I have a table of numbers that look like this:
2 8 4 0
3 1 0 9
1 2 3 4
5 4 14 2
I put all the numbers in an array { 2,8,4,0,3,1... }. Is there a way to sort it by the first column only using a 1D array so that it ends up like this:
1 2 3 4
2 8 4 0
3 1 0 9
5 4 14 2
I know there's a way of doing it with a 2D array, but, assuming I know the number of columns, is it possible with only a 1D array?
I'd create an array of indexes into your data, and then sort those indexes; this will save a decent number of the copies.
Your sort would then examine the value of the number at the given index.
ie for your example - indexes would be 1,2,3,4
and then sorted would read 3,1,2,4
edit: this was 1 based; the code 0 based. Makes no difference.
Essentially converting your 1d array into 2. Since the bulk of your data is still contiguous (especially for large numbers of columns) reading should still be fast.
Example code:
std::vector<int> getSortedIndexes(std::vector<int> data, int size) {
int count = data.size() / size;
std::vector<int> indexes(count);
// fill in indexes from 0 to "count" since that's the size of our vector
std::iota(indexes.begin(), indexes.end(), 0);
// don't write your own sorting implementation .... really; don't.
std::sort(indexes.begin(), indexes.end(), [data, size](int indexA, int indexB) {
return data[indexA*size] < data[indexB*size];
});
return indexes;
}
For arrays of non-user defined types it is easy to do the task using the standard C function qsort.
Here is a demonstrative program.
#include <iostream>
#include <cstdlib>
int cmp( const void *a, const void *b )
{
const int *left = static_cast<const int *>( a );
const int *right = static_cast<const int *>( b );
return ( *right < *left ) - ( *left < *right );
}
int main()
{
const size_t N = 4;
int a[N * N] =
{
2, 8, 4, 0, 3, 1, 0, 9, 1, 2, 3, 4, 5, 4, 14, 2
};
for ( size_t i = 0; i < N; i++ )
{
for ( size_t j = 0; j < N; j++ )
{
std::cout << a[N * i + j] << ' ';
}
std::cout << '\n';
}
std::cout << '\n';
std::qsort( a, N, sizeof( int[N] ), cmp );
for ( size_t i = 0; i < N; i++ )
{
for ( size_t j = 0; j < N; j++ )
{
std::cout << a[N * i + j] << ' ';
}
std::cout << '\n';
}
std::cout << '\n';
}
The program output is
2 8 4 0
3 1 0 9
1 2 3 4
5 4 14 2
1 2 3 4
2 8 4 0
3 1 0 9
5 4 14 2
So all you need is to write the function
int cmp( const void *a, const void *b )
{
const int *left = static_cast<const int *>( a );
const int *right = static_cast<const int *>( b );
return ( *right < *left ) - ( *left < *right );
}
and add just one line in your program
std::qsort( a, N, sizeof( int[N] ), cmp );
You can use bubblesort:
void sort_by_name(int* ValueArray, int NrOfValues, int RowWidth)
{
int CycleCount = NrOfValues / RowWidth;
int temp;
(int j = 0; j < CycleCount ; j++)
{
for (int i = 1; i < (CycleCount - j); i++)
{
if(ValueArray[((i-1)*RowWidth)] > ValueArray[(i*RowWidth)])
{
for(int k = 0; k<RowWidth; k++)
{
temp = ValueArray[(i*RowWidth)+k]
ValueArray[(i*RowWidth)+k] = ValueArray[((i-1)*RowWidth)+k];
ValueArray[((i-1)*RowWidth)+k] = temp;
}
}
}
}
}
keep in mind that simply making your array 2D will be a MUCH BETTER solution
edit: variable naming
I am currently doing Matric in C++, I want to create a matrix that can store non_zero_elements. What I want to do is to insert an element in specific position and further print it out the matrix with non_zero_element and zero_element. This is my initialised matrix :
int A[4][4] =
{
{ 0 , 0 , 0 , 0 },
{ 0 , 0 , 0 , 0 },
{ 0 , 0 , 0 , 0 },
{ 0 , 0 , 0 , 0 }
};
Below are my codes
void SM::readElement(int row, int column, int value)
{
m = row;
n = column;
for (int i = 0; i < m; i++) {
cout << "\t\t";
for (int j = 0; j < n; j++) {
if (i == row && j == column)
{
A[i][j] = value;
}
}
}
}
void SM::printMatrix()
{
for (i = 0; i < 4; i++)
{
cout << "\n";
for (j = 0; j < 4; j++)
cout << A[i][j];
}
}
The driver :
int main()
{
SM sm;
int choice, column, row, value;
do {
sm.Menu();
cin >> choice;
switch (choice)
{
case 1:
do
{
cout << "Enter row -> ";
cin >> row;
} while (row < 0 || row >= 11);
do
{
cout << "Enter column -> ";
cin >> column;
} while (column < 0 || column >= 11);
do {
cout << "Enter value -> ";
cin >> value;
} while (value <= 0);
sm.readElement(row, column, value);
}
I want to read the elements and insert it into specific row and column and print out a whole matrix with updated matrix after inserting the element. I need somebody point out the errors. Thank you very much.
