An excerpt from Puzzle Laboratory's A Guide to Twisting Puzzles

[This was our first published solution, which appeared in the inaugural issue of WGR in November of 1983. It has been considerably edited for this page.]

Pyraminx is a tetrahedral twisting puzzle devised by a German inventor, Uwe Mèffert, and originally manufactured by Tomy Corporation. Though conceived earlier, it was only manufactured in the wake of Rubik's Cube. Pyraminx is composed of 14 visible parts: four small corners (sometimes called tips), four large corners, and six edges (see photos above). It is a much easier puzzle: the Cube can be twisted into over 43 billion billion positions, while Pyraminx has only 75 million [75,582,720 = 2^5x3^8x6!/2], and once the small corners are (trivially) twisted into their correct positions, there are less than a million possible positions [933,120]. The object is to take a scrambled Pyraminx and return it to its original condition (called Start), in which each of the four faces is a single color.  There is now a truncated version with no small corners, sold under the name Tetraminx.

The diagram above is a flattened aerial view (with the Down face hidden), showing the various parts of the Pyraminx. The large corners are shown in darker shades. Note that, from a vertical view, you can see all three facets of the Up corner and two of the three facets of the Left, Back, and Right corners.   You can see both facets of the Left-Right, Front-Left, and Front-Right edges, and only one facet of the Front-Down, Left-Down, and Right-Down faces.

I believe that the solution we present here is reasonably efficient, and is fairly easy to use. We will use the notation above. The four large corners (from now on simply called corners) are called up, left, right, and back. Edge locations are named by the capitalized initials of the two faces they lie on. Edges are named by the two faces they should lie on when Pyraminx is solved, in lower case initials (fd is the edge which belongs at location FD). The faces are called Front, Left, Right, and Down. Clockwise turns of the four corners are named by capital letters U, L, R, and B. We will also use a turn of the base (the part of the Pyraminx which is not part of the Up corner), holding the Up corner in place. This is called D (Down).  The five corresponding anticlockwise turns are named by lower case letters (u, l, r, b, and d).  We only use the D and d turns during Phase 2, to keep the position of the Up corner (and its two adjacent edges) fixed.  We will only need 8 turns in Phases 3 through 5.

We will solve Pyraminx in five phases. Phase 1 turns the small corners so that their colors match the corners they are attached to. Phase 2 puts the two front edges fl and fr in place. Phase 3 turns the left and right corners, and Phase 4 finishes the front face by placing the edge fd. Phase 5 reorients the Pyraminx so the front face becomes the down face, turns the up corner, and finishes the solution by simultaneously placing the fl, fr, and lr edges.

Phase 1 is the easiest one. Simply turn each tip (small corner), if necessary, in the correct direction, so that its three colors match those of the corner it is connected to. After all four small corners are turned correctly, Phase 1 is finished, and we will not need to turn the small corners again. Phase 1 can take as many as four turns. (I usually don't bother twisting the small corners when scrambling the Pyraminx, since this phase is trivial anyway).

In Phase 2, we place the correct edges at FL and FR so that their colors match the up corner. We will assume that the Up corner is correctly placed. The color of the Up corner which shows on the Front face will be called the front color. For the rest of the solution, we will assume that the front color is blue, but of course you may choose any of the four colors in an actual solution. First find fl, the edge which belongs at FL (check the colors of the Up corner on the Front and Left faces -- blue and orange in our example -- and find the edge which has those colors).  If fl is already at FL and it shows blue on the Front face, continue with the edge fr. If not, we still have work to do.

We need to get fl to one of the two target positions (above left) from which it can be placed correctly by turning the Left corner. We need to get fl either to FD with its blue side on the Down face (from where an l turn puts it in place), or to LD with its blue side on the left face (from where an L turn puts it in place).  See the diagram above. If fl is at LR or FR, or already at FL but with the colors flipped, make a turn (B from LR, r from FR, or L from FL) which puts it at FD or LD.   Half of the time, B from LR or r from FR will put it in a good position directly.  Once fl is in the bottom layer, turn the base if needed so that fl goes into one of the two good positions described above.   The base turn is always necessary always when fl starts at FR, and is also needed when fl is at FD or LD, but with blue on the wrong face.  Once fl is in the correct position, make the Left turn that places fl in FL. The process of placing fl may take as many as three turns.

The edge fr is now placed in the same way. The two key positions to aim for (above right) are FD with its blue side on the Down face (R now puts fr in place), and RD with its blue side on the Right face (r now puts fr in place).   If fr is at LR, turn r to put it at FR (where either r or dr puts it in place).   If it is at FR but flipped, rdR fixes it. If it is already in the Down layer, turn that layer if needed to put it in a good position and then turn R or r.  Placing fr can also take up to three turns. Phase 2 has now been completed, taking a maximum of six turns. The Front face is now all blue except for the bottom row.



In Phase 3, we turn the Left and Right corners so that they also show blue on the Front face. But we must find which face is really the Front face. Look at the Left, Right, and Back corners. Two of these will have blue on them. The face these two corners share is the correct front face! For example, if the Left and Back corners both have blue on them, then the Left face (shared by the Left and Back corners) should really be the Front face.  (Back and Right blue corners means the Right face should really be the Front face).  If the Left and Right corners both have blue on them, then we already have the correct Front face.  

