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At the last step of solving a big cube I've been told to solve the cube "like a 3x3" I don't really understand that. Do I do OLL? PLL? F2L? What is it about big cube's last step that correlates to 3x3? All the other steps like edge pairing, OLL parity, and PLL parity (4x4 PLL parity and OLL parity) were all simple.
 
At the last step of solving a big cube I've been told to solve the cube "like a 3x3" I don't really understand that. Do I do OLL? PLL? F2L? What is it about big cube's last step that correlates to 3x3? All the other steps like edge pairing, OLL parity, and PLL parity (4x4 PLL parity and OLL parity) were all simple.
It should speak for itself, but if you don't understand, watching a big cube example solve should suffice.
 
At the last step of solving a big cube I've been told to solve the cube "like a 3x3" I don't really understand that. Do I do OLL? PLL? F2L? What is it about big cube's last step that correlates to 3x3? All the other steps like edge pairing, OLL parity, and PLL parity (4x4 PLL parity and OLL parity) were all simple.

Basically, the goal of modern big cube methods is to reduce the cube to a 3x3. If you look at a 3x3, it has 8 corners, 12 edges, and 6 centers. Big cubes have 8 corners, but obviously way more edges and centers. However, after doing all the other steps (building centers and pairing edges), you should be able to see that the giant block of 4 white center pieces on your 4x4 can exactly correspond with the white center of a 3x3. This applies for all centers (this is why “getting your color scheme right” is so important). For edges, once they have all been paired, you can stop thinking of each individual edge piece, but instead think of the two paired red-blue edges as one red-blue edge. If you do that, you’ll see that there are now 12 edges, 6 centers, and 8 corners, just like a 3x3. Then you can solve the rest of the puzzle the same way you’d solve a scrambled 3x3, with Cross/F2L/OLL/PLL, using only outer layer turns (this will prevent edges/centers from being messed up)

Tl;dr: Think of each big paired edge piece and built center piece as one edge / one center on a 3x3. Then solve the 4x4 the same way you’d solve a 3x3, with only outer layer turns
 
Basically, the goal of modern big cube methods is to reduce the cube to a 3x3. If you look at a 3x3, it has 8 corners, 12 edges, and 6 centers. Big cubes have 8 corners, but obviously way more edges and centers. However, after doing all the other steps (building centers and pairing edges), you should be able to see that the giant block of 4 white center pieces on your 4x4 can exactly correspond with the white center of a 3x3. This applies for all centers (this is why “getting your color scheme right” is so important). For edges, once they have all been paired, you can stop thinking of each individual edge piece, but instead think of the two paired red-blue edges as one red-blue edge. If you do that, you’ll see that there are now 12 edges, 6 centers, and 8 corners, just like a 3x3. Then you can solve the rest of the puzzle the same way you’d solve a scrambled 3x3, with Cross/F2L/OLL/PLL, using only outer layer turns (this will prevent edges/centers from being messed up)

Tl;dr: Think of each big paired edge piece and built center piece as one edge / one center on a 3x3. Then solve the 4x4 the same way you’d solve a 3x3, with only outer layer turns
Ohh, okay. Watching example solves really didn't suffice. I just finished my 4x4 in like a minute afterwards. Same with my 5x5. So all algorithms for OLL and PLL work as well. Thank you for explaining that. Wouldn't have figured it out otherwise.
 
What's the best 5x5 I can get on the market right now? The MGC doesn't really suit me and I'm looking for a better 5x5, any suggestions? The price does not matter.
 
What is the best way to solve these cases (for 6x6, and 7x7)? Any case where for the last 2 centers, the opposite color pieces are in the exact same place. Is there a trick to solving these cases? I attached screenshots on a 7x7 of what I'm talking about. I'm currently stuck on a 7x7 with the first picture case.
1671984589208.png
1671984631240.png


Edit: I figured out how to solve the first case, but a method/strategy that is consistent in working for this scenario no matter where the pieces are would help.
 
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He said so; he didn’t understand what 3x3 stage was and how to “solve the rest like a 3x3”
And I want to know what blockade about"solve it like a 3x3" they had. Most likely they didn't understand to only turn the outer layers but I wanna know exactly.
 
What is the best way to solve these cases (for 6x6, and 7x7)? Any case where for the last 2 centers, the opposite color pieces are in the exact same place. Is there a trick to solving these cases? I attached screenshots on a 7x7 of what I'm talking about. I'm currently stuck on a 7x7 with the first picture case.
View attachment 21517View attachment 21518

Edit: I figured out how to solve the first case, but a method/strategy that is consistent in working for this scenario no matter where the pieces are would help.
The best way is to probably use comutators. I know how they work, but I'm bad at explaining things. Maybe someone else can explain them. @abunickabhi you help?
 
