336, Flag = MF
Fragment 5: Length = 1040 bytes, ID = 333, Offset = 448, Flag = Last Fragment (LF)
Each fragment has a length of 1120 bytes, except for the last fragment, which has a remaining size of 1040 bytes. The ID of all fragments is 333, indicating they belong to the same original segment. The offset value specifies the position of each fragment within the original segment. The first fragment has an offset of 0, and each subsequent fragment increments the offset by the MTU size. The "More Fragments" (MF) flag is set for all fragments except the last one, which has the "Last Fragment" (LF) flag.
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Explain the reason for moving from stop and wai (ARQ protocol to the Gezbackay ARO peotsced (2 points) 2. Define briefly the following: ( 6 points) - Data link control - Framing and the reason for its need - Controlled access protocols 3. Define piggybacking and is usefuiness (2 points):
Gezbackay ARO offers higher efficiency and selective repeat ARQ, while Stop-and-Wait has limitations in efficiency and error handling.
The move from Stop-and-Wait (ARQ) protocol to the Gezbackay ARO protocol can be attributed to the following reasons:Improved Efficiency: The Stop-and-Wait protocol is a simple and reliable method for error detection and correction. However, it suffers from low efficiency as it requires the sender to wait for an acknowledgment before sending the next data frame.
This leads to significant delays in the transmission process. The Gezbackay ARO protocol, on the other hand, employs an Automatic Repeat Request (ARQ) mechanism that allows for continuous data transmission without waiting for acknowledgments. This results in higher throughput and improved efficiency.
Error Handling: Stop-and-Wait ARQ protocol handles errors by retransmitting the entire frame when an error is detected. This approach is inefficient for large frames and high-error rate channels.
The Gezbackay ARO protocol utilizes selective repeat ARQ, where only the damaged or lost frames are retransmitted, reducing the overhead and improving the overall error handling capability.
Definitions:Data Link Control (DLC): Data Link Control refers to the protocols and mechanisms used to control the flow of data between two network nodes connected by a physical link.
It ensures reliable and error-free transmission of data over the link, taking care of issues such as framing, error detection and correction, flow control, and access control.
Framing: Framing is the process of dividing a stream of data bits into manageable units called frames. Frames consist of a header, data payload, and sometimes a trailer.
The header contains control information, such as source and destination addresses, sequence numbers, and error detection codes. Framing is necessary to delineate the boundaries of each frame so that the receiver can correctly interpret the data.
Controlled Access Protocols: Controlled Access Protocols are used in computer networks to manage and regulate access to a shared communication medium. These protocols ensure fair and efficient sharing of the medium among multiple network nodes.
They can be categorized into two types: contention-based protocols (e.g., CSMA/CD) and reservation-based protocols (e.g., token passing). Controlled access protocols help avoid data collisions and optimize the utilization of the communication channel.
Piggybacking is a technique used in networking where additional information is included within a data frame or packet that is already being transmitted. This additional information may be unrelated to the original data but is included to make more efficient use of the communication medium.The usefulness of piggybacking can be understood in the context of acknowledgement messages in a network.
Instead of sending a separate acknowledgment frame for each received data frame, the receiver can piggyback the acknowledgment onto the next outgoing data frame. This approach reduces the overhead of transmission and improves efficiency by utilizing the available bandwidth more effectively.
Piggybacking is particularly beneficial in scenarios where network resources are limited or when the transmission medium has constraints on the number of messages that can be sent.
By combining data and acknowledgments in a single frame, piggybacking optimizes the utilization of the network and reduces the overall latency in the communication process.
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Translate the following C-code into RISC-V assembly.
Please leave comments next to the instructions.
Consider the following C source code.
int D[100];
int main(int argc, char *argv[])
{
return foo(10);
}
int foo(int a)
{
for (int i=0; i < a; i++) {
bar(i, i);
}
}
void bar(int x, int y)
{
D[x] = y;
}
The RISC-V assembly code for the provided C-code is as follows:
``` .text
.align 2
.globl main
main:
addi sp,sp,-16
sw ra,12(sp)
sw s0,8(sp)
addi s0,sp,16
li a0,10
jal foo
lw ra,12(sp)
lw s0,8(sp)
addi sp,sp,16
jr ra
.align 2
.globl foo
foo:
addi sp,sp,-16
sw ra,12(sp)
sw s0,8(sp)
addi s0,sp,16
li t0,0
mv t1,a0
loop:
beq t0,t1,exit
jal bar
addi t0,t0,1
j loop
exit:
lw ra,12(sp)
lw s0,8(sp)
addi sp,sp,16
jr ra
.align 2
.globl bar
bar:
addi sp,sp,-16
sw ra,12(sp)
sw s0,8(sp)
addi s0,sp,16
sw a1,0(a0) # Stores the value of 'y' in the D[x] array
lw ra,12(sp)
lw s0,8(sp)
addi sp,sp,16
jr ra
```
Comments next to instructions are as follows:-
First, it declares the memory for D.```int D[100];```
- It starts the main function.```int main(int argc, char *argv[])```
- The function 'foo' is called with argument 10.```return foo(10);```
- The 'foo' function starts here.```int foo(int a)```
- Initializes register t0 to 0 and moves the value of register a0 to t1.```li t0,0
mv t1,a0```
- Loops through values of i using register t0 and t1.
If the value of t0 is equal to t1, the loop ends.```loop:
beq t0,t1,exit
jal bar
addi t0,t0,1
j loop
exit:```
- The 'bar' function starts here.```void bar(int x, int y)```- The value of register a1 is stored in the array D[x].```sw a1,0(a0)```
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The ________ statement executes one block of statements if a test condition is true, and another block if the condition is false.
trailing else
if/else
if/else if
if
switch
The if/else statement executes one block of statements if a test condition is true, and another block if the condition is false.
This statement is used to create conditional programming, meaning a program that can act differently based on the input or other factors.To put it simply, if/else is used when we want to execute a statement or a block of statements when a certain condition is true, and another statement/block when that condition is false.An if/else statement is an important construct in programming languages, as it allows the program to make choices based on certain conditions. Using conditional statements such as if/else helps create code that is more flexible, interactive, and easier to read and understand.
:Thus, the if/else statement is used to execute a block of statements if a test condition is true, and another block if the condition is false.
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Let’s say a program has 1010 bytes and will be loaded into page frames of 256 bytes each, (assuming the job begins loading at the first page (Page 0) in memory), and the instruction to be used is at Byte 577, answer the following question:
Compute the page number and exact displacement for the byte addresses where the data is stored.
Please give a detailed explanation as I am confused.
The program has 1010 bytes and will be loaded into page frames of 256 bytes each. The instruction to be used is at Byte 577. Find the page number and the exact displacement for the byte addresses where the data is stored.