First let us handle the case where the current front face is correct. Now we need to turn the left and right corners so that they show blue on the front face, but without displacing the edges from FL and FR.  We don't care yet which edge is the third edge on the Up corner; we show this edge in gray in the examples above.  We must turn the Up corner so that this gray edge is on the Left or Right corner which is being fixed, so that turning the Left or Right corner does not disrupt one of the two edges already placed correctly on the Up corner.  If the right corner needs to be turned, make turn U first (getting fr out of the way), and turn R or r (whichever brings the blue side of the right corner to the front side).  Now if the left corner need to be turned, turn U again (getting fr out of the way again), turn L or l, and turn U a third time to finish phase 3.   The diagram above left shows an example.  If only one (Left or Right) corner originally needed to be turned, only two turns of the Up corner are needed -- the first away from the corner needing to be fixed, and the second in the opposite direction (uLU or ulU to fix the Left corner, Uru or URu to fix the Right corner).   [When you need to fix both corners, you can actually fix either one first; the upper left diagram can also be solved by the sequence uluru.]

turn Left and Right corners to show the front color (blue in our example) on that face, moving the Up corner out of the way as necessary.  The Front face should have eight (or nine if we are lucky) of its facets showing the same color.  Phase 3 can take up to five turns.  In one case out of eight, we are lucky and FD is correct too, and we can go directly to the last step, Phase 5.  If not, we go to Phase 4 to place fd at FD. 

In Phase 4, we need to find and place edge fd.  If fd is at LD or RD with blue on the Down face, turn b or B to get it to LR. If fd is at LD with blue on the Left face, place it with the sequence of turns rBR. If fd is at RD with blue on the Right face, place it with sequence Lbl (the mirror image of rBR).  If fd started at LR, or you moved it there as instructed, you can place it with LBl if its blue side is on the Left face, or with rbR if its blue side is on the Right face. 

If fd is at FD but flipped (i.e., with blue on the Down face), fix it by turning rBRLBl.  The diagrams above summarizes these sequences.  Phase 4 is completed in a maximum of six turns. 

Phase 5

Finally we come to Phase 5. The Front face is all blue, so we are finished with it. Turn the entire Pyraminx so that the blue face becomes the new Down face. Now turn the new Up corner so that its colors match those of the base. We are faced with four possibilities. We may be finished solving (an 11-to-1 longshot).  If so, mix it up and solve it again -- practice makes perfect.  If we are not so lucky, we need to find out whether we need to move the last three edges (FL, FR, LR) cyclically, and whether two of the edges need to be flipped.   The easiest way to tell whether an edge needs to be flipped is to observe whether either of its facelets match the color of the adjacent Up corner facelet.  An edge in the wrong position needs to be flipped if one of the colors match; an edge which does not need to be flipped does not match on either facelet.   (If an edge is in the correct position, both colors match, of course, if it does not need to be flipped; the colors are reversed if it needs to be flipped).  

If one of the three edges is completely correct, the other two need to be flipped.  If none of the three edges is correct, then a cyclic exchange of the three edges is needed (we call this a tricycle). Decide which direction this cycle must be in by visualizing whether the clockwise turn U or the anticlockwise turn u will put at least one of the edges in the correct place with respect to the base. If it is U, you need a clockwise tricycle. If it is u, you need an anticlockwise one. If all three edges are put in the right place by your trial turn, you do not need to flip any of them. If only one edge is put right, you need to flip the other two. If two of the edges need to be flipped (whether you need a tricycle or not), hold the Pyraminx so that they are at FL and FR.


Now we need five sequences. The first sequence flips the edges at FL and FR without a tricycle. This sequence is rUluLuRU. This is a somewhat longer sequence than we have seen so far, but it is not hard to memorize, since it has a natural rhythm similar to the corner tricycle used in solving Rubik's Cube. It is actually another example of an isoflip. The second and third sequences perform tricycles without flipping edges. The clockwise tricycle is ruRuruR. The anticlockwise one is rURUrUR. Note that these sequences are identical except for the direction of the Up turn, and you don't even have to remember which is which; just make the first Up turn in the direction which will cause the second Right turn to put an edge in its correct slot, and keep turning the Up corner in the same direction; each Right turn will place an edge correctly. These three crucial sequences are shown in the diagrams above. In fact, you now have enough sequences to solve the whole Pyraminx; the rest of the solution consists of shortcuts.


The fourth and fifth sequences are not absolutely necessary, since they handle the cases which can be be solved using a tricycle followed by a double edge flip. But they are short and not too hard to remember, and considerably shorten the solution when you need to perform a tricycle and flip the edges FL and FR simultaneously (which happens about half the time). When you are in one of these situations, hold the Pyraminx so that the edge which does not need to be flipped is at LR (note again that the edges which need to be flipped will each have one of their facets the same color as the face it is on). The clockwise tricycle is LURurl; The anticlockwise is RBUbur. Note that both sequences have three clockwise turns followed by three anticlockwise turns, in slightly different orders. When you have chosen and performed the correct sequence from these five, the Pyraminx is solved! Phase 5 takes a maximum of nine turns. The total maximum length of the basic method is therefore 4+6+5+6+9, or 30 turns!   The average length, not counting the Phase 1 tip turns, is about 17 turns.  Practice solving Pyraminx until you can do it fairly easily;