找到（x，y）坐标之间的最大距离

Question

Im trying to calculate the maximum manhattan distance of a large 2D input , the inputs are consisting of (x, y)s and what I want to do is to calculate the maximum distance between those coordinates In less than O(n^2) time , I can do it in O(n^2) by going through all of elements sth like :
*(Manhattan distance between two points (X1,Y1) and (X2,Y2) is : |X1-X2| + |Y1-Y2|)

for ( 0 -> n )  
   for ( 0-> n )   
       { // here i calculate |Xi - Xj| + |Yi - Yj| which is maximum }  

but It won't work efficiently for very large inputs :(
anyone has any idea for a better algorithm ?

Answer 1

There are only two cases to consider, if we only consider results such that Xi <= Xj .

• If Yi <= Yj , then the distance is (Xj + Yj) - (Xi + Yi)
• Otherwise, the distance is (Xj - Yj) - (Xi - Yi)

By breaking it down into these cases, I have gotten rid of the absolute value function making it much easier to reason about the distances.

So, we simply pick points with minimum and maximum x+y , and compute the distance. Then pick points with minimum and maximum xy , and compute the distance. One of those two distances is your maximum.

This can be done in O(n) , which is asymptotically optimal.

Answer 2

It is fairly simple and can be calculated in O(n)

Let x1>x2 and y1>y2

max(|x1-x2|+|y1-y2|) = max(x1-x2+y1-y2) = max(x1+y1) - min(x2+y2)

Let x1>x2 and y1<y2

max(|x1-x2|+|y1-y2|) = max(x1-x2-y1+y2) = max(x1-y1) - min(x2-y2)

Now change x1 with x2 and you take the same results.

So in general your solution is

max ( (max(xi+yi)-min(xi+yi)), (max(xi-yi) - min(xi-yi)) ) 

Answer 3

The best thing to do with questions like this is try to establish some small results that will help you with the overall problem.

For example, it is not too hard to determine that for any three points, A, B and C, which have the condition that B is between (more on this in a second) A and C, B will never be further from a fourth point D than one of A and C. With the standard Euclidean metric of distance, a point is between two other points if it lies on the segment joining them. For Manhattan measurements it is not so simple - partly because the concept of a segment is not as well understood.

A more general way of describing 'between' is this (using the notation that the distance from A to B is |AB|): A point B is between two points A, C if |AB| + |BC| = |AC|

You can see that in Euclidean distance this means that B lies on the segment joining A and C.

In Manhattan distance, this means that the point B is contained in the rectangle defined by A and C (which of course could be a straight segment if AC is parallel one of the axis).

This result means that for any point, if it lies between two existing points, it can be no further from any new points that are added to the set than the two which surround it.

Now, this information does not solve the issue for you, but it does let you throw away many potential future calculations. Once you have determined that a point is between two others, there is no point in tracking it.

So, you can solve this problem by only tracking the outermost points, and disregarding any that fall within.

An interesting exercise for the casual observer

Prove that you can have no more than 4 distinct points such that none of the points are between two of the others, in the Manhattan sense.

With this second result it becomes clear that you will only ever need to track up to 4 points.

Some of the other methods already presented are probably faster, but this way is more fun!

Extra Credit

Generalise these ideas to n dimensions

Answer 4

the first big improvement would be:

for ( X: 0 -> n )
    for ( Y: X -> n )
        { compute the distance between X and Y }

since the distance between X and Y is the same as the distance between Y and X. that would cut your computation by a half...

Answer 5

Maximum distance will be between most far from each other points. So you just have find dot with maximum X and maximum Y and then find dot with minimum X and minimum Y and calculate distance between them. There can be a lot of dots which will match criteria.. but at least yu'll have a much less amount of points to check