Input row = 2
Input column = 2
Input value = 5
Real Output :
int A [4][4] =
{
{ 5 , 5 , 0 , 0 },
{ 5 , 5 , 0 , 0 },
{ 0 , 0 , 0 , 0 },
{ 0 , 0 , 0 , 0 }
};
Expected Output :
int A [4][4] =
{
{ 0 , 0 , 0 , 0 },
{ 0 , 5 , 0 , 0 },
{ 0 , 0 , 0 , 0 },
{ 0 , 0 , 0 , 0 }
};
And I found that my codes will not accept row/column =0, it only starting with row/column >= 1.
int A[4][4] =
{
{ 0 , 0 , 0 , 0 },
{ 0 , 0 , 0 , 0 },
{ 0 , 0 , 0 , 0 },
{ 0 , 0 , 0 , 0 }
};
is the correct syntax for declaring and initializing a 2d array. You should probably be using vectors though.
In order to insert an element, say int value, into position [row,column] of a matrix A it suffices to do
A[row][column] = value;
In order to print the new matrix just call the function printMatrix() you wrote after the insertion
There is a problem where I need to fill an array with zeros, with the following assumptions:
in the array there can only be 0 and 1
we can only change 0 to 1 and 1 to 0
when we meet 1 in array, we have to change it to 0, such that its neighbours are also changed, for instance, for the array like the one below:
1 0 1
1 1 1
0 1 0
When we change element at (1,1), we then got the array like this:
1 1 1
0 0 0
0 0 0
We can't change the first row
We can only change the elements that are in the array
The final result is the number of times we have to change 1 to 0 to zero out the array
1) First example, array is like this one below:
0 1 0
1 1 1
0 1 0
the answer is 1.
2) Second example, array is like this one below:
0 1 0 0 0 0 0 0
1 1 1 0 1 0 1 0
0 0 1 1 0 1 1 1
1 1 0 1 1 1 0 0
1 0 1 1 1 0 1 0
0 1 0 1 0 1 0 0
The answer is 10.
There also can be situations that its impossible to zero out the array, then the answer should be "impossible".
Somehow I can't get this working: for the first example, I got the right answer (1) but for the second example, program says impossible instead of 10.
Any ideas what's wrong in my code?
#include <iostream>
using namespace std;
int main(int argc, char **argv)
{
int n,m;
cin >> n >> m;
bool tab[n][m];
for(int i=0; i<n; i++)
for(int j=0; j<m; j++)
cin >> tab[i][j];
int counter = 0;
for(int i=0; i<n-1; i++)
{
for(int j=0; j<m-1; j++)
{
if(tab[i][j] == 1 && i > 0 && j > 0)
{
tab[i-1][j] = !tab[i-1][j];
tab[i+1][j] = !tab[i+1][j];
tab[i][j+1] = !tab[i][j+1];
tab[i][j-1] = !tab[i][j-1];
tab[i][j] = !tab[i][j];
counter ++;
}
}
}
bool impossible = 0;
for(int i=0; i<n; i++)
{
for(int j=0; j<m; j++)
{
if(tab[i][j] == 1)
{
cout << "impossible\n";
impossible = 1;
break;
}
}
if(impossible)
break;
}
if(!impossible)
cout << counter << "\n";
return 0;
}
I believe that the reason your program was returning impossible in the 6x8 matrix is because you have been traversing in a left to right / top to bottom fashion, replacing every instance of 1 you encountered with 0. Although this might have seemed as the right solution, all it did was scatter the 1s and 0s around the matrix by modifying it's neighboring values. I think that the way to approach this problem is to start from bottom to top/ right to left and push the 1s towards the first row. In a way cornering (trapping) them until they can get eliminated.
