Arimaa Forum - 2006 World Championship

Welcome, Guest. Please Login or Register.
May 7^th, 2024, 6:34pm

Home

Help

Members

Arimaa Forum « 2006 World Championship »

   Arimaa Forum
   Arimaa
   Events (Moderator: supersamu)
   2006 World Championship

« Previous topic | Next topic »

Pages: 1 2 3 4 ... 6

Notify of replies

Send Topic

Author

Topic: 2006 World Championship (Read 5935 times)

Fritzlein
Forum Guru

Arimaa player #706

Gender:

Posts: 5928

Re: 2006 World Championship
« Reply #15 on: Nov 15^th, 2005, 9:03am »

Quote

Modify

on Nov 15^th, 2005, 2:17am, Ryan_Cable wrote:

As seeds 1 and 2, I believe that 99of9 and Fritzlein can potentially play each other at only two specific times: The first opportunity is in round 4 in a critical battle to be the last remaining undefeated player. The only other time they can meet is at the end of the tournament playing for the win.

I hadn't considered the significance of this. The seeding does keep me away from 99of9, and I couldn't be happier about that. Yet if I'm not mistaken, we can meet twice without playing for the title. Consider the "Achilles heel" scenario: In round four 99of9 plays me for the for the privelege of the round five bye, and and wins. Going into round six, we could have 99of9 undefeated and Adanac and myself with one loss. I would get the round six bye, but then if Adanac beats 99of9 in round six, I would have to play 99of9 in round seven. It could happen that 99of9 beats me a second time, and then loses a second time to Adanac in round eight, and Adanac is the World Champion!

I have definitely been hoping that someone else knocks 99of9 out of the tournament, because it sure worked to my advantage last year. Until now I hadn't considered that someone might knock him out only after he has knocked me out!

Generally speaking, it isn't so unlikely that 99of9 and I will deal each other two losses in rounds 4 and 7, so only two upsets would be necessary, specifically in rounds 6 and 8. However, if we are each upset in the first three rounds, 99of9 and I couldn't be paired against each other ever unless we met in round 8 for the Championship. If this happens, I plan to claim it was my strategy all along.

It is interesting to hear speculation that I might be underrated at 2273. I remember that when I crossed 2300 for the first time there were generous comments that I was poised to rise straight to 2400. I made a bet then that I was overrated from botbashing, and would visit 2200 again before seeing 2400. Since that time my rating has been as high as 2342 and as low as 2206, but not crossed either marker. Today I would still bet that I cross 2200 before 2400, not as a 50-50 speculation, but as a moneymaker.

IP Logged

Fritzlein
Forum Guru

Arimaa player #706

Gender:

Posts: 5928

Re: 2006 World Championship
« Reply #16 on: Nov 15^th, 2005, 9:27am »

Quote

Modify

By the way, my wife pointed out that the color determination for the first round happens to be alphabetical by username. By my calculation there's only a 1 in 128 chance that could happen randomly. 'fess up, Omar!

IP Logged

Ryan_Cable
Forum Guru

Arimaa player #951

Gender: male

Posts: 138

Re: 2006 World Championship
« Reply #17 on: Nov 15^th, 2005, 5:16pm »

Quote

Modify

on Nov 15^th, 2005, 9:03am, Fritzlein wrote:

Yet if I'm not mistaken, we can meet twice without playing for the title. Consider the "Achilles heel" scenario: In round four 99of9 plays me for the for the privelege of the round five bye, and and wins. Going into round six, we could have 99of9 undefeated and Adanac and myself with one loss. I would get the round six bye, but then if Adanac beats 99of9 in round six, I would have to play 99of9 in round seven. It could happen that 99of9 beats me a second time, and then loses a second time to Adanac in round eight, and Adanac is the World Champion!

