Mathematrec has moved to: https://richholmes.xyz/mathematrec.
Imagine me finding myself a new blogging platform
As of today, anything on wordpress.com can be assumed to be false, AI-generated garbage.
I’ll let you know when I figure out where to move this blog to.
A brick
Continuing on the subject of cube (etc.) dissections into polycubes.
Subjectively:
- It seems to me dissections that use multiple copies of the same piece tend to be easier than ones that use different pieces.
- Unique solutions are not necessarily harder to find
- Multiple solutions are of some interest in the way they’re related to each other
After thinking about other dissections of the 3x3x3 cube I thought I would take a look at dissections of the 4x3x3 brick. One set of distinct pieces that — according to my solver program — gives a moderate number of solutions (125) is:
I call this (for no particular reason) Qbrick. “Cube brick” I guess?
Cube counting
For Qbrick, cube counting seems not too useful.
There are four odd-size pieces with checkerboard excess/deficit of 1 each and two even-size pieces with excess/deficit of 2 each so there are, I think, fourteen ways to get zero total excess/deficit, not very constraining:
| a | b | e | f | g | h |
|---|---|---|---|---|---|
| 2 | 2 | -1 | -1 | -1 | -1 |
| 2 | -2 | 1 | 1 | -1 | -1 |
| 2 | -2 | 1 | -1 | 1 | -1 |
| 2 | -2 | 1 | -1 | -1 | 1 |
| 2 | -2 | -1 | 1 | 1 | -1 |
| 2 | -2 | -1 | 1 | -1 | 1 |
| 2 | -2 | -1 | -1 | 1 | -1 |
| -2 | 2 | 1 | 1 | -1 | -1 |
| -2 | 2 | 1 | -1 | 1 | -1 |
| -2 | 2 | 1 | -1 | -1 | 1 |
| -2 | 2 | -1 | 1 | 1 | -1 |
| -2 | 2 | -1 | 1 | -1 | 1 |
| -2 | 2 | -1 | -1 | 1 | 1 |
| -2 | -2 | 1 | 1 | 1 | 1 |
The sum of the maximum number of vertex cubes occupied per piece is 11, and we need 8; no piece can occupy 3 vertex cubes, so between 2 and 3 pieces must be vertex-deficient. Also not very constraining. (A little searching suggests e cannot be doubly deficient, but that doesn’t help much even if it’s correct.)
All pieces except a must occupy at least 1 face, and up to 3; we need 10, so betwen 2 and 3 pieces must be face-excessive. Also not very constraining.
So:
- There are 8 pieces vs. 7 for Soma
- There are 36 cubes vs. 27 for Soma
- There is no evident way to rule out any position for any piece (other than perhaps doubly deficient positions for e) as there is for several pieces in Soma
which means the combinatorics for this dissection — or almost any similar dissection of the same brick — are much, much worse. There are fewer solutions but many more cases to investigate. If there’s a way to make this tractible for human investigation I haven’t figured it out.
Non methodical searching
I made a Qbrick set out of Artec blocks, which were called to my attention by mathart4all.
Just recently someone posted 3D printing files for a similar sort of linking cubes over on Printables, which might be worth a look. Unfortunately I don’t have enough different colors of Artec blocks to make each piece a different color, and not enough of any color to make every piece be a single color. But for exploration it beats trying to 3D print a full set of pieces.
I’ve spent some unmetered amount of time over the past week or so looking for solutions without trying to be systematic about it — just putting pieces together and seeing what I can make work. In some cases I’ve found a solution and then realized a small-ish number of pieces can be rearranged to make one different solution. In one case I found a cycle of four solutions linked by 2- and 3-piece moves. Altogether I’ve found 18 solutions so far.
That’s another difference with Soma, in which 239 out of 240 solutions connect to each other via sequences of 2- or 3-piece moves. Find one solution and look for these moves and you can find them (almost) all. The Qbrick solutions I’ve found mostly appear to be solitary or in clusters of 2 or 4 under moves of less than 5 pieces, unless I’m overlooking non-obvious 4 piece moves (which would not surprise me at all). Finding one solution does not easily lead you to finding many others.
Between the worse combinatorics with no piece positions ruled out, and the apparent lack of simple connectivity between solutions, getting a handle on finding solutions other than by luck (or by computer!) has been elusive.
