need to find the min distance and make a route - c++

Question

The graph:
and how the points are connected:
the file how points are connected are the output i have . The expected output should be 1<->216 , 23<->157 115<->157 and then 115<->216 . The order doesn't matter but those points should be connected this way
Let me ask you a question.
I have the following code:
for (int i = 0; i < values.size()-1;i++)
{
int result = 0;
double Min2 = DBL_MAX;
int x = 0;
for (int j = i+1; j < values.size(); j++)
{
/*if (visited[j] == false)*/
{
distance = sqrt(pow((values[i].x2 - values[j].x2), 2) + pow((values[i].y2 - values[j].y2), 2));
if (distance < Min2 && distance != 0)
{
Min2 = distance;
x = j;
}
}
}
If you look at the graph the left first node is 1 then goes 216 then 115 then 157 and then 23. I have the right connection except for 216-115 but instead of it connects 216-157. Why does it ignore or min distance won't help. I tried to use flag( like if visited or not)the same result. All nodes work , just this one doesn't want to be connected right.

Answer 1

The logic of the code seems a bit odd...
for (int i = 0; i < values.size()-1;i++) {
int result = 0;
double Min2 = DBL_MAX;
int x = 0;
for (int j = i+1; j < values.size(); j++) {
iterating till size()-1 seems like you assume that the relation "closest neighbour" is commutative, but it is not. Consider this example:
a b c
a has b as closest node and (by chance) also b has a as closest node, but c has b as closest node and this connection you would miss.
Further, starting the second loop at i+1 seems rather odd, as it assumes that the points in the vector are already ordered (if this would be the case, then whats the point of searching the closest nodes?).
If you know already, that the nodes are along a route, then I would do it like this:
1) start with the first node n0 in the route (its a bit more complicated if you dont know this node)
2) find the closest node to n0 call it n1
3) continue to find the closest node to ni among the nodes that are not yet connected to the route, call it ni+1
4) repeat 3) until you reach the end of the route.
I have the impression, that you dont understand the difference between those steps and your code. I dont like to give up, so I will try to explain again in a less pseudo way:
int currentNode = startNode; // as mentioned before, if you dont know
// this its a bit more involved
std::vector<int> route;
route.push_back(currentNode);
while (currentNode != endNode) { // also knowing the last node of the route helps...
double minDistance = std::numeric_limits<double>::max();
int nextNode = -1;
for (int i=0;i<nodes.size();i++) {
if (std::find(route.begin(), route.end(), i) == route.end()){
double distance = getDistance(currentNode,i);
if (distance < minDistance) {
minDistance = distance;
nextNode = i;
}
}
}
route.push_back(nextNode);
currentNode = nextNode;
}
Your outer loop shouldnt iterate through all nodes, if you actually want to find only that one route. Also you have to remember nodes that are already in the route (in case each node should appear only once and you dont want a closed loop).

Answer 2

for (int i = 0; i < values.size();i++)
{
int result = 0;
double Min2 = DBL_MAX;
int x = 0;
for (int j =0; j < values.size(); j++)
{
/*if (visited[j] == false)*/
{
distance = sqrt(pow((values[i].x2 - values[j].x2), 2) + pow((values[i].y2 - values[j].y2), 2));
if (distance < Min2 && distance != 0)
{
Min2 = distance;
x = j;
}
}
}
/*visited[i] = true;*/