# What is the size of a maximum matching in a cycle of length $2k + 1$

What is the size of a maximum matching in a cycle of length $2k + 1$:

$$\large v_1, v_2,..v_{2k+1}$$

What is the size of a minimum vertex cover?

I know that If $G$ is a bipartite graph, then the maximum size of a matching in $G$ equals to the minimum size of a vertex cover in $G$. I am not sure on how to go on after this for a cycle of length $2k+1$

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What have you tried? For instance, have you tried it for small graphs by hand $k = 1, 2, 3$ etc.? – ShreevatsaR Jan 23 '13 at 16:45
An important aspect of the assignment is to work individually.. Bruce – user59886 Jan 27 '13 at 23:28
@B.Shepherd: if you are the instructor of a course and you think the asker is one of your students cheating on their homework, you can consider the discussion about a similar incident before: One of my students may be getting math.SE to do their homework. Unfortunately, posting your comments here is likely to result in them simply being deleted. – Rahul Jan 28 '13 at 0:10
@B.Shepherd: you may be interested in this meta thread: meta.math.stackexchange.com/questions/1277/… – rschwieb Jan 28 '13 at 0:10
@B.Shepherd: I have converted this to a comment to prevent it from being deleted. – robjohn Jan 28 '13 at 0:32

The size of a maximum matching is $k$.
On one hand, it's easy to find a matching with size $k$.
On the other hand, if you find a matching with size bigger than $k$, then the matching has vertexes not less than $2k+2$ as no vertex are in two edges of the matching. But the cycle has $2k+1$ vertexes only. The contradictory occurs.
The size of a minimum vertex cover is $k+1$.
On one hand, it's easy to find a vertex cover with size $k+1$.
On the other hand, if you find a vertex cover with size smaller than $k+1$, then it covers at most $2k$ edges as one vertex covers at most $2$ edges. But the cycle has $2k+1$ edges. The contradictory occurs.