Homework 13 SOLUTIONS

In-class Exercise 46. (a) Show that K2,3 , K5,2 , K5 − e for any edge e, and W6 are all planar by giving planar representations of them.

(b) Explain why K2,n is planar for any n. (Draw a picture)

1

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(c) Argue that K5 is not planar by doing a regional analysis like we did for K3,3 .

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In-class Exercise 47.

(a) Determine which of the following are planar. For those that are not, explain why. For those that are, draw a planar representation and verify Euler’s formula. You will want to label the vertices to aid in your explanations and/or drawings.

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(b) Suppose that a connected bipartite planar simple graph has m edges and n ≥ 3 vertices. Show that m ≤ 2n − 4.

Proof. In a bipartite graph, there are no odd cycles. So every face of a bipartite planar graph has at least 4 edges along its boundary. So the handshake theorem for faces, followed by Euler’s formula, says

2m =

X

deg(f ) ≥ 4|F | = 4(2 + m − n) = 8 + 4m − 4n.

f ∈F

So 2m ≤ 4n − 8, which is

m ≤ 2n − 4.



(c) Suppose that a planar graph has k connected components, m edges, and n vertices. Also suppose that the plane is divided into f faces by a planar representation of the graph. Find a formula for f in terms of m, n, and k.

Consider the graph with k isolated vertices (i.e. no edges). This graph has 1 face and k vertices. So

|F | − |E| + |V | = k + 1.

(∗)

Now, just as we did for Euler’s formula, we can prove that this formula holds (for a fixed k) in general by induction on the number of vertices (where we just did the base case. So assume that any planar graph with exactly k connected components and n ≥ k vertices has (∗). Now let G be a graph with k connected components and n + 1 vertices. Since there are more vertices than connected components, there is at least one connected component with more than one vertex. Let v be a vertex in that component, and consider G − v. By our inductive hypothesis, we have that G − v satisfies (∗). Then adding v to G, just as in the proof of Euler’s formula, preserves |F | − |E| + |V |. So (∗) holds.

25. graph. Find a formula for r in terms of e, v, and k. c 20. 20. 19. Which of these nonplanar graphs have the property that a b a the removal of any vertex and all edges incident with that vertex produces a planar graph? a e b) K6 c) K3,3 d) eK3,4 a) K5

ected components, e that the plane is presentation of the f e, v, and k.

24.

b

a

f c

c

i

This is not planar, because G − {ad, bi, df, eg, f h} (all the curvy edges) is a subdivision of K5 .

(ii) h

d e

g f

25.

