OK, I guess it's that time that a thread get started for all those who are ashamed to admit they are stuck with the professor pyraminx.

I used intuition to get as far as having the entire puzzle solved, except for the four "middle" prices (i.e. the very middle of each face, and the three pieces that touch it - or the Triforce if you will...)

Was it a bad idea to wait and solve these pieces last? Is there a better way? I found an algorithm that is a three cycle from the front face to the bottom face, but it also cycles three pieces around the L, B, and R faces too - making it very difficult to cycle everything around to where it needs to go. (basically the alg is taking the slices that make an "X" including the middle piece, and doing a [R L' R' L]x2

Any suggestions? I thought about trying to do these pieces first, but then solving for the "edges" and "centers" becomes much less intuitive.

Thanks!

It is normal to leave these pieces for the end. Solvingswise they are identical to the edge pieces of FTO or Master Skewb.

My method is showed in the following picture.

If 4 are broken than two pairs need to be swapped with each other. I'm not sure what you mean by R and L but assume that they are face moves, then the slice just below the face would be r and l. Do [r l r' l']x3 for a total of 12 moves.

Here it is in Gelatinbrain's notation [D&2, A&2, D'&2, A'&2]x3 which you can paste into the applet (5.1.10) to see what it does.

Awesome - this is what I was looking for. Thanks!

Or, in Gelatinbrain notation, it would be this:

/*000000*/ADC&2,
/*000001*/A'&2,
/*000002*/D&2,
/*000003*/A&2,
/*000004*/D'&2,
/*000005*/A'&2,
/*000006*/D&2,
/*000007*/A&2,
/*000008*/D'&2,
/*000009*/ADC'&2,
/*000010*/B',
/*000011*/A'&2,
/*000012*/D&2,
/*000013*/A&2,
/*000014*/D'&2,
/*000015*/A'&2,
/*000016*/D&2,
/*000017*/A&2,
/*000018*/D'&2,
/*000019*/B,

Here is a (4,1) commutator that does a 3-cycle of the triangles:
(b, d‘, b‘, d ) rbd (d‘, b, d, b‘, rbd‘)
all turns are inner layer turns.
In Gelatinbrain Notation:
C&2,D'&2,C'&2,D&2,BCD&2,D'&2,C&2,D&2,C'&2,BCD'&2,

See the explanation of my notation below!


In Gelatinbrain notation: C&2,D'&2,C'&2,D&2, BCD&2, D'&2,C&2,D&2,C'&2, BCD'&2,

Here are the two notations:

I use two types of turns
- face turns
- vertex turns
In addition to that, I use inner layer turns, under a face and under a corner.
I use capital letters for faces, (L=Left, R=Right, B=Back, D=Down), three letters for a corner using the three face letters the corner sits in (a corner is the entity of 4 pieces sitting under the trivial tips), small letters for the second layer (e.g. rbd is the layer below corner RBD, l is the layer below face L).

Here is a picture describing the moves by example:


The above mentioned 20 move algorithm repeated:

My notation: lbd, (l‘, d, l, d’)*2, lbd‘, R‘, (l‘, d, l, d‘) *2, R
Gelatinbrain notation:
ADC&2, [A'&2,D&2,A&2, D'&2]x2, ADC'&2, B', [A'&2,D&2,A&2,D'&2]x2, B,

EDIT: I have tried to improve the description of my notation
EDIT2: I have corrected a typo in the picture showing the moves by example.

I thought the easiest solution was to reduce it to a Jing's Pyraminx: Centers, Edges and then The Jing Pyraminx.
I guess I'm not the only one who is solving it like this. It's kinda similar to a 5x5x5 compared to the 3x3x3.

That's how I did my one and only solve so far.

Paired the edges, fixed the centres, solved the centre edge pieces like a normal pyraminx, solved the result as a jings pyraminx.

It seems to me like it should all be intuitive if you use reduction or an "inside-out" method. My only question is as follows: you have the "triforce" and the 2 edge pieces that directly touch the 2 outer sides of any "triforce" piece, and each of these edge pieces touches 2 "triforce" pieces. Once those are all solved, you can solve the middle edge pieces by slotting and the outer edge pieces like a mosaic cube or platypus. The question is what should I solve first: the "triforce" pieces or the aforementioned edge pieces? It seems like solving one will affect the other because of the nature of the slice moves needed to place the pieces...any algorithms needed for this? If so, it is funny because normally what you need is algorithms at the end and not the beginning. I just need algorithms for moving those edge pieces without disturbing the "triforce" pieces...

