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Prove, using mathematical induction: if H is a bipartite simple graph on n vertices then H has ≤ $n^2/4$ edges if n is even.

Not terribly good at inductive proofs, especially in the context of graph theory. This is a representative problem from the chapter, so a breakdown would be much appreciated.

J. Sanders
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    Hint: Can you prove that, if $x+y=n,$ then $xy\le n^2/4$? – bof Aug 01 '17 at 03:05
  • Where does the x+y come in in this case? – J. Sanders Aug 01 '17 at 10:52
  • $H$ is a bipartite graph. It has $x$ vertices in one part, $y$ vertices in the other part. – bof Aug 01 '17 at 11:03
  • Ah, okay! Thanks for that clarification. Now tackling the proof, algebraically how do we go from addition to multiplication? – J. Sanders Aug 01 '17 at 11:34
  • Finding the maximum value of $xy$ given the constraint $x+y=n$ is a standard calculus problem. It's the same as maximizing the area of a rectangle with a perimeter of $2n.$ – bof Aug 01 '17 at 12:04
  • Algebraically: $$xy=\frac{(x+y)^2-(x-y)^2}4\le\frac{(x+y)^2}4$$ – bof Aug 01 '17 at 12:15
  • It's the case $n=2$ of the famous AM-GM inequality $$(x_1x_2\cdots x_n)^{1/n}\le\frac{x_1+x_2+\cdots+x_n}n$$ for positive $x_1,x_2,\dots,x_n;$ for a simple proof see this answer. – bof Aug 01 '17 at 12:25
  • Okay, so I'm starting to see where we're going from here. But where does induction come into play? Is it even necessary at this point? – J. Sanders Aug 01 '17 at 13:08
  • I've been ignoring the part that says "using mathematical induction". I don't know why one would use induction on this problem. – bof Aug 01 '17 at 13:14
  • Yeah, that is where I am having difficulty. I also thought of approaching it this way on pretty early, but I do not know where induction would come in. – J. Sanders Aug 01 '17 at 13:21
  • Does the above proof still apply since we are looking for the number of edges, not the area? – J. Sanders Aug 01 '17 at 13:38
  • A simple bipartite graph, with $x$ vertices in one part and $y$ vertices in the other part, has at most $xy$ edges; this maximum is attained by the complete bipartite graph $K_{x,y}.$ – bof Aug 01 '17 at 21:34
  • The mathematics of maximizing $f(x,y)=xy$ subject to the constraing $g(x,y)=x+y=c$ is the same, regardless of whether $xy$ is the area of a square with height $x$ and width $y$ or the number of edges in the complete bipartite graph $K_{x,y}.$ – bof Aug 02 '17 at 00:20

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Just giving an inductive solution since OP asked for one, in addition to the direct counting ones in the comments above:

Induct on $k = \frac{n}{2}$. Note that then $\frac{n^2}{4} = \frac{(2k)^2}{4} = k^2$.

Base case: $k = 1$; the graph can have at most one edge between these two vertices, and $1 \leq 1^2$.

Induction Step: Assume true for $k$, now WTS that a bipartite graph with $2k+2$ vertices has at most $(k+1)^2$ edges. Consider an arbitrary bipartite graph $G = (U,V,E)$ on $2k+2$ vertices. WLOG, let $|U| \leq |V|$.

Case 1: $|U| = 1$. Then every edge has to be between this vertex and one in $V$, giving us at most $2k +1$ choices. Since $k \geq 1$, $2k+1 < 2(k+1) \leq (k+1)^2$ and we are finished.

Case 2: $|U| > 1$. Remove one vertex $u,v$, from each of $U$ and $V$, respectively. Note that this induced subgraph $H$ is still bipartite and has $2k+2-2 = 2k$ vertices; by the induction hypothesis, it has at most $k^2$ edges. Now replace the vertices added. Note that each of the $2k$ vertices in $H$ can have an edge to at most one of either $u$ or $v$ (since our original graph was bipartite). Finally, there can be at most one edge between the new vertices. This gives a total of at most $k^2 + 2k + 1 = (k+1)^2$ edges, and we are done.

platty
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