Given that the page frames are 256 bytes, it is necessary to calculate the number of page frames that are needed to store the program. This can be computed using the following formula:Number of Page Frames = Size of Program / Size of Page Frame= 1010/256= 3.945 ≈ 4 page framesFor the instruction that will be used, the byte address is 577.
Therefore, the page number is given by the formula:Page Number = Byte Address / Size of Page Frame= 577/256 = 2.253 ≈ 2 page framesTo determine the exact displacement, the byte address must be taken modulo the size of the page frame as follows: Displacement = Byte Address modulo Size of Page Frame= 577 modulo 256= 65Therefore, the data is stored in Page 2, and the exact displacement is 65. Hence,Page number is 2, and the exact displacement is 65.
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Write a class Conversion containing the following methods: (i) Constructor: which builds the frame shown on the right side. The frame consists of a text field for inputting a WON amount, a label with 10 spaces for an equivalent WON amount in USD, and a button to start the calculation. Declare any necessary attributes in the class and add appropriate action listeners for future use. Copy the class, including import statement(s), as the answers to this part. (ii) actionPerformed() : which performs the calculation and puts the result on the label when the button is pressed. You can assume one WON is equivalent to 0.00077 USD. You can assume a valid real number is entered in the textfield. Copy the method as the answers to this part. (iii) main( ) : which creates a Conversion object and sets it visible for testing. Copy the method as the answers to this part.
The question involves creating a class called "Conversion" with methods for constructing a frame, performing currency conversion, and setting up the main method for testing.
How can we create a class called "Conversion" with a constructor for building a frame, an actionPerformed() method for performing currency conversion, and a main() method for testing?We create a class called "Conversion" that contains a constructor for building a frame. The frame consists of a text field for inputting a WON amount, a label for displaying the equivalent amount in USD, and a button for initiating the calculation. We declare necessary attributes in the class and add appropriate action listeners for future use.
The actionPerformed() method is implemented to perform the calculation when the button is pressed. Assuming one WON is equivalent to 0.00077 USD, the method retrieves the input from the text field, performs the conversion calculation, and displays the result on the label. It assumes a valid real number is entered in the text field.
The main() method is set up to create an object of the Conversion class and make it visible for testing purposes. This allows us to verify the functionality of the frame and the currency conversion process.
By creating the Conversion class with the necessary methods and implementing the currency conversion logic, we can construct a frame for currency conversion and perform calculations based on the provided conversion rate.
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. imagine a program that processes 5000 input values in 10 seconds. about how long would it take the program to process 20,000 input values if the algorithm implemented by the program is: a. a o(n) algorithm. b. a o(n^2) algorithm
It would take approximately 40 seconds for an O(n) algorithm and approximately 400 seconds for an O(n^2) algorithm to process 20,000 input values.
How long would it take to process 20,000 input values with O(n) and O(n^2) algorithms?An O(n) algorithm has a linear time complexity, which means that the time taken to process input values increases linearly with the number of values. In this case, the program processes 5000 values in 10 seconds.
Therefore, the time taken to process 20,000 values can be estimated by calculating the ratio of the input values:
Time taken for 20,000 values = (20,000 / 5000) * 10 seconds = 40 seconds.
An O(n^2) algorithm has a quadratic time complexity, which means that the time taken to process input values increases quadratically with the number of values. In this case, the program processes 5000 values in 10 seconds.
Therefore, the time taken to process 20,000 values can be estimated by calculating the ratio of the input values:
Time taken for 20,000 values = (20,000 / 5000)^2 * 10 seconds = 400 seconds.
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Given a binary tree using the BinaryTree class in chapter 7.5 of your online textbook, write a function CheckBST(btree) that checks if it is a binary search tree, where btree is an instance of the BinaryTree class. Question 2 In the lecture, we introduced the implementation of binary heap as a min heap. For this question, implement a binary heap as a Maxheap class that contains at least three member functions: - insert (k) adds a new item to the heap. - EindMax() returns the item with the maximum key value, leaving item in the heap.
Here is the Python code implementation of the CheckBST function and MaxHeap class: Function to Check if a Binary Tree is a Binary Search Tree
def CheckBST(btree):
def CheckBSTHelper(node, min_val, max_val):
if node is None:
return True
if node.key < min_val or node.key > max_val:
return False
return (CheckBSTHelper(node.left, min_val, node.key - 1) and
CheckBSTHelper(node.right, node.key + 1, max_val))
return CheckBSTHelper(btree.root, float("-inf"), float("inf"))```
Class for MaxHeap```python
class MaxHeap:
def __init__(self):
self.heap_list = [0]
self.size = 0
def percolate_up(self, i):
while i // 2 > 0:
if self.heap_list[i] > self.heap_list[i // 2]:
self.heap_list[i], self.heap_list[i // 2] = \
self.heap_list[i // 2], self.heap_list[i]
i //= 2
def insert(self, k):
self.heap_list.append(k)
self.size += 1
self.percolate_up(self.size)
def percolate_down(self, i):
while (i * 2) <= self.size:
mc = self.max_child(i)
if self.heap_list[i] < self.heap_list[mc]:
self.heap_list[i], self.heap_list[mc] = \
self.heap_list[mc], self.heap_list[i]
i = mc
def max_child(self, i):
if (i * 2) + 1 > self.size:
return i * 2
else:
if self.heap_list[i * 2] > self.heap_list[(i * 2) + 1]:
return i * 2
else:
return (i * 2) + 1
def find_max(self):
if self.size > 0:
return self.heap_list[1]
else:
return None
def del_max(self):
if self.size == 0:
return None
max_val = self.heap_list[1]
self.heap_list[1] = self.heap_list[self.size]
self.size -= 1
self.heap_list.pop()
self.percolate_down(1)
return max_val
A binary tree can be checked if it is a binary search tree or not by traversing through all the nodes of the tree and checking whether it satisfies the properties of binary search tree or not.
Binary Heap can be implemented as MaxHeap and the methods that it can include are insert(k), find_max(), and del_max() which add new item to heap, return the maximum key value item and delete the maximum item respectively.
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A local variable (automatic variable) exists for the life of the program and can be accessed anywhere in the program file. True False When using the C math library, which of the following flags are required to link the math library? A. -o B. −1 m C. −c D. -math-library A break statement can be used to exit a while loop. True False When generating random numbers, which of the following functions "seeds" the random number generator? A. srand() B. time() C. rand () D. sqrt() Given the following declaration and initialization: double result =104/5; What is the value stored in the variable result? A. 19 B. 20 C. 20.0 D. 20.8 int result =104.0/5.0; What is the value stored in the variable result? A. 20.0 B. 21 C. 20.8 D. 20
The flag required to link the math library in C is "-1 m". The srand function is used to seed the random number generator. The value stored in the variable "result" is 20.