You’re right, I missed this possibility. This makes it significantly less of a lock if a titan gets the final undefeated spot. Probably adds another 3 or 4 percentage points to the field’s chances, since a titan being undefeated after round 4 is quite likely. That brings my prediction into line with the simulations 10%.

on Nov 15^th, 2005, 9:03am, Fritzlein wrote:

I have definitely been hoping that someone else knocks 99of9 out of the tournament, because it sure worked to my advantage last year. Until now I hadn't considered that someone might knock him out only after he has knocked me out!

As a spectator, I am hoping to see you and 99of9 meet in round 4, and then again for the title, which together almost forms a best of 3 mini-match. There is probably a >10% chance of seeing 3 battle of the titans games in this tournament.

on Nov 15^th, 2005, 9:03am, Fritzlein wrote:

Your bet is almost certainly a winner unless you intentionally inflate your rating. I think 2273 would make a good best guess for your rating with 99of9 at 2200. Looking at your record vs. the sub-titans centered around 2025, it is reasonable to think you outrank them by 250 to 300 points. If adjustments were made to the record to remove learning games, blitz games, etc, the advantage might fall near the bottom of this range, but I doubt it falls much below it. Likewise, I think 99of9’s record suggests he is 125 to 200 points above them. This puts you somewhere around 2300 and 99of9 around 2175. Then comes the question of how to deal with the fact that 99of9 seems to be roughly equal to you at the board. Mathematically, I think this calls for simulated annealing to get around the intransitivity in the ratings, but I don’t think that would produce a result much more meaningful than my just cutting 25 points from your rating and giving it to 99of9.

Philosophically, it comes down to the nature of your relationship with 99of9. It could be that there is a 100 to perhaps even 200 point gap in your true-ratings, and 99of9 just outperforms against you like he has by going undefeated against PMertens. Alternatively, it could be that your true-ratings are about equal and 99of9 has just been unlucky/underperformed in a few sub-titan games. The only way to determine which is correct is if another or better yet several other players of 99of9’s strength join Arimaa (or rise from the sub-titan ranks). Short of this all we have is opinions and guesses. Mine is that you are 50 to 125 points above 99of9 in true-rating.

Of course, there is also the details of different ratings at different time controls. I have heard rumors that you are undefeated at postal, which would put you 400+ above everyone else.

on Nov 15^th, 2005, 9:27am, Fritzlein wrote:

99of9 finds another strategic advantage. Cheesy

IP Logged

omar
Forum Guru

Arimaa player #2

Gender: male

Posts: 1003

Re: 2006 World Championship
« Reply #18 on: Nov 15^th, 2005, 5:50pm »

Quote

Modify

on Nov 15^th, 2005, 9:27am, Fritzlein wrote:

You had me scared there for a minute. I just double checked the pairing program and it seems OK. I ran it again a few times and got pairings like this:

pick Fritzlein MrBrain
pick acheron 99of9
pick jdb Ryan_Cable
pick PMertens grey_0x2A
pick megamau Adanac
pick robinson naveed
pick Belbo BlackKnight
pick Paul omar

pick MrBrain Fritzlein
pick acheron 99of9
pick jdb Ryan_Cable
pick grey_0x2A PMertens
pick megamau Adanac
pick naveed robinson
pick BlackKnight Belbo
pick Paul omar

pick MrBrain Fritzlein
pick 99of9 acheron
pick jdb Ryan_Cable
pick PMertens grey_0x2A
pick Adanac megamau
pick naveed robinson
pick BlackKnight Belbo
pick Paul omar

So I guess it was just a fluke.

Im also going to post the pairing script here so we have an official record of what we used. This is the same script that was used in my simulations of FDE.

#!/usr/bin/ruby

if ARGV[0] == nil
print <<EOM;

Floating Double Elimination Format
variant to format proposed by Karl Juhnke (Fritzlein)
see the Arimaa Forum for more details
search for "World Championship tournament format"

1) Everyone plays until they have lost two times.
At the end of each round, any player who has lost
for the second time is eliminated.

