Check if two or more objects are side by side

Programming Questions

knuppel

Check if two or more objects are side by side

in Programming Questions • 3 months ago

I made a little sketch where you have to move to boxes. The sketch should check if the two boxes are side by side or on top of each other. Each box is an object, and the check-function works god with this two objects. But when i want to have three or more boxes it will get complicated. Do you have any suggestions to simplify this?

box box1;

box box2;

int[] xxx= {

1, 2, 3, 4, 5, 6, 7, 8, 9, 10

};

int[] yyy= {

1, 2, 3, 4, 5, 6, 7, 8, 9, 10

};

float[] xpos=new float[10];

float[] ypos=new float[10];

int boxh;

float weite, breite;

int posx;

int posy;

boolean blockaktiv, blockaktiv1;

void setup() {

size(displayWidth, displayHeight);

weite=width/15;

breite=height/20;

boxh=width/13;

for (int i=0;i<=9;i++)

{

ypos[i]=breite;

breite=breite+boxh;

}

for (int i=0;i<=9;i++)

{

xpos[i]=weite;

weite=weite+boxh;

}

box1=new box(color(124, 32, 211), 1, 1, 1, 1);

box2=new box(color(124, 32, 211), 1, 3, 1, 2);

}

void draw() {

background(0);

//x++

box1.display();

box2.display();

//rect(posx,posy,boxh,boxh);

println(width+" "+height);

println(box1.posx+" "+box2.posy);

if (box1.posx==box2.posx-boxh

&&box1.posy==box2.posy)text("BANG!!!!", 10, 10);

if (box2.posx==box1.posx-boxh

&&box1.posy==box2.posy)text("BANG!!!!", 10, 10);

if (box1.posy==box2.posy-boxh

&&box2.posx==box1.posx)text("BANG!!!!", 10, 10);

if (box2.posy==box1.posy-boxh

&&box1.posx==box2.posx)text("BANG!!!!", 10, 10);

}

class box {

color c;

int xx;

int yy;

int[] xa=new int[10];

int[] ya=new int[10];

float xspeed;

boolean aktiv;

float posx, posy;

// The Constructor is defined with arguments.

box(color tempC, int tempXpos, int tempYpos, int tempXa, int tempYa ) {

c = tempC;

xx = tempXpos;

xx = tempYpos;

}

void display() {

stroke(0);

fill(c);

if (mousePressed==false) {

aktiv=false;

}

if (mousePressed==true

&& mouseX>=xpos[xx]

&& mouseX<=xpos[xx]+boxh

&& mouseY>=ypos[yy]

&& mouseY<=ypos[yy]+boxh)

aktiv=true;

if (

mousePressed==true

&& mouseX>=xpos[xx] +boxh

&& mouseX<=xpos[xx] +boxh+boxh

&& mouseY>=ypos[yy]

&& mouseY<=ypos[yy]+boxh+boxh

&& aktiv==true) {

xx=xx+1;

if (xx>=9)xx=9;

if (xx<=0)xx=0;

for (int i=0;i<=9;i++)

{

xa[i]=0;

}

xa[xx]=1;

}

//x--

if (

mousePressed==true

&& mouseX>=xpos[xx] -boxh

&& mouseX<=xpos[xx]

&& mouseY>=ypos[yy]

&& mouseY<=ypos[yy]+boxh+boxh

&& aktiv==true) {

xx=xx-1;

if (xx>=9)xx=9;

if (xx<=0)xx=0;

for (int i=0;i<=9;i++)

{

xa[i]=0;

}

xa[xx]=1;

}

//y++

if (

mousePressed==true

&& mouseX>=xpos[xx]

&& mouseX<=xpos[xx] +boxh

&& mouseY>=ypos[yy] +boxh

&& mouseY<=ypos[yy] +boxh+boxh

&& aktiv==true) {

yy=yy+1;

if (yy>=9)yy=9;

if (yy<=0)yy=0;

for (int i=0;i<=9;i++)

{

ya[i]=0;

}

ya[yy]=1;

}

//y--

if (

mousePressed==true

&& mouseX>=xpos[xx]

&& mouseX<=xpos[xx] +boxh

&& mouseY>=ypos[yy]-boxh

&& mouseY<=ypos[yy]

&& aktiv==true) {

yy=yy-1;

if (yy>=9)yy=9;

if (yy<=0)yy=0;

for (int i=0;i<=9;i++)

{

ya[i]=0;

}

ya[yy]=1;

}

rect(xpos[xx], ypos[yy], boxh, boxh);

posx=xpos[xx];

posy=ypos[yy];

}

Replies(12)

benj.n.su..

