Loading in 5 sec....

Separator Theorems for Planar GraphsPowerPoint Presentation

Separator Theorems for Planar Graphs

- 124 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' Separator Theorems for Planar Graphs' - mikaia

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

- S – a class of graphs.
- An f(n)-separator theorem for S is a theorem of the following form:
- There exist constants α<1, β>0, such that, if G is any n-vertex graph in S, the vertices of G can be partitioned into three sets A, B, C, such that no edge joins a vertex in A with a vertex in B, neither A nor B contains more than αn vertices, and C contains no more than βf(n) vertices.

Theorem 1 (Lipton & Tarjan)

Let G be any n-vertex planar graph.

The vertices of G can be partitioned into three sets A, B, C such that:

- No edge joins a vertex in A with a vertex in B.
- Neither A nor B contains more than 2n/3 vertices.
- C contains no more than vertices.

Facts

- Any n-vertex binary tree can be separated into two subtrees, each with no more than 2n/3 vertices, by removing a single edge.
- Any n-vertex tree can be divided into two parts, each with no more than 2n/3 vertices, by removing a single vertex.
- A -separator theorem holds for the class of grid graphs.

Most Graphs do not have a good Separator Theorem

- Theorem: for every ε>0, there exists a positive constant c=c(ε) such that almost all graphs G with n=(2+ε)k vertices and ck edges have the property that, after the omission of any k vertices, a connected component of at least k vertices remains.
- Sparsity is not enough, but planarity is!!

Some facts about planarity

- Let C be any simple closed curve in the plane. Removal of C divides the plane into exactly two connected regions, the “inside” and the “outside” of C.
- Any n-vertex planar graph with n≥3 contains no more than 3n-6 edges.
- A graph is planar iff it contains neither a complete graph on 5 vertices nor a complete bipartite graph on two sets of 3 vertices as a generalized subgraph.
4. Any planar graph can be triangulated.

Lemma 1

- Let G be any planar graph. Shrinking any edge of G to a single vertex, we preserve planarity.
- Proof:

Corollary 1

- Let G be any planar graph. Shrinking any connected subgraph of G to a single vertex preserves planarity.
- Proof: Immediate from Lemma 1 by induction on the number of vertices in the subgraph to be shrunk.

Lemma 2

Let G be any planar graph with:

- Nonnegative vertex costs summing to at most 1.
- A spanning tree of radius r.
Then the vertices of G can be partitioned into three sets A, B, C, such that:

- No edge joins a vertex in A with a vertex in B.
- Neither A nor B has total cost exceeding 2/3.
- C contains no more than 2r+1 vertices, one the root of the tree.

Lemma 2 – Proof

Assume no vertex has cost exceeding 1/3. Otherwise, this vertex is C and the rest of the graph is B.

Make each face a triangle by adding additional edges.

Any nontree edge forms a simple cycle with some of the tree edges.

This cycle is of length at most 2r+1 or 2r-1. The cycle divides the plane (and the graph) into two parts.

Lemma 2 – Proof – Cont.

- Claim: at least one cycle separates the graph so that neither the inside nor the outside contains vertices whose total cost exceeds 2/3.
- Proof: Let (x,z) be the nontree edge whose cycle minimizes the maximum cost either inside or outside the cycle.
Break ties by choosing the nontree edge whose cycle has the smallest number of faces on the same side as the maximum cost.

Lemma 2 – Proof – Cont.

- Suppose without l.o.g that the graph is embedded so that the cost inside the (x,z) cycle is ≥ the cost outside the cycle.
- If the vertices inside the cycle have total cost ≤ 2/3, the claim is true.
- Suppose the vertices inside the cycle have total cost > 2/3.
- We show by case analysis that this contradicts the choice of (x,z).

Consider the face which has (x,z) as a boundary edge and lies inside the cycle.

Lemma 3 - Definitions lies inside the cycle.

- Let G be any n-vertex connected planar graph having nonnegative vertex costs summing to no more than 1.
- The vertices of G are partitioned into levels according to their distance from some vertex v.
- L(l) denotes the number of vertices at level l.
- r - the maximum distance of any vertex from v.
- Let r+1 be an additional level containing no vertices.

Lemma 3 lies inside the cycle.

Given any two levels l1 and l2 such that:

- Levels 0 through l1-1 have total cost not exceeding 2/3
- Levels l2+1 through r+1 have total cost not exceeding 2/3.
It is possible to find a partition A, B, C of the vertices of G such that:

- No edge joins a vertex in A with a vertex in B.
- Neither A nor B has total cost exceeding 2/3.
- C contains no more than L(l1)+L(l2)+max{0,2(l2-l1-1)} vertices.

Lemma 3 - Proof lies inside the cycle.

If l1≥l2,

let A be all vertices on levels 0 through l1-1,

B all vertices on levels l1+1 through r,

and C all vertices on level l1.

Then the Lemma is true.

0

l2

A

l1-1

l1

C

l1+1

B

r

Proof – Cont. lies inside the cycle.

