stoic
Premium Member
Dayan Bermuda is a series of twelve puzzles, roughly analogous to 3x3x3, with some of the faces featuring triangular bandaging. There are eight "core" puzzles, all named after planets and cubic when solved, as well as four "bonus" puzzles which are all shapemods. I've always admired them, and recently I was lucky enough to get my hands on a set. Information online appears to be patchy at best, and so I wanted to post my completed solutions here.
I'll put my solutions to the planet cubes in the order I completed them, which is roughly increasing in difficulty. The shapemods could be solved at any time in the sequence, as they only feature bandaging on one face.
It's worth pointing out that there are other approaches and methods which will solve each of these puzzles. Each one poses its own unique challenges, and asks many questions of the solver. Multiple solving methods are required to complete the set.
I'd be interested to hear if there are any other Bermuda solvers here, and how you've gone about them. They really are a terrific set, and I hope somebody finds this useful!
I'll put my solutions to the planet cubes in the order I completed them, which is roughly increasing in difficulty. The shapemods could be solved at any time in the sequence, as they only feature bandaging on one face.
There's a useful thread on TP about these puzzles, and I am indebted to the "triangular pioneers" therein for the guidance which I have incorporated into some of these solutions. I should also mention my debt to chareaves whose "Ultimate Solution" has been of great help to me when solving all sorts of puzzles. While I'm proud of the effectiveness of my solutions, I'm aware that I'm standing on the shoulders of giants.
I understand from the above thread that there was a variation introduced at the factory which manufactured these. Some of the early Bermuda cubes were assembled with ordinary 3x3x3 cores, and the later ones with Fisher cube cores. This means that there can be a variance in approach required. I'm going to give as full a description as possible of the cubes I'm using.
All my puzzles are black plastic and have the standard "BOY" colour scheme. That is, with white on U and yellow on D, clockwise from above: green, orange, blue, red. White plastic variants are also available, which substitute black stickers in place of white.
Another note: when I solve puzzles beginner-style, I use the alg L U' R' U L' U' R (U2) to swap UBR and UFR corners. I follow Lars Petrus and use the term "Niklas" to describe this alg.
Lastly: due to shapeshifting, some of the puzzles require unusual adjustments, so where necessary I'm going to adopt the convention of using "+" to mean a 45 degree clockwise face turn, and "-" for 45 degrees anticlockwise. e.g. F+ D-
I understand from the above thread that there was a variation introduced at the factory which manufactured these. Some of the early Bermuda cubes were assembled with ordinary 3x3x3 cores, and the later ones with Fisher cube cores. This means that there can be a variance in approach required. I'm going to give as full a description as possible of the cubes I'm using.
All my puzzles are black plastic and have the standard "BOY" colour scheme. That is, with white on U and yellow on D, clockwise from above: green, orange, blue, red. White plastic variants are also available, which substitute black stickers in place of white.
Another note: when I solve puzzles beginner-style, I use the alg L U' R' U L' U' R (U2) to swap UBR and UFR corners. I follow Lars Petrus and use the term "Niklas" to describe this alg.
Lastly: due to shapeshifting, some of the puzzles require unusual adjustments, so where necessary I'm going to adopt the convention of using "+" to mean a 45 degree clockwise face turn, and "-" for 45 degrees anticlockwise. e.g. F+ D-
(Triangle in one face, Fisher style)
Start with the triangle on D, and orient the (2-colour) E-layer centres. (You'll have to ensure that you maintain their orientation throughout everything that follows.)
Next, place the single pentagonal D-layer piece across the hypotenuse of the triangle. Add the two small triangles on either side, and incorporate their matching edges (the squares which look confusingly like centres) into the E-layer. Expand this out by blockbuilding into a Petrus-style 3x2x2 (I can't think of a better way to describe this, but it involves ten solved pieces plus the triangle). I prefer to finish this phase with the triangle/pentagon block at BL, leaving just the R- and U-layers unsolved but giving the freedom to do moves on F in the next phase.
At this point, continue with the Petrus-style approach. If you start from a cubic shape and perform half a Uw (aka u+ or Uw+) move to align the layers, you can solve EO, finish F2L and use Sune/doublesune for CO in a pretty straightforward way, all in <R,U,F>. NB the square edges don't appear to have any orientation, so bear that in mind during EO if it appears that you have an odd number of bad edges.
Finally, solve PLL using something both supercube-safe and <R,U,F> (see list below). Note that, because of the Fisher style, the edges have three stickers, while the corners only have two. This takes a bit of getting used to!
There's a very helpful post on this topic here.
