Closest pair of four points c program - c

Question

I need to find closest pair of four points C program. This code for three points. I need this solution for four point.
I tried this. This solution for three input.
When I entering the three points then I will get the closest but I need the closest point of four points.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
struct Point
{
int x, y ;
};
double getDistanceAB(struct Point a, struct Point b)
{
double distanceAB;
distanceAB = sqrt((a.x - b.x) * (a.x - b.x) + (a.y-b.y) *(a.y-b.y));
return distanceAB;
}
double getDistanceBC(struct Point b, struct Point c)
{
double distanceBC;
distanceBC = sqrt((b.x - c.x) * (b.x - c.x) + (b.y-c.y) *(b.y-c.y));
return distanceBC;
}
double getDistanceAC(struct Point a, struct Point c)
{
double distanceAC;
distanceAC = sqrt((a.x - c.x) * (a.x - c.x) + (a.y-c.y) *(a.y-c.y));
return distanceAC;
}
int main()
{
struct Point a, b, c;
printf("Enter coordinate of points a: ");
scanf("%d %d", &a.x, &a.y);
printf("Enter coordinate of points b: ");
scanf("%d %d", &b.x, &b.y);
printf("Enter coordinate of points c: ");
scanf("%d %d", &c.x, &c.y);
if((getDistanceAB(a,b))>(getDistanceBC(b,c)) && (getDistanceAB(a,b))>(getDistanceBC(a,c)))
{
printf("Point A and B are closest.");
}
else if((getDistanceBC(b,c))>(getDistanceAC(a,c)) && (getDistanceBC(b,c))>(getDistanceAC(a,b)))
{
printf("Point B and C are closest.");
}
else if((getDistanceBC(a,c))>(getDistanceAC(a,b)) && (getDistanceBC(a,c))>(getDistanceAC(b,c)))
{
printf("Point A and C are closest.");
}
else
{
printf("All point are same.");
}
}

Answer 1

First, change this:
double getDistanceAB(struct Point a, struct Point b)
{
double distanceAB;
distanceAB = sqrt((a.x - b.x) * (a.x - b.x) + (a.y-b.y) *(a.y-b.y));
return distanceAB;
}
double getDistanceBC(struct Point b, struct Point c)
{
double distanceBC;
distanceBC = sqrt((b.x - c.x) * (b.x - c.x) + (b.y-c.y) *(b.y-c.y));
return distanceBC;
}
double getDistanceAC(struct Point a, struct Point c)
{
double distanceAC;
distanceAC = sqrt((a.x - c.x) * (a.x - c.x) + (a.y-c.y) *(a.y-c.y));
return distanceAC;
}
to just this:
double getDistance(struct Point a, struct Point b)
{
double distance;
distance = sqrt((a.x - b.x) * (a.x - b.x) + (a.y-b.y) * (a.y-b.y));
return distance;
}
One of the main points of functions is that you don't have to repeat code.
Now all you have to do is create your four points by adding one more scan for the fourth point and add that to the decision tree.
Keep in mind this for the decision tree that you made... If you check if point 'a' is not the closest using the same logic you used in your original post, you don't have to compare point 'a' again.

Answer 2

I would reduce the number of functions to just double getDistance(struct Point p, struct Point o)
and keep your points in a list so you can allow the program to run through the points dynamically instead of programming each condition.
Once you have your points in a list, you can run a loop that checks each pair in the list for their distance and check that against the currently shortest distance; and if the distance of the pair checked is closer you change the currently shortest distance to the checked pair and which pair of points have that distance.
That way you can expand it to work for arbitrarily large number of points.
I'm not used to the syntax of C, but for the checking of points in the list you'll need a double for loop, in which the first goes through each point in the list, and the second checks the distance from/to that first point to all points later in the list.
for i = 0, i++, length(listOfPoints) {
for j = i+1, j++, length(listOfPoints) {
getDistance(listOfPoints[i], listOfPoints[j]
}
}
Hope this helps some.

Answer 3

This is how I'd solve that,
#include <stdio.h>
typedef struct
{
int x;
int y;
} Point;
int square(int x) { return x * x; }
int distanceSq(Point *a, Point *b)
{
return square(a->x - b->x) + square(a->y - b->y);
}
int main(int argc, char const *argv[])
{
int n = 4;
Point a[4];
for (int i = 0; i < n; i++)
{
printf("Enter Point %d <as x y>: ", i + 1);
scanf("%d %d", &a[i].x, &a[i].y);
}
int distance = __INT_MAX__;
int p1 = -1, p2 = -1;
for (int i = 0; i < n - 1; i++)
for (int j = i + 1; j < n; j++)
{
int current = distanceSq(&a[i], &a[j]);
if (current < distance)
{
p1 = i;
p2 = j;
distance = current;
}
}
printf("The closest points are [%d %d] and [%d %d]", a[p1].x, a[p1].y, a[p2].x, a[p2].y);
return 0;
}
Note:
This can be extended for n number of points
Gives us the first pair closest points
we do not need to take square roots since if the square is large the square root will be proportionally large( in case of a large number(n) of points it might save computation time)

Answer 4

Here you go, a solution for any number of points.
Just change MAX_POINTS to anything you might need.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>
#define MAX_POINTS (4U)
struct Point
{
int x;
int y;
};
struct PointPair
{
struct Point a;
struct Point b;
};
double getDistance(const struct PointPair pair)
{
return sqrt((pair.a.x - pair.b.x) * (pair.a.x - pair.b.x) +
(pair.a.y - pair.b.y) * (pair.a.y - pair.b.y));
}
void readPoints(struct Point points[const])
{
for (unsigned i = 0; i < MAX_POINTS; i++)
{
printf("Enter coordinate of point %u: ", i);
scanf("%d %d", &(points[i].x), &(points[i].y));
}
}
bool checkForShorterDistance(const struct PointPair pair, double *const p_minDistance)
{
double tempDistance = getDistance(pair);
if (tempDistance < *p_minDistance)
{
*p_minDistance = tempDistance;
return true;
}
return false;
}
struct PointPair getClosestPair(const struct Point points[const])
{
struct PointPair result =
{
.a = points[0],
.b = points[1]
};
double minDistance = getDistance(result);
struct PointPair tempPair;
unsigned i, j;
for (i = 0; i < MAX_POINTS; i++)
{
tempPair.a = points[i];
for (j = 0; j < MAX_POINTS; j++)
{
if (i == j)
{
continue;
}
tempPair.b = points[j];
if (checkForShorterDistance(tempPair, &minDistance))
{
result = tempPair;
}
}
}
return result;
}
int main(void)
{
struct Point points[MAX_POINTS];
readPoints(points);
struct PointPair pair = getClosestPair(points);
printf("Closest pair is (%d, %d) and (%d, %d)\n",
pair.a.x,
pair.a.y,
pair.b.x,
pair.b.y);
return 0;
}