Description
- For the cube graph ππ, the distance π·(π, π) between two vertices
π = (π1, π2, β¦ , ππ) and π = (π1, π2, β¦ , ππ)
is called the βHamming distance.βΒ This is the number of positions where π and π differ.Β For instance, the Hamming distance between (0,0,1,0) and (1,1,0,0) is 3 because these two vertices differ in three positions.Β In each of the parts A,B below, π·(π±, π²) is the Hamming distance in ππ:
- Show that if π·(π, π) and π·(π, π) have the same parity (i.e., are both even or are both odd), then π·(π, π) must be even.
- Show that if π·(π, π) and π·(π, π) have different parity, then π·(π, π) must be odd.
- Consider πΎ4 as drawn and labeled below:
Since this graph is simple, we can specify a walk by listing only the vertices.Β For instance, πΆ: 1,2,3,4,1 is a 4-cycle; this can be abbreviated as β12341β.Β List all of the cycles (abbreviated style is fine) that start and end at vertex 1 in this drawing of πΎ4.
- For 1 β€ π β€ 11, let πΊπ be the graph with vertex set
π = {0,1,2,3,4,5,6,7,8,9,10,11}
and where vertices π’ and π€ are adjacent iff π€ β π’ = π modulo 12 or π’ β π€ = π modulo 12.Β We observe that πΊ1 = πΆ12, a twelve-cycle. A. Β For what values of π is πΊπ connected?
- What are the possible numbers of components of πΊπ?
- Let πΊ be a graph and let π1 and π2 be edges. Show that if deleting π1 and π2 disconnects vertices π’ and π£, then any cycle containing both π’ and π£ must contain both π1 and π2.Β One approach:Β You could apply to the graphs πΊ β π1 and πΊ β π2 the fact that if π is a bridge whose removal disconnects π’ and π£, then any path connecting π’ and π£ must contain π.




