Faster way to find the closest pair of points - performance

Question

I'm trying to compare all elements of an array to each other. I can do it with nested loops but it's a very inefficient algorithm and I can tell it's not the right way to do it. Here's what I'm doing right now.
PER answers below I've changed this code and I'm expanding on the question.
// Point from java.awt.Point;
private static void findShortestDistance(Point[] pt) {
ArrayList<Double> distance = new ArrayList<Double>(1000);
for(int i=0; i<pt.length; i++) {
for(int j=i+1; j<pt.length; j++) {
double tmp = pt[i].distance(pt[j]);
distance.add(tmp);
}
}
double min = distance.get(0);
for(Double d : distance) {
if(d < min) { min = d; }
}
}
There's the full code for the method I have so far. I'm trying to find the shortest distance between two points in the given array.

Answer 1

Have a look at wikipedia.
http://en.wikipedia.org/wiki/Closest_pair_of_points
And this question seems to be the same as
Shortest distance between points algorithm

Answer 2

As long as you say the above code is exactly what you want the following will be the implementation you want:
for(int i=0; i<pt.length; i++) {
for(int j = i + 1; j<pt.length; j++) {
/* do stuff */
}
}
However, I have a god feeling you are actually interested in comparing the array values or am I wrong?

Answer 3

for(int i=0; i<pt.length; i++) {
for(int j=i+1; j<pt.length; j++) {
if(pt[i] != pt[j]) { /* do stuff */ }
}
}

Answer 4

for(int i=0; i<pt.length; i++) {
for(int j=i; j--; ) {
{ /* do stuff */ }
}
}
And, for completeness:
for(int i=pt.length; i--; ) {
for(int j=i; j--; ) {
{ /* do stuff */ }
}
}

Answer 5

Initializing j to i+1 will skip the redundant comparisons.
for(int i = 0; i < pt.length; i++) {
for(int j = i + 1; j < pt.length; j++) {
if(pt[i] != pt[j]) { /* do stuff */ }
}
}
(BTW, changed the if-statement, assuming you meant to use i and j as indexes into an array.)
That's about the best you can get for the most basic case where you need to get a cross product of the array with itself, minus all the elements {x, y} | x == y. (i.e. an exhaustive list of unordered pairs of differing elements.) If you need ordered pairs of differing elements, then your code is best.