The statement "A local variable (automatic variable) exists for the life of the program and can be accessed anywhere in the program file" is false. Local variables have a limited scope and exist only within the block of code where they are declared.
When using the C math library, the flag "-lm" is required to link the math library functions. This flag instructs the compiler to include the math library during the linking process, enabling the use of mathematical functions in the program.
A break statement can be used to exit a while loop. When encountered, the break statement immediately terminates the loop and control is transferred to the next statement outside the loop.
The srand() function is used to seed the random number generator. It initializes the random number generator with a seed value, allowing for the generation of different sequences of random numbers. The srand() function is typically used with the time() function to seed the generator based on the current time.
In the given declaration and initialization, double result = 104/5, the division operation is performed using integer operands. Since both operands are integers, the division results in the quotient of 20. Therefore, the value stored in the variable result is 20.
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The purpose of this assignment is to demonstrate knowledge of the basic syntax of a SQL query. Specifically, you will be asked to demonstrate: - use of the SELECT clause to specify which fields you want to query. - use of the FROM clause to specify which tables you want to query, and - use of the WHERE clause to specify which conditions the query will use to query rows in a table. These are the basic commands that will make up your foundational knowledge of SQL. There are other clauses besides SELECT, FROM, and WHERE, but by building up your knowledge of these basic clauses, you will have constructed a foundation upon which to base your knowledge of SQL. Tasks 1. Design the following queries, using the lyrics.sql schema: 1. List the Title, UPC and Genre of all CD titles. (Titles table) 2. List all of the information of CD(s) produced by the artist whose ArtistlD is 2. (Titles table) 3. List the First Name, Last Name, HomePhone and Email address of all members. (Members table) 4. List the Member ID of all male members. (Members table) 5. List the Member ID and Country of all members in Canada. (Members table)
The basic syntax of a SQL query involves using various clauses in order to specify what information you want to retrieve from a database. There are three fundamental clauses: SELECT, FROM, and WHERE. The SELECT clause specifies which fields you want to query.
The FROM clause specifies which tables you want to query. The WHERE clause specifies which conditions the query will use to query rows in a table. In order to demonstrate your knowledge of these basic clauses, you have been given five tasks to complete using the lyrics. sql schema. Task 1: List the Title, UPC and Genre of all CD titles. (Titles table)The query for this task is as follows: SELECT Title, UPC, Genre FROM Titles; This query specifies the SELECT and FROM clauses. We are selecting the Title, UPC, and Genre fields from the Titles table. Task 2: List all of the information of CD(s) produced by the artist whose Artist lD is 2. (Titles table)The query for this task is as follows:SELECT *FROM TitlesWHERE ArtistID = 2;This query specifies the SELECT, FROM, and WHERE clauses. '
We are selecting all fields from the Titles table where the ArtistID is equal to 2.Task 3: List the First Name, Last Name, HomePhone and Email address of all members. (Members table)The query for this task is as follows:SELECT FirstName, LastName, HomePhone, EmailFROM Members;This query specifies the SELECT and FROM clauses. We are selecting the FirstName, LastName, HomePhone, and Email fields from the Members table.Task 4: List the Member ID of all male members. (Members table)The query for this task is as follows:SELECT MemberIDFROM MembersWHERE Gender = 'M';This query specifies the SELECT, FROM, and WHERE clauses. We are selecting the MemberID field from the Members table where the Gender is equal to 'M'.
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To get started take a look at the following sites:
Business Insider: What is the Internet of Things (IoT)? (Links to an external site.) (Links to an external site.)
The Guardian: Can we secure the internet of things in time to prevent another cyber-attack? (Links to an external site.)
INFOSEC Institute: How Hackers Violate Privacy and Security of the Smart Home (Links to an external site.)
Where you go next is up to you. Do you have any experience developing IoT devices? Have you had a security issue with an IoT device that you use? If so begin your post by sharing your experience with us. Alternatively, research the area online and tell us about some interesting things that you find there.
In your original post, answer the following:
Draft a response as specified below. Remember to use your textbook and other academic resources to justify your responses [see the note below regarding references].
Option 1: Do you have personal experience with IoT development or security issues? If so begin by sharing your personal experiences with us. Research the particular issue that you experienced and tell us more about it.
Option 2: Research C# IoT development or C# IoT security issues. You want to keep the discussion current, so at least one of your sources must have been published in the last year. [Choose a search engine that allows you to set a date range for your search.]
Yes, I have personal experience with IoT development and security issues.
During my time as an IoT developer, I encountered a security issue related to a smart home device. The device in question was a smart door lock that was connected to the internet, allowing users to control and monitor access to their homes remotely.
However, I discovered a vulnerability in the device's firmware that could potentially be exploited by hackers to gain unauthorized access to the lock.
Upon further investigation, I found that the firmware lacked proper encryption mechanisms and had weak default credentials, making it susceptible to brute-force attacks.
This meant that an attacker could potentially gain access to the lock by guessing the default username and password or by intercepting the communication between the device and the accompanying mobile app.
To address this issue, I immediately reported it to the manufacturer, who acknowledged the vulnerability and released a firmware update to fix the security flaw. They also provided guidelines for users to change the default credentials and recommended enabling two-factor authentication for added security.
This experience highlighted the importance of robust security measures in IoT devices. It is crucial for manufacturers to prioritize security from the design phase, including strong encryption, secure default settings, and regular firmware updates to address vulnerabilities.
Additionally, users must be proactive in implementing security best practices such as changing default credentials and keeping their devices up to date.
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Which of the following terms are often synonymous with or made possible with CIDR? (Select two.)
NAT
OSPF
Classful
VLSM
Classless
The two terms that are often synonymous with or made possible with CIDR include: Classless and VLSM. CIDR (Classless Inter-Domain Routing) is an IP addressing scheme that modifies the traditional IP address structure.
The notation used in CIDR is a suffix attached to the IP address that indicates the number of bits in the address that can be used to identify hosts. It uses Variable Length Subnet Masks (VLSM) that allow for efficient allocation of IP addresses and routing. CIDR replaced the Classful network addressing scheme.
NAT (Network Address Translation) is a technique used in IP addressing that translates IP addresses from one network to another. OSPF (Open Shortest Path First) is a routing protocol that is used for dynamic routing in IP networks. It helps routers to calculate the shortest path to a destination network. Classful is an outdated IP addressing scheme that was used in the early stages of the internet.
