by Jacques Haubrich
Originally published in "Games and Puzzles Journal" Issue 44
Once upon a time, long before the word 'pentomino' and other -omino words were invented, these puzzle pieces were referred to as 3-pieces, 4-pieces, 5-pieces, 6-pieces etcetera and they were numbered sequentially, starting with the monomino as number 1, the domino as number 2, and so on. In those days, for example the I pentomino was known as piece 10. We are talking of the years around World War II and especially of the British journal named The Fairy Chess Review. This fine old journal is a real treasure trove on polyomino problems (among others). Unfortunately, issues of this journal are hard to find and the way in which problems and solutions are described is a real puzzle in itself. Fortunately enough, George P. Jelliss in 1989 deciphered virtually everything and sent copies of his results, in hand-drawn form, to various correspondents .
This article is about one special class of polyomino problems that has been highly under-regarded over the years: the covering of the surface of a given 3D object by 2D polyominoes.
The first problems of this kind that I could find, were problems 6841, 6842 and 6843 in The Fairy Chess Review of June 1946 , , . Problem 6841 was to cover the surface of a 2×2×7 box (with one 2×2 end open) with the 12 5-pieces, overlapping all edges. The problem was posed and one solution (in fact over 1 million are now known to be possible) was given by R. J. French, which looked like this:
fig. 1 - Solution of FCR-problem 6841 by R.J. French
Of course, this particular surface looks a lot less interesting than the surface used in the next two problems: a 3×3×3 cube with 9 squares on each face, 54 in total, to be covered with one tetromino and 10 pentominoes. In problem 6842, the pieces could be folded around the edges without restrictions; in problem 6843, the covering should also cover piece-46, which is one particular hexomino. French's solution as published in FCR August 1946 looked like this:
fig. 2 - Solution of FCR-problem 6842 by R.J. French
fig. 3 - Solution of FCR-problem 6843 by R.J. French
It should be obvious that the 16,308 possible solutions for problem 6843 are just a subset of the millions of solutions for problem 6842.
So far, the polyominoes are placed in a quite straightforward way since the underlying grid on the cube is parallel with the edges of the cube. But suddenly, in October 1946, the same R. J. French poses problem 7002 in FCR  based on a new idea: rotate the grid with respect to the edges of the cube in a 1:2 direction. The resulting grid has 24 squares to be covered, leaving some pyramid-like areas around each corner uncovered (shown in light yellow):
fig. 4 - R.J. French's 24 squares grid on a cube.
Covering 24 squares takes one tetromino and 4 pentominoes and somehow, French must have felt that there are lots of solutions for this (actually, there are 72,871), so French included a second novelty in his problem 7002: to cover not just one such cube, but two of them simultaneously, using 2 different tetrominoes and 8 different pentominoes. In the original problem 7002, the editor forgot to mention which tetrominoes should be used, but the solution, two months later in December 1946, clearly showed his intentions:
fig. 5 - R.J. French's solution for covering two cubes simultaneously.
There was apparently a lot of confusion around this problem: when the problem was first posed (FCR October 1946 page 56), it said that pieces 11, 12, 13 and 15 should be used for one cube and pieces 16, 17, 18 and 20 for the other cube:
fig. 6 - R.J. French's original problem 7002 in FCR.
Translated to modern terminology, one cube should be covered with the L, Y, P and N and the second cube with pieces V, W, T and Z. The tetrominoes were completely forgotten. In the solutions, two moths later, the required tetrominoes were mentioned: pieces 6 and 7, or in modern terms the tetrominoes l and t. Let us now look at the solution itself:
fig. 7 - R.J. French's solution for FCR problem 7002.
And we notice immediately that the solution shows pentomino X instead of pentomino L that was originally mentioned. Quite a difference! And apart from that, pentomino X is completely useless for covering this cube since one of its squares should do something impossible: covering a corner of the cube. It turns out that if we move that particular square of the X one unit down, changing piece X into piece F, the solution suddenly becomes a correct solution. However, piece F was originally denoted as piece 19, which was not mentioned at all. And to complete the confusion, the original set of pieces l, L, Y, P and N also do cover the cube, so a hypothesis might be that French originally found such a solution, which was then lost before the December issue of FCR was assembled, so he had to enter a new solution. The stress and time frame could then have led to the double mistakes.
It is unclear if French ever considered covering not two, but three such cubes simultaneously, using 3 tetrominoes and all 12 pentominoes. Maybe he did and discovered that this would be impossible: it is not really hard to see that pentominoes I and X are useless for covering this particular cube. Also, the only way in which pentomino U could be used is quite interesting, since it has to be folded around a corner so that two of its squares touch each other at their corners. All together, there are 369 different solutions for covering this cube with one tetromino and four pentominoes (of course not counting rotations of these solutions), and this result comes from 239 different sets of 5-pieces, each having from 1 to a maximum of 8 different solutions. Simultaneously covering two cubes with 5+5 different pieces can be done using 1267 different sets of 5+5 pieces and in 695 of these sets, both cubes can only be covered in one unique way. French's original problem would have had 2×2 = 4 different solutions whereas the corrected version (X becomes F) of the published left-most solution has 4 different solutions.
The next step follows in February 1947 when F. Hansson publishes problem 7124 in The Fairy Chess Review : using the twelve pentominoes, cover the outside of two root-five cubes simultaneously:
fig. 8 - F. Hansson's problem 7124.
The mentioning of problem 7002 is quite confusing since in that problem, the cube was not completely covered and one tetromino plus 4 pentominoes are used for each cube. This time, each cube must be completely covered using exactly 6 pentominoes. The first step is clearly thinking carefully about the grid that is to be used. The words 'root-five edge cubes' are a great hint since according to Pythagoras, the square root of five can only be created as the hypotenuse of a triangle with sides 1 and 2 at a square angle. Contrary to problem 7002, the grid lines pass through the corners of the cube, which is the only way to get the cube entirely covered.
fig. 9 - F. Hansson's solution to problem 7124.
