A polyiamond (also polyamond or simply iamond, or sometimes triangular polyomino[1]) is a polyform whose base form is an equilateral triangle. The word polyiamond is a back-formation from diamond, because this word is often used to describe the shape of a pair of equilateral triangles placed base to base, and the initial 'di-' looks like a Greek prefix meaning 'two-' (though diamond actually derives from Greek ἀδάμας - also the basis for the word "adamant"). The name was suggested by recreational mathematics writer Thomas H. O'Beirne in New Scientist 1961 number 1, page 164.
Counting
editThe basic combinatorial question is, How many different polyiamonds exist with a given number of cells? Like polyominoes, polyiamonds may be either free or one-sided. Free polyiamonds are invariant under reflection as well as translation and rotation. One-sided polyiamonds distinguish reflections.
The number of free n-iamonds for n = 1, 2, 3, ... is:
The number of free polyiamonds with holes is given by OEIS: A070764; the number of free polyiamonds without holes is given by OEIS: A070765; the number of fixed polyiamonds is given by OEIS: A001420; the number of one-sided polyiamonds is given by OEIS: A006534.
| Name | Number of forms | Forms | ||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Moniamond | 1 |
| ||||||||||||
| Diamond | 1 |
| ||||||||||||
| Triamond | 1 |
| ||||||||||||
| Tetriamond | 3 |
| ||||||||||||
| Pentiamond | 4 |
| ||||||||||||
| Hexiamond | 12 |
|
Some authors also call the diamond (rhombus with a 60° angle) a calisson after the French sweet of similar shape.[2][3]
Symmetries
editPossible symmetries are mirror symmetry, 2-, 3-, and 6-fold rotational symmetry, and each combined with mirror symmetry.
2-fold rotational symmetry with and without mirror symmetry requires at least 2 and 4 triangles, respectively. 6-fold rotational symmetry with and without mirror symmetry requires at least 6 and 18 triangles, respectively. Asymmetry requires at least 5 triangles. 3-fold rotational symmetry without mirror symmetry requires at least 7 triangles.
In the case of only mirror symmetry we can distinguish having the symmetry axis aligned with the grid or rotated 30° (requires at least 4 and 3 triangles, respectively); ditto for 3-fold rotational symmetry, combined with mirror symmetry (requires at least 18 and 1 triangles, respectively).
Generalizations
editLike polyominoes, but unlike polyhexes, polyiamonds have three-dimensional counterparts, formed by aggregating tetrahedra. However, polytetrahedra do not tile 3-space in the way polyiamonds can tile 2-space.
Tessellations
editEvery polyiamond of order 8 or less tiles the plane, except for the V-heptiamond. [4]
Correspondence with polyhexes
edit
Every polyiamond corresponds to a polyhex, as illustrated at right. Conversely, every polyhex is also a polyiamond, because each hexagonal cell of a polyhex is the union of six adjacent equilateral triangles. Neither correspondence is one-to-one.
In popular culture
editThe set of 22 polyiamonds, from order 1 up to order 6, constitutes the shape of the playing pieces in the board game Blokus Trigon, where players attempt to tile a plane with as many polyiamonds as possible, subject to the game rules.
See also
editExternal links
edit- Weisstein, Eric W. "Polyiamond". MathWorld.
- Polyiamonds at The Poly Pages. Polyiamond tilings.
- VERHEXT — a 1960s puzzle game by Heinz Haber based on hexiamonds (Archived March 3, 2016, at the Wayback Machine)
References
edit- ^ Sloane, N.J.A. (July 9, 2021). "A000577". OEIS. The OEIS Foundation Inc. Retrieved July 9, 2021.
triangular polyominoes (or triangular polyforms, or polyiamonds)
- ^ Alsina, Claudi; Nelsen, Roger B. (31 December 2015). A Mathematical Space Odyssey: Solid Geometry in the 21st Century. ISBN 9781614442165.
- ^ David, Guy; Tomei, Carlos (1989). "The Problem of the Calissons". The American Mathematical Monthly. 96 (5): 429–431. doi:10.1080/00029890.1989.11972212. JSTOR 2325150.
- ^ "All of the polyiamonds of order eight or less, with the exception of one of the heptiamonds will tessellate the plane. The exception is the V-shaped heptiamond. Gardner (6th book p.248) posed the problem of identifying this heptiamond and reproduced an impossibilty proof of Gregory. However, in combination with other heptiamonds or other polyiamonds, tesselations using this V-shaped heptiamond can be achieved."