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Which of the following are categories of encryption algorithms? a. symmetric b. cross-positional c. asymmetric d. transposition e. transactional f. substitution
The following are categories of encryption algorithms:a. symmetricb. asymmetricc. transpositiond. substitution. Encryption algorithms are mainly categorized into two categories as follows:
Symmetric Encryption Asymmetric Encryption Symmetric Encryption: This algorithm involves only one secret key to encrypt and decrypt the information.Both sender and receiver must know the same key. This type of encryption is also known as private-key encryption.Asymmetric Encryption:
This algorithm uses two keys, one for encryption, and another for decryption. This type of encryption is also known as public-key encryption.Transposition and substitution are techniques used in cryptography, which means they are not categories of encryption algorithms.Therefore, a. symmetricb. asymmetricc. transpositiond. substitution
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Question 4 (Modular Arithmetic \& Brute-Force attacks −10 marks) a) If 10 8
computers work in parallel on a brute force attack and each computer can test 10 11
keys per second, how long will it take on average to find a DES 56-bit encryption key? Show working. b) If 10 20
computers work in parallel on a brute force attack and each computer can test 10 15
keys per second, how long will it take on average to find a AES 128-bit encryption key? Show working. c) If 10 11
computers work in parallel on a brute force attack and each computer can test 10 30
keys per second, how long will it take on average to find a TwoFish 256-bit encryption key? Show working. d) If 1,000,000 computers work in parallel on a brute force attack and each computer can test 100,000 keys per second, how long will it take on average to find a 64-bit encryption key? Show working. e) If 10,000,000 computers work in parallel on a brute force attack and each computer can test 1,000,000 keys per second, how long will it take on average to find a 128-bit encryption key? Show working.
a. it will take 7.20576 x 10^(-14) sec. b. it will take 3.40282 x 10^(-18) sec. c. it will take 1.15792 x 10^(-45) sec. d. it will take 36,893.5 sec. e. it will take 3.40282 x 10^13 sec.128-bit encryption has 2^128 combinations.
a) To find the time it takes to find a DES 56-bit encryption key using parallel brute force attacks with computers testing 10^11 keys per second:
Calculate the number of possible keys:
A 56-bit encryption key has 2^56 possible combinations.
Calculate the time it takes to find the key:
Divide the total number of possible keys by the rate at which keys are tested.
Time = Number of keys / Rate of testing
Time = (2^56) / (10^11)
Using scientific notation for convenience:
Time = (2^56) / (10^11)
= (2^56) / (1 x 10^11)
= (2^56) / (1 x 10^11) * (10^(-11) / 10^(-11))
= (2^56 x 10^(-11)) / (1)
= (2^56) x (10^(-11))
≈ 7.20576 x 10^(-14) seconds
Therefore, it will take approximately 7.20576 x 10^(-14) seconds to find a DES 56-bit encryption key using parallel brute force attacks with computers testing 10^11 keys per second.
b) To find the time it takes to find an AES 128-bit encryption key using parallel brute force attacks with 10^20 computers testing 10^15 keys per second:
Calculate the number of possible keys:
A 128-bit encryption key has 2^128 possible combinations.
Calculate the time it takes to find the key:
Divide the total number of possible keys by the rate at which keys are tested.
Time = Number of keys / Rate of testing
Time = (2^128) / (10^20 x 10^15)
Time = (2^128) / (10^35)
Using scientific notation for convenience:
Time = (2^128) / (10^35)
≈ 3.40282 x 10^(-18) seconds
Therefore, it will take approximately 3.40282 x 10^(-18) seconds to find an AES 128-bit encryption key using parallel brute force attacks with 10^20 computers testing 10^15 keys per second.
c) To find the time it takes to find a TwoFish 256-bit encryption key using parallel brute force attacks with 10^11 computers testing 10^30 keys per second:
Calculate the number of possible keys:
A 256-bit encryption key has 2^256 possible combinations.
Calculate the time it takes to find the key:
Divide the total number of possible keys by the rate at which keys are tested.
Time = Number of keys / Rate of testing
Time = (2^256) / (10^11 x 10^30)
Time = (2^256) / (10^41)
Using scientific notation for convenience:
Time = (2^256) / (10^41)
≈ 1.15792 x 10^(-45) seconds
Therefore, it will take approximately 1.15792 x 10^(-45) seconds to find a TwoFish 256-bit encryption key using parallel brute force attacks with 10^11 computers testing 10^30 keys per second.
d) To find the time it takes to find a 64-bit encryption key using parallel brute force attacks with 1,000,000 computers testing 100,000 keys per second:
Calculate the number of possible keys:
A 64-bit encryption key has 2^64 possible combinations.
Calculate the time it takes to find the key:
Divide the total number of possible keys by the rate at which keys are tested.
Time = Number of keys / Rate of testing
Time = (2^64) / (1,000,000 x 100,000)
Using scientific notation for convenience:
Time = (2^64) / (1,000,000 x 100,000)
≈ 3.68935 x 10^4 seconds
Therefore, it will take approximately 36,893.5 seconds (or about 10.25 hours) to find a 64-bit encryption key using parallel brute force attacks with 1,000,000 computers testing 100,000 keys per second.
e) To find the time it takes to find a 128-bit encryption key using parallel brute force attacks with 10,000,000 computers testing 1,000,000 keys per second:
Calculate the number of possible keys:
A 128-bit encryption key has 2^128 possible combinations.
Calculate the time it takes to find the key:
Divide the total number of possible keys by the rate at which keys are tested.
Time = Number of keys / Rate of testing
Time = (2^128) / (10,000,000 x 1,000,000)
Using scientific notation for convenience:
Time = (2^128) / (10,000,000 x 1,000,000)
≈ 3.40282 x 10^13 seconds
Therefore, it will take approximately 3.40282 x 10^13 seconds (or about 1,077,178,865.97 years) to find a 128-bit encryption key using parallel brute force attacks with 10,000,000 computers testing 1,000,000 keys per second.
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Is there any point in keeping old routers?.
There can be several reasons why it might be useful to keep old routers:
1. Backup or Redundancy:
2. Experimental or Learning Purposes:
Keeping old routers can serve as a backup or redundancy option. In case your current router malfunctions or stops working, having an old router can be a lifesaver. You can quickly switch to the old router and continue using the internet until you can replace or repair the new one. This ensures uninterrupted connectivity and avoids any inconvenience caused by a sudden internet outage. Additionally, if you have a large house or office space, using old routers as Wi-Fi extenders can help improve the Wi-Fi coverage in areas where the main router's signal is weak.
Another reason to keep old routers is for experimental or learning purposes. If you are interested in networking or want to gain hands-on experience with routers, having access to old routers can be beneficial. You can experiment with different settings, configurations, and firmware updates without risking the functionality of your primary router. In summary, keeping old routers can be useful for backup or redundancy purposes, providing uninterrupted internet connectivity in case of router failure. Additionally, it can serve as a valuable tool for experimentation and learning about networking concepts.
