[Help Thread]Big Cube Discussion (5x5 / 6x6 / 7x7 / etc)

xyzzy

Member
(transferring to this thread)
(also I say "you" a bunch of times but I mean that generically rather than carcass specifically)
Here is a trick to avoid triple parity on 6x6. Finish all of your edges, and if you have parity, don't do it. Pick a cross color so the parity edge will end up in your last layer. If you aren't color neutral, do a Uw Dw' Flip Uw' Dw or something like that to make the edge yellow. Then, when you get to OLL, you can be sure if you have inner or outer parity.
The value of this "trick" is predicated on triple parity being the worst thing ever, and since that is a cowdung premise, this trick turns out to be not really that useful in practice. Like, I use it in maybe 10% of my solves and it's not completely useless, but its value is way overblown by beginners who think it's some super fancy trick.

So this trick lets you "avoid" "triple parity". But concretely, what's happening here?

1. What is triple parity? It's when you get parity during edge pairing (parity of the inner wings and parity of the outer wings differ), then you get OLL parity (odd parity both for inner wings and for outer wings), and finally PLL parity.

2. What is meant by "avoiding"? The simplest interpretation is just that you won't encounter the above scenario.

Okay, fine, so this trick does allow you to avoid "triple parity". So what is the alternative? To encounter the triple parity scenario, we need these conditions to line up once the centres are solved (**):
(i) Inner wings have odd parity.
(ii) Outer wings have even parity.
(iii) Edge groups and corners together have odd parity.

These events are independent (*) and each happen with probability 1/2 (**), so even without going out of your way to avoid triple parity, it only happens one-eighth of the time anyway. When this does happen, with the basic "beginner" method, you end up doing three parity algs. That sucks, right? (No, it doesn't; you just have an irrational hatred of parity algs because they took you a long time to learn. This is also besides the point.) But going back to my original question: what is the alternative?

(i) You see that there's one edge group left to fix, so you decide to leave it to the end. (No parity alg here!)
(ii) You are restricted in your choice of cross colour. Actually not that big of a deal because it's not like you could choose the best cross in one second anyway, and the restriction can help you look ahead into cross better.
(ii.5) If you're not colour neutral (or at least, quad CN), sometimes you'll get the messed up edge on an inconvenient colour. That sucks. Maybe you'd want to slice-flip-slice to transfer it to a yellow edge group instead. Not as bad as a parity alg, but slice-flip-slice is not a free operation.
(iii) After finishing F2L, you count the number of bad edges and see that you have inner parity. (We're focusing on the situation where you'd otherwise have gotten triple parity. It's impossible in that situation to have outer parity here.) Gosh, the inner parity alg stinks. So many slice moves!
(iv) Do you use OLL parity tricks like using a trigger setup to force OLL skip on the 1-flip L cases? Too bad, you can't use them in conjunction with the inner parity alg.
(v) Oh, and after you've solved inner parity, you still have to deal with PLL parity. No change there.

Congratulations, you replaced two bad parity algs with one super-bad parity alg, and one of the three parity algs you originally had to do didn't even change at all. Granted, inner parity is not as bad as having to do both edge pairing parity and OLL parity, but it's barely any better.

Oh, do you use a dumb version of the inner parity alg where you have to do U2 l r' at the end? Even worse; you just eliminated whatever little benefit this trick had.

(*) Almost, but the deviation from independence is negligible. (If you want to go into the nitty gritty details, you do have to take into account that humans do not solve twisty puzzles by rigidly following a flowchart.)
(**) Assuming you don't use the parity L2E algs. This is actually important! Read on.

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Now, if you just take everything I wrote in the above section at face value, you might come to the conclusion that at least the trick isn't actively harmful if you're using a proper inner parity alg. Which is true, but not the right perspective. The right perspective is that once you start learning the parity L2E algs (here, have one: r U2 r U2 F2 r F2 l' U2 l U2 r2), the value of this trick plummets even further. It's never worth it to force this trick when you have a parity L2E case other than the OLL parity case, and that's also the rarest L2E case. The cost-benefit analysis is a bit more involved here, so bear with me for a moment.

Oh, and in case you haven't noticed, PLL parity is actually irrelevant to the discussion here, so I'll be ignoring that henceforth. (It's kind of funny that that's the case because this trick is always marketed as "triple parity avoidance" and then it turns out one of the three parities is unavoidable.) The following four scenarios are equally likely (1/4 each).

