Example of Dynamic Programming to achieve O(n).
Interview question: N houses, each with weighting on value of goods. How can a burglar maximise profit if they are not allowed to visit neighbouring houses?
Solution: optimum(i) = max( optimum(i-1)-w(i-1)+w(i) , optimum(i-2)+w(i) )
Numerical solution of ordinary differential equations by using Runge-Kutta Method of Order Two and Runge-Kutta Method of Order Four
How to write the C++ codes?
Numerical solution of ordinary differential equations by using Runge-Kutta Method of Order Two and Runge-Kutta Method of Order Four
How to write the C++ codes?
Notes for C++ Programming / Object Oriented C++ Programming for MCA, BCA and ...ssuserd6b1fd
C++ programming language notes for beginners and Collage students. Written for beginners. Colored graphics. Function by Function explanation with complete examples. Well commented examples. Illustrations are made available for data dealing at memory level.
03 of 03 parts
Get Part 1 from https://www.slideshare.net/ArunUmrao/notes-for-c-programming-for-bca-mca-b-sc-msc-be-amp-btech-1st-year-1
Get Part 2 from https://www.slideshare.net/ArunUmrao/notes-for-c-programming-for-bca-mca-b-sc-msc-be-amp-btech-1st-year-2
C is a general-purpose, procedural computer programming language supporting structured programming, lexical variable scope, and recursion, while a static type system prevents unintended operations. C provides constructs that map efficiently to typical machine instructions and has found lasting use in applications previously coded in assembly language. Such applications include operating systems and various application software for computers, from supercomputers to PLCs and embedded system.
In this talk, we will review an experience of rearchitecting and migrating a system that appeared reactive and microservice-based, but was in fact a monolith with RPC calls to a truly reactive architecture.
The migration work had to be done without causing disruption to the current system, and without taking time to rewrite the system. The result is a biometric computer vision system with a distributed domain in Akka / Scala with storage in Apache Cassandra, with the computer vision components in OpenCV in C++, connected with RabbitMQ and with batch analytics code in Apache Spark.
This talk will show the architectural and code smells that were the result of half-harted reactive implementation and the way to address them, but also the impact of the changes on privacy and security of the stored biometric information.
Notes for C Programming for MCA, BCA, B. Tech CSE, ECE and MSC (CS) 4 of 5 by...ssuserd6b1fd
C programming language notes for beginners and Collage students. Written for beginners. Colored graphics. Function by Function explanation with complete examples. Well commented examples. Illustrations are made available for data dealing at memory level.
02 of 03 parts
Get Part 1 from https://www.slideshare.net/ArunUmrao/notes-for-c-programming-for-bca-mca-b-sc-msc-be-amp-btech-1st-year-1
Get Part 3 from https://www.slideshare.net/ArunUmrao/notes-for-c-programming-for-bca-mca-b-sc-msc-be-amp-btech-1st-year-3
C is a general-purpose, procedural computer programming language supporting structured programming, lexical variable scope, and recursion, while a static type system prevents unintended operations. C provides constructs that map efficiently to typical machine instructions and has found lasting use in applications previously coded in assembly language. Such applications include operating systems and various application software for computers, from supercomputers to PLCs and embedded system.
Assg 12 Binary Search Trees COSC 2336assg-12.cppAssg 12 Binary .docxfestockton
Assg 12 Binary Search Trees COSC 2336/assg-12.cppAssg 12 Binary Search Trees COSC 2336/assg-12.cpp/**
* @author Jane Programmer
* @cwid 123 45 678
* @class COSC 2336, Spring 2019
* @ide Visual Studio Community 2017
* @date April 8, 2019
* @assg Assignment 12
*
* @description Assignment 12 Binary Search Trees
*/
#include<cassert>
#include<iostream>
#include"BinaryTree.hpp"
usingnamespace std;
/** main
* The main entry point for this program. Execution of this program
* will begin with this main function.
*
* @param argc The command line argument count which is the number of
* command line arguments provided by user when they started
* the program.
* @param argv The command line arguments, an array of character
* arrays.
*
* @returns An int value indicating program exit status. Usually 0
* is returned to indicate normal exit and a non-zero value
* is returned to indicate an error condition.
