What is the most efficient algorithm to find a straight line that goes through most points?

Question

The problem:

N points are given on a 2-dimensional plane. What is the maximum number of points on the same straight line?

The problem has O(N^2<

Answer 1

Again an O(n^2) solution with pseudo code. Idea is create a hash table with line itself as the key. Line is defined by slope between the two points, point where line cuts x-axis and point where line cuts y-axis.

Solution assumes languages like Java, C# where equals method and hashcode methods of the object are used for hashing function.

Create an Object (call SlopeObject) with 3 fields

1. Slope // Can be Infinity
2. Point of intercept with x-axis -- poix // Will be (Infinity, some y value) or (x value, 0)
3. Count

poix will be a point (x, y) pair. If line crosses x-axis the poix will (some number, 0). If line is parallel to x axis then poix = (Infinity, some number) where y value is where line crosses y axis. Override equals method where 2 objects are equal if Slope and poix are equal.

Hashcode is overridden with a function which provides hashcode based on combination of values of Slope and poix. Some pseudo code below

Hashmap map;
foreach(point in the array a) {
    foeach(every other point b) {
        slope = calculateSlope(a, b);
        poix = calculateXInterception(a, b);
        SlopeObject so = new SlopeObject(slope, poix, 1); // Slope, poix and intial count 1.
        SlopeObject inMapSlopeObj = map.get(so);
        if(inMapSlopeObj == null) {
            inMapSlopeObj.put(so);
        } else {
            inMapSlopeObj.setCount(inMapSlopeObj.getCount() + 1);
        }
    }
}
SlopeObject maxCounted = getObjectWithMaxCount(map);
print("line is through " + maxCounted.poix + " with slope " + maxCounted.slope);