Other Methods

Step 1 - Solve all edges

Solving these pieces first allows them to be completed using only outer face turns, and no slice moves, so they can be solved layer-by-layer with very simple inserts. However this does require knowing the color scheme, or at least checking it with the corners. The edge pieces on an FTO also behave identically to those of the Dino Cube, so solving outwards from a vertex works too.

Step 2 - Solve all corners

These are the remaining pieces that make up the "cage" that ends up surrounding the inner triangles. They can very easily be moved around and put into place with the familiar sledges and hedges. There are only six corner pieces on the FTO to solve, so even permuting them individually can still be done in relatively few moves.

Step 3 - Solve all triangles

Finally the 24 triangles remain to be solved. These pieces are best saved for last because they are the easiest to manipulate independently of the other piece types. The commutator here is fairly straightforward, with a 3-move insert and a slice layer interchange. There is no particular order for solving in this step, but solving one orbit of the triangles before the other can make the recognition easier.

Step 1 - Solve four center groups

A center (highlighted in cyan) consists of three edges and three triangles of the same color being matched up on one face of the puzzle, in the same way as Bencisco. The edge pieces must be placed so as to get the overall color scheme correct. Only one orbit of triangles is used here, and there is more freedom for blockbuilding compared to Bencisco as all of the triples can be ignored for this stage.

Step 2 - Match six triple groups

A triple (highlighted in magenta) consists of one corner and two triangles matched with it across one vertex of the puzzle, again just like in Bencisco. With one of the centers on the top layer, the bottom layer can be rotated to match up the corners with their respective triangles, which can then be inserted into the top layer to be preserved. Some algorithms are needed for certain difficult cases when completing the last two triples. This is the least efficient step of this method and ultimately what renders it obsolete to modern techniques.

Step 3 - Solve the reduced puzzle

Now we have simplified the puzzle into a Tetraminx equivalent - with four "corners" and six "edges". This can be solved with any number of different methods, but one basic approach is to solve three of the triples around one center while orienting the other centers correctly (like the top-first method for a Pyraminx). This leaves three triples left to solve which can be placed with simple sledges and hedges.

Step 1 - Solve one center

This is identical to the first step of reduction/Bencisco; Julian describes it as building a "hexagon" in two halves. Like those methods, it is important to know the color scheme for this step and the others. The center becomes part of the first layer.

Step 2 - Solve two adjacent arrow blocks

These effectively consist of two first-layer triples, and two half-centers (which use edges and triangles that belong in the second layer). There are several ways to go about completing this step, and Julian's way to solve each block is equivalent to forming a triple and then the half-center below it. However, solving a half-center and then inserting the triple also works well, or even using arrow-block techniques like Bencisco to form each of them as single units. The triangles that lie between the edges are technically optional here.

Step 3 - Solve the last half-center and bottom layer corner

This step is a bit like the previous one, as you are solving what would be the last arrow block in the F2L, but intentionally leaving out the two triangles that would also belong in the first layer. This allows it to be completed using only 2-gen <R,U> moves while still being fairly efficient. There is a bit less freedom for this block, so a typical solution consists of matching the bottom corner piece (and triangle) with one of the second-layer edges, and then matching that pair with the last edge. Again, solving the triangle is technically optional.

Step 4 - Solve corners of the last layer

Here the majority of the F2L is now solved, also forcing the last layer edges to be solved, so we now solve the remaining three corners. There are a total of five cases, and a variety of approaches can be taken since we don't care about the triangles for this step. The simple hedge and sledge work the best when both orientation and permutation need to be solved, and basic commutators work for pure 3-cycles. Alternate algorithms may also be useful for cases where it's easy to influence the triangles in the following step.

Step 5 - Solve remaining triangles with commutators

At this point there will now be at most 17 triangle pieces left to solve (14 with complete half-centers). This may seem like only a small reduction compared to just a cage method, but it will often be fewer than that, since some triangles are identical in coloring and will happen to be solved already. The remaining triangles will also be in separate orbits which further reduces the number of commutators actually needed. With that, combined with even basic influencing during step 4, many scrambles can be solved with only four total commutators (and sometimes even three) with the proper choices and setups. Even so, this is still the least efficient step since the ratio of moves to pieces being solved is higher than the others.

Step 1 - Solve the first layer

The first step of this method is fairly self-explanatory; solving the first layer which consists of a center and three triples. The most obvious way is solving the center and then solving the triples next to it, just like in Bencisco, but blockbuilding strategies can also be used as well. This is the quickest and most efficient step.

