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This puzzle was a curious one and I required a helpful tutorial to solve it. Just some quick algorithms to help me solve this cool cube.

Step 1: Return to a cube

First, you’ll get offset centers lined up with an edge piece sloped downward above it. Here, you move the center piece 45° to the right, then perform a R, U’, R’ before returning the center. You then keep repeating this process. If you’re stuck with a flipped edge, just bring it down to make it a center and repeat.

Step 2: Restore centers

Just prep centers. This will swap front and right centers. When you have a pair to exchange, move your center 45° to the right, then R2 and return. Super easy.

Step 3: Solve F2L

Solve the first two layers are you would on a normal 3×3.

Step 4: Last-Layer Parities

If you get the classic 4×4 parity, you can bring down your front edge 45° (an M slice). Then F2, an E (turn left as you look at it), F2, and return the E and return M’.

This will mess up three layer 2 edges which can be fixed easily apparently. Place the proper piece on bottom/back and whip out a: B2 M B2 M. This should fix it.

If you get a parity where your final two edges are swapped, place the flipped edge in front. A M’ here is a 45° upward.

M’, R, U, R’, U’
M’, U, R, U’, R’

I’m finding a problem where this doesn’t necessarily solve it all. But for now, it’s a good start. Perhaps I’ll edit this again in the future.

This puzzle looks cool, turns wonderfully and appears simple enough. While it’s effectively just a 3×3 shape mod, I simply cannot visualize it as such. This tutorial helped me whittle the whole thing down to something digestible.

Step 1: Solve two-colored edges

These act like corners, so just find the proper plane and solve all three two-colored edges. This is an easy step and no image is necessary.

Step 2: Flatten Petals

Now you will be faced with one of three situations. If the two pieces are beside one another like shown here, place them on the right plane. Move this right side down, then move the bottom/left layer (here, the other green petal) down, then back up and up (R’, L, R, L’)

The second scenario is when the two incorrectly-placed pieces are on adjacent faces. Hold the tip of the triangle at you and the upright petals on the top-left and top-right. With the one stickered side facing you positioned on the left, turn this layer down 90°, then the right layer a full 180° and return the first face. I’ve included a video here for easier reference.

A third scenario exists where they’re on adjacent layers but on the same pivot. In this case, move one side away so it mirrors scenario #2. I believe this will be an easy fix.

Step 3: Solve all Petals

This step will involve either a two or a three-cycle. For the latter, hold pyramid tip at you, this swaps far left, top center and far right. Shown here, it’ll exchange the left red piece, the top yellow and the right blue. Do 180° flips between the two layers (DDUU – L2, R2, L2, R2). This is kind of intuitive and while it may involve commutators, for your basic purposes, it’s not difficult.

Now, sometimes you will encounter three that you can’t easily swap. You have two options here. A three-cycle may work (for instance, swapping two blues at once). But if you’re lost, which I often get, you may benefit from some two-cycles instead. Here, bring a face down with a 90° turn. This will look like a person wearing glasses looking at you (see pic). Now, your three-cycle maneuver (clockwise or CCW based on your needs) will cycle these just as you’d hope. This is a really-quick fix as, like before, you can swap the same-colored pieces to maneuver what you want without much thinking.

Step 4: Solve Centers/Corners

Any inverted (jutting out) centers here signify a hidden piece under a pyramid tip. Find it! OK, now leave it hidden in the tip of the pyramid. You are now going to swap this hidden piece with one that is jutting out, making sure the inverted piece is on top, also shown here. You’ll reveal it w/ a 90° right turn upwards (like shown in this image), then the usual algorithm (R, L’, R, L) but you’ll do this three times. This will also swap the two lower-layer centers as well, FYI. Do this as many times as necessary. You may get lucky and solve the puzzle this way. If not, move on to step #5.

Step 5: Swap Centers

Now that everything is flat, we can swap centers. You’re going to use the same algorithm as in step 4, but you don’t want to kick out those inverted pieces. First, find two centers you want to swap. (If you have more, just choose two and then repeat this step). Place these in front of you on the bottom as shown here. In order to prevent bumping out those inverted pieces, turn the top layer 90°. This way, those inner pieces will harmlessly swap. Do the same algorithm as before (again, three times) and your two bottom centers will be exchanged.