*/
int main(int argc,char** argv)
{
// -----------------------------------------------------------------------
cout <<"--------------- testing BinaryTree construction ----------------"<< endl;
BinaryTree t;
cout <<"<constructor> Size of new empty tree: "<< t.size()<< endl;
cout << t << endl;
assert(t.size()==0);
cout << endl;
// -----------------------------------------------------------------------
cout <<"--------------- testing BinaryTree insertion -------------------"<< endl;
//t.insert(10);
//cout << "<insert> Inserted into empty tree, size: " << t.size() << endl;
//cout << t << endl;
//assert(t.size() == 1);
//t.insert(3);
//t.insert(7);
//t.insert(12);
//t.insert(15);
//t.insert(2);
//cout << "<insert> inserted 5 more items, size: " << t.size() << endl;
//cout << t << endl;
//assert(t.size() == 6);
cout << endl;
// -----------------------------------------------------------------------
cout <<"--------------- testing BinaryTree height -------------------"<< endl;
//cout << "<height> Current tree height: " << t.height() << endl;
//assert(t.height() == 3);
// increase height by 2
//t.insert(4);
//t.insert(5);
//cout << "<height> after inserting nodes, height: " << t.height()
// << " size: " << t.size() << endl;
//cout << t << endl;
//assert(t.height() == 5);
//assert(t.size() == 8);
cout << endl;
// -----------------------------------------------------------------------
cout <<"--------------- testing BinaryTree clear -------------------"<< endl;
//t.clear();
//cout << "<clear> after clearing tree, height: " << t.height()
// << " size: " << t.size() << endl;
//cout << t << endl;
//assert(t.size() == 0);
//assert(t.height() == 0);
cout << endl;
// return 0 to indicate successful completion
return0;
}
Assg 12 Binary Search Trees COSC 2336/assg-12.pdf
Assg 12: Binary Search Trees
COSC 2336 Spring 2019
April 10, 2019
Dates:
Due: Sunday April 21, by Midnight
Objectives
� Practice recursive algorithms
� Learn how to construct a binary tree usi ...
Notes for C++ Programming / Object Oriented C++ Programming for MCA, BCA and ...ssuserd6b1fd
C++ programming language notes for beginners and Collage students. Written for beginners. Colored graphics. Function by Function explanation with complete examples. Well commented examples. Illustrations are made available for data dealing at memory level.
03 of 03 parts
Get Part 1 from https://www.slideshare.net/ArunUmrao/notes-for-c-programming-for-bca-mca-b-sc-msc-be-amp-btech-1st-year-1
Get Part 2 from https://www.slideshare.net/ArunUmrao/notes-for-c-programming-for-bca-mca-b-sc-msc-be-amp-btech-1st-year-2
C is a general-purpose, procedural computer programming language supporting structured programming, lexical variable scope, and recursion, while a static type system prevents unintended operations. C provides constructs that map efficiently to typical machine instructions and has found lasting use in applications previously coded in assembly language. Such applications include operating systems and various application software for computers, from supercomputers to PLCs and embedded system.
In this talk, we will review an experience of rearchitecting and migrating a system that appeared reactive and microservice-based, but was in fact a monolith with RPC calls to a truly reactive architecture.
The migration work had to be done without causing disruption to the current system, and without taking time to rewrite the system. The result is a biometric computer vision system with a distributed domain in Akka / Scala with storage in Apache Cassandra, with the computer vision components in OpenCV in C++, connected with RabbitMQ and with batch analytics code in Apache Spark.
This talk will show the architectural and code smells that were the result of half-harted reactive implementation and the way to address them, but also the impact of the changes on privacy and security of the stored biometric information.
Notes for C Programming for MCA, BCA, B. Tech CSE, ECE and MSC (CS) 4 of 5 by...ssuserd6b1fd
C programming language notes for beginners and Collage students. Written for beginners. Colored graphics. Function by Function explanation with complete examples. Well commented examples. Illustrations are made available for data dealing at memory level.
02 of 03 parts
Get Part 1 from https://www.slideshare.net/ArunUmrao/notes-for-c-programming-for-bca-mca-b-sc-msc-be-amp-btech-1st-year-1
Get Part 3 from https://www.slideshare.net/ArunUmrao/notes-for-c-programming-for-bca-mca-b-sc-msc-be-amp-btech-1st-year-3
C is a general-purpose, procedural computer programming language supporting structured programming, lexical variable scope, and recursion, while a static type system prevents unintended operations. C provides constructs that map efficiently to typical machine instructions and has found lasting use in applications previously coded in assembly language. Such applications include operating systems and various application software for computers, from supercomputers to PLCs and embedded system.
Assg 12 Binary Search Trees COSC 2336assg-12.cppAssg 12 Binary .docxfestockton
Assg 12 Binary Search Trees COSC 2336/assg-12.cppAssg 12 Binary Search Trees COSC 2336/assg-12.cpp/**
* @author Jane Programmer
* @cwid 123 45 678
* @class COSC 2336, Spring 2019
* @ide Visual Studio Community 2017
* @date April 8, 2019
* @assg Assignment 12
*
* @description Assignment 12 Binary Search Trees
*/
#include<cassert>
#include<iostream>
#include"BinaryTree.hpp"
usingnamespace std;
/** main
* The main entry point for this program. Execution of this program
* will begin with this main function.
*
* @param argc The command line argument count which is the number of
* command line arguments provided by user when they started
* the program.
* @param argv The command line arguments, an array of character
* arrays.
*
* @returns An int value indicating program exit status. Usually 0
* is returned to indicate normal exit and a non-zero value
* is returned to indicate an error condition.
