Combinatorics and Graph Theory Exam - 28 January 2010 Michael Bushell
[email protected]
December 23, 2012
1
2 (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 ) such that V ⊆ V and E ⊆ E . A spanning tree of G is a subgraph of G which contains all vertices of G and which is also a tree.
(b) See Theorem Theorem 1.1 in lecture lecture notes. notes. (c) Using the method of spanning trees, we have have the following following spanning spanning trees containing the edge sa or bc on the path from s to t. s s s 2
2
2
a 3
a 3
2
b
c
2
b
2
c
b
c 1
t
t
t
s
s
s
2
a
2
a 4
2
b
4
2
1
2 3
c
a 4
3
c
b
2
2
4
c
b 1
t
t
s
s
2
t
2
a
a
3
4
c
b 2
a
1
c
b 2
t
1
t
With respective weights 1/ 1/24, 24, 1/12, 12, 1/16, 16, 1/32, 32, 1/48, 48, 1/24, 24, 1/12, 12, 1/16. The edge sa appears in every tree on the path from s to t, so I sa (1/24+1/ 24+1/12+1/ 12+1/16+1/ 16+1/32+1/ 32+1/48+1/ 48+1/24+1/ 24+1/12+1/ 12+1/16) = 41α/ 41α/96 96 sa = α(1/ The edge bc appears positively in the 2nd tree and negatively in 4th
3 tree, so I bc (1/12 − 1/32) = 5α/ 5α/96 96 bc = α(1/ We are given that I bc 96/5. bc = 1, hence α = 96/ Therefore I sa 41α/96 96 = (41/ (41/96)(96/ 96)(96/5) = 8. 8.2 sa = 41α/
4 (2) (a) Given Given a directed directed graph graph G = (V, E ), ), a flow is a function f : E → such that
R
• ∀(x, y ) ∈ E : 0 ≤ f ( f (x, y ) ≤ c(x, y ), and • ∀x ∈ V \{s, t} we have
f ( f (x, y ) =
xy ∈E
f ( f (y, x)
yx ∈E
(b) Let (S, (S, T ) T ) be a cut of G of G, then v (f ) f ) =
f ( f (s, y ) −
sy ∈E
=
f ( f (y, s)
ys ∈E
f ( f (x, y ) −
x∈S
xy ∈E
f ( f (y, x)
yx ∈E
Because for all terms except when x = s, the value inside the large brackets is equal to 0 by the definition of a flow. We can rewrite this as follows v (f ) f ) =
f ( f (x, y ) −
x∈S,y∈V,xy ∈E
f ( f (y, x)
x∈S,y∈V,yx∈E
For every edge (x, (x, y ) such that x, y ∈ S , the value +f +f ((x, y ) appears precisely once in the left sum and −f ( f (x, y ) appears precisely once in the right sum, cancelling each other. This just laves the edges (x, (x, y ) such that x ∈ S, y ∈ T , T , hence
v (f ) f ) =
f ( f (x, y ) −
x∈S,y ∈T,xy ∈E
≤
f ( f (y, x)
x∈S,y∈T,yx∈E
f ( f (x, y )
x∈S,y ∈T,xy∈E
≤
c(x, y )
x∈S,y ∈T,xy∈E
= c(S, T ) T )
Therefore, the value of any flow is less than or equal to the capacity of any cut. (c) The follow following ing flow can be found found
5 9(10)
a
s
1(1)
2(2)
2(2)
0(3)
9(10)
b
2(2)
c 1(5)
3(3)
e
5(7)
2(2)
d 1(2)
2(2)
4(4)
t
6(10)
f
by starting from an empty flow, and using the augmenting paths and increments increments as follows follows
• • • • • • •
path path path path path path path
s,a,b,t, s,a,b,t, increment = 4, s,a,b,d,t, s,a,b,d,t , increment = 2, s,a,b,d,e,f,t, s,a,b,d,e,f,t, increment = 1, s,a,b,c,d,e,f,t, s,a,b,c,d,e,f,t, increment = 1, s,a,b,c,e,f,t, s,a,b,c,e,f,t, increment = 1, s,c,d,f,t, s,c,d,f,t, increment = 1, s,e,f,t, s,e,f,t, increment = 2.
giving a flow of value 12. We can check by finding the cut (using notation as in the lecture notes) S 0 = {s}, S 1 = {s, a}, S 2 = {s,a,b}, S 3 = {s,a,b} hence (S, T ) T ) = ({s,a,b}, {c,d,e,f,t}) is the required cut whose capacity is c(S, T ) T ) = 1 + 2 + 2 + 3 + 4 = 12
sc
se
bc
bd
bt
as expected (edges labelled under their capacities). (d) See Theorem Theorem 2.6 in lecture lecture notes. notes.
