Given a rectilinear path, find if it has a loop. A rectilinear path is made up of made up of alternating horizontal and vertical segments.
Input => Ordered set of points representing ra ectilinear path. Points have been sanitised to ensure that they represent alternating horizontal and vertical segments. This means there are no two consecutive horizontal (or vertical) segments.
Output => True
if it has loop, False
otherwise.
I could think of two algorithms.
Algorithm 1
For a loop, there must be crossing between horizontal and vertical line segments.
Crossing/Overlap of two horizontal (or vertical) line segments can't lead to loop.
- Break path into horizontal and vertical line segments.
- Check if any horizontal line segment crosses any vertical line segment.
- If crossing found return
True
. Else returnFalse
Algorithm 1 Complexity:
- N points means N-1 line segments.
- Vertical segment count is (N-1)/2. Same for horizontal segment count.
- Checking each pair of horizontal and vertical segments means (N-1)(N-2)/4 check. That gives complexity O(N²).
Algorithm 2
For a loop, there must be crossing between horizontal and vertical line segments.
These line segments shouldn't be consecutive to each other in rectilinear path.
- Break path into horizontal and vertical line segments.
- Sort vertical line segments based on their x-values.
- Iterate over horizontal line segments and check if any vertical line segment falls in its x-range. Use binary search over sorted vertical segments to find candidate vertical segments.
- Check if horizontal line segment cross with any of the vertical line segments.
- If crossing found return
True
. Else returnFalse
Algorithm 2 Complexity:
- Split into horizontal and vertical segments = O(N).
- vertical segment count = (N-1)/2.
- Sorting vertical segments: O(NLogN)
- Iteration over horizontal segment = O(N).
- For each iteration, Binary search O(LogN).
For each iteration, Check for crossing with candidate vertical segments. Worst case (N-1)/4. - Worst case complexity remains O(N²). But number of pairs checked will be less than for Algorithm 1.
Implementation of Algorithm 2
#Code for Algorithm 2
from functools import cmp_to_key
# Represents line_segment which is either horizontal or vertical.
class line_segment:
__start_point = (0, 0)
__end_point = (0, 0)
def __init__(self, start_point, end_point):
if start_point[0] == end_point[0]:
self.__start_point = (start_point, end_point)[start_point[1] > end_point[1]]
self.__end_point = (start_point, end_point)[start_point[1] < end_point[1]]
else:
self.__start_point = (start_point, end_point)[start_point[0] > end_point[0]]
self.__end_point = (start_point, end_point)[start_point[0] < end_point[0]]
def does_intersect(self, target_line_segment):
is_vertical = self.is_segment_vertical()
is_traget_vertical = target_line_segment.is_segment_vertical()
# Check for parallel segments
if is_vertical and is_traget_vertical:
return False
if is_vertical:
return self.__start_point[0] >= target_line_segment.__start_point[0] and \
self.__start_point[0] <= target_line_segment.__end_point[0] and \
target_line_segment.__start_point[1] >= self.__start_point[1] and \
target_line_segment.__start_point[1] <= self.__end_point[1]
else:
return target_line_segment.__start_point[0] >= self.__start_point[0] and \
target_line_segment.__start_point[0] <= self.__end_point[0] and \
self.__start_point[1] >= target_line_segment.__start_point[1] and \
self.__start_point[1] <= target_line_segment.__end_point[1]
def is_segment_vertical(self):
return self.__start_point[0] == self.__end_point[0]
def get_value(self):
if self.is_segment_vertical():
return self.__start_point[0]
else:
return self.__start_point[1]
def get_non_constant_start_coordinate(self):
if self.is_segment_vertical():
return self.__start_point[1]