Scenario A: Even inner parity, even outer parity.
Do edge pairing as usual. No edge parity, no OLL parity.

Scenario B: Odd inner parity, odd outer parity.
Do edge pairing as usual. No edge parity, yes OLL parity.

(These two scenarios above do not involve the trick at all.)

Scenario C: Even inner parity, odd outer parity.
Do edge pairing up to L2E. Pair up the inners too, if they're not already. (***) Depending on your edge pairing strategy for L4E, the probability of getting the OLL parity case will differ (e.g. it's almost zero with AvG-like edge pairing), but it'll be somewhere between almost-0 and 1/6. At least 5/6 of the time you'll be getting something else, which is the interesting subcase.
If you use the trick: You do slice-flip-slice here. Get outer parity when you reach OLL later.
If you don't use the trick: You do the parity L2E alg here. No parity when you reach OLL later.

Scenario D: Odd inner parity, even outer parity.
As in Scenario C, do edge pairing up to L2E, and there'll be a ≥5/6 chance of getting a not-OLL-parity L2E parity case.
If you use the trick: You do slice-flip-slice here. Get inner parity when you reach OLL later.
If you don't use the trick: You do the parity L2E alg here. Get OLL parity when you reach OLL later.

In Scenario C, you're clearly better off not using the trick. Scenario D is a bit fuzzier, but if you look at the average move counts, using the trick costs around 8 (slice-flip-slice) + 22 (inner parity) = 30 moves, while not using the trick costs around 15 (parity L2E) + 17 (OLL parity) = 32 moves. The numbers will vary somewhat depending on your exact choice of algs, but my point here is that even if you do save on move count by using the trick here, it's a tiny saving. You can't quickly distinguish between Scenario C and Scenario D, so conditioned on being in C or D, by using the trick, there's a 50% chance you're wasting moves big time (Scenario C) and a 50% chance you're saving maybe a little bit (Scenario D). The trick turns out to be a net negative here.

(***) Ha ha, would you believe it if I said it can get even more complicated? Ha… ha… (I'm not writing up any more than this today, but hint: inner parity algs other than OLL parity.)

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tl;dr Were you expecting a summary here? Hell no, I spent Time writing it so you should spend Time reading it too. triple parity avoidance trick overrated, do not mindlessly (ab)use

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One Wheel

Member
what should i average on 6x6? I use Redux btw

average under 1:10 on 4x4 (Yau)
and under 2:30 on 5x5 (Redux)
Your times will get better with practice. I average about 1:20/2:20/4:00, Yau for 4x4.

NewCuber257

Member
Hi I’m Taylor. I’m pretty new to the cubing world but I love doing them, even if I’m extremely slow at it! I just figured out how to solve the 8x8 and the 9x9 and I just bought a 10x10 and a 13x13. My question is are all big cubes solved using the same beginners algorithms. For instance can I solve the the 10x10 the same as a 8x8 and can I solve the 13x13 the same as a 9x9. Any advice would help a lot. Thank you!

xyzzy

Member
Hi I’m Taylor. I’m pretty new to the cubing world but I love doing them, even if I’m extremely slow at it! I just figured out how to solve the 8x8 and the 9x9 and I just bought a 10x10 and a 13x13. My question is are all big cubes solved using the same beginners algorithms. For instance can I solve the the 10x10 the same as a 8x8 and can I solve the 13x13 the same as a 9x9. Any advice would help a lot. Thank you!
Yep, you can use pretty much the same method. There isn't really anything new to big cubes past a 6×6×6.

(Unless you're trying to go fast, in which case various optimisation tricks only make sense on certain sizes.)

NewCuber257

Member
Yep, you can use pretty much the same method. There isn't really anything new to big cubes past a 6×6×6.

