Reachability

In graph theory, reachability refers to the ability to get from one vertex to another within a graph. A vertex ${\displaystyle s <semantics><mrow><mstyle><mi>s</mi></mstyle></mrow><annotation>{\displaystyle s}</annotation></semantics></math><img src=$ ${\displaystyle t <semantics><mrow><mstyle><mi>t</mi></mstyle></mrow><annotation>{\displaystyle t}</annotation></semantics></math><img src=$ ${\displaystyle t <semantics><mrow><mstyle><mi>t</mi></mstyle></mrow><annotation>{\displaystyle t}</annotation></semantics></math><img src=$ ${\displaystyle s <semantics><mrow><mstyle><mi>s</mi></mstyle></mrow><annotation>{\displaystyle s}</annotation></semantics></math><img src=$ ${\displaystyle s <semantics><mrow><mstyle><mi>s</mi></mstyle></mrow><annotation>{\displaystyle s}</annotation></semantics></math><img src=$ ${\displaystyle t <semantics><mrow><mstyle><mi>t</mi></mstyle></mrow><annotation>{\displaystyle t}</annotation></semantics></math><img src=$

In an undirected graph, reachability between all pairs of vertices can be determined by identifying the connected components of the graph. Any pair of vertices in such a graph can reach each other if and only if they belong to the same connected component; therefore, in such a graph, reachability is symmetric ( ${\displaystyle s <semantics><mrow><mstyle><mi>s</mi></mstyle></mrow><annotation>{\displaystyle s}</annotation></semantics></math><img src=$ ${\displaystyle t <semantics><mrow><mstyle><mi>t</mi></mstyle></mrow><annotation>{\displaystyle t}</annotation></semantics></math><img src=$ ${\displaystyle t <semantics><mrow><mstyle><mi>t</mi></mstyle></mrow><annotation>{\displaystyle t}</annotation></semantics></math><img src=$ ${\displaystyle s <semantics><mrow><mstyle><mi>s</mi></mstyle></mrow><annotation>{\displaystyle s}</annotation></semantics></math><img src=$

Definition

For a directed graph ${\displaystyle G=(V,E) <semantics><mrow><mstyle><mi>G</mi><mo>=</mo><mo>(</mo><mi>V</mi><mo>,</mo><mi>E</mi><mo>)</mo></mstyle></mrow><annotation>{\displaystyle G=(V,E)}</annotation></semantics></math><img src=$ ${\displaystyle V <semantics><mrow><mstyle><mi>V</mi></mstyle></mrow><annotation>{\displaystyle V}</annotation></semantics></math><img src=$ ${\displaystyle E <semantics><mrow><mstyle><mi>E</mi></mstyle></mrow><annotation>{\displaystyle E}</annotation></semantics></math><img src=$ ${\displaystyle G <semantics><mrow><mstyle><mi>G</mi></mstyle></mrow><annotation>{\displaystyle G}</annotation></semantics></math><img src=$ ${\displaystyle E <semantics><mrow><mstyle><mi>E</mi></mstyle></mrow><annotation>{\displaystyle E}</annotation></semantics></math><img src=$ ${\displaystyle (s,t) <semantics><mrow><mstyle><mo>(</mo><mi>s</mi><mo>,</mo><mi>t</mi><mo>)</mo></mstyle></mrow><annotation>{\displaystyle (s,t)}</annotation></semantics></math><img src=$ ${\displaystyle V <semantics><mrow><mstyle><mi>V</mi></mstyle></mrow><annotation>{\displaystyle V}</annotation></semantics></math><img src=$ ${\displaystyle v_{0}=s,v_{1},v_{2},...,v_{k}=t <semantics><mrow><mstyle><msub><mi>v</mi><mrow><mn>0</mn></mrow></msub><mo>=</mo><mi>s</mi><mo>,</mo><msub><mi>v</mi><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mi>v</mi><mrow><mn>2</mn></mrow></msub><mo>,</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>,</mo><msub><mi>v</mi><mrow><mi>k</mi></mrow></msub><mo>=</mo><mi>t</mi></mstyle></mrow><annotation>{\displaystyle v_{0}=s,v_{1},v_{2},...,v_{k}=t}</annotation></semantics></math><img src=$ ${\displaystyle (v_{i-1},v_{i}) <semantics><mrow><mstyle><mo>(</mo><msub><mi>v</mi><mrow><mi>i</mi><mo>−</mo><mn>1</mn></mrow></msub><mo>,</mo><msub><mi>v</mi><mrow><mi>i</mi></mrow></msub><mo>)</mo></mstyle></mrow><annotation>{\displaystyle (v_{i-1},v_{i})}</annotation></semantics></math><img src=$ ${\displaystyle E <semantics><mrow><mstyle><mi>E</mi></mstyle></mrow><annotation>{\displaystyle E}</annotation></semantics></math><img src=$ ${\displaystyle 1\leq i\leq k <semantics><mrow><mstyle><mn>1</mn><mo>≤</mo><mi>i</mi><mo>≤</mo><mi>k</mi></mstyle></mrow><annotation>{\displaystyle 1\leq i\leq k}</annotation></semantics></math><img src=$