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Answer the following questions: 1. What do you understand by a computer system? Do you think that the information and data are same? Why or why not? Explain with examples.- 2. How is object-oriented programming (OOP) different from procedural programming? Is C\# an OOP language? Name a few OOP languages. 3. What does an interpreter do? How is it different from a compiler? What is an executable statement? How is logic error different from syntax error? Will a program run if the program has logic errors? . How is source code different from object code? Do the programmers write object code? -
1. A computer system is a combination of software and hardware that works together to achieve a specific goal. The software includes the applications, operating system, and programming languages, whereas the hardware includes the processor, memory, input/output devices, and storage.
Information and data are not the same. Data refers to raw facts and figures, such as numbers or characters, that have no meaning on their own. For example, the letters "A," "B," and "C" are data, but they do not provide any information until they are organized into words and sentences.
Information, on the other hand, is a meaningful interpretation of data that has been processed and analyzed. For example, the sentence "John likes apples" provides information because it gives meaning to the data "John" and "apples."
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Select any application used in your organization or any application that you are working on to develop, maintain or test (If you are unable to find one such application, you may choose any application in the public domain). Find out the architecture of the application. You may be able to find the architecture in the supporting documents for the application. Study and understand the architecture and then attempt the following.
Create an architecture diagram for the application. You can use a readymade diagram if you are able to find one.
List the architecturally significant requirements that shaped the architecture of the application.
Discuss how the architecture addresses any three quality attributes defined by the architecturally significant requirements identified in step 2.
Identify architectural patterns used in the architecture
Analyse the architecture for any three failure scenarios and suggest improvements in the architecture to prevent these failures. If you are unable to find any failure scenarios, suggest three generic improvements that could make the architecture better in your opinion.
Analyzing the application's architecture, identifying significant requirements, addressing quality attributes, identifying architectural patterns used, and suggesting improvements for failure scenarios or overall enhancement.
Application: E-commerce Website
1. Architecture Diagram:
Create a visual representation of the application's architecture, including its major components, modules, and their interactions. This diagram can be created using various tools like Lucidchart, draw.io, or any other diagramming tool.2. Architecturally Significant Requirements:
Scalability: The ability of the application to handle increasing user traffic and growing data. Availability: Ensuring the application remains accessible and operational even during hardware or software failures. Security: Protecting user data, preventing unauthorized access, and ensuring secure transactions.3. Addressing Quality Attributes:
Scalability: The architecture may include horizontal scaling by employing load balancers and distributed caching to handle increased traffic. It can use a distributed database to manage data growth. Availability: The architecture can incorporate redundancy and fault-tolerant components such as load balancers, multiple web servers, and database replication. It may also use auto-scaling mechanisms to handle traffic spikes. Security: The architecture may include firewalls, secure communication protocols (e.g., HTTPS), authentication mechanisms, encryption for sensitive data, and regular security audits.4. Architectural Patterns:
Model-View-Controller (MVC): Separating the application into three interconnected components to manage user interactions, data, and presentation. Microservices: Breaking down the application into smaller, loosely coupled services that can be developed, deployed, and scaled independently. Layered Architecture: Organizing the application into layers (e.g., presentation layer, business logic layer, data access layer) to separate concerns and improve maintainability.5. Failure Scenarios and Improvements:
Scenario 1: Database FailureImprovement: Implementing database replication or a distributed database solution to ensure data availability and fault tolerance.
Scenario 2: Network OutageImprovement: Introducing a content delivery network (CDN) to cache static assets and reduce dependency on a single network infrastructure.
Scenario 3: Security BreachImprovement: Implementing a robust authentication and authorization mechanism, performing regular security audits, and staying up-to-date with security patches and protocols.
The above example is generalized, and the actual architecture, requirements, and improvements will vary depending on the specific application being analyzed.
For your own application, you will need to study and understand its architecture, document the architecture diagram, identify architecturally significant requirements, discuss how the architecture addresses quality attributes, identify architectural patterns, and analyze failure scenarios to suggest improvements.
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Instructions Mobile Phone Bill Write a FLOWGORITHM program that will calculate a mobile phone bill based on the customer plan and the data used. The program should perform the following: Prompt for input of for customer name Prompt for input of customer’s mobile plan Prompt for input of number of gigabytes of data used If the plan choice is invalid or gigabytes used is less than zero (0) display a message and terminate program Calculate the monthly bill based on plan & data usage Display customer name, plan and monthly mobile charges Mobile data plans are: Plan A 19.99/month, w/6 gigs of data, additional data $8.50/gig Plan B 29.99/month, w/10 gigs of data, additional data $3.50/gig Plan C 39.99/month, unlimited data Remember the following: declare necessary variables and constants initialize the constants use comment box for your name, date and purpose of program use other comments where appropriate DO NOT "hard code numbers" in calculations, use constants round all real variable calculations to 2 decimals use clear prompts for your input clearly label each output number or name
a FLOWGORITHM program that calculates a mobile phone bill based on customer plan and data used. The program performs the following steps:Prompt for input of customer name.Prompt for input of customer’s mobile plan.Prompt for input of number of gigabytes of data used.
If the plan choice is invalid or gigabytes used is less than zero (0), display a message and terminate program.Calculate the monthly bill based on plan & data usage.Display customer name, plan, and monthly mobile charges.Mobile data plans are:Plan A 19.99/month, w/6 gigs of data, additional data $8.50/gig.Plan B 29.99/month, w/10 gigs of data, additional data $3.50/gig.Plan C 39.99/month, unlimited data.
Declare necessary variables and constants.Initialize the constants.Use a comment box for your name, date, and purpose of the program.Use other comments where appropriate.DO NOT "hard code numbers" in calculations, use constants.
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Consider the following grammar for a language, where a and b are called terminals (there are no rules to replace them) while < Start >, and are called nonterminals (there are rules to replace them). The goal is to generate a sequence of as' and bs'. < Start >→ab
→b∣b
→a∣a
Which of the following sequences are in the language generated by this grammar? You need to explain your answers. (a) bbaab (b) bbaaaaa (c) bbbaab
Therefore, the correct option is (a) and (c). (a) bbaab (c) bbbaab
The following grammar for a language, where a and b are called terminals (there are no rules to replace them) while < Start >, and are called nonterminals (there are rules to replace them).
The goal is to generate a sequence of as' and bs'.→ab→b∣b→a∣aWhich of the following sequences are in the language generated by this grammar?
1. bbaab 2. bbaaaaa 3. bbbaab(a) bbaab: It is in the language generated by the grammar. It can be generated by the following sequence.→b→b→a→a→b(b) bbaaaaa:
It is not in the language generated by the grammar. It starts with bb which can only be generated as→b, but then it follows with aaaaa which cannot be generated by the given grammar.
