# World Championship 2007



## pjk (Sep 19, 2007)

Who is going? Post where you are staying, and yours goals for the competition.

Should be an awesome competition! Unfortunately, I won't be able to make it. I am looking forward to seeing some nice videos afterwards.


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## Erik (Sep 19, 2007)

Erik Akkersdijk, 
Arrival day: Octobre 3rd
Hotel: Novotel
Return day: Monday 
Events: about all
Goals: have (a) good time(s)!


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## Rama (Sep 19, 2007)

Me,
3 October I arrive in Budapest Hotel and I stay with Mr. Akkersdijk.
I am graduatly practising again and stopped playing the Bass-Guitar and stopped riding the BMX, untill were done with the Worlds.

My goals: Having fun with my new friends I am going to make there.


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## Pedro (Sep 19, 2007)

I arrive october 4th and leave the 9th

not sure where I'm going to stay, but probably the Ginkgo hostel

I'm going to do 3x3, OH, bld and 4x4x4 (maybe multi bld, if there's room for me  but I just suck at it...my best ever was 2 cubes)

My goals: Learning Rama's killer instinct for OH cubing 

haha...seriously, meeting a lot of cubers, learning and sharing experiences, and maybe learning Matyas' bld system


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## joey (Sep 19, 2007)

Me.
Staying in the competition hotel, sharing with Joel Van Noort.
I am only doing a few events, cos I have improved alot since I registered, so I didn't register for enough events 
Goals: Just do well, its my first comp after all!



Rama said:


> My goals: Having fun with my new friends I am going to make there.



Yeh, I am sure Erik will introduce us to each other!


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## tim (Sep 19, 2007)

Tim Habermaas,
Arrival day: Thursday (morning)
Ho(s)tel: Ginkgo Hostel
Return day: Monday (morning)
Events: 3x3, 2x2, bld (i'll try to ask for multiple bld)
Goals: getting through the qualification rounds


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## FrankMorris (Sep 20, 2007)

Frank Morris
Novotel Budapest Congress
3x3, 4x4, 5x5

I hope that I don't make a fool of myself, and have a great time hanging out with everyone in case this is my last competition.

I arrive on 9/28 and I will leave on 10/10


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## Pedro (Sep 20, 2007)

Kal El said:


> Frank Morris
> Novotel Budapest Congress
> 3x3, 4x4, 5x5
> 
> ...



why is that?


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## adragast (Sep 20, 2007)

I will come with a friend. I arrive the 3rd and he will arrive the 4th. We stay at Mercure hotel. 
Goals: have fun of course, learn a lot from other cubers, manage a sub-1'30 in OH
Dream: Manage a sub-30 in the normal 2 handed competition (but I don't think I will manage it).


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## gillesvdp (Sep 20, 2007)

I will arrive on Thursday 4th October and leave on Monday 8th October.
I will stay at the Novotel hotel.

My goals are as follow:
- Get into the semi-final for the 3x3 event
- Get into the final for the 3x3 OH event
- Get belgian record for 3x3 BLD event (2-18.xx)

I am really looking forward to it. ^^


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## DanHarris (Sep 20, 2007)

Arrive 3 Oct Depart 10 Oct.
a lot of events

Goals: to win, of course! And to make sure I stay UK#1 at least...

Dan


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## AvGalen (Sep 20, 2007)

I don't know how and when I will get there and how/when I will leave. I might come by car, or by plane. I will stay in the same hotel/room as Bob Burton.

Goals improve/keep my two National Records (mbf, fmc) IF I am allowed to compete in them. I seem to be allowed to compete in every competition except the two for which I have National Records and top-10 rankings.

I think all top-50 ranked people should be allowed to compete in that event (if they come)


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## Mike Hughey (Sep 21, 2007)

AvGalen said:


> Goals improve/keep my two National Records (mbf, fmc) IF I am allowed to compete in them. I seem to be allowed to compete in every competition except the two for which I have National Records and top-10 rankings.
> 
> I think all top-50 ranked people should be allowed to compete in that event (if they come)



I'm very curious about this - you've mentioned it several times. What have you been told that indicates you might not be allowed to compete, and what is the reason behind it? It sounds outrageous that you're not allowed to compete, but I don't have all the facts, so I'm curious what exactly you've been told. The one thing I can see in the rules that would certainly apply has to do with a limit to number of competitors - is that what happened to you?