*/
int main(int argc,char** argv)
{
// -----------------------------------------------------------------------
cout <<"--------------- testing BinaryTree construction ----------------"<< endl;
BinaryTree t;
cout <<"<constructor> Size of new empty tree: "<< t.size()<< endl;
cout << t << endl;
assert(t.size()==0);
cout << endl;
// -----------------------------------------------------------------------
cout <<"--------------- testing BinaryTree insertion -------------------"<< endl;
//t.insert(10);
//cout << "<insert> Inserted into empty tree, size: " << t.size() << endl;
//cout << t << endl;
//assert(t.size() == 1);
//t.insert(3);
//t.insert(7);
//t.insert(12);
//t.insert(15);
//t.insert(2);
//cout << "<insert> inserted 5 more items, size: " << t.size() << endl;
//cout << t << endl;
//assert(t.size() == 6);
cout << endl;
// -----------------------------------------------------------------------
cout <<"--------------- testing BinaryTree height -------------------"<< endl;
//cout << "<height> Current tree height: " << t.height() << endl;
//assert(t.height() == 3);
// increase height by 2
//t.insert(4);
//t.insert(5);
//cout << "<height> after inserting nodes, height: " << t.height()
// << " size: " << t.size() << endl;
//cout << t << endl;
//assert(t.height() == 5);
//assert(t.size() == 8);
cout << endl;
// -----------------------------------------------------------------------
cout <<"--------------- testing BinaryTree clear -------------------"<< endl;
//t.clear();
//cout << "<clear> after clearing tree, height: " << t.height()
// << " size: " << t.size() << endl;
//cout << t << endl;
//assert(t.size() == 0);
//assert(t.height() == 0);
cout << endl;
// return 0 to indicate successful completion
return0;
}
Assg 12 Binary Search Trees COSC 2336/assg-12.pdf
Assg 12: Binary Search Trees
COSC 2336 Spring 2019
April 10, 2019
Dates:
Due: Sunday April 21, by Midnight
Objectives
� Practice recursive algorithms
� Learn how to construct a binary tree usi ...
SaveI need help with this maze gui that I wrote in java, I am tryi.pdfarihantstoneart
Save
I need help with this maze gui that I wrote in java, I am trying to find the shortest path but I dont
know how to implement the FindshortestPath Method.
I was given this as a stragedy to solve the findshortestpathMethod():
Create two LinkedLists of MazeMap objects. One will be the list of map objects for cells at
distance x from the center. The other will be the list of map objects for cells at distance x + 1.
Initialize the list at distance x to contain a MazeMap object which refers to the center cell of the
maze. Since this map object refers to the center cell, it does not need to refer to any other
MazeMap object, so its MazeMap reference can be set to null. Set the center cell’s visited field
to true to indicate that this cell has already been added to the map.
Set up a loop to run until the maze entrance is mapped. This loop does the following:
Set up a loop to process each map object from the list at distance x. This loop does the following:
i.Get the cell from the map object
ii.If this cell is the entrance cell, set the entrance and break out of the loop!
iii.Check each direction from this cell and for an accessible neighbor cell
Get a reference to that neighbor cell
If the neighbor cell has not already been visited
Set the cell’s visited field to true so its only processed once
Create a MazeMap object that refers to this neighbor cell
Make this MazeMap object refer to the current map object being processed – ie the cell one step
closer to center
Add this MazeMap object to the list at distance x + 1
List at distance x has been processed. The list at distance x + 1 is now complete, so make the list
at distance x refer to list at distance x + 1. Then create a new empty list for the list at distance x +
1.
At this point, the entrance points at the map object which refers to cell 1,1. That map object also
refers to a map object which is one step closer to the center. Write a loop that traverses this trail
of references until it reaches the end (a null reference). For each map object:
Get the cell referred to.
Call the drawCircle method with a color, the cell’s row and column, and SMALL.
Call the delay method with a SHORT delay. This slows down the drawing process so you can
watch the path being displayed.
Advance to the next map object in the list.
here is the code so far*********
here is my code: ******mazeApp*******
package mazepackage;
import javafx.application.Application;
import javafx.application.Platform;
import javafx.scene.Group;
import javafx.scene.Scene;
import javafx.scene.canvas.Canvas;
import javafx.scene.canvas.GraphicsContext;
import javafx.scene.control.TextInputDialog;
import javafx.scene.paint.Color;
import javafx.stage.Stage;
public class MazeApp extends Application
{
// default canvas size is DEFAULT_SIZE-by-DEFAULT_SIZE
private static final int DEFAULT_SIZE = 768;
private int width = DEFAULT_SIZE;
private int height = DEFAULT_SIZE;
// The graphics context is needed to enable drawing on the canvas
private Graphics.
Help please!!(Include your modified DList.java source code file in.pdfjyothimuppasani1
Help please!!
(Include your modified DList.java source code file in your homework solution zip archive)
Using
whatever Java IDE you prefer, create a project and add DList.java and DListTest.java to it (these
files are provided
in the Week 6 Source zip archive). Modify DList to implement a method that removes all
occurrences of a specific
integer from the list. Here is the pseudocode:
Method removeAll(In: Integer pData) Returns Nothing
Define index variable i and initialize i to 0
While i < the size of this Dlist Do
If get(i) equals pData Then
remove(i)
Else
Increment i
End If
End While
End Method removeAll
Next, modify DListTest() to add test case 21 which tests that removeAll() works correctly.
