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I am trying to understand the Burnside's lemma in order to use it in an example but all my efforts are in vain. The example is as follows:

Cards are to be constructed from equilateral triangles, divided into 9 congruent sub-triangles by punching holes in two of the small triangles. How many cards can be constructed??

Brian M. Scott
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nikos
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  • I couldn't figure out what exactly is being asked. A picture of a sample card with the divisions and punchings outlined would be nice. Anyway, for a list of on-site applications of Burnside enumeration see my comment to this question. – Jyrki Lahtonen Feb 03 '13 at 13:00
  • Never mind. Reading joriki's answer made me realize that the card itself is also an equilateral triangle. You do say so, but I misread :-) – Jyrki Lahtonen Feb 03 '13 at 13:20
  • here is the image of the triangle.

    http://imageshack.us/photo/my-images/32/picxrp.jpg/

     – nikos Feb 03 '13 at 13:24

2 Answers2

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It's not clear whether the front and back are distinguishable (as they often are with playing cards), so I'll do the counting both ways. In both cases I'll be assuming that whatever else is printed on the cards respects the rotational symmetry of the triangles.

If the front and back are distinguishable, the symmetry group consists of the three rotations through integer multiples of $2\pi/3$. To apply Burnside's lemma, we need to count how many patterns are fixed by each symmetry element. The identity leaves all $\binom92=36$ patterns invariant. For a pattern to be fixed by a rotation through $2\pi/3$ (in either direction), the punch state would have to be the same across each of the three groups of three triangles that are rotated into each other. Since only two holes are punched, this is impossible, so none of the patterns is fixed by such a rotation. Thus by Burnside's lemma the number of inequivalent patterns is $36/3=12$.

If the front and back are indistinguishable, the symmetry group additionally contains the three reflections in the bisectors. For a pattern to be invariant under such a reflection, the two holes need to be punched either in two triangles that are both reflected into themselves, or in a pair of triangles that are reflected into each other. There are $\binom32=3$ possibilities in the first case and $\binom31=3$ in the second case, for a total of $6$. There are three such reflections, so $3\cdot6=18$ is added to the total, yielding $36+18=54$. Thus by Burnside's lemma in this case there are $54/6=9$ inequivalent patterns.

joriki
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This problem may also be solved using Polya counting by constructing the automorphism group of the subdivided triangle, where the small triangles are being permuted. There are three types of permutations. The cycle structure may be read off the diagram in a trivial way without the need to factorize permutations. These are the three types:

  • the identity, having cycle structure $a_1^9$
  • three flips, having cycle structure $a_1^3 a_2^3$
  • two rotations, having cycle structure $a_3^3$

It follows that the cycle index $Z(Q)$ of the permutation group $Q$ acting on the small triangles is $$ Z(Q) = \frac{1}{6} \left( a_1^9 + 3 a_1^3 a_2^3 + 2 a_3^3 \right).$$ Now evaluate at $x+y$ to distinguish between triangles that have a hole punched into them and triangles that don't: $$ Z(Q)(x+y) = \frac{1}{6} \left( (x+y)^9 + 3 (x+y)^3 (x^2+y^2)^3 + 2 (x^3+y^3)^3 \right)$$ which is $${x}^{9}+3\,{x}^{8}y+9\,{x}^{7}{y}^{2}+20\,{x}^{6}{y}^{3}+27\,{x}^{5}{y}^{4}+27 \,{x}^{4}{y}^{5}+20\,{x}^{3}{y}^{6}+9\,{x}^{2}{y}^{7}+3\,x{y}^{8}+{y}^{9}.$$ Finally we have that since $$[x^7 y^2] Z(Q)(x+y) = 9,$$ it follows that there are nine such cards, confirming the result from above.

Marko Riedel
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