Re: Check if two or more objects are side by side

3 months ago

your current box checking isn't generalized. right now its limited only to box1 and box2. in order to generalize it, you have to abstract out the parts that will be the same no matter the box and put it in a function of it's own. also, in order to have more than one box, you'll need to have an array of boxes. The following isn't quite psuedocode, but it hasn't been tested either.

1) Generalization

boolean checkEdges(box1, box2){
if (box1.posx==box2.posx-boxh && box1.posy==box2.posy) ||
(box2.posx==box1.posx-boxh && box1.posy==box2.posy) ||
(box1.posy==box2.posy-boxh && box2.posx==box1.posx) ||
(box2.posy==box1.posy-boxh && box1.posx==box2.posx) {
return true;
} else {
return false;
}
}

2) Checking the edges of multiple boxes using an array of Boxes

for(int i = 0; i < boxes.length; i++){
for(int j = 0; j < boxes.length; j++){
boolean touching = checkEdges(boxes[i], boxes[j]);
if(touching){ //could also skip line 5 and go if(checkEdges(boxes[i], boxes[j]));
text("BANG!!!!", 10, 10);
}
}
}

knuppel

Re: Check if two or more objects are side by side

3 months ago

How can i implent that in my code?

benj.n.su..

Re: Check if two or more objects are side by side

3 months ago

sure thing.

1) in your draw loop, remove the if/thens that test for the edges and prints the text "Bang!", and replace it with the double for loop.

2) take the definition of the function checkEdges(b1 , b2), and put it in between the closing bracket of your draw() loop, and the beginning of the box class. side note: the definition of any functions you've written go anywhere below the draw() loop or on a separate tab.

benj.n.su..

Re: Check if two or more objects are side by side

3 months ago

oh, also. when ever you share code, after you've pasted it into the editor, highlight it, and click the icon in the editor that looks like a document with <> on top. That will give your code line numbers which will help in people telling you what goes where with a greater accuracy.

knuppel

Re: Check if two or more objects are side by side

3 months ago

What do you mean with "double loop"?

knuppel

Re: Check if two or more objects are side by side

3 months ago

Will this work with more than two boxes?

I have to generate the boxes as an array?

