Burnside Lemma Necklace

Question

I'm trying to solve this problem for 4-6-8 necklace.

So far I am at the following:

$\frac{18!}{4!6!8!}$ (different arrangements) + $\frac{9!}{2!3!4!}$ (rotating through 180 degrees).

Now I consider reflections about a line of symmetry passing through the midpoints of opposite sides and there are 9 such axes and the order of the 9 beads on one side could be permuted in $9! = 362880$ ways.

I now have to consider reflections about a line of symmetry through opposite corners. There are 9 such lines of symmetry. The corners must have the same color and it can be chosen in 3 ways? And the other 8 beads can be arranged in 8! ways?

So I come at $(\frac{18!}{4!6!8!}$ + $\frac{9!}{2!3!4!}$ + $9*9!$ + 9*3*8!)/36$ = 376,250 permutations.

But I think this is entirely wrong and not sure how to fix it.

Is this a re-worded re-post of your previous question Orientation-Less Circular Permutation with In-distinguishable Objects which seems to be a re-worded re-post of of Necklace combinations with three group of beads ? — Vepir, May 11 '20 at 16:21
Only the first post is mine, should I have updated it instead of making this new one? Not really regular here :( — Kradec na kysmet, May 11 '20 at 16:47
Usually, you should not post the same question multiple times. Yes, you should've probably edited your own original question (with what seems to be your thoughts and attempts, and a clarification that it is a necklace problem), instead of reposting the question. The second question is not yours, but is it the same problem once again? I noticed you placed a bounty there. — Vepir, May 11 '20 at 17:14
Yeah, I was looking for similar problem with explanation and found out that it's pretty much the same (but that the OP's calculations are pretty far from solving it). — Kradec na kysmet, May 11 '20 at 18:25

score 2 · Accepted Answer · answered May 12 '20 at 08:29

You can check how I approached the problem:

I manage to answer the question and this is the process that I followed:

I consider the 18-bead necklace in the first part of the problem. Here are the eighteen rotations expressed in cycle form where we assume that the slots are numbered from 1 to 18 in clockwise order. The first is the identity (e: no rotation) and the second is the generator g—a rotation by a single position which, when repeated, generates all the elements of the group:

$e = g^0 \text{ = (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16) (17) (18)}$

$g^1 \text{ = (1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18) }$

$g^2 \text{= (1 3 5 7 9 11 13 15 17) (2 4 6 8 10 12 14 16 18)}$

$g^3 \text{= (1 4 7 10 13 16) (2 5 8 11 14 17) (3 6 9 12 15 18)}$

$g^4 \text{= (1 5 9 13 17 3 7 11 15) (2 6 10 14 18 4 8 12 16)}$

$g^5 \text{= (1 6 11 16 3 8 13 18 5 10 15 2 7 12 17 4 9 14)}$

$g^6 \text{= (1 7 13) (2 8 14) (3 9 15) (4 10 16) (5 11 17) (6 12 18)}$

$g^7 \text{= (1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18)} $

$g^8 \text{= (1 9 17 7 15 5 13 3 11) (2 10 18 8 16 6 14 4 12)} $

$g^9 \text{= (1 10) (2 11) (3 12) (4 13) (5 14) (6 15) (7 16) (8 17) (9 18)} $

$g^{10} \text{= (1 11 3 13 5 15 7 17 9) (2 12 4 14 6 16 8 18 10)} $

$g^{11} \text{= (1 12 5 16 9 2 13 6 17 10 3 14 7 18 11 4 15 8)} $

$g^{12} \text{= (1 13 7) (2 14 8) (3 15 9) (4 16 10) (5 17 11) (6 18 12)} $

$g^{13} \text{= (1 14 9 4 17 12 7 2 15 10 5 18 13 8 3 16 11 6)} $

$g^{14} \text{= (1 15 11 7 3 17 13 9 5) (2 16 12 8 4 18 14 1 6)} $

$g^{15} \text{= (1 16 13 10 7 4) (2 17 14 11 8 5) (3 18 15 12 9 6)} $

$g^{16} \text{= (1 17 15 13 11 9 7 5 3) (2 18 16 14 12 10 8 6 4)} $

$g^{17} \text{= (1 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2)} $

After that I found the GCD for all cycle form's with and group them in a table:

