import numpy as np
def simplify_coordinates(coordinates, tolerance):
    coordinates = np.array(coordinates)  # Convert to NumPy array
    if len(coordinates) <= 2:
        return coordinates.tolist()  # Convert back to list
    
    # Find the point with the maximum distance
    max_distance = 0
    index = 0
    end = len(coordinates) - 1
    for i in range(1, end):
        distance = perpendicular_distance(coordinates[i], coordinates[0], coordinates[end])
        if distance > max_distance:
            max_distance = distance
            index = i
    
    # If the maximum distance is greater than the tolerance, recursively simplify
    if max_distance > tolerance:
        simplified1 = simplify_coordinates(coordinates[:index+1], tolerance)
        simplified2 = simplify_coordinates(coordinates[index:], tolerance)
        return simplified1[:-1] + simplified2  # No need to convert back to list
    else:
        return [coordinates[0], coordinates[end]]  # No need to convert back to list

def perpendicular_distance(point, start, end):
    if np.all(start == end):
        return np.linalg.norm(np.array(point) - np.array(start))
    
    line_length = np.linalg.norm(np.array(end) - np.array(start))
    t = np.dot(np.array(point) - np.array(start), np.array(end) - np.array(start)) / (line_length * line_length)
    t = np.clip(t, 0, 1)
    closest_point = np.array(start) + t * (np.array(end) - np.array(start))
    return np.linalg.norm(np.array(point) - closest_point)