Anyway, here's my solution to this problem. I'm not entirely sure if this is what you were going after, but I think it does the job for the three matrices you provided. The code is not very sophisticated and it would be nice to test it with some harder problems to see if it truly works.
#include <iostream>
static unsigned counter = 0;
template<std::size_t M, std::size_t N>
void print( const bool (&mat) [M][N] )
{
for (std::size_t i = 0; i < M; ++i)
{
for (std::size_t j = 0; j < N; ++j)
std::cout<< mat[i][j] << " ";
std::cout<<std::endl;
}
std::cout<<std::endl;
}
template<std::size_t M, std::size_t N>
void flipNeighbours( bool (&mat) [M][N], unsigned i, unsigned j )
{
mat[i][j-1] = !(mat[i][j-1]);
mat[i][j+1] = !(mat[i][j+1]);
mat[i-1][j] = !(mat[i-1][j]);
mat[i+1][j] = !(mat[i+1][j]);
mat[i][j] = !(mat[i][j]);
++counter;
}
template<std::size_t M, std::size_t N>
bool checkCornersForOnes( const bool (&mat) [M][N] )
{
return (mat[0][0] || mat[0][N-1] || mat[M-1][0] || mat[M-1][N-1]);
}
template<std::size_t M, std::size_t N>
bool isBottomTrue( bool (&mat) [M][N], unsigned i, unsigned j )
{
return (mat[i+1][j]);
}
template<std::size_t M, std::size_t N>
bool traverse( bool (&mat) [M][N] )
{
if (checkCornersForOnes(mat))
{
std::cout<< "-Found 1s in the matrix corners." <<std::endl;
return false;
}
for (std::size_t i = M-2; i > 0; --i)
for (std::size_t j = N-2; j > 0; --j)
if (isBottomTrue(mat,i,j))
flipNeighbours(mat,i,j);
std::size_t count_after_traversing = 0;
for (std::size_t i = 0; i < M; ++i)
for (std::size_t j = 0; j < N; ++j)
count_after_traversing += mat[i][j];
if (count_after_traversing > 0)
{
std::cout<< "-Found <"<<count_after_traversing<< "> 1s in the matrix." <<std::endl;
return false;
}
return true;
}
#define MATRIX matrix4
int main()
{
bool matrix1[3][3] = {{1,0,1},
{1,1,1},
{0,1,0}};
bool matrix2[3][3] = {{0,1,0},
{1,1,1},
{0,1,0}};
bool matrix3[5][4] = {{0,1,0,0},
{1,0,1,0},
{1,1,0,1},
{1,1,1,0},
{0,1,1,0}};
bool matrix4[6][8] = {{0,1,0,0,0,0,0,0},
{1,1,1,0,1,0,1,0},
{0,0,1,1,0,1,1,1},
{1,1,0,1,1,1,0,0},
{1,0,1,1,1,0,1,0},
{0,1,0,1,0,1,0,0}};
std::cout<< "-Problem-" <<std::endl;
print(MATRIX);
if (traverse( MATRIX ) )
{
std::cout<< "-Answer-"<<std::endl;
print(MATRIX);
std::cout<< "Num of flips = "<<counter <<std::endl;
}
else
{
std::cout<< "-The Solution is impossible-"<<std::endl;
print(MATRIX);
}
}
Output for matrix1:
-Problem-
1 0 1
1 1 1
0 1 0
-Found 1s in the matrix corners.
-The Solution is impossible-
1 0 1
1 1 1
0 1 0
Output for matrix2:
-Problem-
0 1 0
1 1 1
0 1 0
-Answer-
0 0 0
0 0 0
0 0 0
Num of flips = 1
Output for matrix3:
-Problem-
0 1 0 0
1 0 1 0
1 1 0 1
1 1 1 0
0 1 1 0
-Found <6> 1s in the matrix.
-The Solution is impossible-
0 1 1 0
1 0 1 1
0 0 0 0
0 0 0 1
0 0 0 0
Output for matrix4 (which addresses your original question):
-Problem-
0 1 0 0 0 0 0 0
1 1 1 0 1 0 1 0
0 0 1 1 0 1 1 1
1 1 0 1 1 1 0 0
1 0 1 1 1 0 1 0
0 1 0 1 0 1 0 0
-Answer-
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
Num of flips = 10
Ok, here comes my somewhat different attempt.
Idea
Note: I assume here that "We can't change the first row" means "We can't change the outmost row".