The following algs are supercube-safe:
Aa: U' R U R' U' R' F R U R' U' R' F' R U R2 U' R'
Ab: R U R2 U' R' F R U R U' R' F' R U R U' R'
E: U R U R' U R' U' R F' R U R' U' R' F R2 U' R2 U R
Ga: R U R' U' R' U F R U R U' R' F' U R' U2 R
Gc: (U') R2' F2 R U2 R U2 R' F R U R' U' R' F R2
Gd: (U') R2' F' (R U R U') (R' F' R) (U2' R' U2' R') F2 R2
H: R U2' R' U' R' U' R2 U' R2' U2' R2 U2' R'
Ja: (y U') L' U' L F L' U' L U L F' L2 U L
Jb: R U R' F' R U R' U' R' F R2 U' R'
Ra: F2 R' F' U' F' U F R F' U2 F U2 F'
Rb: R' U2 R U2 R' F (R U R' U') R' F' R2'
Z: R' U' R' F R F' U R U' R' U' F' U F R
The following algs solve PLL, but cause a 90 degree rotation of the U centre only in the process*:
F: (y U') L U' L' U L2 F U F' U' L' F' L F L2
Na: F' R U R' U' R' F R2 F U' R' U' R U F' R'
Nb: R' U R U' R' F' U' F R U R' F R' F' R U' R
T: R U R' U' R' F R2 U' R' U' R U R' F'
Y: F R U' R' U' R U R' F' R U R' U' R' F R F'
*If the resulting U centre needs rotated 180 degrees, it can be solved simply with (R U R' U)5
The following I couldn't find suitable published <R,U,F> algs for, but can be done simply by combinination, for example:
Gb: (Y-perm) U' (Jb-perm)
Ua: (T-perm) U (Ra-perm)
Ub: (T-perm) U2 (Ja-perm)
V: (H-perm) U (Y-perm)
Start with the triangle on D, and orient the (2-colour) E-layer centres. (You'll have to ensure that you maintain their orientation throughout everything that follows.)
Next, place the single pentagonal D-layer piece across the hypotenuse of the triangle. Add the two small triangles on either side, and incorporate their matching edges (the squares which look confusingly like centres) into the E-layer. Expand this out by blockbuilding into a Petrus-style 3x2x2 (I can't think of a better way to describe this, but it involves ten solved pieces plus the triangle). I prefer to finish this phase with the triangle/pentagon block at BL, leaving just the R- and U-layers unsolved but giving the freedom to do moves on F in the next phase.
At this point, continue with the Petrus-style approach. If you start from a cubic shape and perform half a Uw (aka u+ or Uw+) move to align the layers, you can solve EO, finish F2L and use Sune/doublesune for CO in a pretty straightforward way, all in <R,U,F>. NB the square edges don't appear to have any orientation, so bear that in mind during EO if it appears that you have an odd number of bad edges.
Finally, solve PLL using something both supercube-safe and <R,U,F> (see list below). Note that, because of the Fisher style, the edges have three stickers, while the corners only have two. This takes a bit of getting used to!
There's a very helpful post on this topic here.
The following algs are supercube-safe:
Aa: U' R U R' U' R' F R U R' U' R' F' R U R2 U' R'
Ab: R U R2 U' R' F R U R U' R' F' R U R U' R'
E: U R U R' U R' U' R F' R U R' U' R' F R2 U' R2 U R
Ga: R U R' U' R' U F R U R U' R' F' U R' U2 R
Gc: (U') R2' F2 R U2 R U2 R' F R U R' U' R' F R2
Gd: (U') R2' F' (R U R U') (R' F' R) (U2' R' U2' R') F2 R2
H: R U2' R' U' R' U' R2 U' R2' U2' R2 U2' R'
Ja: (y U') L' U' L F L' U' L U L F' L2 U L
Jb: R U R' F' R U R' U' R' F R2 U' R'
Ra: F2 R' F' U' F' U F R F' U2 F U2 F'
Rb: R' U2 R U2 R' F (R U R' U') R' F' R2'
Z: R' U' R' F R F' U R U' R' U' F' U F R
The following algs solve PLL, but cause a 90 degree rotation of the U centre only in the process*:
F: (y U') L U' L' U L2 F U F' U' L' F' L F L2
Na: F' R U R' U' R' F R2 F U' R' U' R U F' R'
Nb: R' U R U' R' F' U' F R U R' F R' F' R U' R
T: R U R' U' R' F R2 U' R' U' R U R' F'
Y: F R U' R' U' R U R' F' R U R' U' R' F R F'
*If the resulting U centre needs rotated 180 degrees, it can be solved simply with (R U R' U)5
The following I couldn't find suitable published <R,U,F> algs for, but can be done simply by combinination, for example:
Gb: (Y-perm) U' (Jb-perm)
Ua: (T-perm) U (Ra-perm)
Ub: (T-perm) U2 (Ja-perm)
V: (H-perm) U (Y-perm)
(Triangles in two opposite faces, Fisher style)
(On mine the triangles are on the red and orange faces, slightly offset from one another)
This can be solved in a fairly straightforward Roux style. I'm not much of a Rouxer, so I present my beginner/Roux-lite solution.