b

e the property that s incident with that

) K3,4

c

e

The crossing numberThe of a s ber of crossings that can occ b ber o d d plane where no three arcs re plane g cross at the same point.cross d 1a 26. Show that K3,3 has26. ∗∗ 27. Find ∗∗ the crossing num 27. 5 In Exercises 20–22 determine whether the given graph is graphs. homeomorphic to K3,3 . f f b) a) K5 f e ? In-class Exercise 48. (a) Which of the following graphsgare homemorphic g to K3,3 h h d) K e) 3,4 c 20. The crossing number of a simple graph is the minimum numa ∗ b b 21. a ber ofccrossings graph drawn in the∗num Findis the crossing b that can occur c when this28. 21.d a 28. ∗∗ edges plane where no three arcs representing are permitted to 29. ∗∗ 29. Show that if m and n ar cross at the same point. is ing number of K m,n g h g h e 26. Show that K3,3 has 1 as its crossing number. (n − 2)/16. [Hint: Pla ∗∗ 27. Find the crossing numbers of each ofthat theynonplanar are equally sp these gin and place n vertice graphs. f equally f K5 e e d b) fK6 c) K7 spaced and sym a) d g h nect d) K e) K f ) Keach 3,4 4,4 5,5 of the m ver (i) (ii) vertices on the y-axis a a b c 22. 22. ∗ 28. Find athe crossing number of the Petersen graph. 21. a Theintegers, thickness a simple The g l theofcrossb ∗∗ 29. l Show that if bm and n are even positive planar subgraphs of G that is less than or equal to mn(m − 2)plana ing number of Km,n g h k c k c has30. 2a 30.along Showthe that K3,3 so (n − 2)/16. [Hint: Place m vertices x-axis ∗ 31. Findabout the thickness of th that they are equally spaced and symmetric the ori-∗ 31. j d j d 32. Show thatthey if G is conn gin and place n vertices along the y-axis so that area 32. e edges, where v ≥ equally spaced and symmetric about theand origin. Now conf e d e vertices on the x-axis i i e leastto⌈e/(3v − the 6)⌉. nect each of the m each of ∗ ∗to vertices on the y-axis and count the crossings.] f 33. Use Exercise 32 f 33.sh h h a 22. least ⌊(n + 7)/6⌋ when g g The thickness of a simple graph G is the smallest number of (iii) l b 34. Show conn planar23–25 subgraphs of that havetheorem G as their union.that if G is a 34. In Exercises 23–25 use Kuratowski’s theorem to G determine In Exercises use Kuratowski’s to determine c k and e edges, where v≥ has 2 as its thickness. 30. that Kis whether the given graph is planar. 3,3planar. whether the Show given graph then the thickness of G ∗ 31. h Find the thickness of thebecause graphs in it Exercise 27. (i) is not because it is planar (pull the edge between a and out). (ii) is not has d j Use Exercise 34 to show b c b d if G is a cconnected simple d 35. 23. a 23. a 32. graph with v vertices 35. 3-cycles. (iii) is, where the edges ki, ka, ab, ce, eg, and gi haveShow beenthatsubdivided. m and n are not both 1 and e edges, where v ≥ 3, then the thickness of G is at whenever m and n are e i least ⌈e/(3v − 6)⌉. ∗ 36. Draw K5 on the∗surfa ∗ 33. Use Exercise 32 to show that the thickness of Kn is at 36. f (b) Which of theh following graphs are planar? Use Kuratowski’s or Wagner’s theorems. solid) so that no edges least ⌊(n + 7)/6⌋ whenever n is a positive integer. g ∗ 37. Draw K3,3 on the∗surfa (i) e f g h e 34. Show fthat if G is a gconnected simple h graph with v vertices 37. In Exercises 23–25 use Kuratowski’s theorem to determine and e edges, where v ≥ 3, and no circuits of length three, whether the given graph is planar. then the thickness of G is at least ⌈e/(2v − 4)⌉. 35. Use Exercise 34 to show that the thickness of Km,n , where a b c d 23. f a b c d a c h m and n are not both 1, is at least ⌈mn/(2m + 2n − 4)⌉ whenever m and n are positive integers. = ∗ 36. Draw Planar! K5 on the surface of a torus (a doughnut-shaped solid) so that no edges cross. e g e g f h b d ∗ 37. Draw K3,3 on the surface of a torus so that no edges cross. e f g h

anar simple graph ≤ 2v − 4 if v ≥ 3.

mple graph with e e circuits of length /3) if v ≥ 4.

b

c

g a

d

he given graph is f

e

The crossing number of a simple graph is the minimum num-

≤ 2v − 4 if v ≥ 3.

i

mple graph with e le circuits of length 0/3) if v ≥ 4.

h

nected components, e that the plane is 6 presentation of the of e, v, and k.

d e

g f

25.

b

c

ve the property that s incident with that

) K3,4

g

(iii)

a

d

This is not planar, because G/{ag, bc} is isomorphic to K5 .

the given graph is

d

f

e

The crossing number of a simple graph is the minimum number of crossings that occurthe when graph is drawn in theand (b) the thickness for each of the following In-class Exercise 49.canFind (a)this crossing number, plane where no three arcs representing edges are permitted to graphs. cross at the same point. (i) K26. The that crossing number is 1, since it can’t be 0 (K5 is not planar) and this drawing has 1 5 : Show 1 as its crossing number. K3,3 has ∗∗crossing: 27. Find the crossing numbers of each of these nonplanar graphs. a) K5 d) K3,4

5

b) K6 e) K4,4

c) K7 f ) K5,5

1

4

∗ 28. Find the crossing number of the Petersen graph. ∗∗ 29. Show that if m and n are even positive integers, the crossing number of Km,n is less than or equal2to mn(m − 32) (n − 2)/16. [Hint: Place m vertices along the x-axis so that they are equally spaced and symmetric about the origin thickness and place n vertices along theKy-axis so that they are The is 2, because planar and 5 is not equally spaced and symmetric about the origin. Now con5 nect each of the m vertices on the x-axis to each of the 5 vertices on the y-axis and count the crossings.] 1