2) At the beginning of each round, if an odd number of
players remain, a bye must be assigned. The bye
goes to the player who has recieved the fewest byes
so far, with ties broken in favor fewest losses so
far, and further ties broken in favor of highest
pre-tournament rating.

3) The players who don't get a bye are paired
iteratively as follows: Select the unpaired player
with the fewest losses (ties broken by highest
pretournament rating), and pair him with the player
in the field he has played the least number of times,
with ties broken in favor of having fewest losses,
and further ties broken in favor of lower
pre-tournament rating.

4) Within each pairing, whoever has played Gold a smaller
percentage of times so far in the tournament should get
to play Gold for that game, with ties broken randomly.

EOM
exit;
end

pa, ga, ra, ph, rn = getTournState(ARGV[0]);

if (pa.length == 0)
print "done\n"
exit
end

# Determine number of loses for each player
# field called 'loses' is set in the player record
for x in 0 ... pa.length
pa[x]['loses'] = 0
#print " #{pa[x]['n']} #{pa[x]['r']} \n"
end
for x in 0 ... ga.length
e = ''
if ga[x]['w'] == ga[x]['1']
e = ga[x]['2']
else
e = ga[x]['1']
end
if e == "BYE"; next; end
pa[ph[e]]['loses'] += 1
end

# Eliminate players who have lost 3 times
pa.delete_if {|x| x['loses'] >= 2}
# also re-generate the player name index
ph.clear
for x in 0 ... pa.length
ph[pa[x]['n']] = x
end

# Used for testing; can be commented out
mpa = pa
mpa.sort! {|x,y| y['r'] <=> x['r']}
print "# player rating loses\n";
for x in 0 ... mpa.length
print "* #{mpa[x]['n']} #{mpa[x]['r']} #{mpa[x]['loses']}\n"
end
print "\n"

# If only one player is left then show the winner
if (pa.length == 1)
if ra[1] != nil
print "done\n"
exit
end
print "result #{pa[0]['n']} 1\n"
exit
end

print "* Round #{rn+1}\n"
print matchPlayers(pa, ga, ph)

exit;

BEGIN{

def val(v=0)
if v==nil; return 0 end
if v==""; return 0 end
return v
end

def matchPlayers(pa=[], ga=[], ph=Hash.new)
m = ""

# Determine how many times each player has played another
# also count number of byes and times played as first player
v = Hash.new
byes = Hash.new
first = Hash.new
gamesplayed = Hash.new
i = 0
for i in 0 ... ga.length
p1 = ga[i]['1']
p2 = ga[i]['2']
v[p1+p2] = val(v[p1+p2]) + 1
v[p2+p1] = val(v[p2+p1]) + 1
first[p1] = val(first[p1]) + 1
if p2 == 'BYE'; byes[p1] = val(byes[p1]) + 1 end
if p2 != 'BYE'
gamesplayed[p1] = val(gamesplayed[p1]) + 1
end
if p1 != 'BYE'
gamesplayed[p2] = val(gamesplayed[p2]) + 1
end
end

# Setup the scores for ordering the players later
# score - is used when pairing the players
# bsc - is used when giving a player a bye
# fsc - is used when deciding which player plays first
# played - is total number of games played by this player
# also mark that they have not been paired
for i in 0 ... pa.length
pa[i]['score'] = sprintf("%03d%08d", 99 - val(pa[i]['loses']), pa[i]['r'])
pa[i]['bsc'] = sprintf("%03d%03d%08d", 99-val(byes[pa[i]['n']]),
99-val(pa[i]['loses']), pa[i]['r'])
pa[i]['fsc'] = val(first[pa[i]['n']])
pa[i]['played'] = val(gamesplayed[pa[i]['n']])
pa[i]['paired'] = 0
end

