Question

We say two colorings c and c' are equivalent if one can be moved to the other by a symmetry π ∈ G = Sym(X), as we showed in class. Formally, we write c' = π(c), where c'(i) = c(π−1(i)). Here π−1 is the inverse symmetry of π, so i' = π−1(i) is the vertex which is moved to i under π. Verify that the rotation ρ1 = (1234) has inverse symmetry ρ1−1 = (4321), and ρ1 moves the coloring c = RBBB to c' = BRBB, by computing c(ρ1−1(i)) for i = 1,2,3,4.

Answer 1

Let X be a square with vertices labelled clockwise by {1,2,3,4}, and having symmetry group G = Sym(X) containing the identity ε = id, rotations ρ1, ρ2, ρ3, and reflections τ1,..., τ4. Consider colorings with k = 2 colors, that is, functions c : {1,2,3,4} → {R,B} which we can write by listing the color of each vertex: for example, BBRB means c(1) = c(2) = c(4) = B and c(3) = R. The set of all colorings is C, with |C| = 24 = 16.

Answer 2

is this graph theory?

Answer 3

yea , combinatorics and graph theory

Answer 4

ughh taking graph theory right now, im really bad at it. Hmm @KingGeorge is good at this if shows up. @amistre64 might be able to help.. Just stick around.

Answer 5

We can picture the square X and the colorings c = RBBB and c' = BRBB:

1 2 R B B R
4 3 B B B B
We have the 1⁄4 clockwise turn ρ1 = (1234), and it is clear visually that ρ1 moves c to c'. We can compute this formally as follows.
The inverse symmetry is ρ1−1 = (4321), and we can compute that ρ1(ρ1−1(i)) = i for i =1,2,3,4. (For example ρ1(ρ1−1(1)) = ρ1(4) = 1.) Then:
c(ρ1−1(1)) = c(4) = B = c'(1),
c(ρ1−1(2)) = c(1) = R = c'(2),
c(ρ1−1(3)) = c(2) = B = c'(3),
c(ρ1−1(4)) = c(3) = B = c'(4).
Hence, ρ1(c) = c'.

Answer 6

If you still need/want help with this, reply at your earliest convenience and I can get back to you tonight (After 8:00-8:30 PM MST).