@Carl Love The time complexity of your code in comparison with the original published Paar's algorithm is valuable and as you said this might be worth publishing. In continue I want to suggest an improvement for reduction of number of required XOR.
Consider the following 8x8 binary matrix
You can check that the number of required XOR for the implementation of the matrix A is 10 XOR when you run the Paar algorithm.
Set X:=[x_1,x_2,.,.x_8]. Then the cost of implementation X.A is equal to the cost of the equation (1) in my question.
Based on my question the cost of f[i] with i=1..7 is equal to 9 XOR. Now the question is that why the Paar algorithm computed one XOR more than the required XOR.
One answer is that in the question I selected columns with the minimum number of 1's and the Paar algorithm starts with columns with the maximum number of 1's.
I suggest you if you edit your code until that the output your edited code for the matrix A be 9 XOR, then your solution is a new improved heuristics for Short Linear Programs.
Although this types of improvement is recently done (you can see the following two papers), the time complexity of your code is amazing
Thanks for everything.