<2/3

- Suppose l1<l2. Delete the vertices at levels l1 and l2 from G. This separates the remaining vertices of G into three parts: vertices at levels 0 – (l1-1), vertices at levels
(l1+1) – (l2-1) and vertices at levels l2+1 and above.

The only part which may have a cost exceeding 2/3 is the middle part.

l1-1

l1

l1+1

l2-1

l2

l2+1

<2/3

Proof – cont. lies inside the cycle.

- If the middle part does not have cost exceeding 2/3,
let A be the most costly part of the three,

let B be consist of the remaining two parts,

and let C be the set of vertices at levels l1 and l2.

Then the lemma is true.

l1-1

l1

l1+1

C

l2-1

l2

l2+1

Proof – cont. lies inside the cycle.

- Suppose the middle part has cost exceeding 2/3.
Delete all vertices at levels l2 and above, and shrink all vertices at levels l1 and below to a single vertex of cost zero.

These operations preserve planarity by Cor. 1.

The new graph has a spanning tree of radius l2-l1-1 whose root corresponds to vertices at levels l1 and below in the original graph.

l1-1

0

l1

l1+1

l2-1

l2

l2+1

Proof – Cont. lies inside the cycle.

- Apply Lemma 2 to the new graph.
Let A*, B*, C* be the resulting vertex partition.

Let A be the set among A* and B* having greater cost.

Let C consist of the vertices at levels l1 and l2 in the original graph and the vertices in C*, excluding the root of the tree, and let B contain the remaining vertices in G.

- By Lemma 2, A has total cost not exceeding 2/3. But AUC* has total cost at least 1/3, so B also has total cost at most 2/3. Furthermore, C contains no more than L(l1)+L(l2)+2(l2-l1-1) vertices. ■

Theorem 4 lies inside the cycle.

Let G be any n-vertex planar graph having nonnegative vertex costs summing to no more than 1.

Then the vertices of G can be partitioned into three sets A, B, C such that:

- No edge joins a vertex in A with a vertex in B,
- Neither A nor B has total cost exceeding 2/3,
- C contains no more than vertices.

Theorem 4 – Proof For a connected G lies inside the cycle.

0

- Partition the vertices into levels according to their distance from some vertex v. Let L(l) be the number of vertices on level l. If r is the maximum distance of any vertex from v, define additional levels -1 and r+1 containing no vertices.
- Let l1 be the level such that the sum of costs in levels 0 through l1-1 is less than ½, but the sum of costs in levels 0 through l1 is at least ½.
- (If no such l1exists, then B=C=ø satisfies the theorem.)

l0

k

l1-1

<½

l1

>½

l2

Proof – Cont. lies inside the cycle.

0

- Let k be the number of vertices at levels 0 through l1. Find a level l0such that l0≤l1 and |L(l0)|+2(l1-l0)≤2√k. Find a level l2 such that l1+1≤l2 and |L(l2)|+2(l2-l1-1)≤2√(n-k).
- If two such levels exist, then by Lemma 3 the vertices of G can be partitioned into three sets A, B, C such that no edge joins a vertex in A with a vertex in B, neither A nor B has cost exceeding 2/3, and C contains no more than 2(√k+√(n-k)) vertices.

l0

k

l1-1

<½

l1

>½

l2

Proof – Cont. lies inside the cycle.

- But,
- Thus, the theorem holds if suitable levels l0 and l2exist.
- Suppose a suitable level l0does not exist. Then, for i≤l1,
. Since L(0)=1, this means

, and .

Thus, and

which is a contradiction.

Proof For a disconnected G lies inside the cycle.

- Let G1, G2,…Gk be the connected components of G, with vertex sets V1, V2,…,Vk.
- If some connected component Gi has total vertex cost between 1/3 and 2/3, let A=Vi, B=G\Vi and C=ø.

Proof For a disconnected G lies inside the cycle.

2. If no connected component has cost >1/3, let i be the minimum index such that the total cost of V1UV2U…UVi exceeds 1/3.

- Let A= V1UV2U…UVi, let B= Vi+1UVi+2U…UVk, and let C=ø.
- Since i is minimal and the cost of vi ≤ 1/3, the cost of A ≤ 2/3.

Proof For a not connected G lies inside the cycle.

- If some connected component (say Gi) has total cost > 2/3, apply the theorem on Gi.
- Let A*, B*, C* be the resulting partition.
- Let A be the set among A* and B* with greater cost.
- C=C*.
- Let B be the remaining vertices of G.

An Algorithm for finding a good partition lies inside the cycle.

- A representation of a planar embedding of a graph:
A list structure – for the edges of the graph, its endpoints and 4 pointers, designating the edges immediately clockwise and counter-clockwise around each of the endpoints of the edge.

Stored with each vertex is some incident edge.

Example of a representation lies inside the cycle.

The algorithm lies inside the cycle.

- Find a planar embedding of G and construct a representation as described before.
- Find the connected components of G and determine the cost of each one. If none has cost >2/3, construct the partition as described in the proof of the theorem.
- Find a BFS tree of the most costly component. Compute the level of each vertex and compute L(l) in each level l.

The algorithm – Cont. lies inside the cycle.