This little puzzle isn’t altogether too difficult, but it’s a good-enough challenge. As always, I need a little help to push myself through these and I will inevitably forget it all. For the sake of posterity, here are the instructions I’ve worked with, alongside some help from this tutorial.

There are only really four major steps to solving this puzzle: two on the Pyraminx portion and two on the inner circles. Step 1 is just matching the tips. This is easy. Step 2 gives me trouble sometimes because, unlike a Pyraminx, you can’t just rotate the tips when need be. Here’s what I do:

Get one face. Once done, you will have either all edges solved, 2 wrong (flipped) or 3 wrong. Usually it’s the latter.

Move the 3rd side to the bottom (requiring two turns) / Then do a standard d/d/u/u / Move the 3rd side back.

If you did it right, all will be properly positioned. However, many times you’ll have two edges flipped. Holding them on the left and right, do L/R’/L’/R – then U’/R/U/R’. This is the normal Pyraminx alg.

1. Situate tri-color tips so they’re all aligned.
2. Solve the 6 two-sided edges to match those tips.
3. Solve small inner-circle triangles.
4. Solve large inner-circle triangles.

The tutorial shows how to swap those large triangles around. It’s not super intuitive for me and this is the hardest step for me. Here’s the timestamp in the video where he discusses this, but he places the swapped large triangles on the top of the front layer, and on the left of the top. You’ll perform this using the right layer of the side facing you. When you do it, this must bring the large triangle you want to swap up with it. If not, something is wrong and it won’t work. In this photo, the triangle is positioned properly.

Up, Circle Right
Down, Circle Left
Up, Circle Right
Down, Circle Right
Up, Circle Left, Down

All the circle rotations are done on the top layer, FYI. Good luck!!!

I grabbed this little guy while in New York and, it being a Skewb variation, I quickly ran into troubles. Not quire sure what’s up with these friggin’ things, but my brain breaks a little bit, particularly this dumb dodecahedron. It didn’t help that after getting a few steps in, I kept getting confused by orientation and the tutorials out there leave a bit to be desired. Not knocking their content – but they’re not too organized. Anyhow, here’s my solution, which will likely require some tweaking before I forget how it’s all done. That’s kinda silly though, as it only really requires the same one algorithm all Skewbs need. R’, L, R, L’. You can reverse it (L, R’, L’, R) for step 2.

Step 1: Solve an X
This isn’t so bad, though sometimes moving an item out of the way takes a second. This should really be intuitive, but worst case, you might have the piece in the correct spot but not oriented.

If so, move it up to the opposite side. If moved to the right, rotate counter-clockwise, if on the left, CW. Then bring it down and fix the initial turn. Note, if the color you want is facing up when starting this, you’ll have to do this step twice. You can situate all four without breaking one another. This is the easiest step and even if you mess it up, it’s easy enough to fix before moving to step 2.

Step 2: Position remaining centers
This will swap the top and front centers as well as the left and the right. You’re going to use the X you’ve just solved as a starting point for the bottom layer: hold that with your thumb on the bottom. As these centers can have a vertical or horizontal black line (between stickers), it’s easy to get the orientation messed up. However, just keep that thumb on the bottom X as a starting point.

Now, if you’re smart enough, you can plan this out. I, on the other hand, just keep bringing the top piece down to its correct spot (and messing up the rest in the process) until everything is right. Don’t worry about orientation – just getting them in the right spot. You’ll know which algorithm to use based on its upper-level orientation. If the piece you need is on the left face, start with a R’ to bring it to its position. If on the right, start with L. This takes a few tries, but it’s easy.