Today I was able to play with my neighbor's (MonkeyZ) Professor Pyraminx (I hadn't ordered one myself- short on money) and was able to solve it the first time. I found it really easy and will explain my method a bit with the help of the following diagrams.
first solve the centers. should be easy
then, pair the edges. (think trignis)
then, solve like a hm-pyraminx
then, solve pyraminx edges (easy)
then, solve the little edges (easy).

If you'll notice the other diagram piece one corresponds to tips on the trignis (albeit with less permutation) and piece two corresponds to edges.

Hmmm...I came across this case:



Is this kind of a parity? Because I don't recall getting something like this on a Jing's Pyraminx. Any idea how to solve this?

If I do the usual center-switch (on any of the faces), I'm always getting a solved face and the other 3 with the messed "centers".

Don't worry, that's easy to solve. Just turn the tip (and by tip I mean the tip and the layer under it) and then resolve as a Jing's Pyraminx.

Lol....right! The only "trouble" is that I had to mess up the edges a bit, by turning he "tip", which is not possible on a normal Jing's Pyraminx. But it was easy to solve. Thanks.

I have had on several solves situations like this:



In the pictures, I have solved it close to the very end, to show better what it looks like. (The invisible faces are completely solved, all blue and yellow).

Seemingly, two middle edges need, to be swapped. Naturally, this is not possible and you have to swap all the outer edges pairs. So, it is not a parity situation, but it looks like one, at the first glance

My outline:
1. centre triangles
2. inner edges (those next to the middle edge piece)
3. middle edges, as on a Pyraminx ( and here I have had the mentioned situations)
4. outer edges (all five pieces building a Jing's Pyraminx edge are now "paired" = grouped)
5. solve it as Jing's Pyraminx

That's the same why I go around it, although I solve the Jing's Pyraminx before the outer and middle tips. I can't see any efficient way of edge pairing, due to the geometry and the seeming parity.

Do you know on the Trignis, where two of the edge-pairs are swapped? Well, this solves exactly in the same way, using the Pyraminx 2edge-swap and some 'depth-three' turns. I'll draw up some pictures if you want.

How on earth do you fix this case? I'm so close to solving this.

Hey Kevin, the trick to fixing that is to pretend those flipped groups are the edges of a standard Pyraminx. Flip them over and that does a 2-2 swap in the centers which you can fix using 12-move sequence I provided above.

Here is the fix in Gelatinbrain's notation ready to be pasted into the applet:
[D, B, A', B, C', D', C', D, A, C', B', A, D, B', C', D] (16 moves).

Those letters are face moves, not vertex moves. Here it is in a different lettering where R = red face, B = blue face, etc:
[B, R, G', R, Y', B', Y', B, G, Y', R', G, B, R', Y', B]

I don't think they're just flipped, but also swapped. If they are just flipped though, I've been wasting a lot of time in PMs!

If it is this case



it should be easy, if you know, how to fix it on a Jing's Pyraminx.
8 moves to flip the edges and then 12 moves to fix the centres.

If you don't know Jing's Pyraminx, congratulations, that you have come so far!!!

EDIT: Mine had been just flipped as on a normal Jing's Pyraminx

The only difference being is that he has the 'parity' like situation too. I think the easiest way to describe how to solve this parity problem is to image the way you solve parity on a Crazy 4x4x4 I. I'll be making a tutorial next weekend, so hopefully that will help some people.

I'm not sure what you're talking about. Those edge groups are flipped, not swapped.

There are no parities on this puzzle because every piece stays in an even permutation.

They are swaped, I know how to flip them but not swap them.

The middle edges can't be swapped. Please show the yellow side. Are the edge groups not formed? That is, are the middle edges in place but flipped and the rest of the edge pieces are swapped?

EDIT: If they are "swapped" then the edge-groups must not be formed. Your puzzle must look like this:It took me 78 moves to put the puzzle into this position. You're going to have to do a lot of individual piece cycling. You should always fully form the edge groups before solving further.

To be honest, the photo Kevin has given us isn't very useful. Both flipping the edges and the parity are relatively easy to solve, but can we see the other sides please?

Finally, I have solved it! Thank you to all who helped me. I will no longer annoy you. Thx Again.

I solved this several times. I think it's a great puzzle! I love it. After several solves I came to the conclusion that this is acutally very similar to a 5x5 cube. I use the same edge pairing algorithms, of course with a bit of analogy, but it's not difficult.

From my point of view, this could be very well even a an official WCA event. (of course, it's not feasible, as it's just a collectible)

Could someone explain what Luke was trying to show here? This is the last step that I am stumped on without any help so far

Thanks,

Chris Hemerich

Edit: Got it! With my own method but could someone still point out to me the most efficient way of doing this? Mine was just awful.

