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Section 3.1 Graph Coloring

Problem 3.1.1.

Your goal is to schedule all the committee meetings using as few time slots as possible. If two committees share a member, they must meet at different times. The committee assignments are below.

  1. Guido, Swen, and Ute

  2. Guido, Swen, and Eva Marie

  3. Eva Marie, Solveig, and Lars

  4. Lars, Solveig, and Wolfgang

  5. Wolfgang, Ute, and Andrea

Solution.

Let's make finding a solution easier by labeling each committee as different sets.

  • Let \(v_1 = \{\text{Guido, Swen, and Ute}\}\) and call it Group 1.

  • Let \(v_2 = \{\text{Guido, Swen, and Eva Marie}\}\) and call it Group 2.

  • Let \(v_3 = \{\text{Eva Marie, Solveig, and Lars}\}\) and call it Group 3.

  • Let \(v_4 = \{\text{Lars, Solveig, and Wolfgang}\}\) and call it Group 4.

  • Let \(v_5 = \{\text{Wolfgang, Ute, and Andrea}\}\) and call it Group 5.

This way we can work systematically by analyzing who belongs to which group and who doesn't belong to each group.

Let's start with Group 1.

Guido and Swen are both in Group 1 and Group 2, so those groups must be in different time slots. Ute is Group 1 and Group 5. So Groups 1, 2, and 5 cannot be in the same time slot

Eva Marie is both Group 2 and Group 3, so those groups cannot meet at the same time. Both Solveig and Lars are in Group 3 and Group 4, so those two groups cannot meet at the same time. So among those three members, we can conclude that Group 3 can share a time slot with Group 1 or Group 5. Let them share with Group 1.

Wolfgang is both Group 4 and Group 5, so those two groups cannot share the same time slot. Andrea is only a member of Group 5. So Group 5 can share a time slot with Group 2 or Group 3. However, since Group 3 shares a time slot with Group 1 and Ute is in Group 1 as well as Group 5, we can conclude that Group 5 will share a time slot with Group 2.

This leaves us with Group 4, which has members Lars, Solveig, and Wolfgang. This means Group 4 cannot meet when Groups 3 or 4 meet. This also means they cannot meet when Groups 1 and 5 meet, since Groups 1 and 5 meet at the same time as Groups 3 and 4, respectively. Ergo, Group 4 will have to meet at a time separate from all the other groups.

Thus, the fewest time slot that can be used is 3.

Problem 3.1.2.

Acme Inc is buying advertising space at each intersection in a downtown area. To minimize cost and maximize buyer interest they want to produce as few advertisements as possible and avoid having the same advertisement at both ends of a block. The graph in FigureĀ 3.1.3 has a vertex for each intersection, and the edges are the blocks.

Solution.

Let us try to meet the conditions with only 2 different advertisements.

This does not work, because if we start in the top lefthand corner and work our way down alternating colors, the bottom righthand corner has the same advertisement at both ends of a block.

Now let us try to meet the conditions with 3 different advertisements. This easily fixes our problem because the bottom-most center point can be replaced by the third advertisement. Now each block has two different advertisements on either end.

Figure 3.1.3. Advertising Plan

Definition 3.1.4.

A vertex coloring of a graph is an assignment of labels to each vertex such that no adjacent vertices share the same label.

Definition 3.1.5.

The minimum number of labels required to color a graph \(G\) is called the chromatic number of \(G\text{.}\) It is denoted \(\chi(G)\text{.}\)

Problem 3.1.6.

For each of the following find the chromatic number.

(d)

\(F8_{n,m}\)

Solution.

Three cases:

  1. When m and n are even, \(\chi(F8_m,n) = 2\)

  2. When m is even and n is odd, and vice versa, \(\chi(F8_m,n) = 3\)

  3. When m and n are odd, \(\chi(F8_m,n) = 3\)

(e)

Bipartite graphs

Solution.

