Redcoders Frenzy
Organiser:
Achille Astolfi (LAchi)
Rules:
| Coresize: | 8000 |
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| Max. processes: | 8000
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| Max. cycles: | 80000
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| Max. length: | 100
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| Min. distance: | 100
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| Rounds: | 4*250
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3 warriors fight: self against 2 known warriors, special scoring rules.
Paper-The-Bully is always teasing your Little Sibling the Stone,
and killing it in the game... now this is your job as a Big
Sibling: kill The Bully and save your Little Sibling!
You are always rewarded if the Little Sibling survives. But to
maximize the happiness of your family, you must both kill the
Bully and survive until the end of the fight!
The opponents:
Paper-The-Bully is acted by the paper in Revenge of the Papers, while your
Little Sibling is acted by Silver
Bullet. Click on the links to download the code.
The Scores:
The following command line will be issued 4 times:
pmars -bk -r 250 bully.red little.red
big.red
(where big.red is your entry). Then for each
round:
You get 6 points if the Bully dies and both your warrior and the
Little Sibling survive.
You get 3 points if only the Little Sibling survive (a little
reward for your sacrifice).
You get 2 points if all the warriors survive (you have another
chance to save your Little Sibling).
You get 1 point if your Little Sibling and the Bully survive (a
minimum reward for your sacrifice).
You get 0 points if your Little Sibling dies (all the others).
How to compute the
score without a modified pMARS:
If you execute the pmars command as above, you get a result like
this:
1176 127 36 8 79
276 17 31 8 194
532 62 7 8 173
the first row is the Bully, the second row is the Little sibling,
the third row is Big, your entry. The first column is the
"standard" score, the second column is how many times the warrior
was the only survivor, the third column is how many times the
warrior was one of two survivors, the fourth column is how many
times the warrior was one of three survivors, the last column is
how many times the warrior did not survive.
Now you know for sure that Little was the only survivor for 17
times, and that all the warriors survived together for 8 times,
but how for the rest? Well, say x is the times of Bully+Little
survived, y of Bully+Big and z of Little+Big; we know from the
third column that:
x+y=36
x+z=31
y+z=7
This system is always resolvable, and the solution is simple to
compute. First you sum the known values and divide by 2:
(36+31+7)/2 = 37; magic: this is the sum of x+y+z! So if you
subtract the known value of a row from the sum, you get how many
times the other two warriors survived together!
In short: calculate half of the sum of the third column
(36+31+7)/2 = 37; then subtract the "Bully" value from the
result, and you get 37-36 = 1: this means that Little and Big
survived together only once! Subtract the "Big" value from the
result: 37-7 = 30: this means that Bully and Little survived
together 30 times!
So the total score in this case is 1*6 + 17*3 + 8*2 + 30*1 =
103.
The Hints:
The kofacoto round 4
is a good start! Anyway the scoring method there was
different.
The results of Bully against Little Sibling on a 1000 rounds
fight are 886 103 11.
A very specific vampire seems a good idea for this round.
The Entries:
2 entries for each participant. The deadline is April 1st, 2004.
I chose this date because that was most liked by A.K.Dewdney
;)
Send your entries to achillu@tin.it
I am going to participate, too. So if you trust in me, send your
entries whenever you want. If you do not trust, please wait until
I have developed mine and sent them to the notary ;) I will give
a public notice on r.g.c. when my entries will be done.
The Results:
I got 13 entries by 9 authors. I expected many evolved entries,
instead there were none! Most entries are a specific
vamp/scanner, even if we had alternative entries such as
vamp/papers, one imp, and one suicide. No P-spacers.
This time we had two "legal cheats". The suicide entry is going
to collect points anyway (exactly the score of Little Sibling
against the Bully: 3 pts for each win and 1 pt for each tie). The
other "legal cheat" is more subtle: the participant sent two
copies of the same warrior. For statistic reasons, one of them
would score more than average.
1000 rounds seemed to be enough: the test entry being the "legal
cheat", the difference in score was just about 1%, so I think
that the results can be reliable.
Here you can find the raw
results; they were computed by this Java program using
Java SDK 1.4.1. Here is the compiled class if
you only have the Java RE 1.4.1.