“…We complete our Hamiltonian cycle through G 1 4 (H) by first taking our path through G b−2 and G b−1 , then setting α b as the coloring in V (G b ) that agrees β b−1 on u and v (such an α b exists because φ b (x) = 1 and φ b (y) = 2 while β b−1 colors u from {2, 3, 4} and colors v from {3, 4}, and β b−1 and α b are adjacent in G 1 4 (H) because they only differ on the vertex of H ′ where φ b−1 and φ b differ), and finally finding a Hamiltonian path through G b (such a path exists: if β b uses a color outside of {3, 4} on u or v, then we selected α b−1 to color u and v from {3, 4}, so there exists a Hamiltonian path through G b from α b−1 to any other vertex; if β b colors u and v from {3, 4}, then we selected α b−1 to color u with 2 and v from {3, 4}, so there exists a Hamiltonian path through G b from α b−1 to any coloring that colors u and v from {3, 4}).…”