Queue is a collection whose elements are added at one end and removed from the other end
What is a Queue
Conceptual View of a Queue
Uses of Queues in Computing
Operations on a Queue
Implementation of a Queue
Applications
Downloadable Resources
Linked Lists: Introduction Linked lists
Representation of linked list
operations on linked list
Comparison of Linked Lists with Arrays and Dynamic Arrays
Types of Linked Lists and operations-Circular Single Linked List, Double Linked List, Circular Double Linked List
Queue is a collection whose elements are added at one end and removed from the other end
What is a Queue
Conceptual View of a Queue
Uses of Queues in Computing
Operations on a Queue
Implementation of a Queue
Applications
Downloadable Resources
Linked Lists: Introduction Linked lists
Representation of linked list
operations on linked list
Comparison of Linked Lists with Arrays and Dynamic Arrays
Types of Linked Lists and operations-Circular Single Linked List, Double Linked List, Circular Double Linked List
Write a program that accepts an arithmetic expression of unsigned in.pdfJUSTSTYLISH3B2MOHALI
Write a program that accepts an arithmetic expression of unsigned integers in postfix notation
and builds the arithmetic expression tree that represents that expression. From that tree, the
corresponding fully parenthesized infix expression should be displayed and a file should be
generated that contains the three address format instructions. This topic is discussed in the week
4 reading in module 2, section II-B. The main class should create the GUI shown below:
Pressing the Construct Tree button should cause the tree to be constructed and using that tree, the
corresponding infix expression should be displayed and the three address instruction file should
be generated. The postfix expression input should not be required to have spaces between every
token. Note in the above example that 9+- are not separated by spaces. The above example
should produce the following output file containing the three address instructions:
Add R0 5 9
Sub R1 3
R0 Mul R2 2 3
Div R3 R1 R2
Inheritance should be used to define the arithmetic expression tree. At a minimum, it should
involve three classes: an abstract class for the tree nodes and two derived classes, one for
operand nodes and another for operator nodes. Other classes should be included as needed to
accomplish good object-oriented design. All instance data must be declared as private.
You may assume that the expression is syntactically correct with regard to the order of operators
and operands, but you should check for invalid tokens, such as characters that are not valid
operators or operands such as 2a, which are not valid integers. If an invalid token is detected a
RuntimeException should be thrown and caught by the main class and an appropriate error
message should be displayed. Three Adddress Generator Enter Postfix Expression 359 2 3
Construct Tree Infix Expression ((3-(5 9))/ (2* 3
Solution
#include
#include
#include
#include
#include
struct TREE //Structure to represent a tree
{
char data;
struct TREE *right,*left,*root;
};
typedef TREE tree; /* Stack Using Linked List */
struct Stack //Structure to represent Stack
{
struct TREE *data;
struct Stack *next,*head,*top; //Pointers to next node,head and top
};
typedef struct Stack node;
node *Nw; //Global Variable to represent node
void initialize(node *&n)
{
n = new node;
n -> next = n -> head = n -> top = NULL;
}
void create(node *n)
{
Nw = new node; //Create a new node
Nw -> next = NULL; //Initialize next pointer field
if(n -> head == NULL) //if first node
{
n -> head = Nw; //Initialize head
n -> top = Nw; //Update top
}
}
void push(node *n,tree *ITEM)
{
node *temp = n -> head;
if(n -> head == NULL) //if First Item is Pushed to Stack
{
create(n); //Create a Node
n -> head -> data = ITEM; //Insert Item to head of List
n -> head = Nw;
return; //Exit from Function
}
create(n); //Create a new Node
Nw -> data = ITEM; //Insert Item
while(temp -> next != NULL)
temp = temp -> next; //Go to Last Node
temp -> next = Nw; //Point New node
n -> top = Nw; //Upda.
Data Structures in C++I am really new to C++, so links are really .pdfrohit219406
Data Structures in C++
I am really new to C++, so links are really hard topic for me. It would be nice if you can provide
explanations of what doubly linked lists are and of some of you steps... Thank you
In this assignment, you will implement a doubly-linked list class, together with some list
operations. To make things easier, you’ll implement a list of int, rather than a template class.
Solution
A variable helps us to identify the data. For ex: int a = 5; Here 5 is identified through variable a.
Now, if we have collection of integers, we need some representation to identify them. We call it
array. For ex: int arr[5]
This array is nothing but a Data Structure.
So, a Data Structure is a way to group the data.
There are many Data Structures available like Arrays, Linked List, Doubly-Linked list, Stack,
Queue, etc.
Doubly-Linked list are the ones where you can traverse from the current node both in left and
right directions.
Why so many different types of Data Structures are required ?
Answer is very simple, grouping of data, storage of data and accessing the data is different.
For example, in case of Arrays we store all the data in contiguous locations.
What if we are not able to store the data in contiguous locations because we have huge data.
Answer is go for Linked List/Doubly-Linked list.
Here we can store the data anywhere and link the data through pointers.
I will try to provide comments for the code you have given. May be this can help you.
#pragma once
/*
dlist.h
Doubly-linked lists of ints
*/
#include
class dlist {
public:
dlist() { }
// Here we are creating a NODE, it has a integer value and two pointers.
// One pointer is to move to next node and other to go back to previous node.
struct node {
int value;
node* next;
node* prev;
};
// To return head pointer, i.e. start of the Doubly-Linked list.
node* head() const { return _head; }
// To return Tail pointer, i.e. end of the Doubly-Linked list.
node* tail() const { return _tail; }
// **** Implement ALL the following methods ****
// Returns the node at a particular index (0 is the head).
node* at(int index){
int cnt = 0;
struct node* tmp = head();
while(tmp!=NULL)
{
if (cnt+1 == index)
return tmp;
tmp = tmp->next;
}
}
// Insert a new value, after an existing one
void insert(node *previous, int value){
// check if the given previous is NULL
if (previous == NULL)
{
printf(\"the given previous node cannot be NULL\");
return;
}
// allocate new node
struct node* new_node =(struct node*) malloc(sizeof(struct node));
// put in the data
new_node->data = new_data;
// Make next of new node as next of previous
new_node->next = previous->next;
// Make the next of previous as new_node
previous->next = new_node;
// Make previous as previous of new_node
new_node->prev = previous;
// Change previous of new_node\'s next node
if (new_node->next != NULL)
new_node->next->prev = new_node;
}
// Delete the given node
void del(node* which){
struct node* head_ref = head();
/* base case */
if(*head_ref == NUL.
