[2012] Multi bld: require at least 2 cubes solved

Discuss the WCA regulations.

[2012] Multi bld: require at least 2 cubes solved

Postby Pedro_S » Wed Feb 08, 2012 3:21 am

I would like to propose a small change to the multi blindfolded rules: requiring at least 2 cubes to be solved for a valid attempt.

Reasons:
- 1 cube solved is not really a "multiple" attempt
- people can cheat, memorising and solving just one cube
- people can cheat as said above and beat someone who did, for example, 5/10, because this other person will much likely take more time

Actually, I've seen it happen. A guy solved just one cube in 9 minutes and got a medal, "beating" someone who actually tried both but got just one, in 15 minutes.
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Re: [2012] Multi bld: require at least 2 cubes solved

Postby ardianto » Wed Feb 08, 2012 6:08 am

It is not a 'cheat', current regulation make it possible to do so.
It is more like a 'strategy', although I don't like it.

I've seen that 1/2 beat 2/4 also in competition I delegated: http://www.worldcubeassociation.org/res ... ngOpen2011
Competitor with 1/2 only memorize 1 cube, and finished third.
Competitors with 2/4 memorize all the cubes, with just only 2 other cubes with twisted corners, of course this took a longer time finished fifth.

I agree that it is not fair, in my opinion.

I would like to propose that treat 0 point as DNF, so 1/2 is DNF and 2/2 is solved. This make more sense than treat 1/2 as DNF and 2/4 as solved. Regarding old results, I don't have any idea to treat them. Opinion?

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Re: [2012] Multi bld: require at least 2 cubes solved

Postby Pedro_S » Wed Feb 08, 2012 12:27 pm

Yes, I know the regulations allow it, and that's the problem.

I meant cheating as in "not doing multiple cubes blindfolded". Some people just try one because of the higher time limit (20 min instead of the usual 10 or less), just for the personal acomplishment of having solved a cube blindfolded officialy...


I don't know what to do with current results...maybe they should stand. They would probably be beaten, and even if they don't, those attempts will fall down the ranking as time passes by.
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Re: [2012] Multi bld: require at least 2 cubes solved

Postby Hippolyte » Wed Feb 08, 2012 3:10 pm

ardianto wrote:I would like to propose that treat 0 point as DNF, so 1/2 is DNF and 2/2 is solved. This make more sense than treat 1/2 as DNF and 2/4 as solved. Regarding old results, I don't have any idea to treat them. Opinion?

I propose something like that:
- An attempt count if at least 50% of tried cubes are solved. All other results are treat as DNF.
AND
- An attempt count if at least 2 cubes are solved when the competitor stop the timer/ask for stopping. All other results are treat as DNF.

So 1/2 will be DNF, and results with just 50% will be counting attempts (even tough their scores are 0).

Once again, sorry for the English, do not hesitate to formulate it in a better way.
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Re: [2012] Multi bld: require at least 2 cubes solved

Postby Sebastien » Fri Feb 10, 2012 4:09 pm

First of all stop talking about "cheating". We know what you mean but this is a word-abuse and absolutely unadequate for people competing in an event according to the rules.

This topic is absolutely not new. there is a 2.5 year old thread about this in which you even posted Pedro:

http://worldcubeassociation.org/forum/viewtopic.php?f=4&t=806

To start my reply I will just quote what Ron said 2 years ago:

Ron wrote:
The problem of cubers "attempting" 2 cubes but only ever intending to even try one.

I do not consider that a problem. The best will still win.


Indeed he is totally right. For me as well going for 1/2 is a totally valid strategy. I would even go further and say, that someone who memorizes one cube and gets 1/2 is better than someone who memorizes 2 cubes and get 1/2. The first person has solved 100% of the cubes he tried, the other person only 50%. If the second person would have solved 100% of these cube, he would have beaten the other.
Btw I still miss valid arguments why going for 1/2 should not be ok. Every explanation I've read so far were personal feelings about this being unfair.
Also where is the matter if someone tries to get 1/2 just for having blindfolded a cube officially? 1/2 is a crappy result compared to a valid result in 3x3x3 BLD. It's everyone's own choice to go for this or not.
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Re: [2012] Multi bld: require at least 2 cubes solved

Postby Pedro_S » Fri Feb 10, 2012 8:19 pm

So you're saying memorising and solving just one cube, and getting a 1/2 is better than memorising 20 and solving 19, because the latter is not 100%?