If ${\displaystyle G <semantics><mrow><mstyle><mi>G</mi></mstyle></mrow><annotation>{\displaystyle G}</annotation></semantics></math><img src=$ ${\displaystyle s <semantics><mrow><mstyle><mi>s</mi></mstyle></mrow><annotation>{\displaystyle s}</annotation></semantics></math><img src=$ ${\displaystyle t <semantics><mrow><mstyle><mi>t</mi></mstyle></mrow><annotation>{\displaystyle t}</annotation></semantics></math><img src=$ ${\displaystyle t <semantics><mrow><mstyle><mi>t</mi></mstyle></mrow><annotation>{\displaystyle t}</annotation></semantics></math><img src=$ cannot reach ${\displaystyle s <semantics><mrow><mstyle><mi>s</mi></mstyle></mrow><annotation>{\displaystyle s}</annotation></semantics></math><img src=$ ${\displaystyle s <semantics><mrow><mstyle><mi>s</mi></mstyle></mrow><annotation>{\displaystyle s}</annotation></semantics></math><img src=$ ${\displaystyle t <semantics><mrow><mstyle><mi>t</mi></mstyle></mrow><annotation>{\displaystyle t}</annotation></semantics></math><img src=$ ${\displaystyle s <semantics><mrow><mstyle><mi>s</mi></mstyle></mrow><annotation>{\displaystyle s}</annotation></semantics></math><img src=$ ${\displaystyle G <semantics><mrow><mstyle><mi>G</mi></mstyle></mrow><annotation>{\displaystyle G}</annotation></semantics></math><img src=$ ${\displaystyle G <semantics><mrow><mstyle><mi>G</mi></mstyle></mrow><annotation>{\displaystyle G}</annotation></semantics></math><img src=$ not acyclic (i.e. it contains at least one cycle), then its reachability relation will correspond to a preorder instead of a partial order.

Algorithms

Algorithms for determining reachability fall into two classes: those that require preprocessing and those that do not.

If you have only one (or a few) queries to make, it may be more efficient to forgo the use of more complex data structures and compute the reachability of the desired pair directly. This can be accomplished in linear time using algorithms such as breadth first search or iterative deepening depth-first search.

If you will be making many queries, then a more sophisticated method may be used; the exact choice of method depends on the nature of the graph being analysed. In exchange for preprocessing time and some extra storage space, we can create a data structure which can then answer reachability queries on any pair of vertices in as low as ${\displaystyle O(1) <semantics><mrow><mstyle><mi>O</mi><mo>(</mo><mn>1</mn><mo>)</mo></mstyle></mrow><annotation>{\displaystyle O(1)}</annotation></semantics></math><img src=$

Floyd–Warshall Algorithm

The Floyd–Warshall algorithm can be used to compute the transitive closure of any directed graph, which gives rise to the reachability relation as in the definition, above.