Some terminology:
With toggling a bit I mean changing it's value from 0 to 1 or 1 to 0.
With flipping a bit I mean toggling said bit and the 4 bits around it.
The act of toggling a bit is commutative. That is, it does not matter in what order we toggle it—the end result will always be the same (this is a trivial statement). This means that flipping is also a commutative action, and we are free to flip bits in any order we like.
The only way to toggle a value on the edge of the matrix is by flipping the bit right next to it an uneven amount of times. As we're looking for the lowest possible flips, we want to flip it a maximum of 1 time. So, in a scenario like the on below, x will need to be flipped exactly once, and y will need to be flipped exactly 0 times.
. .
1 x
0 y
. ,
From this we can draw two conclusions:
A corner of the matrix can never be toggled—if a 1 on the corner is found it is not possible with any number of flips to make the matrix zero. Your first example can thus be discarded without even flipping a single bit.
A bit next to a corner must have the same same value as the bit on the other side. This matrix that you posted in a comment can thus as well be discarded without flipping a single bit (bottom right corner).
Two examples of the conditions above:
0 1 .
0 x .
. . .
Not possible, as x needs to be flipped exactly once and exactly zero times.
0 1 .
1 x .
. . .
Possible, x needs to be flipped exactly once.
Algorithm
We can now make an recursive argument, and I propose the following:
We are given an m by n matrix.
Check the corner conditions above as stated above (i.e. corner != 1, bits next to corner has to be the same value). If either criteria are violated, return impossible.
Go around the edge of the matrix. If a 1 is encountered, flip the closest bit inside, and add 1 to the counter.
Restart now from #1 with a m - 2 by n - 2 matrix (top and bot row removed, left and right column) if either dimension is > 2, otherwise print the counter and quit.
Implementation
Initially I had thought this would turn out nice and pretty, but truth be told it is a little more cumbersome than I originally thought it would be as we have to keep track of a lot of indices. Please ask questions if you're wondering about the implementation, but it is in essence a pure translation of the steps above.
#include <iostream>
#include <vector>
using Matrix = std::vector<std::vector<int>>;
void flip_bit(Matrix& mat, int i, int j, int& counter)
{
mat[i][j] = !mat[i][j];
mat[i - 1][j] = !mat[i - 1][j];
mat[i + 1][j] = !mat[i + 1][j];
mat[i][j - 1] = !mat[i][j - 1];
mat[i][j + 1] = !mat[i][j + 1];
++counter;
}
int flip(Matrix& mat, int n, int m, int p = 0, int counter = 0)
{
// I use p for 'padding', i.e. 0 means the full array, 1 means the outmost edge taken away, 2 the 2 most outmost edges, etc.
// max indices of the sub-array
int np = n - p - 1;
int mp = m - p - 1;
// Checking corners
if (mat[p][p] || mat[np][p] || mat[p][mp] || mat[np][mp] || // condition #1
(mat[p + 1][p] != mat[p][p + 1]) || (mat[np - 1][p] != mat[np][p + 1]) || // condition #2
(mat[p + 1][mp] != mat[p][mp - 1]) || (mat[np - 1][mp] != mat[np][mp - 1]))
return -1;
// We walk over all edge values that are *not* corners and
// flipping the bit that are *inside* the current bit if it's 1
for (int j = p + 1; j < mp; ++j) {
if (mat[p][j]) flip_bit(mat, p + 1, j, counter);
if (mat[np][j]) flip_bit(mat, np - 1, j, counter);
}
for (int i = p + 1; i < np; ++i) {
if (mat[i][p]) flip_bit(mat, i, p + 1, counter);
if (mat[i][mp]) flip_bit(mat, i, mp - 1, counter);
}
// Finished or flip the next sub-array?
if (np == 1 || mp == 1)
return counter;
else
return flip(mat, n, m, p + 1, counter);
}
int main()
{
int n, m;
std::cin >> n >> m;
Matrix mat(n, std::vector<int>(m, 0));
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
std::cin >> mat[i][j];
}
}
int counter = flip(mat, n, m);
if (counter < 0)
std::cout << "impossible" << std::endl;
else
std::cout << counter << std::endl;
}
Output
3 3
1 0 1
1 1 1
0 1 0
impossible
3 3
0 1 0
1 1 1
0 1 0
1
6 8
0 1 0 0 0 0 0 0
1 1 1 0 1 0 1 0
0 0 1 1 0 1 1 1
1 1 0 1 1 1 0 0
1 0 1 1 1 0 1 0
0 1 0 1 0 1 0 0
10
4 6
0 1 0 0
1 0 1 0
1 1 0 1
1 1 1 0
1 1 1 0
impossible
If tab[0][j] is 1, you have to toggle tab[1][j] to clear it. You then cannot toggle row 1 without unclearing row 0. So it seems like a reduction step. You repeat the step until there is one row left. If that last row is not clear by luck, my intuition is that it's the "impossible" case.
