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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.
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