Start by building two "3x2x1" blocks, orange and red, on L and R. Because of the restrictions imposed by the triangles, this is the hardest part of the solve. I tend to leave a straight edge (three pieces) unsolved on the orange face, and a diagonal edge (three pieces) unsolved on the red face. I also find it easier to make part of each block before finishing them both off. The orange triangle has one pentagonal edge and two corners across its hypotenuse, while the red face has pentagons on the two shorter sides, and a straight line (comprising one rectangular edge and two square corners) across the hypotenuse.
Next, solve the LL corners with Niklas and Sune/doublesune. Depending on how you have solved the blocks, a y2 might be helpful.
Finish with L6E. At this point, I find it easiest to place the last red pentagon and continue with four square LL corners. I'm a nub at Roux, so I solve UL and UR <M,U> and then L4E <M,U2>. Pretty intuitive.
The Fisher cube variant which I have has the additional complexity of finally needing the last four centres to be solved. This can be done with (R U R' U)5 - to rotate the U centre 180 degrees - and/or (M' U M U')5 - which rotates the U and F centres 90 degrees.
(On mine the triangles are on the red and orange faces, slightly offset from one another)
This can be solved in a fairly straightforward Roux style. I'm not much of a Rouxer, so I present my beginner/Roux-lite solution.
Start by building two "3x2x1" blocks, orange and red, on L and R. Because of the restrictions imposed by the triangles, this is the hardest part of the solve. I tend to leave a straight edge (three pieces) unsolved on the orange face, and a diagonal edge (three pieces) unsolved on the red face. I also find it easier to make part of each block before finishing them both off. The orange triangle has one pentagonal edge and two corners across its hypotenuse, while the red face has pentagons on the two shorter sides, and a straight line (comprising one rectangular edge and two square corners) across the hypotenuse.
Next, solve the LL corners with Niklas and Sune/doublesune. Depending on how you have solved the blocks, a y2 might be helpful.
Finish with L6E. At this point, I find it easiest to place the last red pentagon and continue with four square LL corners. I'm a nub at Roux, so I solve UL and UR <M,U> and then L4E <M,U2>. Pretty intuitive.
The Fisher cube variant which I have has the additional complexity of finally needing the last four centres to be solved. This can be done with (R U R' U)5 - to rotate the U centre 180 degrees - and/or (M' U M U')5 - which rotates the U and F centres 90 degrees.
(Triangles in two adjacent faces, cubic style)
This is a terrific puzzle, very challenging. I'm very much indebted to Andrea and Burgo at TP for the guidance which forms the basis of the LL solution.
As with a lot of the Bermuda cubes, it's a good strategy to try and get the 45 degree edges solved first, so start by building two "1x2x3" blocks on the green and red faces. That is, on the green face the green/white/red and green/red/yellow pentagons and the three adjacent corners. On the red face the red/white/blue and red/blue/yellow pentagons and the three corners. That takes care of the two shortest sides of both triangles. It should be straightforward to complete one block, but there's quite a bit of challenge involved in getting the second.
I like to make the green block first, then - with green on left - move it out of the way with L2. Make sure that the red pentagons are correctly oriented (if not, run them through a quarter turn on F), and with yellow on F and red on U you should be able to solve the red block in <R,U>. It may help to piece together one section (two or three adjacent pieces) and store it against the hypotenuse while you make the other - then move the two blocks into place.
It's difficult to proceed from here without making some adjustments to the shape of the puzzle, so refer to the notation above.
The next stages are EO and EP, and we're working towards orange being the last layer.
Holding orange on F and blue on U, setup the shape of the puzzle with (F+) D- B-. We're going to proceed by cycling three edges at a time through the U layer using one of two approaches and undoing this setup each time. Turn F as required between cycles.
For EO, (R U2 R' U2)5 cycles three edges UL>UR>FR, while flipping FR and UR.
For EP, use an <R,U> U-perm which cycles UL, UR and UF to manipulate the remaining edges into place. Either R2 U R U R' U' R' U' R' U R' for clockwise or R U' R U R U R U' R' U' R2 for anticlockwise.
If you encounter a "parity" situation where the last two edges need to be swapped, you'll have to rotate the F face through 90 degrees, and resolve the orange edges into these new positions. Alternatively, exchange UL and UR with (F-) D2 B2 U2 B2 D2 F2 (a second pair of edges - FL and FR - also change places) and continue from there.
Now to complete the puzzle using a simple corner series alg. Rotate the position of the cube to blue F, orange U and setup with U- B+ D+.
