MATH39001 - Combinatorics and Graph Theory - Exam - Jan-2010

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Combinatorics and Graph TheoryExam - 28 January 2010

Michael [email protected]

December 23, 2012

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(1) (a) (i) A tree is a connected, acyclic graph.

(ii) Let G = (V, E). A subgraph of G is a graph G = (V, E) suchthat V V and E E. A spanning tree of G is a subgraphof G which contains all vertices of G and which is also a tree.

(b) See Theorem 1.1 in lecture notes.

(c) Using the method of spanning trees, we have the following spanningtrees containing the edge sa or bc on the path from s to t.

s

a

b c

t

2

32

2

s

a

b c

t

2

32

1

s

a

b c

t

2

42

1

s

a

b c

t

2

42

2

s

a

b c

t

2

43

2

s

a

b c

t

2

43

1

s

a

b c

t

2

3

2 1

s

a

b c

t

2

4

2 1

With respective weights 1/24, 1/12, 1/16, 1/32, 1/48, 1/24, 1/12, 1/16.

The edge sa appears in every tree on the path from s to t, so

Isa = (1/24+1/12+1/16+1/32+1/48+1/24+1/12+1/16) = 41/96

The edge bc appears positively in the 2nd tree and negatively in 4th


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tree, so

Ibc = (1/12 1/32) = 5/96

We are given that Ibc = 1, hence = 96/5.

ThereforeIsa = 41/96 = (41/96)(96/5) = 8.2


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(2) (a) Given a directed graph G = (V, E), a flow is a function f : E R

such that

(x, y) E : 0 f(x, y) c(x, y), and

x V\{s, t} we havexyE

f(x, y) =yxE

f(y, x)

(b) Let (S, T) be a cut ofG, then

v(f) = syEf(s, y) ysEf(y, s)=xS

xyE

f(x, y) yxE

f(y, x)

Because for all terms except when x = s, the value inside the largebrackets is equal to 0 by the definition of a flow. We can rewrite thisas follows

v(f) =

xS,yV,xyEf(x, y)

xS,yV,yxEf(y, x)

For every edge (x, y) such that x, y S, the value +f(x, y) appearsprecisely once in the left sum and f(x, y) appears precisely once inthe right sum, cancelling each other.

This just laves the edges (x, y) such that x S, y T, hence

v(f) =

xS,yT,xyE

f(x, y)

xS,yT,yxE

f(y, x)

xS,yT,xyEf(x, y)

xS,yT,xyE

c(x, y)

= c(S, T)

Therefore, the value of any flow is less than or equal to the capacityof any cut.

(c) The following flow can be found


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s

a

c

e

b

d

f

t

9(10)

1(1)

2(2)

9(10)

2(2)3(3) 4(4)0(3)

2(2)

1(5) 2(2) 1(2)

2(2)

5(7)

6(10)

by starting from an empty flow, and using the augmenting paths andincrements as follows

path s,a,b,t, increment = 4,

path s,a,b,d,t, increment = 2,

path s,a,b,d,e,f,t, increment = 1,

path s,a,b,c,d,e,f,t, increment = 1,

path s,a,b,c,e,f,t, increment = 1,

path s,c,d,f,t, increment = 1, path s,e,f,t, increment = 2.

giving a flow of value 12.

We can check by finding the cut (using notation as in the lecturenotes)

S0 = {s}, S1 = {s, a}, S2 = {s,a,b}, S3 = {s,a,b}

hence(S, T) = ({s,a,b}, {c,d,e,f,t})

is the required cut whose capacity is

c(S, T) = 1sc

+ 2se

+ 2bc

+ 3bd

+ 4bt

= 12

as expected (edges labelled under their capacities).

(d) See Theorem 2.6 in lecture notes.


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(3) (a) A sequence {an}n=0 has ordinary power series generating function

given by the formal power series n=0 anxn.(b) Iff(x) =

n0 anx

n ops {an}n0, then

f(x) =n0

nanxn1

hence, working from the definiton

{nan}n0ops

n0

nanxn = x

n0

nanxn1 = xf(x)

as claimed.

