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edited March 2017

I wonder how I can return the walker to its original position once the counter condition is reached, no matter what the new position is.

``````PVector walker;
PVector start;
PVector seek;
float count;

void setup() {
size (900, 900);

walker= new PVector (width/2, height-10);
start= new PVector(0, -1);
seek=new PVector(random(-1, 1), random( 1));

count=0;
}

void draw() {
background(53, 111);

count++;

stroke(0, 0, 255);
rect(width/2, height-10, 10, 10);

noStroke();
fill(255, 0, 0);
ellipse(walker.x, walker.y, 5, 5);

startt();
limits();
println(count);
}

void startt() {
if (walker.y>=150 && count<50) {
} else {
}
}
// once the condition is reached (count=1000) ,  walker "walks back" to its original position.

void limits() {
if (walker.x>width || walker.x <0) {
seek.x = -seek.x;
} else if (walker.y>height || walker.y <0) {
seek.y = -seek.y;
}
}
``````
Tagged:

@jozze -- do you mean that you want to restore the original position? You defined this as:

``````walker= new PVector (width/2, height-10);
``````

...so that would be:

``````if (count==1000) {
walker.set(width/2, height-10); // walker to original position
}
``````

You could also store that original position in a global `PVector origin` and then `walker.set(origin.x, origin.y);`

That concept is similar to a walking ant returning to its anthill, right?

• @jeremydouglass __ the .set method resets the position straight away from the current one. We do not see the object "going back home". Although this is not desired, it pointed out a way I think works out:

``````  if (count==1000) {
current.set(walker.x, walker.y);//store current position in a new vector
current.sub(origin);// measure distance between origin and current position
current.normalize();// this line is added "to see" the way back
walker.sub(current);
}
``````

@GoToLoop __ Somehow yes, I found a similar approach in your AntHill Sketch (lines 118,122)

thx both of you!