(c) bbbaab: It is in the language generated by the grammar. It can be generated by the following sequence.→b→b→b→a→a→b
Therefore, the correct option is (a) and (c).
Answer:Option A and C are the correct
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IN JAVA
A common problem for compilers and text editors is to determine if the parentheses (or other brackets) in a string are balanced and properly nested. For example, the string "((())())()" contains properly nested pairs of parentheses, but the string ")()(" does not; and the string "())" does not contain properly matching parentheses.
(a) Give an algorithm that returns true if a string contains properly nested and balanced parentheses, and false if otherwise. Hint: At no time while scanning a legal string from left to right will you have encountered more right parentheses than left parentheses.
(b) Give an algorithm that returns the position in the string of the first offending parenthesis if the string is not properly nested and balanced. That is, if an excess right parenthesis is found, return its position; if there are too many left parentheses, return the position of the first excess left parenthesis. Return −1 if the string is properly balanced and nested.
(a) Algorithm to check if a string contains properly nested and balanced parentheses or not:
Step 1: Define a stack to store opening brackets
.Step 2: Scan each character of the string from left to right.
Step 3: If a left bracket ( (, [, or { ) is encountered, push it onto the stack.
Step 4: If a right bracket ( ), ], or } ) is encountered, pop the top element from the stack. If the stack is empty, the string contains an excess right bracket and is not properly nested and balanced. If the popped bracket doesn't match with the current bracket, the string contains mismatched brackets and is not properly nested and balanced.
Step 5: If the stack is empty at the end of the string, the string contains properly nested and balanced parentheses; otherwise, it contains excess left brackets and is not properly nested and balanced.
(b) Algorithm to find the position of the first offending parenthesis if the string is not properly nested and balanced:Step 1: Define a stack to store opening brackets and their positions.
Step 2: Scan each character of the string from left to right.
Step 3: If a left bracket ( (, [, or { ) is encountered, push its position onto the stack.
Step 4: If a right bracket ( ), ], or } ) is encountered, pop the top element from the stack. If the stack is empty, return the position of the excess right bracket. If the popped bracket doesn't match with the current bracket, return the position of the mismatched bracket. Repeat this step until the stack is empty.
Step 5: If the stack is empty at the end of the string, return -1 (the string contains properly nested and balanced parentheses); otherwise, return the position of the first excess left bracket.
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11 This program ask the user for an average grade. 11. It prints "You Pass" if the student's average is 60 or higher and 11 prints "You Fail" otherwise. 11 Modify the program to allow the following categories: 11 Invalid data (numbers above 100 and below 0), 'A' category (90âe'100), l1 'B' categoryc(80ấ" 89), 'C' category (70âe"79), 'You Fail' category (0áe'"69). 1/ EXAMPLE 1: 1/. Input your average: −5 1/ Invalid Data 1/ EXAMPLE 2: 1) Input your average: θ // You fail 11 EXAMPLE 3: 1) Input your average: 69 1) You fail 1/ EXAMPLE 4: 11) Input your average: 70 lf you got a C 1) EXAMPLE 5: II Inout vour average: 79 1/ EXAMPLE 6: 1/ Input your average: 80 1f You got a B 1/ EXAMPLE 7: 1/ Input your average: 89 11 You got a 8 1/ EXAMPLE 8: 1/ Input your average: 90 11 You got a A 11 EXAMPLE 9: 11 Input your average: 100 1. You got a A II EXAMPLE 10: 1/. Input your average: 101 If Invalid Data 1/ EXAMPLE 10: 1) Input your average: 101 /1 Invalid Data I/ PLACE YOUR NAME HERE using namespace std; int main() \{ float average; If variable to store the grade average If Ask user to enter the average cout «< "Input your average:" ≫ average; if (average ⟩=60 ) else cout « "You Pass" << end1; cout «< "You Fail" k< endl; return θ;
The modified program for the given requirements is as follows:#includeusing namespace std;int main() { float average; cout << "Input your average: "; cin >> average; if (average < 0 || average > 100) { cout << "Invalid Data" << endl; } else if (average >= 90) { cout << "You got an A" << endl; } else if (average >= 80) { cout << "You got a B" << endl; } else if (average >= 70) { cout << "You got a C" << endl; } else { cout << "You Fail" << endl; } return 0;
}
The program asks the user to enter the average grade of a student and based on the value, the program outputs the grade category or Invalid Data if the entered grade is not in the range [0, 100].Explanation:First, the program takes input from the user of the average grade in the form of a float variable named average.
The if-else-if conditions follow after the input statement to categorize the average grade of the student. Here, average < 0 || average > 100 condition checks whether the entered average is in the range [0, 100] or not.If the entered average is outside of this range, the program outputs Invalid Data.
If the average lies within the range, it checks for the average in different grade categories by using else-if statements:else if (average >= 90) { cout << "You got an A" << endl; }else if (average >= 80) { cout << "You got a B" << endl; }else if (average >= 70) { cout << "You got a C" << endl; }else { cout << "You Fail" << endl; }.
The first else-if condition checks whether the entered average is greater than or equal to 90. If the condition is true, the program outputs "You got an A."If the condition is false, the next else-if condition is checked. It checks whether the average is greater than or equal to 80.
If the condition is true, the program outputs "You got a B."This process continues with the else-if conditions until the last else condition. If none of the above conditions are true, the else part of the last else-if condition executes. The program outputs "You Fail" in this case.
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How to display time & date using code below in visual studio 2022?
1. Displaying the current date and time using a Page_Load event
The current date and time is:
ID="lblServerTime"
runat="server" />
To display the time and date using the given code below in Visual Studio 2022, follow these steps:
1. First, open the project or web page in Visual Studio 2022.
2. Go to the .aspx.cs file, and add the following code to the Page_Load event:
'protected void Page_Load(object sender, EventArgs e){lblServerTime.Text = DateTime.Now.ToString();}`
3. Now, run the project in Visual Studio, and the current date and time will be displayed on the web page. The output will look like the following: The output of the given code can be seen in the following image: Thus, this is how you can display the time and date using the given code below in Visual Studio 2022.
