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minimum matrix values

Nov 18, 2022AustinLeath
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Infection Simulation

Nov 18, 2022AustinLeath

0 likes • 2 views

/*
this program will simulate the spreading of a disease through a
grid of people, starting from a user-defined person. It will count
the number of turns taken before everyone on the grid is immunized
to the disease after having caught it once.
This program will user the SIR model (Susceptible, Infectious, Recovered)
and cellular automata to simulate the people in the grid.
*/
#include <iostream>
using namespace std;
/* Any and all global variables */
const int SIZE = 8; //Size of the square person array
/* Any and all functions */
void gridDefaultify(char[][SIZE], int);
//Purpose: Sets each item in the person array to 's'
//Parameters: A square, two-dimensional array
// The size of that array's bounds
void gridDisplay(char[][SIZE], int);
//Purpose: Formats and prints the information in the person grid
//Parameters: A square, two-dimensional array
// The value of the current day
void nextTurn(char[][SIZE], char[][SIZE], int&);
//Purpose: Updates the grid of people, and the current day
//Parameters: Two square, two-dimensional arrays
// A reference to the current day (so that it can be updated)
int countInfected(char[][SIZE], int);
//Purpose: Counts the number of infectious people on the grid
//Parameters: A square, two-dimensional array
// The size of that array's bounds
int main(){
int currentDay = 0; //Infection begins on day 0, and ends one day after the last person is Recovered
char gridCurrent[SIZE][SIZE]; //Grid of all people
char gridUpdate[SIZE][SIZE]; //Where the user chooses to start the infection
int xToInfect;
int yToInfect; //Set of coordinates for the initial infection position, given by user
//Initializes the grids to all 's'
gridDefaultify(gridCurrent, SIZE);
gridDefaultify(gridUpdate, SIZE);
//The below block gets the initial infection coordinates from the user
cout << "Please enter a location to infect: ";
while(true){
cin >> xToInfect >> yToInfect;
xToInfect--;
yToInfect--;
if(xToInfect < 0 || yToInfect < 0 || xToInfect >= SIZE || yToInfect >= SIZE){
cout << "Those coordinates are outside the bounds of this region." << endl;
cout << "Please enter another location to infect: ";
continue;
} else {
gridCurrent[xToInfect][yToInfect] = 'i';
break;
}
}
//Displays the initial state of the grid
gridDisplay(gridCurrent, currentDay);
//The below block will display and update the grid until the infection is done.
while(true){
nextTurn(gridCurrent, gridUpdate, currentDay);
gridDisplay(gridCurrent, currentDay);
if(countInfected(gridCurrent, SIZE) == 0) break; //Once there are no more infected, the game is done
}
//Displays the number of days taken for the infection to end
cout << "It took " << currentDay + 1 << " days for the outbreak to end";
cout << endl;
return 0;
}
void gridDefaultify(char arr[][SIZE], int arrSize){
for(int x = 0; x < arrSize; x++){
for(int y = 0; y < arrSize; y++){
arr[x][y] = 's'; //Sets all items in the passed-in array to 's'
}
}
return;
}
void gridDisplay(char arr[][SIZE], int day){
cout << "Day " << day << endl; //Prints the current day
for(int x = 0; x < SIZE; x++){
for(int y = 0; y < SIZE; y++){
cout << arr[x][y] <<" "; //Prints the array's contents
}
cout << endl; //Formats with newlines
}
cout << endl; //Some spacing
return;
}
void nextTurn(char today[][SIZE], char update[][SIZE], int& day){
day++; //Updates the day
int xCheck; //X coordinate to be checked
int yCheck; //Y coordinate to be checked
for(int x = 0; x < SIZE; x++){
for(int y = 0; y < SIZE; y++){
//Sets all 'i' to 'r' in the new grid
if(today[x][y] == 'i' || today[x][y] == 'r'){
update[x][y] = 'r'; //Updates all infectious to recovered, and keeps current recovered
}
if(today[x][y] == 's'){ // If the person is susceptible...
for(int xCheck = x-1; xCheck <= x+1; xCheck++){ // Check all x coordinates around the person
for(int yCheck = y-1; yCheck <= y+1; yCheck++){ // Check all y coordinates around the person
if(xCheck == x && yCheck == y){
// Don't check at the person because there is no need to check there
} else {
if(xCheck >= 0 && yCheck >= 0 && xCheck < SIZE && yCheck < SIZE){ // Make sure the checked coordinates are in bounds
if(today[xCheck][yCheck] == 'i'){ //Is the person at the checked coordinates infected?
update[x][y] = 'i'; //If so, update the 's' to 'i' in the new grid
}
}
}
}
}
}
}
}
for(int x = 0; x < SIZE; x++){
for(int y = 0; y < SIZE; y++){
today[x][y] = update[x][y]; //Updates today's grid with the new values
}
}
}
int countInfected(char arr[][SIZE], int arrSize){
int count = 0;
for(int x = 0; x < arrSize; x++){
for(int y = 0; y < arrSize; y++){
if(arr[x][y] == 'i') count++; //Increments count for each infected person in the grid
}
}
return count;
}