At this moment, it is left to the reader to decipher the original solution. Hansson's problem will be analysed in depth in a later section.
The final stages of covering cubes are found in problems 7590 and 7591, published in the February 1948 issue of The Fairy Chess Review , with the solutions published in the April 1948 issue. H.D. Benjamin posed and solved both problems:
fig. 10 - H.D. Benjamin's problems 7590 and 7591.
fig. 11 - H.D. Benjamin's solution for problem 7590.
fig. 12 - H.D. Benjamin's solution for problem 7591.
fig. 13 - A drawing for the solution of problem 7591.
And finally, problem 8848 in The Fairy Chess Review of October 1950  was about covering a rectangular block with edges √8 × √8 × √32, hence with a total surface of 80, using the 5 tetrominoes and the 12 pentominoes together. It was again H.D. Benjamin who posed the problem and gave a solution in the February 1951 issue.
fig. 14 - A drawing for the solution of problem 8848.
So far the ancient history on cube coverings, since after the publication in FCR of the solution for problem 8848 in February 1951, there is hardly anyone noticing that cubes or other solid objects can be covered by polyominoes. The rare exceptions are W. Stead (giving another solution for problem 7591 originating from an unknown solver ), and then Martin Gardner , Solomon Golomb  and George E. Martin , all three doing hardly more than just presenting Stead's solution as a new problem.
Then we suddenly find 4 new articles on covering cubes with hexominoes, all in CFF-30, December 1992, three by Pieter Torbijn  and one by Anton Hanegraaf . Many different cubes are covered by all kinds of subsets of the 35 hexominoes. From a remark in one of these articles, it seems likely that Torbijn and Hanegraaf became triggered by the 1989 republication of the FCR results by G.P. Jelliss .
In 1997, C.J. Bouwkamp starts working on several covering problems. Most of his work and results were published by him in , ,  and , and both his published as well as some unpublished results are integrated in the body of this article. And finally, the history ends in 2002 and 2003 with three publications in CFF by Pieter Torbijn ,  and .
So why this study? There is a combination of reasons for this. One is that several earlier publications are barely accessible; another is that all these publications only cover one small specific aspect of a larger more general problem; there were also unpublished results by Bouwkamp and mistakes in some publications, like ,  and . And an overview of the entire field had never been published at all. All together, this was more than enough reason for me to check earlier investigations by my own computer programs and attempting to give as complete as possible an overview of as many as possible kinds of coverings. As with many other types of polyomino problems, the subject is open-ended, but I do believe that this article at least covers (almost?) all coverings of elementary 3D surfaces with fairly limited numbers of all kinds of simple polyominoes.
Problem 1: Internal edge or 'nice'
The first problem becomes apparent when looking at figure 15, where we see a pentomino on a 3×3×3 cube.
fig. 15. A pentomino on a 3×3×3 cube.
In this example, the pentomino is somehow folded around corner F of the cube, and as a result, one square of the pentomino now touches another of its squares by the edges. Which pentomino is shown in figure 15? Well, it could be piece L, piece N, and also piece Y, depending on how the covering is cut open: along FB, along FE, or along FG respectively. For obvious reasons, a covering in which this happens is not really 'nice': the internal edge is causing confusion. For this reason, we define a covering of any surface to be 'nice' if there are no internal edges. With a nice covering, it is no problem at all to identify which piece is covering any given part of the surface.
Problem 2: Multiple coverings
The second problem is that quite often, the same piece can cover the same area in 2 (or even 3, as happens with one particular hexomino: p in fig. 34) different ways. It is easily seen that the P-pentomino, when folded around a corner, always can cover its own area by turning it upside down. Another example is the Y-pentomino when folded around a corner. Throughout this article, these multiple coverings of the same area by the same piece are not counted different. One reason is that this is a lot easier for programming the computer; the other reason is that the numbers of solutions would be many times larger than they already are.
Problem 3: Counting solutions
The third quite fundamental problem appears when counting solutions: when are two coverings considered the same or different? In too many publications, this problem is not recognised and the given numbers of solutions cannot be interpreted. A cube can fill its own space in 24 different ways, just by rotating the cube in all possible ways. Do we count these 24 ways as different? That depends. Once the faces of the cube are painted in different colours, or decorated in any other way, all 24 different positions at least look different and can be identified. The same holds for a die, though in most games, only 6 of the 24 positions are considered different. For a gambler, the die has 6 different ways to present itself but for a mathematician, it is still the same die which is only rotated. When covering a cube with polyominoes, the decorated cube can present itself in 24 different ways, but most people prefer to see this as just one solution for the problem of covering. However, in many cases, a polyomino-covered cube does not look different in all these 24 possible rotations; there may be some form of symmetry in the covering, resulting in only 2, or 3, or 4, 6, 8 or 12 different views. In this article, detailed information is presented so that the reader has the option to look at his or her preferred way of counting. Finally, once a covering is found, the mirror-image of that covering is a free extra solution. Some people prefer not to count that as a different solution, but sometimes it happens that the mirror image is exactly the same as the original, though sometimes only after a suitable rotation. Again, this article will tell how many solutions are unchanged after a reflection and how many mirror-image pairs are possible.
There are endless numbers of solids that may be completely covered by polyominoes without overlaps and (preferably) without gaps, or even by other forms of polyshapes. In this article, basically only cubes and double-cubes (1×1×2 rectangular blocks) are considered, but two other solids are added to the list. Throughout this study, cubes are denoted by the letter C and blocks by the letter B, in both cases followed by the number of squares to be covered.
It is obvious that for a complete covering, each corner of the solid must always be on a gridline of the covering, but the direction of the covering is not necessarily parallel to the edges of the solid.
The solids that are discussed in this article are ordered by the number of squares on their surfaces. They are: B10, C12, B16, B20, P20, C24, C24R, C30, B40, B50, C54R, C60 and in that order, they are discussed in the following subsections.
