I have a matrix and I need to find a pattern inside this matrix. Matrix is:

1 0 0 1 1 1 0 0 0 1
0 0 0 1 1 0 1 0 0 1
0 1 1 1 0 0 0 1 0 1
1 0 1 0 0 1 1 0 1 0
1 1 1 0 0 0 1 1 0 1
0 1 0 0 1 1 0 1 0 1
1 1 1 0 0 0 1 0 0 1
1 0 0 1 0 1 1 1 0 1

Rules:

1. We choose one number from every row.
2. The next choosen number from second row must be an opposite of the precedent.
3. Positions of the numbers choosed by the 1 and 2 rules, must be a precise pattern.

So the question would be:

Find the best pattern that respect the 3 rules. Example from the matrix shown:

1. Choosed a number: 0(2) //what is in "()" represents the position of the value..position start from 1 to 10 on rows.
2. 1(4)
3. For the positions 2 and 4 to be a pattern must support rules 1 and 2 for the rest of the matrix.

So we go further on the 3rd row and we check 2nd position:1. We go 4th row, we check 4th position:0. Seems to respect the rules. There are opposite numbers on 2nd and 4th position, so we continue: 5th row, 2nd position:, and so on, but you will see on 7th row 2nd position:1 and 8th row 4th position:1; so the pattern of positions 2-4 is not good.

How could I make an algorithm based on these rules?

• algorithm
• math
• statistics
• pattern-matching

Maybe this will help (motivated by the comment to your question). This is a C++ sort of answer. This answer assumes 0 is always the number you pick, but you can easily edit this to allow 1 to be first.

int firstPos, secondPos;

for(int i = 0; i < 10; ++i)
    if(matrix[0][i] == 0)
        firstPos = i;

for(int i = 0; i < 10; ++i)
    if(matrix[0][i] == 1)
        secondPos= i;

bool success = true;

for(int i = 0; i < 10/2; ++i)
    if(matrix[2*i][firstPos] == matrix[2*i][secondPos])
        success == false;

if(success)
    cout << "success" << endl;
else 
    cout << "failure" << endl;

I would define a pattern by index of the first item (F) and index of the second item (S). I'll also assume that indices begin with 0 (instead of 1 as in your example). Both F and S can take a value between 0 and 9. Solution is simple. Have a double nested loop that runs F and S from 0 to 9, and in third innermost loop just verify that current F and S form a pattern.