## 7 Pure Conditions
### Assur Graphs and Assur Decompositions
Each vertex can belong to only one strongly connected component (which may consist of only a single vertex), so the strongly connected components form a partition of the vertex set. There is a fast combinatorial algorithm for partitioning a directed graph into strongly connected components [106]. This is the basis for the Assur decomposition of a graph in [106]. **Theorem 10** (Shai, Sevatius, Sljoka and Whiteley, [80,106]). : _Assume \(G=(I,P;E)\) is a pinned isostatic graph in \(\mathbb{P}^{2}\). Then the following are equivalent:_  
1. \(G=(I,P;E)\) _is an Assur graph._
2. _If the set_ \(P\) _is contracted to a single vertex_ \(p\)_, then the resulting contracted graph is a rigidity circuit._
3. _Either the graph has a single inner vertex of degree 2 or each time we delete a vertex, the resulting pinned graph has a motion with non-zero velocity at all inner vertices (in generic position)._
4. _Deletion of any edge from_ \(G\) _results in a pinned graph that has a motion with non-zero velocity at all inner vertices (in generic position)._
5. _Any_ \(2\)_-directed orientation of_ \(G\) _is strongly connected._  
**Example 23**. : _Consider the pinned framework in Figure 33a. It has a \(2\)-directed orientation (\(b\)) so it is pinned isostatic. The circle in (\(c\)) highlights a set of directed edges in one direction which disconnects the pinned graph into two components, so it is not strongly connected and therefore is not Assur. The subgraph outside the circle is Assur. If the four directed edges crossing the circle are pinned, the subgraph inside the circle, with these edges pinned, also forms an Assur graph. In general, identifying all the pins identifies a (sub)-circuit-a subgraph which, with pins identified becomes a circuit-as an Assur graph: in this case with vertices \(V^{\prime}=\{a_{1},a_{2},a_{3},a_{4},p\}\)._  
**Example 24**. : _Consider the example in Figure 34a. This is isostatic, as the \(2\)-directed orientation in (\(b\)) confirms. As a \(2\)-directed graph, the orientation is strongly connected. When the pins are all identified, there is a single circuit (\(c\)) which includes all vertices and all edges. There is a (non-generic) singular realisation (\(d\)), where there is a non-trivial infinitesimal motion fixing the pins, which is represented visually with a parallel drawing of all the inner vertices (red arrows). Parallel drawing is discussed in much more detail in our companion paper [27]. (\(c\)) and (\(d\)) are illustrations of Theorems 10 and 11._  
**Theorem 11** (Servatius, Shai, Whiteley [105]). : _A pinned graph \(G\) is an Assur graph in \(\mathbb{P}^{2}\), if and only if it has a realisation \(p\) in \(\mathbb{P}^{2}\) such that_  
1. \((G,p)\) _has a unique (up to scalar) equilibrium stress which is non-zero on all edges; and_
2. \((G,p)\) _has a unique (up to scalar) infinitesimal motion, and this is non-zero on all inner vertices._  
These special positions \(p\) will be preserved by projective transformations. **Conjecture 2** ([105]). : _Let \(G\) be an Assur graph and let \((G,p)\) be a framework in \(\mathbb{P}^{2}\) with a single equilibrium stress which is non-zero on all edges. Then there is a unique (up to scaling) non-trivial infinitesimal motion that is non-zero on at least one end of each bar._  
The converse does not hold. If we pin a triangle with three bars to pinned vertices, it is Assur and the pinned pure condition has the triangle factor times the factor for the three pinning edges. If the triangle is collinear, then the stress is zero on some edges.