#include <memory>
template <typename Elem>
class Arr_2d
{
public:
Arr_2d(unsigned r, unsigned c)
: rows_(r), columns_(c), data(new Elem[rows_ * columns_]) { }
Elem * operator [] (unsigned row_idx)
{ return(data.get() + (row_idx * columns_)); }
unsigned rows() const { return(rows_); }
unsigned columns() const { return(columns_); }
private:
const unsigned rows_, columns_;
std::unique_ptr<Elem []> data;
};
inline void toggle_one(bool &b) { b = !b; }
void toggle(Arr_2d<bool> &tab, unsigned row, unsigned column)
{
toggle_one(tab[row][column]);
if (column > 0)
toggle_one(tab[row][column - 1]);
if (row > 0)
toggle_one(tab[row - 1][column]);
if (column < (tab.columns() - 1))
toggle_one(tab[row][column + 1]);
if (row < (tab.rows() - 1))
toggle_one(tab[row + 1][column]);
}
int solve(Arr_2d<bool> &tab)
{
int count = 0;
unsigned i = 0;
for ( ; i < (tab.rows() - 1); ++i)
for (unsigned j = 0; j < tab.columns(); ++j)
if (tab[i][j])
{
toggle(tab, i + 1, j);
++count;
}
for (unsigned j = 0; j < tab.columns(); ++j)
if (tab[i][j])
// Impossible.
return(-count);
return(count);
}
unsigned ex1[] = {
0, 1, 0,
1, 1, 1,
0, 1, 0
};
unsigned ex2[] = {
0, 1, 0, 0, 0, 0, 0, 0,
1, 1, 1, 0, 1, 0, 1, 0,
0, 0, 1, 1, 0, 1, 1, 1,
1, 1, 0, 1, 1, 1, 0, 0,
1, 0, 1, 1, 1, 0, 1, 0,
0, 1, 0, 1, 0, 1, 0, 0
};
Arr_2d<bool> load(unsigned rows, unsigned columns, const unsigned *data)
{
Arr_2d<bool> res(rows, columns);
for (unsigned i = 0; i < rows; ++i)
for (unsigned j = 0; j < columns; ++j)
res[i][j] = !!*(data++);
return(res);
}
#include <iostream>
int main()
{
{
Arr_2d<bool> tab = load(3, 3, ex1);
std::cout << solve(tab) << '\n';
}
{
Arr_2d<bool> tab = load(6, 8, ex2);
std::cout << solve(tab) << '\n';
}
return(0);
}
The problem is stated like this:
y
yxy If you flip x, then you have to flip all the ys
y
But it's easy if you think about it like this:
x
yyy If you flip x, then you have to flip all the ys
y
It's the same thing, but now the solution is obvious -- You must flip all the 1s in row 0, which will flip some bits in rows 1 and 2, then you must flip all the 1s in row 1, etc, until you get to the end.
If this is indeed the Lights Out game, then there are plenty of resources that detail how to solve the game. It is also quite likely that this is a duplicate of Lights out game algorithm, as has already been mentioned by other posters.
Let's see if we can't solve the first sample puzzle provided, however, and at least present a concrete description of an algorithm.
The initial puzzle appears to be solvable:
1 0 1
1 1 1
0 1 0
The trick is that you can clear 1's in the top row by changing the values in the row underneath them. I'll provide coordinates by row and column, using a 1-based offset, meaning that the top left value is (1, 1) and the bottom right value is (3, 3).
Change (2, 1), then (2, 3), then (3, 2). I'll show the intermediate states of the board with the * for the cell being changed in the next step.
1 0 1 (2,1) 0 0 1 (2,3) 0 0 0 (3, 2) 0 0 0
* 1 1 ------> 0 0 * ------> 0 1 0 ------> 0 0 0
0 1 0 1 1 0 1 * 1 0 0 0
This board can be solved, and the number of moves appears to be 3.