My question is pretty simple, I just want to know how to call my ope.pdfjeetumordhani
My question is pretty simple, I just want to know how to call my operator== function in
Stack.cpp using a list function.
Here is what I meant. This is my Stack.h file:
class Stack {
public:
/**constructors and destructors*/
Stack();
Stack(const Stack &S);
~Stack();
void pop();
void push(int data);
bool operator==(const Stack &S);
[ .......]
private:
List stack;
};
Here is my List.h file:
template
class List {
private:
struct Node {
listitem data;
Node* next;
Node* previous;
Node(listitem data) :
next(NULL), previous(NULL), data(data) {
}
};
typedef struct Node* NodePtr;
NodePtr start;
NodePtr end;
NodePtr cursor;
int length;
public:
........
bool operator==(const List &list);
[.......] }
[........]
template
bool List::operator==(const List& list) {
if (length != list.length)
return false;
NodePtr temp1 = start;
NodePtr temp2 = list.start;
while (temp1 != NULL) {
if (temp1->data != temp2->data)
return false;
temp1 = temp1->next;
temp2 = temp2->next;
}
return true;
}
[......]
So I already define my functions in List.h and I only need to call my function from List again in
my Stack.cpp file. For example from my Stack.cpp:
Stack::Stack(const Stack &S) :
stack(S.stack) {
}
void Stack::pop() {
stack.remove_start();
}
[......]
And I don\'t know how to call my operator== function in Stack.cpp from the List.h like the way
above. This is what my guess so far but did not work:
bool Stack::operator==(const Stack &S) {
stack.operator==(S.stack);
}
Thank you
Solution
When we wish to make a comparison, such as in an if statement, we use the double equals sign
(==).
1)bool operator==(const Stack &S); [ .......]
It means whatever the boolean operator it should be belongs to stack or equal to.
2) bool operator==(const List &list);
Likewise link list also compare with the constructor link list.
and constructor need not to call explicitly. it will call itself when program called first..
Need Help!! C++ #include-iostream- #include-linkedlist-h- using namesp.pdfEdwardw5nSlaterl
Need Help!! C++
#include<iostream>
#include"linkedlist.h"
using namespace std;
/**
* @brief Destructor to destroy all nodes and release memory
*/
LinkedList::~LinkedList() {
//TODO: Add code here. Make sure memory is released properly.
}
/**
* @brief Purpose: Checks if the list is empty
* @return true if the list is empty, false otherwise
*/
bool LinkedList::isEmpty() const {
// TODO: Add code here
return count == 0;
}
/**
* @brief Get the number of nodes in the list
* @return int The number of nodes in the list
*/
int LinkedList::length() const{
//TODO: Add code here
}
/**
* @brief Convert the list to a string
*
*/
string LinkedList::toString() {
string str = "[";
Node *ptr = front;
if (ptr != nullptr) {
// Head node is not preceded by separator
str += to_string(ptr->val);
ptr = ptr->next;
}
while (ptr != nullptr) {
str += ", " + to_string(ptr->val);
ptr = ptr->next;
}
str += "]";
return str;
}
/**
* @brief Displays the contents of the list
*/
void LinkedList::displayAll() {
cout << toString() << endl;
}
//TODO: Add comments
void LinkedList::addRear(T val) {
// TODO: Add code here
// consider the two cases of whether the list was empty
}
//TODO: Add comments
void LinkedList::addFront(T val) {
// TODO: Add code here
// consider the two cases of whether the list was empty
}
//TODO: Add comments
bool LinkedList::deleteFront(T &OldNum) {
// TODO: Add code here
// consider if the list was empty and return false if the list is empty
// consider the special case of deleting the only node in the list
}
//TODO: Add comments
bool LinkedList::deleteRear(T &OldNum) {
// TODO: Add code here
// consider if the list was empty and return false if the list is empty
// consider the special case of deleting the only node in the list
}
/* --- harder ones for test 2 and 3 -- */
/**
* @brief Delete a node at a given position from the list. The
* node at position pos is deleted and the value of the deleted node is returned in val.
* The valid range of pos is 1 to count. pos = 1 is the first node, and pos = count is the last node.
* @param pos: position of the node to be deleted
* @param val: it is set to the value of the node to be deleted
* @return true: if the node was deleted successfully
* @return false: if the node was not deleted successfully because the position was out of range
*/
bool LinkedList::deleteAt(int pos, T &val) {
//TODO: Add code here
// check if the pos is valid first, then move the ptr to the rigth positon
// consider the special case of deleting the first node and the last node
// Do not forget to set value.
}
/**
* @brief Insert a value at a specified position in the list. The valid pos is in the range of 1 to
count+1.
* The value will be inserted before the node at the specified position. if pos = 1, the value will be
inserted
* at the front of the list. if pos = count+1, the value will be inserted at the rear of the list.
* @param pos: position to insert the value at.
* @param val: value to insert.
* @return true: if the value was inserte.
This assignment and the next (#5) involve design and development of a.pdfEricvtJFraserr
This assignment and the next (#5) involve design and development of a sequential
non contiguous and dynamic datastructure called LinkedList. A linked list object is
a container consisting of connected ListNode objects. As before, we are not going
to use pre-fabricated classes from the c++ library, but construct the LinkedList
ADT from scratch.
The first step is construction and testing of the ListNode class. A ListNode object
contains a data field and a pointer to the next ListNode object (note the recursive
definition).
