AMS 303 Second Test and Solutions

AMS 303 TEST 1A Spring 2021 56 pts

1. a) (5 pts) Show a minimal face coloring of this graph? Explain why fewer colors will not suffice.

b) (5 pts) Show a minimal edge coloring of this graph? Explain why fewer colors will not suffice.

ANS a) Because all vertices have even degree, graph is face 2-colorable by Theorem 5.8

b) Trial-and error using 4 colors for edges (=maximum degree). See above

2. (4 pts) Re-state the following statement about a planar graph G in the dual: If G is a simple, planar graph with every vertex of degree at least 5, then G has a face with bdy ≤ 3. (JUST RESTATE FOR DUAL)

ANS: If G* is a planar graph with deg ≥ 3 and face boundaries at least 5, then G* has a vertex of deg ≤ 3.

NOTE ‘simple’ means bdg ≥3)

3. Find the chromatic polynomial of the following graph.

(6 pts) ANS: k(k-1)5(k-2)6(k-3)2

4. Let G be a simple, connected plane graph. Explain the effect of the following operations on the number of vertices, edges and faces in the dual of G .

a) ( 3pts) Delete a vertex of degree 4; b) (3pts) Contract a triangular face.

ANS: a) m* -4, n*-3, f*-1 b) m*-3, n*-1, f*-2

5. (10 pts) A bridge in a graph is an edge whose removal disconnects the graph. A graph G is critical k-chromatic if the minimal vertex coloring number is k and when any vertex x is removed, the minimal vertex coloring number of G-x will be k-1. E.g., an odd-length circuit is critical 3-chromatic. Show that if G is critical k-chromatic, then G cannot contain a bridge. Hint: draw a picture.

ANS: bridge

e

Removal of bridge e produces a disconnected graph G-e with components G1 and G2. Observe that (G), the chromatic number of G, will be larger of the chromatic no.s of (G1) and (G2). Suppose (G1) ≥ (G2).

Then the removal of any vertex of G2 will not reduce (G1), and hence not reduce (G). Not critical.

NOTE: There is actually a mistake in this problem. If you had a graph with just 2 vertices and 1 edge (the edge being a bridge), this graph is critical 2-chromatic.

6. (8 pts) Show if a planar graph G has all triangular faces and is vertex 3-colorable, then every vertex has even degree.

ANS: Restate result in dual: Show that if planar graph G* has all vertices deg. 3 (cubic) and is face 3-colorable, then every face F has even boundary. Cubic means the faces neighboring any face F are consecutively adjacent and form a circuit of faces around F. With one color for F, the circuit of surrounding faces must be 2 colorable and hence of even length. Then F has even bdy. (See left diagram)

x

Can also be proved in G. Triangular faces means the neighbors of a vertex x are successively adjacent forming a circuit around x. With one color for any x, the circuit of neighbors must be 2-colorable; hence the circuit must be even length implying that x has an even no. of neighbors, that is, even degree.

7. (12 pts) Let G be a connected simple graph with all deg ≥4 and all faces have boundaries of 3 or 5. Prove that there are at least 8 more faces of boundary 3 than of boundary 5. Hint: Write f =f3 + f5.

ANS: Because the result concerns faces, we use the fact that deg ≥ 4 along with Euler’s formula to get an upper bound on m in terms of f:

4n ≤ sum(deg) = 2m 4m – 4f + 8 = 4n ≤ 2m 2m – 4f + 8 ≤ 0 2m ≤ 4f - 8. (*)

Next we use f3(3) + f5(5) = sum(bdy) = 2m.

With (*), we have f3(3)+ f5(5)=2m ≤ 4f-8 =4f3+4f5 – 8 or 3f3 + 5f5 ≤ 4f3 + 4f5 – 8.

Subtracting 3f3 and 4f5 from each side we have f5 ≤ f3 – 8 or f5 + 8 ≤ f3.

AMS 303 Test 2 Fall, 2020 54 points

1a) (5 pts)What is the face chromatic number of the graph below? Why can’t it be smaller?

b) (5 pts) What is the edge chromatic number of the graph below? Why can’t it be smaller?

SCORING: a) 3 pts coloring; 2 pts, why minimum b) 4 pts coloring, 1 pt why minimum

ANS: a) 4 colors required- 4 mutually adjacent faces in middle. Easy to 4-color remaining 3 faces (including unbounded face) after the middle 4 faces are 4-colored. Score

b) max deg of 4 = min edge coloring (see above)

2. (6 pt) Find the chromatic poly for this graph.

k*(k-1)^3*(k-2)^4*(k-3)^2 NOTE: vertex 7 must be colored before vertex 8; otherwise non-adjacent neighbors of 7- 5 and 8- are colored before 7. SCORING: -1 pt for each wrong exponent

3. (6 pts) Let G be a simple, connected plane graph. How do the numbers of vertices, edges, and faces in the dual G* change when one performs the following operations in G:

a) delete an edge of G; b) contract a face of G with boundary 5; c) delete a vertex of degree 3 in G.