The pseudo-algorithm is as follows:
flipCount = 0
for each row _below_ the top row:
for each element in the current row:
if the element in the row above is 1, toggle the element in this row:
increment flipCount
if the board is clear, output flipCount
if the board isnt clear, output "Impossible"
I hope this helps; I can elaborate further if required but this is the core of the standard lights out solution. BTW, it is related to Gaussian Elimination; linear algebra crops up in some odd situations :)
Finally, in terms of what is wrong with your code, it appears to be the following loop:
for(int i=0; i<n-1; i++)
{
for(int j=0; j<m-1; j++)
{
if(tab[i][j] == 1 && i > 0 && j > 0)
{
tab[i-1][j] = !tab[i-1][j];
tab[i+1][j] = !tab[i+1][j];
tab[i][j+1] = !tab[i][j+1];
tab[i][j-1] = !tab[i][j-1];
tab[i][j] = !tab[i][j];
counter ++;
}
}
}
Several issues occur to me, but first assumptions again:
i refers to the ith row and there are n rows
j refers to the jth column and there are m columns
I'm now referring to indices that start from 0 instead of 1
If this is the case, then the following is observed:
You could run your for i loop from 1 instead of 0, which means you no longer have to check whether i > 0 in the if statement
You should drop the for j > 0 in the if statement; that check means that you can't flip anything in the first column
You need to change the n-1 in the for i loop as you need to run this for the final row
You need to change the m-1 in the for j loop as you need to run this for the final column (see point 2 also)
You need to check the cell in the row above the current row, so you you should be checking tab[i-1][j] == 1
Now you need to add bounds tests for j-1, j+1 and i+1 to avoid reading outside valid ranges of the matrix
Put these together and you have:
for(int i=1; i<n; i++)
{
for(int j=0; j<m; j++)
{
if(tab[i-1][j] == 1)
{
tab[i-1][j] = !tab[i-1][j];
if (i+1 < n)
tab[i+1][j] = !tab[i+1][j];
if (j+1 < m)
tab[i][j+1] = !tab[i][j+1];
if (j > 0)
tab[i][j-1] = !tab[i][j-1];
tab[i][j] = !tab[i][j];
counter ++;
}
}
}
A little class that can take as a input file or test all possible combination for first row with only zeros, on 6,5 matrix:
#include <iostream>
#include <fstream>
#include <vector>
#include <string>
#include <cstdlib>
#include <ctime>
typedef std::vector< std::vector<int> > Matrix;
class MatrixCleaner
{
public:
void swapElement(int row, int col)
{
if (row >= 0 && row < (int)matrix.size() && col >= 0 && col < (int)matrix[row].size())
matrix[row][col] = !matrix[row][col];
}
void swapElements(int row, int col)
{
swapElement(row - 1, col);
swapElement(row, col - 1);
swapElement(row, col);
swapElement(row, col + 1);
swapElement(row + 1, col);
}
void printMatrix()
{
for (auto &v : matrix)
{
for (auto &val : v)
{
std::cout << val << " ";
}
std::cout << "\n";
}
}
void loadMatrix(std::string path)
{
std::ifstream fileStream;
fileStream.open(path);
matrix.resize(1);
bool enconteredNumber = false;
bool skipLine = false;
bool skipBlock = false;
for (char c; fileStream.get(c);)
{
if (skipLine)
{
if (c != '*')
skipBlock = true;
if (c != '\n')
continue;
else
skipLine = false;
}
if (skipBlock)
{
if (c == '*')
skipBlock = false;
continue;
}
switch (c)
{
case '0':
matrix.back().push_back(0);
enconteredNumber = true;
break;
case '1':
matrix.back().push_back(1);
enconteredNumber = true;
break;
case '\n':
if (enconteredNumber)
{
matrix.resize(matrix.size() + 1);
enconteredNumber = false;
}
break;
case '#':
if(!skipBlock)
skipLine = true;
break;
case '*':
skipBlock = true;
break;
default:
break;
}
}
while (matrix.size() > 0 && matrix.back().empty())
matrix.pop_back();
fileStream.close();
}
void loadRandomValidMatrix(int seed = -1)
{
//Default matrix
matrix = {
{ 0,0,0,0,0 },
{ 0,0,0,0,0 },
{ 0,0,0,0,0 },
{ 0,0,0,0,0 },
{ 0,0,0,0,0 },
{ 0,0,0,0,0 },
};
int setNum = seed;
if(seed < 0)
if(seed < -1)
setNum = std::rand() % -seed;
else
setNum = std::rand() % 33554432;
for (size_t r = 1; r < matrix.size(); r++)
for (size_t c = 0; c < matrix[r].size(); c++)
{
if (setNum & 1)
swapElements(r, c);
setNum >>= 1;
}
}
bool test()
{
bool retVal = true;
for (int i = 0; i < 33554432; i++)
{
loadRandomValidMatrix(i);
if( (i % 1000000) == 0 )
std::cout << "i= " << i << "\n";
if (clean() < 0)
{
// std::cout << "x";
std::cout << "\n" << i << "\n";
retVal = false;
break;
}
else
{
// std::cout << ".";
}
}
return retVal;
}
int clean()
{
int numOfSwaps = 0;
try
{
for (size_t r = 1; r < matrix.size(); r++)
{
for (size_t c = 0; c < matrix[r].size(); c++)
{
if (matrix.at(r - 1).at(c))
{
swapElements(r, c);
numOfSwaps++;
}
}
}
}
catch (...)