| Cycle length | Permutations | GCD with 18 |

| 1 | $g^0$ | GCD(0, 18)=18 |

| 2 | $g^9$ | GCD(9, 18)=9 |

| 3 | $g^6$, $g^{12}$ | GCD(6, 18)=GCD(12, 18)=6 |

| 6 | $g^3$, $g^{15}$ | GCD(3, 18)=GCD(15, 18)=3 |

| 9 | $g^2$, $g^4$, $g^8$, $g^{10}$, $g^{14}$, $g^{16}$ | GCD(2, 18)=GCD(4, 18)=GCD(8, 18)=GCD(10, 18)=GCD(14, 18)=GCD(16, 18)=2 |

| 18 | $g^1$, $g^5$, $g^7$, $g^{11}$, $g^{13}$, $g^{17}$ | GCD(1, 18)=GCD(5, 18)=GCD(7, 18)=GCD(11, 18)=GCD(13, 18)=GCD(17, 18)=1 |

We have 18 permutations for rotation and lets name cycle 1 with $f_1$, cycle 2 with $f_2$ .. cycle n with $f_n$

Than the formula for Cycling index is:

$$\frac{f_1^{18} + f_2^9 + 2f_3^6 + 2f_6^3 + 6f_9^2 + 6f_{18}^1}{18}$$

If we solve all the possible necklaces with three colors the result should be (for the moment we do not solve for the three colors with 4, 6 and 8 beads in respective groups):

$$\frac{3^{18} + 3^9 + 2*3^6 + 2*3^3 + 6*3^2 + 6*3^1}{18} = \text{21 524 542}$$

From here because turn around is allowed we need to add and the necklace(bracelet if we follow the right terms) is with even beads we should add the symethric turn arounds.

$$\frac{f_1^{18} + f_2^9 + 2f_3^6 + 2f_6^3 + 6f_9^2 + 6f_{18}^1 + 9f_1^2f_2^8 + 9f_2^9}{2 * 18}$$

and again for three colors without including the different weight:

$$\frac{3^{18} + 3^9 + 2*3^6 + 2*3^3 + 6*3^2 + 6*3^1 + 9 * 3^8 + 9 * 3^9}{2 * 18} = \text{10 781 954}$$

For the moment we have all the possible necklaces and bracelets with three colors. However, in order to find the necklaces and bracelets with three colors (4 reds, 6 greens and 8 blues) we need to replace:

$$f_1 = (x + y + z)$$

$$f_2 = (x^2 + y^2 + z^2)$$

$$f_3 = (x^3 + y^3 + z^3)$$

$$f_6 = (x^6 + y^6 + z^6)$$

$$f_9 = (x^9 + y^9 + z^9)$$

$$f_{18} = (x^{18} + y^{18} + z^{18})$$

and if we replace in the formula it becomes:

$$\frac{(x + y + z)^{18} + (x^2 + y^2 + z^2)^9 + 2(x^3 + y^3 + z^3)^6 + 2(x^6 + y^6 + z^6)^3 + 6(x^9 + y^9 + z^9)^2 + 6(x^{18} + y^{18} + z^{18}) + 9(x + y + z)^2(x^2 + y^2 + z^2)^8 + 9(x^2 + y^2 + z^2)^9}{36}$$

Then we need to find which expressions can expand to $x^4y^6z^8$.

After then by using multinominal coeficient I managed to calculate the following results

9 189 180

1260

11 340

Then I sum all of them and divide them to 36. This gives me the answer of 255 920 which is the answer of the question. We can create 255 920 bracelets with 4 red 6 green and 8 blue beads.

Burnside Lemma Necklace

1 Answers1