(c) Iff(x)ops

{an}n0 and g(x)ops

{bn}n0, then

f(x)g(x) =

n0

anxn

n0

bnxn

=n0

n

k=0

akbnk

xn

ops

n

k=0akbnk

n0

by definition.

(d) Using standard results of geometric series

{1}n0ops

n0

xn =1

1 x

then by applying (b) twice

{n}n0ops

xd

dx

11 x

=

x

(1 x)2

{n2}n0ops

xd

dx

x

(1 x)2

=

x(1 + x)

(1 x)3


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so by definition

{n2 n + 1}n0ops

n0

(n2 n + 1)xn

=n0

n2xn n0

nxn +n0

xn

=1

1 x

x

(1 x)2+

x(1 + x)

(1 x)3

=1 + 2x 3x2

(1 x)3

(e) Write

{an}n0ops

n0

anxn = A(x)

then

{an+k}n0ops

n0

an+kxn =

A(x) a0 a1x ak1xk1

xk

so using this and the recurrence relation gives

A(x) a0 a1xx2

= 6 A(x) a0x

8A(x)

hence

A(x) =2x + 1

(1 2x)(1 4x)

=3

1 4x

2

1 2x

= 3

n0(4x)n 2

n0(2x)n

=n0

(3 4n 2 2n)xn

ops {3 4n 2 2n}n0

thereforean = 3 4

n 2 2n

since the sequence uniquely determines the generating function.


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(f) Define

an = (1)n2010

n

, bn =

2010 + n2010

for n 0, then

{an}n0ops

n0

2010

n

(1)nxn

=2010n=0

2010

n

(x)n

=(1 x)2010

where summation is over 0 n 2010 since the binomial coefficientsare 0 for n > 2010, and the final step follows from the binomialtheorem.

We also have

{bn}n0ops

n0

2010 + n

2010

xn =

1

(1 x)2011

using the identity given in the question.

Hence, using the product formula in (c), we haven

k=0

akbnk

ops

(1 x)2010 1

(1 x)2011=

1

1 x

ops {1}n0

Therefore, since the sequence uniquely determines the generatingfunction, we have

1 =n

k=0akbnk =

n

k=0(1)k

2010

k

2010 + n k

2010

as claimed.


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(4) (a) A sequence {an}n=0 has exponential generating function given by the

formal power series n=0 anxn/n!.(b) Write {an}n0

egf f(x) =

n0 anx

n/n!, then

df(x)

dx=n1

anxn1

(n 1)!=n0

an+1xn

n!

egf {an+1}n0

and induction on k gives the result

dkf(x)

dxkegf

{an+k}n0

as required.

(c) Suppose {an}n0egf

f(x) and {bn}n0egf

g(x), then

f(x)g(x) =

n0

anxn

n!

n0

bnxn

n!

=n0

n

k=0

ank!

bnk(n k)!

xn

=n0

nk=0

nk

anbnk xn

n!

egf

n

k=0

n

k

anbnk

n0

as claimed.

(d) Let {an}n0egf

n0 anxn/n! = A(x), then

{nan}n0egf

n0

nanxn

n!=

n0

anxn

(n 1)!= xA(x)

As {1}n0egf

ex, it follows that

{n}n0egf

xex and {n2}n0egf

x(x + 1)ex

using (b) and the recurrence relation gives

A(x) = xA(x) x(x + 1)ex + 2ex


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hence

A(X) = (2 + x)ex and A(x) = (x + 1)ex + c1

for some constant c1, as 1 = a0 = A(0) = 1 + c1 we get c1 = 0, thus

A(x) = xex + exegf

{n + 1}n0

thereforean = n + 1

is the solution.

(e) Suppose {bn}n0egf

B(x).

Now by the recurrence relation, using part (b) and the product for-

mula from (c) with {an = 1}n0egf

ex, we have

B(x) = exB(x)

hence B(x)

B(x)dx =

ex dx

and soln |B(x)| = ex + c1 and B(x) = c2e

ex

for some constant c2. As 1 = b0 = B(0) = c2e, we get c2 = e1 giving

B(x) = eex1

as the solution.

MATH39001 - Combinatorics and Graph Theory - Exam - Jan-2010

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