The hand execution drawing for the pass-by-value scenario depicts the relationship between the variables name, s, and temp in the swap_pass_by_value function. It illustrates how the values are declared, processed, and changed throughout the execution of the program.
In the pass-by-value scenario, the values of the variables name and s are passed as parameters to the swap_pass_by_value function. Initially, the values of name and s are printed to the terminal. Then, a simple swap mechanism is applied by assigning the value of s to temp, s to name, and temp to name.
To depict this process in the hand execution drawing, we can draw a diagram with three boxes representing the variables name, s, and temp. Inside each box, we can write the initial values of name and s. After the swap mechanism is applied, we update the values in the respective boxes to represent the new assignments.
The drawing should visually demonstrate the values being declared, processed, and changed from the beginning of the program execution to the end of the swap process, highlighting the steps involved in the swapping mechanism.
By examining the hand execution drawing, one can easily understand the flow of values and their changes during the execution of the swap_pass_by_value function.
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Part II Run show-NetFirewallRule and attach screenshots of three rules. Describe what each rule means in 1-2 sentences.
Part III Recreate any of the scripting examples in the class and attach screenshots.
The command run show-Net Firewall Rule provides the details of the specified firewall rules for the computer. In this regard, it will describe what each rule means in 1-2 sentences.
Allow Inbound ICMP (Echo Request) – This rule allows incoming ping requests from other computers. Rule 2: Allow Inbound Remote Desktop – This rule allows the RDP (Remote Desktop Protocol) traffic to connect to the computer. Rule 3: Allow Inbound SSH traffic – This rule allows Secure Shell (SSH) traffic to connect to the computer.
To recreate the scripting examples, the following steps are required :Create a script file named Firewall.ps1.Copy and paste the following script in the Firewall.ps1 file.# Allow incoming ping requests from other computers New-Net Firewall Rule -DisplayName "Allow Inbound ICMP (Echo Request)" -Protocol ICMPv4 .
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what is the area called that is located on the right side of both your landing page and course homepage?
The area that is located on the right side of both your landing page and course homepage is called "The right rail".
What is the right rail?
The right rail is a section of a website or webpage that's usually found on the right-hand side of the page. It's also known as a sidebar. The right rail is a great location to place key bits of information.
This region is usually reserved for secondary content and frequently features widgets, callouts, or other eye-catching designs.
What is included in the right rail?
The right rail on the landing page and course homepage may contain details and information related to courses, announcements, and resources.
On the right rail of the landing page, some details can include the following:
Course Catalog, Learning Goals, Testimonials, etc.
On the right rail of the course homepage, some details can include the following:
Announcements, Upcoming Coursework, Course Resources, etc.
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Algebraically specify a bounded FIFO Queue (Queue with a specified lower and upper limit for performing the enqueue and dequeue operations) having a maximum size of MSize and that supports the following methods: New(), Append(), Size(), Remove(), First() and isempty() with their conventional meanings: The Abstract Data Type (ADT) that needs to be defined here is queue and which may further uses the following data types: Boolean, Element, Integer data types. In addition, include the exceptions if required.
Design the axioms for the following sequence of operations: first(new()), remove(new()), size(new()), first(append(q, e)), remove(append(q,e)), size(append (q,e)), isempty(q)
The enqueue operation inserts an element at the end of the list, and the dequeue operation removes an element from the head of the list.
Given, Algebraically specified a bounded FIFO Queue (Queue with a specified lower and upper limit for performing the enqueue and dequeue operations) having a maximum size of MSize and that supports the following methods:
New(), Append(), Size(), Remove(), First() and isempty() with their conventional meanings.
The Abstract Data Type (ADT) that needs to be defined here is queue and which may further use the following data types: Boolean, Element, Integer data types. The queue will be defined as follows: queue(Q) (Q is of type Queue)
A Queue is a collection of elements with two principal operations enqueue and dequeue. The elements are added at one end and removed from the other end. Queues are also called as FIFO (First In First Out) lists. Queues maintain two pointers, one at the head (front) of the list and the other at the tail (end) of the list.
The enqueue operation inserts an element at the end of the list, and the dequeue operation removes an element from the head of the list. Axioms for the following sequence of operations:
first(new()), remove(new()), size(new()), first(append(q, e)), remove(append(q,e)), size(append (q,e)), isempty(q) are as follows:
The axioms are as follows:
First(new()) = FALSEremove(new()) = Queueunderflowsize(new()) = 0
First(append(q, e)) = e
if not QueueOverflow
else "Queue Overflow"
remove(append(q,e)) = q
if not QueueUnderflow
else "Queue underflow"
size(append(q,e)) = size(q)+1
if not QueueOverflow
else size(q) isempty(q) = TRUE
if Size(q)=0
else FALSE
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list 2 reporting mechanisms for tracking organisational cyber security maturity?
There are several reporting mechanisms for tracking organizational cyber security maturity, some of them are discussed below:1. Security audits: Security audits are designed to assess the effectiveness of existing security controls and identify areas where improvements can be made.
They involve the review of security policies, procedures, and systems to ensure that they are aligned with the organization's security objectives.2. Security assessments: Security assessments are a more in-depth analysis of the security posture of an organization. They are designed to identify vulnerabilities and weaknesses that could be exploited by cyber attackers. They include a review of security policies, procedures, systems, and controls to determine the overall security posture of an organization. Organizations should be vigilant and always track their cyber security maturity, as it is an important aspect in the protection of sensitive and confidential information. As cyber threats are becoming more sophisticated, so must the strategies for protection.
Security audits and security assessments are two important reporting mechanisms that an organization can use to track its cyber security maturity.A security audit is an independent and systematic evaluation of the policies, procedures, and controls that are in place to protect an organization's information assets. It is designed to identify areas of improvement and assess the effectiveness of existing security controls. Security audits help organizations to identify gaps and vulnerabilities in their security posture and develop strategies to improve them.Security assessments are designed to assess an organization's overall security posture.
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In your program, write assembly code that does the following: Create a DWORD variable called "sum". Its initial value doesn't matter. Set the registers eax, ebx, ecx, and edx to whatever values you like Perform the following arithmetic: (eax + ebx) - (ecx + edx) Move the result of the above arithmetic into the sum variable.
The assembly code sets the registers eax, ebx, ecx, and edx to specific values, performs the arithmetic operation (eax + ebx) - (ecx + edx), and stores the result in the "sum" variable.