Step 3: Orient top-layer X
Still using your bottom facing down, you now have 4 corners correctly positioned and oriented and now your goal is the remaining 4. This can be a pain. Figure out the top-layer colors (here they’re pink and green) and you want there to be two on one face. In this photo these stickers are close, but it could be on the other side where they are farther apart. (Close = they’re adjacent to a horizontal face, far is if they’re adjacent to a vertical face) These can be any combination of those colors too. G,G/P,P or how it is here, one of each. You’ll put these on the left side and rotate the puzzle clockwise (if looking from the right) so that the bottom piece (thumb still on it!) is now the crossroads for this algorigm.
Doubled algorithm this time: R’, L, R, L’ (x2).

If you have no doubles, which is likely, find one sticker you want. If it’s on the left, start with the R’ version of the algorithm and vice versa. I believe this should consistently give you a usable pair.

Step 4: Orient remaining centers
The very same algorithm will be used to rotate centers. This will flip four centers: U, F, L and R. If you only have two, you’ll be doing this algorithm twice – by fixing one and breaking three others (3+1=4).

Get the four centers U, F, L and R as mentioned and rotate up slightly. You will be performing this algorithm on a properly-oriented side as shown here. Make sure the top and front centers are part of the algorithm (sometimes I do this and my F layer is on the bottom – ensure that it’s part of your algorithm!) Do the same algorithm 6x. Every so often you will have four mis-oriented centers in a row and not a plus pattern. Despite having four, your goal, you still have to perform this step repeatedly. Use your intuition to figure out which to fix so to prep yourself for a proper final step.

Good luck!

When I first moved to Sweden, I purchased the new flat-edged Master Pyraminx. The puzzle looks great and I solved it a number of times. Then, I forgot how.

It’s strange; I can still solve the rest of my puzzles, or at least I think I can. In recent months, I’ve returned to the puzzle each time frustrated by my lack of memory and the unclear tutorials online. Today I checked again and hobbled together a solution. As always, for my own purposes, here are the steps I used to finally remember this solve!

Step 1: Solve One Face

Well, of course, get tips matching their adjacent sides first and get all three corners properly aligned. Then, pick one face to solve. This requires some intuitive positioning but it shouldn’t prove too difficult. Get those three center edges and then fill in the remaining six middle-edge pieces.

Step 2: Flip Middle Edges

Sometimes you’ll have all these middle pieces placed correctly. If not, however, two will be placed correctly, but oriented in reverse. In this case, move the correct piece to the back. Then you perform this algorithm, using intuition for replacing the bottom layer.

LD, RD, LU, RU
U’, L’, U, L

Step 3: Finish Second Layer

Here we bring the red piece down, not vice versa. Again we break the bottom layer, but that’s OK, it’s easy to fix! Of course, you can figure out the opposite if mirrored, so use intuition when figuring out what goes where here.

l, R’, l’, R’
Then get the U out of the way (u’)
L’, u, L, u’

Step 4: Centers

There are three possibilities here. Centers are all solved, none are solved or three are out of place. You could solve centers earlier or at the end, but if you have three centers misplaced, the algorithm will make you repeat the final step, so here’s a good time to do this algorithm.

To solve four centers, place opposite centers on top and bottom. It should be an easy one here:

LD, RD, LU, RU (x3).

As I mentioned, when three centers are out, it messes with things. This is sometimes referred to as parity. For this case, place the one properly-placed center on the left side and do the following:

R, U, R’, U (x2).

If this doesn’t swap your centers correctly, do it again. Done!

Step 5: Last-Layer Edges

All that should should remain are last-layer edges. Either these are solved or they need to be permuted. These could go clockwise or counter-clockwise. The algorithm here works opposite of that rotation. But it’s easy enough and can be done twice to accomplish the same result.

R, U, R’, U, R, U

Change that to U’ for a clockwise last-layer spin.

Another quick algorithm update. I saw I had left my awesome MoYu AoSu 4×4 Windmill Cube on the shelf unsolved. That last-layer edge parity (either opposite or adjacent) can’t be fixed by the normal algorithm, as it messes up your centers. “Oh, what was the solution again?” I pondered…To the YouTube!

Ah, that’s simple! OK, so first, rotate the U layer either way and then drop down the white center. Move back the top layer so your parity is properly located for your algorithm and the white/yellow centers are front and back. Do the awesome LL parity algorithm and then just reset the white and yellow faces. Since those two don’t have any orientation, it’s OK if you flip them around.