# Order the players by fewest loses and highest ratings
pa.sort! {|x,y| y['score'] <=> x['score']}

# also re-generate the player name index
ph.clear
for i in 0 ... pa.length
ph[pa[i]['n']] = i
end

# Give one of the players a bye if there are odd number of players
if pa.length % 2 == 1
bpa = pa.sort {|x,y| y['bsc'] <=> x['bsc']}
p1 = bpa[0]['n']
m = m + "pick #{p1} BYE\n"
pa[ph[p1]]['paired'] = 1;
end

# Now pair the players
for i in 0 ... pa.length
p1 = pa[i]
if p1['paired'] != 0; next end
op = []
for j in i+1 ... pa.length
p2 = pa[j]
if p2['paired'] != 0; next end
p2['nsc'] = sprintf("%03d%03d%08d", 99 - val(v[p1['n']+p2['n']]),
99 - val(p2['loses']), 9999999 - p2['r'])
op.push(p2)
end
op.sort! {|x,y| y['nsc'] <=> x['nsc']}
p2 = op[0]
p1['paired'] = 1
p2['paired'] = 1
if (p1['fsc']/(p1['played']+1)) < (p2['fsc']/(p2['played']+1))
m = m + "pick #{p1['n']} #{p2['n']}\n"
elsif p2['fsc'] < p1['fsc']
m = m + "pick #{p2['n']} #{p1['n']}\n"
else
if rand(2)>0
m = m + "pick #{p1['n']} #{p2['n']}\n"
else
m = m + "pick #{p2['n']} #{p1['n']}\n"
end
end
end
return m
end

def shuffle!(a=[])
x = a.length
if x < 2
return
end
for n in 1 .. 5*x
p1 = rand(x)
p2 = rand(x)
a[p1], a[p2] = a[p2], a[p1]
end
end

# General tournament routines

def getTournState(tf)
p, g, w = [], [], []
i = Hash.new
round = 0

l = IO.readlines(tf)
l.each do |x|
# read the player
if x =~ /^player +(\w+) +(\d+)/
r = Hash.new
r['n'], r['r'] = $1, $2.to_i
p.push(r)
i[$1] = p.length - 1
end
# read the game
if x =~ /^(pair|pick) +(\w+) +(\w+) winner +(\w+) *(\w*)/
r = Hash.new
r['1'], r['2'], r['w'] = $2, $3, $4
if $5 != nil
r['d'] = [$4,$5].join(" ")
end
g.push(r)
end
# read the result
if x =~ /^result +(\w+) *(\w*)/
w[$2.to_i] = $1
end
if x =~ /^done$/
break;
end
# read the player properites we set
if x =~ /^\* +set +player +(\w+) +(\w+) +(\w+)/
if i[$1] != nil
p[i[$1]][$2] = $3
end
end
# read the round number
if x =~ /^\* Round +(\d+)/i
round = $1.to_i
end
end
return p, g, w, i, round
end

def printPlayerProp(p, k, v)
print "* set player #{p} #{k} #{v}\n"
end

def printPlayersProp(p, k)
p.each do |r|
print "* set player #{r['n']} #{k} #{r[k]}\n"
end
end

}

exit

IP Logged

99of9
Forum Guru

Gnobby's creator (player #314)

Gender:

Posts: 1413

Re: 2006 World Championship
« Reply #19 on: Nov 15^th, 2005, 6:37pm »

Quote

Modify

on Nov 15^th, 2005, 5:16pm, Ryan_Cable wrote:

99of9 finds another strategic advantage. Cheesy

Actually I was disappointed to get gold in the first round, because it means I get less gold later.

But I do like having my name at the top of the players online list

... except when 6sense is around.

IP Logged

Fritzlein
Forum Guru

Arimaa player #706

Gender:

Posts: 5928

Re: 2006 World Championship
« Reply #20 on: Nov 15^th, 2005, 7:14pm »

Quote

Modify

on Nov 15^th, 2005, 5:50pm, omar wrote:

You had me scared there for a minute. I just double checked the pairing program and it seems OK. [...] So I guess it was just a fluke.