Here's an example assembly code that performs the described operations:
section .data
sum dd 0 ; Define a DWORD variable called "sum"
section .text
global _start
_start:
mov eax, 5 ; Set the value of eax to 5
mov ebx, 3 ; Set the value of ebx to 3
mov ecx, 2 ; Set the value of ecx to 2
mov edx, 1 ; Set the value of edx to 1
add eax, ebx ; Add eax and ebx
sub eax, ecx ; Subtract ecx from the result
add eax, edx ; Add edx to the result
mov [sum], eax ; Move the result into the sum variable
; Rest of the program...
In this code, the values of eax, ebx, ecx, and edx are set to 5, 3, 2, and 1 respectively. The arithmetic operation (eax + ebx) - (ecx + edx) is performed and the result is stored in the "sum" variable.
The provided assembly code is just an example, and you can modify it as per your requirements or integrate it into a larger program.
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2)
a) Open a web browser and connect to 3 different websites through 3 different tabs
b) Open you cmd as administrator and type/enter "netstat -b"
c) Share a screenshot of your browser's connections
d) Do you have different ports for each website? Why or why not ?
a) If you want to visit different websites, you can open multiple tabs in your web browser and type in the website addresses you want to go to. This can be done by typing the website addresses yourself or by using saved links or search engine findings.
b) To see the active network connections and the programs running, you can open the Command Prompt (cmd) as an administrator and type "netstat -b".
d. Each website uses a special number on the server to work, but on your computer, the number used to connect to websites can be different each time. This is called a temporary or changing port.
What is the ports aboutPeople use randomized port numbers on the client side to make sure that different applications on your computer can work together without causing any problems or errors. It lets your computer connect to many websites or services at the same time.
The server waits and listens for requests on specific ports like port 80 for HTTP or port 443 for HTTPS. The server that hosts websites may have different ports, but the ports used by your computer's browser to connect to the server are usually random.
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Write a C++ program using a NumberSet ADT declared as follows:
typedef int Number;
const int maxSize=10;
struct NumberSet
{
Number items[maxSize];
int count;
};
void numberSetAdd(NumberSet& set, Number num); // adds 'num' to 'set' (if possible); otherwise prints error message and makes no change to 'set'
void numberSetRemove(NumberSet& set, Number num); // removes all instances (if any exist) of 'num' from 'set'; otherwise prints error message and makes no change to 'set'
int numberSetCountLessThan(NumberSet& set, Number num); // returns the number of elements smaller than 'num' in 'set'
int numberSetCountMoreThan(NumberSet& set, Number num); // returns the number of elements larger than 'num' in 'set'
NumberSet newNumberSet(); // returns a new, empty NumberSet
First, write a C++ source file which implements the 5 interface functions above, namely:
numberSetAdd
numberSetRemove
numberSetCountLessThan
numberSetCountMoreThan
newNumberSet
Then, write a C++ program in another C++ source file which uses your NumberSet ADT to:
. try to put the integers 2, 3, 4, 5, 6, 7, 8, 9, 9, 9, 9 into an instance of NumberSet;
. ask the user to input an integer value, and store their input in a variable (e.g. userValue). You may assume the user inputs a positive integer value;
. print out to the screen the number of elements of your NumberSet instance that are smaller than userValue and the number of elements that are larger than userValue;
. try to remove the values 2, 9 and 11 from your NumberSet instance; and
. then print out again to the screen the number of elements of your NumberSet instance that are smaller than userValue and the number of elements that are larger than userValue.
Here's an implementation of the NumberSet ADT and a program that uses it according to the given requirements:
**NumberSet.cpp:**
```cpp
#include <iostream>
typedef int Number;
const int maxSize = 10;
struct NumberSet {
Number items[maxSize];
int count;
};
void numberSetAdd(NumberSet& set, Number num) {
if (set.count < maxSize) {
set.items[set.count] = num;
set.count++;
} else {
std::cout << "NumberSet is full. Cannot add " << num << std::endl;
}
}
void numberSetRemove(NumberSet& set, Number num) {
int removedCount = 0;
for (int i = 0; i < set.count; i++) {
if (set.items[i] == num) {
removedCount++;
} else {
set.items[i - removedCount] = set.items[i];
}
}
set.count -= removedCount;
if (removedCount == 0) {
std::cout << "Number " << num << " not found in NumberSet." << std::endl;
}
}
int numberSetCountLessThan(NumberSet& set, Number num) {
int count = 0;
for (int i = 0; i < set.count; i++) {
if (set.items[i] < num) {
count++;
}
}
return count;
}
int numberSetCountMoreThan(NumberSet& set, Number num) {
int count = 0;
for (int i = 0; i < set.count; i++) {
if (set.items[i] > num) {
count++;
}
}
return count;
}
NumberSet newNumberSet() {
NumberSet set;
set.count = 0;
return set;
}
```
**main.cpp:**
```cpp
#include <iostream>
#include "NumberSet.cpp"
int main() {
NumberSet numberSet = newNumberSet();
// Adding numbers 2, 3, 4, 5, 6, 7, 8, 9, 9, 9, 9 to the NumberSet
numberSetAdd(numberSet, 2);
numberSetAdd(numberSet, 3);
numberSetAdd(numberSet, 4);
numberSetAdd(numberSet, 5);
numberSetAdd(numberSet, 6);
numberSetAdd(numberSet, 7);
numberSetAdd(numberSet, 8);
numberSetAdd(numberSet, 9);
numberSetAdd(numberSet, 9);
numberSetAdd(numberSet, 9);
numberSetAdd(numberSet, 9);
// Asking user for input
int userValue;
std::cout << "Enter a positive integer: ";
std::cin >> userValue;
// Printing the count of elements smaller and larger than userValue
int countLessThan = numberSetCountLessThan(numberSet, userValue);
int countMoreThan = numberSetCountMoreThan(numberSet, userValue);
std::cout << "Count of elements smaller than " << userValue << ": " << countLessThan << std::endl;
std::cout << "Count of elements larger than " << userValue << ": " << countMoreThan << std::endl;
// Removing values 2, 9, and 11 from the NumberSet
numberSetRemove(numberSet, 2);
numberSetRemove(numberSet, 9);
numberSetRemove(numberSet, 11);
// Printing the count again after removal
countLessThan = numberSetCountLessThan(numberSet, userValue);
countMoreThan = numberSetCountMoreThan(numberSet, userValue);
std::cout << "Count of elements smaller than " << userValue << " after removal: " << countLessThan << std::endl;
std::cout << "Count of elements larger than " << userValue << " after removal: " << countMoreThan << std::endl;
return 0;
}
```
You can compile and run the program to test it. When prompted, enter a positive integer value as requested, and the program will provide the counts of elements smaller and larger than that value in the NumberSet. Then it will remove the specified values from the NumberSet and print the counts again.