(Unless you're trying to go fast, in which case various optimisation tricks only make sense on certain sizes.)
Thank you so much for the help

SlimJimTx

Member
Can anyone help me swap these 2 middle pieces on this 8x8. It would really make a kid happy on Easter. Thanks in advance!

qwr

BenChristman1

Member
Hold red on top and white in the front, and do:

4R2 U’ 4L2 U 4R2 U’ 4L2 U2 4R2 U’ 4L2 U 4R2 U’ 4L2

4R2 = turn the 4th layer from the right 180 degrees
4L2 = turn the 4th layer from the left 180 degrees

SlimJimTx

Member
Hold red on top and white in the front, and do:

4R2 U’ 4L2 U 4R2 U’ 4L2 U2 4R2 U’ 4L2 U 4R2 U’ 4L2

4R2 = turn the 4th layer from the right 180 degrees
4L2 = turn the 4th layer from the left 180 degrees
Hi - hmmm I just tried but ended up in the same situation...

BenChristman1

Member
Try going to 5:10 in this video. He uses a 6x6, but it works with any big cube. It’s just doing 2 commutators to do 2 2-swaps, which is the algorithm that I gave you. It seems to work fine for me, but maybe I typed it wrong. Also, you’ll have to turn the layer twice, since your swapped centers are on opposite faces, but his are on adjacent faces. Hope this helps!

SlimJimTx

Member
Try going to 5:10 in this video. He uses a 6x6, but it works with any big cube. It’s just doing 2 commutators to do 2 2-swaps, which is the algorithm that I gave you. It seems to work fine for me, but maybe I typed it wrong. Also, you’ll have to turn the layer twice, since your swapped centers are on opposite faces, but his are on adjacent faces. Hope this helps!

Thanks for your help. It still seems to be just reverting back to where it started. I'll keep working at it.

Christopher Mowla

Premium Member
Hold red on top and white in the front, and do:

4R2 U’ 4L2 U 4R2 U’ 4L2 U2 4R2 U’ 4L2 U 4R2 U’ 4L2

4R2 = turn the 4th layer from the right 180 degrees
4L2 = turn the 4th layer from the left 180 degrees
This algorithm doesn't solve the case he was looking to solve unless you do a B2 setup.

But here is a much shorter algorithm to solve the exact case @SlimJimTx was looking for.
4L' 4R' 4U2 4L 4R 4U2

LukasDikic

Member
I have a really bad 6x6 (Yj Yushi), and I ordered the MGC recently, and on my bad cube my best is 3:22, but my average is around 4:00. I average 50 on 4x4 (Yau), 1:38 on 5x5 (Redux), and around 5:30 on 7x7 (Redux), although it keeps dropping everyday because of the MGC 7x7. I don't really practice any of these events more than once of twice a week so would my average on 6x6 be with some regular practice and a better cube?

One Wheel

Member
I have a really bad 6x6 (Yj Yushi), and I ordered the MGC recently, and on my bad cube my best is 3:22, but my average is around 4:00. I average 50 on 4x4 (Yau), 1:38 on 5x5 (Redux), and around 5:30 on 7x7 (Redux), although it keeps dropping everyday because of the MGC 7x7. I don't really practice any of these events more than once of twice a week so would my average on 6x6 be with some regular practice and a better cube?
Hard to say exactly, but I average about 30 seconds slower than you on 4x4, 35 seconds slower on 5x5, and 30 or 40 seconds slower on 7x7, and I am just a few seconds faster than you on my MGC 6x6. You've got a good chance of dropping pretty significantly quickly.

robdawg421

Member
Can't find 7x7 algorithm for last edge parity

Hello.. I've recently been learning big cubes and had a question about 7x7. JPerm's guides on youtube stop at 6x6.

For 4x4... jperm uses for OLL parity:

Rw U2 x Rw U2 Rw U2 Rw' U2 Lw U2 Rw' U2 Rw U2 Rw' U2 Rw'

For 5x5... he uses this for edge parity:

Rw U2 x Rw U2 Rw U2 3Rw' U2 Lw U2 Rw' U2 Rw U2 Rw' U2 Rw'

And for 6x6... he uses this for edge parity:

Rw U2 x Rw U2 Rw U2 Rw' U2 Lw U2 4Rw' U2 Rw U2 Rw' U2 Rw'

Additionally, for 6x6 OLL parity he uses this:

3Rw U2 x 3Rw U2 3Rw U2 3Rw' U2 3Lw U2 3Rw' U2 3Rw U2 3Rw' U2 3Rw'

Now... what I like about these is they are fairly easy to remember and they all are basically the same algorithm except one move might use a different big slice... or all of the moves use a particular big slice.