6 (3) (a) A sequence {an }n∞=0 has ordinar ordinary y power series series generating generating function function ∞ n given by the formal power series n=0 an x . (b) If f If f ((x) =
ops
n n≥0 an x ←→ {an }n≥0 , then
f (x) =
nan xn−1
n≥0
hence, working from the definiton ops
{nan }n≥0 ←→
n
nan x = x
n≥0
nan xn−1 = xf (x)
n≥0
as claimed. ops
ops
(c) If f If f ((x) ←→ {an }n≥0 and g(x) ←→ {bn }n≥0 , then f ( f (x)g (x) =
an xn
bn xn
n≥0
n≥0
n
=
xn
ak bn−k
n≥0
k=0
n
ops
←→
ak bn−k
k=0
n≥0
by definition. definition. (d) Using standard standard results of geometric series ops
{1}n≥0 ←→
xn =
n≥0
1 1−x
then by applying (b) twice
{n}n≥0
d ←→ x dx ops
ops
{n2 }n≥0 ←→ x
d dx
1 1−x
x (1 − x)2
=
x (1 − x)2
=
x(1 + x) (1 − x)3
7 so by definition
ops
2
(n2 − n + 1)x 1)xn
{n − n + 1 }n≥0 ←→
n≥0
n2 xn −
=
n≥0
nxn +
n≥0
xn
n≥0
1 x x(1 + x) + − 1 − x (1 − x)2 (1 − x)3 2 x − 3x2 −1 + 2x = (1 − x)3 =
(e) Write ops
{an }n≥0 ←→
an xn = A(x)
n≥0
then
{an+k }n≥0
A(x) − a0 − a1 x − · · · − ak−1 xk−1 ←→ an+k x = xk n≥0 ops
n
so using this and the recurrence relation gives A(x) − a0 − a1 x A(x) − a0 = 6 − 8A(x) x2 x hence 2x + 1 (1 − 2x)(1 − 4x) 3 2 − = 1 − 4x 1 − 2x =3 (4x (4x)n − 2 (2x (2x)n
A(x) =
n≥0
=
n≥0
(3 × 4n − 2 × 2n )xn
n≥0
ops
←→ {3 × 4n − 2 × 2n }n≥0 therefore an = 3 × 4n − 2 × 2n since the sequence uniquely determines the generating function.
8 (f) Define an = (−1)
n
for n ≥ 0, then
ops
{an }n≥0 ←→
2010 (−1)n xn n
n≥0
2010
=
2010 + n , bn = 2010
2010 n
2010 (−x)n n
n=0
=(1 − x)2010
where summation is over 0 ≤ n ≤ 2010 since the binomial coefficients are 0 for n > 2010, and the final step follows from the binomial theorem. We also have ops
{bn }n≥0 ←→
n≥0
2010 + n n 1 x = 2010 (1 − x)2011
using the identity given in the question. Hence, using the product formula in (c), we have
n
ak bn−k
1 1 ops ← → {1}n≥0 = (1 − x)2011 1−x
ops
←→ (1 − x)2010 ×
k=0
Therefo Therefore, re, since since the sequenc sequencee uniquely uniquely determi determines nes the generati generating ng function, we have n
1=
k=0
as claimed.
n
ak bn−k =
k
(−1)
k=0
2010 k
2010 + n − k 2010
9 (4) (a) A sequence {an }n∞=0 has exponential generating function given by the formal power series n∞=0 an xn /n!. /n!.
egf
(b) Write {an }n≥0 ←→ f ( f (x) = df ( f (x) = dx
n≥0
an xn /n!, /n!, then
xn−1 an = − ( n 1)! n≥1
xn egf ←→ {an+1 }n≥0 an+1 n ! n≥0
and induction on k gives the result dk f ( f (x) egf ←→ {an+k }n≥0 dxk as required. egf
egf
(c) Suppo Suppose se {an }n≥0 ←→ f ( f (x) and {bn }n≥0 ←→ g (x), then f ( f (x)g (x) =
xn an n! n≥0 n
=
n≥0
k=0 n
=
n≥0
k=0
n
egf
←→
k=0
xn bn n! n≥0
an bn−k k! (n − k)!
xn
n an bn−k k
xn n!
n an bn−k k
n≥0
as claimed. egf
(d) Let {an }n≥0 ←→
{nan }n≥0
n≥0
an xn /n! /n! = A(x), then
xn nan = ←→ n ! n≥0 egf
xn an = xA (x) (n − 1)! n≥0
egf
As {1}n≥0 ←→ ex , it follows that egf
egf
{n}n≥0 ←→ xex and {n2 }n≥0 ←→ x(x + 1)e 1)ex using (b) and the recurrence relation gives A (x) = xA (x) − x(x + 1)e 1)ex + 2e 2ex
10 hence A (X ) = (2 + x)ex and A(x) = (x + 1)e 1)ex + c1 for some constant c1 , as 1 = a0 = A(0) = 1 + c1 we get c1 = 0, thus egf
A(x) = xex + ex ←→ {n + 1}n≥0 therefore an = n + 1 is the solution. egf
(e) Suppose Suppose {bn }n≥0 ←→ B (x). Now by the recurrence relation, using part (b) and the product foregf mula from (c) with {an = 1}n≥0 ←→ ex, we have B (x) = ex B (x) hence
and so
B (x) dx = B (x)
ex dx
ln |B (x)| = ex + c1 and B (x) = c2 ee
x
for some constant constant c2 . As 1 = b0 = B (0) = c2 e, we get c2 = e−1 giving B (x) = ee as the solution.
x
−1