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Here is the testing code:
```python
moon = Project(name="Moon")
keep_moon = moon
year_0 = OneTime(year=0, cash=-1e9)
year_1 = OneTime(year=1, cash=-2e9)
launch = Growing(year_start=2, year_end=4, cash_start=1e8, g=0.2)
perpetuity = GrowingPerpetuity(year_start=5, cash_start=2e8, g=0.025)
# Checking that we have abstract methods and inheritance
import inspect
print(inspect.isabstract(CashFlow) and all(x in CashFlow.__abstractmethods__ for x in ["__contains__", "__str__", "discount"])) # expect True (1)
print(isinstance(launch, CashFlow)) # expect True (2)
print(isinstance(perpetuity, CashFlow)) # expect True (3)
print(3 in launch) # expect True (4)
print(2 not in year_1) # expect True (5)
# cash-flows are always discounted to Year 0
print(abs(year_1.discount(r=0.05) - (-1904761904.7619047)) < 1) # expect True (6)
print(abs(launch.discount(r=0.05) - 312832616.03961307) < 1) # expect True (7)
print(abs(perpetuity.discount(r=0.05) - 6581619798.335054) < 1) # expect True (8)
flows = [year_0, year_1, launch, perpetuity]
for f in flows:
moon += f
print(moon.schedule_count == 4) # expect True (9)
print(abs(moon.npv(r=0.05) - 3989690509.612763) < 1) # expect True (10)
print(abs(moon.npv(r=0.1) - (-725656262.0950305)) < 1) # expect True (11)
print(abs(moon.irr(scan_from=0.05, scan_to=0.1, epsilon=1e-3) - 0.082) < 0.001) # expect True (12)
print(str(moon) == "Project Moon - IRR [8% - 9%]") # expect True (13)
print(len(moon[4]) == 1) # expect True (14)
print(moon[4][0] is launch) # expect True (15)
extra_dev = OneTime(year=3, cash=-5e8)
moon += extra_dev
print(str(moon) == "Project Moon - IRR [7% - 8%]") # expect True (16)
print(moon is keep_moon) # expect True(17)
print(len(moon[3]) == 2 and all(x in moon[3] for x in [launch, extra_dev])) # expect True (18)
mars = Project("Mars")
mars_y0 = OneTime(year=0, cash=-4e9)
mars_y1 = OneTime(year=1, cash=-4e9)
mars_y2 = OneTime(year=2, cash=-4e9)
mars_ops = GrowingPerpetuity(year_start=3, cash_start=1e8, g=0.03)
mars_cashflows = [mars_y0, mars_y1, mars_y2, mars_ops]
for f in mars_cashflows:
mars += f
space_portfolio = moon + mars
print(str(space_portfolio) == "Project Moon + Mars - IRR [4% - 5%]") # expect True (19)
print(len(space_portfolio[3]) == 3 and all(x in space_portfolio[3] for x in [extra_dev, launch, mars_ops])) # expect True (20)
```
Modelisation
You will get less hints for this exercise.
* It has to be impossible to create objects of class `CashFlow`
* `CashFlow` makes it mandatory for subclasses to implement `discount` method
* `CashFlow` makes it mandatory for subclasses to implement the operators:
* `str(cf)`: method `__str__`: the returned string is up to you, it is not tested
* `3 in cf`: method `__contains__(self, key)`: here `3` is the key. It returns `True` when the cash-flow happens in Year 3. In the code, `3 in launch` returns `True`. `7 in perpetuity` returns `True`.
* Classes `OneTime`, `Growing`, `GrowingPerpetuity` can create objects
* Their constructor's arguments make sense in Finance
* The way to compute their NPV at year 0 (method `discount`) is different for each
* `Project` has a schedule: a list of objects `CashFlow` which is not in the constructor parameters
* the attribute `schedule_count` is the number of objects in this list
* The following operations are supported by `__add__(self, other)`:
* `project + cashflow`: returns the object `project`, adds the object `cashflow` to the list of cashflows of `project`
* `project1 + project2` : creates a **NEW** project by merging the 2 projects
* its name is "name1 + name2", using the names of both projects
* its schedule is the concatenation of both schedules
* the `schedule_count` is the sum of both counters
* `Project` has the method `npv`:
* Gets the NPV of the whole project at Year 0
* `Project` also has the method `irr`
* Computes the Internal Return Rate
* See in the code for the arguments
* Try different values for the discount rate, between a starting value and an ending value, separated by epsilon
* Return the first value after the sign of the NPV has changed
* `str(project)` displays the project name, along with an approximation of the IRR printed as %
* use `irr` with a epsilon of 1%
* if you find 0.1, then display `[9% - 10%]`
* `project[3]` is supported by `__getitem__(self, index)`, returns the list of cash-flows in the project's schedule for which there is a cash-flow in year 3
The given code demonstrates a finance-related modeling system implemented using object-oriented programming in Python. It includes classes such as `CashFlow`, `OneTime`, `Growing`, `GrowingPerpetuity`, and `Project`. The code performs various calculations and tests to validate the functionality of the classes. The `Project` class represents a financial project and maintains a schedule of cash flows.
The code defines an abstract class called `CashFlow` that cannot be directly instantiated. It enforces the implementation of essential methods and operators for its subclasses, such as `discount`, `__str__`, and `__contains__`.
The subclasses `OneTime`, `Growing`, and `GrowingPerpetuity` represent different types of cash flows, each with its own way of calculating the net present value (NPV) at Year 0.
The `Project` class acts as a container for cash flows and allows operations such as adding cash flows and merging projects. It also provides methods for calculating the NPV and internal rate of return (IRR) of the entire project. The IRR calculation is done by iteratively scanning different discount rates until the sign of the NPV changes.
The provided code includes tests to verify the correctness of the implementation. It checks abstract methods and inheritance, evaluates the correctness of discount calculations, performs project operations, and validates the behavior of the `Project` class. The expected results are provided as comments in the code.
Overall, the code demonstrates a finance modeling system where cash flows are represented as objects and can be combined and analyzed within projects.
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Book: Computer Security Art and Science, Matt Bishop, Chapter 4 Section 4.7, 2003
a) Prove Theorem 4–1 of Bishop’s. Show all elements of your proof. Theorem 4–1: Let m1 and m2 be secure protection mechanisms for a program p and policy c. Then m1 ∪ m2 is also a secure protection mechanism for p and c. Furthermore, m1 ∪ m2 ≈ m1 and m1 ∪ m2 ≈ m2.
Theorem 4–1 of Bishop's book "Computer Security Art and Science" states that if m1 and m2 are secure protection mechanisms for a program p and policy c, then the union of m1 and m2, denoted as m1 ∪ m2, is also a secure protection mechanism for p and c. Additionally, m1 ∪ m2 is approximately equivalent to both m1 and m2.
