Can the same point appear more than once as input to methods in Point?
slopeTo() method, this requirement is explicitly stated in the API;
for the comparison methods, this requirement is implicit in the
The reference solution outputs a line segment in the order p→q but my solution outputs it in the reverse order q→p. Is that ok? Yes, there are two valid ways to output a line segment.
The reference solution outputs the line segments in a different order than my solution. Is that OK? Yes, if there are k line segments, then there are k! different possible ways to output them.
How do I sort a subarray in Java?
Arrays.sort(a, lo, hi) sorts the subarray
a[hi-1] according to the natural
a. You can use a
Comparator as the fourth
argument to sort according to an alternate order.
Where can I see examples of Comparable and Comparator? See the lecture slides.
My program fails only on (some) vertical line segments. What could be going wrong?
Are you dividing by zero? With integers, this produces a run-time
exception. With floating-point numbers, 1.0/0.0 is positive infinity
and -1.0/0.0 is negative infinity. You may also use the constants
What does it mean for slopeTo() to return positive zero? Java (and the IEEE 754 floating-point standard) define two representations of zero: negative zero and positive zero.
Note that whiledouble a = 1.0; double x = (a - a) / a; // positive zero ( 0.0) double y = (a - a) / -a; // negative zero (-0.0)
(x == y)is guaranteed to be true, Arrays.sort() treats negative zero as strictly less than positive zero. Thus, to make the specification precise, we require you to return positive zero for horizontal line segments. Unless your program casts to the wrapper type
Double(either explicitly or via autoboxing), you probably will not notice any difference in behavior; but, if your program does cast to the wrapper type and fails only on (some) horizontal line segments, this may be the cause.
Is it OK to compare two floating-point numbers a and b for exactly equality?
In general, it is hazardous to compare a and b for equality if either is susceptible to
floating-point roundoff error.
However, in this case, you are computing b/a, where a and b are integers between -32,767 and 32,767.
In Java (and the IEEE 754 floating-point standard), the result of
a floating-point operation (such as division) is the nearest representable value. Thus, for example,
it is guaranteed that
(9.0/7.0 == 45.0/35.0).
In other words, it's sometimes OK to compare floating-point numbers for exact equality
(but only when you know exactly what you are doing!)
Note also that it is possible to implement
using only integer arithmetic (but you are not required to do so).
I'm having trouble avoiding subsegments Fast.java when there are 5 or more points on a line segment. Any advice? Not handling the 5-or-more case is a bit tricky, so don't kill yourself over it.
I created a nested Comparator class within Point. Within the nested Comparator class,
this refers to the Comparator object. How do I refer to the Point instance
of the outer class?
Point.this instead of
Note that you can refer directly to instance methods of the outer class (such as
with proper design, you shouldn't need this awkward notation.
Sample data files.
The zip file
contains some sample input files in the specified format.
Associated with some of the input .txt files are output .png files that contains the
desired graphical output.
Thanks to Jesse Levinson '05 for the remarkable input file
rs1423.txt; feel free
to create your own and share with us in the Discussion Forums.
slopeTo()method. Be sure to consider a variety of corner cases, including horizontal, vertical, and degenerate line segments.
Hint: don't waste time micro-optimizing the brute-force solution. Though, there are two easy opportunities. First, you can iterate through all combinations of 4 points (N choose 4) instead of all 4 tuples (N^4), saving a factor of 4! = 24. Second, you don't need to consider whether 4 points are collinear if you already know that the first 3 are not collinear; this can save you a factor of N on typical inputs.
Point. The complicating issue is that the comparator needed to compare the slopes that two points
rmake with a third point
p, which changes from sort to sort. To do this, create a private nested (non-static) class
SlopeOrderthat implements the
Comparator<Point>interface. This class has a single method
compare(q1, q2)that compares the slopes that
q2make with the invoking object
slopeOrder()method should create an instance of this nested class and return it.
Can the problem be solved in quadratic time and linear space? Yes, but the only compare-based algorithm I know of that guarantees quadratic time in the worst case is quite sophisticated. It involves converting the points to their dual line segments and topologically sweeping the arrangement of lines by Edelsbrunner and Guibas.
Can the decision version of the problem be solved in subquadratic time? The original version of the problem cannot be solved in subquadratic time because there might be a quadratic number of line segments to output. (See next question.) The decision version asks whether there exists a set of 4 collinear points. This version of the problem belongs to a group of problems that are known as 3SUM-hard. A famous unresolved conjecture is that such problems have no subquadratic algorithms. Thus, the sorting algorithm presented above is about the best we can hope for (unless the conjecture is wrong). Under a restricted decision tree model of computation, Erickson proved that the conjecture is true.
What's the maximum number of (maximal) collinear sets of points in a set of n points in the plane? It can grow quadratically as a function of N. Consider the n points of the form: (x, y) for x = 0, 1, 2, and 3 and y = 0, 1, 2, ..., n / 4. This means that if you store all of the (maximal) collinear sets of points, you will need quadratic space in the worst case.