B10 is a 1×1×2 block, or two unit cubes joined together. After C6, the elementary cube for which the possible coverings are well-know and easily found, this is the next smallest object and the first one that is still interesting enough. The surface of 10 squares asks for being covered by 2 pentominoes. To do so, we have a choice of using either 2 different pentominoes, or twice the same pentomino.
For the last case, the following table lists the results:
|different views||self |
Since this is the first table of this kind, ample explanation is due. The first column mentions all possible pentominoes that might be used to cover the cube, in this case two of each are required. With pentomino F, there are 21 different coverings of B10. The heading 'sols bruto' means that this is the number of solutions when different looking rotations are counting as different. The next column, headed 'sols netto' gives the number of solutions when rotations of the covered object are no longer counted as different. For the F pentomino, there are 6 different ways to cover B10. Then follow four columns about 'different views'. The first of these columns, headed by the number 1, tells the reader that among the net total of 6 solutions, there is one that has only 1 different view, meaning that whatever rotation is applied to the covered object, it always looks exactly the same. And then, there are 5 more solutions where each have 4 different appearances, resulting in a count of 20 in the bruto number of solutions. None of the solutions with twice piece F appear in only 2 different views and also none appears in exactly 8 different views. And finally, 2 of the 6 solutions are self-symmetric, meaning that when viewed in a mirror, the pattern looks unchanged (maybe after a suitable rotation). This implies that the remaining 4 solutions can be seen as 2 pairs of mirror-image solutions and some people even prefer to say that the F pentomino yields only 4 different solutions. It is a matter of what one prefers to count as different, as discussed in 'Problem 3' above.
Since B10 is a relatively small object, pentominoes must be
folded over the edges of the block and in many cases they also fold
around a corner, resulting in a solution that is not nice. Nice
solutions, solutions without internal edges, are very rare in this case:
• there is only 1 with piece I and that one is its own mirror image;
• there is 1 with piece T, also its own mirror image;
• and there is 1 pair of mirror-image solutions with piece V.
All 34 other solutions show internal edges. Easily found are those solutions where each pentomino covers one end of the block, or in other words covers 5 faces of one of the two cubes that are glued together to form this block. The remaining solutions are not too hard to find.
Block B10 can also be covered using 2 different pentominoes. In this case, we find 442 solutions which number reduces to 71 after elimination of rotational invariances. Among these 71 are 31 solutions that are their own mirror image and the remaining 40 hence form 20 mirror-image pairs. No less than 21 of the 31 self-symmetric solutions are formed by one piece covering one half of the block and the other piece covering the other half. It is easily seen that pentominoes F, L, T, W, X, Y and Z can be folded this way and a simple calculation of (7C2) = 21 confirms that there should be 21 such pairs.
Among the net number of 71 solutions, we find 21 solutions that have only 2 different appearances, and 50 that have exactly 8 different appearances.
There is only one nice solution, formed by the pieces I and U. This solution may appear in 8 differently looking rotations. It is also the only solution where piece I is one of the two pentominoes.
The frequency in which each of the pentominoes appears in solutions is shown in the following table:
If the edges of a cube have length √2 then the cube has a total surface of 12. With a grid rotated 45 degrees with respect to the edges of the cube, 12 squares appear, each of them with an edge of the cube as a diagonal.
C12 could be covered by 4 trominoes, by 3 tetrominoes, by 2 hexominoes or by 1 dodeca-omino. The last case was never attacked.
Covering C12 by 4 trominoes could be done using 4 identical trominoes and also using 4 different trominoes. Using 4 identical trominoes, we obtain (before eliminating rotations) 18 solutions with the straight tromino, and 24 with the L-shaped tromino. Table 3 lists the results in detail:
All solutions with piece I are nice and one of them is its own mirror image; only one of the L-solutions is nice and that one is also its own mirror image.
The number 1 with the I tromino in the column with heading '12' means that there is 1 (netto = modulo rotations) solution with piece I that has 12 different views. Since 24:12 = 2, the pattern has obviously a 2-fold symmetry (180 degrees rotational symmetry). The other numbers have similar meanings. The 5 (of 8 in total) solutions that remain when only 1 of each mirror-image pair is shown in figure 16, using colours for the different pieces. It is not too difficult to find these solutions without the help of a computer.
fig. 16 - Some solutions for C12 with 4 identical trominoes
Covering C12 with 3 tetrominoes is more difficult. The 5 tetrominoes are named according to their shape I, L, N, S and T, where 'S' stands for 'square'. Fig. 4 shows them all:
fig. 17 - The five tetrominoes and their 'names'
Using 3 identical tetrominoes, we obtain the following results:
It is impossible to cover C12 with 3 pieces I or with 3 pieces S. Some solutions (column 24) have no rotational symmetry, and all other solutions have a 3-fold symmetry so that rotation over 120 degrees around a body diagonal shows the same covering. The last column ('mip') in table 4 shows the number of mirror-image pairs of solutions.
fig. 18 - Some coverings for C12 with 4 tetrominoes
Figure 18 shows 1 covering for each of the 5 mirror-image pairs. Very peculiar because the drawings for pieces N and T are identical. This is caused by the fact that these two pieces can each be folded around 2 corners of the cube so that they cover exactly the same squares! Piece N can even be folded in two different ways and still cover the same 4 squares. This is what happens when a covering is not nice!
And then, the same C12 cube cannot only be covered by 3 identical tetrominoes, but also by 3 different tetrominoes, though only by 7 of the 10 possible triplets. Before removing rotations, there are 624 solutions so there are 624/24 = 26 different coverings modulo rotations. None of these is its own mirror image, so there are 13 mirror-image pairs; 11 of these show 2 internal edges, 2 of them even have 3 internal edges. The triplets that do not cover C12 are ILS, INS and IST. The triplets LNS, LST and NST have a common property in that each has only 1 solution (plus the mirror image) whereas triplet LNT is the most productive with 4 different solutions (plus their mirror images).