Step 2 - Solve the second layer

This part of the method can be broken down into three stages, solving the edges, lower triangles and then upper triangles.  The edges that belong in this layer can be brought in from the third layer using 5-move inserts that preserve the first-layer triples,  including  the lower triangles between them where possible. Any remaining lower triangles can be solved by setting up to a sledge, or using 3-cycle commutators when it is efficient. Lastly, the three upper triangles can be solved with just sledges and hedges, with three required for the worst case. Completing this step will also solve the remaining edges and top triangles in the third layer for the final step.

Step 3 - Solve the third layer

Here we are left with the pieces in the third layer, which amounts to just three corners and six triangles. There are multiple ways to approach this step, but a simple yet still efficient one is to solve the corners first (which can always be done with one algorithm in at most 7 moves) and then the triangles (which can always be solved with at most two commutators). Alternate algorithms can also be used when solving the corners to influence better cases for the remaining triangles.

Step 1 - Solve one vertex

A vertex consists of one corner piece, the four triangles which surround it, and the four edges which fit in between those triangles. This step can be completed intuitively, borrowing techniques from Bencisco such as building a large trap and extending it to the completed vertex, or building an arrow block and bringing in the remaining edges and triangles.

Step 2 - Solve an adjacent vertex

With a second vertex sharing an edge with the first one, we now have a large section of solved pieces with two triangles left out. This vertex can again be solved intuitively, but a consistent approach I have found is to solve an edge-triangle pair next to the first vertex, and then add the remaining arrow block to complete it . Due to there being four options for this vertex, a solver has lots of flexibility and greater potential for efficiency. Solving the pieces onto the left side of the puzzle helps with transitioning to the next step.

Step 3 - Solve the remaining edges and two bottom triangles

At this stage, the key insight is that the remaining unsolved pieces are really two sets of four triples in two separate orbits, along with the edges. This step is solving those edges while also solving the triangles that make up a bottom triple.  It is best to use 2-gen <R,U> moves when these triangles are readily available, but in some instances they may be stuck in the middle slice, in which case 4-move triple inserts are useful. Since there are only four pieces necessary to solve, this step can be very efficient and quick.

Step 4 - Solve the last six triangles

We now have six triangles left to solve in one of the orbits, similar to the six triangles in the last 3 triples stage of Bencisco. Solving them can again be accomplished using only sledges/hedges and top layer moves, but ignoring the corners makes this step more efficient than L3T. 3-cycle commutators can also be helpful in some situations which would normally require many uses of sledges.

Step 5 - Solve the last four triples

Finally we are left with four triples to solve, and this step can be done just like in Bencisco. There isn't much else to say for this part, but since the triangles in this orbit are not affected during step 4, there are unique opportunities for lookahead compared to Bencisco.

Step 1 - Solve the first block

The first block here is the exact same as in Bencisco, but it now becomes part of the first two layers. This block can again be solved by building a center and then solving two triples next to it, but techniques such as building arrow blocks can also be used for greater efficiency. The common color shared by the two triples becomes the bottom face, so a solver will need to be comfortable with their color scheme to take full advantage of available options.

Step 2 - Solve two centers

After the first block, we can now solve the two center groups that belong in the first 2 layers. This works the same way as Bencisco, by using <Rw,R,U> moves to preserve the first block. Once these centers are solved, the center group that belongs in the top layer will also become automatically solved.

Step 3 - Solve the bottom triple

This step is where the method begins to truly deviate from Bencisco; solving the bottom triple here will now complete two entire layers of pieces on the FTO.  The basic strategy is to first solve the bottom corner and bottom triangle together with one algorithm, and then solve the remaining middle layer triangle above them with a second algorithm. A full algorithm set also exists for solving all three pieces at once. Avoiding the bottom corner from ending up in the bottom slot during step 2 helps to improve the potential cases for this step.

Step 4 - Solve the last layer

Here we again have a last layer as the final step, and this can be solved in a number of ways. The basic approach is to first solve the six side triangles in at most two algorithms, by solving a small trapezoid on one side and then the remaining two trapezoids on the other sides. The remaining three corners can then be finished off with another algorithm. Again, a full algorithm set also exists for solving all six triangles at once. Due to corners having only two orientations each, the recognition for this step only relies on checking the sides of the last layer, similar to a PLL recognition system.