For CP, we'll use (R F' R' F)3. If you try this on an ordinary 3x3x3, you'll note it swaps UFR<>DFR and also DLF<>DBR. It's the first of these swaps that interests us, as we can exchange out the blue/red corner and use this position to "shoot" corners to where we want them to go to by rotating the U layer (undoing and redoing setups as we go). The other two corners will just shoot themselves back and forth; blindsolvers will be very familiar with this concept.
Finally, CO is done in the same manner, using (R F' R' F)2 or - depending on the corner orientation - (R F' R' F)4. Be careful at this final stage, as the F2L will appear scrambled until you have all the corners correctly oriented. It's easy to go wrong!
This is a terrific puzzle, very challenging. I'm very much indebted to Andrea and Burgo at TP for the guidance which forms the basis of the LL solution.
As with a lot of the Bermuda cubes, it's a good strategy to try and get the 45 degree edges solved first, so start by building two "1x2x3" blocks on the green and red faces. That is, on the green face the green/white/red and green/red/yellow pentagons and the three adjacent corners. On the red face the red/white/blue and red/blue/yellow pentagons and the three corners. That takes care of the two shortest sides of both triangles. It should be straightforward to complete one block, but there's quite a bit of challenge involved in getting the second.
I like to make the green block first, then - with green on left - move it out of the way with L2. Make sure that the red pentagons are correctly oriented (if not, run them through a quarter turn on F), and with yellow on F and red on U you should be able to solve the red block in <R,U>. It may help to piece together one section (two or three adjacent pieces) and store it against the hypotenuse while you make the other - then move the two blocks into place.
It's difficult to proceed from here without making some adjustments to the shape of the puzzle, so refer to the notation above.
The next stages are EO and EP, and we're working towards orange being the last layer.
Holding orange on F and blue on U, setup the shape of the puzzle with (F+) D- B-. We're going to proceed by cycling three edges at a time through the U layer using one of two approaches and undoing this setup each time. Turn F as required between cycles.
For EO, (R U2 R' U2)5 cycles three edges UL>UR>FR, while flipping FR and UR.
For EP, use an <R,U> U-perm which cycles UL, UR and UF to manipulate the remaining edges into place. Either R2 U R U R' U' R' U' R' U R' for clockwise or R U' R U R U R U' R' U' R2 for anticlockwise.
If you encounter a "parity" situation where the last two edges need to be swapped, you'll have to rotate the F face through 90 degrees, and resolve the orange edges into these new positions. Alternatively, exchange UL and UR with (F-) D2 B2 U2 B2 D2 F2 (a second pair of edges - FL and FR - also change places) and continue from there.
Now to complete the puzzle using a simple corner series alg. Rotate the position of the cube to blue F, orange U and setup with U- B+ D+.
For CP, we'll use (R F' R' F)3. If you try this on an ordinary 3x3x3, you'll note it swaps UFR<>DFR and also DLF<>DBR. It's the first of these swaps that interests us, as we can exchange out the blue/red corner and use this position to "shoot" corners to where we want them to go to by rotating the U layer (undoing and redoing setups as we go). The other two corners will just shoot themselves back and forth; blindsolvers will be very familiar with this concept.
Finally, CO is done in the same manner, using (R F' R' F)2 or - depending on the corner orientation - (R F' R' F)4. Be careful at this final stage, as the F2L will appear scrambled until you have all the corners correctly oriented. It's easy to go wrong!
(Triangles in three adjacent faces, cubic style)
This is one of the key Bermuda cubes, as the strategies below will serve us well for several other puzzles in the series.
Examining the solved cube and considering the position and "direction" of the three triangles led me to decide quite quickly how to approach this puzzle. That is, begin by building a 2x2x2 block along the yellow/green/red axis and work towards ultimately leaving a finish of four corners and three edges along the opposite white/orange/blue axis. It involves a whole lot of blockbuilding!
After the 2x2x2 is made, we need to build three separate blocks: a group of three involving the pentagons in each of the white, orange and blue faces, inserted against the hypotenuses of the triangles.
I hold white F, orange U, blue R and solve the white block into position first.
(The first time I did this, I proceeded by orienting the pentagons, and building all of the blocks in <R,U>. It's doable, but far from ideal.)
I solve blue next and store it in place, although I find it can be useful to move the blue block around U whilst solving the last orange pieces. (Or, if you get the orange block solved but placed against the wrong side of the triangle, cycle out part of the red face while you insert the block correctly.)
The last pieces to solve are along the white/orange/blue axis, and we're going to be rotating the cube around this corner axis as required to enable us to affect the right pieces at the right time. If you've solved a rex cube you'll be familiar with this idea.
First, solve edge permutation intuitively in <F,R>.
There's a "parity" situation you might encounter at this point, whereby two edges and two corners need to be swapped, and I'm grateful to Andrea for the neat fix below (although I think my edges-first recognition as described is a slight improvement).
