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VikaD [51]
3 years ago
5

This project was used to drive the VR Robot to number 25 on the Number Grid

Computers and Technology
1 answer:
Ludmilka [50]3 years ago
3 0

Answer:

B Wait until position y in millimeter should be greater than 500 at least im guessing because you didnt add what the map looked like, making my job MUCH harder

Explanation:

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g Write a program to sort an array of 100,000 random elements using quicksort as follows: Sort the arrays using pivot as the mid
shtirl [24]

Answer:

header.h->function bodies and header files.

#include<iostream>

#include<cstdlib>

#include<ctime>

using namespace std;

/* Partitioning the array on the basis of piv value. */

int Partition(int a[], int bot, int top,string opt)

{

int piv, ind=bot, i, swp;

/*Finding the piv value according to string opt*/

if(opt=="Type1 sort" || opt=="Type3 sort")

{

piv=(top+bot)/2;

}

else if(opt=="Type2 sort" || opt=="Type4 sort")

{

piv=(top+bot)/2;

if((a[top]>=a[piv] && a[top]<=a[bot]) || (a[top]>=a[bot] && a[top]<=a[piv]))

piv=top;

else if((a[bot]>=a[piv] && a[bot]<=a[top]) || (a[bot]>=a[top] && a[bot]<=a[piv]))

piv=bot;

}

swp=a[piv];

a[piv]=a[top];

a[top]=swp;

piv=top;

/*Getting ind of the piv.*/

for(i=bot; i < top; i++)

{

if(a[i] < a[piv])

{

swp=a[i];

a[i]=a[ind];

a[ind]=swp;

ind++;

}

}

swp=a[piv];

a[piv]=a[ind];

a[ind]=swp;

return ind;

}

void QuickSort(int a[], int bot, int top, string opt)

{

int pindex;

if((opt=="Type3 sort" || opt=="Type4 sort") && top-bot<19)

{

/*then insertion sort*/

int swp,ind;

for(int i=bot+1;i<=top;i++){

swp=a[i];

ind=i;

for(int j=i-1;j>=bot;j--){

if(swp<a[j]){

a[j+1]=a[j];

ind=j;

}

else

break;

}

a[ind]=swp;

}

}

else if(bot < top)

{

/* Partitioning the array*/

pindex =Partition(a, bot, top,opt);

/* Recursively implementing QuickSort.*/

QuickSort(a, bot, pindex-1,opt);

QuickSort(a, pindex+1, top,opt);

}

return ;

}

main.cpp->main driver file

#include "header.h"

int main()

{

int n=100000, i;

/*creating randomized array of 100000 numbers between 0 and 100001*/

int arr[n];

int b[n];

for(i = 0; i < n; i++)

arr[i]=(rand() % 100000) + 1;

clock_t t1,t2;

t1=clock();

QuickSort(arr, 0, n-1,"Type1 sort");

t2=clock();

for( i=0;i<n;i++)

arr[i]=b[i];

cout<<"Quick sort time, with pivot middle element:"<<(t2 - t1)*1000/ ( CLOCKS_PER_SEC );

cout<<"\n";

t1=clock();

QuickSort(arr, 0, n-1,"Type2 sort");

t2=clock();

for( i=0;i<n;i++)

arr[i]=b[i];

cout<<"Quick sort time, with pivot median element:"<<(t2 - t1)*1000/ ( CLOCKS_PER_SEC );

cout<<"\n";

t1=clock();

QuickSort(arr, 0, n-1,"Type3 sort");

t2=clock();

for( i=0;i<n;i++)

arr[i]=b[i];

cout<<"Quick sort time and insertion sort time, with pivot middle element:"<<(t2 - t1)*1000/ ( CLOCKS_PER_SEC );

cout<<"\n";

t1=clock();

QuickSort(arr, 0, n-1,"Type4 sort");

t2=clock();

cout<<"Quick sort time and insertion sort time, with pivot median element:"<<(t2 - t1)*1000/ ( CLOCKS_PER_SEC );

return 0;

}

Explanation:

Change the value of n in the main file for different array size. Output is in the same format as mentioned, time is shown in milliseconds.

7 0
3 years ago
Write a function in MATLAB that takes as an input a matrix of coefficients for a system of linear equations (A), the solution ve
Illusion [34]
Answer is A all the way to the right
7 0
3 years ago
The number of pixels displayed on the screen is known as ​
puteri [66]

Answer: Resolution

Explanation: The total number of pixels that can be displayed on the screen at a time is called the resolution of the screen. This resolution is normally described in the pair of numbers, such as 2560 x 1440. This means, the computer screen is 2560 pixels wide and 1440 pixels tall.

4 0
3 years ago
Describe a sub routine​
Maksim231197 [3]

Explanation:

A routine or subroutine, also referred to as a function procedure and sub program is code called and executed anywhere in a program. FOr example a routine may be used to save a file or display the time.

4 0
3 years ago
How to get amount after a percentage increase in excel​
Tresset [83]

Answer:

First: work out the difference (increase) between the two numbers you are comparing.

Increase = New Number - Original Number.

Then: divide the increase by the original number and multiply the answer by 100.

% increase = Increase ÷ Original Number × 100.

Explanation:

3 0
2 years ago
Read 2 more answers
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