#This assignment requires you to
1. Read the Assignment 4 Notes
2. Watch the Assignment 4 Support video
3. Implement the following methods of the ListNode class
-custom constructor
-setters for next pointer and data
4. Implement the insert and remove method in the main program
5. Scan the given template to find the above //TODO and implement the code
needed
//TODO in ListNodecpp.h file
public: ListNode(T idata, ListNode<T> * newNext);
public: void setNext(ListNode<T> * newNext);
public: void setData(T newData);
// TODO in main program
void remove(ListNode<int> * &front,int value)
void insert(ListNode<int> * &front,int value)
# The driver is given ListNodeMain.cpp is given to you that does the following
tests
1. Declares a pointer called front to point to a ListNode of datatype integer
2. Constructs four ListNodes with data 1,2,4 and adds them to form a linked
list.
3. Inserts ListNode with data 3 to the list
4. Removes node 1 and adds it back to test removing and adding the first
element
5. Removes node 3 to test removing a middle node
6. Removes node 4 to test removing the last node
7. Attempt to remove a non existent node
8. Remove all existing nodes to empty the list
9. Insert node 4 and then node 1 to test if insertions preserve order
10.Print the list
Main.cpp
#include <iostream>
#include "ListNodecpp.h"
// REMEMBER each ListNode has two parts : a data field
// and an address field. The address is either null or points to the next node
//in the chain
//Requires: integer value for searching, address of front
//Effects: traverses the list node chain starting from front until the end comparing search value
with listnode getData. Returns the original search value if found, if not adds +1 to indicate not
found
//Modifies: Nothing
int search(ListNode<int> * front, int value);
//Requires: integer value for inserting, address of front
//Effects: creates a new ListNode with value and inserts in proper position (increasing order)in
the chain. If chain is empty, adds to the beginning
//Modifies: front, if node is added at the beginning.
//Also changes the next pointer of the previous node to point to the
//newly inserted list node. the next pointer of the newly inserted pointer
//points to what was the next of the previous node.
//This way both previous and current links are adjusted
//******** NOTE the use of & in passing pointer to front as parameter -
// Why do you think this is needed ?**********
void insert(ListNode<int> * &fr.
pleaase I want manual solution forData Structures and Algorithm An.pdfwasemanivytreenrco51
pleaase I want manual solution for
Data Structures and Algorithm Analysis in C++, Third Edition
By Clifford A. Shaffer
because my question from this book.page270 ,question 7.4
The implementation for Mergesort given in Section 7.4 takes an array as input and sorts that
array. At the beginning of Section 7.4 there is a simple pseudocode implementation for sorting a
linked list using Mergesort. Implement both a linked list-based version of Mergesort and the
array-based version of Mergesort, and compare and analyze their running times.
Solution
Linked list:
#include
#include
struct node
{
int data;
struct node* next;
};
struct node* SortedMerge(struct node* a, struct node* b);
void FrontBackSplit(struct node* source,
struct node** frontRef, struct node** backRef);
void MergeSort(struct node** headRef)
{
struct node* head = *headRef;
struct node* a;
struct node* b;
if ((head == NULL) || (head->next == NULL))
{
return;
}
FrontBackSplit(head, &a, &b);
MergeSort(&a);
MergeSort(&b);
*headRef = SortedMerge(a, b);
}
struct node* SortedMerge(struct node* a, struct node* b)
{
struct node* result = NULL;
if (a == NULL)
return(b);
else if (b==NULL)
return(a);
if (a->data <= b->data)
{
result = a;
result->next = SortedMerge(a->next, b);
}
else
{
result = b;
result->next = SortedMerge(a, b->next);
}
return(result);
}
void FrontBackSplit(struct node* source,
struct node** frontRef, struct node** backRef)
{
struct node* fast;
struct node* slow;
if (source==NULL || source->next==NULL)
{
*frontRef = source;
*backRef = NULL;
}
else
{
slow = source;
fast = source->next;
while (fast != NULL)
{
fast = fast->next;
if (fast != NULL)
{
slow = slow->next;
fast = fast->next;
}
}
*frontRef = source;
*backRef = slow->next;
slow->next = NULL;
}
}
void printList(struct node *node)
{
while(node!=NULL)
{
printf(\"%d \", node->data);
node = node->next;
}
}
void push(struct node** head_ref, int new_data)
{
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
int main()
{
struct node* res = NULL;
struct node* a = NULL;
push(&a, 15);
push(&a, 10);
push(&a, 5);
push(&a, 20);
push(&a, 3);
push(&a, 2);
MergeSort(&a);
printf(\"\ Sorted Linked List is: \ \");
printList(a);
getchar();
return 0;
}
Time complexity:o(nlogn)
Array :
#include
#include
void merge(int arr[], int l, int m, int r)
{
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;
int L[n1], R[n2];
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1+ j];
i = 0; // Initial index of first subarray
j = 0;
k = l;
while (i < n1 && j < n2)
{
if (L[i] <= R[j])
{
arr[k] = L[i];
i++;
}
else
{
arr[k] = R[j];
j++;
}
k++;
}
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
void mergeSort(int arr[], int l, int r)
{
if (l < r)
{
int m = l+(r-l)/2;
// Sort first and second halves
mergeSort(arr, l, m);
mergeSort(arr, m+1, r);
merge(arr, l, m, r);
}
}
void printArray(int A[], i.
Accelerate your Kubernetes clusters with Varnish CachingThijs Feryn
A presentation about the usage and availability of Varnish on Kubernetes. This talk explores the capabilities of Varnish caching and shows how to use the Varnish Helm chart to deploy it to Kubernetes.
This presentation was delivered at K8SUG Singapore. See https://feryn.eu/presentations/accelerate-your-kubernetes-clusters-with-varnish-caching-k8sug-singapore-28-2024 for more details.
GraphRAG is All You need? LLM & Knowledge GraphGuy Korland
Guy Korland, CEO and Co-founder of FalkorDB, will review two articles on the integration of language models with knowledge graphs.