To correct parity: hold the puzzle with the triangles at F, R and D, and rotate it around the corner until the relevant two edges are at FR and DR. This parity fix swaps (and flips) the FR and DR edges, and also swaps the DRB and DFR corners. Ensure that DFR is the "point" of the 3-way axis between the triangles on F, R and D.
Then perform:
R F' U //setup (note the U is done on a non-triangle face)
(R' F R F')3 U'
(R' F R F')3 U'
(R' F R F')3 U' //modified corner piece series (CPS)
(y) L' U L U L' U2 L (y') //Left Sune
R U R' U R U2 R' //Sune
U' F R' //undo setup
(The two Sunes combine to make a U-perm which corrects the position of the edges in U once the corners have exchanged.)
Now, resolve the edges and, if needed, flip two edges placed at UR and BR (and holding the triangles at U, B and R) with (R' U R U' z x')9. It's handy to remember that the (z x') is an anticlockwise rotation around the axis we're working on.
Next, permute the corners with either a double 2-cycle or a 3-cycle.
By holding the triangles at U, R and F you can swap diagonal corners and adjacent corners (UFL<>UBR and URF<>FRD) with a triple CPS: (R' F R F')3. Note that you might have to do this a couple of times, from different angles.
To cycle three corners clockwise, place the triangles at F, R and D and rotate so that the corners are on the front face at URF, DFR and DLF then perform:
R F' U'
( (R' F R F')3 U ) 6
(R' F R F')3
U' F R'
Be careful; it's long but it does work. Note that you can convert this to an anticlockwise cycle by simply reversing the direction of all the U moves.
Finally, solve CO: again holding the triangles at U, B and R, rotate three corners in the R face (RFU, RUB and RBD) clockwise with (R' U R U' z x')6. It might take a bit of thought to orient all four corners by this method - especially if two are solved - but it is effective.
And just for fun: I've seen an alternate method proposed, where the triangular blocks are solved first, and then the 2x2x2 considered later, along with the requisite adjustments made to the red/yellow layers using Sunes. This may well be a superior approach which I might return to; although the method described above transfers better to the harder puzzles as we progress through the series.
This is one of the key Bermuda cubes, as the strategies below will serve us well for several other puzzles in the series.
Examining the solved cube and considering the position and "direction" of the three triangles led me to decide quite quickly how to approach this puzzle. That is, begin by building a 2x2x2 block along the yellow/green/red axis and work towards ultimately leaving a finish of four corners and three edges along the opposite white/orange/blue axis. It involves a whole lot of blockbuilding!
After the 2x2x2 is made, we need to build three separate blocks: a group of three involving the pentagons in each of the white, orange and blue faces, inserted against the hypotenuses of the triangles.
I hold white F, orange U, blue R and solve the white block into position first.
(The first time I did this, I proceeded by orienting the pentagons, and building all of the blocks in <R,U>. It's doable, but far from ideal.)
I solve blue next and store it in place, although I find it can be useful to move the blue block around U whilst solving the last orange pieces. (Or, if you get the orange block solved but placed against the wrong side of the triangle, cycle out part of the red face while you insert the block correctly.)
The last pieces to solve are along the white/orange/blue axis, and we're going to be rotating the cube around this corner axis as required to enable us to affect the right pieces at the right time. If you've solved a rex cube you'll be familiar with this idea.
First, solve edge permutation intuitively in <F,R>.
There's a "parity" situation you might encounter at this point, whereby two edges and two corners need to be swapped, and I'm grateful to Andrea for the neat fix below (although I think my edges-first recognition as described is a slight improvement).
To correct parity: hold the puzzle with the triangles at F, R and D, and rotate it around the corner until the relevant two edges are at FR and DR. This parity fix swaps (and flips) the FR and DR edges, and also swaps the DRB and DFR corners. Ensure that DFR is the "point" of the 3-way axis between the triangles on F, R and D.
Then perform:
R F' U //setup (note the U is done on a non-triangle face)
(R' F R F')3 U'
(R' F R F')3 U'
(R' F R F')3 U' //modified corner piece series (CPS)
(y) L' U L U L' U2 L (y') //Left Sune
R U R' U R U2 R' //Sune
U' F R' //undo setup
(The two Sunes combine to make a U-perm which corrects the position of the edges in U once the corners have exchanged.)
Now, resolve the edges and, if needed, flip two edges placed at UR and BR (and holding the triangles at U, B and R) with (R' U R U' z x')9. It's handy to remember that the (z x') is an anticlockwise rotation around the axis we're working on.
Next, permute the corners with either a double 2-cycle or a 3-cycle.
By holding the triangles at U, R and F you can swap diagonal corners and adjacent corners (UFL<>UBR and URF<>FRD) with a triple CPS: (R' F R F')3. Note that you might have to do this a couple of times, from different angles.
