= Solution
In the crystallographic convention for this course, an <admissible Coxeter graph> comes from linearly independent unit <vectors> $v_1,\ldots,v_r$ in a <Euclidean space>, such that for $i\ne j$,
$$
(v_i,v_j)\leq0,\qquad n_{ij}=4(v_i,v_j)^2\in\{0,1,2,3\}.
$$
The vertices $i,j$ have no connection when $n_{ij}=0$, and a single, double or triple connection when $n_{ij}=1,2,3$. Equivalently, the symmetric <Gram matrix>
$$
\boxed{G_{ii}=1,\qquad G_{ij}=-\frac{\sqrt{n_{ij}}}{2}\ (i\ne j)}
$$
must be <positive-definite>. This equivalence follows because a positive definite <Gram matrix> can be realized by linearly independent <vectors>, while such vectors necessarily have a positive definite Gram matrix. In the edge-labelled <Coxeter graph> convention, these connections have labels $3,4,6$, since the corresponding inner products are $-\cos(\pi/3)$, $-\cos(\pi/4)$ and $-\cos(\pi/6)$.
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