1. Unifying Large Language Models and Knowledge Graphs: A Roadmap.
https://arxiv.org/abs/2306.08302
2. Microsoft Research's GraphRAG paper and a review paper on various uses of knowledge graphs:
https://www.microsoft.com/en-us/research/blog/graphrag-unlocking-llm-discovery-on-narrative-private-data/
Essentials of Automations: Optimizing FME Workflows with ParametersSafe Software
Are you looking to streamline your workflows and boost your projects’ efficiency? Do you find yourself searching for ways to add flexibility and control over your FME workflows? If so, you’re in the right place.
Join us for an insightful dive into the world of FME parameters, a critical element in optimizing workflow efficiency. This webinar marks the beginning of our three-part “Essentials of Automation” series. This first webinar is designed to equip you with the knowledge and skills to utilize parameters effectively: enhancing the flexibility, maintainability, and user control of your FME projects.
Here’s what you’ll gain:
- Essentials of FME Parameters: Understand the pivotal role of parameters, including Reader/Writer, Transformer, User, and FME Flow categories. Discover how they are the key to unlocking automation and optimization within your workflows.
- Practical Applications in FME Form: Delve into key user parameter types including choice, connections, and file URLs. Allow users to control how a workflow runs, making your workflows more reusable. Learn to import values and deliver the best user experience for your workflows while enhancing accuracy.
- Optimization Strategies in FME Flow: Explore the creation and strategic deployment of parameters in FME Flow, including the use of deployment and geometry parameters, to maximize workflow efficiency.
- Pro Tips for Success: Gain insights on parameterizing connections and leveraging new features like Conditional Visibility for clarity and simplicity.
We’ll wrap up with a glimpse into future webinars, followed by a Q&A session to address your specific questions surrounding this topic.
Don’t miss this opportunity to elevate your FME expertise and drive your projects to new heights of efficiency.
Smart TV Buyer Insights Survey 2024 by 91mobiles.pdf91mobiles
91mobiles recently conducted a Smart TV Buyer Insights Survey in which we asked over 3,000 respondents about the TV they own, aspects they look at on a new TV, and their TV buying preferences.
UiPath Test Automation using UiPath Test Suite series, part 4DianaGray10
Welcome to UiPath Test Automation using UiPath Test Suite series part 4. In this session, we will cover Test Manager overview along with SAP heatmap.
The UiPath Test Manager overview with SAP heatmap webinar offers a concise yet comprehensive exploration of the role of a Test Manager within SAP environments, coupled with the utilization of heatmaps for effective testing strategies.
Participants will gain insights into the responsibilities, challenges, and best practices associated with test management in SAP projects. Additionally, the webinar delves into the significance of heatmaps as a visual aid for identifying testing priorities, areas of risk, and resource allocation within SAP landscapes. Through this session, attendees can expect to enhance their understanding of test management principles while learning practical approaches to optimize testing processes in SAP environments using heatmap visualization techniques
What will you get from this session?
1. Insights into SAP testing best practices
2. Heatmap utilization for testing
3. Optimization of testing processes
4. Demo
Topics covered:
Execution from the test manager
Orchestrator execution result
Defect reporting
SAP heatmap example with demo
Speaker:
Deepak Rai, Automation Practice Lead, Boundaryless Group and UiPath MVP
Encryption in Microsoft 365 - ExpertsLive Netherlands 2024Albert Hoitingh
In this session I delve into the encryption technology used in Microsoft 365 and Microsoft Purview. Including the concepts of Customer Key and Double Key Encryption.
Builder.ai Founder Sachin Dev Duggal's Strategic Approach to Create an Innova...Ramesh Iyer
In today's fast-changing business world, Companies that adapt and embrace new ideas often need help to keep up with the competition. However, fostering a culture of innovation takes much work. It takes vision, leadership and willingness to take risks in the right proportion. Sachin Dev Duggal, co-founder of Builder.ai, has perfected the art of this balance, creating a company culture where creativity and growth are nurtured at each stage.
Software Delivery At the Speed of AI: Inflectra Invests In AI-Powered QualityInflectra
In this insightful webinar, Inflectra explores how artificial intelligence (AI) is transforming software development and testing. Discover how AI-powered tools are revolutionizing every stage of the software development lifecycle (SDLC), from design and prototyping to testing, deployment, and monitoring.
Learn about:
• The Future of Testing: How AI is shifting testing towards verification, analysis, and higher-level skills, while reducing repetitive tasks.
• Test Automation: How AI-powered test case generation, optimization, and self-healing tests are making testing more efficient and effective.
• Visual Testing: Explore the emerging capabilities of AI in visual testing and how it's set to revolutionize UI verification.
• Inflectra's AI Solutions: See demonstrations of Inflectra's cutting-edge AI tools like the ChatGPT plugin and Azure Open AI platform, designed to streamline your testing process.
Whether you're a developer, tester, or QA professional, this webinar will give you valuable insights into how AI is shaping the future of software delivery.
Kubernetes & AI - Beauty and the Beast !?! @KCD Istanbul 2024Tobias Schneck
As AI technology is pushing into IT I was wondering myself, as an “infrastructure container kubernetes guy”, how get this fancy AI technology get managed from an infrastructure operational view? Is it possible to apply our lovely cloud native principals as well? What benefit’s both technologies could bring to each other?
Let me take this questions and provide you a short journey through existing deployment models and use cases for AI software. On practical examples, we discuss what cloud/on-premise strategy we may need for applying it to our own infrastructure to get it to work from an enterprise perspective. I want to give an overview about infrastructure requirements and technologies, what could be beneficial or limiting your AI use cases in an enterprise environment. An interactive Demo will give you some insides, what approaches I got already working for real.
Epistemic Interaction - tuning interfaces to provide information for AI supportAlan Dix
Paper presented at SYNERGY workshop at AVI 2024, Genoa, Italy. 3rd June 2024
https://alandix.com/academic/papers/synergy2024-epistemic/
As machine learning integrates deeper into human-computer interactions, the concept of epistemic interaction emerges, aiming to refine these interactions to enhance system adaptability. This approach encourages minor, intentional adjustments in user behaviour to enrich the data available for system learning. This paper introduces epistemic interaction within the context of human-system communication, illustrating how deliberate interaction design can improve system understanding and adaptation. Through concrete examples, we demonstrate the potential of epistemic interaction to significantly advance human-computer interaction by leveraging intuitive human communication strategies to inform system design and functionality, offering a novel pathway for enriching user-system engagements.