To cycle three corners clockwise, place the triangles at F, R and D and rotate so that the corners are on the front face at URF, DFR and DLF then perform:
R F' U'
( (R' F R F')3 U ) 6
(R' F R F')3
U' F R'
Be careful; it's long but it does work. Note that you can convert this to an anticlockwise cycle by simply reversing the direction of all the U moves.
Finally, solve CO: again holding the triangles at U, B and R, rotate three corners in the R face (RFU, RUB and RBD) clockwise with (R' U R U' z x')6. It might take a bit of thought to orient all four corners by this method - especially if two are solved - but it is effective.
And just for fun: I've seen an alternate method proposed, where the triangular blocks are solved first, and then the 2x2x2 considered later, along with the requisite adjustments made to the red/yellow layers using Sunes. This may well be a superior approach which I might return to; although the method described above transfers better to the harder puzzles as we progress through the series.
(Triangles in three faces on same layer, cubic style)
My first thoughts when I examined Mars were to solve the three pentagon blocks and leave myself one F2L slot to solve along with an orange last layer. Happily, this turned out to be pretty straightforward. (Note that the pentagons when solved all point towards FRD, when looking at them face on with orange U.)
I solve them in order yellow>blue>white, mostly using the red/green/white slot to manipulate the pieces.
Next, orient the last five edges. We can't do fruruf but we can flip two edges by moving them to the blue face: rotate the white and then blue faces in order 90 degrees clockwise and utilise the slot created at red/blue/yellow to orient them.
Return the blue and white triangles to their correct positions, and move back to working with the red/green/white slot in FR. At this point we are going to tackle the last five corners and edges together rather than leaving a LL. It's not necessary but I like to solve the F2L slot now anyway for reference. As I'm primarily a CFOP solver for 3x3x3, it's simple and intuitive in <R,U>.
Freedom of movement is limited with green F, so switch to white F, orange U and perform a single F move. Note that the slot we have just completed is going to be adversely affected, but also note that from this position we have the freedom to work across three faces in the movegroup <R,U,F> which is going to be very useful.
There are a couple of sequences adapted from the Neptune solution which can come into play now as we tackle CP.
Firstly: we can swap diagonal corners and adjacent corners (UFL<>UBR and URF<>FRD) with a triple CPS: (R' F R F')3.
Secondly, this long sequence:
( (R' F R F')3 U ) 6 (R' F R F')3 U2
cycles ULB>URF>DFR.
Some setups may be required, but these two algorithms are sufficient to permute all corners.
Next, I like to resolve the orientation of the red/green/white corner if required with Sune before returning the white triangle block to its correct position, and solving CO in the remaining four corners using Sune with green F, orange U.
Finally, EP; this is also effected using the white/blue/red slot, so perform a (y) move to get back to white as F, and we can again work with algorithms we've seen before.
A clockwise U-perm, cycling UL>UR>UF, is achieved with a combination of Sunes:
R U R' U R U2 R' (y) L' U L U L' U2 L (y')
Reversing the order of the Sunes gives an anticlockwise U-perm, cycling UB>UF>UR.
( (R' F R F')3 U ) 6 is an H-perm
Setting up to a combination of these sequences should permute edges as required and complete the solution once setups are undone and the white triangle is restored.
My first thoughts when I examined Mars were to solve the three pentagon blocks and leave myself one F2L slot to solve along with an orange last layer. Happily, this turned out to be pretty straightforward. (Note that the pentagons when solved all point towards FRD, when looking at them face on with orange U.)
I solve them in order yellow>blue>white, mostly using the red/green/white slot to manipulate the pieces.
Next, orient the last five edges. We can't do fruruf but we can flip two edges by moving them to the blue face: rotate the white and then blue faces in order 90 degrees clockwise and utilise the slot created at red/blue/yellow to orient them.
Return the blue and white triangles to their correct positions, and move back to working with the red/green/white slot in FR. At this point we are going to tackle the last five corners and edges together rather than leaving a LL. It's not necessary but I like to solve the F2L slot now anyway for reference. As I'm primarily a CFOP solver for 3x3x3, it's simple and intuitive in <R,U>.
Freedom of movement is limited with green F, so switch to white F, orange U and perform a single F move. Note that the slot we have just completed is going to be adversely affected, but also note that from this position we have the freedom to work across three faces in the movegroup <R,U,F> which is going to be very useful.
There are a couple of sequences adapted from the Neptune solution which can come into play now as we tackle CP.
Firstly: we can swap diagonal corners and adjacent corners (UFL<>UBR and URF<>FRD) with a triple CPS: (R' F R F')3.
Secondly, this long sequence:
( (R' F R F')3 U ) 6 (R' F R F')3 U2
cycles ULB>URF>DFR.