Is there an equivalent for 7x7? Most algorithms I am able to see in youtube vids (jperm doesnt have a 7x7 vid) or in guides online are using completely different algorithms that use B, L or Fs... and at least after learning the above one seems like heading back to the drawing board.

Does anyone know off-hand how to adjust the above algorithms to work to solve the final edge parity on 7x7? I am assuming maybe I have to repeat the algorithm twice sometimes but not really sure. Originally was hoping it would be a single move change like 5x5 and 6x6 seemed to do.

Edit: fyi I am solid with 4x4 and 5x5.. the others are coming in the mail.

Just found alg.cubing.net... answers all. So glad one base algo can be used for edges in bigger cubes.

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abunickabhi

Member
Yes just one algorithm works for any NxN, you just have to adjust/change the layers of the alg, according to the number of layers on the cube.

tsmosher

Member
jperm does have a couple of videos on 7x7 BTW. Big Cube Fundamentals is one of them, I think.

In learning this alg, there's 2 tricks I use to remember how to scale it up for larger cubes:

I have it written down like below. (Note the M move in the middle.)

[Rw U2] (x) [Rw U2]2 (M) [Rw' U2] [Lw U2] [Rw' U2 Rw] U2 [Rw' U2 Rw] (Rw2)

That can still get confusing though, so the way that I intuitively think about it:

Rw U2 x Rw U2 Rw U2 Rw' U2 Lw U2 Rw' U2 Rw U2 Rw' U2 Rw

All of the Rw moves except the bolded/underlined one will use the same number of layers.
(All of the layers up to but not including the flipped edges.)
The Lw move will use this same number of layers as well (just mirrored onto the left side).

If you've been turning 2 layers for your Rw moves on a 7x7 cube, the Rw' move will be (7 - 2) = 5 layers coming backward.
If you've been turning 3 layers for your Rw moves on a 9x9 cube, the Rw' move will be (9 - 3) = 6 layers coming backward.
The Rw' move basically turns: (# total layers - # layers you've been turning thus far)

Hope this helps.

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xyzzy

Member
That can still get confusing though, so the way that I intuitively think about it:

Rw U2 x Rw U2 Rw U2 Rw' U2 Lw U2 Rw' U2 Rw U2 Rw' U2 Rw

All of the Rw moves except the bolded/underlined one will use the same number of layers.
(All of the layers up to but not including the flipped edges.)
The Lw move will use this same number of layers as well (just mirrored onto the left side).
I feel like the least confusing rendition of this alg (to me! ymmv!) is
Rw U2 Rw F2 Rw F2 Lw' U2 Lw U2 Rw' U2 Rw U2 Rw' U2 Rw'
=
Rw U2 Rw (x U2 Rw U2 x') Lw' U2 Lw U2 Rw' U2 Rw U2 Rw' U2 Rw'

This "use rotations to execute F2s easier" trick also features in literally the only 555 L2E parity alg everyone should learn:
Rw U2 Rw U2 F2 Rw F2 Lw' U2 Lw U2 Rw2
=
Rw U2 Rw U2 (x U2 Rw U2 x') Lw' U2 Lw U2 Rw2
(and of course it does, look how similar the algs are)

This specific rendition of Lucas parity is also easily modified to a variant for solving only inner parity for 666 and 777 (and even bigger cubes):
3R U2 3R x U2 3R U2 (x' 3L') U2 3L U2 3R' U2 3R U2 3R' U2 3R'

PikachuPlayz_MC

Member
Just found alg.cubing.net... answers all. So glad one base algo can be used for edges in bigger cubes.
I was going to say that, it is the same alg for all the NxN big cube, just change what layer you ned to do it.
Yes just one algorithm works for any NxN, you just have to adjust/change the layers of the alg, according to the number of layers on the cube.

robdawg421

Member
Yeah.. my main issue was i have 4x4 and 5x5.. but am still waiting on a 6x6 and 7x7 to arrive so can't experiment (until I found the algo site).

I suspected could move diff big slices for larger cubes but didn't see a vid explaining how to adjust pre-existing algorithms. Also threw me off that for 6x6, JPerm changed which move he changed the layer on (didn't realize could do in different spots).

All of this is prob from me being lazy and just following algorithms instead of thinking how they are working as I use them. And perhaps that the 4x4 algo page on this site is a little overwhelming (but in retrospect is great).

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