The algorithm requires ${\displaystyle O(|V|^{3}) <semantics><mrow><mstyle><mi>O</mi><mo>(</mo><mrow><mo>|</mo></mrow><mi>V</mi><msup><mrow><mo>|</mo></mrow><mrow><mn>3</mn></mrow></msup><mo>)</mo></mstyle></mrow><annotation>{\displaystyle O(|V|^{3})}</annotation></semantics></math><img src=$ ${\displaystyle O(|V|^{2}) <semantics><mrow><mstyle><mi>O</mi><mo>(</mo><mrow><mo>|</mo></mrow><mi>V</mi><msup><mrow><mo>|</mo></mrow><mrow><mn>2</mn></mrow></msup><mo>)</mo></mstyle></mrow><annotation>{\displaystyle O(|V|^{2})}</annotation></semantics></math><img src=$

Thorup's Algorithm

For planar digraphs, a much faster method is available, as described by Mikkel Thorup in 2004. This method can answer reachability queries on a planar graph in ${\displaystyle O(1) <semantics><mrow><mstyle><mi>O</mi><mo>(</mo><mn>1</mn><mo>)</mo></mstyle></mrow><annotation>{\displaystyle O(1)}</annotation></semantics></math><img src=$ ${\displaystyle O(n\log {n}) <semantics><mrow><mstyle><mi>O</mi><mo>(</mo><mi>n</mi><mi>log</mi><mo>⁡</mo><mrow><mi>n</mi></mrow><mo>)</mo></mstyle></mrow><annotation>{\displaystyle O(n\log {n})}</annotation></semantics></math><img src=$ ${\displaystyle O(n\log {n}) <semantics><mrow><mstyle><mi>O</mi><mo>(</mo><mi>n</mi><mi>log</mi><mo>⁡</mo><mrow><mi>n</mi></mrow><mo>)</mo></mstyle></mrow><annotation>{\displaystyle O(n\log {n})}</annotation></semantics></math><img src=$

The overall approach is to associate with each vertex a relatively small set of so-called separator paths such that any path from a vertex ${\displaystyle v <semantics><mrow><mstyle><mi>v</mi></mstyle></mrow><annotation>{\displaystyle v}</annotation></semantics></math><img src=$ ${\displaystyle w <semantics><mrow><mstyle><mi>w</mi></mstyle></mrow><annotation>{\displaystyle w}</annotation></semantics></math><img src=$ ${\displaystyle v <semantics><mrow><mstyle><mi>v</mi></mstyle></mrow><annotation>{\displaystyle v}</annotation></semantics></math><img src=$ ${\displaystyle w <semantics><mrow><mstyle><mi>w</mi></mstyle></mrow><annotation>{\displaystyle w}</annotation></semantics></math><img src=$

Given a graph ${\displaystyle G <semantics><mrow><mstyle><mi>G</mi></mstyle></mrow><annotation>{\displaystyle G}</annotation></semantics></math><img src=$ ${\displaystyle v_{0 <semantics><mrow><mstyle><msub><mi>v</mi><mrow><mn>0</mn></mrow></msub></mstyle></mrow><annotation>{\displaystyle v_{0}}</annotation></semantics></math><img src=$ from the previous step (starting with just ${\displaystyle v_{0 <semantics><mrow><mstyle><msub><mi>v</mi><mrow><mn>0</mn></mrow></msub></mstyle></mrow><annotation>{\displaystyle v_{0}}</annotation></semantics></math><img src=$ to the previous step until all vertices have been assigned to a layer. By construction of the layers, every vertex appears at most two layers, and every directed path, or dipath, in ${\displaystyle G <semantics><mrow><mstyle><mi>G</mi></mstyle></mrow><annotation>{\displaystyle G}</annotation></semantics></math><img src=$ ${\displaystyle L_{i <semantics><mrow><mstyle><msub><mi>L</mi><mrow><mi>i</mi></mrow></msub></mstyle></mrow><annotation>{\displaystyle L_{i}}</annotation></semantics></math><img src=$ ${\displaystyle L_{i+1 <semantics><mrow><mstyle><msub><mi>L</mi><mrow><mi>i</mi><mo>+</mo><mn>1</mn></mrow></msub></mstyle></mrow><annotation>{\displaystyle L_{i+1}}</annotation></semantics></math><img src=$ ${\displaystyle k <semantics><mrow><mstyle><mi>k</mi></mstyle></mrow><annotation>{\displaystyle k}</annotation></semantics></math><img src=$ ${\displaystyle k <semantics><mrow><mstyle><mi>k</mi></mstyle></mrow><annotation>{\displaystyle k}</annotation></semantics></math><img src=$ ${\displaystyle \bigcup _{i=0}^{k}L_{i}=V <semantics><mrow><mstyle><munderover><mo>⋃</mo><mrow><mi>i</mi><mo>=</mo><mn>0</mn></mrow><mrow><mi>k</mi></mrow></munderover><msub><mi>L</mi><mrow><mi>i</mi></mrow></msub><mo>=</mo><mi>V</mi></mstyle></mrow><annotation>{\displaystyle \bigcup _{i=0}^{k}L_{i}=V}</annotation></semantics></math><img src=$