WHat more do i need to provide to get help on this???
Solution
/* DListNode.java */
/**
* A DListNode is a node in a DList (doubly-linked list).
*/
class DListNode {
/**
* item references the item stored in the current node.
* prev references the previous node in the DList.
* next references the next node in the DList.
*
* DO NOT CHANGE THE FOLLOWING FIELD DECLARATIONS.
*/
public Object item;
protected DListNode prev;
protected DListNode next;
/**
* DListNode() constructor.
* @param i the item to store in the node.
* @param p the node previous to this node.
* @param n the node following this node.
*/
DListNode(Object i, DListNode p, DListNode n) {
item = i;
prev = p;
next = n;
}
}
/* DList.java */
/**
* A DList is a mutable doubly-linked list ADT. Its implementation is
* circularly-linked and employs a sentinel (dummy) node at the head
* of the list.
*
* DO NOT CHANGE ANY METHOD PROTOTYPES IN THIS FILE.
*/
class DList {
/**
* head references the sentinel node.
* size is the number of items in the list. (The sentinel node does not
* store an item.)
*
* DO NOT CHANGE THE FOLLOWING FIELD DECLARATIONS.
*/
protected DListNode head;
protected int size;
/* DList invariants:
* 1) head != null.
* 2) For any DListNode x in a DList, x.next != null.
* 3) For any DListNode x in a DList, x.prev != null.
* 4) For any DListNode x in a DList, if x.next == y, then y.prev == x.
* 5) For any DListNode x in a DList, if x.prev == y, then y.next == x.
* 6) size is the number of DListNodes, NOT COUNTING the sentinel,
* that can be accessed from the sentinel (head) by a sequence of
* \"next\" references.
*/
/**
* newNode() calls the DListNode constructor. Use this class to allocate
* new DListNodes rather than calling the DListNode constructor directly.
* That way, only this method needs to be overridden if a subclass of DList
* wants to use a different kind of node.
* @param item the item to store in the node.
* @param prev the node previous to this node.
* @param next the node following this node.
*/
protected DListNode newNode(Object item, DListNode prev, DListNode next) {
return new DListNode(item, prev, next);
}
/**
* DList() constructor for an empty DList.
*/
public DList() {
// Your solution here.
head = newNode(null, head, head);
head.prev = head;
head.next = head;
si.
need help with code I wrote. This code is a maze gui, and i need hel.pdfarcotstarsports
need help with code I wrote. This code is a maze gui, and i need help with the method
shorterstpath(); here is my code:
******mazeApp*******
package mazepackage;
import javafx.application.Application;
import javafx.application.Platform;
import javafx.scene.Group;
import javafx.scene.Scene;
import javafx.scene.canvas.Canvas;
import javafx.scene.canvas.GraphicsContext;
import javafx.scene.control.TextInputDialog;
import javafx.scene.paint.Color;
import javafx.stage.Stage;
public class MazeApp extends Application
{
// default canvas size is DEFAULT_SIZE-by-DEFAULT_SIZE
private static final int DEFAULT_SIZE = 768;
private int width = DEFAULT_SIZE;
private int height = DEFAULT_SIZE;
// The graphics context is needed to enable drawing on the canvas
private GraphicsContext gc;
// boundary of drawing canvas, 0% border
// private static final double BORDER = 0.05;
private static final double BORDER = 0.00;
private double xmin, ymin, xmax, ymax;
public static void main(String[] args)
{
launch(args);
}
@Override
public void start(Stage primaryStage)
{
Group root = new Group();
Canvas canvas = new Canvas(width, height);
gc = canvas.getGraphicsContext2D();
gc.setLineWidth(2);
gc.setFill(Color.WHITE);
gc.fillRect(0, 0, width, height);
root.getChildren().add(canvas);
TextInputDialog tid = new TextInputDialog();
tid.setTitle(\"Maze Size\");
tid.setHeaderText(\"Enter maze size between 10 and 50\");
tid.showAndWait();
int size = Integer.parseInt(tid.getResult());
if (size > 50)
size = 50;
if (size < 10)
size = 10;
primaryStage.setTitle(\"Maze Application\");
primaryStage.setScene(new Scene(root));
primaryStage.setResizable(false);
// Make sure that the application goes away when then window is closed
primaryStage.setOnCloseRequest(e -> System.exit(0));
primaryStage.show();
Maze maze = new Maze(this, size);
// Must solve the maze in a separate thread or else
// the GUI wont update until the end.....
Thread solver = new Thread(
new Runnable () {
public void run()
{
while(true)
{
maze.buildAndDrawMaze();
maze.findShortestPath();
try
{
Thread.sleep(5000);
}
catch(Exception e) { }
gc.setFill(Color.WHITE);
gc.fillRect(0, 0, width, height);
}
}
});
solver.start();
}
/**
* Sets the pen color to the specified color.