Finally, C12 could be covered using 2 hexominoes. This was investigated in 1992 by Anton Hanegraaf . Using two copies of the same hexomino, there are 308 solutions which number reduces to 23 after eliminating rotations and reflections. There are two different coverings for a pair of one particular hexomino, which was not detected by Hanegraaf, so in his publication, he finds only 22 solutions. That particular piece is piece b in the following list:
fig. 19 - The 35 hexominoes
The hexominoes are shown in the same order in which they were arranged by T.R. Dawson and W.E. Lester, as published in The Fairy Chess Review, April 1937. and since they do not have widely used designations, I used the numbers 1-9 and letters a-z to refer to each. Apart from the two different solutions for piece b, there is one solution for a pair of each of the following pieces:
2, 3, 6, 7, 8, 9, c, d, e, g, h, i, j, k, l, m, n, p, t, u and v and two of the remaining 13 pieces do not cover C12, which is quite obvious for the pieces 1 and a. All solutions have 2 or 4 internal edges.
Using two different hexominoes, then there are 3900 solutions which reduces to 115 after elimination of rotations and reflections. Pieces 2, 9, d and u only appear in 2 of the 115 solutions and piece i is the winner (in this respect) since it appears in 17 different solutions.
B16 is a 1×1×2 block, or 4 cubes glued together. The grid is parallel to the edges, so there are 4 squares on top and on the bottom, and 2 on each of the four sides. Since it has a surface of 16, a set of 4 tetrominoes looks like an obvious set to cover the surface.
Again, figure 17 shows the five tetrominoes and the letter used to identify each of them, as used throughout this study.
Using 4 copies of the same tetromino, the following table shows the numbers of solutions, including how many appear in how many different rotations and also how many nice solutions were found amidst the netto set of solutions:
Using 4 different tetrominoes, there are 752 solutions, or 47 modulo rotations and reflections. There is only one nice solution, using the pieces I, L, S and T. That solution, and one for each other set of 4 out of the 5 available pieces, is shown in Figure 20:
fig. 20 - B16 covered with 4 different tetrominoes in 5 different ways
Reading the solutions is far from easy, since most pieces look quite distorted, even more when they are folded around a corner in a non-nice way. For that reason, each piece is in a fixed colour: green for I, yellow for L, red for N, cyan for S, and magenta for T. Besides, the missing piece is mentioned with each drawing and so is the number of internal edges.
Block B20 is a kind of duplication of cube C12. With its surface of 20 squares, it invites us to cover it with either 5 tetrominoes or 4 pentominoes.
fig. 21 - Block B20
5 tetrominoes on B20
Covering B20 with 5 identical tetrominoes (see fig. 17) is only possible using piece L, N or T. With piece L, there are 56 solutions, there is only 1 with piece N and T yields 2 solutions. None of these 59 solutions shows any symmetry.
Among the 56 with piece L are the only 4 nice solutions. Figure 22 shows one of these four, also the N-solution and one of the two solutions with piece T.
fig. 22 - 5 tetrominoes covering B20
Block B20 can also be covered with (the) 5 different tetrominoes. The 1136 solutions each appear in 8 different rotations, resulting in 142 solutions modulo rotation. These 142 in turn form 71 mirror-image pairs. Only one of these 71 is nice (also shown in figure 22); all others have 1, 2, 3 or even 4 internal edges.
4 identical pentominoes on B20
This block can be covered using 4 pentominoes. Though this was completely investigated in the mid 1990s by C.J. Bouwkamp, I could not find any publication of his results. Covering B20 with 4 identical pentominoes can be done in 340 different ways, which number reduces to 96 after elimination of rotations. Among these 96, 22 are their own mirror image, leaving 37 mirror-image pairs of solutions. Only 13 of the 96 solutions are nice, 5 of them are their own mirror image and the remaining 8 form 4 pairs. Table 4 lists all the details and figure 23 shows some solutions.
fig. 23 - Identical pentominoes on B20
As can be seen in table 6, only 10 of the 96 solutions (column '8') have no internal rotational symmetry and in contrast to this, there are 2 mirror-image pairs solutions (one with piece I and one with Z) where rotations have no effect at all on the visible pattern. There are 3 solutions, all three with piece L, where the covering shows exactly 1 internal edge. In all other cases, there are 0, 2 or 4 internal edges. Figure 23 shows two nice coverings, one with 4 copies of piece V and one with piece L.
4 different pentominoes on B20
Covering B20 with 4 different pentominoes is another nice problem. Since there are 12 pentominoes, there are (12C4) = 495 sets of 4 different pentominoes. In 446 of these cases, there is at least one possible covering; the remaining 49 sets do not cover block B20. These 495 sets generate 65,072 solutions which number reduces to 4,071 after elimination of rotations and reflections. This is not exactly divided by 8 for rotations and 2 for reflections! The reason for this is that 64 of the 65,072 cases have an internal reflection symmetry. These 64 reduce to 8 cases and the remaining 65,008 reduce to 4,063 cases.
Among all these 4,071 coverings, only 2 are nice, consisting of the pieces LVXZ in one case and LNYZ in the other case. The set LVXZ has two other solutions (with at least 1 internal edge) and LNYZ even has 18 other solutions. Some pieces are more useful in covering B20 than others. Piece X appears in only 507 of the 4071 solutions whereas piece P appears in 1986 cases.
Simultaneously covering three B20 blocks with 12 different pentominoes
Since almost every pentomino quadruplet has at least one solution for covering B20, it is quite easy to split the set of 12 pentominoes into three quadruplets and use each set of four to cover a B20 block. There are a staggering 858,088 solutions for simultaneously covering three B20 blocks and there is no nice simultaneous covering. Using FTVZ-ILNY-PUWX, there are 14×13×7 = 1274 simultaneous coverings. At the other end of the scale, we find 2 sets of quadruplets, namely FTWX-ILPY-NUVZ and FUVZ-ILPY-NTWX each with only 1 simultaneous solution.