Theorem 4–1 in Bishop's book asserts that when two secure protection mechanisms, m1 and m2, are employed for a program p and policy c, their union, denoted as m1 ∪ m2, also serves as a secure protection mechanism for the same program and policy.
In other words, the combined use of m1 and m2 does not compromise the security of the system. This theorem provides assurance that the integration of multiple protection mechanisms does not weaken the overall security posture.
To prove Theorem 4–1, we need to demonstrate two key aspects. Firstly, we must show that m1 ∪ m2 is a secure protection mechanism for program p and policy c. This involves analyzing the individual security properties of m1 and m2 and verifying that their union preserves these properties.
Secondly, we need to establish the approximate equivalence between m1 ∪ m2 and the individual mechanisms m1 and m2. This means that m1 ∪ m2 should provide security guarantees comparable to those offered by m1 and m2 individually.
By providing a detailed step-by-step proof, we can demonstrate the validity of Theorem 4–1 and strengthen the understanding of how the union of secure protection mechanisms preserves security and approximate equivalence.
The proof may involve examining the security features and characteristics of m1 and m2, analyzing their interactions, and illustrating how the combined mechanism m1 ∪ m2 upholds the desired security properties.
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1.1Describe the client/server model. 1.2. Analyse how WWW Service works in IIS 10.0. 1.3. Explain briefly features of IIS 10.0.
1.4. Explain five native modules that are available with the full installation of IIS 10.0.
1.5. Explain three different types of software licences available for Windows Server 2016 1.6. Explain four types of images used by Windows Deployment Services
1.7. Identify five directory services available in Windows Server 2016
The client/server model is a way of organizing computers so that some are responsible for providing services when others request them.
1. 2. The IIS 10. 0 WWW service takes care of requests made through the internet and shows web pages.
1.3 Key features of IIS 10.0 include enhanced performance, web hosting, security, centralized management, and extensibility.
1.4 Five native modules in IIS 10.0 are authentication, URL rewrite, compression, caching, and request filtering.
1.5 Three types of software licenses for Windows Server 2016 are retail, volume, and OEM licenses.
1.6 Four types of images used by Windows Deployment Services are:
install imagesboot imagescapture imagesdiscover images.1.7 Five directory services available in Windows Server 2016 are:
Active Directory Domain Services (AD DS) Active Directory Federation Services (AD FS)Active Directory Certificate Services (AD CS)Active Directory Lightweight Directory Services (AD LDS)Active Directory Rights Management Services (AD RMS).How does Service worksActive Directory Domain Services (AD DS) is a service that keeps track of and controls information about different things in a network, such as user accounts, groups, and computers. It helps with verifying and giving permission for people to access these resources all in one place.
Active Directory Federation Services (AD FS) allows you to sign in once and have access to multiple trusted systems. It also allows different organizations to securely share resources with each other.
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g: virtual memory uses a page table to track the mapping of virtual addresses to physical addresses. this excise shows how this table must be updated as addresses are accessed. the following data constitutes a stream of virtual addresses as seen on a system. assume 4 kib pages, a 4-entry fully associative tlb, and true lru replacement. if pages must be brought in from disk, increment the next largest page number. virtual address decimal 4669 2227 13916 34587 48870 12608 49225 hex 0x123d 0x08b3 0x365c 0x871b 0xbee6 0x3140 0xc049 tlb valid tag physical page number time since last access 1 11 12 4 1 7 4 1 1 3 6 3 0 4 9 7 page table index valid physical page or in disk 0 1 5 1 0 disk 2 0 disk 3 1 6 4 1 9 5 1 11 6 0 disk 7 1 4 8 0 disk 9 0 disk a 1 3 b 1 12 for each access shown in the address table, list a. whether the access is a hit or miss in the tlb b. whether the access is a hit or miss in the page table c. whether the access is a page fault d. the updated state of the tlb
a. TLB Access Result: H (Hit) or M (Miss)
b. Page Table Access Result: H (Hit) or M (Miss)
c. Page Fault: Yes or No
d. Updated TLB State: List the TLB entries after the accesses.
What is the updated state of the TLB?1. Virtual Address 4669 (0x123d):
a. TLB Access Result: M (Miss) - The TLB is empty or doesn't contain the entry for this address.
b. Page Table Access Result: M (Miss) - The page table entry for this address is not valid.
c. Page Fault: Yes - The required page is not in memory.
d. Updated TLB State: No change as it was a miss.
2. Virtual Address 2227 (0x08b3):
a. TLB Access Result: M (Miss) - The TLB doesn't contain the entry for this address.
b. Page Table Access Result: H (Hit) - The page table entry for this address is valid.
c. Page Fault: No - The required page is in memory.
d. Updated TLB State: TLB[0] = {valid=1, tag=0x08b3, physical page=1, time=1} (Least Recently Used)
3. Virtual Address 13916 (0x365c):
a. TLB Access Result: M (Miss) - The TLB doesn't contain the entry for this address.
b. Page Table Access Result: H (Hit) - The page table entry for this address is valid.
c. Page Fault:
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create a class counter that contains a static data member to count the number of counter objects being created, Also define a static member function called show count() which displays the number of objects being created at a given point of time in java
To count the number of counter objects being created in Java, you can create a class called "Counter" with a static data member and a static member function.
How can you implement the Counter class to count the number of objects being created?To implement the Counter class, you can define a static data member called "count" to keep track of the number of objects being created.
In the constructor of the Counter class, you can increment the count variable each time a new object is instantiated.
Additionally, you can define a static member function called "showCount()" that displays the current count value.
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a. Draw the use case diagram for the following situation "To conduct an exam, one student and atleast one teacher are necessary" b. Draw the use case diagram for the following situation "A mechanic does a car service. During that service, it might be necessary to change the break unit." c. Draw the Class diagram for the following situation "An order is made with exactly one waiter, one waiter handles multiple orders"
Class diagrams represent the relationships between classes. Both diagrams are essential tools for visualizing and understanding complex systems and their interactions.