Get Coefficient

Nov 18, 2022AustinLeath

0 likes • 4 views

#include <iostream>
using namespace std;
/* Function: get_coeff
Parameters: double& coeff, int pos passed from bb_4ac
Return: type is void so no return, but does ask for user to input data that establishes what a b and c are.
*/
void get_coeff(double& coeff, int pos) {
char position;
if(pos == 1) {
position = 'a';
} else if(pos == 2) { //a simple system to determine what coefficient the program is asking for.
position = 'b';
} else {
position = 'c';
}
cout << "Enter the co-efficient " << position << ":"; //prompt to input coeff
coeff = 5; //input coeff
}
/* Function: bb_4ac
Parameters: no parameters passed from main, but 3 params established in function, double a, b, c.
Return: b * b - 4 * a * c
*/
double bb_4ac() {
double a, b, c; //coefficients of a quadratic equation
get_coeff(a, 1); // call function 1st time
get_coeff(b, 2); // call function 2nd time
get_coeff(c, 3); // call function 3rd time
return b * b - 4 * a * c; //return b * b - 4 * a * c
}
int main() {
cout << "Function to calculate the discriminant of the equation. . . " << endl;
double determinate = bb_4ac(); //assign double determinate to bb_4ac function
cout << "The discriminant for given values is: " << determinate << endl; //output the determinate!
}

2D Array Chessboard Pattern

Nov 18, 2022AustinLeath

0 likes • 14 views

#include<iostream>
using namespace std;
const int rows = 8;
const int cols = 8;
char chessboard[rows][cols];
void setBoard(char chessboard[][cols]);
void printBoard(char chessboard[][cols]);
void setBoard(char chessboard[][cols]) {
for(int i = 0; i < rows; i++) {
for(int j = 0; j < cols; j++) {
if(i % 2 == 0 && j % 2 == 0) {
chessboard[i][j] = 'x';
} else {
if(i % 2 != 0 && j % 2 == 1) {
chessboard[i][j] = 'x';
} else {
chessboard[i][j] = '-';
}
}
}
}
return;
}
void printBoard(char chessboard[][cols]) {
for(int i = 0; i < rows; i++) {
for(int j = 0; j < cols; j++) {
cout << chessboard[i][j] << " ";
}
cout << endl;
}
return;
}
int main(int argc, char const *argv[])
{
setBoard(chessboard);
printBoard(chessboard);
return 0;
}

Big O(n^2) Ascending Sort

Nov 18, 2022AustinLeath

1 like • 6 views

#include <iostream>
using namespace std;
int main() {
int arr[5];
for(int i = 0; i < 5; i++) {
arr[i] = i;
}
for(int i = 0; i < 5; i++) {
cout << "Outputting array info at position " << i + 1 << ": " << arr[i] << endl;
}
for(int i=0;i<5;i++)
{
for(int j=i+1;j<5;j++)
{
if(arr[i]>arr[j])
{
int temp=arr[i];
arr[i]=arr[j];
arr[j]=temp;
}
}
}
cout << endl;
for(int i = 0; i < 5; i++) {
cout << "Outputting sorted array info at position " << i + 1 << ": " << arr[i] << endl;
}
return 0;
}