The graph is then re-expressed as a series of digraphs ${\displaystyle G_{0},G_{1},\ldots ,G_{k-1 <semantics><mrow><mstyle><msub><mi>G</mi><mrow><mn>0</mn></mrow></msub><mo>,</mo><msub><mi>G</mi><mrow><mn>1</mn></mrow></msub><mo>,</mo><mo>…</mo><mo>,</mo><msub><mi>G</mi><mrow><mi>k</mi><mo>−</mo><mn>1</mn></mrow></msub></mstyle></mrow><annotation>{\displaystyle G_{0},G_{1},\ldots ,G_{k-1}}</annotation></semantics></math><img src=$ ${\displaystyle G_{i}=r_{i}\cup L_{i}\cup L_{i+1 <semantics><mrow><mstyle><msub><mi>G</mi><mrow><mi>i</mi></mrow></msub><mo>=</mo><msub><mi>r</mi><mrow><mi>i</mi></mrow></msub><mo>∪</mo><msub><mi>L</mi><mrow><mi>i</mi></mrow></msub><mo>∪</mo><msub><mi>L</mi><mrow><mi>i</mi><mo>+</mo><mn>1</mn></mrow></msub></mstyle></mrow><annotation>{\displaystyle G_{i}=r_{i}\cup L_{i}\cup L_{i+1}}</annotation></semantics></math><img src=$ ${\displaystyle r_{i <semantics><mrow><mstyle><msub><mi>r</mi><mrow><mi>i</mi></mrow></msub></mstyle></mrow><annotation>{\displaystyle r_{i}}</annotation></semantics></math><img src=$ ${\displaystyle L_{0}\ldots L_{i-1 <semantics><mrow><mstyle><msub><mi>L</mi><mrow><mn>0</mn></mrow></msub><mo>…</mo><msub><mi>L</mi><mrow><mi>i</mi><mo>−</mo><mn>1</mn></mrow></msub></mstyle></mrow><annotation>{\displaystyle L_{0}\ldots L_{i-1}}</annotation></semantics></math><img src=$ ${\displaystyle G_{i <semantics><mrow><mstyle><msub><mi>G</mi><mrow><mi>i</mi></mrow></msub></mstyle></mrow><annotation>{\displaystyle G_{i}}</annotation></semantics></math><img src=$ ${\displaystyle G <semantics><mrow><mstyle><mi>G</mi></mstyle></mrow><annotation>{\displaystyle G}</annotation></semantics></math><img src=$ ${\displaystyle G_{i <semantics><mrow><mstyle><msub><mi>G</mi><mrow><mi>i</mi></mrow></msub></mstyle></mrow><annotation>{\displaystyle G_{i}}</annotation></semantics></math><img src=$

