"Tayssir John Gabbour" <ta*********@yahoo.com> wrote:

I don't quite understand this sentence, perhaps I'm not thinking

straight: "If a stone is hit, the speed vectors are devided by 2

(integer devision without fraction) as long as no stone is hit (which

can result in a speed vector of (0, 0))."

the sentence before is important:

| After this the speed vector is added to the coordinate, but only, if

| the new coordinate does not hit a stone after adding the speed.

So, suppose when step 20 occurs:

* You're 2 pixels left of a stone.

* Your velocity-x will be 2 pixels/step on step 20 and (if possible)

21.

* velocity-y will be arbitrary.

What happens? Are you able to go partly through a stone?

no, the robot is always stone-free. If you are 2 pixels left of a stone

and your velocity-x is 2, the next turn you are 0 pixels left of a stone.

Then adding the velocity will result in hitting a stone, so the new

velocity is (1, velocity-y/2) and because this will result in a hit, too,

again the velocity is devided, which will result in a velocity-x of 0 and

velocity-y devided by 2 again.

--

Frank Buß,

fb@frank-buss.de http://www.frank-buss.de,

http://www.it4-systems.de