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## Harris Chan (Sep 21, 2007)

Arrive: Oct 4th (Leave Canada at Oct 3)

Depart: Oct 8th

Hotel: Novotel (that official hotel)

Events: Just all the 3x3 (speedsolve, OH, BLD (I didn't register for the big cubes because it was too late...and at the time of the registration I didn't even have any XD)

Goal: -Have a successful BLD (I don't care how slow I just want to make a successful one lol)
-Get into semi-finals or even finals of 3x3 speedsolve
-And of course, meet all the people I talk to on the internet in real life! (And have fun!)


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## AvGalen (Sep 21, 2007)

Your new events are:
= Rubik's Cube : Qualification round (Friday)
= 4x4x4 Cube : Qualification round (Friday)
= 5x5x5 Cube : First round (Saturday)
= Rubik's Cube Blindfolded : Qualification round (Friday)
= Rubik's Cube One-handed : Qualification round (Friday)
= 2x2x2 Cube : Combined final (Friday)
= Rubik's Magic : Final (Saturday)
= Rubik's Master Magic : Final (Saturday)
= Rubik's Clock : Combined final (Sunday)
= Square-1 : Combined final (Saturday)
= Megaminx : Combined final (Saturday)
= Pyraminx : Combined final (Friday)
= Rubik's Cube Multiple blindfolded : Waiting list
= Rubik's Cube Fewest moves : Final (Sunday)


This was the latest mail that I received today. Someone cancelled on Fewest Moves and I was next on the list


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## Stefan (Sep 21, 2007)

Arrive Oct 3, leave Oct 9
Hostel: Ginkgo
Events: Most
Goal: Beat Matyas at something. Megaminx, I guess.


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## Mike Hughey (Sep 21, 2007)

AvGalen said:


> Your new events are:
> = Rubik's Cube Multiple blindfolded : Waiting list
> = Rubik's Cube Fewest moves : Final (Sunday)
> 
> ...



Oh, I see. I'm glad you got into Fewest Moves - you REALLY deserve to be in that one!

Does some nice person want to cancel on Multiple Blindfolded so Arnaud can go for the new world record?


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## gillesvdp (Sep 21, 2007)

Hey, I'm still up for the 2 cubes Multiple Blindfolded Speed WR ^^


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## krnballerzzz (Sep 22, 2007)

I plan to attend. 
My goals are meet as many speedcubers as possible, get an autograph from Erno Rubik (if he attends xP), and win the 3x3 main competition.


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## gillesvdp (Sep 22, 2007)

krnballerzzz said:


> I plan to attend.
> My goals are meet as many speedcubers as possible, get an autograph from Erno Rubik (if he attends xP), and win the 3x3 main competition.



I'm not sure he will sign (m)any cubes as he only signs 2 cubes per year.

PS: here are the events I am competing in:
= Rubik's Cube : First round (Saturday)
= 4x4x4 Cube : First round (Saturday)
= 5x5x5 Cube : First round (Saturday)
= Rubik's Cube Blindfolded : First round (Sunday)
= Rubik's Cube One-handed : First round (Saturday)
= Rubik's Magic : Final (Saturday)
= Rubik's Clock : Combined final (Sunday)
= Square-1 : Combined final (Saturday)
= Rubik's Cube Multiple blindfolded : Waiting list


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## Ron (Sep 22, 2007)

Check out http://www.worldcubeassociation.org/results/c.php?allResults=All+Results&competitionId=WC2005
and see which events you could have won if you were as good as you are now.
Especially 2x2 and 5x5 would be piece of cake.


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## AvGalen (Sep 22, 2007)

Nice first post on this forum Ron! I hope we will see you here often.



> Does some nice person want to cancel on Multiple Blindfolded so Arnaud can go for the new world record?


 I plan on doing 5/5 or 6/6 so there is no way that I could beat the world record. However I might just win if the real experts all do 10/11 or 19/20 or so. I don't think that is entirely impossible. In some multiple blindfold competitions it would have been enough to:

1) Ask for 2 cubes
2) Only study 1 cube
3) Only solve 1 cube
4) Win the event with 1/2 cubes in just a couple of minutes.