For each ${\displaystyle G_{i <semantics><mrow><mstyle><msub><mi>G</mi><mrow><mi>i</mi></mrow></msub></mstyle></mrow><annotation>{\displaystyle G_{i}}</annotation></semantics></math><img src=$ ${\displaystyle 1/2 <semantics><mrow><mstyle><mn>1</mn><mrow><mo>/</mo></mrow><mn>2</mn></mstyle></mrow><annotation>{\displaystyle 1/2}</annotation></semantics></math><img src=$ ${\displaystyle G_{i <semantics><mrow><mstyle><msub><mi>G</mi><mrow><mi>i</mi></mrow></msub></mstyle></mrow><annotation>{\displaystyle G_{i}}</annotation></semantics></math><img src=$ ${\displaystyle S <semantics><mrow><mstyle><mi>S</mi></mstyle></mrow><annotation>{\displaystyle S}</annotation></semantics></math><img src=$

For each ${\displaystyle Q\in S <semantics><mrow><mstyle><mi>Q</mi><mo>∈</mo><mi>S</mi></mstyle></mrow><annotation>{\displaystyle Q\in S}</annotation></semantics></math><img src=$ ${\displaystyle Q <semantics><mrow><mstyle><mi>Q</mi></mstyle></mrow><annotation>{\displaystyle Q}</annotation></semantics></math><img src=$ ${\displaystyle v <semantics><mrow><mstyle><mi>v</mi></mstyle></mrow><annotation>{\displaystyle v}</annotation></semantics></math><img src=$ ${\displaystyle G_{i <semantics><mrow><mstyle><msub><mi>G</mi><mrow><mi>i</mi></mrow></msub></mstyle></mrow><annotation>{\displaystyle G_{i}}</annotation></semantics></math><img src=$ ${\displaystyle Q <semantics><mrow><mstyle><mi>Q</mi></mstyle></mrow><annotation>{\displaystyle Q}</annotation></semantics></math><img src=$ ${\displaystyle v <semantics><mrow><mstyle><mi>v</mi></mstyle></mrow><annotation>{\displaystyle v}</annotation></semantics></math><img src=$ ${\displaystyle Q <semantics><mrow><mstyle><mi>Q</mi></mstyle></mrow><annotation>{\displaystyle Q}</annotation></semantics></math><img src=$ ${\displaystyle v <semantics><mrow><mstyle><mi>v</mi></mstyle></mrow><annotation>{\displaystyle v}</annotation></semantics></math><img src=$ ${\displaystyle Q <semantics><mrow><mstyle><mi>Q</mi></mstyle></mrow><annotation>{\displaystyle Q}</annotation></semantics></math><img src=$ ${\displaystyle v <semantics><mrow><mstyle><mi>v</mi></mstyle></mrow><annotation>{\displaystyle v}</annotation></semantics></math><img src=$ ${\displaystyle Q <semantics><mrow><mstyle><mi>Q</mi></mstyle></mrow><annotation>{\displaystyle Q}</annotation></semantics></math><img src=$ ${\displaystyle v <semantics><mrow><mstyle><mi>v</mi></mstyle></mrow><annotation>{\displaystyle v}</annotation></semantics></math><img src=$ ${\displaystyle v <semantics><mrow><mstyle><mi>v</mi></mstyle></mrow><annotation>{\displaystyle v}</annotation></semantics></math><img src=$ ${\displaystyle u <semantics><mrow><mstyle><mi>u</mi></mstyle></mrow><annotation>{\displaystyle u}</annotation></semantics></math><img src=$ ${\displaystyle w <semantics><mrow><mstyle><mi>w</mi></mstyle></mrow><annotation>{\displaystyle w}</annotation></semantics></math><img src=$ ${\displaystyle u <semantics><mrow><mstyle><mi>u</mi></mstyle></mrow><annotation>{\displaystyle u}</annotation></semantics></math><img src=$ ${\displaystyle w <semantics><mrow><mstyle><mi>w</mi></mstyle></mrow><annotation>{\displaystyle w}</annotation></semantics></math><img src=$ via ${\displaystyle Q <semantics><mrow><mstyle><mi>Q</mi></mstyle></mrow><annotation>{\displaystyle Q}</annotation></semantics></math><img src=$ ${\displaystyle u <semantics><mrow><mstyle><mi>u</mi></mstyle></mrow><annotation>{\displaystyle u}</annotation></semantics></math><img src=$ ${\displaystyle Q <semantics><mrow><mstyle><mi>Q</mi></mstyle></mrow><annotation>{\displaystyle Q}</annotation></semantics></math><img src=$ ${\displaystyle w <semantics><mrow><mstyle><mi>w</mi></mstyle></mrow><annotation>{\displaystyle w}</annotation></semantics></math><img src=$ ${\displaystyle Q <semantics><mrow><mstyle><mi>Q</mi></mstyle></mrow><annotation>{\displaystyle Q}</annotation></semantics></math><img src=$