Figure 24 shows one of the simultaneous coverings with a unique solution and also one of the two nice coverings:
fig. 24 - Simultaneous coverings of three B20 blocks
P20 is a special case within this study. It is nothing else than a P-pentomino made out of 5 cubes, usually called the P-pentacube. The total surface of this particular object is 20 and it happens to be the only pentacube with a surface of 20. That surface could be covered by 5 tetrominoes (not investigated) or by 4 pentominoes. Using four identical pentominoes, my computer found 104 different solutions (reflections not counting different), none of them consisting of the piece I, U or X. Only 1 of these 104 solutions is nice, and this happens when 4 copies of piece V are used. The other extreme is a solution with 9 internal edges and exactly that same pattern emerges with 3 different pieces: L, N and Y.
fig. 25 - Four special solutions for covering P20
C24R is a cube with 24 squares on its surface. The letter R means 'reduced' since this is the covering invented by R. J. French as mentioned in the introduction and shown in figure 4, now again in figure 26; the corners of the cube remain uncovered.
fig. 26 - A three-dimensional view of C24R
Covering the squares on this surface could be done using one tetromino and four pentominoes. After removal of rotations, 369 solutions remain and they are the result of 239 different sets of 5 pieces. In most cases, namely in 166 cases, the set of 5 pieces covers this cube in a unique way. Then there are 50 sets that each give 2 solutions, 5 sets that each give 3 solutions, 9 with 4 solutions, 6 with 5, 2 with 7 and 1 with 8 different solutions. This last set is composed of the L-tetromino and the pentominoes F, N, P and U.
In R. J. French's original problem, two of these cubes should be covered using 10 different pieces. Doing so is not a very hard problem since there are 1267 different sets that can do so, and still 695 sets that do so in a unique way for both cubes. Also, there is one set that yields 2×7 = 14 different coverings for a pair of cubes. Figure 25 shows one of the unique solutions for the double covering and one of the 14 solutions for the sets of pieces T+PUVZ and Y+LTWY.
fig. 27 - Two solutions for two simultaneous C24R cubes
C24R can also be covered using 4 hexominoes
Using four different hexominoes, there are 206 solutions (modulo rotations). In some solutions, piece h folds around a corner, forming a closed ring, which is a very nice way to be not nice. Pieces 1, 2, 3, 4, 8, a, g, n, y do not appear in any solution, which is hardly amazing since none of these would fit anywhere on the grid.
Using four copies of the same hexomino, the number of 68 solutions reduces to only 13 solutions after removing rotations, and solutions appear only with pieces 5, 6, 9, b, c, e, h, j, o, p and w, usually only one solution for each, but there are two different solutions with pieces b and e. Since each solutions involves only 4 relatively large pieces, these solutions are quite easily found. The solutions with pieces c and x are special since they both show a threefold symmetry and the solution with piece h even has a 2-fold symmetry.
C24 is a cube with 4 squares on each face, parallel to the edges of the cube. Figure 28 shows what it looks like:
fig. 28 - A view of C24
The grid on C24 is very straightforward and its 24 squares could be covered by either 6 tetrominoes, or 1 tetromino and 4 pentominoes, or by 4 hexominoes. Despite all this, I could not find any publication about coverings of C24. As usual in all computer-assisted research, there is never 100% certainty that the results are complete and correct, so I am looking forward to anyone who is interested in writing his or her own programs to check my results, not only for C24 but also for all other cases that I investigated.
Covering C24 with 6 tetrominoes
Using 6 copies of the same tetromino, the results are shown in the following table 7:
When counting nice solutions only, the numbers reduce to:
Just in case anyone thinks that I forgot the covering of C24 using 6 different tetrominoes: there were only 5 different tetrominoes in my box with polyominoes.
Covering C24 with 1 tetromino and 4 pentominoes
After reduction for rotations and reflections, there are 54,503 different solutions using 1 tetromino and 4 different pentominoes. Only 689 of these are nice. Picking 4 pentominoes from the 12 available ones can be done in (12C4) = 495 different ways. All of these except for 6 sets appear with at least one of the 5 possible tetrominoes. Limiting ourselves to nice solutions only, there are 164 different sets of 4 pentominoes that can be used.
It is even possible to cover three C24 cubes simultaneously,
using 3 different tetrominoes and all 12 different pentominoes. And not
just possible, there are zillions such sets and there are still no less
than 291 sets of 3 times a tetromino and 4 pentominoes that cover
simultaneously three C24 cubes with the following restrictions:
- each of the three set of 5 pieces has only one unique solution to cover C24;
- in 2 of the 3 cases the solution is nice;
- in the third case, it is only the X-pentomino that unavoidably has an internal edge.
One such triple solution is shown in figure 29:
fig. 29 - Simultaneous covering of three C24 cubes with maximum restrictions.
Covering C24 with 4 identical hexominoes
The gross number of 1946 solutions reduces to only 138 different solutions after removing all rotations and reflections. Further details are given in the following table:
|number of views|
It should be clear from this table that only hexominoes d and x (4 copies of each) cannot cover this cube. Hexomino p is the only one that has a covering with only 4 different views (though with 4 internal edges) and 16 different views is limited to pieces p (again with 4 internal edges). Pieces l and m each can cover the cube in such a way that all 48 possible views (including mirror images) look different (with 4 and 3 internal edges respectively). So these four solutions are shown in figure 30, along with one nice solution for piece 8:
fig. 30 - Coverings of C24 with four identical hexominoes
Covering C24 with four different hexominoes
C24 can also be covered using 4 different hexominoes. In this case, after removal of rotations, there are 77 solutions that are their own mirror image and 67276 mirror-image pairs. Only 218 mirror image pairs are completely nice, all other solutions have from 1 to 8 corners with internal edges. One example of a nice solution is easily found using the pieces 1-2-5-6. The same four pieces can also cover C24 with 3 internal edges, and that solution is even easier found than the one without internal edges.