Other brick dissections?
It might be possible to engineer a brick dissection that does have meaningful constraints from cube counting. For example, I think there are 12 pentacubes that can occupy no more than 1 vertex cube, including 1 that can occupy only none, as well as 5 tetracubes that can occupy no more than 1. In the Qbrick there are 3 pieces that can occupy 2 vertices and the rest can occupy 1. But if you examined dissections in which 0 or 1 pieces could occupy 2 vertices and the rest 1 (or 2 occupying up to 2 vertices and 1 occupying only 0), the summed maximum vertices would be 8 or 9 and there might be useful constraints from that. As for checkerboard coloring constraints, I think there are 2 pentacubes whose checkerboard coloring has an excess/deficit of 3 instead of 1 (1 of which occupies up to 2 vertices but the other occupies none). If both were present, and neither of the 2-excess tetracubes, I think they’d have to have opposite excesses. Maybe there’s an interesting dissection along these lines.
But this sort of just punts. It doesn’t do anything to aid in coming to grips with what seem likely to be more typical brick dissections. And maybe nothing can, but maybe there’s something I just haven’t hit upon.
7 pieces, 1 solution
(This is a rewrite of, and replacement for, an earlier post.)
In his 1972 Scientific American article (or rather the book chapter based on it) on polycubes, Martin Gardner wrote:
Is there a seven-piece dissection [of the 3x3x3 cube into polycubes] with a unique solution? If so, it has not come to my attention.
As 0xDE on the Fediverse points out, the answer is trivially yes:
These fit together in only one way (up to rotations) to form a big cube. The solution is symmetric under mirror reversal, so you don’t get a second solution that way. There are lots of other similar dissections: knock six of the corners off a cube and that gives you another, for instance. Of course this isn’t the sort of thing Gardner had in mind, but how would you rephrase his question to eliminate these and allow only the kinds of dissections he did have in mind?
You could object that monocubes are not “real” polycubes and insist on polycubes of order 2 or more… but right away I can see a 7-piece dissection into six dicubes and a 15-cube with a unique solution. Okay, tricubes or higher? Sorry, but this has what seems to me a fairly uninteresting unique solution:
Nonconvex polycubes, then? Maybe. Nothing as obvious as the above comes to mind.
Then again, there’s this:
There’s only one solution with these. Since pieces C through F are asymmetric, and we don’t have their chiral partners, the mirror solution is not possible. Piece A is convex, but I think this dissection is nevertheless the kind of thing Gardner had in mind — I don’t know how hard a time you’d have finding it, but I don’t think you’d find it obvious.
You can do cube counting to narrow down the search space. A checkerboard coloring argument shows piece A must have vertex, edge, face, central counts [0, 2, 1, 0]. Then all the remaining pieces must occupy their maximum number of vertex cubes: B has to be [2, 2, 0, 0] and G must be [2, 1, 1, 0]. One of C–F must be central and nondeficient — [1, 1, 1, 1] — and the other three must be noncentral and nondeficient, [1, 2, 1, 0]. That tells us the positions of all the pieces. Armed with that, maybe the solution isn’t hard to find, but surely it’s not trivial.
So I think there are both “interesting” and “uninteresting” 7-piece dissections with unique solutions, but I don’t know how to define “interesting”.
And maybe the lesson is that having a unique solution is not a particularly meaningful attribute. Certainly it doesn’t correlate with difficulty.
Cube O
Yet another cube dissection. This is one I designed; I call it Cube O — Oh as in orange.
Here are the pieces.
Like the Mikusiński Cube, it has six pieces, three pentacubes and three tetracubes. One piece is asymmetric, so there are no mirror image solutions. In fact, there is a grand total of one solution. Definitely no map for this puzzle.
I’m curious to see others’ opinions as to how hard this one is to solve. I know how long it took my solver program to get it, but it’s not at all clear how or if that maps onto human-perceived difficulty:
| Puzzle | Time for 1 solution (seconds) | Time for all solutions (seconds) |
| Soma | 0.05 | 41.8 |
| Diabolical | 0.04 | 4.2 |
| Mikusiński | 0.11 | 3.8 |
| Cube O | 0.18 | 3.0 |
From my experience it seems Cube O is considerably easier than the Mikusiński Cube. Or did I just get lucky? I don’t think so, but it’s hard to tell.