Every vertex is labelled as above for each step of the recursion which builds ${\displaystyle G_{0}\ldots ,G_{k <semantics><mrow><mstyle><msub><mi>G</mi><mrow><mn>0</mn></mrow></msub><mo>…</mo><mo>,</mo><msub><mi>G</mi><mrow><mi>k</mi></mrow></msub></mstyle></mrow><annotation>{\displaystyle G_{0}\ldots ,G_{k}}</annotation></semantics></math><img src=$ ${\displaystyle O(\log {n}) <semantics><mrow><mstyle><mi>O</mi><mo>(</mo><mi>log</mi><mo>⁡</mo><mrow><mi>n</mi></mrow><mo>)</mo></mstyle></mrow><annotation>{\displaystyle O(\log {n})}</annotation></semantics></math><img src=$ ${\displaystyle Q <semantics><mrow><mstyle><mi>Q</mi></mstyle></mrow><annotation>{\displaystyle Q}</annotation></semantics></math><img src=$ ${\displaystyle O(1) <semantics><mrow><mstyle><mi>O</mi><mo>(</mo><mn>1</mn><mo>)</mo></mstyle></mrow><annotation>{\displaystyle O(1)}</annotation></semantics></math><img src=$

In summarizing the analysis of this method, first consider that the layering approach partitions the vertices so that each vertex is considered only ${\displaystyle O(1) <semantics><mrow><mstyle><mi>O</mi><mo>(</mo><mn>1</mn><mo>)</mo></mstyle></mrow><annotation>{\displaystyle O(1)}</annotation></semantics></math><img src=$ ${\displaystyle 1/2 <semantics><mrow><mstyle><mn>1</mn><mrow><mo>/</mo></mrow><mn>2</mn></mstyle></mrow><annotation>{\displaystyle 1/2}</annotation></semantics></math><img src=$ ${\displaystyle O(n\log n) <semantics><mrow><mstyle><mi>O</mi><mo>(</mo><mi>n</mi><mi>log</mi><mo>⁡</mo><mi>n</mi><mo>)</mo></mstyle></mrow><annotation>{\displaystyle O(n\log n)}</annotation></semantics></math><img src=$ ${\displaystyle O(\log {n}) <semantics><mrow><mstyle><mi>O</mi><mo>(</mo><mi>log</mi><mo>⁡</mo><mrow><mi>n</mi></mrow><mo>)</mo></mstyle></mrow><annotation>{\displaystyle O(\log {n})}</annotation></semantics></math><img src=$

Kameda's Algorithm

An even faster method for pre-processing, due to T. Kameda in 1975, can be used if the graph is planar, acyclic, and also exhibits the following additional properties: all 0-indegree and all 0-outdegree vertices appear on the same face (often assumed to be the outer face), and it is possible to partition the boundary of that face into two parts such that all 0-indegree vertices appear on one part, and all 0-outdegree vertices appear on the other (i.e. the two types of vertices do not alternate).