- Back to slides 22-23.
- Construct a new vertex x to represent all vertices on levels 0 through l0.
Construct a Boolean table with one entry per vertex. Initialize to true the entry for each vertex on levels 0-l0 and initialize to false the entry for each vertex on levels l0+1 through l2-1.

The algorithm – Cont. lies inside the cycle.

The vertices on levels 0-l0 correspond to a subtree of the BFS generated before.

Scan the edges incident to this tree clockwise around the tree.

When scanning an edge (v,w) with v in the tree, check the table entry for w. If it is true – delete edge (v,w). If it is false – change it to true, create (x,w) and delete (v,w).

We get a planar representation of the shrunken graph, to which Lemma 2 is to be applied.

Planar representation for the shrunken graph - Example lies inside the cycle.

The algorithm – Cont. lies inside the cycle.

- Construct a BFS for the shrunken graph.
- Find an appropriate cycle, with cost ≤ 2/3 in the inside of the cycle.
- Use the found cycle and levels l0 and l2 (found in step 4) to construct a satisfactory vertex partition as described in the proof of Lemma 3. Extend this partition from the connected component chosen in step 2, to the entire graph as described in the proof of Theorem 4.
→ All steps require O(n) time.

Finding an appropriate cycle lies inside the cycle.

Application of the theorem lies inside the cycle.

- Divide-and-conquer in combination with the theorem can be used to rapidly find good approximate solutions to certain NP-complete problems on planar graph.
- The maximum independent set problem.

The maximum independent set problem. lies inside the cycle.

- Theorem 5: Let G be an n-vertex planar graph with nonnegative vertex costs summing to no more than 1, and let 0 ≤ ε≤ 1. Then there is some set C of O(√(n/ε)) vertices whose removal leaves G with no connected component of cost exceeding ε.
- C can be found in O(n log n) time.

Proof for theorem 5 lies inside the cycle.

- If ε ≤ 1/√n, let C contain all the vertices of G.
- Otherwise, apply the following algorithm:
Init: Let C=ø.

Step: Find some connectedcomponent K in G\C with cost > ε.

Apply Theorem 4 to K, producing a partition A1, B1, C1 of its vertices.

Let C=CUC1.

Proof – Cont. lies inside the cycle.

- Repeat the step until G\C has no component with cost > ε.
- Consider all components arising during the course of the algorithm. Assign a level to each component.

Level 0

Level 1

Level 2

Level 3

Complexity lies inside the cycle.

- Any two components at the same level are vertex-disjoint.
- Each level 1 component has cost > ε.
- For i≥1, each level i component has cost ≥ .
- The total cost of G is ≤ 1.
- Therefore, the total number of components of level i is at most .
- The maximum level k satisfy

Complexity – Cont. lies inside the cycle.

- The time to split a component is linear in its number of vertices.
- Any two components at the same level are vertex-disjoint.
→ The total running time of the algorithm is O(n log n).

The size of the set C lies inside the cycle.

- Let K1, K2, …, Kl, of sizes n1, n2, …, nl, respectively, be the components of some level i≥1.
- The number of vertices added to C by splitting K1, …, Kl is at most .
- and .

The size of C – Cont. lies inside the cycle.

- We have
which leads to

Algorithm for IS lies inside the cycle.

- Apply Theorem 5 to G with ε=k(n)/n and each vertex having cost 1/n.
Thus we find a set of vertices C of size O(n/√k(n)) whose removal leaves no connected component with more than k(n) vertices.

2. In each connected component of G\C, find a max IS by checking every subset of vertices. Form I as a union of max ISs, one from each component.

Correctness lies inside the cycle.

- Let I* be a maximum IS of G.
- I*\I cannot have vertices from G\C. (all possibilities were checked.)
→I*\I can have vertices only from C.

→|I*|-|I|=O(n/√k(n)).

- G is planar → G is 4-colorable→ |I*| ≥ n/4.
→

Correctness – Cont. lies inside the cycle.

- Thus, the relative error in the size of I → 0 as n → ∞ if K(n) → ∞ as n → ∞.

Complexity lies inside the cycle.

- Step 1 of the algorithm requires O(n log n) by Theorem 5.
- Step 2 requires time on a connected component of ni vertices.
- The total time required by step 2:

Complexity – Cont. lies inside the cycle.

- Hence the entire algorithm requires
time.

- If we choose k(n)=log n, we get an O(n²)-time algorithm with O(1/√(log n)) relative error.
- If we choose k(n)=log log n, we get an O(n log n)-time algorithm with O(1/√(log log n)) relative error.

Better Results lies inside the cycle.

- Alon, Seymour and Thomas gave a shorter proof to the theorem of Lipton & Tarjan.
- They also proved a better theorem, with a smaller constant.

The new Separator Theorem lies inside the cycle.

Let G be any n-vertex planar graph.

The vertices of G can be partitioned into three sets A, B, C such that:

- No edge joins a vertex in A with a vertex in B.
- Neither A nor B contains more than 2n/3 vertices.
- C contains no more than vertices.
This theorem also holds for weighted graphs.

Download Presentation

Connecting to Server..