[Math] Size of minimum vertex cover on complete graph

graph theory

I know that the maximum size of an independent vertex set, also known as the independence number and denoted by $\alpha$, plus the the size of the minimum vertex cover, $\tau$, is equal to the number of vertices in the graph, $n$.

$$\alpha(G) + \tau(G) = |G|$$

However, for a complete graph I don't understand how this is true. Take $K_4$, it seems to have $\alpha = \tau = 1$ but in this case $n = 4$.

Best Answer

Given the complete graph $K_n$, it is clear that $\alpha(K_n)=1$ as any two vertices are adjacent and thus any set with more than one vertex is not independent. Similarly, any subset of less than $n-1$ vertices cannot be a vertex cover, as it would not include an edge in the graph, so $\tau(K_n)=n-1$. So the formula is valid: $$\alpha(K_n) + \tau(K_n) = 1 + (n-1) = n. $$