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only thing slowing him down is that his customers always want to know what the monthly payment is going to be. Therefore, Louie has asked you to write a program for him to do just that. This formula will come in handy: P∗(1+r)∗−1r(1+r)nP= principal (amount borrowed) r= interest rate /12n= term (number of payments) A couple of notes to consider: - The user will have three inputs: (1) the principal, (2) the interest rate, (3) the length of the loan. - The interest rate entered will be an annual percentage (i.e., 6.5). You will need to convert it to a monthly rate (divide it by 12) and convert it to a decimal number (divide it by 100). - The length of the loan will be input in number of years. You will need to convert this to months (multiply by 12). - The monthly payment should be output using 2 decimals with a dollar sign immediately before the payment amount. - The user is prompted after each loan calculation to see if he would like to go again. - When the user is finished doing calculations, the program should output the total number of loans processed. - The screen interaction will look something like this: - Implement the solution to your problem as a Python program. - Make sure your program uses meaningful variable names. - Be sure your program has at least 4 lines of comments at the top.
The program will prompt the user to enter the principal, interest rate, and length of the loan. It will then convert the interest rate to a monthly rate and the length of the loan to months.
Finally, it will calculate the monthly payment using the formula provided and output the result with 2 decimals and a dollar sign before the payment amount. The program will then prompt the user to go again or exit and output the total number of loans processed.```# Program to calculate monthly payment for a loanimport math# Initialize loop variablekeep_going = True# Initialize loan countloan_count = 0# Loop until user wants to exitwhile keep_going:# Get user input for principal, interest rate, and length of loanprincipal = float(input("Enter the principal amount: "))interest_rate = float(input("Enter the annual interest rate: "))loan_length = int(input("Enter the length of the loan in years: "))# Convert annual interest rate to monthly rate and loan length to monthsmonthly_rate = interest_rate / 1200months = loan_length * 12#
Calculate monthly payment payment = (principal * (monthly_rate * ((1 + monthly_rate) ** months))) / (((1 + monthly_rate) ** months) - 1)# Output monthly paymentprint("Monthly payment: ${:.2f}".format(payment))# Increment loan countloan_count += 1# Prompt user to go again or exitresponse = input("Do you want to go again? (Y/N): ")if response.upper() != "Y":keep_going = False# Output total number of loans processed print("Total number of loans processed:", loan_count)```
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00000110b in ASCII stands for End of Transmission. Select one: True False
00000110b in ASCII stands for End of Transmission.The correct option is True.
In ASCII, 00000110b represents the End of Transmission (EOT) character. This character is used to indicate the end of a transmission or message and is commonly used in telecommunications and computer networking.ASCII is a character encoding scheme that represents text in computers and other devices. It assigns unique binary codes to each character in the standard ASCII character set, which includes letters, numbers, and symbols.ASCII codes are widely used in computing, telecommunications, and other fields where data needs to be transmitted and processed electronically.
Therefore, the given statement is true.
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5.14 lab: convert to reverse binary write a program that takes in a positive integer as input, and outputs a string of 1's and 0's representing the integer in reverse binary. for an integer x, the algorithm is: as long as x is greater than 0 output x modulo 2 (remainder is either 0 or 1) assign x with x divided by 2 note: the above algorithm outputs the 0's and 1's in reverse order. ex: if the input is: 6 the output is: 011 6 in binary is 110; the algorithm outputs the bits in reverse. 428030.2135884.qx3zqy7 lab activity 5.14.1: lab: convert to reverse binary
To convert a positive integer to reverse binary, write a program that uses the given algorithm: output the remainder (0 or 1) of the integer modulo 2 while it is greater than 0, and then divide the integer by 2. The resulting string will represent the integer in reverse binary.
How can we implement the algorithm to convert a positive integer to reverse binary?To convert a positive integer to reverse binary, we can use the following algorithm:
1. Take the positive integer as input.
2. While the integer is greater than 0, perform the following steps:
a. Output the remainder (0 or 1) of the integer modulo 2.
b. Divide the integer by 2 and assign the quotient back to the integer.
3. The above steps will output the bits in reverse order, representing the integer in reverse binary.
For example, if the input is 6, the algorithm will output "011." In binary, 6 is represented as 110, but the algorithm outputs the bits in reverse.
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Arrow networks have another name; what is it? What is the reason for this name?
Arrow networks are also known as Directed Acyclic Graphs (DAGs) due to their structure.
Arrow networks, commonly referred to as Directed Acyclic Graphs (DAGs), are a type of data structure used in computer science and mathematics. The name "Directed Acyclic Graphs" provides a concise description of the key characteristics of these networks.
Firstly, the term "Directed" indicates that the connections between nodes in the network have a specific direction. In other words, the relationship between the nodes is one-way, where each node can have multiple outgoing connections, but no incoming connections. This directed nature of the connections allows for the representation of cause-and-effect relationships or dependencies between different nodes in the network.
Secondly, the term "Acyclic" signifies that the network does not contain any cycles or loops. In other words, it is impossible to start at any node in the network and traverse the connections in a way that eventually brings you back to the starting node. This acyclic property is essential in various applications, such as task scheduling or dependency management, where cycles can lead to infinite loops or undefined behaviors.
The name "Directed Acyclic Graphs" accurately captures the fundamental characteristics of these networks and provides a clear understanding of their structure and behavior. By using this name, researchers, programmers, and mathematicians can easily communicate and refer to these specific types of networks in their respective fields.
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by using visual studio code
Create 5 variables relating to a hobby you have, make sure at least 1 is a string, 2 are numbers, and 1 is a boolean
Write a comment above your variables saying what hobby they represent
Here are the five variables relating to a hobby that consists of 1 string, 2 numbers, and 1 boolean.let hobbyName = 'Cycling';let durationInMinutes = 60;let distanceInKm
The variable is Completed is a boolean variable that can hold either true or false value.We have also written a comment above the variables that describe the hobby of cycling. Comments are used to provide additional information about the code and can help other developers to understand the code better.
In this program, we declare five variables named hobbyName, durationInMinutes, distanceInKm, isCompleted, and caloriesBurnt. These variables represent the hobby of cycling. We have assigned a string value to the variable hobbyName and numerical values to durationI n Minutes, distanceInKm, and caloriesBurnt.
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list 5 testing tools that provides the least decision code
coverage.
The five testing tools that provide the least decision code coverage are: JUnit, Selenium WebDriver, Mockito, Cucumber, and Postman.
JUnit is a popular unit testing framework for Java applications. While it is effective for testing individual units of code, such as methods or classes, it does not provide extensive coverage of decision-based code paths. It focuses more on verifying the correctness of individual units rather than exploring different decision outcomes.
Selenium WebDriver is a widely used tool for automating web browsers. It is primarily used for functional testing of web applications by interacting with elements on web pages. However, its purpose is not to specifically target decision-based code coverage. Instead, it focuses on simulating user interactions and validating expected behaviors.
Mockito is a mocking framework for Java that allows developers to create test doubles of dependencies. It is commonly used in unit testing to isolate and test individual components. While Mockito helps in isolating dependencies, it does not explicitly address decision coverage. Its main goal is to facilitate unit testing by providing a means to stub or mock dependencies.