{
return -2;
}
if (!matrix.empty())
for (auto &val : matrix.back())
{
if (val == 1)
{
numOfSwaps = -1;
break;
}
}
return numOfSwaps;
}
Matrix matrix;
};
int main(int argc, char **argv)
{
std::srand(std::time(NULL));
MatrixCleaner matrixSwaper;
if (argc > 1)
{
matrixSwaper.loadMatrix(argv[argc - 1]);
std::cout << "intput:\n";
matrixSwaper.printMatrix();
int numOfSwaps = matrixSwaper.clean();
std::cout << "\noutput:\n";
matrixSwaper.printMatrix();
if (numOfSwaps > 0)
std::cout << "\nresult = " << numOfSwaps << " matrix is clean now " << std::endl;
else if (numOfSwaps == 0)
std::cout << "\nresult = " << numOfSwaps << " nothing to clean " << std::endl;
else
std::cout << "\nresult = " << numOfSwaps << " matrix cannot be clean " << std::endl;
}
else
{
std::cout << "Testing ";
if (matrixSwaper.test())
std::cout << " PASS\n";
else
std::cout << " FAIL\n";
}
std::cin.ignore();
return 0;
}
Say I've got an N-dimensional boost::multi_array (of type int for simplicity), where N is known at compile time but can vary (i.e. is a non-type template parameter). Let's assume that all dimensions have equal size m.
typedef boost::multi_array<int, N> tDataArray;
boost::array<tDataArray::index, N> shape;
shape.fill(m);
tDataArray A(shape);
Now I would like to loop over all entries in A, e.g. to print them. If N was 2 for example I think I would write something like this
boost::array<tDataArray::index, 2> index;
for ( int i = 0; i < m; i++ )
{
for ( int j = 0; j < m; j++ )
{
index = {{ i, j }};
cout << A ( index ) << endl;
}
}
I've used an index object to access the elements as I think this is more flexible than the []-operator here.
But how could I write this without knowing the number of dimensions N. Is there any built-in way? The documentation of multi_array is not very clear on which types of iterators exist, etc.
Or would I have to resort to some custom method with custom pointers, computing indices from the pointers, etc.? If so - any suggestions how such an algorithm could look like?
Ok, based on the Google groups discussion already mentioned in one of the comments and on one of the examples from the library itself, here is a possible solution that lets you iterate over all values in the multi-array in a single loop and offers a way to retrieve the index for each of these elements (in case this is needed for some other stuff, as in my scenario).
