I found this challenge problem which states the following :
Suppose that there are n rectangles on the XY plane. Write a program to calculate the maximum possible number of rectangles that can be crossed with a single straight line drawn on this plane.
I have been brainstorming for quite a time but couldn't find any solution. Maybe at some stage, we use dynamic programming steps but couldn't figure out how to start.
Here is a sketch of an O(n^2 log n) solution.
First, the preliminaries shared with other answers. When we have a line passing through some rectangles, we can translate it to any of the two sides until it passes through a corner of some rectangle. After that, we fix that corner as the center of rotation and rotate the line to any of the two sides until it passes through another corner. During the whole process, all points of intersection between our line and rectangle sides stayed on these sides, so the number of intersections stayed the same, as did the number of rectangles crossed by the line. As a result, we can consider only lines which pass through two rectangle corners, which is capped by O(n^2), and is a welcome improvement compared to the infinite space of arbitrary lines.
So, how do we efficiently check all these lines? First, let us have an outer loop which fixes one point A and then considers all lines passing through A. There are O(n) choices of A.
Now, we have one point A fixed, and want to consider all lines AB passing through all other corners B. In order to do that, first sort all other corners B according to the polar angle of AB, or, in other words, angle between axis Ox and vector AB. Angles are measured from -PI to +PI or from 0 to 2 PI or otherwise, the point in which we cut the circle to sort angles can be arbitrary. The sorting is done in O(n log n).
Now, we have points B1, B2, ..., Bk sorted by the polar angle around point A (their number k is something like 4n-4, all corners of all rectangles except the one where point A is a corner). First, look at the line AB1 and count the number of rectangles crossed by that line in O(n). After that, consider rotating AB1 to AB2, then AB2 to AB3, all the way to ABk. The events which happen during the rotation are as follows:
When we rotate to ABi, and Bi is the first corner of some rectangle in our order, the number of rectangles crossed increases by 1 as soon as the rotating line hits Bi.
When we rotate to ABj, and Bj is the last corner of some rectangle in our order, the number of rectangles crossed decreases by 1 as soon as the line rotates past Bj.
Which corners are first and last can be established with some O(n) preprocessing, after the sort, but before considering the ordered events.
In short, we can rotate to the next such event and update the number of rectangles crossed in O(1). And there are k = O(n) events in total. What's left to do is to track the global maximum of this quantity throughout the whole algorithm. The answer is just this maximum.
The whole algorithm runs in O(n * (n log n + n + n)), which is O(n^2 log n), just as advertised.
(Edit of my earlier answer that considered rotating the plane.)
Here's sketch of the O(n^2) algorithm, which combines Gassa's idea with Evgeny Kluev's reference to dual line arrangements as sorted angular sequences.
We start out with a doubly connected edge list or similar structure, allowing us to split an edge in O(1) time, and a method to traverse the faces we create as we populate a 2-dimensional plane. For simplicity, let's use just three of the twelve corners on the rectangles below:
9| (5,9)___(7,9)
8| | |
7| (4,6)| |
6| ___C | |
5| | | | |
4| |___| | |
3| ___ |___|(7,3)
2| | | B (5,3)
1|A|___|(1,1)
|_ _ _ _ _ _ _ _
1 2 3 4 5 6 7
We insert the three points (corners) in the dual plane according to the following transformation:
point p => line p* as a*p_x - p_y
line l as ax + b => point l* as (a, -b)
Let's enter the points in order A, B, C. We first enter A => y = x - 1. Since there is only one edge so far, we insert B => y = 5x - 3, which creates the vertex, (1/2, -1/2) and splits our edge. (One elegant aspect of this solution is that each vertex (point) in the dual plane is actually the dual point of the line passing through the rectangles' corners. Observe 1 = 1/2*1 + 1/2 and 3 = 1/2*5 + 1/2, points (1,1) and (5,3).)
Entering the last point, C => y = 4x - 6, we now look for the leftmost face (could be an incomplete face) where it will intersect. This search is O(n) time since we have to try each face. We find and create the vertex (-3, -18), splitting the lower edge of 5x - 3 and traverse up the edges to split the right half of x - 1 at vertex (5/3, 2/3). Each insertion has O(n) time since we must first find the leftmost face, then traverse each face to split edges and mark the vertices (intersection points for the line).
In the dual plane we now have:
After constructing the line arrangement, we begin our iteration on our three example points (rectangle corners). Part of the magic in reconstructing a sorted angular sequence in relation to one point is partitioning the angles (each corresponding with an ordered line intersection in the dual plane) into those corresponding with a point on the right (with a greater x-coordinate) and those on the left and concatenating the two sequences to get an ordered sequence from -90 deg to -270 degrees. (The points on the right transform to lines with positive slopes in relation to the fixed point; the ones on left, with negative slopes. Rotate your sevice/screen clockwise until the line for (C*) 4x - 6 becomes horizontal and you'll see that B* now has a positive slope and A* negative.)
