def branch_solve(items, max_cost):
    """
    Finds the optimal value possible from a combination of Items, given a cost constraint
    Parameters:
        items, a list of Items to consider
        max_cost, the cost/weight constraint
    Returns:
        the list of optimal Items and the total value attained
    """

    # If there are no items left to choose from or no cost left to "spend"
    if not items or max_cost <= 0:
        return ([], 0) # Return an empty list and zero value

    # Examine the branch that doesn't include taking this item
    without_item, without_item_val = branch_solve(items[1:], max_cost)

    with_item_val = 0
    # If we can afford current item, examine branches that include it
    if items[0].cost <= max_cost: # Can afford current item
        # Explore options that include this item
        next_item = items[0]
        # Recursively examine all options with this item
        with_item, with_item_val = branch_solve(items[1:], max_cost - next_item.cost)
        with_item_val += next_item.value # Add value of next_item to total value

    # Choose better "branch"; i.e. determine if we should include this item in the solution
    if with_item_val > without_item_val:
        # If value with item is higher, include the item
        result = (with_item + [next_item], with_item_val)
    else:
        # If value without item is higher, do not include the item
        result = (without_item, without_item_val)

    # Return the list of items and the value they yield
    return result