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.
Hence, the maximum possible number of squares that can be formed using 12 straight lines is 55.
The maximum number of rectangles that can be counted in a shape constructed using all six sticks is 9 (Option C).
Number of rectangles are =m(m+1)n(n+1)/4=2×4×3×5/4=30.
The shape of a rectangle has four lines, two of which are vertical lines. The other two lines are horizontal lines.
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.
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
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