Please explain me for this case.
I still cannot solve it.
For this case, it's only 3 centers swap.
But Jing Pyraminx case is 4 centers swap.
Thanks.

1. Move the middle two layers so that the centres are correct relative to the corners.
2. Moving like a Jing pyraminx (i.e. 4 layers at a time) do a 3-cycle of the edge blocks (that consist of 5 pieces each)
3. Do a 3-cycle on the three remaining incorrect wing edge pieces.

Many thanks Mr. Jaap.
Now, I can solve it.
Your website is the one of my favorite pages.

Thanks again.

If anybody's interested, I've made a video showing how to fix parity for the edges (since I don't have a Prof. Pyraminx I used a trignis, but as stated before it's the same pieces)
It was requested by a member, but I felt to post it here so others can see as well.
here it is!

You forgot to mention one important thing: on your picture "D" means rotating ALL levels under the top to the right, not just ONE level under the top.

It is a little bit confusing, but after understanding this I could fix the parity.

In my case I had situation when 2 edges were swapped. Not flipped but swapped. And I made them not-swapped only by breaking center triangles. So, the task was later to fix those center triangles, and this combination (on the picture of kwsjack054) helped.

Hi,
Sorry for being a bit off topic but, my `help` is that mine is a bit loose and when I turn it faster, sometimes a piece falls out of the edge. I can't find where to adjust it. Can I adjust (tighten) it? It only requires a tweak I think. I didn't want to start a new thread for this small thing.
Cheers,
Burgo.

I've had no problems or complaints with the tension on either my Professor Pyraminx or the Vulcano puzzle. I'm not sure of any way to remove the caps. You can try the "Pull and Twist" method to tighten or loosen the screws. Basically, twist the corner cap sideways to grip it and pull with firm, steady pressure while rotating the cap. This will provide the necessary torsion forces on the screw to allow it to rotate.

While the thread is open again, I need some help with the last touches of the solving process. The twin edge pieces can be solved relatively easily at the beginning. As well, so can the centers, but I can't seem to initially solve the centers without messing them up while solving the twin edges. The rest of the puzzle (positioning of the outer edge pieces, corners, etc is relatively intuitive, so when I finish solving everything else, I am left with the centers all messed up. I use the techniques that I have learned from the Master Pyraminx to solve the face centers, but then I get stuck with the entire puzzle solved except for the "Triforce" pieces (classic Zelda reference as so brilliantly coined by Jebeck). It's been sitting on my desk in this nearly solved state, taunting me for weeks now:The "twin edges" behind the center edge piece are not so equivalent to the "twin edges" on the Vulcano (as Monopoly proposed earlier), as the vulcano twin edge pieces are lined up on either side of the edge and equidistant to the corners, whereas the Professor twin edges are directly inline with the edge and equidistant to the adjacent face centers. The Vulcano is actually like a six-layer pyraminx with edges removed, or a "babyface" Pyraminx with extended rotatable faces. I have solved the Vulcano several times, including the dreaded edge parities, but have not yet come to grips with a solution for the Professor's remaining "triforce" pieces that doesn't scramble the rest of the puzzle.

Kwsjack054 and Konrad, I am having difficulty interpreting your notations, algorithms, and diagrams. Though I generally solve the Pyraminx as a Corner-Turning puzzle, I realize it can just as easily be solved as a face-turning puzzle as well. I even view the Vulcano as having two corner-turning layers and one face-turning layer, but I feel like the Professor Pyraminx needs to be either one or the other. Slice turns are fine too btw

Radu, I have encountered the three centers parity several times on my Master Pyraminx. Simply rotate the big tip to match the centers, then use a normal Pyraminx algorithm to rotate the center corner. Now the Jing's centers will either be correct, or an easily fixable double swaped position. Use your normal edge swapping technique to commute the final edge wings about the corner piece and you have it solved.

I have tried to improve the description of my notation in my earlier post.
I use two types of turns
- face turns
- vertex turns
In addition to that, I use inner layer turns, under a face and under a corner.
I use capital letters for faces, (L=Left, R=Right, B=Back, D=Down), three letters for a corner using the three face letters the corner sits in (a corner is the entity of 4 pieces sitting under the trivial tips), small letters for the second layer (e.g. rbd is the layer below corner RBD, l is the layer below face L).