knuppel

Re: Check if two or more objects are side by side

3 months ago

box[] boxes=new box[3];
int[] xxx= {
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
};
int[] yyy= {
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
};
float[] xpos=new float[10];
float[] ypos=new float[10];
int boxh;
float weite, breite;
int posx;
int posy;
boolean blockaktiv, blockaktiv1;
void setup() {
orientation(PORTRAIT);
size(displayWidth, displayHeight);
weite=width/15;
breite=height/20;
boxh=width/11;
for (int i=0;i<=9;i++)
{
ypos[i]=breite;
breite=breite+boxh;
}
for (int i=0;i<=9;i++)
{
xpos[i]=weite;
weite=weite+boxh;
}
boxes[0]=new box(color(124, 32, 211), 1, 1, 1, 1);
boxes[1]=new box(color(124, 32, 211), 1, 3, 1, 2);
boxes[2]=new box(color(124, 132, 211), 9, 9, 1, 2);
}
void draw() {
background(0);
for (int i = 0; i < boxes.length; i ++ ) {
boxes[i].display();
}
for (int i = 0; i < boxes.length; i++) {
for (int j = 0; j < boxes.length; j++) {
boolean touching = checkEdges(boxes[i], boxes[j]);
if (touching) { //could also skip line 5 and go if(checkEdges(boxes[i], boxes[j]));
text("BANG!!!!", 10, 10);
}
}
}
}
boolean checkEdges(boxes1, boxes2) {
if ((boxes1.posx==boxes2.posx-boxh && boxes1.posy==boxes2.posy) ||
(boxes2.posx==box1.posx-boxh && boxes1.posy==boxes2.posy) ||
(boxes1.posy==box2.posy-boxh && boxes2.posx==boxes1.posx) ||
(boxes2.posy==boxes1.posy-boxh && boxes1.posx==boxes1.posx)) {
return true;
}
else {
return false;
}
}
class box {
color c;
int xx;
int yy;
int[] xa=new int[10];
int[] ya=new int[10];
float xspeed;
boolean aktiv;
float posx, posy;
// The Constructor is defined with arguments.
box(color tempC, int tempXpos, int tempYpos, int tempXa, int tempYa ) {
c = tempC;
xx = tempXpos;
xx = tempYpos;
}
void display() {
stroke(0);
fill(c);
if (mousePressed==false) {
aktiv=false;
}
if (mousePressed==true
&& mouseX>=xpos[xx]
&& mouseX<=xpos[xx]+boxh
&& mouseY>=ypos[yy]
&& mouseY<=ypos[yy]+boxh)
aktiv=true;
if (
mousePressed==true
&& mouseX>=xpos[xx] +boxh
&& mouseX<=xpos[xx] +boxh+boxh
&& mouseY>=ypos[yy]
&& mouseY<=ypos[yy]+boxh+boxh
&& aktiv==true) {
xx=xx+1;
if (xx>=9)xx=9;
if (xx<=0)xx=0;
for (int i=0;i<=9;i++)
{
xa[i]=0;
}
xa[xx]=1;
}
//x--
if (
mousePressed==true
&& mouseX>=xpos[xx] -boxh
&& mouseX<=xpos[xx]
&& mouseY>=ypos[yy]
&& mouseY<=ypos[yy]+boxh+boxh
&& aktiv==true) {
xx=xx-1;
if (xx>=9)xx=9;
if (xx<=0)xx=0;
for (int i=0;i<=9;i++)
{
xa[i]=0;
}
xa[xx]=1;
}
//y++
if (
mousePressed==true
&& mouseX>=xpos[xx]
&& mouseX<=xpos[xx] +boxh
&& mouseY>=ypos[yy] +boxh
&& mouseY<=ypos[yy] +boxh+boxh
&& aktiv==true) {
yy=yy+1;
if (yy>=9)yy=9;
if (yy<=0)yy=0;
for (int i=0;i<=9;i++)
{
ya[i]=0;
}
ya[yy]=1;
}
//y--
if (
mousePressed==true
&& mouseX>=xpos[xx]
&& mouseX<=xpos[xx] +boxh
&& mouseY>=ypos[yy]-boxh
&& mouseY<=ypos[yy]
&& aktiv==true) {
yy=yy-1;
if (yy>=9)yy=9;
if (yy<=0)yy=0;
for (int i=0;i<=9;i++)
{
ya[i]=0;
}
ya[yy]=1;
}
rect(xpos[xx], ypos[yy], boxh, boxh);
posx=xpos[xx];
posy=ypos[yy];
}
}

The transfer to the function does not work.

knuppel

Re: Check if two or more objects are side by side

3 months ago

Ok, i did it like this:

boolean checkEdges(int i, int j) {
if ((boxes[i].posx==boxes[j].posx-boxh && boxes[i].posy==boxes[j].posy) ||
(boxes[j].posx==boxes[i].posx-boxh && boxes[i].posy==boxes[j].posy) ||
(boxes[i].posy==boxes[j].posy-boxh && boxes[j].posx==boxes[i].posx) ||
(boxes[j].posy==boxes[i].posy-boxh && boxes[i].posx==boxes[j].posx)) {
return true;
}
else {
return false;
}}

and:

boolean touching = checkEdges( i, j);

When two or three boxes are side by side the text is displayed.

How can i solve it that the text is only shown when all boxes are side by side?

benj.n.su..

Re: Check if two or more objects are side by side

3 months ago

i've been putting some thought to this and i think it requires a bit more clarification:

will all the following work or just some?

[][][][]

[]

[][][]

[][]

(please say they all work)

benj.n.su..

Re: Check if two or more objects are side by side

3 months ago

or say it will always be no more than three boxes.

knuppel

Re: Check if two or more objects are side by side

3 months ago

It will not be more than three boxes.

benj.n.su..

Re: Check if two or more objects are side by side

3 months ago

hmmm...well. it seems to me that in any configuration of three, you are going to have one of the three having two sides touching. therefore you could set up an int variable in your box class called numSidesTouching which you increment whenever a side is touching. then in the check edges part of your draw() loop you add something like:

if(boxes[i].numSidesTouching || boxes[j].numSidesTouching == 2) {

//do some stuff

}

The part that I have not yet figured out is how to decrement numSidesTouching when the sides are no longer touching. Sorry. Still need to think on it. Hopefully someone else can chime in, too.

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