But I didn’t just go in by trial and error or by exhaustive unconstrained search. Here once again cube counting tells you a lot. I’ll put my analysis behind a spoiler warning in case you want to work it out for yourself — or to try solving the puzzle without it.
3D print files are here: https://www.printables.com/model/463446-cube-o-puzzle
The possible positions — listing the number of occupied [vertex, edge, face, and central] cubes, are:
A: [3, 2, 0, 0], [0, 3, 2, 0]
B: [1, 2, 1, 1], [1, 3, 1, 0], [1, 1, 2, 1], [0, 2, 2, 1], [0, 1, 3, 1], [1, 2, 2, 0]
C: [2, 1, 1, 1], [0, 3, 2, 0], [0, 1, 3, 1], [0, 2, 2, 1]
D: [1, 2, 1, 0], [1, 1, 1, 1], [0, 2, 2, 0], [0, 1, 2, 1]
E: [1, 2, 1, 0], [0, 1, 2, 1]
F: [1, 3, 0, 0], [1, 1, 2, 0], [0, 2, 1, 1], [0, 0, 3, 1]
Piece A can occupy up to 3 vertex cubes, C can occupy up to 2, and the other four can occupy a maximum of 1 each. 3+2+1+1+1+1 = 9, so yet again there must be one and only one singly deficient piece. Or let me say singly vertex-deficient. Neither piece A nor piece C can be singly vertex-deficient, and we can cross their multiply-deficient positions off our list of possibilities. And C’s only non vertex-deficient position is central, so C must be the central piece.
But B, E, and F have no vertex-deficient non central positions. So D has to be the vertex-deficient piece.
The positions we’re now left with are
A: [3, 2, 0, 0]
B: [1, 3, 1, 0], [1, 2, 2, 0]
C: [2, 1, 1, 1]
D: [0, 2, 2, 0]
E: [1, 2, 1, 0]
F: [1, 3, 0, 0], [1, 1, 2, 0]
Pieces A, C, D, and E occupy 4 face cubes out of 6, and B and F could occupy another 4 for a total of 8, so there must be either two singly face-deficient pieces or one doubly face-deficient. But piece F cannot be singly face-deficient, and B cannot be doubly, so the only possibility is that F is doubly face-deficient and B is non face-deficient. We’re left with only one possible set of positions:
A: [3, 2, 0, 0]
B: [1, 2, 2, 0]
C: [2, 1, 1, 1]
D: [0, 2, 2, 0]
E: [1, 2, 1, 0]
F: [1, 3, 0, 0]And these do in fact add up to 8 vertex, 12 edge, 6 face, and 1 central cubes.
So there’s one starting position for pieces A and E, three for each of pieces C, D, and F, and five for piece B that may look plausible but are in fact dead ends! Without doing the cube counting, you wouldn’t know that, and I can imagine you might spend a lot of time trying to work from such bad starts and not getting anywhere. But by restricting myself to the above positions I was able to find the solution fairly quickly. And I stopped there, my solver program having told me there were no other solutions — can you convince yourself of that?
Another cube
Another cube dissection…
The Mikusiński Cube was invented by J. G. Mikusiński sometime before 1950. It’s almost the anti-Diabolical Cube. Both have six polycube pieces but the Diabolical’s all are planar; five of the Mikusiński’s are nonplanar. Those five also are asymmetric, and all six are nonconvex.
There are three pentacubes and three tetracubes. The latter are pieces 2, 5, and 6 from the Soma cube — one symmetric piece and a chiral pair of asymmetric ones. But the pentacubes are three distinct asymmetric shapes; their chiral partners are not used. This means mirror image solutions are not possible.
Here are the shapes along with the letters I identify them by:
It’s useful to think about these pieces and do some cube counting before trying to make a large cube out of them. Pieces b, c, and e can occupy two vertex cubes, or one, or none. Pieces a, d, and f can occupy one or none. The sum of the maximum number of vertex cubes is 9 — just like for the Soma Cube, and leading to the same conclusion: One and only one piece must be deficient, and no more than singly deficient. That rules out any positions for b, c, and e that occupy no vertex cubes. Those are doubly deficient positions. Piece e’s one central position is one of these, so e cannot be central.