*
* @param color the color to make the pen
*/
public void setPenColor(Color color) {
gc.setStroke(color);
}
/**
* Sets the pen color to the specified color.
*
* @param color the color to make the pen
*/
public void setFillColor(Color color) {
gc.setFill(color);
}
/**
* Sets the x-scale to the specified range.
*
* @param min the minimum value of the x-scale
* @param max the maximum value of the x-scale
* @throws IllegalArgumentException if {@code (max == min)}
*/
public void setXscale(double min, double max) {
double size = max - min;
if (size == 0.0) {
throw new IllegalArgumentException(\"the min and max are the same\");
}
xmin = min - BORDER * size;
xmax = max + BORDER * size;
}
/**
* Sets the y-scale to the specified range.
*
* @param min the minimum valu.
This is the user guide for the quantum simulator I have developed. It includes tutorials on advanced quantum mechanics, quantum algorithms and geometric algebra, together with examples of using the simulator do quantum mechanics, Special Relativity, geometric algebra and quantum computing.
O(n) in time, O(1) in space substr-search algorithm, using O(1)-updatable hash derived using Group Theory. Hash function is greyed out, to prevent IP-theft. Algorithm is far simpler than trie or Knuth-Morris-Pratt. Very surprised it has not been discovered by others.
O(n) in time, O(1) in space substr-search algorithm, using O(1)-updatable hash derived using Group Theory. Hash function is greyed out, to prevent IP-theft. Algorithm is far simpler than trie or Knuth-Morris-Pratt. Very surprised it has not been discovered by others.
O(n) in time, O(1) in space substr-search algorithm, using O(1)-updatable hash derived using Group Theory. Hash function is greyed out, to prevent IP-theft. Algorithm is far simpler than trie or Knuth-Morris-Pratt. Very surprised it has not been discovered by others.
An O(n+m) in time and O(1) in space search-algorithm for substr, length m, in target string, length n. Simpler than trie or Knuth-Morris-Pratt. Can be reduced to O(n) using Group Theoretic O(1)-updatable hash that replaces the cumulative sum, which requires an O(m) verification step because "+" is commutative and cannot distinguish between anagrams: a+b+c = a+c+b = c+a+b etc. I am not uploading the hash, because it means giving away R&D I do in quantum simulations.
This is a roughly O(n) algorithm that generates the kth lexicographically ordered permutation of an n-element array from the integer k. Example, for a three-element
array:
0 --> 0 1 2
1 --> 0 2 1
2 --> 1 0 2
3 --> 1 2 0
4 --> 2 0 1
5 --> 2 1 0
This is a roughly O(n) algorithm that generates the kth lexicographically ordered permutation of an n-element array from the integer k. Example, for a three-element array:
0 --> 0 1 2
1 --> 0 2 1
2 --> 1 0 2
3 --> 1 2 0
4 --> 2 0 1
5 --> 2 1 0
A class that automates conversion from a C++ recursive function to an iterative function. It allow the recursive function to preserve its structure by reproducing the "call stack" on an std::stack. The examples use combinatorics to illustrate usage.
Dirac demo (quantum mechanics with C++). Please note: There is a problem with...Russell Childs
Simple demo of a framework I wrote for doing quantum mechanics in C++. It uses Dirac bras, kets, inner & tensor products, operators and so forth for the linear algebra of QM.
UML design for C++11 written to solve a problem at interview, please also see "Interview C++11 code". The UML design can be zoomed to render it more legible.
Globus Compute wth IRI Workflows - GlobusWorld 2024Globus
As part of the DOE Integrated Research Infrastructure (IRI) program, NERSC at Lawrence Berkeley National Lab and ALCF at Argonne National Lab are working closely with General Atomics on accelerating the computing requirements of the DIII-D experiment. As part of the work the team is investigating ways to speedup the time to solution for many different parts of the DIII-D workflow including how they run jobs on HPC systems. One of these routes is looking at Globus Compute as a way to replace the current method for managing tasks and we describe a brief proof of concept showing how Globus Compute could help to schedule jobs and be a tool to connect compute at different facilities.
Field Employee Tracking System| MiTrack App| Best Employee Tracking Solution|...informapgpstrackings
Keep tabs on your field staff effortlessly with Informap Technology Centre LLC. Real-time tracking, task assignment, and smart features for efficient management. Request a live demo today!