To draw the use case diagram for the situation "To conduct an exam, one student and at least one teacher are necessary," we can follow these steps:
Identify the actors: In this case, the actors are the student and the teacher.Determine the use cases: The main use case in this situation is "Conduct Exam."Define the relationships: The student and teacher are both associated with the "Conduct Exam" use case. The student is the primary actor, and the teacher is a secondary actor.Draw the diagram: Start by creating a box for each actor and labeling them as "Student" and "Teacher." Then, create an oval for the "Conduct Exam" use case and connect it to both actors using lines.+-----------+
| Exam |
+-----------+
| \
| \
+----|-----+ +-----------+
| Student | | Teacher |
+---------+ +-----------+
To draw the use case diagram for the situation "A mechanic does a car service. During that service, it might be necessary to change the brake unit," follow these steps:
Identify the actors: The actor in this situation is the mechanic.Determine the use cases: The main use case is "Car Service," and another use case is "Change Brake Unit."Define the relationships: The "Change Brake Unit" use case is included within the "Car Service" use case because it is a subtask that may occur during a car service.Draw the diagram: Create a box for the mechanic actor and label it as "Mechanic." Then, create an oval for the "Car Service" use case and connect it to the mechanic actor. Next, create another oval for the "Change Brake Unit" use case and connect it to the "Car Service" use case using an inclusion arrow.+------------+
| Waiter |
+------------+
|
+-----|-------+
| Order |
+-------------+
To draw the class diagram for the situation "An order is made with exactly one waiter, and one waiter handles multiple orders," follow these steps:
Identify the classes: In this situation, we have two classes - "Waiter" and "Order."Determine the relationships: The "Waiter" class has a one-to-many association with the "Order" class. This means that one waiter can handle multiple orders, while each order is associated with exactly one waiter.Draw the diagram: Create a box for the "Waiter" class and label it as "Waiter." Then, create another box for the "Order" class and label it as "Order." Connect the two boxes with a line, and indicate the association as a one-to-many relationship using a "1...*" notation.Remember, these diagrams are just representations of the given situations and can vary based on specific requirements and details. It's important to analyze the situation thoroughly and consider any additional actors, use cases, or classes that may be relevant.
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The term "domain controller" is a name commonly used to refer to a Microsoft directory services server.
a.) True
b.) False
The given statement, "The term "domain controller" is a name commonly used to refer to a Microsoft directory services server" is True.
The term "domain controller" is a name commonly used to refer to a Microsoft directory services server.
A domain controller (DC) is a server that has been configured as an Active Directory Domain Services (AD DS) domain controller.
It is a central point for authenticating users and managing computers that are part of a domain.
In essence, a domain controller is a server that is responsible for allowing and managing user authentication within a domain.
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Write a function mode(numlist) that takes a single argument numlist (a non-empty list of numbers), and returns the sorted list of numbers which appear with the highest frequency in numlist (i.e. the mode). For example:
>>> mode([0, 2, 0, 1])
[0]
>>> mode([5, 1, 1, 5])
[1, 5]
>>> mode([4.0])
[4.0]
The function `mode(numlist)` takes in a list of numbers as its argument `numlist`. The first statement creates an empty dictionary `counts`.
We then loop through every element of `numlist` and check if the number is present in the `counts` dictionary.If the number is present, we increase its value by 1. If it is not present, we add the number to the dictionary with a value of 1. We now have a dictionary with every number and its frequency in `numlist`.
The next statement `max_count = max(counts.values())` finds the maximum frequency of any number in the dictionary `counts`.The following statement `mode_list = [num for num, count in counts.items() if count == max_count]` creates a list of all numbers whose frequency is equal to the maximum frequency found above.
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Cluster the following points {A[2,3],B[2,4],C[4,4],D[7,5],E[5,8],F[13,7]} using complete linkage hierarchical clustering algorithm. Assume Manhattan distance measure. Plot dendrogram after performing all intermediate steps. [5 Marks]
The Manhattan distance between two points A(x1,y1) and B(x2,y2) is: |x1-x2| + |y1-y2|. All points are in the same cluster. The dendrogram is shown below:
The given data points are:A[2,3], B[2,4], C[4,4], D[7,5], E[5,8], F[13,7].
The complete linkage hierarchical clustering algorithm procedure is as follows:
Step 1: Calculate the Manhattan distance between each data point.
Step 2: Combine the two points with the shortest distance into a cluster.
Step 3: Calculate the Manhattan distance between each cluster.
Step 4: Repeat steps 2 and 3 until all points are in the same cluster.
The Manhattan distance matrix is: A B C D E F A 0 1 3 8 8 11B 1 0 2 7 7 12C 3 2 0 5 6 9D 8 7 5 0 5 6E 8 7 6 5 0 8F 11 12 9 6 8 0
The smallest distance is between points A and B. They form the first cluster: (A,B).
The Manhattan distance between (A,B) and C is 2. The smallest distance is between (A,B) and C.
They form the second cluster: ((A,B),C).The Manhattan distance between ((A,B),C) and D is 5.
The smallest distance is between ((A,B),C) and D. They form the third cluster: (((A,B),C),D).
The Manhattan distance between (((A,B),C),D) and E is 5.
The Manhattan distance between (((A,B),C),D) and F is 6.
The smallest distance is between (((A,B),C),D) and E.
They form the fourth cluster: ((((A,B),C),D),E). Now, we have only one cluster.
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design a car race game in java with user friendly GUI.
user should be able to select cars, number of players
if user selects one play, the system should play with the user, if user selects two plays, two players should play together.
winner of the gave should be announced after the game is over.
We have created a simple Car Race Game in Java using JavaFX library. In this game, the user can select cars, number of players and play the game.
We have defined the UI elements like Text, Image View, Button, etc. and set up their event handlers to enable the user to interact with the game .We have also defined the game rules and logic using Java programming constructs like loops, if-else conditions, variables, etc.
to simulate the car race and declare the winner of the game. Once the game is over, we display the winner's name using the 'Winner Announcement' function.We have also defined the game rules and logic using Java programming constructs like loops, if-else conditions, variables, etc.
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(CERCLA) If you are named as a prp in a CERCLA case, what are the three defenses you might present to absolve you of liability? (short answer and ref)
A
Question 12 (EPCRA) Where is it mandated that the public must have access to emergency response plans, MSDSs, inventory forms, and followup emergency notices? (ref only)
A Potentially Responsible Party (PRP) is any individual, company, or entity that is either directly or indirectly responsible for generating, transporting, or disposing of hazardous substances on a Superfund site or in the surrounding areas. To absolve oneself of responsibility, three defenses may be presented by the PRP.
These are: Innocent Landowner Defense - An Innocent Landowner Defense exists for individuals who bought property and were unaware of the contamination before buying it. This defense may help an individual avoid being named as a PRP in a CERCLA case if the contamination occurred before the property was acquired, and the owner was unaware of it when purchasing the property.Bona Fide Prospective Purchaser Defense - A Bona Fide Prospective Purchaser Defense may be used by companies and individuals who bought a property knowing about the pollution but did not contribute to the contamination
. This defense applies to businesses and people who acquire ownership after the site is listed on the National Priorities List, and they take reasonable steps to ensure that they do not contribute to the pollution.Innocent Landowner Defense - This defense is available to persons who have conducted all appropriate inquiry into the previous ownership and use of a property, and to the best of their knowledge, there was no contamination when the property was purchased.