Le prétraitement effectue les étapes suivantes. Nous ajoutons un nouveau sommet. ${\displaystyle s <semantics><mrow><mstyle><mi>s</mi></mstyle></mrow><annotation>{\displaystyle s}</annotation></semantics></math><img src=$ ${\displaystyle t <semantics><mrow><mstyle><mi>t</mi></mstyle></mrow><annotation>{\displaystyle t}</annotation></semantics></math><img src=$ ${\displaystyle G <semantics><mrow><mstyle><mi>G</mi></mstyle></mrow><annotation>{\displaystyle G}</annotation></semantics></math><img src=$ ${\displaystyle i=n+1 <semantics><mrow><mstyle><mi>je</mi><mo>=</mo><mi>n</mi><mo>+</mo><mn>1</mn></mstyle></mrow><annotation>{\displaystyle i=n+1}</annotation></semantics></math><img src=$ ${\displaystyle s <semantics><mrow><mstyle><mi>s</mi></mstyle></mrow><annotation>{\displaystyle s}</annotation></semantics></math><img src=$ ${\displaystyle i <semantics><mrow><mstyle><mi>je</mi></mstyle></mrow><annotation>{\displaystyle i}</annotation></semantics></math><img src=$ ${\displaystyle i <semantics><mrow><mstyle><mi>je</mi></mstyle></mrow><annotation>{\displaystyle i}</annotation></semantics></math><img src=$ ${\displaystyle t <semantics><mrow><mstyle><mi>t</mi></mstyle></mrow><annotation>{\displaystyle t}</annotation></semantics></math><img src=$ ${\displaystyle n+1 <semantics><mrow><mstyle><mi>n</mi><mo>+</mo><mn>1</mn></mstyle></mrow><annotation>{\displaystyle n+1}</annotation></semantics></math><img src=$ ${\displaystyle s <semantics><mrow><mstyle><mi>s</mi></mstyle></mrow><annotation>{\displaystyle s}</annotation></semantics></math><img src=$ ${\displaystyle 0 <semantics><mrow><mstyle><mn>0</mn></mstyle></mrow><annotation>{\displaystyle 0}</annotation></semantics></math><img src=$

Une fois terminé, ${\displaystyle s <semantics><mrow><mstyle><mi>s</mi></mstyle></mrow><annotation>{\displaystyle s}</annotation></semantics></math><img src=$ ${\displaystyle t <semantics><mrow><mstyle><mi>t</mi></mstyle></mrow><annotation>{\displaystyle t}</annotation></semantics></math><img src=$ ${\displaystyle 1 <semantics><mrow><mstyle><mn>1</mn></mstyle></mrow><annotation>{\displaystyle 1}</annotation></semantics></math><img src=$ ${\displaystyle n <semantics><mrow><mstyle><mi>n</mi></mstyle></mrow><annotation>{\displaystyle n}</annotation></semantics></math><img src=$ ${\displaystyle u <semantics><mrow><mstyle><mi>vous</mi></mstyle></mrow><annotation>{\displaystyle u}</annotation></semantics></math><img src=$ ${\displaystyle v <semantics><mrow><mstyle><mi>v</mi></mstyle></mrow><annotation>{\displaystyle v}</annotation></semantics></math><img src=$ ${\displaystyle L(u)=(a_{1},a_{2}) <semantics><mrow><mstyle><mi>L</mi><mo>(</mo><mi>vous</mi><mo>)</mo><mo>=</mo><mo>(</mo><msub><mi>un</mi><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mi>un</mi><mrow><mn>2</mn></mrow></msub><mo>)</mo></mstyle></mrow><annotation>{\displaystyle L(u)=(a_{1},a_{2})}</annotation></semantics></math><img src=$ ${\displaystyle L(v)=(b_{1},b_{2}) <semantics><mrow><mstyle><mi>L</mi><mo>(</mo><mi>v</mi><mo>)</mo><mo>=</mo><mo>(</mo><msub><mi>b</mi><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mi>b</mi><mrow><mn>2</mn></mrow></msub><mo>)</mo></mstyle></mrow><annotation>{\displaystyle L(v)=(b_{1},b_{2})}</annotation></semantics></math><img src=$ ${\displaystyle L(u)<L(v) <semantics><mrow><mstyle><mi>L</mi><mo>(</mo><mi>vous</mi><mo>)</mo><mo><</mo><mi>L</mi><mo>(</mo><mi>v</mi><mo>)</mo></mstyle></mrow><annotation>{\displaystyle L(u)<L(v)}</annotation></semantics></math><img src=$ ${\displaystyle a_{1}\leq b_{1 <semantics><mrow><mstyle><msub><mi>un</mi><mrow><mn>1</mn></mrow></msub><mo>≤</mo><msub><mi>b</mi><mrow><mn>1</mn></mrow></msub></mstyle></mrow><annotation>{\displaystyle a_{1}\leq b_{1}}</annotation></semantics></math><img src=$ ${\displaystyle a_{2}\leq b_{2 <semantics><mrow><mstyle><msub><mi>un</mi><mrow><mn>2</mn></mrow></msub><mo>≤</mo><msub><mi>b</mi><mrow><mn>2</mn></mrow></msub></mstyle></mrow><annotation>{\displaystyle a_{2}\leq b_{2}}</annotation></semantics></math><img src=$ ${\displaystyle a_{1 <semantics><mrow><mstyle><msub><mi>un</mi><mrow><mn>1</mn></mrow></msub></mstyle></mrow><annotation>{\displaystyle a_{1}}</annotation></semantics></math><img src=$ ${\displaystyle a_{2 <semantics><mrow><mstyle><msub><mi>un</mi><mrow><mn>2</mn></mrow></msub></mstyle></mrow><annotation>{\displaystyle a_{2}}</annotation></semantics></math><img src=$ ${\displaystyle b_{1 <semantics><mrow><mstyle><msub><mi>b</mi><mrow><mn>1</mn></mrow></msub></mstyle></mrow><annotation>{\displaystyle b_{1}}</annotation></semantics></math><img src=$ ${\displaystyle b_{2 <semantics><mrow><mstyle><msub><mi>b</mi><mrow><mn>2</mn></mrow></msub></mstyle></mrow><annotation>{\displaystyle b_{2}}</annotation></semantics></math><img src=$

