How do I create a permutation of a 2D string array? - c++

Question

I have the following code. I set each box equal to 1. Now I want to set 3 boxes at a time to 0. How do I do that without manually setting each of them to 1?
Is there a permutation formula that will set 3 at the time to 1?
#include <iostream>
using namespace std;
int main()
{
int array[2][2];
for (int x = 0; x<2; x++)
{
for (int y = 0; y<2; y++)
array[x][y] = 1;
}
// display all cells to see that all of them are set to zero
cout << "diplaying" << endl;
for (int x = 0; x<2; x++)
{
for (int y = 0; y<2; y++)
cout << array[x][y] << " " ;
cout << endl;
}
Printing this would look something like.
1 1
1 1
Now how do I get to print
0 1
0 0
and
1 0
0 0
and
0 0
1 0
and
0 0
0 1
without having to set them individually that way?

Answer 1

Personally, I would store the array as a 1D std::vector<int> of size n*n. Then you could call std::next_permutation() on it very simply. (It's worth noting that you don't have to use a std::vector; as long as it is contiguous in memory, you should be able to use std::next_permutation() properly)
The only thing you have to do that makes your permutation logic "2D" is the act of printing it out. However, your loop as-is should handle that properly, so no problems there either.
EDIT: Upon re-reading your code, you could not use this as-is. Instead, you should initialize your 1D std::vector to be 0 everywhere, except 1 at position 0. THEN, permutations of that would yield the output you want.
Furthermore, your printing loop would not print out the array properly. You probably want:
for (int i = 0; i < 2; ++i) {
for (int j = 0; j < 2; ++j) {
std::cout << vector[i*2+j] << " " ;
}
std::cout << std::endl;
}

Answer 2

After reading carefully, I interpreted your question this way:
You want to go from this
1 1
1 1
to this
0 1
0 0
using the for clause.
if you simply want to leave a cell unchanged.. you could just save it and restore it after filling the array with 0.
memo = array[0][1];
for (int x = 0; x<2; x++) {
for (int y = 0; y<2; y++) {
array[x][y] = 0;
}
}
array[0][1] = memo;
If that's what you want to do.
Where's the "string array" by the way..?

Answer 3

I would write a function that prints all zeros except where it needs to print the actual value, then call this function with different indexes for desired effect.
#include <iostream>
using namespace std;
#define SZ 2
void printValue(int a [SZ][SZ], int x, int y)
{
for(int i=0; i<SZ; ++i )
{
for(int j=0; j<SZ; ++j)
{
if(i==x && j==y) cout<<a[i][j];
else cout<<"0";
cout<<" ";
}
cout<<endl;
}
}
Now you can make use of this function in a for loop
int main()
{
int arr[SZ][SZ];
for (int x = 0; x<SZ; x++)
{
for (int y = 0; y<SZ; y++)
arr[x][y] = 1;
}
// display all cells to see that all of them are set to zero
cout << "diplaying" << endl;
for (int x = 0; x<SZ; x++)
{
for (int y = 0; y<SZ; y++)
cout << arr[x][y] << " " ;
cout << endl;
}
//now display all combos:
for(int i=0; i<SZ; ++i)
{
for(int j=0; j<SZ; ++j)
{
printValue(arr, i,j);
}
cout<<"\n\n";
}
}