Neuro-symbolic is not enough, we need neuro-*semantic*Frank van Harmelen
Neuro-symbolic (NeSy) AI is on the rise. However, simply machine learning on just any symbolic structure is not sufficient to really harvest the gains of NeSy. These will only be gained when the symbolic structures have an actual semantics. I give an operational definition of semantics as “predictable inference”.
All of this illustrated with link prediction over knowledge graphs, but the argument is general.
JMeter webinar - integration with InfluxDB and GrafanaRTTS
Watch this recorded webinar about real-time monitoring of application performance. See how to integrate Apache JMeter, the open-source leader in performance testing, with InfluxDB, the open-source time-series database, and Grafana, the open-source analytics and visualization application.
In this webinar, we will review the benefits of leveraging InfluxDB and Grafana when executing load tests and demonstrate how these tools are used to visualize performance metrics.
Length: 30 minutes
Session Overview
-------------------------------------------
During this webinar, we will cover the following topics while demonstrating the integrations of JMeter, InfluxDB and Grafana:
- What out-of-the-box solutions are available for real-time monitoring JMeter tests?
- What are the benefits of integrating InfluxDB and Grafana into the load testing stack?
- Which features are provided by Grafana?
- Demonstration of InfluxDB and Grafana using a practice web application
To view the webinar recording, go to:
https://www.rttsweb.com/jmeter-integration-webinar
JMeter webinar - integration with InfluxDB and Grafana
Lists
1. CHAPTER 4
LISTS
All the programs in this file are selected from
Ellis Horowitz, Sartaj Sahni, and Susan Anderson-Freed
“Fundamentals of Data Structures in C”,
Computer Science Press, 1992.
CHAPTER 4 1
2. Introduction
Array
successive items locate a fixed distance
disadvantage
• data movements during insertion and deletion
• waste space in storing n ordered lists of varying
size
possible solution
• linked list
CHAPTER 4 2
3. Pointer Review (1)
Pointer Can Be Dangerous
int i, *pi;
1000 2000
i ? pi ?
pi = &i;
i 1000 2000
*pi ? pi 1000
i = 10 or *pi = 10
i 1000 2000
*pi 10 pi 1000
CHAPTER 4 3
5. 4.1.2 Using Dynamically Allocated Storage
int i, *pi;
float f, *pf;
pi = (int *) malloc(sizeof(int));
request memory
pf = (float *) malloc (sizeof(float));
*pi =1024;
*pf =3.14;
printf(”an integer = %d, a float = %fn”, *pi, *pf);
free(pi);
return memory
free(pf);
*Program4.1:Allocation and deallocation of pointers (p.138)
CHAPTER 4 5
6. 4.2 SINGLY LINKED LISTS
bat cat sat vat NULL
*Figure 4.1: Usual way to draw a linked list (p.139)
CHAPTER 4 6
7. Insertion
bat cat sat vat NULL
mat
*Figure 4.2: Insert mat after cat (p.140)
CHAPTER 4 7
8. bat cat mat sat vat NULL
dangling
reference
*Figure 4.3: Delete mat from list (p.140)
CHAPTER 4 8
9. Example 4.1: create a linked list of words
Declaration
typedef struct list_node, *list_pointer;
typedef struct list_node {
char data [4];
list_pointer link;
};
Creation
list_pointer ptr =NULL;
Testing
#define IS_EMPTY(ptr) (!(ptr))
Allocation
ptr=(list_pointer) malloc (sizeof(list_node));
CHAPTER 4 9
10. e -> name (*e).name
strcpy(ptr -> data, “bat”);
ptr -> link = NULL;
address of ptr data ptr link
first node
b a t 0 NULL
ptr
*Figure 4.4:Referencing the fields of a node(p.142)
CHAPTER 4 10
11. Example: create a two-node list
ptr
10 20 NULL
typedef struct list_node *list_pointer;
typedef struct list_node {
int data;
list_pointer link;
};
list_pointer ptr =NULL
Example 4.2: (p.142)
CHAPTER 4 11
12. list_pointer create2( )
{
/* create a linked list with two nodes */
list_pointer first, second;
first = (list_pointer) malloc(sizeof(list_node));
second = ( list_pointer) malloc(sizeof(list_node));
second -> link = NULL;
second -> data = 20;
ptr
first -> data = 10;
first ->link = second;
return first; 10 20 NULL
} *Program 4.2:Create a tow-node list
(p.143)
CHAPTER 4 12
13. List Insertion:
Insert a node after a specific node
void insert(list_pointer *ptr, list_pointer node)
{
/* insert a new node with data = 50 into the list ptr after node */
list_pointer temp;
temp = (list_pointer) malloc(sizeof(list_node));
if (IS_FULL(temp)){
fprintf(stderr, “The memory is fulln”);
exit (1);
}
CHAPTER 4 13
14. temp->data = 50;
if (*ptr) { noempty list
temp->link =node ->link;
node->link = temp;
ptr
}
else { empty list 10 20 NULL
temp->link = NULL; node
*ptr =temp;
} 50
}
temp
*Program 4.3:Simple insert into front of list (p.144)
CHAPTER 4 14
15. List Deletion
Delete the first node.
ptr trail node ptr
10 50 20 NULL 50 20 NULL
(a) before deletion (b)after deletion
Delete node other than the first node.