For more details, visit us : https://informapuae.com/field-staff-tracking/
How to Position Your Globus Data Portal for Success Ten Good PracticesGlobus
Science gateways allow science and engineering communities to access shared data, software, computing services, and instruments. Science gateways have gained a lot of traction in the last twenty years, as evidenced by projects such as the Science Gateways Community Institute (SGCI) and the Center of Excellence on Science Gateways (SGX3) in the US, The Australian Research Data Commons (ARDC) and its platforms in Australia, and the projects around Virtual Research Environments in Europe. A few mature frameworks have evolved with their different strengths and foci and have been taken up by a larger community such as the Globus Data Portal, Hubzero, Tapis, and Galaxy. However, even when gateways are built on successful frameworks, they continue to face the challenges of ongoing maintenance costs and how to meet the ever-expanding needs of the community they serve with enhanced features. It is not uncommon that gateways with compelling use cases are nonetheless unable to get past the prototype phase and become a full production service, or if they do, they don't survive more than a couple of years. While there is no guaranteed pathway to success, it seems likely that for any gateway there is a need for a strong community and/or solid funding streams to create and sustain its success. With over twenty years of examples to draw from, this presentation goes into detail for ten factors common to successful and enduring gateways that effectively serve as best practices for any new or developing gateway.
Paketo Buildpacks : la meilleure façon de construire des images OCI? DevopsDa...Anthony Dahanne
Les Buildpacks existent depuis plus de 10 ans ! D’abord, ils étaient utilisés pour détecter et construire une application avant de la déployer sur certains PaaS. Ensuite, nous avons pu créer des images Docker (OCI) avec leur dernière génération, les Cloud Native Buildpacks (CNCF en incubation). Sont-ils une bonne alternative au Dockerfile ? Que sont les buildpacks Paketo ? Quelles communautés les soutiennent et comment ?
Venez le découvrir lors de cette session ignite
AI Pilot Review: The World’s First Virtual Assistant Marketing SuiteGoogle
AI Pilot Review: The World’s First Virtual Assistant Marketing Suite
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AI Pilot Review: Key Features
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✅More than 85 AI features are included in the AI pilot.
✅No setup or configuration; use your voice (like Siri) to do whatever you want.
✅You Can Use AI Pilot To Create your version of AI Pilot And Charge People For It…
✅ZERO Manual Work With AI Pilot. Never write, Design, Or Code Again.
✅ZERO Limits On Features Or Usages
✅Use Our AI-powered Traffic To Get Hundreds Of Customers
✅No Complicated Setup: Get Up And Running In 2 Minutes
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Dynamic programming burglar_problem
1. /**
* @mainpage
* @anchor mainpage
* @brief
* @details
* @copyright Russell John Childs, PhD, 2016
* @author Russell John Childs, PhD
* @date 2016-03-06
*
* This file contains classes: BurglarProblem
* Problem statement: A burglar wishses to maximise the value of goods stolen
* from N houses, subject to the constraint that they can only visit houses
* that are not neighbours.
*
* Solution: Dyanmic programming with complexity O(n) through the fololowing
* recurrence relation:
*
* optimum(i) = max(optimum(i-1)-w(i-1)+w(i), optimum(i-2)+w(i)), where w(i) is
* the weighting for house i and optimum(i) is the optimum solution for the 1st
* i houses.
*
* Algorithm tested against exhaustive search using binary number 1 to 2^(N-1)
* where 1-bts are houses visited, 0-bits are houses skipped and neighbours are
* illegal: E.g. 10010, but not 100110
*
* Compiled and verified under Visual Studio 13
*
* Documentation: Doxygen comments for interfaces, normal for impl.
*
* Usage: After compiling, run. The binary will send:
* n
* (1) test results to stdout
*
* The file unit_test.hpp is required for the tests and must be requested from
* author.
*
* @file dynamic_programming_burglar_problem.cpp
* @see
* @ref mainpage
*/
#include <vector>
#include <utility>
#include <random>
#include <bitset>
#include<algorithm>
#include "unit_test.hpp"
//Unit test framework is written for Linux. This #define ports it to Visual Studio 2013
#define __PRETTY_FUNCTION__ __FUNCSIG__
/**
* addtogroup BurglarProblem
* @{
*/
namespace BurglarProblem
{
class Burglar
{
/**
* This class solves the following problem:
* A burglar wishses to maximise the value of goods stolen
* from N houses, subject to the constraint that they can only visit houses
* that are not neighbours.
*
* It uses Dynamic Programming to achieve O(n) complexity using the recurrence
* relation:
*
* optimum(i) = max(optimum(i-1)-w(i-1)+w(i), optimum(i-2)+w(i)), where w(i) is
* the weighting for house i and optimum(i) is the optimum solution for the 1st
* i houses.
*
*/
private:
2. /**
* Initialiser function that makes all weights positive, sets optimum(0) and
* optimum(1), house list for optimum(0) and house list for optimum(1)
*/
void init(void)
{
//Make all weights positive
unsigned i = 0;
for (auto weight : m_weights)
{
m_weights[i] = std::abs(m_weights[i]);
++i;
}
//Store current and previous max for recurrence relation
auto min_max = std::minmax(m_weights[0], m_weights[1]);
m_max_found[0] = min_max.first;
m_max_found[1] = min_max.second;
//Store current and previous best list of houses for recurrence relation
m_houses[0].clear();
m_houses[0].push_back(m_weights[0] > m_weights[1]);
m_houses[1].clear();
m_houses[1].push_back(m_weights[1] > m_weights[0]);
}
public:
/**
* @param weights {const std::vector<double>&} - new list of weights
*/
Burglar(const std::vector<double>& weights) :
m_weights(weights),
m_which_list(false)
{
init();
}
/**
* dtor
*/
~Burglar(void)
{
}
/**
* This function resets the list of weights used.
