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[Max White]

Wait, I'm not so sure about this...

return (this.Right > r.Left && this.Left < r.Right && this.Top < r.Bottom && this.Bottom > r.Top);

In the above example, there is clearly an overlap, but assuming the red box is this then it's bottom is less than the green ones top (Remembering that top left is 0,0) and there for not all four conditions are filled.

This only checks for containment, not interception.

[Virex]

Screw me, I didn't take into account that you can't just negate the comparison operators if you're dealing with something that isn't a point. Forget what I said please.

[Max White]

See that is why I needed to bust out a pencil in the first place... Finding if things overlap is a bitch to do! Much easier to find if they don't, and reverse it.

[MagmaMcFry]

Wait, I'm not so sure about this...

return (this.Right > r.Left && this.Left < r.Right && this.Top < r.Bottom && this.Bottom > r.Top);

In the above example, there is clearly an overlap, but assuming the red box is this then it's bottom is less than the green ones top (Remembering that top left is 0,0) and there for not all four conditions are filled.

This only checks for containment, not interception.

Wait, all four conditions are actually fulfilled. Or am I seeing something wrong?

[Max White]

I think you might be reading it upside down.

[Siquo]

Last one isn't (assuming topleft = 0,0).

[Virex]

If topleft = 0.0 then bottomleft = n.0 > 0.0. So if A is below B, then A > B, which corresponds to the given situation, no? You would have been right if bottomleft = 0.0, but down is higher, not lower in this case. Unless top is actually bottom in your image (that is, top has the highest Y coordinate, not the lowest)

[Nadaka]

The intersection of two grid aligned squares in pseudocode:

(((B.maxX > A.maxX > B.minX) OR (B.maxX > A.minX > B.minX)) AND ((B.maxY > A.maxY > B.minY) OR (B.maxY > A.minY > B.minY))) OR (((A.maxX > B.maxX > A.minX) OR (A.maxX > B.minX > A.minX)) AND ((A.maxY > B.maxY > A.minY) OR (A.maxY > B.minY > A.minY)))

You can trade some efficiency for readability assuming a method that determines if a point is in a square and methods that return the corners of a square:

B.containsPoint(A.topleft) | B.containsPoint(A.topright) | B.containsPoint(A.bottomleft) | B.containsPoint(A.bottomright) | A.containsPoint(B.topleft) | A.containsPoint(B.topright) | A.containsPoint(B.bottomleft) | A.containsPoint(B.bottomright)

[Levi]

Anything math related like that I just steal from elsewhere.  I stole mine from the libtcod python tutorial:

Code: [Select]

def intersect(self, other):
        #returns true if this rectangle intersects with another one
        return (self.x1 <= other.x2 and self.x2 >= other.x1 and
                self.y1 <= other.y2 and self.y2 >= other.y1)


[Max White]

Ok, well I'm wiling to go with that.
Libtcod has been pretty well worked over.

[Stargrasper]

You guys are giving me nightmares of CS projects past...

...do you still need help with collision detection or do you collectively have it down now...?

Also, that didn't look like nearly as much snow until I started shoveling it...that took forever and I'm still thawing...

[Max White]

Well while your offering...
Knowing if there is a collision is nice, but how would you go about finding firstly the depth of the collision, and secondly given an x and y speed, the place and time of collision?

And you can have it on my desk by Monday.

[Nadaka]

After 1.5 minutes of watching 30 rock on comedy central, my method is longer than necessary as there are dead paths.

(((B.maxX > A.maxX > B.minX) OR (B.maxX > A.minX > B.minX)) AND ((B.maxY > A.maxY > B.minY) OR (B.maxY > A.minY > B.minY))) OR ((A.maxX > B.maxX > A.minX) AND (A.maxY > B.maxY > A.minY))

You can trade some efficiency for readability assuming a method that determines if a point is in a square and methods that return the corners of a square:

B.containsPoint(A.topleft) | B.containsPoint(A.topright) | B.containsPoint(A.bottomleft) | B.containsPoint(A.bottomright) | A.containsPoint(B.topleft)

The reason for this is that if no corners of square A are inside square B, then either all or no corners of B will be inside A.

[Max White]

You... Ok there Nadaka?
I mean the entire reason we are going over this is because it is simpler than having a negator, so lets make sure it stays simpler than having a negator.

[Nadaka]

Calling 1 method 4 times on 1 object and once on another isn't simple enough?