关于最小度为4的图中3-环数量的良好上界技术问询

Great question! Let's walk through the tightest upper bounds we can get for the number of triangles (3-cycles) in your graph (G) (with (v) vertices, (e) edges, and minimum degree (\delta=4)):

1. The Absolute Maximum: Complete Graphs #

When (G) is a complete graph (K_v) (which satisfies minimum degree 4 as long as (v≥5)), we hit the theoretical maximum number of triangles. Every triple of vertices forms a triangle, so:
[
t = \binom{v}{3} = \frac{v(v-1)(v-2)}{6}
]
This is the largest possible number of triangles you can have in any graph with (v) vertices, and it's achievable when (e = \binom{v}{2}) (all possible edges exist).

2. Tight Upper Bound for General Cases #

For graphs that aren't complete, we can use double-counting of edges in vertex neighborhoods to derive a tight bound:

• Every triangle is counted three times in the sum of edges across all vertex neighborhoods (once for each vertex in the triangle). So if (t) is the number of triangles, we have (3t = \sum_{u \in V} \text{edges in } N(u)), where (N(u)) is the neighborhood of (u).
• For any vertex (u), the maximum number of edges in (N(u)) is (\binom{d(u)}{2}) (when the neighborhood is a complete subgraph), and since (d(u)≥4), this is at least (\binom{4}{2}=6).

Putting these together, we get our first upper bound:
[
t ≤ \frac{1}{3} \sum_{u \in V} \binom{d(u)}{2}
]

Simplified Practical Bound #

If you want a more usable bound that only depends on (v) and (e), we can leverage the handshaking lemma ((\sum d(u) = 2e)) and the convexity of the function (\binom{x}{2}). For minimum degree 4 graphs, (e≥2v) (since (2e = \sum d(u) ≥4v)).

When the graph is as "regular as possible" (degrees are balanced), this simplifies to:
[
t ≤ \frac{e(2e - v)}{3v}
]
This bound is tight for 4-regular graphs (where (e=2v)): plugging in (e=2v) gives (t ≤2v), which matches the triangle count of disjoint unions of (K_5) (each (K_5) has 10 triangles across 5 vertices, so total (t=2v)).

More Precise Bound (For Degree-Distributed Graphs) #

If you have information about the degree distribution of your graph, you can get an even tighter bound. Suppose (k) vertices have maximum degree (\Delta), and the remaining (v-k) vertices have the minimum degree 4. Using the handshaking lemma to solve for (\Delta = \frac{2e -4(v-k)}{k}), the bound becomes:
[
t ≤ \frac{1}{3} \left( \frac{e(2e -k)}{k} + 6(v -k) \right)
]
This works when (\Delta ≥4) (which is guaranteed since (e≥2v)).

3. Edge-Centered Perspective #

Another way to think about it: each edge can be part of at most (\min(d(u)-1, d(v)-1)) triangles (where (u) and (v) are the edge's endpoints). For minimum degree 4, this means each edge could be in up to (v-2) triangles (if one endpoint is connected to every other vertex), but this bound is much looser than the neighborhood-based ones above.

To wrap up:

• The absolute maximum is (\binom{v}{3}) (complete graph).
• For most practical cases, (\frac{e(2e -v)}{3v}) is a tight, easy-to-use upper bound.
• If you know the degree distribution, the sum-based bound (\frac{1}{3}\sum \binom{d(u)}{2}) is the most precise.

内容的提问来源于stack exchange，提问作者Brian Nugent