1. Planarity Testing

2. Overview Definitions Menger’s Theorem Ear Decomposition Lemma Pieces
Planarity Testing Algorithm
Complexity O(n )
Connected – otherwise can show for each connected component. 3
* We’ll only discuss connected undirected graphs

3. Definitions Cut vertex Cut edge Block
Two blocks share at most one vertex – cut-vertex
Connected Components: 3 1
Connected Components: 2 1
Cut vertex\edge – whose deletion increases the number of connected components.
Block – maximal connected sub-graph with no cut-vertex
Why? Suppose 2 or more shared vertices, then deleting one still leaves a connected graph – contradicts the maximally property of a block.
Why cut vertex? If it’s the only connection, then removing it will disconnect the two blocks, adding to the number of connected components.

4. DFS Depth First Search Use the DFS tree to find Blocks G DFS tree 3 45 7 6 1 2 G
Undirected graph – only tree or back edges.
Block – if no w BELOW v has an ancestor above u – which is v’s parent, then the sub tree of v along with u is a block.
DFS tree

5. Menger’s Theorem (1927) G is k-connected 
k-connected – deleting (k-1) vertices cannot disconnect it (|V(G)| >= k).
Corollary (Dirac 1960) –
G is 2-connected  G consists of a single block
G is k-connected 
Any pair u, v V(G) can be connected by k vertex-disjoint paths.
G is k-connected
Any k vertices of G
lie on a simple cycle

6. ST-Ordering Numbering - v1, v2, ... , vn of V(G) so that
(v1, vn) E(G)
vi has a neighbor vj where j < i
vi also has a neighbor vk where i < k (for every i , 1 < i < n)
G admits an st-ordering  G is 2-connected
Note that finding ST ordering isn’t trivial (though there are algorithms).
Explain 2-connected  single block.
If G is 2-connected then any pair vi, vj that are adjacent can be chosen as V1 and Vn.
We’ll prove it soon

7. Ear Decomposition Lemma
Every 2-connected graph can be obtained from a cycle by adding paths
For each path:
Both end points are on the current graph
But otherwise it is disjoint from it
Proof
Gi G, Gi ≠ G the current graph.
Pick u V(Gi) , v V(Gi) such that (u, v) E(G) and connect v to Gi by a shortest path.

8. G admits an st-ordering  G is 2-connected
Use induction on |E(G)|
Show G admits st-ordering with
v1 = u, v2, ... , vn = v (u, v) E(G)
Pick a cycle C through (u, v)
Base:
G = C
trivial
(v1) u v
(v5)
G is 2-connected, so there’s such cycle.
Base is trivial because the ordering will be determined by a walk from u to v on the cycle (without using (u,v)). v2 v4 v3

9. Continued
C ≠ G
(v1) u v
(v5)
G0 = C G

10. Continued C ≠ G Add a path to it as in the ear decomposition
Number the path’s vertices so they form an increasing chain connecting its endpoints
(v1) u v
(v5)
G0 = C G1
We might need to adjust our numbering later v4 v3
Shortest path

11. Continued C ≠ G Add a path to it as in the ear decomposition
Number the path’s vertices so they form an increasing chain connecting its endpoints
(v1) u v
(v5) G2 G1 v2 v4 v3

12. Pieces G is 2-connected C is a cycle in G Pieces C is Separating P2 P1
The remaining red vertices are called “Attachments”.
Separating – gives rise to at least 2 pieces.
Finding the pieces is linear in the number of edges P3

13. Pieces
Pieces can be drawn on either side of C P2 P1 P3

14. Pieces
Pieces can be drawn on either side of C P2 P1 P3

15. Pieces
Pieces can be drawn on either side of C P2 P1 P3

16. Pieces
Pieces can be drawn on either side of C P2 P1 P3

17. Pieces
Two pieces Interlace or Conflict if they cannot be drawn on the same side of C without crossing edges
When does this happen? G G’ P2 a1
Cyclic order: a1, b1, a3, b3 P1 a2 b1 b3 P3 a3 b2

18. Pieces G’ might be a planar graph P2 G’ P1 P3
Because p3 can be drawn on the outside. P3

19. Pieces G’ might be a planar graph
If one of the conflicting pieces can be drawn on the other side of C P2 G’ P1 P3

20. Our Goal Find 2 sets (S1, S2) of pieces so that: P2 G P1 P3
No two pieces in the same set conflict
S S2 C = G ∩ ∩ P2 G
S1 = {P1, P2}
S2 = {P3} P1 P3

21. Interlacement Graph Vertex set – the set of pieces (with respect to C)
Two vertices are connected  the pieces interlace
Interlacement(G, C): P2 G P1 P3

22. Is G planar? G 2-connected graph with cycle C G is planar 
For each piece P (with respect to C)
P C is a planar graph
The Interlacement graph is bipartite (2-colorable) ∩

23. Why? P2
Interlacement(G, C): G S1 P1 S2
If Interlacement(G, C) is bipartite, we can divide the pieces to 2 sets - like we wanted P3

24. Why? Each piece combined with C is a planar graph
And we can assemble all the pieces together so that they don’t interlace
So G is planar

25. Planarity Testing Recursive Algorithm
G 2-connected graph with
n vertices
O(n) edges
C is a cycle in G
The Algorithm:
Find pieces with respect to C
Build interlace graph
Check if it is bipartite
Check planarity for C’s pieces (recursively) ( )
O(n)
O(n) edges – from Euler’s Theorem corollary : E <= 3*V – 6.
The number of pieces (V in the interlace graph) <= n.
That’s also why there’ll be O(n) recursive calls at most (totall).
Check for bipartite:
Perform BFS (Broad First Search), and add the following check: Grey node (that was already discovered) v that we encounter from a new node u most be even if u is odd, or odd if u is even for the graph to be bipartite. + *
O(n ) 2 +
O(n ) 2
O(n)
Total Cost = O(n ) 3

26. Questions?