A maximum of 8 cubes can be covered simultaneously using 32 different pieces. If only nice solutions are allowed, then no more than 6 cubes can be covered simultaneously. Sets that give simultaneous unique nice solutions are -125a-3jmr-48bi-6cdx-9egh-fluw-.
The grid on cube C30 is oriented in a 1:2 direction. This can obviously be done in 2 different ways that are each others mirror image. Since both cases have exactly the same solutions, only one of these cases needs to be investigated. A total of 30 squares asks for being covered either by 6 pentominoes, or by 5 hexominoes. These can again be identical pieces or different pieces. In the case of 6 different pentominoes, a further step is looking for two C30 cubes simultaneously covered by all 12 pentominoes.
Covering C30 with 6 identical pentominoes
There are 1821 ways to cover C30 with 6 identical pentominoes. In 219 of these cases, the covering is nice. These numbers reduce to 137 and 27 respectively when counting modulo rotations.
Interesting are maybe the two solutions with piece X, the nice one easily found without computer. The other X-solution has 6 internal edges. Piece Z is the only one with a solution that has only 2 different appearances (rotations), so this solution has a lot of rotational symmetry, both around axes parallel to the cube's edges as well as around diagonal axes. Diagonal rotational axes are quite rare: column '8' only counts 8 of such coverings. Maybe slightly unexpected is the fact that 5 copies of the I-pentomino cannot cover the C30 cube.
Covering C30 with 6 different pentominoes
This problem was earlier investigated by C. J. Bouwkamp and some of his results were published in  and . Later, Torbijn and Götz investigated the same problem again and published their findings in ,  and , unfortunately in the two last cases with some incorrect numbers of solutions. My own findings are the same as Bouwkamp's, who was renowned for his extreme accuracy, so it seems that these are indeed correct results. Choosing 6 different pentominoes from the set of 12 can be done in (12C6) = 924 ways. Each of these 924 sets of six pentominoes can cover the C30 cube, though in a vastly different number of ways. Sets with the least number of solutions are ITUVWX (2 solutions), ITUVXZ (2 solutions) and ILUVXZ (3 solutions) while FLNPWY has 1780 solutions and FLNPVY even 1794 solutions. If we limit ourselves to the 3090 nice coverings only, then there are still 160 sets of 6 pentominoes with only 1 (nice) covering of which set FLNPUY holds the record with 56 nice coverings. Remarkably, the winner in nice solutions only is not the same set as the winner in all solutions.
The gross number of solutions is 4,304,064. After eliminating 23 of every 24 rotations, 179,336 coverings remain. Less than 2% of these are nice.
The following table shows how many solutions have how many internal edges:
|nr of internal edges||0||1||2||3||4||5||6|
|nr of solutions||3090||14047||32531||45631||44246||31190||8601|
In how many solutions appear each of the 12 pentominoes? With 179,336 solutions, each containing 6 pentominoes, each pentomino should appear in an average of 89,668 solutions. Only piece W more or less fulfils this prophecy. The topper is the P pentomino which manages to be part of almost 120,000 solutions and the most hideous one is the X which is found in only 37,322 solutions, or hardly more than 20% of all solutions instead of the expected 50%. The table shows the frequency for each piece and also the frequency in appearing in nice coverings:
Covering two C30 cubes simultaneously
Simultaneously covering two C30 cubes was first mentioned in 1947 by F. Hansson in . He also gave a solution . The way in which his solution was coded, is a kind of puzzle in itself and it even contains a typing error:
fig. 31 - Hansson's solution to problem 7124 in The Fairy Chess Review
The solution to this 'interpretation puzzle' was published in 1989 by G.P. Jelliss . A few years later, C.J. Bouwkamp started working on the problem, found the typo and published his results in . One of his discoveries was that covering this cube with the 12 pentominoes can be done in 5,626,985 different ways (modulo rotations). In only 19 of these cases, both coverings are nice. Bouwkamp called them the "19 beauties of Hansson's problem". As the interested reader will be able to find out, Hansson's solution was not nice. One of the not-nice 'double six' sets is FNPUVX-ILTWYZ with 126×67 = 8442 solutions. The interesting aspect in this double set of 6 pentominoes is that they not only simultaneously cover two C30 cubes, but they also cover simultaneously two 5×6 rectangles (though in only 2 ways). Bouwkamp was the first to discover this and published this result in .
Figure 32 shows one of the 'beauties' and figure 33 shows a view of one of the 534,964 possible nice coverings of two C30 cubes that are hinged to form a flexible object by pentomino connections. Finding the 'invisible' part of the covering is left as an exercise for the interested reader.
fig. 32 - One of Hansson's 'beauties'
fig. 33 - A nice covering of two hinged C30 cubes
Covering C30 with 5 identical hexominoes
This problem was first considered by Torbijn and Götz and published in . Unfortunately, their numbers of solutions ( and ) were not confirmed by my findings. My results are:
and I found no nice solutions at all. Only 12 of the 35 hexominoes have solutions for this problem, which was also the case in the earlier results by Torbijn and Götz.
(The hexominoes and the way they are labelled were already shown in fig. 19.)
Most of the 19 solutions can be rotated into 24 different views, and only one has a 3-fold rotational symmetry over a diagonal. This very special case is caused by the fact that hexomino p can be folded around a corner in such a way that the folded piece itself has a 3-fold symmetry (though with an internal edge). This solution, as well as 1 symmetric solution each for the pieces q, r and z is shown in figure 34.
fig. 34 - A special solution with hexomino p
Covering C30 with 5 different hexominoes
Using 5 different hexominoes, my computer found 3,841,783 different coverings, 19,833 of them being nice (both numbers are modulo rotations). The nice coverings are formed by 16,535 different sets of 5 hexominoes, with the quintuplet 5fhjm being the most fruitful one with 14 different (nice) coverings. Among the 324,632 possible sets of 5 hexominoes, there are only 14,065 different quintuplets (5.1%) with a unique solution for the covering. So most of them do not cover a C30 cube in a nice way.