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## pjk (Sep 24, 2007)

I am guessing the Multi-BLD event winner will do 15/15  Just a guess....


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## AvGalen (Sep 24, 2007)

I think the real multi-blind-experts will want to tell how many they are attempting in secret. Otherwise Row might just hear Mátyás will do 12 and then he will do 13.


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## Mátyás Kuti(shaipo) (Sep 24, 2007)

AvGalen said:


> I think the real multi-blind-experts will want to tell how many they are attempting in secret. Otherwise Row might just hear Mátyás will do 12 and then he will do 13.



I'll do 18. Rowe will do (maximum) 17. 
We have already discussed it. xD


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## tim (Sep 24, 2007)

Mátyás Kuti(shaipo) said:


> AvGalen said:
> 
> 
> > I think the real multi-blind-experts will want to tell how many they are attempting in secret. Otherwise Row might just hear Mátyás will do 12 and then he will do 13.
> ...



And you guys can trust each other? 

I thought 10 - 13 would be enough to win this event. 18 cubes are crazy . If one of you will get this, i'll stop practicing for the multi-blind world record :/.


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## AvGalen (Sep 24, 2007)

What if you only get 1 attempt? Would you guys try 17/18 cubes also and risk me winning with a 2/2?


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## AvGalen (Sep 24, 2007)

After Ron's post about the WC2005 I digged a little and found this story of the EC2006 by Erik. You will be amazed how much has (and hasn't) changed in just 1 year

http://erikku.er.funpic.org/rubik/euro2006.html


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## Stefan (Sep 24, 2007)

AvGalen said:


> What if you only get 1 attempt?


What do you mean, "if" ?


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## AvGalen (Sep 24, 2007)

That was my subtle way of pointing out to everyone that you only get one attempt. That means that for 17 cubes at roughly 200 moves per cube you cannot make a single wrong move out of 3400 moves or else you could get beaten by a beginner (like me) that chooses to do 2/2 and does it correctly in 29:59.99.


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## Erik (Sep 24, 2007)

AvGalen said:


> After Ron's post about the WC2005 I digged a little and found this story of the EC2006 by Erik. You will be amazed how much has (and hasn't) changed in just 1 year
> 
> http://erikku.er.funpic.org/rubik/euro2006.html


I found the following sentences quite funny now:
'Gilles has became very fast in OH cubing. Best avg below 30 sec!'
'I started fairly good with a 17.66' (3x3)
'the 3rd was better 2:04.88NR' (minx)
'I manage a sub-4min, but I got a 4:27.78' (5x5)
'ER average of 29.09!' (gilles on OH)
'Anssi won with an amazing time of 1:25.63' (3x3 feet)
'won with 14.97 ' (Joel on 3x3 finals)
also what was the solution to this riddle of Guus again? 'you have the digits 1, 3, 4 and 6. By dividing, multiplying adding and the other thing of which I don't know the english name whe had to make 24'
I forgot


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## Stefan (Sep 24, 2007)

AvGalen said:


> That means that for 17 cubes at roughly 200 moves per cube you cannot make a single wrong move out of 3400 moves


And Matyas did make a mistake before, in one of the 52 cubes he attempted in competitions so far.


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## Stefan (Sep 24, 2007)

Erik said:


> also what was the solution to this riddle of Guus again? 'you have the digits 1, 3, 4 and 6. By dividing, multiplying adding and the other thing of which I don't know the english name whe had to make 24


It's called "subtract". And there are at least two answers. I asked it at the US Open and Tyson found the answer I knew, while Shelley and Toby found another one.


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## DanHarris (Sep 24, 2007)

Erik said:


> By dividing, multiplying adding and the other thing of which I don't know the english name whe had to make 24'
> I forgot



Subtracting!


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## DanHarris (Sep 24, 2007)

sheesh, stefan got his post in inbetween me viewing this page and me refreshing this page to see my post. Git.


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## AvGalen (Sep 24, 2007)

I think this should be correct: 6 / (1-(3/4))
1-3/4 = 0.25
6/0.25 = 24


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## Stefan (Sep 24, 2007)

I just noticed I presented the riddle with the numbers 1, 4, 5, 6. I don't see a second solution with 1, 3, 4, 6.