Le principal résultat de cette méthode est donc que ${\displaystyle v <semantics><mrow><mstyle><mi>v</mi></mstyle></mrow><annotation>{\displaystyle v}</annotation></semantics></math><img src=$ ${\displaystyle u <semantics><mrow><mstyle><mi>vous</mi></mstyle></mrow><annotation>{\displaystyle u}</annotation></semantics></math><img src=$ ${\displaystyle L(u)<L(v) <semantics><mrow><mstyle><mi>L</mi><mo>(</mo><mi>vous</mi><mo>)</mo><mo><</mo><mi>L</mi><mo>(</mo><mi>v</mi><mo>)</mo></mstyle></mrow><annotation>{\displaystyle L(u)<L(v)}</annotation></semantics></math><img src=$ ${\displaystyle O(1) <semantics><mrow><mstyle><mi>O</mi><mo>(</mo><mn>1</mn><mo>)</mo></mstyle></mrow><annotation>{\displaystyle O(1)}</annotation></semantics></math><img src=$

Problèmes connexes

Un problème connexe consiste à résoudre les requêtes d'accessibilité avec un certain nombre ${\displaystyle k <semantics><mrow><mstyle><mi>k</mi></mstyle></mrow><annotation>{\displaystyle k}</annotation></semantics></math><img src=$ ${\displaystyle u <semantics><mrow><mstyle><mi>vous</mi></mstyle></mrow><annotation>{\displaystyle u}</annotation></semantics></math><img src=$ ${\displaystyle v <semantics><mrow><mstyle><mi>v</mi></mstyle></mrow><annotation>{\displaystyle v}</annotation></semantics></math><img src=$ ${\displaystyle s_{1},s_{2},...,s_{k <semantics><mrow><mstyle><msub><mi>s</mi><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mi>s</mi><mrow><mn>2</mn></mrow></msub><mo>,</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>,</mo><msub><mi>s</mi><mrow><mi>k</mi></mrow></msub></mstyle></mrow><annotation>{\displaystyle s_{1},s_{2},...,s_{k}}</annotation></semantics></math><img src=$

Un autre problème lié aux requêtes d'accessibilité concerne le recalcul rapide des relations d'accessibilité lorsqu'une partie du graphe est modifiée. Par exemple, cela pose un problème important pour le ramasse-miettes , qui doit trouver un équilibre entre la récupération de la mémoire (afin qu'elle puisse être réallouée) et les impératifs de performance de l'application en cours d'exécution.