4

1

4

The thickness of a simple graph G is the smallest number of 4 ∪ 1 planar subgraphs of G that have G as = their union. 30. Show that K3,3 has 2 as its thickness. ∗ 31. Find the thickness of the graphs in Exercise 27. 2 3 2 3 32. Show that if G is a connected simple graph with v vertices and e edges, where v ≥ 3, then the thickness of G is at least ⌈e/(3v − 6)⌉. (ii) ∗K6 : The crossing number has to be more than 2: 33. Use Exercise 32 to show that the thickness of Kn is at leastSuppose ⌊(n + 7)/6⌋ whenever n is a positive Proof. there’s a drawing of K6integer. with only one crossing. Then deleting a vertex incident 34. Show that if G is a connected simple graph with v vertices to one of the edges in that crossing would leave a planar graph. But K6 − v = K5 for any orem to determine and e edges, where v ≥ 3, and no circuits of length three, vertex in thickness K6 . Similarly, you⌈e/(2v had just thenv the of G is atifleast − 4)⌉.2 crossings, each crossing involves 4 distinct vertices (the two incident to each edge in the 35. Use Exercise 34 to show that the thickness of crossing). Km,n , where So since there are only 6 vertices, the two d crossing would have to have a vertex in So deleting that vertex would eliminate m and n are not both 1, is at least ⌈mn/(2m + common. 2n − 4)⌉ m and n are positive integers. bothwhenever crossings, leaving a planar representation of K5 again.  ∗ 36. Draw K5 on the surface of a torus (a doughnut-shaped But thesofollowing drawing solid) that no edges cross. of K6 has 3 crossings, so K6 has crossing number 3: ∗ 37. Draw K3,3 on the surface of a torus so that no edges cross. h

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The thickness is 2, because K6 is not planar, but is the union of 5 2

1

and

3

1

2

3

4

.

6

4

(iii) K7 There is the following drawing of K − 7 with 9 crossings, which shows that the crossing number is no more than 9:

It turns out that there is no drawing of K7 with fewer crossings. The thickness of K7 is 2 since it is not planar, but is the union of 5

7

1

4

and 2

1

4

2

3

5

.

3 6

(iv) K3,4 : The crossing number cannot be 1, since any crossing involves an edge incident to one of the vertices on the 4 side, and if there were only 1 crossing, then deleting one such vertex on the 4 side would yield a planar drawing of K3,3 . Thus the crossing number of K3,4 is 2 since the following is a drawing of K3,4 with 2 crossings:

Further, its thickness is 2 because it is the union of

and

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(v) K4,4 : Similarly as above, the crossing number of K4,4 bust be larger than 2 because we could eliminate a crossing with the removal of some vertex, and K4,4 − v = K3,4 . Also, it can’t be 3 because some two crossings would have to share a special vertex in common (3 ∗ 4 > 8) whose removal would eliminate two crossings at a time. So it must be at least 4, and the following drawing of K4,4 has 4 crossings:

The thickness of K4,4 is then 2 because it’s not planar, but is the union of

and

(vi) K4 : Since K4 is planar, it has crossing number 0 and thickness 1. In-class Exercise 50. Let A, B, C, D, G, and H be the graphs a

b

b

g

a A=

d

d

f

e

a a

d

b

b e

c

C= b

B= c

D=

a

c

G= e

c

a

b

e

c

d

f

H=

f c

d

d

(a) Choose a planar drawing of each of A, B, C, D, G, and H, and draw the corresponding dual graph.

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Planar drawing: a b

Dual graph:

R

A:

R

S

S c

d

Planar drawing: b c

S

a

B:

R g

Q

T

Dual graph: R S

Q

T

V

d

V U e

f

U

Planar drawing: T

Dual graph: T

a C: R

b

c

S

R

S

d

Planar drawing: Dual graph: U a

P

P

b U

R D: Q

e

S

N

f

M

M c

S T

T N

R Q

d

732

10 / Graphs

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1 Red 1 2 Blue

Brown 7 7

2 Planar drawing: d R b 3 a c T S e

G:

6

5

Dual graph: R Red 6

Time Period I II III IV

3 Green

T

S 5 Green

4

4 Brown

FIGURE 8 The Graph FIGURE 9 Using a Coloring to Schedule Final Exam Planar drawing: Representing the Scheduling d Dual graph: of Final Exams. R a R b e H: T Now consider an application to the assignment of television channels. T f S S EXAMPLE 6 Frequency Assignments Television channels 2 through 13 are assigned to stations c America so that no two stations within 150 miles can operate on the same channel. How assignment of channels be modeled by graph coloring? (b) Pick another planar representation of H whose dual graph is not isomorphic to the graph you drew in part (a). Solution: Construct a graph by assigning a vertex to each station. Two vertices are conn an edge if they are located within 150 miles of each other. An assignment of channels cor to a coloring of the graph, where each color represents a different channel. Planar drawing: d An application of graph coloring to compilers is considered in Example 7.