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## Pedro (Sep 24, 2007)

AvGalen said:


> That was my subtle way of pointing out to everyone that you only get one attempt. That means that for 17 cubes at roughly 200 moves per cube you cannot make a single wrong move out of 3400 moves or else you could get beaten by a beginner (like me) that chooses to do 2/2 and does it correctly in 29:59.99.



that's what I don't understand...

if someone goes for, say, 10 cubes, and finish 9, why should someone who did 2/2 win???


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## joey (Sep 24, 2007)

Pedro said:


> that's what I don't understand...
> 
> if someone goes for, say, 10 cubes, and finish 9, why should someone who did 2/2 win???



100% perecent accuracy?

Is 2/2 better than 9/10? Yes because of accuracy, but the 9/10 person still got 9 cubes solved.


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## Erik (Sep 24, 2007)

then I could ask for 1000 cubes and pick the 10 easiest and forget about the rest


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## Pedro (Sep 25, 2007)

nah...makes no sense to me that someone who makes 2/2 wins when someone does like 19/20...


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## AvGalen (Sep 25, 2007)

2/2 > 19/20. It is the easiest math there is. (1 > 0.95)


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## normanbim (Sep 25, 2007)

that will be very hard stefan. he made 16 world records just this year. it took macky three years just to get 15 world records. i hope macky is coming. he is one of the reasons i started speedsolving and blindsolving. hail macky!


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## AvGalen (Sep 25, 2007)

Stefan1,4,5,6 is just the same idea as 1,3,4,6: Don't do 6*4, but do (6/(1/4)). Don't do 4*6, but do ((4/(1/6)).
Solution: 4 / (1-(5/6))
Which one of the two has 2 solutions?


> I just noticed I presented the riddle with the numbers 1, 4, 5, 6. I don't see a second solution with 1, 3, 4, 6.


 
normanbim: What were you responding to when you said this?


> that will be very hard stefan


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## Mark (Sep 25, 2007)

1,4,5,6 has two solutions: 4/(1-5/6) and 6/(5/4-1)


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## AvGalen (Sep 25, 2007)

OK, same idea again (use 3 numbers to create 1/4 or 1/6)


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## hdskull (Sep 25, 2007)

normanbim said:


> that will be very hard stefan. he made 16 world records just this year. it took macky three years just to get 15 world records. i hope macky is coming. he is one of the reasons i started speedsolving and blindsolving. hail macky!



that's because matyas competes in like 10 more events than macky does. all macky competes in is 3x3, OH, BLD, 2x2, 4x4, that's 5 events total.

EDIT: i just checked, matyas competes in 17.


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## Stefan (Sep 25, 2007)

hdskull said:


> [that's because matyas competes in like 10 more events than macky does.


On the other hand, Matyas started much later, when the records were already a whole lot better than when Macky collected so many. Macky's last record is from 2005, Matyas only got one in 2006 and all others in 2007. I'm in the same boat, I got most of my records somewhat for free, because there was little serious competition back then. In retrospect I regret this, because I was a lot more motivated a while ago and now that the overall level has risen and I could try to improve accordingly (for example by learning the better methods for the magics), I feel a little fatigued, like I'm too old to try to get to the top again. I've already given up on the magics quite a while ago.


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## Pedro (Sep 25, 2007)

AvGalen said:


> 2/2 > 19/20. It is the easiest math there is. (1 > 0.95)



I know the % is 'better' if you do 2/2 instead of 19/20, but that just doesn't seem fair for me...

now imagine some attempts 100 and gets 99 correct...that's some thousands of correct moves...

if someone does 2/2, that's something like 200-some moves...

isn't making 99/100 *MUCH* harder than 2/2? shouldn't someone who does that win?


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## AvGalen (Sep 25, 2007)

I think you have to create a pretty complicated formula Pedro.

The following are ordered according to the current rules. Could you order them according to your ideas and then find a formula that matches that idea? (I have left the time factor out of the examples, that would complicated matters even more!)



