3 Squares, 11 Pieces

Here are the 11 pieces made from squares of size 40mm x 40mm x 3mm.

To construct stable constructions I added a quadratic frame for 9 squares, which can be used as a bottom of a figure. It also prevents further vertical squares from falling over. Now you can place 9 squares on this bottom and with the remaining 24 pieces you can make a grid of vertical squares. This way you get kind of a typesetting box or display box. Other figures are possible if you swap squares at the bottom with missing squares at the top. This is a nice puzzle, which can be solved by hand in about 10 minutes. Solutions are here.

There are also some figures with a bottom shaped like an octomino and 33-8=25 squares left for a grid onto this octomino. The picture shows some examples with the bottom upside. Here are the layers.

For figures with more than one layer it seems to be neccessary to use pieces possible to stick together. I looked at a P-Pentomino shaped grid without a bottom and added a V-Tromino shaped grid in the second layer. Now we have 15+5 squares for the first layer and 10+3 squares for the second one, so that all pieces are used. Here are the layers.

An open box of size 3x2 has 6 horizontal and 17 vertical squares. Adding two single cubes with one open face we have 2*5 additional squares and a toal of 6+17+10=33 squares, just enough for constructions with the whole set.

In the above mentioned open 3x3x1 box one 3x3 face is missing. A construction with two 3x3 faces and three missing 3x1 faces is also possible. Here is this construction and another one with height 3.


1, 2 or 3 Squares, 14 Pieces

There is one single square and two pieces with two squares. Adding the 11 pieces with three squares we get 14 pieces with a total of 38 squares. Now we can construct a box of size 4x2x1 with no open faces. Here is a solution placed in a 4x2 frame. The construction is rather unstable, because there is no possibility to connect different pieces . Especially the single square should be on the bottom, otherwise it's likely to fall down.

Here is the above solution with bottom and top faces swapped.

If we start with a 3x3x1 box, open at the top, we can add a single cube at three different positions. This single cube has one missing face and can't be made with the pieces of order 1 and 2. Therefore we need a new solution for each position. The picture shows the solutions, and the layers are also given.

Here are three other symmetric constructions and their layers.


4 Squares, 80 Pieces

Here are the 80 Tetromonoids with a total of 320 squares. 16 pieces are red color, indicating that the centers of their squares lie in a single plane and the squares are orthogonal to these planes. The pieces can be used as kind of tetrasticks in a quadratic grid.

I noticed, that a box of size 10x10x2 with a missing 10x10 face has 10*11*2=320 squares. Is a construction possible with this set? Unfortunately not! 10*11*2=220 vertical squares are needed but the maximum number of vertical pieces in best optional position and orientation of the pieces is only 199.

Are there other open boxes with one missing face and a total of 320 squares possible? We have to look for solutions of a*b*c+(a+1)*b*c+a*(b+1)*c=320. We get four solutions: (10,10,1), (1,1,64), (2,2,20) and (12,4,2). Due to a small cross section a construction of the open 1x1x64 box isn't possible. An open 12x4x2 box is shown below with the missing face at the bottom, the layers are here.

You can get the open 2x2x20 box with a missing 2x2 face if you divide it into four open boxes of size 2x2x5. This is shown in the title and the layers are given here.

You can also combine the four open boxes into a ring of size 6x6x2 with a hole of 3x3x2.

Boxes with no missing faces are also possible.We have to solve a*b*(c+1)+(a+1)*b*c+a*(b+1)*c=320. Then we can look for constructions of size 7x6x2 and 15x5x1. Both boxes are possible Click the pictures to see the layers.

The 15x5x1 box has a cross section of 5x1, which is the shape of the I-pentomino. Other bars with length 15 and a cross section like a pentomino can also be made with the whole set. An exception is the shape of the P-pentomino because only 16*5 +15*15=305 squares are needed.

The bar with the X-shaped cross section is a construction with a jagged border, so I looked for other ones and found a prism with a 9x2 jagged cross section and a height of 3.

At last some figures with rotational symmetry are shown. Boxes with a quadratic cross section like 4x4x4, 4x4x3, 4x5x3 or 9x9x1 are the starting point. Then we can add or subtract single cubes or boxes preserving the symmetry. The tower is made from a 4x4x4 cube and a 2x2x5 open box using the remaining pieces. Click the pictures to see the layers.

The subset of the red marked pieces has a parity issue. Puzzler has solved many figures with this set leaving one piece out. My physical pieces can also be used as tetrasticks. In this case you can stabilize a construction with clamps.


3 or 4 Squares, 93 Pieces

If we take the 11 pieces of order 3 and the 80 pieces of order 4, we have 93 pieces with a total of 33+320=252 squares. Solving (a+1)*b*c+a*(b+1)*c+a*b*(c+1)=353 leads to a box of size 39x2x1. The cross section is very small and I wondered, if such a box is possible to construct. But it worked. You can see the long bar made from physical pieces displayed on the floor. In the picture, showing the layers, I split the construction into two parts.

The only option for an open box is 35x3x1. But you can't get a solution. We have 248 vertical squares in this box but only 199+25 vertical squares are available, even if the pieces are best oriented.

The subset of pieces, where all squares can be vertically positioned, has no parity issue, and I constructed two figures with two axis of symmetry.

If we add the pieces of order 1 and 2, we have a set of pieces equivalent to polysticks of order 1 up to 4. Only two examples are shown below. A lot of other figures is given by Puzzler.


5 Squares, 780 Pieces

Because there are too many pieces I never thought about physical pieces of order 5. There might be further problems by putting those pieces together if they aren't made from elastic material. A picture of the pieces is here.

But we can try to construct some virtual boxes with the pieces. Solving (a+1)*b*c+a*(b+1)*c+a*b*(c+1)=780*5=3900 we can look for a quadratic prism of size 13x13x7. If you fill a box plane by plane the cross section must be small to allow backtracking to the previous plane after few placements. Therefore a cross section of 13x13 or 13x7 is much too large, and I divided the box into multiple blocks of size 3x3x10, 4x3x10, 4x2x10 and 3x3x7. You must ensure, that the missing faces match given faces of the neighboring block. Filling a small box a thousand times can help to mark pieces with good tileability and save them for the last blocks. Here is a solution and the assoziated layers.

An open box has one missing face, and we must solve (a+1)*b*c+a*(b+1)*c+a*b*c=3900. Solutions are: 195x2x4, 51x6x4, 44x7x4 and 20x5x12. For the last box with height 12 I divided the whole box into smaller ones and got the solution below. The layers are here.

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