Piece a’s three possible positions all occupy two face cubes. So the other five pieces between them can occupy only four. Piece e is the only one that can occupy zero face cubes, and the other four must then occupy only one each. Piece b’s only central position occupies two faces, so b cannot be central either.
What you are left with is one central position for pieces c, d, and f; one non central position for pieces a, c, d, e, and f; two central positions for a (one deficient and one non); and two non central positions for b (also one deficient and one non).
Suppose a is central. If it is in its deficient central position then b must be in its non deficient non central position; if a is non deficient and central then b must be deficient and non central. Either way, once we know a is central and whether it’s deficient, we know the positions of all the pieces, and the numbers of vertex, edge, and face cubes add up to the required 8, 12, and 6.
c’s central position and a’s non central position both are deficient, which means c cannot be central.
d’s central position is non deficient, though, so a can be deficient provided b is not. Again, when d is central, all the other pieces’ positions are determined and the number of each kind of cube ends up being correct.
The same reasoning holds for piece f, since it’s the mirror image of d.
Here’s a summary of the four possible position combinations. The sets of four numbers indicate how many vertex, edge, face, and central cubes the pieces occupy:
| Piece | a deficient and central | b deficient a central | a deficient d central | a deficient f central |
| a | [0, 2, 2, 1] | [1, 1, 2, 1] | [0, 3, 2, 0] | [0, 3, 2, 0] |
| b | [2, 2, 1, 0] | [1, 3, 1, 0] | [2, 2, 1, 0] | [2, 2, 1, 0] |
| c | [2, 2, 1, 0] | [2, 2, 1, 0] | [2, 2, 1, 0] | [2, 2, 1, 0] |
| d | [1, 2, 1, 0] | [1, 2, 1, 0] | [1, 1, 1, 1] | [1, 2, 1, 0] |
| e | [2, 2, 0, 0] | [2, 2, 0, 0] | [2, 2, 0, 0] | [2, 2, 0, 0] |
| f | [1, 2, 1, 0] | [1, 2, 1, 0] | [1, 2, 1, 0] | [1, 1, 1, 1] |
Note c and e must both have their three cubes in a row along an edge of the big cube. We’ve eliminated three out of four positions for each, along with two out of four for b, d, and f. A nice reduction in the search space!
As for dexterity, piece e is the only planar piece, so it can — in fact must — lie in a face of the big cube, and can be a forward or backward L. But we can’t say which until we find solutions.
Which I found is… not easy, even knowing the above. There are just two solutions and I found both, but it took a long time. Obviously this means not all the above “possible” position combinations in fact occur. I’ll say which one or ones do at the end of this post, after a spoiler alert.
Some models designed for the Soma Cube are also possible, including the Tower, the Staircase, and the Canal from the Parker Brothers Soma booklet — according to my solver program, anyway. The Tower is easy, but I haven’t solved the other two yet myself. the other two are harder but I’ve solved them.
Files for 3D printing this puzzle are here: https://www.printables.com/model/461220-mikusinski-cube-puzzle
Both solutions have piece a central and piece b deficient. In one, piece e is a forward L (dexterity=1) and in the other it’s backward (dexterity=0).
You can get from one solution to the other via a 3-piece move. With even more abject apologies to Conway and Guy, here is the Mikumap:
Diabolical models
I made an upgrade to ssoma.py; now the input file can have multiple models and the program will attempt to solve them all. I threw a file with a bunch of Soma figures at it to solve them with the Diabolical Cube pieces instead. Not too surprisingly most of them were impossible. The very large pieces #7 and #6 and the awkward #5 make it hard to produce interesting shapes, or at least interesting shapes of the sort you can do with Soma. There were 11 it could solve, though, and turns out so could I.
These all are possible with the Diabolical Cube pieces. (Most are from the Soma Addict.)
Added: Here’s two more.
Diabolical cube counting
My arguments for why pieces #2 and #3 in the Diabolical Cube cannot be central rest on #5 not being deficient, and I think examination of the positions shows that, but it’s not the most satisfactory approach. And I had no argument at all other than exhaustive search for why #3 cannot be in the positions with 1 vertex and 2 edge cubes or 1 edge and 2 face cubes.
Some progress on that front:
If either #2 or #3 is central it must supply at least 1 face cube.
If #4, #6, #7 are non central they must supply 7 vertex cubes and 3 face cubes.