> 9f16) For the 3x3x3 Cube: Multiple Blindfolded event the order in the results is based on:
> Highest in the ranking are the competitors who solved all puzzles, among these competitors a larger number means better. For equal results the lower total time means better.
> Next in the ranking are the competitors who solved at least one of the puzzles, among these competitors a larger number of solved puzzles means better, and after this a lower number of attempted puzzles means better. For equal results the lower total time means better.
> Next in the ranking are the competitors who did not solve at least one puzzle, these competitors all finish at the worst position.


 
1000/1000
100/100
10/10
2/2
499/1000
99/100
99/1000
50/100
9/10
9/100
9/1000
8/9
6/7
4/5
2/3
1/2
1/10
1/100
1/1000
0/2, 0/10, 0/100, 0/1000


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## Johannes91 (Sep 25, 2007)

AvGalen said:


> 499/100


That's quite good.


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## AvGalen (Sep 25, 2007)

Edited. I include this because I want to know if 50/100 or 499/1000 is considered better.

The only type of formula I can personally think of (besides the current rules which I think are fine) would be somethink like:
n points for each solved cube
-m points for each unsolved cube
Most points wins.

But it will be very hard/arbitrary to determine n and m and some unwanted effects might arise.

I think the current rules are good because they force you to do well if you want to be sure that you will win. 2/2 might sound easier than 9/10, but the current system is already pretty motivating to attempt a lot of cubes
1) If I attempt 2 and get 2/2 than my ranking still depends on someone else failure on 3/3 or 10/10
2) If I make 1 mistake and do 1/2 (50%) I will lose to someone that does 2/10 (20%)
3) If you attempt 100 cubes and all others attempt 5 you can take a gamble and pick the 5 easiest cubes (5%), memorize and solve only those and beat everyone that gets 4/5 or worse.
4) If you use trick 3 you can have 6*15 + 94*10 = All day  to memorize and solve those 5 easieast cubes


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## Pedro (Sep 25, 2007)

AvGalen said:


> I think you have to create a pretty complicated formula Pedro.
> 
> The following are ordered according to the current rules. Could you order them according to your ideas and then find a formula that matches that idea? (I have left the time factor out of the examples, that would complicated matters even more!)
> 
> ...



well, for me it would be:

1000/1000 (of course )
499/1000
100/100
99/1000
99/100
50/100
10/10
9/10
9/100
9/1000
8/9
6/7
4/5
2/3
2/2
1/2
1/100
1/1000
0/2, 0/10, 0/100, 0/1000 (meaning all of them are at the last position )

no need for complicated formulas...

who solves more cubes wins 
if 2 persons solve the same number, then the one who tried more wins...(99/1000>99/100;9/10>9/100>9/1000)

well, that's just what I think...

I don't think is fair in any way to give victory to someone who did 2/2 when another one did 999/1000...


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## Stefan (Sep 25, 2007)

If you don't like the WCA rules, talk to the WCA. They have a forum for that.

Here's my suggestion from a while ago, inspired by the hour number discipline of the memorizers:

a) Allow a fixed amout of total time (e.g. one hour) 
b) Count the number of correctly solved cubes, not the attempted ones. 
c) Require fixed success rate (e.g. 2/3 of given cubes solved).

The current rules requiring 100% success rate stem from an era when three cubes was considered awesome. I actually talked with Ralf about two years ago and he considered blindsolving even one cube to be hard, which I think we now all agree is just wrong, as blindcubing is rather very easy. The higher the number of cubes gets, the more ridiculous it is to keep those old rules. Even the professional memorizers don't require perfection, in case of mistakes they just don't get full points (see the Scoring section in the page linked above).


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## Pedro (Sep 25, 2007)

already posted there 

(and I knew the wca forum, btw )


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## gillesvdp (Sep 25, 2007)

The number of competitors for Multible Blindfolded has been raised...

WR for 2 cubes, here I come !!!


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## Pedro (Sep 25, 2007)

nope...I'm coming for it hahaha


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## gillesvdp (Sep 25, 2007)

haaa noooooo
I will do it sub2 (Matyas can do it  )


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## Pedro (Sep 25, 2007)

sub-2? 

then you'll beat me...I just did 2 cubes once...was like 4:39


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## tim (Sep 25, 2007)

gillesvdp said:


> The number of competitors for Multible Blindfolded has been raised...



oh, really? i wish i would have registered for multi bld .