Some setups may be required, but these two algorithms are sufficient to permute all corners.
Next, I like to resolve the orientation of the red/green/white corner if required with Sune before returning the white triangle block to its correct position, and solving CO in the remaining four corners using Sune with green F, orange U.
Finally, EP; this is also effected using the white/blue/red slot, so perform a (y) move to get back to white as F, and we can again work with algorithms we've seen before.
A clockwise U-perm, cycling UL>UR>UF, is achieved with a combination of Sunes:
R U R' U R U2 R' (y) L' U L U L' U2 L (y')
Reversing the order of the Sunes gives an anticlockwise U-perm, cycling UB>UF>UR.
( (R' F R F')3 U ) 6 is an H-perm
Setting up to a combination of these sequences should permute edges as required and complete the solution once setups are undone and the white triangle is restored.
(Triangles in four faces, cubic style)
This puzzle is so heavily bandaged, it's not easy to get it scrambled! Note that it's not possible to turn either of the red or green faces, so the red/green piece is always fixed in place. However, there's not here much we haven't seen before so the solution is quite quick to describe.
First, solve the white pentagon block (three pieces). I find it best to hold yellow U, orange F and bring the pieces to the yellow/blue/orange axis. This is because we have free movement of R, U and F when the triangles are correctly positioned with URF the "point" of the 3-way axis between the triangles. Once this block is made and twisted into position across the white hypotenuse, we have effectively built a red/white/green "2x2x2" and we can pretty much forget about white for the rest of the solve while we concentrate on the other three triangles.
I'd note at this juncture that, because of the additional bandaging, it's not possible to involve the D face while placing the last block after it's made. This means that the last block has to be manufactured in such a way that it fits directly onto the hypotenuse as it's completed, and this increases the difficulty level quite significantly. I haven't found a better way to do this, so I somewhat chance to luck with broadly a 1 in 3 chance of getting it right. Otherwise, I break it up and go again...
From here, proceed exactly as for Neptune with a finish on the yellow/blue/orange axis.
This puzzle is so heavily bandaged, it's not easy to get it scrambled! Note that it's not possible to turn either of the red or green faces, so the red/green piece is always fixed in place. However, there's not here much we haven't seen before so the solution is quite quick to describe.
First, solve the white pentagon block (three pieces). I find it best to hold yellow U, orange F and bring the pieces to the yellow/blue/orange axis. This is because we have free movement of R, U and F when the triangles are correctly positioned with URF the "point" of the 3-way axis between the triangles. Once this block is made and twisted into position across the white hypotenuse, we have effectively built a red/white/green "2x2x2" and we can pretty much forget about white for the rest of the solve while we concentrate on the other three triangles.
I'd note at this juncture that, because of the additional bandaging, it's not possible to involve the D face while placing the last block after it's made. This means that the last block has to be manufactured in such a way that it fits directly onto the hypotenuse as it's completed, and this increases the difficulty level quite significantly. I haven't found a better way to do this, so I somewhat chance to luck with broadly a 1 in 3 chance of getting it right. Otherwise, I break it up and go again...
From here, proceed exactly as for Neptune with a finish on the yellow/blue/orange axis.
(Triangles in three adjacent sides, cubic style)
The solution to Jupiter is similar to that for Neptune, albeit with one main difference. As we don't have a 3-way axis anywhere on the puzzle to work with, we'll need to manufacture one. This can be done relatively simply with a setup as follows: hold red U, blue F and perform R2 U' U-.
Follow the Neptune solution from here, starting with the 2x2x2 block in yellow/orange/green placed at DBL, manufacturing three blocks (three pieces on each of blue and white based around a pentagon, and one straight edge based on white/orange - NB not white/green!) and then finishing with four corners and three edges - two of which are red pentagons - around the blue/red/white axis.
Note that the orange stickers on the straight edge match up with red - it's easy to orient this block the wrong way. Also be aware that the white/blue/red pentagon points to white/green/orange because of the setup moves.
Undo setup to reveal a solved cube.
One further note: I did encounter a situation on this puzzle that I didn't see on Neptune or Uranus; that is, I wasn't able to orient all of the corners at the last stage as I kept ending up with either a pair unsolved, or only the central corner solved (which doesn't lend itself to a 3-cycle on one face). It's likely to be a function of three of the corners being triangles, rather than "true" corners (they look more like edges when solved), and due to the initial half-move setup applied. I fixed this by using (R' F R F')3 to switch out the solved central corner, orienting the other three corners and replacing the centre with (R' F R F')3.
The solution to Jupiter is similar to that for Neptune, albeit with one main difference. As we don't have a 3-way axis anywhere on the puzzle to work with, we'll need to manufacture one. This can be done relatively simply with a setup as follows: hold red U, blue F and perform R2 U' U-.