ptr trail node ptr
10 50 20 NULL 10 20 NULL
CHAPTER 4 15
16. void delete(list_pointer *ptr, list_pointer trail, list_pointer
node)
{
/* delete node from the list, trail is the preceding node
ptr is the head of the list */
trail node
if (trail)
trail->link = node->link; 10 50 20 NULL
else
*ptr = (*ptr) ->link; 10 20 NULL
free(node);
}
ptr node ptr
10 50 20 NULL 50 20 NULL
CHAPTER 4 16
17. Print out a list (traverse a list)
void print_list(list_pointer ptr)
{
printf(“The list ocntains: “);
for ( ; ptr; ptr = ptr->link)
printf(“%4d”, ptr->data);
printf(“n”);
}
*Program 4.5: Printing a list (p.146)
CHAPTER 4 17
18. 4.3 DYNAMICALLY LINKED STACKS AND QUEUES
top
element link
NULL
(a) Linked Stack
front rear
element link
NULL
(b) Linked queue
*Figure 4.10: Linked Stack and queue (p.147)
CHAPTER 4 18
19. Represent n stacks
#define MAX_STACKS 10 /* maximum number of
stacks */
typedef struct {
int key;
/* other fields */
} element;
typedef struct stack *stack_pointer;
typedef struct stack {
element item;
stack_pointer link;
};
stack_pointer top[MAX_STACKS];
CHAPTER 4 19
20. Represent n queues
#define MAX_QUEUES 10 /* maximum number of
queues */
typedef struct queue *queue_pointer;
typedef struct queue {
element item;
queue_pointer link;
};
queue_pointer front[MAX_QUEUE],
rear[MAX_QUEUES];
CHAPTER 4 20
21. Push in the linked stack
void add(stack_pointer *top, element item)
{
/* add an element to the top of the stack */
stack_pointer temp =
(stack_pointer) malloc (sizeof (stack));
if (IS_FULL(temp)) {
fprintf(stderr, “ The memory is fulln”);
exit(1);
}
temp->item = item;
temp->link = *top;
*top= temp; * Program 4.6:Add to a linked stack
(p.149)
} CHAPTER 4 21
22. pop from the linked stack
element delete(stack_pointer *top) {
/* delete an element from the stack */
stack_pointer temp = *top;
element item;
if (IS_EMPTY(temp)) {
fprintf(stderr, “The stack is emptyn”);
exit(1);
}
item = temp->item;
*top = temp->link;
free(temp);
return item;
}
*Program 4.7: Delete from a linked stack (p.149)
CHAPTER 4 22
23. enqueue in the linked queue
void addq(queue_pointer *front, queue_pointer *rear, element item)
{ /* add an element to the rear of the queue */
queue_pointer temp =
(queue_pointer) malloc(sizeof (queue));
if (IS_FULL(temp)) {
fprintf(stderr, “ The memory is fulln”);
exit(1);
}
temp->item = item;
temp->link = NULL;
if (*front) (*rear) -> link = temp;
else *front = temp;
*rear = temp; } CHAPTER 4 23
24. dequeue from the linked queue (similar to push)
element deleteq(queue_pointer *front) {
/* delete an element from the queue */
queue_pointer temp = *front;
element item;
if (IS_EMPTY(*front)) {
fprintf(stderr, “The queue is emptyn”);
exit(1);
}
item = temp->item;
*front = temp->link;
free(temp);
return item;
}
CHAPTER 4 24
25. Polynomials
A( x ) =a m − x e
1
m−1
+ m− x e
a 2 m−2
+ + 0xe
... a 0
Representation
typedef struct poly_node *poly_pointer;
typedef struct poly_node {
int coef;
int expon;
poly_pointer link;
};
poly_pointer a, b, c;
coef expon link
CHAPTER 4 25
26. Examples
a = 3x + 2 x +1
14 8
a
3 14 2 8 1 0 null
b = 8 x 14 − 3x 10 +10 x 6
b
8 14 -3 10 10 6 null
CHAPTER 4 26
27. Adding Polynomials
3 14 2 8 1 0
a
8 14 -3 10 10 6
b
11 14 a->expon == b->expon
d
3 14 2 8 1 0
a
8 14 -3 10 10 6
b
11 14 -3 10 a->expon < b->expon
d
CHAPTER 4 27
29. Alogrithm for Adding Polynomials
poly_pointer padd(poly_pointer a, poly_pointer b)
{
poly_pointer front, rear, temp;
int sum;
rear =(poly_pointer)malloc(sizeof(poly_node));
if (IS_FULL(rear)) {
fprintf(stderr, “The memory is fulln”);
exit(1);
}
front = rear;
while (a && b) {
switch (COMPARE(a->expon, b->expon)) {
CHAPTER 4 29
30. case -1: /* a->expon < b->expon */
attach(b->coef, b->expon, &rear);
b= b->link;
break;
case 0: /* a->expon == b->expon */
sum = a->coef + b->coef;
if (sum) attach(sum,a->expon,&rear);
a = a->link; b = b->link;
break;
case 1: /* a->expon > b->expon */
attach(a->coef, a->expon, &rear);
a = a->link;
}
}
for (; a; a = a->link)
attach(a->coef, a->expon, &rear);
for (; b; b=b->link)
attach(b->coef, b->expon, &rear);
rear->link = NULL;
temp = front; front = front->link; free(temp);
return front;
}
Delete extra initial node.
CHAPTER 4 30
31. Analysis
(1) coefficient additions
0 ≤ additions ≤ min(m, n)
where m (n) denotes the number of terms in A (B).
(2) exponent comparisons
extreme case
em-1 > fm-1 > em-2 > fm-2 > … > e0 > f0
m+n-1 comparisons
(3) creation of new nodes
extreme case
m + n new nodes
summary O(m+n)
CHAPTER 4 31
32. Attach a Term
void attach(float coefficient, int exponent,
poly_pointer *ptr)
{
/* create a new node attaching to the node pointed to
by ptr. ptr is updated to point to this new node. */
poly_pointer temp;
temp = (poly_pointer) malloc(sizeof(poly_node));
if (IS_FULL(temp)) {
fprintf(stderr, “The memory is fulln”);
exit(1);
}
temp->coef = coefficient;
temp->expon = exponent;
(*ptr)->link = temp;
*ptr = temp;
}
CHAPTER 4 32
33. A Suite for Polynomials
e(x) = a(x) * b(x) + d(x)
poly_pointer a, b, d, e; read_poly()
...
a = read_poly(); print_poly()
b = read_poly(); padd()
d = read_poly();
temp = pmult(a, b); psub()
e = padd(temp, d); pmult()
print_poly(e);
temp is used to hold a partial result.