Why does it work? If a point p in the original plane is transformed into a line p* in the dual plane, then traversing that dual line from left to right corresponds with rotating a line around p in the original plane that also passes through p. The dual line marks all the slopes of this rotating line by the x-coordinate from negative infinity (vertical) to zero (horizontal) to infinity (vertical again).
(Let's summarize the rectangle-count-logic, updating the count_array for the current rectangle while iterating through the angular sequence: if it's 1, increment the current intersection count; if it's 4 and the line is not directly on a corner, set it to 0 and decrement the current intersection count.)
Pick A, lookup A*
=> x - 1.
Obtain the concatenated sequence by traversing the edges in O(n)
=> [(B*) 5x - 3, (C*) 4x - 6] ++ [No points left of A]
Initialise an empty counter array, count_array of length n-1
Initialise a pointer, ptr, to track rectangle corners passed in
the opposite direction of the current vector.
Iterate:
vertex (1/2, -1/2)
=> line y = 1/2x + 1/2 (AB)
perform rectangle-count-logic
if the slope is positive (1/2 is positive):
while the point at ptr is higher than the line:
perform rectangle-count-logic
else if the slope is negative:
while the point at ptr is lower than the line:
perform rectangle-count-logic
=> ptr passes through the rest of the points up to the corner
across from C, so intersection count is unchanged
vertex (5/3, 2/3)
=> line y = 5/3x - 2/3 (AC)
We can see that (5,9) is above the line through AC (y = 5/3x - 2/3), which means at this point we would have counted the intersection with the rightmost rectangle and not yet reset the count for it, totaling 3 rectangles for this line.
We can also see in the graph of the dual plane, the other angular sequences:
for point B => B* => 5x - 3: [No points right of B] ++ [(C*) 4x - 6, (A*) x - 1]
for point C => C* => 4x - 6: [(B*) 5x - 3] ++ [(A*) x - 1]
(note that we start at -90 deg up to -270 deg)
In the space of all lines in the graph, the lines which pass by a corner are exactly the ones where the number or intersections is about to decrease. In other words, they each form a local maximum.
And for every line which passes by at least one corner, there exist an associated line that passes by two corners that has the same number of intersections.
The conclusion is that we only need to check the lines formed by two rectangle corners as they form a set that fully represents the local maxima of our problem. From those we pick the one which has the most intersections.
This solution first needs to recovers all lines that pass by two corners. The number of such line is O(n^2).
We then need to count the number of intersections between a given line and a rectangle. This can obviously be done in O(n) by comparing to each rectangles.
There might be a more efficient way to proceed, but we know that this algorithm is then at most O(n^3).
Here is a Python implementation of this algorithm. I oriented it more toward readability than efficiency, but it does exactly what the above defines.
def get_best_line(rectangles):
"""
Given a set of rectangles, return a line which intersects the most rectangles.
"""
# Recover all corners from all rectangles
corners = set()
for rectangle in rectangles:
corners |= set(rectangle.corners)
corners = list(corners)
# Recover all lines passing by two corners
lines = get_all_lines(corners)
# Return the one which has the highest number of intersections with rectangles
return max(
((line, count_intersections(rectangles, line)) for line in lines),
key=lambda x: x[1])
This implementation uses the following helpers.
def get_all_lines(points):
"""
Return a generator providing all lines generated
by a combination of two points out of 'points'
"""
for i in range(len(points)):
for j in range(i, len(points)):
yield Line(points[i], points[j])
def count_intersections(rectangles, line):
"""
Return the number of intersections with rectangles
"""
count = 0
for rectangle in rectangles:
if line in rectangle:
count += 1
return count
And here are the class definition that serve as data structure for rectangles and lines.
import itertools
from decimal import Decimal
class Rectangle:
def __init__(self, x_range, y_range):
"""
a rectangle is defined as a range in x and a range in y.
By example, the rectangle (0, 0), (0, 1), (1, 0), (1, 1) is given by
Rectangle((0, 1), (0, 1))
"""
self.x_range = sorted(x_range)
self.y_range = sorted(y_range)
def __contains__(self, line):
"""
Return whether 'line' intersects the rectangle.
