Suppose we have a positive integer $n\geq 4$. We call a graph $G=(V,E)$ an *$K^{(n)}\_{n-1}$-graph* if


1. there are $n$ subsets $S\_1,\ldots, S\_n$, each consisting of $n-1$ points, and each are cliques;
2. for positive integers $i<j\leq n$ we have $|S\_i\cap S\_j| = 1$, and
3. for all $v\in V$ there are positive integers $i<j\leq n$ such that $S\_i \cap S\_j = \{v\}$.


If $G\_1, G\_2$ are both $K^{(n)}\_{n-1}$-graphs, do we have $G\_1\cong G\_2$?



A graph consisting of cliques pairwise intersecting in 1 point
The answer is yes (in a certain sense; see below). This proof seems a bit long... so perhaps someone can come up with an easier argument!


The assumption that the sets form cliques in a graph seems to be irrelevant here. In fact, this problem isn't really a "graph problem" at all, so let's rephrase it in terms of hypergraphs. We want to show that there is unique (up to isomorphism) $(n-1)$-uniform hypergraph $\mathcal{H}$ such that


1. $\mathcal{H}$ has exactly $n$ edges,
2. any two edges of $\mathcal{H}$ intersect on exactly one vertex, and
3. for every vertex $v$, there exists a pair $e,e'$ of hyperedges with $e\cap e'=\{v\}$.


**Claim 1:** $\mathcal{H}$ has at most $\binom{n}{2}$ vertices.


*Proof:* Consider the function which maps each vertex $v$ to a pair of hyperedges whose intersection is $\{v\}$. This map is injective. $\square$


Okay, now order the edges of $\mathcal{H}$ by $e\_1,\dots,e\_n$ in an arbitrary order and let $\mathcal{H}\_k$ be the subhypergraph of $\mathcal{H}$ induced by the hyperedges $e\_1,\dots,e\_k$. Clearly, $\mathcal{H}\_1$ consists of $n-1$ vertices contained in a single hyperedge.


**Claim 2:** For $k\geq2$ we have $|V(\mathcal{H}\_k)|\geq |V(\mathcal{H}\_{k-1})| + n-k$.


*Proof:* By assumption, $e\_k$ intersects each of the edges $e\_1,\dots,e\_{k-1}$ on exactly one vertex. Therefore, there must be at least $(n-1)-(k-1) = n-k$ vertices of $e\_k$ not contained in any of the edges $e\_1,\dots,e\_{k-1}$. This completes the proof. $\square$


So, by Claim 2,


$$|V(\mathcal{H})| \geq |V(\mathcal{H}\_1)| + (|V(\mathcal{H}\_2)|-|V(\mathcal{H}\_1)|) + \dots +(|V(\mathcal{H}\_n)|-|V(\mathcal{H}\_{n-1})|)$$
$$= (n-1) + (n-2)+\dots+1+0 = \binom{n}{2}.$$
Combining this with Claim 1, we see that we must have $|V(\mathcal{H})|=\binom{n}{2}$. Moreover, it must be the case that 
$$|V(\mathcal{H}\_k)| = \binom{n}{2} - \binom{n-k}{2}$$
for $0\leq k\leq n$ (otherwise, we could apply Claim 2 to obtain $|V(\mathcal{H})|>\binom{n}{2}$, contradicting Claim 1). 


In particular, we get that any vertex $v$ is contained in exactly two hyperedges of $\mathcal{H}$ (otherwise, if we let $e\_1,e\_2,e\_3$ be the three hyperedges containing $v$, then this contradicts the fact that $|V(\mathcal{H}\_3)|=3n-6$). 


Now, let us show that $\mathcal{H}\_k$ is uniquely determined (up to isomorphism) for each $k$ by induction. The result is trivial for $k=1$. Now for $k\geq2$, we can assume (by induction) that $\mathcal{H}\_{k-1}$ contains no vertex of degree greater than two and that each hyperedge contains exactly $n-k+1$ vertices of degree one. By the above arguments, the graph $\mathcal{H}\_k$ must be constructed from $\mathcal{H}\_{k-1}$ by 


(a) adding a set $A$ of $n-k$ new vertices, and


(b) adding a new hyperedge $e\_k$ containing $A$ and containing exactly one vertex $v\_i$ of degree one from the hyperedge $e\_i$ for $1\leq i\leq k-1$.


All that remains is to show that any such choice yields the same hypergraph, up to isomorphism. For $1\leq i\leq k-1$, let $u\_i$ and $v\_i$ be (possibly different) vertices of $e\_i$ of degree one. There is an isomorphism from the hypergraph obtained by adding the edge $A\cup \{u\_1,\dots,u\_{k-1}\}$ and the hypergraph obtained by adding the edge $A\cup \{v\_1,\dots,v\_{k-1}\}$ where $v\_i\mapsto u\_i$, $u\_i\mapsto v\_i$ and every other vertex simply maps to itself. This completes the proof.


**Edit:** As others have pointed out, what I have proved is that these graphs are unique under the additional assumption that *every edge of $G$ is contained in one of the cliques $S\_1,...,S\_n$.* Without this assumption, the class of graphs is closed under adding edges, and therefore we cannot have uniqueness. I originally interpreted the problem as having this assumption, but I do agree that it was not included in the statement.


