Mars Rescue Mission Challenge

[Frank Buss]

A new challenge:

http://www.frank-buss.de/marsrescue/index.html

Have fun! Now you can win real prices.

 

[Howard]

A new challenge:

http://www.frank-buss.de/marsrescue/index.html

Have fun! Now you can win real prices.

Would those be retail prices or wholesale prices?

Oh, never mind. It's off-topic here.

 

[Frank Buss]

Would those be retail prices or wholesale prices?

you are right, I meant prize, not price

 

[Tayssir John Gabbour]

A new challenge:

http://www.frank-buss.de/marsrescue/index.html

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))."

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?


MfG,
Tayssir

 

[Neo-LISPer]

A new challenge:

http://www.frank-buss.de/marsrescue/index.html

Have fun! Now you can win real prices.

Terribly similar to ICFP 2003. No 1 or 3-day time constraint. What's the fun
in that?

 

[Frank Buss]

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]

Terribly similar to ICFP 2003. No 1 or 3-day time constraint. What's
the fun in that?

I know this challenge (to the other readers: you can see it at
http://www.dtek.chalmers.se/groups/icfpcontest/ ) and I liked the idea,
this is one of the reason for this challenge. There might be many people
who don't know the ICFP challenge, and I hope they'll have fun with it.

One difference in my challenge is that you don't need to simulate fixed-
point arithmetic or complicated definitions of sin and cos. This helps to
concentrate on a good algorithm.

Another difference is the time limit. For my challenge you have much time,
so you can think and test a lot and perhaps there will be some interesting
ideas or variations of the challenge submitted.

 

[Neo-LISPer]

I know this challenge (to the other readers: you can see it at
http://www.dtek.chalmers.se/groups/icfpcontest/ ) and I liked the idea,
this is one of the reason for this challenge.

You might want to give proper credit then. Otherwise, some people might call
it plagiarism (sponsored by DrDobbs ?)

 

[Surendra Singhi]

A new challenge:

http://www.frank-buss.de/marsrescue/index.html

Have fun! Now you can win real prices.

Please correct the typos in the spelling of "devided" and "devision". It
should be "divided" and "division".