To do so we check if the line intersects one of the diagonals of the rectangle
"""
c1, c2, c3, c4 = self.corners
x1 = line.intersect(Line(c1, c4))
x2 = line.intersect(Line(c2, c3))
if x1 is True or x2 is True \
or x1 is not None and self.x_range[0] <= x1 <= self.x_range[1] \
or x2 is not None and self.x_range[0] <= x2 <= self.x_range[1]:
return True
else:
return False
@property
def corners(self):
"""Return the corners of the rectangle sorted in dictionary order"""
return sorted(itertools.product(self.x_range, self.y_range))
class Line:
def __init__(self, point1, point2):
"""A line is defined by two points in the graph"""
x1, y1 = Decimal(point1[0]), Decimal(point1[1])
x2, y2 = Decimal(point2[0]), Decimal(point2[1])
self.point1 = (x1, y1)
self.point2 = (x2, y2)
def __str__(self):
"""Allows to print the equation of the line"""
if self.slope == float('inf'):
return "y = {}".format(self.point1[0])
else:
return "y = {} * x + {}".format(round(self.slope, 2), round(self.origin, 2))
@property
def slope(self):
"""Return the slope of the line, returning inf if it is a vertical line"""
x1, y1, x2, y2 = *self.point1, *self.point2
return (y2 - y1) / (x2 - x1) if x1 != x2 else float('inf')
@property
def origin(self):
"""Return the origin of the line, returning None if it is a vertical line"""
x, y = self.point1
return y - x * self.slope if self.slope != float('inf') else None
def intersect(self, other):
"""
Checks if two lines intersect.
Case where they intersect: return the x coordinate of the intersection
Case where they do not intersect: return None
Case where they are superposed: return True
"""
if self.slope == other.slope:
if self.origin != other.origin:
return None
else:
return True
elif self.slope == float('inf'):
return self.point1[0]
elif other.slope == float('inf'):
return other.point1[0]
elif self.slope == 0:
return other.slope * self.origin + other.origin
elif other.slope == 0:
return self.slope * other.origin + self.origin
else:
return (other.origin - self.origin) / (self.slope - other.slope)
Here is a working example of the above code.
rectangles = [
Rectangle([0.5, 1], [0, 1]),
Rectangle([0, 1], [1, 2]),
Rectangle([0, 1], [2, 3]),
Rectangle([2, 4], [2, 3]),
]
# Which represents the following rectangles (not quite to scale)
#
# *
# *
#
# ** **
# ** **
#
# **
# **
We can clearly see that an optimal solution should find a line that passes by three rectangles and that is indeed what it outputs.
print('{} with {} intersections'.format(*get_best_line(rectangles)))
# prints: y = 0.50 * x + -5.00 with 3 intersections
How about the following algorithm:
RES = 0 // maximum number of intersections
CORNERS[] // all rectangles corners listed as (x, y) points
for A in CORNERS
for B in CORNERS // optimization: starting from corner next to A
RES = max(RES, CountIntersectionsWithLine(A.x, A.y, B.x, B.y))
return RES
In other words, start drawing lines from each rectangle corner to each other rectangle corner and find the maximum number of intersections. As suggested by @weston, we can avoid calculating same line twice by starting inner loop from the corner next to A.
If you consider a rotating line at angle Θ and if you project all rectangles onto this line, you obtain N line segments. The maximum number of rectangles crossed by a perpendicular to this line is easily obtained by sorting the endpoints by increasing abscissa and keeping a count of the intervals met from left to right (keep a trace of whether an endpoint is a start or an end). This is shown in green.
Now two rectangles are intersected by all the lines at an angle comprised between the two internal tangents [example in red], so that all "event" angles to be considered (i.e. all angles for which a change of count can be observed) are these N(N-1) angles.
Then the brute force resolution scheme is
for all limit angles (O(N²) of them),
project the rectangles on the rotating line (O(N) operations),
count the overlaps and keep the largest (O(N Log N) to sort, then O(N) to count).
This takes in total O(N³Log N) operations.
Assuming that the sorts needn't be re-done in full for every angle if we can do them incrementally, we can hope for a complexity lowered to O(N³). This needs to be checked.
Note:
The solutions that restrict the lines to pass through the corner of one rectangle are wrong. If you draw wedges from the four corners of a rectangle to the whole extent of another, there will remain empty space in which can lie a whole rectangle that won't be touched, even though there exists a line through the three of them.
We can have an O(n^2 (log n + m)) dynamic-programming method by adapting Andriy Berestovskyy's idea of iterating over the corners slightly to insert the relationship of the current corner vis a vis all the other rectangles into an interval tree for each of our 4n iteration cycles.
A new tree will be created for the corner we are trying. For each rectangle's four corners we'll iterate over each of the other rectangles. What we'll insert will be the angles marking the arc the paired-rectangle's farthest corners create in relation to the current fixed corner.
In the example directly below, for the fixed lower rectangle's corner R when inserting the record for the middle rectangle, we would insert the angles marking the arc from p2 to p1 in relation to R (about (37 deg, 58 deg)). Then when we check the high rectangle in relation to R, we'll insert the interval of angles marking the arc from p4 to p3 in relation to R (about (50 deg, 62 deg)).
When we insert the next arc record, we'll check it against all intersecting intervals and keep a record of the most intersections.
(Note that because any arc on a 360 degree circle for our purpose has a counterpart rotated 180 degrees, we may need to make an arbitrary cutoff (any alternative insights would be welcome). For example, this means that an arc from 45 degrees to 315 degrees would split into two: [0, 45] and [135, 180]. Any non-split arc could only intersect with one or the other but either way, we may need an extra hash to make sure rectangles are not double-counted.)