#5 cannot be non deficient since this would bring the vertex cube count up to 9. But it cannot be deficient since in that case it would supply 3 face cubes, bringing the face cube count up to at least 7.
Therefore neither #2 nor #3 can be central.
With #4 central: #4, #6, #7 supply 6 vertex cubes, 4 face cubes. #5 cannot be deficient since it would supply 3 more face cubes. Then #2 and #3 must be deficient (so #3 is not in the 1 vertex, 2 edges position). Each must supply exactly one face cube (1 edge, 2 faces is ruled out for #3) so #5 must be in the position with 2 vertex cubes, 3 edge cubes, and no face cubes. This adds up to the right numbers of vertex, edge, face, and center cubes.
With #6 central all the rest must be non deficient. Then #4, #6, and #7 supply 4 face cubes, and #2 supplies none. #3 must be in the position with 1 vertex, 1 edge, and 1 face cubes, and #5 must supply 2 vertices, 2 edges, 1 face.
With #7 central, #4, #6, and #7 supply 5 face cubes. Either #3 or #5 could supply the remaining face (#3 cannot be in the 1 edge, 2 faces position) and the correct number of edges: there are two combinations for them that would add up.
So cube counting can tell us #2 and #3 cannot be central, and can tell us what position every piece must be in if #4 or #6 is central. But it cannot rule out #5 being central, or #3 being in the 2 edges, 1 face position when #7 is central.
More on the Diabolical solutions
I’ve been categorizing the Diabolical Cube solutions. For the Soma Cube, Conway and Guy characterized solutions three ways:
- Which piece is central? That is, which occupies the central cube? (The big cube is made up of 8 vertex cubes — the ones on the corners; 12 edge cubes — orthogonally in between the vertex cubes; 6 face cubes — in the centers of the big cube’s faces; and the central cube — the one in the center of the big cube.)
- Which piece is deficient? There are 8 vertex cubes and between them the 7 Soma pieces can fill a maximum of 9 vertices, so one and only one piece must be filling one less vertex than it could; that’s the deficient piece.
- What’s the dexterity? Pieces 1, 2, and 4 can look like an L, an L, or a Z if on a face of the big cube, or like a backward L or a backward Z. Add the numbers of the pieces that are on faces and look like forward letters and that’s the dexterity.
Knowing the central and deficient pieces and the dexterity tells you the position of every piece, up to rotation, with respect to the big cube.
For the Diabolical Cube things are a little different.
Since every piece is planar, it follows that the central piece must be deficient. The six pieces can provide up to 11 vertex cubes, so more than one piece can be deficient. But pieces 6 and 7 must provide either 0 or 3 vertices, so if either is deficient, no other piece can be.
Piece 5 can provide 0 or 2 vertices. If it’s 0 that means the cube-void in the middle — I call it the notch — must be occupied by piece 2, 3, or 4. It can’t be 6 or 7 unless that piece is central, hence deficient, and then 5 can’t be deficient. But if it’s 2 or 3 then pieces 6 and 7 can’t fit around it and still leave room for piece 4, and if it’s 4 then pieces 6 and 7 can’t fit. So piece 5 cannot be deficient, and if it can’t be deficient it can’t be central.
If piece 4 is not central it’s not deficient, so if piece 2 or 3 is central, pieces 4, 6, and 7 are contributing a total of 7 vertices. Then piece 5 has to be deficient, but it can’t be. So pieces 2 and 3 cannot be central.
If piece 4 is central it’s deficient, and pieces 6 and 7 are contributing 6 vertices. 5 can’t be deficient, so 2 and 3 both must be.
The conclusion is we can have only the following:
- 4 central, with 2, 3, 4 deficient
- 6 central and deficient
- 7 central and deficient
The deficient pieces are determined by the central piece.
As for dexterity, any of the six pieces can appear on a face but only two can be forward or backward: 3 and 7. Whether 3 is forward or backward is ambiguous if it’s occupying two edge cubes. In the Soma cube the piece of the same shape can’t do that. In the Diabolical cube it can, though it can’t occupy a vertex and two edges — for reasons I can’t find a simple proof of, but that position doesn’t occur in any solution. If it occupies a face and two edges, it’s deficient. If it’s occupying one edge, one face, and one vertex, it’s not deficient, and the cube can be rotated to put that edge at the bottom of the front face; then the piece is either a forward L or backward. Piece 7 is unambiguous, it’s either a forward L (with a bold width upright) or not. If we say piece 3’s dexterity contribution is 0 if it’s deficient or backward and 1 if it’s forward, and piece 7’s is 0 if it’s central or backward and 2 if it’s forward, then we can add them and get a dexterity in the range 0 to 3 — in principle. In fact there are no solutions where both pieces are forward, so the dexterity is 0 to 2. Nor are there solutions where both are deficient, so the mirror image solution always has a different dexterity.