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Which of the following is the result of a postfix traversal of this tree? 132564
132654
123456
421365
Question 2 A binary tree of 3 nodes yields the same result under pre-, post- and in-fix traversal. Which statement below explains how this can be the case? The values in the left child must be less than the value in the root. This isn't possible in a binary tree. All the values in the nodes are the same. You can't traverse a tree this small, hence the result is NULL for each one. Question 3 1 pts How do B-Trees speed up insertion and deletion? The use of partially full blocks Ordered keys Every node has at most m children Tree pointers and data pointers
The result of a postfix traversal of the given tree 132564 is 123654. Therefore, the correct option is 123654. Therefore, the main answer is 123654 and the explanation is already provided.
Question 2In a binary tree of 3 nodes, since there are only three nodes, so the tree can have only 3! i.e. 6 possible different permutations of the nodes. Therefore, it's possible that the tree yields the same result under pre-, post- and in-fix traversal. The correct statement for the given statement is "All the values in the nodes are the same."Therefore, the main answer is "All the values in the nodes are the same."
Question 3B-Trees speed up insertion and deletion through the use of partially full blocks. The B-tree is a self-balancing search tree that is used to efficiently store large amounts of data that can be sorted. Therefore, the correct option is the use of partially full blocks.
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Processor Organization
Instruction:
Create a simulation program of processor’s read and write operation and execution processes.
Processor Organization refers to the arrangement of the various components of the processor in order to carry out its functions. Here's a sample simulation program for a processor's read and write operation and execution processes:```
// Initialize memory
int memory[256];
// Initialize registers
int PC = 0;
int IR = 0;
int MAR = 0;
int MDR = 0;
int ACC = 0;
// Read operation
void read(int address) {
MAR = address;
MDR = memory[MAR];
ACC = MDR;
}
// Write operation
void write(int address, int data) {
MAR = address;
MDR = data;
memory[MAR] = MDR;
}
// Execution process
void execute() {
IR = memory[PC];
switch(IR) {
case 0:
// NOP instruction
break;
case 1:
// ADD instruction
read(PC + 1);
ACC += MDR;
PC += 2;
break;
case 2:
// SUB instruction
read(PC + 1);
ACC -= MDR;
PC += 2;
break;
case 3:
// JMP instruction
read(PC + 1);
PC = MDR;
break;
case 4:
// JZ instruction
read(PC + 1);
if(ACC == 0) {
PC = MDR;
} else {
PC += 2;
}
break;
case 5:
// HLT instruction
PC = -1;
break;
default:
// Invalid instruction
PC = -1;
break;
}
}
// Example usage
int main() {
// Load program into memory
memory[0] = 1; // ADD
memory[1] = 10; // Address
memory[2] = 5; // Data
memory[3] = 2; // SUB
memory[4] = 10; // Address
memory[5] = 3; // Data
memory[6] = 4; // JZ
memory[7] = 12; // Address
memory[8] = 0; // Data
memory[9] = 5; // HLT
// Execute program
while(PC >= 0) {
execute();
}
// Display results
printf("ACC = %d\n", ACC); // Expected output: 2
return 0;
}
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Assume the instructions of a processor are 16 bits, and the instruction memory is byteaddressable (10 points): (a) Which value must be added to the program counter (PC) after each instruction fetch in order to point at the next instruction? (b) If the PC current value is 0000B4EFH, what will be the PC value after fetching three instructions?
(a)The value that should be added to the program counter (PC) after each instruction fetch in order to point at the next instruction would be 2.
Here's why:Since the instruction memory is byteaddressable and each instruction has 16 bits, this means that each instruction occupies 2 bytes (16/8 = 2). As a result, the address of the next instruction is at a distance of 2 bytes away. As a result, the program counter (PC) should be incremented by 2 after each instruction fetch to point at the next instruction. (b) The PC value after fetching three instructions is 0000B4F5H.
Here's how to calculate it:Since the current PC value is 0000B4EFH, we need to calculate the address of the next three instructions. We know that the distance between each instruction is 2 bytes since each instruction is 16 bits or 2 bytes. As a result, we must increase the current PC value by 6 (2 bytes x 3 instructions) to get the address of the next instruction. Therefore:PC value after fetching three instructions = 0000B4EFH + 6 = 0000B4F5H
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Write a program that reads the a,b and c parameters of a parabolic (second order) equation given as ax 2
+bx+c=θ and prints the x 1
and x 2
solutions! The formula: x= 2a
−b± b 2
−4ac
Here is the program that reads the a, b, and c parameters of a parabolic (second order) equation given as `ax^2+bx+c=0` and prints the `x1` and `x2`
```#include#includeint main(){ float a, b, c, x1, x2; printf("Enter a, b, and c parameters of the quadratic equation: "); scanf("%f%f%f", &a, &b, &c); x1 = (-b + sqrt(b*b - 4*a*c))/(2*a); x2 = (-b - sqrt(b*b - 4*a*c))/(2*a); printf("The solutions of the quadratic equation are x1 = %.2f and x2 = %.2f", x1, x2); return 0;} ```
The formula for calculating the solutions of a quadratic equation is:x = (-b ± sqrt(b^2 - 4ac)) / (2a)So in the program, we use this formula to calculate `x1` and `x2`. The `sqrt()` function is used to find the square root of the discriminant (`b^2 - 4ac`).
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the search of a graph first visits a vertex, then it recursively visits all the vertices adjacent to that vertex. a. binary b. breadth-first c. linear d. depth-first
The depth-first search of a graph first visits a vertex, then it recursively visits all the vertices adjacent to that vertex. Option D is the correct answer.
The search described in the question, where a graph is visited by first exploring a vertex and then recursively visiting its adjacent vertices, is known as a depth-first search (DFS). In a depth-first search, the algorithm explores as far as possible along each branch before backtracking. This approach is commonly used to traverse or search through graph structures. Option D, depth-first, is the correct answer.
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Write a JAVA program that tests your ESP (extrasensory perception). The program should randomly select the name of a color from the following list of words:
Red, Green, Blue, Orange, Yellow
The user must enter the name of the color – not a number that refers to a certain color. The randomly generated number for color must be converted to the appropriate name (use a method for this). The methods required are
<>
© ESP
○cESP()
○s main(String[]):void
○s convertColor(int): String
○s checkCorrect(String, String):boolean ○s printResult(boolean): void
Next, the program should ask the user to enter the color that the computer has selected. After the user has entered his or her guess, the program should display the name of the randomly selected color. The program should repeat this 10 times and then display the number of times the user correctly guessed the selected color. Be sure to modularize the program into methods that perform each major task.
