Polyforms derived from combinations of other n-forms.
The only rectangle possible with this set is the 6x23 shown above. If we use one piece twice than we can form a 12x12 square. All 23 problems are possible.
Roel Huismann has found a number of simultaneous rectangles with the 40 one-sided tridominoes.
There are 211 tetradominoes including three with internal holes.
The 208 pieces without holes are shown here in eight 8x26 rectangles made by Roel Huismann and also in sixteen 8x13 rectangles made by Patrick Hamlyn. Patrick's solution can be coloured with just three colours.
We can also look at the solid tridominoes and form a number of three dimensional constructions the last of which shows a construction with eight internal holes.
One box is possible with this set a 2x3x23 as shown below.
Polytrominoes are combinations of trominoes. The figure below shows the 29 ditrominoes firstly in the only possible rectangle the 6x29.
The one-sided set has 51 pieces but, since there is an odd number of unbalanced pieces, no rectangle can be made with the set.
Notice that there are two types of tromino used and other, smaller sets could be produced if we were to look at specific combinations - two trominoes of the same type or one of each type as in this construction.
Ditetrominoes
If we join two L tetrominoes then we get a set of 64 pieces one of which has a hole. Patrick Hamlyn has managed to pack the remaining 63 pieces into three 8x21 rectangles.
Polydiamonds are sets formed from the joining of diamonds (two joined equilateral triangles). The n-diamonds are subsets of the 2n-iamonds. For constructions with these see Miroslav Vicher's site.
Brendan Owen has made some constructions with the one sided tridiamonds and the tetradiamonds.
If, instead of diamonds, we join triamonds (trapezia) together then we get subsets of the 3n-iamonds which we can call polytraps. There are 9 ditraps shown in a number of constructions below. The 15 piece one-sided set can also make a number of figures.
The following table is due to Brendan Owen.
The 94 tritraps include one piece with a hole. The 93 pieces without a hole can form a trapezium as shown below.
Beolw are two constructions, one with the full set and the other without the piece with a hole.
The full set can be used to make 11-fold copies of an heptiamond with a hole.
The 178 one-sided tritraps without a hole are shown here.
Constructions with the chequered sets are also possible.
Brendan Owen has suggested forming figures from the combination of pairs of hexagons. The following table is due to him.
The figure below shows the six didohexes and the nine one-sided didohexes.
There are 74 tridohexes shown here in a number of constructions.
The diagram below shows two constructions made with the 73 tridohexes without a hole.
There are 134 one-sided tridohexes including one with a hole. The pieces without a hole can form seven 6x19 parallelograms.
Pairs of congruent figures are also possible with the set.
Brendan Owen has found a set of 19 congruent shapes made with this set (omitting the piece with a hole).
If we join dicubes we form subsets of the 2n-cubes. There are six didicubes with a total volume of 24 unit cubes (see the figure below which also shows how they can for a 2x3x4 box). Joining three dicubes gives us the 103 tridicubes made up of solid versions of the 23 tridominoes and a further 80 three dimensional pieces giving a total volume of 618 unit cubes. If we use one of each mirror pair then we get a set of 67 pieces. Since both 103 and 67 are primes the only possible boxes would be 2x3x103 or 2x3x67.
The 2x3x103 is possible by combining the 2x3x23 box made with the solid tridominoes and this set of 16 2x3x5 boxes found by Brendan Owen.
Note: With all the of the above there is a problem of choice as to whether we consider different combinations producing the same shape as different on not (see diagram below for examples).
Mostly we should consider these to be the same but they could be considered as distinct. If we were to consider these as different for the tridominoes we should get 26 pieces as shown below in two 6x13 rectangles. Also shown here is the one-sided set of 45 pieces in five 6x9 rectangles and other sets of rectangles.
Similarly with the ditrominoes there are two pieces which can be formed in two different ways. The resulting set of 31 pieces has an unbalanced colouring so no rectangle is possible. The set can, however, form a number of symmetrical shapes.
With the one-sided set we get 54 pieces but these also can only form unbalanced figures.