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## AvGalen (Sep 26, 2007)

Pedro:


> if 2 persons solve the same number, then the one who tried more wins...(99/1000>99/100;9/10>9/100>9/1000)


Where is the logic in that? You actually think 2/3 is better then 2/2? I don't think you thought about


> 3) If you attempt 100 cubes and all others attempt 5 you can take a gamble and pick the 5 easiest cubes (5%), memorize and solve only those and beat everyone that gets 4/5 or worse.


enough.

Stefan:


> If you don't like the WCA rules, talk to the WCA. They have a forum for that.


 I think the WCA IS the place to talk about this, but only after you have finalized an idea. I think this forum is a great place to discuss general ideas first, work out a few small oversights and then post at the WCA.
Although I personnaly like the current rules (I think blindfolded should be based on perfection), I think your "1 hour" plan is a good idea. It makes multi_blind more into a speedcubing event, more managable for an organisation, more competitive and more inline with fewest moves. I also see a big disadvantage: Doing > 30 cubes might not be possible with the new rules.

And I just received a messages telling me that I will be able to compete in multi-blind at Worlds


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## gillesvdp (Sep 26, 2007)

Pedro said:


> sub-2?
> 
> then you'll beat me...I just did 2 cubes once...was like 4:39



Ok that's fast ^^

I will just go for sub10 I think. So don't worry about me. ;-)


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## Stefan (Sep 26, 2007)

AvGalen said:


> Although I personnaly like the current rules (I think blindfolded should be based on perfection), I think your "1 hour" plan is a good idea. It makes multi_blind more into a speedcubing event, more managable for an organisation, more competitive and more inline with fewest moves. I also see a big disadvantage: Doing > 30 cubes might not be possible with the new rules.


I also think blindcubing should be based on perfection. Just on a finer level now that we've in my opinion clearly outgrown the original rules. Like mentioned before, those rules were made with a handful of cubes in mind, not dozens. I want to still require each single cube to be solved perfectly in order to count, but not require all cubes to be solved perfectly.

Consider the "hour numbers" memorization again. There they count each row (containing 40 digits) individually. Make one mistake in a row and you get half the points for that row, make two mistakes and you lose the entire row. This however doesn't make you lose any of the other rows, your total score is the sum of your individual row scores. I'd like us to treat our cubes like their rows, except that each cube must be solved perfectly, not still getting a half point when you made a small mistake. Um wait, actually I'd like to give a half point in case of situations that would just cause a two seconds penalty in speedsolving, e.g., when you're a quarter turn away from solved.

I don't think it's a disadvantage we couldn't solve more than 30 cubes this way. The current rules allow 5 hours and 30 minutes for 30 cubes. That's way too long to handle in most competitions. Usually competitions do limit the total time for this event already (I think) anyway. Thus how many cubes you can do depends on the competition you go to. Not good. Also, given enough time, even 100 cubes or more are possible. But I'd say competitions are not the right place for this. Compare again to the memorizers. They do have "hour cards" in competitions (see here) and they have larger record attempts outside of the competitions (see here and here).

I just noticed the hour cards discipline is actually a better analogy to blindcubing, because "a deck of cards" seems more like "a cube" than "a row of 40 digits" does.. You get 52 points for a correct deck of cards, 26 if you make one mistake, 0 for more mistakes, and then your total score is the sum of all individual deck scores.


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## AvGalen (Sep 26, 2007)

I really like this idea!
A solved cube is +1 point
A +2 penalty solved cube is +0.5 points.
An unsolved cube is 0 points.
2/3 of the attempted cubes must be solved (to prevent my "trick 3")
Most points win.

Some questions:
1) How/when do you determine how many cubes you are going to do?
2) Can you decide to do more/less cubes during memorisation?
3) How does adding/removing cubes affect the 2/3 solved rule?
4) How would the ranking look with these results (would they all share 1st place?)
--4/4 in 54:05
--4/5 in 54:05
--4/4 in 60:00
--4/5 in 60:00


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## guusrs (Sep 26, 2007)

*original riddle*



AvGalen said:


> OK, same idea again (use 3 numbers to create 1/4 or 1/6)



Hello guys,

Do not %$#@* my riddle. The original riddels was: used all digits 1,3,4,6 (exact 1 time) to create 24 by dividing, multiplying, adding or substracting. The same for 3,3,8,8.

