Constructing Identity Matrix [closed] - c++

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Closed 10 years ago.
I have tried to solve an algorithm problem, I'm newbie and I'm trying to practice a lot in programming problems. So I wanted to construct an Identity matrix n*n. I came up with a stupid solution, that worked for a 4*4 matrix, but it didn't work with 5*5. I know that its weird solution and the solution to the problem is really easy when I looked at it. I need to know what did I do wrong so that I can learn, and is my solution is really stupid and I will be better in the future after working much in solving those kind of problems ?
#include <iostream>
#include <vector>
#include <sstream>
#include <iomanip> // for setw, setfill
using namespace std;
int binary(int number);
int main()
{
vector<vector<int> > matrix;
cout<<"Please enter the size of the identity matrix"<<endl;
int n;
cin>>n;
matrix.resize(n);
for (int i=0; i<n;i++)
{
matrix[i].resize(n);
}
int steps = 1<<n-1;
int bin = binary(steps);
ostringstream binString;
binString <<bin;
if(binString.str().size()<n)
{
std::string dest = binString.str();
int nPaddings = n-binString.str().size();
if (nPaddings==0) nPaddings=1;
dest = std::string( nPaddings, '0').append( binString.str());
binString.str("");
binString<<dest;
}
for (int col = 0; col<n; col++)
{
if(col>=1)
{
steps= (int)steps/2;
int bin = binary(steps);
binString.str("");
binString << bin;
if(binString.str().size()<n)
{
std::string dest = binString.str();
int nPaddings = n-steps;
if (nPaddings==0) nPaddings=1;
dest = std::string( nPaddings, '0').append( binString.str());
binString.str("");
binString<<dest;
}
}
for (int row=0; row<n; row++)
{
matrix[col][row] =binString.str().at(row)-'0';
}
}
return 0;
}
int binary(int number) {
long rem,i=1,sum=0;
do
{
rem=number%2;
sum=sum + (i*rem);
number=number/2;
i=i*10;
}while(number>0);
return sum;
}

Answer 1

There is a much simpler way to do it.
First, you should allocate your matrix with the specified size. Then, you know that only the diagonal is 1s:
vector<vector<int> > matrix;
int n;
cout << "Please enter the size of the identity matrix" << endl;
cin >> n;
// Initialize the matrix as a n x n array of 0.
matrix = vector<vector<int> >(n, vector<int>(n,0));
// Set the diagonal to be 1s
for(unsigned int t = 0; t < n; t++)
matrix[t][t] = 1;
You can see a live example here.
Edit:
Your error comes from this line:
int nPaddings = n-steps;
In fact, you're not using the size of dest to compute the padding, which is not correct. See here, I added some debug printfs to see the state of the variables. You can see that nPaddings == -3, hence the errors.
The idea you have:
for each column
get the representation of the column as a string
set the i-th value of the column as the i-th character of the string
So, here is a simpler program using your idea. Separating the code in several functions helps a lot. Also, std::ostringstream and std::string is just pure overkill here.
#include <iostream>
#include <vector>
#include <iomanip> // for setw, setfill
using namespace std;
std::string binStr(unsigned int exponent, unsigned int size);
int main()
{
vector<vector<int> > matrix;
cout<<"Please enter the size of the identity matrix"<<endl;
int n;
cin>>n;
// Initialize the matrix
matrix.resize(n);
for (int i=0; i<n;i++)
matrix[i].resize(n);
// Fill the matrix
for (int col = 0; col<n; col++)
{
std::string bin = binStr(n-col,n);
for (int row=0; row<n; row++)
matrix[col][row] = bin[row]-'0';
}
// Print the matrix and return
for(unsigned int y = 0; y < n; y++)
{
for(unsigned int x = 0; x < n; x++)
cout << "\t" << matrix[y][x];
cout << "\n";
}
return 0;
}
std::string binStr(unsigned int exponent, unsigned int size)
{
// You do not need a string stream (which is like using a bazooka to kill a fly...)
// Instead, just create a string of the required length
// 'str' will contain the binary representation of 2^exponent
std::string str(size,'0');
if(exponent <= size && exponent > 0)
str[size - exponent] = '1';
return str;
}
You can see it in action here.

Answer 2

#include <iostream>
#include <vector>
using namespace std;
vector<vector<int> > make_idty_matrix( int n )
{
vector<vector<int> > idty( n, vector<int>( n, 0 ));
for( int i = 0; i < n; ++i )
idty[i][i] = 1;
return idty;
}
int main()
{
vector<vector<int> > matrix = make_idty_matrix( 5 );
// your code here
// ...
return 0;
}