A bipartite graph is a graph whose vertices can be divided into two disjoint sets M and N. Each set of vertices in M and N are not adjacent to each other, so you only need two colors for the graph. All vertices in M are one color, and all vertices in N are another color.

(f)

Tripartite graphs

Solution.

Similarly, for a tripartite graph, you need three colors, per the same logic as above.

Problem 3.1.8.

Describe a graph with \(\chi(G)=1\text{.}\)

Solution.

Any graph with at least one vertex and no edges will have a chromatic colouring of one, since no vertex is adjacent to any other.

Problem 3.1.9.

Describe a graph with \(\chi(G)=n\) where \(n\) is the number of vertices.

Solution.

Suppose \(G\) is a graph such that \(\chi(G)=n\) where \(n\) is the number of vertices in \(G\text{.}\) We know from TheoremĀ 3.1.7 that the largest clique in this graph must be of size \(n\text{.}\) However, for \(\chi(G)\) to equal the number of vertices, \(G\) must be a complete graph (even if \(G=K_1\text{,}\) the trivial case). By DefinitionĀ 1.3.4, every vertex in a complete graph is adjacent to every other vertex present in the graph. Thus, if \(G=K_n\) then every vertex must receive a unique label to prevent any vertex from being adjacent to a vertex with the same label and \(\chi(G)=n\text{.}\)

Definition 3.1.10.

Kiva(G) is a function to make a graph using graph G where:.

G has nodes {G1, G2, ... Gk} with \(k \gt 0\) nodes.

Kiva(G) has nodes {A1, A2, ... An, B1, B2, ... Bn, C1, C2, ... Cn, D1, D2, ... Dn, and E1, E2, ... En}

Kiva(G) contains edges {Akn, Akm} for all n,m where G has edge {Gkn, Gkm}

Kiva(G) contains edges {Bkn, Bkm} for all n,m where G has edge {Gkn, Gkm}

Kiva(G) contains edges {Ckn, Ckm} for all n,m where G has edge {Gkn, Gkm}

Kiva(G) contains edges {Dkn, Dkm} for all n,m where G has edge {Gkn, Gkm}

Kiva(G) contains edges {Ekn, Ekm} for all n,m where G has edge {Gkn, Gkm}

Kiva(G) contains edges {An, Bm} for all n,m where \(0 \lt n \lt k\text{,}\) and \(0 \lt m \lt k\text{.}\)

Kiva(G) contains edges {Bn, Cm} for all n,m where \(0 \lt n \lt k\text{,}\) and \(0 \lt m \lt k\text{.}\)

Kiva(G) contains edges {Cn, Dm} for all n,m where \(0 \lt n \lt k\text{,}\) and \(0 \lt m \lt k\text{.}\)

Kiva(G) contains edges {Dn, Em} for all n,m where \(0 \lt n \lt k\text{,}\) and \(0 \lt m \lt k\text{.}\)

Kiva(G) contains edges {En, Fm} for all n,m where \(0 \lt n \lt k\text{,}\) and \(0 \lt m \lt k\text{.}\)

Kiva(G) contains no other edges.

note: Kiva(G) contains no edges between any {A,B},{B,C},{C,D},{D,E} or {E,A}.

note: in Kiva(G) A, B, C, D, E can be relabeled by rotating in alphabetical order, or reversing the order of A-E labeling.

note: the subset containing only nodes labeled A, is isomorphic to G.

Proof.

Direct Proof:

Take Kiva(G) for some graph G

Let k1 be equal to G's maximum clique.

Let A, B, C, D, E represent the subset of Kiva(G) containing nodes labeled An, Bn, Cn, Dn, En.

The maximum clique in {A} is equal to k1 because they are isomorphic.

As Kiva(G) can be relabeled this applies to B,C,D,E also.

By definition All nodes in A are adjacent to all nodes in B and E

By definition no edges from B to E exits.

By definition no edges from A to C or D exist.