See you in Budapest


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## guusrs (Sep 26, 2007)

Hello Guys,

I'm arriving at october 4 (around noon) and leaving october 8 (afternoon).
Staying at Kings hotel

Competing in speed 3x3x3, 2x2x2 and fewest moves

Goals:
Having fun!
2x2x2 best sub 6 seconds
3x3x3 best sub 16 seconds (would I then be in the semi-finals?)
3x3x3 average sub 18 seconds
fewest moves: beating the Dutch record (33 moves by Arnaud)

See ya 

Gus


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## Erik (Sep 26, 2007)

Guus! It will be so great to meet you again  
You will be at dutch open both days?


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## AvGalen (Sep 26, 2007)

> Guus! It will be so great to meet you again


 and even better if you don't break my record 

But seriously, I have tipped you as the winner with a new WR. I am not bad, it is just my bad luck that I live in the Netherlands where everyone seems to be better. I never thought I would get a NR and now I have 2.


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## Pedro (Sep 26, 2007)

yeah, maybe solving x% necessary is the best way...

then the point counting


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## hait2 (Sep 26, 2007)

i dunno, why don't you just do this:
determine what number of cubes you will solve beforehand. once determined, it cannot change.
+1 point for a solved cube, -1 point for an unsolved cube
highest points wins. in ties, lower time wins. 0.5 points for a penalty cube if you like
i can't see how this would be abused..

the only question is whether you agree that stuff like
14/17 is better than 10/10, equal to 11/11 and worse than 12/12++
seems okay to me


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## AvGalen (Sep 26, 2007)

I think 5/5 is much better than 7/9, so I don't agree with this system. Now what do we do?


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## hait2 (Sep 27, 2007)

in my system they are equivalent so 5/5 would win since it would take less time
i don't see an issue
and i think 4/4 is worse than 7/9 so the line is drawn nicely, no?


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## AvGalen (Sep 28, 2007)

I am not so sure that 5/5 would take less time than 7/9.

And I think 4/4 is better than 7/9


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## Stefan (Sep 28, 2007)

AvGalen said:


> And I think 4/4 is better than 7/9


Even if they both took the same time?


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## hait2 (Sep 28, 2007)

i don't know arnaud.. i doubt many people will agree that 4/4 is better than 7/9
i mean in the 7/9 case you're solving almost double the cubes bld with fairly good accuracy rate (almost 80%)

and 5/5 has to take less time than 7/9 simply because you're memorizing and solving less, by quite a bit

well it was just an idea anyhow, everyone will of course have different opinions on how it should work


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## Mike Hughey (Sep 28, 2007)

hait2 said:


> i don't know arnaud.. i doubt many people will agree that 4/4 is better than 7/9


I will agree with Arnaud - I think 4/4 is better than 7/9. If you did 7/9, and thought you could do better with 4 cubes, you should have chosen to solve 4 instead.


hait2 said:


> well it was just an idea anyhow, everyone will of course have different opinions on how it should work


Very true. I'm sure lots of people agree with you. But I agree with Arnaud. I think accuracy is very important, and being able to judge your ability is also very important.


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## Stefan (Sep 28, 2007)

Mike Hughey said:


> I will agree with Arnaud - I think 4/4 is better than 7/9.


Then I ask you, too: Even if they both took the same time?


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## Mike Hughey (Sep 28, 2007)

StefanPochmann said:


> Mike Hughey said:
> 
> 
> > I will agree with Arnaud - I think 4/4 is better than 7/9.
> ...


Absolutely! If the person who did 7/9 was capable of doing 4/4 (as opposed to, say, 3/4, which would have been almost the same accuracy as 7/9), then that person SHOULD have chosen to solve 4 cubes instead of 9. They would undoubtedly have then gotten a better time than the person who did do 4/4 (unless they have to REALLY slow down for the accuracy). So they chose wrong on the number of cubes to do - that was their fatal mistake. And there is a chance that the 7/9 person might have gone slower than the 4/4 person if only attempting 4 cubes, but going for 100% accuracy.