So far we know piece 2 has only two positions, corresponding to its deficiency; so do 4 and 6; piece 3 has three, corresponding to its deficiency and dexterity; so does 7.
Piece 5 can’t be deficient and has no dexterity, but it still has two possible positions: One where the notch is an edge cube, and one where it’s a face cube. It might seem we need another parameter if we want to encode 5’s position. But it turns out the notch is an edge cube if and only if 6 is central.
In summary, if we know the central piece and the dexterity, we know the position of every piece.
And that means we’re ready for — with apologies to Conway and Guy — the Diabolimap!
Like the Somap, this doesn’t tell you how to build any solution, but it does tell you how to get from any solution to any other once you find your place on the map.
The solution designators are of the form CDx where C is the central piece (4, 6 or 7), D is the dexterity (0, 1, or 2), and x is a lowercase letter to distinguish solutions with the same CD.
There are 13 solutions up to rotations and reflections. The 42x solutions under reflection become 40x; the 70x become 71x; the 61x become 62x. The map shows only the 42x, 70x, and 61x solutions, plus one 71x solution; the rest of the reflected solutions can be imagined as a second half of the map. a mirror image of the first.
Solid lines connect solutions that are related by two-piece moves: remove the two indicated pieces from one solution and put them back in different positions and you get the other. Unlike with Soma, you can’t get from solutions with one central piece to solutions with another using 2- or 3-piece moves; you need one of the 4-piece moves shown. (Other 4-piece moves not affecting which piece is central are not shown. And no, unlike Conway and Guy, I did not do this by hand, I used a computer to find them.) But there are numerous ways to get from this side of the map to the mirror image side with 3-piece moves. One such move, from 70b to 71d, is shown. You can look for others, but this is enough to allow you to get from any of the 26 solutions to any other via a sequence of 2-, 3-, and 4-piece moves.
Diabolical Cube
I printed a Diabolical Cube, a sort of 19th century precursor to the Soma Cube. The earliest known reference is an 1893 book by “Professor Hoffmann”. There are six polycube pieces composed of from two to seven small cubes, and you have to put them together to form a 3x3x3 cube.
It’s easier to find a solution than for the Soma Cube. There are 13 solutions in all (up to rotations and reflections). I found all of them. I did try to apply some of the Soma Cube tricks — using checkerboard coloring and vertex cube counting, you can rule out some of the positions for some of the Soma pieces. But that’s because Soma is more tightly constrained by those things; you need to fill eight vertex cube positions, for instance, but the sum of the maximum number of vertex cubes for each piece is only nine, so one and only one piece must be “deficient” by one. The Diabolical Cube pieces’ maximum number of vertex cubes, on the other hand, sums to eleven.
I did work out that if piece #2 or #3 (named after the number of cubes in each) is central — occupying the cube at the center of the big cube — then #5 has to be doubly deficient, and that means its awkward notch must straddle the central cube. But then even if there’s a place to put #7 and #6, you’re left with no place to put #4. Similarly with #4 central if #5 is doubly deficient, but in that case #2 and #3 can be deficient instead. If either #6 or #7 is central then none of the other pieces can be deficient.
With #7 central, #6 can go perpendicular or parallel to it. If perpendicular you quickly find four solutions, in pairs linked by a swap of two pieces. If #6 is parallel you end up unable to place #4 and #5.
With #7 not central and #6 perpendicular there’s one pair of solutions, again with a two piece swap linking them.
Then if #6 is parallel and central, there are three solutions linked by 2-piece swaps, plus another pair.
Finally if #6 is parallel and noncentral, there is one more solution pair.
And that’s it for the cube. You can try for other figures too, but the W Wall is still impossible…
You can download my model files here: https://www.printables.com/model/449349-diabolical-cube-puzzle
MathematRec 