Evaluating triangle creation in bipartite graph

For a complete bipartite graph with parts K_{⌊n/2⌋, ⌈n/2⌉} plus an extra edge inside, the number of triangles formed is approximately the size of the opposite part. I’ve concluded that in the case of K_{a,b}, the triangle count will be floor(n/2) since each vertex in part A can form a triangle with the new edge's endpoints in part B.

Proving triangle count for dense graphs

For odd n, parts have sizes (a,b) where $a = \frac{n-1}{2}$ and $b = \frac{n+1}{2}$. Adding an edge inside the larger part B gives $a$ triangles. For even n, adding an edge inside either part results in exactly n/2 triangles.

To prove, we can use Mantel’s theorem or a probabilistic approach. We can also revert to degree-sum techniques for estimation of triangle count in dense graphs.

Refining triangle count argument

I wonder if removing an edge from a graph can reduce the number of triangles. If we take an edge that, when removed, reduces the total number of edges to $\lfloor n^2/4 \rfloor$, it might still leave at least $\lfloor n/2 \rfloor$ triangles, but it's not guaranteed. Rademacher’s theorem says that there will be at least $\lfloor n/2 \rfloor$ triangles in total without focusing on specific edges.

Optimizing partition for minimal triangles

To prove Mantel's theorem, we consider a partition that maximizes cross-cut edges. This guarantees that at least $\lfloor n^2/4 \rfloor$ edges exist between the parts. For a graph exceeding the Mantel bound by 1, the number of edges inside parts A and B must sum to at least 1, which helps minimize triangles.

Minimizing triangles with extra edges

To minimize triangles in a graph, adding edges within the larger partition yields exactly $\lfloor n/2 \rfloor$ triangles. To prove that $m = \lfloor n^2/4 \rfloor + 1$ edges guarantees this many triangles, we can use a partition maximizing cross edges and apply various counting techniques, including the max-cut approach. More detailed inequalities can help bound the number of triangles created.

Thinking through triangle minimization

When considering how to minimize triangles by adding edges, a key approach is always using a max-cut partition. Here, a triangle relies on having at least one inside edge, meaning we need to count these edges carefully. One option involves Zykov symmetrization to avoid triangle increase. Another alternative is using Goodman's bound for a lower triangle bound in bipartite-like graphs.

Rethinking triangle minimization proof

When exploring ways to minimize triangles in a graph, I'm considering a lot of steps. One option is focusing on maximum cuts, which can prevent triangles from occurring entirely within a single partition. Alternatively, using Mantel’s theorem, a triangle-free graph can provide a baseline, and choosing a graph with minimal triangles gives some insights into the structure of edges and partitions.

Exploring minimal triangles in graphs

I’m working through proving that a graph with edges greater than Turán’s number must contain triangles. Through various strategies—like using symmetrization and focusing on vertex degree—I’m interested in showing that the minimum number of triangles achievable is ⌊n/2⌋. My goal is to refine the argument and find a clean, rigorous proof using induction or known inequalities.

Exploring triangle bounds in graphs

I’m refining a proof using supersaturation to show that graphs with more edges than Turán’s number must have at least a certain number of triangles. I’m calculating the degree sums over edges, exploring Cauchy-Schwarz inequalities, and testing boundaries to find the minimal triangle count for large graphs, especially focusing on cases with $m = \lfloor n^2/4 \rfloor + 1$.

Revisiting triangle bounds

I’ve overcounted somewhere. The earlier inequality for summing triangle counts leads to inconsistencies — it suggested $\Theta(n^3)$, but that's too large. The corrected approach, involving inclusion-exclusion and taking codegrees into account, suggests the inequality’s bounds can’t hold. The sum over edges gives too high a value — I’ll refine further.

Revisiting triangle bounds with the Rademacher proof

So earlier, we had issues with the inequality — trying to reason through the bounds gave confusion. Perhaps I should present the classical Rademacher proof, which provides a clean argument using the maximal bipartite cut. The key here is to examine maximum triangle-free subgraphs and their edge counts for a sharper bound, which simplifies the problem.

Simplifying edge-switching proof strategy

I’ve been overcomplicating things with triangle counting. The crucial insight lies in edge-switching, where adding a cross-edge doesn't increase the number of triangles. A few conditions must hold, like ensuring careful selection of cross edges and avoiding internal triangles within parts. For a minimal triangle count, the structure simplifies to a complete bipartite graph with a single additional edge.

Providing a proof outline

For this problem, I'll present a concise proof based on Rademacher’s theorem, focusing on "edge shifting" operations that preserve edge count while not increasing the number of triangles. By applying symmetrization, we get the extremal configuration as a complete multipartite graph. I’ll finish by counting the triangles in this structure and showing that using cross edges does not introduce new triangles.

Thought for 5m 9s