ANS: a) m* -1, f* unchanged, n*-1; b) m* -5, f* -4, n* -1; c) m* -3, f* -1, n* -2. SCORING: 2 pts each part

4. (4 pts) Restate the following statement in terms of the dual graph (DO NOT PROVE): If G is a simple planar graph with all degrees ≤4, then G has a triangular face.

ANS: If G* is a planar graph with boundaries ≥ 4 and deg >=3, then G* has a vertex of deg 3.

SCORING: -2 pts for each mistake

5. ( 8 pts) Let G be a connected graph and let a and b be two distinct vertices in G. If every path between a and b must include the vertex c, show that G-c (that is, removing c and its edges from G) is not connected. Use the definition of a connected graph that there is a path between every pair of vertices in the graph. SCORING: Variable partial credit

ANS: Since all paths in G between a and b must pass through c, if c is removed, then G-c has no path between a and b. Hence G-c is not connected.

NOTE: One cannot assume that G-c has two components. That is what you are supposed to prove.

6. (8pts) Show that if a planar graph G is cubic (all degrees = 3) and if its dual G* is vertex 3-colorable, then every face of G has an even boundary. You may not cite any theorem.

ANS: G* vertex 3-colorable means G is face 3-colorable.

Surrounding faces 2 colorable -> even circuit of faces

-> even boundary

SCORING: +2 pt, use dual vertex 3-colorable -> G is face 3-colorable.

+5 pt, draw picture of face surrounded by a 2-colored circuit of faces

+1 pts, even-length surrounding circuit implies even boundary of center face

7. (12 pts) If G is a simple, planar graph with all vertices of degree 5 or 6. Prove that G has at least 12 vertices of degree 5.

ANS: Let n = n5 + n6 . Simple -> bdy ≥ 3. So 3f ≤ sum(bdy) = 2m or f ≤ (2/3)m

Then m – n + 2 = f ≤ (2/3)m -> (1/3)m – n + 2 ≤ 0 -> m ≤ 3n – 6

n5(5) + n6(6) = sum(deg) = 2m ≤ 2(3n-6) = 6n – 12 = 6n5 + 6n6 -12

Subtracting left-side terms from right-side terms, we have

0 ≤ n5 -12

SCORING Step A m ≤ 3n-6 - 5 points (must be proved): +3 pts: use bdy >=3 --> f ≤ (2/3)m;

+3 pts: use m – n + 2 = f ≤ (2/3)m to derive m <= 3n-6

Step B- 7 points: +1 pt: Let n = n5 + n6

+2 pts: n5(5) + n6(6) = sum(deg) = 2m

+3 pts: 2m ≤ 2(3n-6) = 6n – 12 = 6n5 + 6n6 -12

+1 pt: use above to conclude 0 ≤ n5 – 12.

Spring 2020 Test 2 Solutions 55 points  1. (10 pts- 5,5): a) faces- 4 colors; reason: unbounded face and the 3 faces incident to it are mutually adjacent.b) edges- 4 edge colors = max degree (start at one of the deg-4 vertices)

2. (6 pts): k*(k-1)5 *(k-2)6 *(k-3)2.

3. (4 pts): Restate the following theorem in terms of the dual graph (DO NOT PROVE): If G is a simple cubic (degree=3) planar graph, then G has a face with boundary ≤ 5.

If G* is a planar graph with deg ≥ 3 and bdy=3, then G* has a vertex of deg ≤5.

4. (4 pts) . Let G be a simple, connected plane graph. How do the numbers of vertices, edges, and faces in the dual G* change when one performs the following operations in G:

a) contract a face of G with boundary 4; b) delete a vertex of degree 6 in G.

a) m* changes by -4, n* by -1, f* by -3; b) m* by -6, n* by -5, f* by -1.

5. (10 pts): Let G be any connected (undirected) graph. If C is a set of edges forming a circuit in G and K is a set of edges forming a cutset, show that C and K must have an even number of edges in common (possibly zero). Reminder: K is a cutset if K’s removal disconnects G, that is, G-K is disconnected, but no subset K’ of K disconnects G.

Proof: Removing the cutset (in blue) splits the graph into two components. Each time a circuit (in red) crosses from the first component to the second component, it must cross back so that it finishes in the same component where it started.

6. (8 pts): Show that if the vertices of a planar graph G are 2-colorable, then the faces of G have even boundaries. You may not cite any theorems

Proof: vertex 2-colorable implies face 2-colorable in dual. If the circuit of faces around any vertex is face 2-colorable, circuit must be of even length; that is, the vertex has an even number of edges incident to it. In original graph, this means any face has even boundary.

Problem can also be solved in primal graph: vertex 2-colorable implies all circuits are vertex 2-colorable, that is, implies even length. Boundaries of faces are circuits and hence of even length.

7. (13 pts) Let G be a planar graph in which every vertex has deg >=3 and every face has bdy =>4. Prove that G has at least one vertex of deg = 3.

Given bdy ≥4. If conclusion false, deg ≥ 4.

Pf: 4f ≤ sum(bdy)= 2m --> m-n+2=f ≤ m/2 --> m ≤ 2n-4.

4n ≤ sum(deg)= 2m --> 2n ≤ m, then 2n = m ≤ 2n-4. Impossible