Follow the Neptune solution from here, starting with the 2x2x2 block in yellow/orange/green placed at DBL, manufacturing three blocks (three pieces on each of blue and white based around a pentagon, and one straight edge based on white/orange - NB not white/green!) and then finishing with four corners and three edges - two of which are red pentagons - around the blue/red/white axis.
Note that the orange stickers on the straight edge match up with red - it's easy to orient this block the wrong way. Also be aware that the white/blue/red pentagon points to white/green/orange because of the setup moves.
Undo setup to reveal a solved cube.
One further note: I did encounter a situation on this puzzle that I didn't see on Neptune or Uranus; that is, I wasn't able to orient all of the corners at the last stage as I kept ending up with either a pair unsolved, or only the central corner solved (which doesn't lend itself to a 3-cycle on one face). It's likely to be a function of three of the corners being triangles, rather than "true" corners (they look more like edges when solved), and due to the initial half-move setup applied. I fixed this by using (R' F R F')3 to switch out the solved central corner, orienting the other three corners and replacing the centre with (R' F R F')3.
(Triangles in four faces, cubic style)
This is the hardest of all the Bermuda cubes in my opinion.
The first thing to note is that the blue/white piece is fixed in place and that neither of these faces can turn, so we'll start by building the three-piece block around the orange pentagon (it goes along the hypotenuse of the orange triangle). This gives us our now-familiar "2x2x2" placed at DBL.
Next, holding green F and red U adjust the shape of the puzzle with the setup (F' F-) (U U+) to give us the triple axis we need to continue.
From here, we can follow the Neptune/Uranus solution; however, because of this setup it's crucial to note that we are proceeding with a couple of key differences. First, we're going to solve with the final, important, axis being green/red/yellow, and these three coloured triangles "pointing" towards this corner.
The green/*yellow*/orange pentagon block needs to be built misoriented against the red hypotenuse (careful: note that the green/*white*/orange pentagon will be solved later as one of the final edges), and the green hypotenuse has a straight green/red edge against it, with regular-shaped corners. (Yellow block solves in a straightforward fashion into DRB.)
The other main difference is corner permutation, as the 3-cycle given for Neptune won't work; instead we can use the following sequence to cycle the three corners on F (UFL, URF and FRD) clockwise. (Thanks to Burgo for the idea.)
(R' R-) U' (R R+) U
(R' F R F')3
U' (R' R-) U (R R+)
(R' F R F')3
(R' R-) U' (R R+) U
(R U' R' U)3
(R' F R F')3
U' (R' R-) U (R R+)
Keep in mind that we're interested in manipulating the green/white corner and the three red corners during the last parts of the solve. It's easy to get confused at this delicate stage.
Finally, we can undo the setup we did at the beginning to reveal a solved cube and the series is complete!
This is the hardest of all the Bermuda cubes in my opinion.
The first thing to note is that the blue/white piece is fixed in place and that neither of these faces can turn, so we'll start by building the three-piece block around the orange pentagon (it goes along the hypotenuse of the orange triangle). This gives us our now-familiar "2x2x2" placed at DBL.
Next, holding green F and red U adjust the shape of the puzzle with the setup (F' F-) (U U+) to give us the triple axis we need to continue.
From here, we can follow the Neptune/Uranus solution; however, because of this setup it's crucial to note that we are proceeding with a couple of key differences. First, we're going to solve with the final, important, axis being green/red/yellow, and these three coloured triangles "pointing" towards this corner.
The green/*yellow*/orange pentagon block needs to be built misoriented against the red hypotenuse (careful: note that the green/*white*/orange pentagon will be solved later as one of the final edges), and the green hypotenuse has a straight green/red edge against it, with regular-shaped corners. (Yellow block solves in a straightforward fashion into DRB.)
The other main difference is corner permutation, as the 3-cycle given for Neptune won't work; instead we can use the following sequence to cycle the three corners on F (UFL, URF and FRD) clockwise. (Thanks to Burgo for the idea.)
(R' R-) U' (R R+) U
(R' F R F')3
U' (R' R-) U (R R+)
(R' F R F')3
(R' R-) U' (R R+) U
(R U' R' U)3
(R' F R F')3
U' (R' R-) U (R R+)
Keep in mind that we're interested in manipulating the green/white corner and the three red corners during the last parts of the solve. It's easy to get confused at this delicate stage.
Finally, we can undo the setup we did at the beginning to reveal a solved cube and the series is complete!
It's worth pointing out that there are other approaches and methods which will solve each of these puzzles. Each one poses its own unique challenges, and asks many questions of the solver. Multiple solving methods are required to complete the set.
I'd be interested to hear if there are any other Bermuda solvers here, and how you've gone about them. They really are a terrific set, and I hope somebody finds this useful!
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