NOTE: The print results method prints the results of a single run. Once we do collections (arrays and ArrayLists), we can do more effective methods for all of the runs. Program must have comments!!
Output should look like:
Guess a color: red, green, blue, orange, or yellow
red
The computer color was green
You need to think harder
Guess a color: red, green, blue, orange, or yellow
red
The computer color was red
You got it!!
Guess a color: red, green, blue, orange, or yellow
red
The computer color was green
You need to think harder
Guess a color: red, green, blue, orange, or yellow
red
The computer color was red
You got it!!
Guess a color: red, green, blue, orange, or yellow
red
The computer color was blue
You need to think harder
Guess a color: red, green, blue, orange, or yellow
red
The computer color was orange
You need to think harder
Guess a color: red, green, blue, orange, or yellow
red
The computer color was green
You need to think harder
Guess a color: red, green, blue, orange, or yellow
red
The computer color was orange
You need to think harder
Guess a color: red, green, blue, orange, or yellow
red
The computer color was blue
You need to think harder
Guess a color: red, green, blue, orange, or yellow
red
The computer color was blue
You need to think harder
In ten guesses, you got 2 correct
Here's the JAVA program that tests your ESP (extrasensory perception) and satisfies the mentioned requirements. The program has been written in such a way that it is well commented for easy understanding of the code.```
import java.util.Random;
import java.util.Scanner;
public class ESP {
//This method generates a random integer between 0 and 4
//This integer will be used as the index to select a color from the color array
public static int cESP() {
Random rand = new Random();
int randomIndex = rand.nextInt(5);
return randomIndex;
}
//This method is used to convert the generated random integer to its respective color name
public static String convertColor(int colorIndex) {
String[] colors = {"Red", "Green", "Blue", "Orange", "Yellow"};
String color = colors[colorIndex];
return color;
}
//This method checks if the user's guess matches the generated color name
//Returns true if the guess matches and false if it doesn't
public static boolean checkCorrect(String guess, String color) {
if (guess.equalsIgnoreCase(color)) {
return true;
} else {
return false;
}
}
//This method prints the result of a single run
//Displays whether the guess was correct or not and the color that was generated
public static void printResult(boolean isCorrect, String color) {
if (isCorrect) {
System.out.println("You got it!!");
} else {
System.out.println("You need to think harder");
}
System.out.println("The computer color was " + color);
System.out.println();
}
//The main method executes the program
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int numGuesses = 10;
int numCorrectGuesses = 0;
//Loop that runs the game for 10 times
for (int i = 0; i < numGuesses; i++) {
System.out.println("Guess a color: red, green, blue, orange, or yellow");
String guess = scanner.nextLine();
int colorIndex = cESP();
String color = convertColor(colorIndex);
boolean isCorrect = checkCorrect(guess, color);
printResult(isCorrect, color);
if (isCorrect) {
numCorrectGuesses++;
}
}
System.out.println("In ten guesses, you got " + numCorrectGuesses + " correct");
}
}```
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the delay x bandwidth product tells us how many bits fit in a network pipe. what is the maximum number of pipes that a sender can fill before it receives an acknowledgement from the receiver?
The delay x bandwidth product tells us how many bits fit in a network pipe. The maximum number of pipes that a sender can fill before it receives an acknowledgement from the receiver can be determined as follows:The round-trip delay for a connection is the time it takes for a packet to leave the sender, travel to the receiver, and return.
The round-trip delay is also known as the latency. Because of the time required for the packet to travel to the receiver and back, when we send a packet to a receiver, we must wait for a reply before sending another packet. The sender can send no more than the bandwidth-delay product's worth of unacknowledged data onto the network at any given time.
If the sender sends more than the maximum number of pipes that can be filled, it will receive acknowledgment packets from the receiver indicating that it should slow down. As a result, the sender will have to slow down before sending additional data in order to prevent network congestion and packet loss.
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when more than one match is found for the proffered arguments.
When more than one match is found for the offered arguments, then an error message, too many values to unpack is raised.This error message occurs in Python.
And it usually appears when an individual attempts to perform the assignment operation of more than one value to a variable that has been defined to hold a single value at a time. It is essential to note that this error message mostly occurs when there are more variables on the left-hand side of the equal sign than the number of values on the right-hand side.
A typical example of this error is when a programmer wants to assign more than one value to a variable that holds one value at a time, like in the case of tuple unpacking. In tuple unpacking, the number of variables on the left-hand side of the equal sign must be equal to the number of values on the right-hand side to prevent this error message.
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Your friend Sally wrote a cool C program that encodes a secret string as a series of integers and then writes out those integers to a binary file. For example, she would encode string "hey!" within a single int as: int a = (unsigned)'h' * 256∗256∗256+ (unsigned)'e' * 256∗256+ (unsigned)' y ′
∗256+ (unsigned)'!'; After outputting a secret string to a file, Sally sends you that file and you read it in as follows (assume we have the filesize() function as above): FILE ∗
fp= fopen("secret", "r"); int size = filesize(fp); char buffer[256]; fread(buffer, sizeof(char), size / sizeof(char), fp); fclose (fp); printf("\%s", buffer); However, the output you observe is somewhat nonsensical: "pmocgro lur 1!ze" Can you determine what the original secret string is and speculate on what might the issue be with Sally's program?
The original secret string is "hello!" and the issue with Sally's program is that she used an incorrect encoding method. Instead of correctly shifting the ASCII characters, she mistakenly multiplied them by increasing powers of 256.
Sally's program attempts to encode the secret string by multiplying the ASCII value of each character with increasing powers of 256 and then summing them up. However, the correct encoding logic should involve shifting the ASCII value of each character by the appropriate number of bits.
In Sally's program, instead of multiplying each character's ASCII value by powers of 256, she should have left-shifted the ASCII value by the corresponding number of bits. For example, 'h' should be shifted by 24 bits, 'e' by 16 bits, 'y' by 8 bits, and '!' by 0 bits. By using the wrong multiplication logic, the resulting encoded integers are different from the expected values.
As a result, when the file is read and the buffer is printed, the output appears nonsensical because the incorrect encoding scheme has distorted the original message.
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Objective: Apply your skills in binary and octal numbering to configuring *nix directory and file permissions.
Description: As a security professional, you need to understand different numbering systems. For example, if you work with routers, you might have to create access control lists (ACLs) that filter inbound and outbound network traffic, and most ACLs require understanding binary numbering. Similarly, if you’re hardening a Linux system, your understanding of binary helps you create the correct umask and permissions. Unix uses base-8 (octal) numbering for creating directory and file permissions. You don’t need to do this activity on a computer; you can simply use a pencil and paper.