That's funny. I guess there was also a 1 in 128 chance it would appear to be ordered by rating, and a 1 in 128 it would appear to be ordered by nationality, and a 1 in 128 chances it would appear to be orded by (fill in the blank). When you consider all the possible correlations one could look for, maybe it isn't so improbable one of them would be true.

IP Logged

Fritzlein
Forum Guru

Arimaa player #706

Gender:

Posts: 5928

Re: 2006 World Championship
« Reply #21 on: Nov 15^th, 2005, 7:26pm »

Quote

Modify

on Nov 15^th, 2005, 5:16pm, Ryan_Cable wrote:

Mathematically, I think this calls for simulated annealing to get around the intransitivity in the ratings[...]
Philosophically, it comes down to the nature of your relationship with 99of9. It could be that there is a 100 to perhaps even 200 point gap in your true-ratings, and 99of9 just outperforms against you [...]

Aha, another convert to the non-transitivity of ratings! It really opens things up when you stop believing in the mathematical model behind the ratings. What then can a rating mean? What then should a rating mean? Let us open a ratings philosophy thread so we can get it all hashed out and ready for Omar to implement a new system next year.

One thing about ratings seems clear from experience: the more the contestant mix around who they play against (e.g. when pairings are determined by tournaments rather than by self-selection) the more accurate ratings become. That's yet another reason broad participation in Arimaa tournaments is good for Arimaa.

IP Logged

omar
Forum Guru

Arimaa player #2

Gender: male

Posts: 1003

Re: 2006 World Championship
« Reply #22 on: Nov 16^th, 2005, 5:18pm »

Quote

Modify

on Nov 15^th, 2005, 7:14pm, Fritzlein wrote:

Ordered by rating is intentional. By nationality is a coincidence.

IP Logged

Fritzlein
Forum Guru

Arimaa player #706

Gender:

Posts: 5928

Re: 2006 World Championship
« Reply #23 on: Nov 20^th, 2005, 7:56pm »

Quote

Modify

OK, fifteen folks showed up for their games, so it is now official that this WC has the most participation ever. Also, many of the games were interesting and intense already in the first round.

Just seeing how things are going in this tournament so far, I think it is definitely a good idea to have winners play winners and losers play losers. If we didn't pair that way, but instead had something like 1 and 2 seeds reversing and playing 15 and 16 in the second round, it would be less fun for the prediction contest, less fun for the players involved, and simply less fun all around. Whatever way we decide to tweak the pairing system for next year, we should leave this desirable feature intact.

We're still on track for this to be the best WC ever, by a wide margin.

« Last Edit: Nov 20^th, 2005, 7:56pm by Fritzlein »

IP Logged

Fritzlein
Forum Guru

Arimaa player #706

Gender:

Posts: 5928

Re: 2006 World Championship
« Reply #24 on: Nov 20^th, 2005, 9:31pm »

Quote

Modify

Ryan Cable noticed something with regard to color assignments. In round two Grey_0x2A plays Megamau. In round one Megamau had silver. It isn't clear whether that should count or not, since he forfeited, but either way Megamau has played gold zero times. Meanwhile Grey_0x2A has played gold once, i.e. more than Megamau, so either way Grey_0x2A should get silver in round two.

All of this is mostly moot if Megamau forfeits again, but it should inspire one to take another look at the pairing code.

IP Logged

99of9
Forum Guru

Gnobby's creator (player #314)

Gender:

Posts: 1413

Re: 2006 World Championship
« Reply #25 on: Nov 20^th, 2005, 10:58pm »

Quote

Modify

Well spotted Ryan.

on Nov 20^th, 2005, 9:31pm, Fritzlein wrote:

In round one Megamau had silver. It isn't clear whether that should count or not, since he forfeited, but either way Megamau has played gold zero times.