When considering all 3,841,783 solutions, there are 296,278 different sets of 5 hexominoes (91,3%) that appear in solutions, 24,015 of them with a unique solution and with the set 56bem being the most 'popular' since it yields no less than 192 different solutions.
The 35 hexominoes can be arranged in 7 sets of 5 each. So it might be possible to simultaneously cover 7 copies of C30 with all 35 hexominoes. It turns out that there are even many thousands of solutions to this problem, most of them having only 1 solution for each of the 7 cubes. But I did find a set of 7 quintuplets that has no less than 1152 different (all nice!) solutions:
6kmow - 19ace - 23fhu - 4qrsx - 57bnv - 8dgyz - ijlpt : 8×2×3×2×2×3×2 = 1152.
Finding at least one solution for each of these 7 coverings, might be considered a feasible challenge for many readers.
I have never seen any publication about covering block B40. Apart from 8 pentominoes, two full sets of the 5 tetrominoes could also be used, but this was not investigated. Using 8 different pentominoes, my computer found the following results:
The X pentomino is apparently useless for covering this B40 block and the U pentomino does the job in one unique way that is even completely symmetric under all rotations and reflection. It may be rewarding to think about how each of these pieces can cover the B40 block. Anyway, figure 35 shows a choice of solutions.
fig. 35 - Some solutions for B40 covered with 8 pentominoes
For the problem of covering B40 with 8 different pentominoes, I found 11,292,596 solutions and 92,678 of these are nice (both numbers are modulo rotations - divide by 2 to obtain the number of mirror-image pairs). Three such blocks can simultaneously be covered by 2 full sets of pentominoes with 8 different pentominoes on each block.
The edges of B50 have lengths √10 × √10 × √40 and the grid is rotated in a 1:3 direction with respect to the edges of the block:
fig. 36 - Block B50 and the grid for 50 squares
Using 10 pentominoes, it should be possible to cover the 50 squares of the grid on the B50 block.
For 10 identical pentominoes, the number of solutions are given in the table. From this table, it is obvious that the number of solutions is extremely dependent on the pentomino chosen.
Using 10 different pentominoes, there are 7,901,856 solutions that are completely nice, which number can be divided by 8 to exclude rotations. Then 987,732 nice solutions remain. Every possible set (there are 66 such sets) of 10 pentominoes has plenty of solutions. The least prolific set is the set without N and P, in which case only 1051 solutions exist. But when I and X are left unused, no less than 112,869 different solutions emerge. Piece X is relatively difficult to use since without piece X, there are 540,357 solutions while leaving piece P out results in 'only' 36,461 solutions. C. J. Bouwkamp investigated solutions with internal edges and he found that with pieces F and I omitted, there are 864 solutions with 8 internal edges. This result was never published by him.
The total number of solutions, both nice and with one or more internal edges, is not (yet?) known. The unique solution with piece X is easily found. C.J. Bouwkamp did not investigate this problem completely; he only looked for the nice solutions with the Y pentomino. His unpublished results are in accordance with mine in the table above.
Cube C54R is highly related to cube C24R, though, in a way, it is rather its 'big brother':
fig. 37 - A view of cube C54R
This covering scheme was never mentioned in any publication, not even in the good old Fairy Chess Review.
The surface of 54 units suggests coverings by either 9 hexominoes or 1 tetromino plus 10 pentominoes. Amazingly, covering the surface with 9 copies of the same hexomino never gives a solution, except with hexomino c in which case 2 different solutions were found, both with 4 different rotations:
fig. 38 - the two coverings with hexomino c
Using 9 different hexominoes, the surface of C54R is easily covered. Even with nice coverings, there are hundreds of thousands of solutions; the exact number is not known.
Using 1 tetromino and 10 (different) pentominoes, the total number of solutions is 7,890,708 (modulo rotations). With so many solutions, it should be fairly easy to find one without electronic assistance. Perhaps even in the case of the S-tetromino and all but the L and P pentominoes, in which case there are 'only' 1292 solutions.
Covering C60 with 12 different pentominoes
Covering of C60 by the set of all 12 pentominoes was first mentioned and one solution was given in 1948 by Herbert Daniel Benjamin (1899-1950) in , later in 1954 by W. Stead , both in The Fairy Chess Review. The problem was also briefly mentioned by Martin Gardner , Solomon Golomb  and George Martin . Then in 1994, C. J. Bouwkamp started solving this problem by computer and he found 26,358,584 solutions (modulo rotations and reflections). Only 284,402 (1.1%) of these are nice solutions. Bouwkamp published his method and results in .
Covering C60 with 12 identical pentominoes
This too was first investigated by C.J. Bouwkamp for nice coverings only and published in 1997 . My computer recently found exactly the same numbers of solutions. The problem was also described in , unfortunately with incorrect numbers of solutions.
The results for all coverings, including those that are not nice, are:
Limiting ourselves to nice coverings, the results become:
Bouwkamp checked lots of these solutions very carefully and built hundreds of them using thin cardboard. Unpublished by him were cases in which a set of 5 identical 15-ominoes, formed by joining 3 identical pentominoes, cover C60. Five of these 15-ominoes are shown in figure 39. The construction of the actual covering is left as a nice exercise for the interested reader, and so is finding possible other of these special 15-ominoes.
fig. 39 - five 15-ominoes that cover C60
In the same way, it looks probable that there exist 20-ominoes such that 3 copies of such a 20-omino cover C60 exactly. It would be nice if each of these 20-ominoes is built-up of 4 different pentominoes, such that together, they are made of all 12 pentominoes. I was able to find several such coverings, but all with two 'flaws': one is that none of the coverings are nice and the second is that at least one of the three 20-ominoes can never be folded flat to form three identically shaped planar 20-ominoes.