By returning the nodes of temp, we
may use it to hold other polynomials
CHAPTER 4 33
34. Erase Polynomials
void earse(poly_pointer *ptr)
{
/* erase the polynomial pointed to by ptr */
poly_pointer temp;
while (*ptr) {
temp = *ptr;
*ptr = (*ptr)->link;
free(temp);
}
}
O(n)
CHAPTER 4 34
36. Maintain an Available List
poly_pointer get_node(void)
{
poly_pointer node;
if (avail) {
node = avail;
avail = avail->link:
}
else {
node = (poly_pointer)malloc(sizeof(poly_node));
if (IS_FULL(node)) {
printf(stderr, “The memory is fulln”);
exit(1);
}
}
return node;
}
CHAPTER 4 36
37. Maintain an Available List (Continued)
Insert ptr to the front of this list
void ret_node(poly_pointer ptr)
{
ptr->link = avail;
avail = ptr;
} Erase a circular list (see next page)
void cerase(poly_pointer *ptr)
{
poly_pointer temp;
if (*ptr) {
temp = (*ptr)->link;
(*ptr)->link = avail; (1)
avail = temp; (2)
*ptr = NULL;
}
}
Independent of # of nodes in a list O(1) constant time
CHAPTER 4 37
38. 4.4.4 Representing Polynomials As Circularly Linked Lists
avail
(2)
ptr
(1) temp
NULL
avail
*Figure 4.14: Returning a circular list to the avail list (p.159)
CHAPTER 4 38
39. Head Node
Represent polynomial as circular list.
(1) zero
a -1
Zero polynomial
(2) others
a
-1 3 14 2 8 1 0
a = 3x + 2 x +1
14 8
CHAPTER 4 39
40. Another Padd
poly_pointer cpadd(poly_pointer a, poly_pointer b)
{
poly_pointer starta, d, lastd;
int sum, done = FALSE;
starta = a;
a = a->link; Set expon field of head node to -1.
b = b->link;
d = get_node();
d->expon = -1; lastd = d;
do {
switch (COMPARE(a->expon, b->expon)) {
case -1: attach(b->coef, b->expon, &lastd);
b = b->link;
break;
CHAPTER 4 40
41. Another Padd (Continued)
case 0: if (starta == a) done = TRUE;
else {
sum = a->coef + b->coef;
if (sum) attach(sum,a->expon,&lastd);
a = a->link; b = b->link;
}
break;
case 1: attach(a->coef,a->expon,&lastd);
a = a->link;
}
} while (!done);
lastd->link = d;
return d; Link last node to
} first
CHAPTER 4 41
42. Additional List Operations
typedef struct list_node *list_pointer;
typedef struct list_node {
char data;
list_pointer link;
};
Invert single linked lists
Concatenate two linked lists
CHAPTER 4 42
43. Invert Single Linked Lists
Use two extra pointers: middle and trail.
list_pointer invert(list_pointer lead)
{
list_pointer middle, trail;
middle = NULL;
while (lead) {
trail = middle;
middle = lead;
lead = lead->link;
middle->link = trail;
}
return middle;
}
0: null
1: lead ...
≥2: lead
CHAPTER 4 43
44. Concatenate Two Lists
list_pointer concatenate(list_pointer
ptr1, list_pointer ptr2)
{
list_pointer temp;
if (IS_EMPTY(ptr1)) return ptr2;
else {
if (!IS_EMPTY(ptr2)) {
for (temp=ptr1;temp->link;temp=temp->link);
temp->link = ptr2;
}
return ptr1;
}
}
O(m) where m is # of elements in the first list
CHAPTER 4 44
45. 4.5.2 Operations For Circularly Linked List
What happens when we insert a node to the front of a circular
linked list?
X1 X2 X3
a
Problem: move down the whole list.
*Figure 4.16: Example circular list (p.165)
CHAPTER 4 45
46. A possible solution:
X1 X2 X3
a
Note a pointer points to the last node.
*Figure 4.17: Pointing to the last node of a circular list (p.165)
CHAPTER 4 46
48. Length of Linked List
int length(list_pointer ptr)
{
list_pointer temp;
int count = 0;
if (ptr) {
temp = ptr;
do {
count++;
temp = temp->link;
} while (temp!=ptr);
}
return count;
}
CHAPTER 4 48
49. Equivalence Relations
A relation over a set, S, is said to be an equivalence relation over S
iff it is symmertric( 对称 ), reflexive( 自反 ), and transitive( 传递 )
over S.
reflexive, x=x
symmetric, if x=y, then y=x
transitive, if x=y and y=z, then x=z
CHAPTER 4 49
51. A Rough Algorithm to
Find Equivalence Classes
void equivalenec()
{
initialize;
while (there are more pairs) {
Phase 1
read the next pair <i,j>;
process this pair;
}
initialize the output;
do {
output a new equivalence class;
Phase 2
} while (not done);
}
What kinds of data structures are adopted?