Compute Volume of Cylinder

Nov 18, 2022AustinLeath

0 likes • 0 views

/*
Algorithm:
Step 1: Get radius of the cylinder from the user and store in variable r
Step 2: Get height of the cylinder from the user and store in variable h
Step 3: Multiply radius * radius * height * pi and store in v
Step 4: Display the volume
*/
#include <iostream>
using namespace std;
int main()
{
float r; //define variable for radius
float h; //define variable for height
float v;
float pi;
pi=3.1416;
cout<<"Enter radius:";
cin>>r;
cout<<"Enter height:";
cin>>h;
v=r*r*h*pi; //compute volume
cout<<"Radius:"<<r<<"\tHeight:"<<h<<endl; //display radius and height
cout<<"\n************************\n";
cout<<"Volume:"<<v<<endl;//display volume
return 0;
}

Daily: Cutting a Wall

Dec 20, 2021aedrarian

0 likes • 0 views

/*
Good morning! Here's your coding interview problem for today.
This problem was asked by LinkedIn.
A wall consists of several rows of bricks of various integer lengths and uniform height. Your goal is to find a vertical line going from the top to the bottom of the wall that cuts through the fewest number of bricks. If the line goes through the edge between two bricks, this does not count as a cut.
For example, suppose the input is as follows, where values in each row represent the lengths of bricks in that row:
[[3, 5, 1, 1],
[2, 3, 3, 2],
[5, 5],
[4, 4, 2],
[1, 3, 3, 3],
[1, 1, 6, 1, 1]]
The best we can we do here is to draw a line after the eighth brick, which will only require cutting through the bricks in the third and fifth row.
Given an input consisting of brick lengths for each row such as the one above, return the fewest number of bricks that must be cut to create a vertical line.
AUTHORS NOTE:
Makes following assumptions:
- Each row is same length
- Data is in file called "data.dat" and formatted in space-separated rows
- The cuts at the beginning and end of the wall are not solutions
This requires the following file named data.dat that is a space separated file, or similar formatted file:
----START FILE----
3 5 1 1
2 3 3 2
5 5
4 4 2
1 3 3 3
1 1 6 1 1
----END FILE----
*/
#include <algorithm>
#include <iostream>
#include <fstream>
#include <map>
#include <sstream>
#include <string>
#include <vector>
using namespace std;
int main()
{
vector<vector<int>> wall;
ifstream in;
in.open("data.dat");
if(!in.good())
{
cout << "ERROR: File failed to open properly.\n";
}
/* Get input from space separated file */
string line;
while(!in.eof())
{
getline(in, line);
int i;
vector<int> currv;
stringstream strs(line);
while(strs >> i)
currv.push_back(i);
wall.push_back(currv);
}
/* Convert each value from "length of brick" to "position at end of brick" */
for(int y = 0; y < wall.size(); y++)
{
wall.at(y).pop_back(); //Delet last val
for(int x = 1; x < wall.at(y).size(); x++) //Skip the first bc data doesn't need change
wall.at(y).at(x) += wall.at(y).at(x-1);
}
/* Check output. COMMENT OUT */
// for(auto row : wall)
// {
// for(int pos : row)
// cout << pos << " ";
// cout << endl;
// }
/* Determine which ending position is most common, and cut there */
//Exclude final position, which will be the size of the wall
int mode = -1;
int amt = -1;
vector<int> tried;
for(auto row : wall)
{
for(int pos : row) //For each pos in the wall
{
//Guard. If pos is contained in the list, skip pos
if(find(tried.begin(), tried.end(), pos) != tried.end())
continue;
tried.push_back(pos);
/* Cycle through each row to see if it contains the pos */
int curramt = 0;
for(auto currrow : wall)
{
if( find( currrow.begin(), currrow.end(), pos ) != currrow.end() )
curramt++;
}
//cout << pos << " " << curramt << endl;
if(curramt > amt)
{
amt = curramt;
mode = pos;
}
}
}
cout << "Please cut at position " << mode << endl;
cout << "This will cut through " << (wall.size() - amt) << " bricks." << endl;
return 0;
}