Since the current regular 3x3x3 BLD event puts so little emphasis on accuracy, it seems only appropriate that the multi-BLD event put more emphasis on accuracy. I'd still rather see valid averages required for 3x3x3 BLD events, but I can live with it the way it is.

Another way to look at it: you and i were both at the US Open and saw Rowe Hessler's 5/8 attempt. I'm sure you were as impressed as I was with that. But don't you think that the average person in the audience there would have been more impressed with 3/3 than with 5/8? Even if they were the same time? And certainly at least as impressed by 2/2 as by 5/8. The average person doesn't understand why, if you can do it sometimes, you miss sometimes. So 100% accuracy is much more impressive to the average person. At least, that's my opinion - I haven't asked enough to know that for sure - I'll admit it's a guess on my part. And since multi-BLD is largely an exhibition anyway, I'd think the most important thing is to appeal to the average person.


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## hait2 (Sep 28, 2007)

Mike Hughey said:


> StefanPochmann said:
> 
> 
> > Mike Hughey said:
> ...



well i guess if you place accuracy above all, then yes, of course my system wouldn't work. i thought it provided a nice balance between accuracy and amount (both which are important imo) without being overly complicated. 

also ironically, 5/8 under my system would indeed be equal to 2/2 (although would lose since it would probably take more time) and worse than 3/3 ^_^;

i think quite a lot of us can agree that current system isn't too good though (is 99/100 really worse than 2/2?)
im just throwing ideas out, and i encourage everyone to do the same ^^;


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## Stefan (Sep 28, 2007)

Mike Hughey said:


> But don't you think that the average person in the audience there would have been more impressed with 3/3 than with 5/8? Even if they were the same time? And certainly at least as impressed by 2/2 as by 5/8.


No I don't. I think the average audience person would think 5 cubes is better than 2 or 3.



Mike Hughey said:


> The average person doesn't understand why, if you can do it sometimes, you miss sometimes. So 100% accuracy is much more impressive to the average person. At least, that's my opinion - I haven't asked enough to know that for sure - I'll admit it's a guess on my part.


How many people did you ask? I'll ask my colleagues next Monday and report the answer here.


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## ExoCorsair (Sep 28, 2007)

StefanPochmann said:


> Mike Hughey said:
> 
> 
> > The average person doesn't understand why, if you can do it sometimes, you miss sometimes. So 100% accuracy is much more impressive to the average person. At least, that's my opinion - I haven't asked enough to know that for sure - I'll admit it's a guess on my part.
> ...



I think I'll do the same; results should be interesting.


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## Pedro (Sep 28, 2007)

StefanPochmann said:


> Mike Hughey said:
> 
> 
> > I will agree with Arnaud - I think 4/4 is better than 7/9.
> ...



I think if they took the same time, more cubes should win...



StefanPochmann said:


> Mike Hughey said:
> 
> 
> > But don't you think that the average person in the audience there would have been more impressed with 3/3 than with 5/8? Even if they were the same time? And certainly at least as impressed by 2/2 as by 5/8.
> ...



I agree that average audience would be more impressed with 5...

and...which opinion is more important? the average audience or the "cubing audience"?


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## Mike Hughey (Sep 28, 2007)

Pedro said:


> I agree that average audience would be more impressed with 5...
> 
> and...which opinion is more important? the average audience or the "cubing audience"?



A valid question. I'll certainly admit I might be focusing on the interests of the wrong audience.

For those who are going to ask their colleagues - I hope you'll try your best to ask the question in a way that will affect their answers; try to word it in an answer-neutral way. I'll ask around myself and see what I get. Maybe I was guessing completely wrong; I'll be interested in finding out.


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## adragast (Sep 29, 2007)

sorry to disturb you but this debate seems to go completely outside the topic, isn't it ?
I think that it would be better to discuss about this in a new topic for people interested in multi-blindfold and not interested in the world championship for example....


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## DavidCalvo (Sep 30, 2007)

Hello you guys!

I'm arriving on 4th and leaving on 8th.

I'm staying on the Mercure Hotel.

I have quite a long scale on Munich both on 4th and on 8th. Do you know any speedcuber attending to the WC who lives there?

See you within a few days!

David


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