Cucumber is a behavior-driven development (BDD) tool that enables collaboration between developers, testers, and business stakeholders. It uses plain-text specifications written in a language called Gherkin to define test scenarios. Although Cucumber encourages specifying scenarios with different inputs and expected outcomes, it does not inherently focus on decision-based code coverage.
Postman is an API testing tool that allows developers to send HTTP requests and analyze responses. It is primarily used for functional testing and integration testing of APIs. While it helps in verifying the correctness of API endpoints, it does not specifically target decision coverage within the codebase.
In summary, while these testing tools are valuable in their respective areas, they do not provide extensive support for achieving high decision code coverage. It's important to use additional testing techniques and tools to ensure comprehensive coverage of decision paths in software applications.
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Create a program with the following functions for FizzBuzz4:
Create a program which implements instructions from FizzBuzz1 up to FizzBuzz3
FizzBuzz4 should be able to do the following:
Create a separate arrays for Fizz values, Buzz values, FizzBuzz values and those with no comments.
Create a function which accepts a parameter which is a string showSpecificArr(String check).
your function should return the values from the selected array.
Note that you need to modify the insertIntoArray function.
Sample Simulation
showSpecificArr("Fizz");
// Should return and display all numbers from that particular array
Submit the java file and make sure to add comments for areas of your code which will require explanation.
the code below is the fizzbuzz 1-3
import java.util.Scanner;
public class fizz1 {
public static int listoffizz[] = new int [100];
public static void main (String[]args){
Scanner x= new Scanner(System.in);
System.out.print("Enter a start: ");
int num1 = x.nextInt();
System.out.print("Enter a end: ");
int num2 = x.nextInt();
fizziterate(num1, num2);
showArray();
}
public static String fizzCheck(int num){
String result = (num%3==0 && num%5==0)?
"FIZZBUZZ": (num%3==0)?
"FIZZ": (num%5==0)?"BUZZ": "";
return result;
}
public static void fizziterate(int start, int end){
for (int i=start; i<=end; i++)
{
System.out.println(i+" - " + fizzCheck(i));
insertIntoFizz(fizzCheck(i), i);
}
}
public static void insertIntoFizz(String comment,int num){
if(comment.equals("FIZZ")|| comment.equals("FIZZBUZZ") || comment.equals("BUZZ"))
{
for(int i=0;i
{
if(listoffizz[i]==0)
{
listoffizz[i]=num;
i=listoffizz.length;
break;
}
}
}
}
public static void showArray()
{
System.out.println("THESE ARE THE FIZZBUZZ VALUES");
for(int i=0;i
{
if(listoffizz[i]==0)
{
break;
}
System.out.print(listoffizz[i]+",");
}
}
}
Fizz Buzz is a popular interview question that is often used to assess a programmer's coding skills. Fizz Buzz is a problem that requires you to output the numbers 1 to 100, but with a twist:
If the number is divisible by three, print Fizz instead of the number; if the number is divisible by five, print Buzz instead of the number; and if the number is divisible by both three and five, print Fizz Buzz instead of the number. The task is to modify the provided Fizz Buzz1-3 code by creating Fizz Buzz4, which can accomplish all of the tasks from FizzBuzz1-3, but with the addition of new features.
The FizzBuzz4 code should be modified to include the following functions:
Create an array for Fizz values, an array for Buzz values, an array for Fizz Buzz values, and an array for values with no comments. Create a function that takes a parameter that is a string, show Specific Array (String check), which should return the values from the selected array. Modify the insert In to Array function.
In Java programming language, the modified FizzBuzz1-3 code is given as follows:
import java.util.
Scanner; public class FizzBuzz1_4 {public static int list off i buzz
= new in to [100];
public static int listofbuzz[] = new int [100];public static int listoffizzbuzz[] = new int [100];
public static int listofnoncomments[] = new int [100];
public static void main (String[]args){Scanner x= new Scanner(System.in);
System. out. print("Enter a start: ");
in t num1 = x. next In to;
System. out. print ("Enter a end: ");
int num2 = x.next Int;
fizziterate(num1, num2);
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c define a function outputvals() that takes two integer parameters and outputs all integers starting with the first and ending with the second parameter in reverse order, each followed by a newline. the function does not return any value. ex: if the input is 3 7, then the output is: 7 6 5 4 3
Below is a C function called outputvals() that takes two integer parameters and outputs the integers in reverseorder.
#include <stdio.h>
void outputvals(int start, int end) {
int i;
for (i = end; i >= start; i--) {
printf("%d\n", i);
}
}
int main() {
int first = 3;
int last =7;
outputvals(first, last);
return0;
}
How does it work?Whenyou run this program, it will output the integers from end to start in reverse order, each followed by a newline. In this example, the output will be -
7
6
5
4
3
You can modify the values of first and last in the main() function to specify differentstarting and ending parameters for the outputvals() function.
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In a data transfer instruction the effective address will be given by: The immediate field in the instruction The base register multiplied by the immediate field The sum of the base register and the immediate field None of the above QUESTION 4 MIPS uses the following addressing modes: Register Indirect Mode, Base or displacement addressing, Immediate addressing, PC-relative addressing, and Pseudodirect addressing. Register Mode, Based plus scaled index addressing, Immediate addressing, PC-relative addressing, and Pseudodirect addressing. Register Mode, Base or displacement addressing, Immediate addressing, PC-relative addressing, and Pseudodirect addressing. None of the above
In a data transfer instruction the effective address will be given by the sum of the base register and the immediate field. The effective address is defined as the address produced by adding a register's content or an immediate value to a memory address during the execution of an instruction.
The addressing modes used by MIPS are:
Register Mode, Base or displacement addressing, Immediate addressing, PC-relative addressing, and Pseudodirect addressing.
Explanation:
In computer programming, the addressing mode specifies how the effective address of an operand is calculated from its logical address. The effective address is defined as the address produced by adding a register's content or an immediate value to a memory address during the execution of an instruction.There are four main types of addressing modes that are used by a processor.
These include:
Immediate addressing mode: In this mode, the operand is specified within the instruction, rather than being loaded from memory. This is also called a constant mode because it is used to provide constant data to the program.
Register addressing mode: In this mode, the operand is specified in a register. It is usually faster than other addressing modes, as it avoids accessing memory.
Base or displacement addressing mode: In this mode, the effective address is calculated by adding a constant value (called the displacement) to the value in a register. It is also called an offset mode because it is used to access data that is located at a specific offset from a base address.
PC-relative addressing mode: In this mode, the effective address is calculated by adding a constant value to the program counter. This mode is useful for accessing data that is located close to the current instruction.
Pseudodirect addressing mode: This mode is a variation of the base or displacement mode, where the offset is stored in a register.
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