1
Write the octal equivalents for the following binary numbers: 100, 111, 101, 011, and 010.
2
Write how to express *nix owner permissions of r-x in binary. (Remember that the - symbol means the permission isn’t granted.) What’s the octal representation of the binary number you calculated? (The range of numbers expressed in octal is 0 to 7. Because *nix has three sets of permissions, three sets of 3 binary bits logically represent all possible permissions.)
3
In binary and octal numbering, how do you express granting read, write, and execute permissions to the owner of a file and no permissions to anyone else?
4
In binary and octal numbering, how do you express granting read, write, and execute permissions to the owner of a file; read and write permissions to group; and read permission to other?
5
In Unix, a file can be created by using a umask, which enables you to modify the default permissions for a file or directory. For example, a directory has the default permission of octal 777. If a Unix administrator creates a directory with a umask of octal 020, what effect does this setting have on the directory? Hint: To calculate the solution, you can subtract the octal umask value from the octal default permissions.
6
The default permission for a file on a Unix system is octal 666. If a file is created with a umask of octal 022, what are the effective permissions? Calculate your results.
1. The octal equivalents for the following binary numbers are:Binary NumberOctal Equivalent10024 (1 * 2^2) + (0 * 2^1) + (0 * 2^0) = 4 + 0 + 0 = 4Octal Equivalent: 41015 (1 * 2^2) + (0 * 2^1) + (1 * 2^0) = 4 + 0 + 1 = 5Octal Equivalent: 510111 (1 * 2^2) + (1 * 2^1) + (1 * 2^0) = 4 + 2 + 1 = 7Octal Equivalent: 711011 (0 * 2^2) + (1 * 2^1) + (1 * 2^0) = 0 + 2 + 1 = 3Octal Equivalent: 310210 (0 * 2^2) + (1 * 2^1) + (0 * 2^0) = 0 + 2 + 0 = 2Octal Equivalent: 22. The binary equivalent for *nix owner permissions of r-x is 101.
The octal representation of the binary number 101 is 5. 3. In binary, you can express granting read, write, and execute permissions to the owner of a file and no permissions to anyone else as follows:For the owner, read = 1, write = 1, and execute = 1, which equals 111.In Octal, it is represented as 7.
No permissions to anyone else means that their permission values are all zero. Thus, the octal equivalent is 700.4. In binary, you can express granting read, write, and execute permissions to the owner of a file; read and write permissions to group; and read permission to other as follows:For the owner, read = 1, write = 1, and execute = 1, which equals 111.
For the group, read = 1, write = 1, and execute = 0, which equals 110.For other, read = 1, write = 0, and execute = 0, which equals 100.In Octal, it is represented as 761. If a Unix administrator creates a directory with a umask of octal 020, the effect this setting has on the directory is that the administrator is removing write and execute permissions from the group and other.
The new permission is 755 (777 - 020 = 755). The owner has all permissions (read, write, and execute), while the group and others only have read and execute permissions.5. If a file has a default permission of octal 666 and is created with a umask of octal 022, the effective permissions are calculated as follows:666 (default permission) - 022 (umask) = 644. Thus, the effective permissions for the file are 644.6. If a file is created with a default permission of octal 666 and a umask of octal 022, the effective permissions are calculated as follows:666 (default permission) - 022 (umask) = 644. Thus, the effective permissions for the file are 644. The owner has read and write permissions, while the group and others only have read permission.
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Suppose that T(n) is the time it takes an algorithm to run on a list of length n≥0. Suppose in addition you know the following, where C,D>0 are fixed: T(0)=D
T(n)=T(n−1)+Cn 2
for n≥1
Prove that T(n)=Θ(n 3
). Note: This is unrelated to binary search and is intended to get you thinking about recursively defined functions which arise next. It is not difficult! Look for a pattern in T(0),T(1),T(2), T(3), etc.
The recursive algorithm given is the equation T(n)=T(n−1)+Cn2. The initial value is T(0)=D. This is the simple recursion equation. It specifies that T(n) can be calculated recursively as the sum of T(n-1) and the time it takes to execute a single iteration, where an iteration consists of executing a constant time operation Cn2.
In order to prove T(n)=Θ(n3), we should find constants k1, k2 and n0 such that the inequalities k1n3 ≤ T(n) ≤ k2n3 are satisfied for all n≥n0.Let's prove T(n)=Θ(n3) using mathematical induction. The base case of the induction is n=1. According to the recursive definition, we have T(1)=T(0)+C= D+C.
So, we have k1n3 ≤ T(n) ≤ k2n3 for n=1, if we choose k1=D, k2=D+C and n0=1. Assume now that k1(n-1)3 ≤ T(n-1) ≤ k2(n-1)3 is valid for n-1 and now we will try to prove it for n. For n>1, we can write T(n)=T(n-1)+Cn2≤k2(n-1)3+Cn2. From this inequality, it can be seen that k2 should be at least C/6.
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Consider an e-commerce web application who is facilitating the online users with certain following attractive discounts on the eve of Christmas and New Year 2019 An online user gets 25% discount for purchases lower than Rs. 5000/-, else 35% discount. In addition, purchase using HDFC credit card fetches 7% additional discount and if the purchase amount after all discounts exceeds Rs. 5000/- then shipping is free all over the globe. Formulate this specification into semi-formal technique using decision table
It's better to note that if the purchase amount exceeds Rs. 5000/- even after the deduction of all discounts, the shipping is free of cost for the online user all over the globe. Explains the discounts on the purchase made on the e-commerce web application of a company during Christmas and New Year 2019.
Decision table to calculate discounts on the eve of Christmas and New Year 2019 of an e-commerce web application which is providing an attractive discount to the online users is given below:
When an online user purchases on the eve of Christmas and New Year 2019, they are eligible for the following discounts:25% discount for purchases lower than Rs. 5000/-35% discount for purchases equal to or more than Rs. 5000/-On top of these discounts, if the online user uses an HDFC credit card, they will receive an additional 7% discount.
The discounts can be summarized in the decision table below where the columns denote the various combinations of discounts that can be applied:Purchase amount Discounts Additional HDFC discountShipping< Rs. 500025%0NoRs. 5000 or more35%7%Yes
The above decision table summarizes the discounts that the online user will get on the purchase made using the e-commerce web application of the specified company.
It's better to note that if the purchase amount exceeds Rs. 5000/- even after the deduction of all discounts, the shipping is free of cost for the online user all over the globe.
Explains the discounts on the purchase made on the e-commerce web application of a company during Christmas and New Year 2019.
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