It's very clear to me that it should count as him having played once with silver. That is the opportunity he had - just because he timed out on move 1 doesn't change the colour he had.

IP Logged

Ryan_Cable
Forum Guru

Arimaa player #951

Gender: male

Posts: 138

Re: 2006 World Championship
« Reply #26 on: Nov 21^st, 2005, 12:53am »

Quote

Modify

on Nov 20^th, 2005, 10:58pm, 99of9 wrote:

It's very clear to me that it should count as him having played once with silver. That is the opportunity he had - just because he timed out on move 1 doesn't change the colour he had.

Yes, I think forfeits should be treated exactly like other games. The person who forfeits doesn’t deserve any favors, and the person receiving the forfeit already gets a big privilege in the free advance to the next round. Mostly, I want this because the rules for losses are predictable up to the coin flips, and handling forfeits is otherwise unspecified. The rules state:

The director will also make the final decision on matters of pairing and color assignment.

So it is David Fotland’s opinion that matters most. If forfeits are not going to be treated like other games, then I think David should at least state in plain English the algorithm that is being used to handle them. (The ruby script is Greek to me; I can’t even find where it is assigning colors. Even if I understood the code, it would probably still be ambiguous, because it is not clear what the inputs are for a forfeit.) Even if megamau forfeits again, it is important to have this worked out to remove the ambiguity and unpredictability in the color assignment for Adanac and grey_0x2A in round 3+.

Whatever rule for handling forfeits is chosen, I can’t think of any good reason for megamau to have silver. I think David should override the script and give megamau gold as soon as possible, so grey_0x2A knows what color to prepare for in case megamau does show.

PS BlackKnight vs. acheron confirms that gold is not being assigned by alphabetical order, when the percent of previous gold is equal.

IP Logged

Fritzlein
Forum Guru

Arimaa player #706

Gender:

Posts: 5928

Re: 2006 World Championship
« Reply #27 on: Nov 21^st, 2005, 9:49am »

Quote

Modify

Speaking of color assignments, I notice that if I play PMertens in round three, one of us will be assigned gold for the third time in a row. However, it doesn't appear possible for someone to be assigned gold for the fourth time in a row in round four. For that to happen, all eight people who played gold in the first round would have had to win.

IP Logged

omar
Forum Guru

Arimaa player #2

Gender: male

Posts: 1003

Re: 2006 World Championship
« Reply #28 on: Nov 22^nd, 2005, 5:57am »

Quote

Modify

on Nov 20^th, 2005, 9:31pm, Fritzlein wrote:

Thanks for pointing this out guys. It was due to a bug in the code. I've fixed the code.

I've also changed the color assignement for Grey_0x2A and Megamau and sent an email to the players. No need to consult David on this since it is not an arbitrary choice or a dispute.

Here is the change to the code:

if (p1['fsc']/(p1['played']+1)) < (p2['fsc']/(p2['played']+1))

changed to:

if (p1['fsc']/(p1['played']+1.0)) < (p2['fsc']/(p2['played']+1.0))

It was using interger division but I was expecting float. By changing the 1 to 1.0 we cause it to use floating point calculations.

« Last Edit: Nov 22^nd, 2005, 8:49am by omar »

IP Logged

Fritzlein
Forum Guru

Arimaa player #706

Gender:

Posts: 5928

Re: 2006 World Championship
« Reply #29 on: Nov 22^nd, 2005, 8:39am »

Quote

Modify

on Nov 22^nd, 2005, 5:57am, omar wrote:

Thanks for pointing this out guys. It was due to a bug in the code. I've fixed the code.

Thank you, Omar, for fixing that so quickly, and for running the tournament so smoothly in general. You have worked out a fine system, and you stay on top of any problems, which makes it easy for everyone to have a good time.

IP Logged

Pages: 1 2 3 4 ... 6

Notify of replies

Send Topic


« Previous topic \| Next topic »