* @param weights {const std::vector<double>&} - new list of weights
*
* @return {Burglar& } - *this
*
* Usage my_burgalr_object(new_list).solve();
*/
Burglar& reset(const std::vector<double>& weights)
{
m_weights = weights;
init();
return *this;
}
/**
* This function finds the optimum set of houses to visit, using:
*
* optimum(i) = max(optimum(i-1)-w(i-1)+w(i), optimum(i-2)+w(i)), where w(i) is
* the weighting for house i and optimum(i) is the optimum solution for the 1st
* i houses.
*/
void solve(void)
{
auto size = m_weights.size();
//Loop over houses
for (unsigned i = 2; i < size; ++i)
{
//Get last house in current optimum list
unsigned last = m_houses[1].back();
3. //Check house has value
if (m_weights[i] > 0)
{
//If is not a neighbour of current house
if (i > (last + 1))
{
//Add house value to current value
m_max_found[1] += m_weights[i];
//set prev optimum to current
m_houses[0] = m_houses[1];
//Add house to current optimum
m_houses[1].push_back(i);
}
else
{
//house is a neighbour, find best way to add it:
//(Skip, replace current house, add to prev optimum)
enum Choose{ do_nothing = 0, replace = 1, add = 2 } choose
= do_nothing;
double tmp_max = m_max_found[1];
//Get value if we replace current house in optimum(i-1)
double tmp = m_max_found[1] + (m_weights[i] - m_weights[i - 1]);
if (tmp > tmp_max)
{
tmp_max = tmp;
choose = replace;
}
//Get value if we add house to previous optimum(i-2)
tmp = m_max_found[0] + m_weights[i];
if ((tmp > tmp_max) && (i != (m_houses[0].back()+1)))
{
tmp_max = tmp;
choose = add;
}
//Set new vals for optimum(i-1), optimum(i)
m_max_found[0] = m_max_found[1];
m_max_found[1] = tmp_max;
//Replace optimum(i) house with new house
if (choose == replace)
{
m_houses[0] = m_houses[1];
m_houses[1].back() = i;
}
//Add new house to optimum(i-1)
else if (choose == add)
{
std::vector<unsigned> tmp_list = m_houses[0];
m_houses[0] = m_houses[1];
m_houses[1] = tmp_list;
m_houses[1].push_back(i);
}
}
}
}
}
/**
* This function return optimum value of goods stolen and houses to visit
*
* @return {std::pair<double,std::reference_wrapper<std::vector<unsigned>>>}
* - {optimum value, list of houses}
*/
std::pair<double, std::reference_wrapper<std::vector<unsigned>>>
get_result(void)
{
//Return optimum value and corresponding house list
return std::pair<double, std::reference_wrapper<std::vector<unsigned>>>
(m_max_found[1], m_houses[1]);
}
4. //private:
std::vector<double> m_weights;
bool m_which_list;
double m_max_found[2];
std::vector<unsigned> m_houses[2];
};
}
/**
* @}
*/
/**
* addtogroup Tests
* @{
*/
namespace Tests
{
/**
* Wrapper class for std::vector converting {a, b, c, ...} to "a b c ..."
*/
struct PrintVector
{
PrintVector(const std::vector<unsigned>& vec) :
m_vec(vec)
{
}
std::string str()
{
std::stringstream ss;
for (auto elem : m_vec)
{
ss << elem << " ";
}
return ss.str();
}
std::vector<unsigned> m_vec;
};
/**
* This function compares algorithm against exhaustive search for best
* solution using: binary number 1 to 2^(N-1)
* where 1-bts are houses visited, 0-bits are houses skipped and neighbours are
* illegal: E.g. 10010, but not 100110
*/
void tests(void)
{
using namespace UnitTest;
using namespace BurglarProblem;
//Print test banner
Verify<> banner("Burglar solver");
typedef std::vector<unsigned> uv;
//Two houses with 1st being the optimal choice
Burglar test(std::vector<double>{3, 1});
test.solve();
std::string msg_arg("std::vector<double>{3, 1}");
VERIFY(std::string("Input weights=") + msg_arg,
test.get_result().first) == 3;
VERIFY(std::string("Input weights=") +
msg_arg, PrintVector(test.get_result().second.get()).str()) ==
PrintVector(uv{ 0 }).str();
//Two houses with 2nd being the optimal choice
test.reset(std::vector<double>{2, 4}).solve();
msg_arg = "std::vector<double>{2, 4})";
VERIFY(std::string("Input weights=") + msg_arg,
test.get_result().first) == 4;
VERIFY(std::string("Input weights=") +
msg_arg, PrintVector(test.get_result().second.get()).str()) ==
PrintVector(uv{ 1 }).str();
//Three houses with 1st and 3rd being the optimal choice
5. test.reset(std::vector<double>{4, 9, 6}).solve();