CHAPTER 4 51
52. First Refinement
#include <stdio.h>
#include <alloc.h>
#define MAX_SIZE 24
#define IS_FULL(ptr) (!(ptr))
#define FALSE 0
#define TRUE 1
void equivalence()
{
initialize seq to NULL and out to TRUE
while (there are more pairs) {
read the next pair, <i,j>;
put j on the seq[i] list;
put i on the seq[j] list;
}
for (i=0; i<n; i++)
if (out[i]) { direct equivalence
out[i]= FALSE;
output this equivalence class;
}
} Compute indirect equivalence
using transitivity
CHAPTER 4 52
54. Final Version for
Finding Equivalence Classes
void main(void)
{
short int out[MAX_SIZE];
node_pointer seq[MAX_SIZE];
node_pointer x, y, top;
int i, j, n;
printf(“Enter the size (<= %d) “, MAX_SIZE);
scanf(“%d”, &n);
for (i=0; i<n; i++) {
out[i]= TRUE; seq[i]= NULL;
}
printf(“Enter a pair of numbers (-1 -1 to quit): “);
scanf(“%d%d”, &i, &j);
Phase 1: input the equivalence pairs:
CHAPTER 4 54
55. while (i>=0) {
x = (node_pointer) malloc(sizeof(node));
if (IS_FULL(x))
fprintf(stderr, “memory is fulln”);
exit(1);
} Insert x to the top of lists seq[i]
x->data= j; x->link= seq[i]; seq[i]= x;
if (IS_FULL(x))
fprintf(stderr, “memory is fulln”);
exit(1);
} Insert x to the top of lists seq[j]
x->data= i; x->link= seq[j]; seq[j]= x;
printf(“Enter a pair of numbers (-1 -1 to
quit): “);
scanf(“%d%d”, &i, &j);
}
CHAPTER 4 55
56. Phase 2: output the equivalence classes
for (i=0; i<n; i++) {
if (out[i]) {
printf(“nNew class: %5d”, i);
out[i]= FALSE;
x = seq[i]; top = NULL;
for (;;) {
while (x) {
j = x->data;
if (out[j]) {
printf(“%5d”, j); push
out[j] = FALSE;
y = x->link; x->link = top;
top = x; x = y;
}
else x = x->link;
}
if (!top) break; pop
x = seq[top->data]; top = top->link;
}
}
}
CHAPTER 4 56
57. 4.7 Sparse Matrices
0 0 11 0
12 0 0 0
0 −4 0 0
0 0 0 − 15
inadequates of sequential schemes
(1) # of nonzero terms will vary after some matrix computation
(2) matrix just represents intermediate results
new scheme
Each column (row): a circular linked list with a head node
CHAPTER 4 57
58. Revisit Sparse Matrices
# of head nodes = max{# of rows, # of columns}
連 down head right 連同一列元素
head node
同
next
一
行 down entry row col right
entry node
元
value
素
entry i j
aij
aij
CHAPTER 4 58
63. Read in a Matrix
matrix_pointer mread(void)
{
/* read in a matrix and set up its linked
list. An global array hdnode is used */
int num_rows, num_cols, num_terms;
int num_heads, i;
int row, col, value, current_row;
matrix_pointer temp, last, node;
printf(“Enter the number of rows, columns
and number of nonzero terms: “);
CHAPTER 4 63
64. scanf(“%d%d%d”, &num_rows, &num_cols,
&num_terms);
num_heads =
(num_cols>num_rows)? num_cols : num_rows;
/* set up head node for the list of head
nodes */
node = new_node(); node->tag = entry;
node->u.entry.row = num_rows;
node->u.entry.col = num_cols;
if (!num_heads) node->right = node;
else { /* initialize the head nodes */
for (i=0; i<num_heads; i++) {
term= new_node();
hdnode[i] = temp;
hdnode[i]->tag = head; O(max(n,m))
hdnode[i]->right = temp;
hdnode[i]->u.next = temp;
}
CHAPTER 4 64
65. current_row= 0; last= hdnode[0];
for (i=0; i<num_terms; i++) {
printf(“Enter row, column and value:”);
scanf(“%d%d%d”, &row, &col, &value);
if (row>current_row) {
last->right= hdnode[current_row];
current_row= row; last=hdnode[row];
} ...
temp = new_node();
temp->tag=entry; temp->u.entry.row=row;
temp->u.entry.col = col;
temp->u.entry.value = value;
last->right = temp;/*link to row list */
last= temp;
/* link to column list */
hdnode[col]->u.next->down = temp;
hdnode[col]=>u.next = temp;
}
利用 next field 存放 column 的 last node
CHAPTER 4 65
66. /*close last row */
last->right = hdnode[current_row];
/* close all column lists */
for (i=0; i<num_cols; i++)
hdnode[i]->u.next->down = hdnode[i];
/* link all head nodes together */
for (i=0; i<num_heads-1; i++)
hdnode[i]->u.next = hdnode[i+1];
hdnode[num_heads-1]->u.next= node;
node->right = hdnode[0];
}
return node;
}
O(max{#_rows, #_cols}+#_terms)
CHAPTER 4 66
67. Write out a Matrix
void mwrite(matrix_pointer node)
{ /* print out the matrix in row major form */
int i;
matrix_pointer temp, head = node->right;
printf(“n num_rows = %d, num_cols= %dn”,
node->u.entry.row,node->u.entry.col);
printf(“The matrix by row, column, and
value:nn”); O(#_rows+#_terms)
for (i=0; i<node->u.entry.row; i++) {
for (temp=head->right;temp!=head;temp=temp->right)
printf(“%5d%5d%5dn”, temp->u.entry.row,
temp->u.entry.col, temp->u.entry.value);
head= head->u.next; /* next row */
}
}
CHAPTER 4 67
68. Free the entry and head nodes by row.
Erase a Matrix
void merase(matrix_pointer *node)
{
int i, num_heads;
matrix_pointer x, y, head = (*node)->right;
for (i=0; i<(*node)->u.entry.row; i++) {
y=head->right;
while (y!=head) {
x = y; y = y->right; free(x);
}
x= head; head= head->u.next; free(x);
}
y = head;
while (y!=*node) {
x = y; y = y->u.next; free(x);
}
free(*node); *node = NULL;
}
O(#_rows+#_cols+#_terms)
Alternative: 利用 Fig 4.14 的技巧,把一列資料 erase (constant time)
CHAPTER 4 68
69. 4.8 Doubly Linked List
Move in forward and backward direction.
Singly linked list (in one direction only)
How to get the preceding node during deletion or insertion?
Using 2 pointers
Node in doubly linked list consists of:
1. left link field (llink)
2. data field (item)
3. right link field (rlink)
CHAPTER 4 69