else:
return self.__start_point[0]
def get_non_constant_end_coordinate(self):
if self.is_segment_vertical():
return self.__end_point[1]
else:
return self.__end_point[0]
# Line segment comparator
def compare(item_1, item_2):
return item_1[0].get_value() - item_2[0].get_value()
def binary_serach_comparator(segment, search_value):
return segment[0].get_value() - search_value
def binary_serach(sorted_collection, serach_value, comparator):
high = len(sorted_collection) - 1
low = 0
index = -1
mid = 0
while(low <= high):
mid = int((low + high)/2)
comparator_value = comparator(sorted_collection[mid], serach_value)
if comparator_value < 0:
low = mid + 1
elif comparator_value > 0:
high = mid - 1
else:
index = mid
break
return (index, low, high)
def split_path_in_segments(path_points):
vertical_segment_start_index = (0, 1) [path_points[0][0] == path_points[1][0]]
vertical_segments = [(line_segment(path_points[index], path_points[index + 1]), index)\
for index in range(vertical_segment_start_index, len(path_points) - 1, 2)]
horizontal_segments = [(line_segment(path_points[index], path_points[index + 1]), index)\
for index in range(int(not(vertical_segment_start_index)), len(path_points) - 1, 2)]
return vertical_segments, horizontal_segments
def find_segments_in_range(segments, range_start, range_end):
(start_index, start_low, start_high) = binary_serach(segments, range_start, binary_serach_comparator)
(end_index, end_low, end_high) = binary_serach(segments, range_end, binary_serach_comparator)
return (start_low, end_high)
# Input: Ordered set of points representing rectilinear paths
# which is made up of alternating horizontal and vertical segments
def check_loop(path_points):
# For loop we need 4 or more segments. Hence more than 5 points
if len(path_points) <= 4:
return False
vertical_segments, horizontal_segments = split_path_in_segments(path_points)
# Sort vertical segmnets for easy serach
vertical_segments = sorted(vertical_segments, key=cmp_to_key(compare))
# Iterate through horizontal segments, find vertical segments
# which fall in rane of horizontal segment and check for intersection
for horizontal_counter in range(len(horizontal_segments)):
horizontal_segment = horizontal_segments[horizontal_counter][0]
horizontal_segment_index = horizontal_segments[horizontal_counter][1]
(start, end) = find_segments_in_range(vertical_segments,\
horizontal_segment.get_non_constant_start_coordinate(),\
horizontal_segment.get_non_constant_end_coordinate())
for vertical_counter in range(start, end + 1):
vertical_segment = vertical_segments[vertical_counter][0]
vertical_segment_index = vertical_segments[vertical_counter][1]
# Avoid adjacent segments. They will always have one endpoint in common
if abs(horizontal_segment_index - vertical_segment_index) <= 1:
continue
if horizontal_segment.does_intersect(vertical_segment):
return True
return False
print(check_loop([(0,0), (5,0), (5, 5)])) # False
print(check_loop([(0,0), (5,0), (5, 5), (0, 5), (0, 0)])) # True
print(check_loop([(0,0), (5,0), (5, 5), (4, 5)])) # False
print(check_loop([(0,0), (5,0), (5, 5), (4, 5), (4, 2)])) # False
print(check_loop([(0,0), (5,0), (5, 5), (4, 5), (4, -1)])) # True
print(check_loop([(0,0), (5,0), (5, 5), (4, 5), (4, 0)])) # True
print(check_loop([(0,0), (5,0), (5, 5), (8, 5), (8, 2), (10, 2)]))# False
print(check_loop([(0,0), (5,0), (5, 5), (8, 5), (8, 2), (10, 2), (11, 2), (11, 1), (-5, 1), (-5, 15)]))# False
print(check_loop([(0,0), (5,0), (5, 5), (8, 5), (8, 2), (10, 2), (10, -1), (2, -1), (2, 15)]))# True
Review request
- Algorithm improvements.
- Functional correctness of implemented algorithm.
- Boundary and error cases.
- Python-specific feedback.
binary_serach
? You'll make your code more maintainable if you correct the spellings. \$\endgroup\$