EXAMPLE 7 H:

R b a Registers Index In efficient compilers the execution of loops is speeded up when fr Dual graph: in index registers in the central processing uni T used variables are stored temporarily S e memory. For a given R loop, how many index registers are needed? This of in regular can be addressed using a graph coloring model. To set up the model, let each vertex o S T represent f a variable in the loop. There is an edge between two vertices if the varia represent must be stored in index registers at the same time during the execution of Thus, the chromatic number of the graph gives the number of index registers needed different registers must be assigned to variables when the vertices representing these are adjacent in the graph. c

(c) Give an example of a plane graph whose dual is simple.

Exercises

See D from part (a). In Exercises 1–4 construct the dual graph for the map shown.

2.

(d) For each map, draw the the corresponding dual graph. Then find number of colors needed to color the map so that no two adjacent regions have the same color.

C

1.

B

B A A

C

D

E

A D E

B

C

D

▲

different registers must be assigned to variables when the vertices representing these variables are adjacent in the graph.

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struct the dual graph for the map shown. r of colors needed to color the map so regions have the same color.

2.

B A D

B

C

D

E

A

B

C

3.

11.

B F

A

10.8 Graph C

D

A

e

i

h

C j

a

F

B

E

C

d

D

n

D

E

b

g

f

c

k

l

Foran theexample graphs in of Exercises 5–11, decide 4. (e) Calculate the chromatic numbers for A, B, C, D, G, and H. For each,12. give a possible to decrease the chromatic number vertex coloring of the corresponding graph using exactly χ colors. a single vertex and all edges incident with D B

F A All of these graphs except for G has a K3 in it, so must have chromatic overhave 2. a chromatic number of 13. number Which graphs E C

a

b

b

In Exercises 5–11afind the chromatic number d of the given g graph.

A:

B= 5. a

c

b

e

f

χ=3

c g

a

c b C = 15. What is the chromatic number of Wn ?

16. Show that a simple graph that has a circui number d of vertices in it cannot be colored u ors.

d

χ=3

χ=3 c

d

c

a

8.

b

b

a

e f

e

f

a

7.

a D=

b

6.

d

14. What is the least number of colors needed t of thea United States? Do not consider ad that meet only at a corner. Suppose that one region. Consider the vertices representin Hawaii as isolated vertices.

c

b

G=

d

e

b

c

e

f

c

d

d

17. Schedule the final exams for Math 115 Math 185, Math 195, CS 101, CS 102, CS 473, using the fewest number of differe ifathere are nob students taking e both Math 11 both Math 116 and CS 473, both Math 195 both Math 195 and CS 102, both Math 115 a H = both Math 115 and Math 185, and both M Math 195, but there are students in every courses.

d

f

18. How many different channels are needed fo χ=2 located at the distances shown in the tabl c χ=3 tions cannot use the same channel when th 150 mileswill of each other? (f) Which of A, B, C,9.D, G, and H have the property that removing a single vertex reduce a b c

d

the chromatic number?

1 to isolated 2 3 4 None of them have a vertex that would totally disconnect the graph (reduce it vertices). So none of them can have their chromatic number reduced to 1. 1So, — in particular, 85 175 200 e c G’s chromatic number cannot be reduced by removing a single vertex. However, removing b 2 85 — 125 175 from A, g from B, a from C, and b from H will remove all triangles and drop the chromatic d a

10. i

b

5

6

50

100

100

160

3

175

125

—

100

200

250

4

200

175

100

—

210

220

5

50

100

200

210

—

100

6

100

160

250

220

100

—

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number to 2. Finally, D has two disjoint 3-cycles, so removing any one vertex will not drop the chromatic number. (g) Classify all graphs with chromatic number (i) 1, and (ii) 2. If the chromatic number of a graph is 1, then that graph cannot have any edges. So the graphs with chromatic number 1 are those that are a collection of isolated vertices. If the chromatic number is 2, then the graph is bipartite. So a graph has chromatic number 2 if and only if it has at least one edge, and has no odd cycles. (h) What are the chromatic numbers of (i) Kn : χ(Kn ) = n since every vertex in incident to every other vertex. (ii) Km,n : χ(Km,n ) = 2 since it is bipartite. (Unless m or n is 0, in which case the chromatic number is 1.) (iii) Cn : χ(Cn ) = 2 if n is even, and 3 if n is odd. (iv) Wn : χ(Wn ) = 3 if n is even (since the cycle is 2-colorable, but the middle vertex is adjacent to everything), and 4 if n is odd (similarly since the cycle is 3-colorable). (v) Qn : χ(Qn ) = 2 since it is bipartite. (vi) The Petersen graph: χ = 3, it cannot be 2 as it contains odd cycles, but can be 3-colored as follows: