[Unofficial] Maskow: Multiblindfold 125/150

[Leo123]

Not really, you just have different types of pieces.

Probably in about 6 to 10 attempts

 

[moralsh]

Wouldn't an 11x11 be too hard to memorise or could he do it in about 10 attempts?

Big cubes are a bit harder to track pieces while doing memo and a lot harder to execute.

Maskow, awesome, just awesome.

 

[Rocky0701]

Not really, you just have different types of pieces.

Well the thing about an 11x11 solve is that 1. It would be about like doing a 30/30 MBLD solve in terms of amount memorized, however have you ever held a cube like bigger than a 9x9? They are hard to hold and turn just because of they're size. Even if they are smooth you can't get much more than like 1 TPS on them. 2. Even just finding the layer that you were at would take up time during centers, you would probably have to count to a certain layer quite a bit just so you don't mess up since you can't see the cube.
3. Because of these reason, execution would take at least 5 times as long as if it were a 30/30 MBLD. It would take Maskow about 15 minutes to execute 30 cubes, and so that would make the 11x11 execution an hour and 15 minutes long. 4. Because of the turning of the cube it would be much easier to make an error. 5. Because the execution would be like 5 times longer, he would have to retain his memo for 5 times as long which would result in error.

Even though he can handle the memo, it would still comoletely different execution. I do think that it would be possible after a ton of attempts, but you can't make an equal comparison.

 

[tseitsei]

No, one cube definitely isn't 20 letters. Every cube has 20 pieces - that's right. But we have some funny things like more cycles, twisted pieces and parity [all three things are terrible]. With that system the average is ~11,5 words per cube, not 10. I counted it on memo of few real MBLD attempts and it's always ~11,5

I see no reason to compare it with 2-digits system only because we use 2-pieces system for cube. We don't use 3-pieces system because it will have almost 9000 words [it's almost as many as you need for 4-digits system!] and will be harder to solve with our 2-pieces solving methods. So I compare it to 3-digits system. 11,5 x 150 x 3 = 5175. I cheated a little before because I used 12, not 11,5

Yeah I knew one cube is more than 20 letters. I also said it in my post ( I just didn't know by how much...) and I am very well aware of how blindsolving works ( I have done 21/21 mbld myself)

And yes I think it makes more sense to compare to 3digit system. I was simply trying to explain why you and qqwreff had different numbers because he had obviously used 2 digit system

 

[MatsBergsten]

A tremendous effort!

125 solved cubes are totally amazing, who cares about the 25 dnfs anyway.

Just curious, do you have any idea of those 25 dnfs, where most of them memory recall errors
or execution errors? (Sheer fatigue should account for a lot of dnfs).

 

[Stefan]

qqwreff had different numbers because he had obviously used 2 digit system

No he obviously didn't. Try again? Hint: his "2945 decimal digits" is pretty exact.

 

[Berd]

No he obviously didn't. Try again? Hint: his "2945 decimal digits" is pretty exact.

Stefan layin down da law [emoji8]

 

[tseitsei]

No he obviously didn't. Try again? Hint: his "2945 decimal digits" is pretty exact.

Well close enough for me

2945 is almost 3200

 

[Stefan]

Quitter.

2945

 

[Mike Hughey]

So nice to see you try something like this. I'm looking forward to seeing your next attempt!

 

[tseitsei]

Quitter.

2945

Oh I see what he did now

That's one way to look at it yes...

 

[qqwref]

I have no idea how did you count it. IMO it's more like ~5400 digits or ~18000 binary digits. And you must also remember that "reading" words from cube is much more harder than from sheet of paper. Revisions are also terrible on cubes.

Yeah, I just mean information-theoretically. log(43252003274489856000^150)/log(10) ~= 2495. Of course in real life memorizing something like a cube is definitely not as efficient or easy as memorizing raw decimal digits.

This is also about the same amount of information as a 28x28x28 cube.

In terms of raw information memorized, multiBLD has been far ahead of big cube BLD for a long time. So this is definitely the most information ever memorized in a BLD cubing attempt.

 

[Zhanchi1]

Holy ****. I can't even blind solve a single 2x2!

 

[cashis]

Yeah, I just mean information-theoretically. log(43252003274489856000^150)/log(10) ~= 2495. Of course in real life memorizing something like a cube is definitely not as efficient or easy as memorizing raw decimal digits.

This is also about the same amount of information as a 28x28x28 cube.

In terms of raw information memorized, multiBLD has been far ahead of big cube BLD for a long time. So this is definitely the most information ever memorized in a BLD cubing attempt.

and he's disappointed, LOL