Quite simple really…until I forget next time, of course!

This ought to be unnecessary. The Gear Cube can be solved with just one algorithm and this has two. However, they’re so simple that I forget them all the time. It’s dumb. And it’s easy to fix! Just make a quick blog entry for it to go along all those other algorithm entries. OK, let’s get to it: First get the corners, which is really quite simple. Then worry about those edges. This algorithm swaps the top-front center and the top-back center. I don’t usually forget that, but the R2, U, R2, U sometimes eludes me. That is enough to solve the Gear Cube, but not the Gear Ball which needs something to move around those pesky inner edges. Here you can move around eight at a time. If all are awry, just do this algorithm twice. If not, you should have one stripe of correctly-placed inner edges. Put that in your M layer, going from left to right and do a R, U, R, U, R, U algorithm. Then of course, just rotating those final edge pieces is a cinch. So, there you go! Easy to remember!

I love me those odd-numbered cubes. Those even-numbered ones are a PITA sometimes and their parity is obnoxious. However, a nice 5×5 is a relaxing, fun jaunt. And relaxing is the word to use…as it’s not difficult and there’s only this parity to know. It’s an easy algorithm but I’d forgotten it along the way. Muscle memory only goes so far when you haven’t used it in a year plus. Here’s a quick-to-look-up algorithm. Placing the mismatched singletons on the right just do the following algorithm for an easy fix!

(Ll)’ / U² / (Ll)’ / U²
F² / (Ll)’ / F² / (Rr)
U² / (Rr)’ / U² / (Ll)²

I remember buying my Skewb when I was just getting into cubing and realized…it’s not that cool. There’s really no true challenge and therefore it sat on my shelf, rarely touched. In the years since, I’ve purchased other octahedral puzzles: the skewb diamond, the rainbow cube, a face-turning octahedron, the flower rex cube. And it turns out…nah, they’re not that cool either.

Among my collection, those are collectively my least-favorite bunch of cubes.

However, I should at least know how to solve them, and I only recently realized I never actually solved my Skewb Diamond. So, I went out to figure it out and I was beyond frustrated. Tutorials weren’t clear enough (I plan on making my own) and it just annoyed me. But now, I think I have it down, and so as usual I’m tossing up some algorithms for future reference as I will inevitably put these away and forget about them…again!

Let’s get to it!

Skewb: (R’ / L / R / L’)(Alg will turn front center piece)

1. Solve White Face
2. Match Centers
3. Top-Layer Corners

The second step will swap the Front and Top centers as well as Left and Right centers.

The last maneuver requires the algorithm to be performed twice.

4 Incorrect

• Place yellow headlights on left (facing left)
• Place yellow on front right facing front
• Place yellow on back right facing back

Bar – Algorithm 2x

• Yellow in Front Right – Facing Right
• Yellow in Back Left – Facing Back

Skewb Diamond

Step One
The first step is to solve one side, not including the single-sticker center.

Then find one other correctly-placed corner to make a second complete side.

Step Two
Now, with a four-color piece facing you, place that new, correctly-placed piece in the back left and rotate the puzzle forward. The correct piece is now in the front-left position and the original face is on the left in your left palm. Perform this algorithm:

Solve All Corners
R / U’ / R’ / U / F

Step Three
You will face one of two situations. You’ll have to swap a three-cycle of faces or you’ll have four.

Four-Cycle
To swap four, place a four-sticker piece facing you. This algorithm swaps top & below 4-sticker piece as well as swapping top right back & top left back. The algorithm is:

Four-Cycle
R / U / R’ / U’ (x3)

Three Cycle
Find the base side (around which the three cycle goes) and put that in left palm.

Colors of the three-cycle around this face will be on bottom, back-left and top-left.

Using the opposite side of base side (same color) as your front face, the single-sticker triangle looking right at you with a flat top. Remember, one ill-placed center will be on the bottom.

Three-Cycle
Clockwise: U / R / U / R’ (x2)
Counter-CW: R / U’ / R’ / U’ (x2)