My point is that: suppose at a competition everyone doing multi bld is trying 2 cubes. Person A takes 15 mins and ends up with 2 corners twisted in one cube. Person B takes 10 mins, 3 edges off. Person C takes 6 mins, 2 edges flipped. I memorise and solve just one and get 1:20. Should I win?

I mean, by the rules, I'll win, but it's not fair.

Now let's say everyone memorises and solves just one of the cubes, because someone is probably doing it already. Is that really the spirit we want?
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Re: [2012] Multi bld: require at least 2 cubes solved

Postby Sebastien » Sat Feb 11, 2012 1:29 am

Pedro_S wrote:So you're saying memorising and solving just one cube, and getting a 1/2 is better than memorising 20 and solving 19, because the latter is not 100%?


No, I'm not saying this.

Pedro_S wrote:My point is that: suppose at a competition everyone doing multi bld is trying 2 cubes. Person A takes 15 mins and ends up with 2 corners twisted in one cube. Person B takes 10 mins, 3 edges off. Person C takes 6 mins, 2 edges flipped. I memorise and solve just one and get 1:20. Should I win?

I mean, by the rules, I'll win, but it's not fair.


Yes, you should win. You still have not provided any argument why this should be unfair. In this scenario you would have chosen the best strategy, so obviously you deserve the win.

If you are happy with winning an event (which likely gives you no benefit) and getting a crappy official MBF result is another point though.

Pedro_S wrote:Now let's say everyone memorises and solves just one of the cubes, because someone is probably doing it already. Is that really the spirit we want?


You are asking random questions. But yes, if everyone wants to go for a bad results, then let them do. Facing realitiy though and a decent competition level those people will not really succeed with this strategy.
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Re: [2012] Multi bld: require at least 2 cubes solved

Postby Pedro_S » Sat Feb 11, 2012 1:19 pm

SebastienAuroux wrote:
Pedro_S wrote:So you're saying memorising and solving just one cube, and getting a 1/2 is better than memorising 20 and solving 19, because the latter is not 100%?


No, I'm not saying this.

Then I'm confused. What does this mean:
I would even go further and say, that someone who memorizes one cube and gets 1/2 is better than someone who memorizes 2 cubes and get 1/2. The first person has solved 100% of the cubes he tried, the other person only 50%. If the second person would have solved 100% of these cube, he would have beaten the other.

?

Pedro_S wrote:My point is that: suppose at a competition everyone doing multi bld is trying 2 cubes. Person A takes 15 mins and ends up with 2 corners twisted in one cube. Person B takes 10 mins, 3 edges off. Person C takes 6 mins, 2 edges flipped. I memorise and solve just one and get 1:20. Should I win?

I mean, by the rules, I'll win, but it's not fair.


Yes, you should win. You still have not provided any argument why this should be unfair. In this scenario you would have chosen the best strategy, so obviously you deserve the win.

If you are happy with winning an event (which likely gives you no benefit) and getting a crappy official MBF result is another point though.

This is unfair because I wouldn't be doing MULTIPLE cubes attempt, which is the event's name and purpose. Or am I wrong?

Pedro_S wrote:Now let's say everyone memorises and solves just one of the cubes, because someone is probably doing it already. Is that really the spirit we want?


You are asking random questions. But yes, if everyone wants to go for a bad results, then let them do. Facing realitiy though and a decent competition level those people will not really succeed with this strategy.

It wasn't a random question, it was a situation example.


Let me put this straight: I don't think people doing this kind of thing is the actualy problem. They're allowed to do so by the rules. The problem is the rules allowing this. If our rules allow someone to NOT do a multiple cubes attempt and get a valid result in a "multiple cubes" event, I guess they have a flaw, right?
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Re: [2012] Multi bld: require at least 2 cubes solved

Postby Sebastien » Tue Feb 14, 2012 6:54 pm

Pedro_S wrote:
SebastienAuroux wrote:Then I'm confused. What does this mean:
I would even go further and say, that someone who memorizes one cube and gets 1/2 is better than someone who memorizes 2 cubes and get 1/2. The first person has solved 100% of the cubes he tried, the other person only 50%. If the second person would have solved 100% of these cube, he would have beaten the other.