Since all nodes within A's Maximum clique are adjacent to B's maximum clique, those two cliques form a clique size \(2 \cdot k1\text{.}\)

Since any node in B were not part of B's maximum clique they can not be added to form a bigger clique.

Since all nodes in B are not adjacent to E, and all nodes in C or D are not adjacent to A, no nodes from C, D, or E can be added to form a largest clique.

Therefor we cannot add any more nodes to increase the clique.

Since Kiva(G) can be relabeled, this applies for any B,C,D,E also.

Since the maximum clique from A was started with, and all other nodes were considered, we have formed the maximum clique of size \(2 \cdot k1\text{.}\)

This proves: Kiva(G)'s maximum clique is 2*G's maximum clique.

Proof.

Take Kiva(G) for some graph G

Let k2 be equal to G's maximum clique.

Let A, B, C, D, E represent the subset of Kiva(G) containing nodes labeled An, Bn, Cn, Dn, En.

The coloring number of A is equal to k2 because they are isomorphic.

As Kiva(G) can be relabeled this applies to B,C,D,E also.

A cannot share a coloring with B as all nodes in A are adjacent to all nodes in B

as Kiva(G) can be relabeled this applies to A B, B C, C D, D E or E A

Therefor A can share a coloring with C,D

Therefor B can share a coloring with D,E

Therefor C can share a coloring with E,A

Therefor D can share a coloring with F,A

Therefor E can share a coloring with B,C

This forms a odd cycle, which from previous proofs three different colors on the nodes

Coloring (A)(BD)(CE) is one such coloring

Since Each subgraph needs k2 colors, the coloring of Kiva(G) is \(3 \cdot k2\)

The coloring number of Kiva(G) is 3*G's coloring number.

Proof.

Notation Kivan(G) represents Kiva(Kiva(Kiva(G))) where Kiva() appears n times.

Example: Kiva2(G) represents Kiva(Kiva(G))

Example: Kiva1(G) represents Kiva(G)

From previous proof Kiva(G)'s coloring Ratio is G's coloring * 3.

From previous proof Kiva(G)'s maximum clique is G's maximum clique * 2

Kivan(k1) has \(3 \cdot 1\) coloring and \(2 \cdot 1\) maximum clique, and (3^1)/(2^1) coloring number to maximum clique ratio.

By induction, Kivan(k1) has \(3^n\) coloring and \(2^n\) maximum clique, and \((3/2)^n\) coloring number to maximum clique ratio.

K1 has 1 coloring and 1 maximum clique.

Base: n=1. Kiva1(K1) has \(3 \cdot 1\) coloring and \(2 \cdot 1\) maximum clique, and \((3^1)/(2^1)\) coloring number to maximum clique ratio.

Inductive: Kiva(Kivan(k1)) has \(3^(n+1)\) coloring number, \(2^(n+1)\) maximum clique, and \((3^(n+1))/(2^(n+1))\) coloring number to maximum clique ratio.

Kivan(k1) has \(3^n\) coloring, \(2^n+1\) maximum clique, and \((3^1)/(2^1)\) coloring number to maximum clique ratio.

From previous, Kiva(Kivan*k1)) has \(3 \cdot (3^n)\) coloring, \(2 \cdot (2^n)\) maximum clique, and thus \((3 \cdot (3^n))/(2 \cdot (2^1))\) coloring number to maximum clique ratio.

Reducing, Kiva(Kivan*k1)) has \(3^(n+1)\) coloring number, \(2^(n+1)\) maximum clique, and \((3^(n+1))/(2^(n+1))\) coloring number to maximum clique ratio.

Thus: Kivan(k1) has \(3^n\) coloring and \(2^n\) maximum clique, and \((3/2)^n\) coloring number to maximum clique ratio.

As Kivan(k1) has \((3/2)^n\) coloring number to maximum clique ratio for any n, and \((3/2)^n\) grows arbitrarily large as n grows arbitrarily large.

A graph can be constructed with an arbitrarily large ratio of coloring number to maximum clique.