N人公司满足三人组约束的一一相识关系数量求解

Hey there! Let's break down this problem using graph theory, since your initial approach misses some critical cases that make it incorrect.

First, let's translate the problem into graph terms:

• Treat each person as a vertex in a simple graph.
• Draw an edge between two vertices if the corresponding people know each other; no edge means they don't.
• The requirement "every trio has both acquaintances and strangers" translates to two key graph properties:
  1. No triangles (clique of size 3, meaning three people all know each other).
  2. No 3-independent sets (three people who all don't know each other).

Why your initial approach is wrong #

Your idea of splitting the group into disjoint trios and counting 6 valid configurations per trio only works if those trios are completely isolated from each other—but that's not allowed! Any three people from different trios could form a forbidden 3-independent set (if none of them know each other) or a triangle (if all three know each other), violating the problem's rules. For example, if you split 6 people into two trios, picking one stranger from each trio could create a trio of mutual strangers, which is not permitted.

Structure of valid graphs #

Valid graphs (called $(K_3, \overline{K_3})$-free graphs) have strict structural constraints:
Every connected component of the graph must be one of these:

• A single isolated vertex (but you can have at most 2 of these total—three isolated vertices would form a forbidden 3-independent set).
• A single edge (two people who know each other).
• A 3-vertex path (one person knows two others, who don't know each other).
• A 4-vertex path.
• A 5-vertex cycle (the only odd cycle allowed, since longer odd cycles have 3-independent sets).
• No larger components, because they would either contain a triangle or a 3-independent set.

Counting valid graphs #

There's no simple closed-form formula for this count, but we can use exponential generating functions (EGF) to calculate the number for a given $N$:

1. Define EGFs for each valid connected component:
   • Isolated vertices: $1 + x + \frac{x^2}{2!}$ (covers 0, 1, or 2 isolated vertices; divided by $2!$ because two isolated vertices form an identical graph regardless of order).
   • Edges (K₂): $\frac{x^2}{2!}$ (counts all ways to pick two connected vertices).
   • 3-vertex paths (P₃): $\frac{x^3}{2}$ (counts all undirected 3-vertex paths).
   • 4-vertex paths (P₄): $\frac{x^4}{2}$ (counts all undirected 4-vertex paths).
   • 5-vertex cycles (C₅): $\frac{x^5}{10}$ (counts all undirected 5-vertex cycles).
2. Combine these into a total EGF:
   $$G(x) = \left(1 + x + \frac{x^2}{2}\right) \times \exp\left(\frac{x^2}{2} + \frac{x^3}{2} + \frac{x^4}{2} + \frac{x^5}{10}\right)$$
   The $\exp(\dots)$ term accounts for any combination of non-isolated connected components, and the leading factor adds the allowed isolated vertices.
3. For a given $N$, the number of valid graphs is $N! \times [x^N]G(x)$, where $[x^N]G(x)$ is the coefficient of $x^N$ in the expanded generating function.

Example verification #

• For $N=3$: Your initial approach gives $6^{3/3}=6$, which matches the valid count (all 3-vertex graphs except the triangle and empty graph). The generating function confirms this: $3! \times [x^3]G(x) = 6 \times \frac{1}{2} = 6$.
• For $N=4$: Your approach would give a non-integer result, which is impossible. Using the generating function, we calculate $4! \times [x^4]G(x) = 24 \times \frac{11}{8} = 33$, which counts all valid 4-vertex graphs (including pairs of edges, 4-vertex paths, 4-vertex cycles, and 3-vertex paths plus an isolated vertex).

内容的提问来源于stack exchange，提问作者Vladok AC