def flippingMatrix(matrix: list[list[int]]) -> int:
    """
    Since any number of column/row flips are possible,
    The maximum values must come from rotations around the center.
    Also, since the problem doesn't require the resultant matrix,
    we can simply determine the largest values for each position
    then return their sum.
    
    q=4
    n=q/2
    m=[
      [ 1, 2, 3, 4],
      [ 5, 6, 7, 8],
      [ 9,10,11,12],
      [13,14,15,16],
    ]
    Top left quad must consist of max for each rotation:
    0,0:max([1,4,13,16]) 0,1:max([2,3,14,15])
    1,0:max([5,8,9,12])  1,1:max([6,7,10,11])
    
    Generalized, these can be represented as follows:
    r,c: max([m[r][c], m[r][q-1-c], m[q-1-r][c], m[q-1-r][q-1-c]])
    
    find the max value for each r,c set
    return the sum of max values
    """
    
    q = len(matrix)
    n = q//2
    m = matrix
    
    # get quad value arrays
    quad_values = [
        [
            max([m[r][c], m[r][q-1-c], m[q-1-r][c], m[q-1-r][q-1-c]])
            for c in range(n)
        ]
        for r in range(n)
    ]
    print(quad_values)
    return sum([sum([c for c in r]) for r in quad_values])