Covering C60 with 10 identical hexominoes
This problem was studied before by Torbijn and Götz in . However, I found completely different numbers of solutions. Unfortunately, Bouwkamp did not work on this problem, so until the problem is tacked by someone independent, it remains unsure which numbers are correct. Figure 19 above shows the hexominoes and their designations as used in this study; only the hexominoes that have solutions appear in the following table:
Only 14 nice solutions were found. The maximum number of internal edges (8) appears in 6 of the 141 solutions, all 6 being solutions with piece l (lima). Some solutions are shown in figure 40:
fig. 40 - some solutions for C60 covered with 10 identical hexominoes
Covering C60 with 10 different hexominoes
This problem has not (yet?) been investigated. One nice solution was shown in figure 40.
Problem 6841 (Fairy Chess Review June 1946) is about covering a 2×2×7 surface with one 2×2 end open with the 12 pentominoes. This would make a nice vase for (puzzle?) flowers (actually 1,000,838 different nice vases).
FCR-problems 6842 and 6843 were about covering a 3×3×3 cube (surface 54 units) with 1 tetromino and 10 pentominoes.
Problem 8848 (FCR October 1950) asks for a covering a 2×2×1 block with edges √8 and 2×√8 using all the tetrominoes and hexominoes. This block would be named B80 and one solution was shown in figure 14.
Most likely, there are other surfaces that can be covered using limited amounts of the many available polyominoes.
Many coverings of rectangular blocks by polyominoes were investigated. Important contributions were made by the mathematician and puzzle enthusiast Prof. Dr. C.J. Bouwkamp (1915-2003) and many of his results were also published by him. His remaining so far unpublished results in this field are all given in this article. His work, repeated by my computer using my own programs, was confirmed by finding the same results. Several additional cases were investigated and some of the results, though believed to be correct, are waiting for confirmation by an independent programmer.
Covering C30 and C60 with 10-ominoes was investigated in  and many more coverings could be researched using larger or less regular blocks, using a combination of two or even more types of polyominoes, using other pieces than the trominoes, tetrominoes, pentominoes and hexominoes as considered in this article, and even depending on the shape of the surface to be covered using other polyforms, like polyiamonds (poly-triangles), polyhexes and perhaps even other shapes.
1. H.D. Benjamin, Problem/Solution No. 5820, The Fairy Chess Review, Feb./April 1944
2. R.J. French, Problem/Solution No. 6841, The Fairy Chess Review, June/August 1946
3. R.J. French, Problem/Solution No. 6842, The Fairy Chess Review, June/August 1946
4. R.J. French, Problem/Solution No. 6843, The Fairy Chess Review, June/August 1946
5. R.J. French, Problem/Solution No. 7002, The Fairy Chess Review, October/December 1946, pp.56/66
6. F. Hansson, Problem/Solution No. 7124, The Fairy Chess Review, Feb./April 1947, pp. 71/81
7. H.D. Benjamin, Problem/Solution No. 7590, The Fairy Chess Review, Feb./April 1948, pp.
8. H.D. Benjamin, Problem/Solution No. 7591, The Fairy Chess Review, Feb./April 1948, pp.
9. H.D. Benjamin, Problem/Solution No. 8848, The Fairy Chess Review, Oct. 1950/Feb. 1951, pp.
10. W. Stead, Solution 7591, The Fairy Chess Review, Dec. 1954 pp. 2-3
11. Solomon W. Golomb, Polyominoes, Charles Scribner's Sons, 1965, p. 136
12. G.P. Jelliss, Dissection Problems in PFCS/FCR, manuscript, St Leonards on Sea, UK, 1989.
(Some photocopies of these notes were sent to a few interested correspondents, it was not 'published' commercially.)
13. Martin Gardner, Covering a Cube with Congruent Polygons, CFF-25, Dec. 1990, p. 13.
(also: Martin Gardner, The Scientific American Book of Mathematical Puzzles & Diversions, 1959, pp. 133-134.)
14. George E. Martin, Polyominoes, The Mathematical Association of America, 1991, pp.72- 73
15. P.J. Torbijn, Six Six-by-Six's, Problem 1711, Journal of Recreational Mathematics 23 (1991) pp. 304-305
16. Pieter Torbijn, Cubic Hexomino Cubes, CFF-30, December 1992, pp. 18-20
17. Pieter Torbijn, General Hexomino Cubes, CFF-30, December 1992, pp. 21-26
18. Anton Hanegraaf, Covering a Cube by 2 Hexominoes, CFF-30, Dec. 1992, pp. 27-29
19. Pieter Torbijn, Covering a Cube by 36 Hexominoes, CFF-30, Dec. 1992, p. 33
20. C.J. Bouwkamp, An old pentomino problem revisited, Simplex Sigillum Veri, Technische Universiteit Eindhoven (1995), pp. 87-96 ISBN 90-386-0197-2
21. C.J. Bouwkamp, An Unusual Pentomino Problem, CFF-41, Oct. 1996, pp. 39-40
22. C.J. Bouwkamp, On Benjamin's Pentomino Cube, EUT Report 97-WSK-01, Eindhoven, Dec. 1997
23. C.J. Bouwkamp, Tiling the surface of the cube by 12 identical pentominoes, EUT Report 98-WSK-02, Eindhoven, Nov. 1998
24. Pieter Torbijn, Covering a Cube with Congruent Polyominoes, CFF-58, July 2002, pp. 18- 20
25. Pieter Torbijn, Covering a Cube with Congruent Polyominoes (2), CFF-59, November 2002, p. 14
26. Pieter Torbijn, Covering a Cube with Congruent Polyominoes (3), CFF-61, July 2003, pp. 12-16
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