[Unofficial] Maskow: Multiblindfold 125/150

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What exactly is a "memory palace"?
I assume that due to the fact that he uses a Journey/ Roman rooms type method, the amount of rooms that he uses for 50 cubes creat his "memory palace". Once he goes past 50 cubes, his palace becomes to big to effectively house the amount of rooms he is using. I wonder how many rooms he has for 150 cubes? Amazing job Marcin! Even though you had some DNFs, 100 points is beyond belief.
 
damn. soon organizers are going to have to bring tablecloths to competitions to cover mbld attempts
 
What can I say? I'm definitely not happy about this result. Too long memorisation phase, too many DNFs. My concentration doesn't work over 1 hour. Also my memory palace doesn't work over50 cubes. I will work on these both things and improve this result because it's definitely not my last attempt on 150 cubes. It was completely crazy but I want to do it again and do it faster and better ;)

1. I don't know how you're stuffing so many images on the same locations (If you're still using 100 locations). I'm guessing this would be easier for you if you made 200 more loci. Am I missing something? or are you placing 3 piece sets(corners/edges) in each location?

2. My concentration can't even get to 1 hour yet. It's a bit of an issue for hour digits. I read somewhere that Marcell Endrey suggested taking scheduled breaks during memo to increase focus later in the memo. I have done some 2 hour digit attempts to test the idea and I take a 2 minute break every 1/2 hour. The 6 minutes of break time are not lost because it makes my memo faster since I don't burnout as much. I'd wholeheartedly suggest giving it a try.

The fact that you memorized for 8 hours straight is super human. Maskow, You're certainly the best of the best!!
 
It's starting to seem like the only thing restricting Maskow with the size of his MBLD attempts is the number of cubes he has...
 
Incredible! It blows my mind that you can remember this much information. It's the equivalent of 2945 decimal digits or 9784 binary digits.

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.

Try /75 in 2 hrs.

Is should definitely try it.

Now where was that BLD tutorial again?

There is a 3-hours videos of full BLD tutorial, including Old Pochmann corners/edges, TuRBo edges and commutators. Of course it's in Polish ;)

1. I don't know how you're stuffing so many images on the same locations (If you're still using 100 locations). I'm guessing this would be easier for you if you made 200 more loci. Am I missing something? or are you placing 3 piece sets(corners/edges) in each location?

My base were my places for 50 cubes, I don't have more. Then I improvised rest of it based on these places. I did it in the same way on 100 cubes. And it wasn't a good idea. It was a very bad idea. I wasted probably about ~4h because of it XD Before my next big old style MBLD attempt I should definitely make more places for saving a lot of time and do all memo in 4 hours ^ _ ^

Heh, remember when people were doing ~30 cubes old style and they had to add a one hour limit because of how crazy it was to scramble/judge/organize? :)

I'm happy about this time limit. Doing something like that on competition is pointless ; P
 
I have no idea how did you count it. IMO it's more like ~5400 digits or ~18000 binary digits.
I would say it is roughly 4800 or 3200 digits. Depends on how you memo. See below for explanation.

One cube is roughly 20 letters. 150 cubes is roughly 20*150 letters=3000 letters and since most people encode two letters per word = 1500words. And one cube is a little more than 20 letters actually I think since in MBLD I can't just visually remember flipped edges and parity. So let's say ~3200 letters and 1600 words.
For digits I would use a memo method that encodes 2 digits per word so 1600 words would mean ~3200 digits. Someone who is actually practising memoing digits actively would probably use a method that encodes 3 digits per word and that would mean 3*1600=4800 digits.

So that probably explains the difference in your numbers at least partially... They still don't quite match what I calculated here but closer already :)

I haven't looked that much into memoing binary yet so I won't comment on that.
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.

And this I have noticed myself also while memoing some cards lately. SO MUCH easier to read the same amount of info from a deck of cards than from cubes...
 
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Amazing! How does the amount of memo compare to say a 9bld?
9BLD would have 3 sets of x-centers 3 + centers and 6 oblique centers sets. About 20 letters per set means 20*12=240 letters for centers.

Also 3 sets of wings and 1 set of midges and one set of corners. About 24 letters for each wing set means 24*3=72. 12 letters for midges and 8 for corners.
So all together 240+72+12+8=332 letters.

That means this attempt equals, in terms of memorized info, about 10 times 9BLD :D
 
I would say it is roughly 4800 or 3200 digits. Depends on how you memo. See below for explanation.

One cube is roughly 20 letters. 150 cubes is roughly 20*150 letters=3000 letters and since most people encode two letters per word = 1500words.

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 ;)
 
Will you ever be able to do 50 cubes blindfolded officially. Is memorising it with pictures easier or is letters. I guess using pictures is more advanced but quicker.
 
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