msg_arg = "std::vector<double>{4, 9, 6}";
VERIFY(std::string("Input weights=") + msg_arg,
test.get_result().first) == 10;
VERIFY(std::string("Input weights=") +
msg_arg, PrintVector(test.get_result().second.get()).str()) ==
PrintVector(uv{ 0,2 }).str();
//Three houses with 2nd being the optimal choice
test.reset(std::vector<double>{5, 13, 7});
msg_arg = "std::vector<double>{5, 13, 7}";
VERIFY(std::string("Input weights=") + msg_arg,
test.get_result().first) == 13;
VERIFY(std::string("Input weights=") +
msg_arg, PrintVector(test.get_result().second.get()).str()) ==
PrintVector(uv{ 1 }).str();
//Test proof of recurrence relation. Fourth house not optimal
test.reset(std::vector<double>{5, 10, 7, 1}).solve();
msg_arg = "std::vector<double>{5, 10, 7, 1}";
VERIFY(std::string("Input weights=") + msg_arg,
test.get_result().first) == 12;
VERIFY(std::string("Input weights=") +
msg_arg, PrintVector(test.get_result().second.get()).str()) ==
PrintVector(uv{ 0, 2 }).str();
//Test proof of recurrence relation. Fourth house optimal
test.reset(std::vector<double>{5, 10, 7, 3}).solve();
msg_arg = "std::vector<double>{5, 10, 7, 3}";
VERIFY(std::string("Input weights=") + msg_arg,
test.get_result().first) == 13;
VERIFY(std::string("Input weights=") +
msg_arg, PrintVector(test.get_result().second.get()).str()) ==
PrintVector(uv{ 1, 3 }).str();
unsigned run = 0;
auto exhaustive = [&]()
{
//Test by exhaustive enumeration. Enumeration subject to constraints
//took me a while to deduce. Idea is to use a binary number from
//1 to 2^N, where "1" means visit house and "0" means skip. Discard
//any binary number that violates the constraints (visits 2 neighbours)
//I am not 100% confident this is right and welcome a code review.
//(1) 20 houses with random weights (if you change this, ensure n<64)
const unsigned n = 20;
std::vector<double> random_weights;
std::random_device dev;
//std::mt19937 generator(dev());
std::mt19937 generator(run);
std::uniform_int_distribution<> distribution(1, 1000000);
for (int i = 0; i < n; ++i)
{
random_weights.push_back(distribution(generator));
}
//(2) Generate a binary number from 1 to 2^n
//and treat as house1=visit/skip, house2=visit/skip ...
double exhaustive_search_max = 0;
std::vector<unsigned> exahustive_search_house_list;
for (long long unsigned bin = 0; bin < ((1 << n) - 1); ++bin)
{
//Reverse to represent house1=0/1, house2=0/1, ...
//Happily bitset does this anyway.
std::bitset<n> houses(bin);
//Loop over bits until we find two neighbours
unsigned prev_pos = 0;
bool is_neighbour = false;
unsigned j = 0;
while ((is_neighbour == false) && j < n)
{
is_neighbour = (j != 0) && (houses[prev_pos] && houses[j]);
prev_pos = j;
++j;
}
//if we haven't found any neighbours, we have a legal permutation
if (is_neighbour == false)
6. {
//Get global maximum and houses corresponding to global max
double tmp = 0;
unsigned house_number = 0;
for (auto weight : random_weights)
{
tmp += (weight * houses[house_number]);
++house_number;
}
//If new global max found
if (tmp > exhaustive_search_max)
{
//Update global max
exhaustive_search_max = tmp;
//update house list
exahustive_search_house_list.clear();
for (unsigned index = 0; index < n; ++index)
{
if (houses[index])
{
exahustive_search_house_list.push_back(index);
}
}
}
}
}
//(3) Validate algorithm agaisnt exhaustive search
test.reset(random_weights).solve();
msg_arg = "Exhaustive search with randomw weights: ";
VERIFY(msg_arg, test.get_result().first) == exhaustive_search_max;
VERIFY(msg_arg, PrintVector(uv(test.get_result().second.get())).str()) ==
PrintVector(exahustive_search_house_list).str();
};
//Run 20 exhasutive checks
for (unsigned i = 0; i < 20; ++i)
{
run = i;
std::cout << std::endl << "Run=" << run << std::endl;
exhaustive();
}
}
}
/**
* @}
*/
int main(void)
{
using namespace Tests;
using namespace UnitTest;
//This struct pauses at the end of tests to print out results
struct BreakPointAfterMainExits
{
BreakPointAfterMainExits(void)
{
static BreakPointAfterMainExits tmp;
}
~BreakPointAfterMainExits(void)
{
unsigned set_bp_here_for_test_results = 0;
}
} dummy;
//Run tests
tests();
Verify<Results> results;
}