?


This is a reasoning for 2 cubes. I would not extend this to an arbitrary amount.

The event's name is "Rubik's Cube: Multiple Blindfolded". For me this indicates, that this is an event where multiple cubes are blindfolded, but not more. The regulations allowing a valid score with only one cube solved is still ok for me.
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Re: [2012] Multi bld: require at least 2 cubes solved

Postby Pedro_S » Tue Feb 14, 2012 7:54 pm

SebastienAuroux wrote:The event's name is "Rubik's Cube: Multiple Blindfolded". For me this indicates, that this is an event where multiple cubes are blindfolded, but not more. The regulations allowing a valid score with only one cube solved is still ok for me.

What I'm saying is that the name means that multiple cubes are to be attempted, which is not always true under current regulations.
The problem is not allowing a valid score with only one cube solved. The problem is allowing a valid score with just one cube attempted. And since we can't make sure people are actually attempting just one or not, the rules should require 2 cubes solved for a valid score, I think.
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Re: [2012] Multi bld: require at least 2 cubes solved

Postby Sebastien » Tue Feb 14, 2012 11:51 pm

I disagree. You attempt as many cubes as you hand in before. So if you hand in 2 cubes, you attempt 2 cubes, no matter if you try to solve both of them or not. this is why the result is 1/2 and not 1/1.
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Re: [2012] Multi bld: require at least 2 cubes solved

Postby Pedro_S » Wed Feb 15, 2012 12:53 am

I think you understood what I meant with "attempt", but let me rephrase it.

The problem is allowing a valid score with just one cube tried. And since we can't make sure people are actually trying just one or not, the rules should require 2 cubes solved for a valid score, I think.
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Re: [2012] Multi bld: require at least 2 cubes solved

Postby Clement Gallet » Wed Feb 15, 2012 8:08 am

Pedro_S wrote:The problem is...

Why is it a problem ?
Pedro_S wrote:And since we can't make sure people are actually trying just one or not, the rules should require 2 cubes solved for a valid score, I think.

So you prefer to assume people are "guilty" than "innocent" ?
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Re: [2012] Multi bld: require at least 2 cubes solved

Postby Pedro_S » Wed Feb 15, 2012 1:54 pm

Clement Gallet wrote:
Pedro_S wrote:The problem is...

Why is it a problem ?

It is a problem because it allows people to not do the event they're supposed to, which is "memorise and (try to) solve more than one cube blindfolded".
Pedro_S wrote:And since we can't make sure people are actually trying just one or not, the rules should require 2 cubes solved for a valid score, I think.

So you prefer to assume people are "guilty" than "innocent" ?

I don't prefer to assume anything, I'm just saying we can't know wheter the person actually tried both cubes or not. In that case, we should not allow just one cube solved, because it may be just one cube tried, and that's not the purpose of the event.
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Re: [2012] Multi bld: require at least 2 cubes solved

Postby Sebastien » Fri Feb 17, 2012 9:34 am

Pedro_S wrote:which is "memorise and (try to) solve more than one cube blindfolded".


Where do you take this from?
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Re: [2012] Multi bld: require at least 2 cubes solved

Postby Pedro_S » Fri Feb 17, 2012 1:52 pm

H1a) Competitor must tell before the start of the event how many puzzles (>1) he wants to solve blindfolded for each attempt.
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Re: [2012] Multi bld: require at least 2 cubes solved

Postby Dene » Sun Feb 19, 2012 4:44 am

SebastienAuroux wrote:Yes, you should win. You still have not provided any argument why this should be unfair. In this scenario you would have chosen the best strategy, so obviously you deserve the win.


Making a mockery of the WCA by abusing the regulations should not be an allowable strategy. There should only be one strategy in this event, and that is to solve as many of the attempted cubes as possible. How one goes about that is where any other strategic decisions should go on.

I'm with Pedro on this issue; I have seen instances of this rule being abused and it is not in the spirit of the event, it's just a waste of my time, and the time of every other competitor at the competition.
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