Here is a picture describing the moves by example:

Here is a picture describing the (4,1) commutator in detail. I have produced the pictures in Gelatinbrain. Two pictures in a row describe the complete Professor Pyraminx, left part Right and Left faces - to the right Back and Down faces. Each pair descibes the four faces after the corresponding move that is written in between the two halves of a pair. The 10 inner layer moves are (b, d,' b,' d) rbd (d', b, d, b') rbd' - b the inner layer next to B, d inner layer next to D, rbd inner layer under corner RBD
Gelatinbrain notation is: C&2,D'&2,C'&2,D&2,BCD&2,D'&2,C&2,D&2,C'&2,BCD'&2,

The result is the following 3-cycle:
I combination of face and vertex turns seems highly logical, in this case. I assume that your problem with mixing the solved centres while pairing the edges, can be avoided if you think in face moves.
With the algorithm above it should be easy to repair the last few centres.

EDIT: Typo corrected

Does anyone have a good algorithm for the double edge swap? I solve the edges first and I don't particularly care if the algorithm messes everything else up. So far I've managed to flub my way past this problem a few times, but I haven't come up with a consistent way of fixing this.

Konrad, thanks. It seems I can follow your notations now. I'll give it a go when I get back to the dorm. It seems this will be the only algorithm I'll need since I can solve the rest of the pieces by myself.

I want to draw you attention to this discussion in the Gelatibrain thread (5.1.10 is the Professor Pyraminx without trivial tips):
Coudos to Elwyn and Brandon (he is number two in the Gelatinbrain record list 5.1.10 "Fewest Moves Ranking") for setting such high standards

Who needs Gelatin Brian when you can hold the real puzzle in your hands

Gelatinbrain's puzzles feel just as real to me as physically holding the puzzle. That combined with macros, undo, the ability to instantly reset the puzzle, compete against others, etc. make Gelatinbrain's puzzles a far superior solving experience for me.

I think once you get over the hurdle of using a mouse instead of your hands and using visual memory rather than muscle memory Gelatinbrain's program is great and acquires a natural feel.

It is Elwyn who always pushes the envelope. His solve order for reduction is far superior to mine. I just used his order and solved in 74 moves, first try. I know I could cut out at least one 8 move routine in the HM pyramid phase. With practice and luck I think 60 moves is possible.

I solve it using a reduction variation: centers, pair middle and inner edges, solve like Jing's pyraminx with the added freedom of being able to twist corners in place, then solve outer edges which are easy. Then I realised the texture of the stickers means you can solve center orientation too, so I decided to solve them without disturbing the rest of the solved puzzle. Took a little bit of thinking, but then spotted an 8-move commutator (with several possible variations) which cycles 3 triforce pieces (glad I'm not the only one who thought of Zelda) and rotates two centers. Super-solving the centers like this is fun for an added challenge, I recommend it to anyone who has solved it normally. Btw, my commutator in Gelatinbrain notation is:

/*000000*/C&2,
/*000001*/A,
/*000002*/C'&2,
/*000003*/A'&2,
/*000004*/C&2,
/*000005*/A',
/*000006*/C'&2,
/*000007*/A&2,

When trying to make edge pairs, I've gotten a parity where two of the center edge pieces are swapped horizontally: (green-yellow and blue-green). Is there an algorithm to fix this?

Are you referring to the situation describe early in this thread (this post)?

You should solve the edge pairs as though the center edge pieces are correct. If you have created the edge groups such that two center edge triangles are out of place you should instead view that as 4 edge wing pieces being out of place. You need to cycle the out-of-place edge wings to match the center edge triangles.

I asked for a Professor Pyraminx and not a Jing's Pyraminx for my birthday (it's too late to change it now, oh no!), from this thread I've gathered that this is a bad idea

Should I scramble it like a Jing's Pyraminx the first few times until I know the method, then fully scramble it?

Scrambling a puzzle is never a bad idea. Despite what you may gather from this thread, the Professor Pyraminx is not a hard puzzle.

Ah, good. I meant "should I emulate a Jing's Pyraminx the first few times, learn to solve a Jing's Pyraminx, then fully scramble it?" (just saying that for clarity).

Sure you might as well. Jing's Pyraminx is simple enough though that it can be solved with intuition. The only trouble you'll run into is a 2-2 swap or 3-cycle in the centers (requires a corner twist if my memory serves me). Repeating pretty much any 2-gen sequence enough times will resolve it though.

I solve it from the bottom up. no jing's pyraminx algs. needed. the only hard part is the centers. (which I solve last.) it's a very fun method actually.

Actually? I always wondered why there aren't many non-reduction methods for solving big cubes, it would be fun to see it build by layer rather than piece type (edge pairing is more like sorting than solving to me).

I could probably apply some 5x5x5 algs to it (my edge pairing algorithm is the best, I can probably adapt it to the professor pyraminx if I put some serious thought into it) if I wanted to, as well.

I just noticed there's no way to reduce a Professor Pyraminx to a Pyraminx. Is there any size Pyraminx that can be reduced like that? (I'm thinking only multiples of 3 can, so the Royal may be able to)