#include <iostream>
#include <boost/multi_array.hpp>
#include <boost/array.hpp>
const unsigned short int DIM = 3;
typedef double tValue;
typedef boost::multi_array<tValue,DIM> tArray;
typedef tArray::index tIndex;
typedef boost::array<tIndex, DIM> tIndexArray;
tIndex getIndex(const tArray& m, const tValue* requestedElement, const unsigned short int direction)
{
int offset = requestedElement - m.origin();
return(offset / m.strides()[direction] % m.shape()[direction] + m.index_bases()[direction]);
}
tIndexArray getIndexArray( const tArray& m, const tValue* requestedElement )
{
tIndexArray _index;
for ( unsigned int dir = 0; dir < DIM; dir++ )
{
_index[dir] = getIndex( m, requestedElement, dir );
}
return _index;
}
int main()
{
double* exampleData = new double[24];
for ( int i = 0; i < 24; i++ ) { exampleData[i] = i; }
tArray A( boost::extents[2][3][4] );
A.assign(exampleData,exampleData+24);
tValue* p = A.data();
tIndexArray index;
for ( int i = 0; i < A.num_elements(); i++ )
{
index = getIndexArray( A, p );
std::cout << index[0] << " " << index[1] << " " << index[2] << " value = " << A(index) << " check = " << *p << std::endl;
++p;
}
return 0;
}
The output should be
0 0 0 value = 0 check = 0
0 0 1 value = 1 check = 1
0 0 2 value = 2 check = 2
0 0 3 value = 3 check = 3
0 1 0 value = 4 check = 4
0 1 1 value = 5 check = 5
0 1 2 value = 6 check = 6
0 1 3 value = 7 check = 7
0 2 0 value = 8 check = 8
0 2 1 value = 9 check = 9
0 2 2 value = 10 check = 10
0 2 3 value = 11 check = 11
1 0 0 value = 12 check = 12
1 0 1 value = 13 check = 13
1 0 2 value = 14 check = 14
1 0 3 value = 15 check = 15
1 1 0 value = 16 check = 16
1 1 1 value = 17 check = 17
1 1 2 value = 18 check = 18
1 1 3 value = 19 check = 19
1 2 0 value = 20 check = 20
1 2 1 value = 21 check = 21
1 2 2 value = 22 check = 22
1 2 3 value = 23 check = 23
so the memory layout goes from the outer to the inner indices. Note that the getIndex function relies on the default memory layout provided by boost::multi_array. In case the array base or the storage ordering are changed, this would have to be adjusted.
There is a lack of simple boost multi array examples. So here is a very simple example of how to fill a boost multi array using indexes and how to read all the entries using a single pointer.
typedef boost::multi_array<double, 2> array_type;
typedef array_type::index index;
array_type A(boost::extents[3][2]);
// ------> x
// | 0 2 4
// | 1 3 5
// v
// y
double value = 0;
for(index x = 0; x < 3; ++x)
for(index y = 0; y < 2; ++y)
A[x][y] = value++;
double* it = A.origin();
double* end = A.origin() + A.num_elements();
for(; it != end; ++it){
std::cout << *it << " ";
}
// -> 0 1 2 3 4 5
If you don't need the index, you can simply do:
for (unsigned int i = 0; i < A.num_elements(); i++ )
{
tValue item = A.data()[i];
std::cout << item << std::endl;
}
Based on the answers before I produced this nice overloaded version of the insertion operator for boost::multi_arrays
using namespace std;
using namespace boost::detail::multi_array;
template <typename T , unsigned long K>
ostream &operator<<( ostream &os , const boost::multi_array<T , K> &A )
{
const T* p = A.data();
for( boost::multi_array_types::size_type i = A.num_elements() ; i-- ; ++p )
{
os << "[ ";
for( boost::multi_array_types::size_type k = 0 ; k < K ; ) {
os << ( p - A.origin() ) / A.strides()[ k ] % A.shape()[ k ]
+ A.index_bases()[ k ];
if( ++k < K )
os << ", ";
}
os << " ] = " << *p << endl;
}
return os;
}
It's just a streamlined version of answer 1, except it should work with any type T that has a working operator<<. I tested like
typedef boost::multi_array<double, 3> array_type;
typedef array_type::index index;
index x = 3;
index y = 2;
index z = 3;
array_type A( boost::extents[ x ][ y ][ z ] );
// Assign values to the elements
int values = 0;
for( index i = 0 ; i < x ; ++i )
for( index j = 0 ; j < y ; ++j )
for( index k = 0 ; k < z ; ++k )
A[ i ][ j ][ k ] = values++;
// print the results
cout << A << endl;
and it seems to work:
[ 0, 0, 0 ] = 0
[ 0, 0, 1 ] = 1
[ 0, 0, 2 ] = 2
[ 0, 1, 0 ] = 3
[ 0, 1, 1 ] = 4
[ 0, 1, 2 ] = 5
[ 1, 0, 0 ] = 6
[ 1, 0, 1 ] = 7
[ 1, 0, 2 ] = 8
[ 1, 1, 0 ] = 9
[ 1, 1, 1 ] = 10
[ 1, 1, 2 ] = 11
[ 2, 0, 0 ] = 12
[ 2, 0, 1 ] = 13
[ 2, 0, 2 ] = 14
[ 2, 1, 0 ] = 15
[ 2, 1, 1 ] = 16
[ 2, 1, 2 ] = 17
Hope this is useful to somebody, and thanks a lot for the original answers: it was very useful to me.