Personally I think triple parity is just a construct to use in order to talk 6x6, 8x8 and the like as a bad puzzle.
This part I agree, but…

What is all this talk about triple parity on 6x6? For my last 2 edges if I get edge parity I make sure the parity edge is yellow (LL colour) and see which edges need to be oriented/ misoriented in order to get an even orientation. If I recognise I have outer edges misoriented I do the normal edge parity but if I have inner I do an inner slice move variant that directly fixes parity. In other words, I force myself out of full oll parity when I see that I have edge parity. So why isn’t this done more especially for Yau users?
If you have to go out of your way to force the one-edge-parity case during edge pairing, this is generally not worth it. The numbers just don't work out. Slice parity is much slower than edge parity or OLL parity; it's faster than having to do both, but not by much, and you lose all of that advantage if you need to do an extra slice-flip-slice.
 
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And I want to know what blockade about"solve it like a 3x3" they had. Most likely they didn't understand to only turn the outer layers but I wanna know exactly.
The problem I had is I didn't get why and how a 5x5 cube would have any similarity with a 3x3. Therefore I couldn't "solve like a 3x3." Oh btw happy new year.

So this meant I didn't get that I could just use usual OLL, PLL, and F2L to solve like I'd normally solve a 3x3. The usual stuff I'd hear is "solve like a 3x3" and no more added context.
He said so; he didn’t understand what 3x3 stage was and how to “solve the rest like a 3x3”
Yes precisely. I didn't get that the edges were aligned so that the cube is more like a 3x3 than a 5x5. Saying "solve like a 3x3" is very vague
 
The problem I had is I didn't get why and how a 5x5 cube would have any similarity with a 3x3. Therefore I couldn't "solve like a 3x3." Oh btw happy new year.

So this meant I didn't get that I could just use usual OLL, PLL, and F2L to solve like I'd normally solve a 3x3. The usual stuff I'd hear is "solve like a 3x3" and no more added context.

Yes precisely. I didn't get that the edges were aligned so that the cube is more like a 3x3 than a 5x5. Saying "solve like a 3x3" is very vague
When people say "solve like a 3x3" they mean that once you have all the centers and edges formed, use only outer moves to solve the puzzle. In this way, all the centers are preserved (like a 3x3), all the edges are preserved (like a 3x3) and all the corners are also preserved (you guessed it, like a 3x3!)

It is more difficult to recognize typical 3x3 cases since the edges are made of multiple edge pieces and are therefore longer, but overall it shouldn't be too difficult to figure out. Let me know if you have extra questions.
 
What is the best way to solve these cases (for 6x6, and 7x7)? Any case where for the last 2 centers, the opposite color pieces are in the exact same place. Is there a trick to solving these cases? I attached screenshots on a 7x7 of what I'm talking about. I'm currently stuck on a 7x7 with the first picture case.
View attachment 21517View attachment 21518

Edit: I figured out how to solve the first case, but a method/strategy that is consistent in working for this scenario no matter where the pieces are would help.
I think the best thing to do in each of these cases is (like @Abram Grimsley said) to use commutators, but there's a specific one that's quite useful in big cube center solving especially. In the image you provided, the center pieces on the blue and red sides need to switch. We can place the blue side on top and the red side in front and do a variation of the algorithm R U' L' U R' U' L U. This algorithm is commonly known as a Niklas, and although it isn't technically a commutator per sé because of the way that the U moves in the second half of the algorithm are, it is the most useful algorithm. For the first case, we must modify this algorithm so that we put the blue pieces in position (using 2 middle layers, so 2r) then do U' normally, then do an l' move and a normal U move, then reverse the 2r, do a U', then l U.

For the second case, it's the same thing but we only put one blue piece in place, then we do a U' as usual, but this time we have to turn the right-hand slice closest to the middle upwards as if it were an L' move. Then the rest of the algorithm goes much the same way.

If I was too confusing, please let me know... I hope this helps!
 
I average sub 8:00 for 6x6. I have a comp in June with a 3:30 cutoff. Does anyone have any tips for me to get remotely close to that. ( Even just sub 5:00) Edge Pairing is were I have the most trouble. Should I learn Freeslice?
Yes, learn freeslice. And make sure you're using a good cube, like the MGC or, if you can, the Aoshi WRM.
 
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