Combinatorics and Graph Theory Exam - 24 January 2011 Michael Bushell michael.bushell@student.manchester.ac.uk
December 19, 2012
1
2 (1) (a)
(i) A walk is an alternating sequence x0 , e1 , x1 , e2 , x2 , . . . , xn−1 , en , xn of vertices xi ∈ V and edges ei = xi−1 xi ∈ E . A path is a walk with no repeated vertices. (ii) A graph is connected if there exists a path between every pair of vertices. vertices.
(b) See Theorem Theorem 1.1 in lecture lecture notes. notes. (c) We have the following following spanning trees containing containing the edge st or ab on the path from s to t. s s s a
a
b
b
t
t
t
s
s
s
a
a
b
b
t
t
t
s
s
s
a
a
b
b
t
t
t
s
s
s
a t
a
b
b
t
a
b
a
b
a
b
a
b
t
with respective weights 1/24, 1/16,1/16, 1/24, 1/4, 1/6,1/4, 1/6, 1/8, 1/12, 1/8, 1/8. By the method of spanning trees I st (1/2 4 + 1/ 1/1 6 + 1/ 1/1 6 + 1/ 1/2 4 + 1/ 1/8 + 1/1 2 + 1/ 1/8 + 1/8) = 2α/ 2α/33 st = α(1/ and I ab (1/4 − 1/6) = α/ α/12 12 ab = α(1/ We knowI knowI st st = 4, therefore I ab α/12 12 = 4(3/ 4(3/2)/ 2)/12 = 1/ 1/2 = 0.5 ab = α/
3 (2) (a) A cut in G is a pair (S, (S, T ) T ) where S, T ⊆ V , V , S ∪ T = V , V , S ∩ T = ∅ and s ∈ S , t ∈ T . T . The capacity of a cut is defined to be
c(S, T ) T ) =
c(x, y )
x∈S,y∈T,xy ∈E
(b) Let (S, (S, T ) T ) be an arbitrary cut of G of G and f an arbitrary flow, 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. If G has a maximum flow, then this maxiumum value is equal to the minimum of the capacities of the cuts. (c) The follow following ing flow can be found found
4 4(4)
a 4(4)
b
4(5)
c
7(7)
10(12) 0(5)
s
1(6) 0(5) 6(8)
3(4)
d
2(2)
e
t
2(2)
0(2)
2(5)
f
3(3)
by starting from an empty flow, and using the augmenting paths and increments increments as follows follows
• • • • •
path path path path path
s,a,b,t, s,a,b,t, increment = 4, s,a,c,b,t, s,a,c,b,t , increment = 3, s,e,a,c,b,t, s,e,a,c,b,t , increment = 1, s,e,d,f,b,t, s,e,d,f,b,t, increment = 2, s,e,f,t, s,e,f,t, increment = 3,
giving a flow of value 13. We can check by finding the cut (using notation as in the lecture notes) S 0 = {s}, S 1 = {s, e}, S 2 = {s,a,e}, S 3 = {s,a,e} hence (S, T ) T ) = ({s,a,e}, {b,c,d,f,t}) is the required cut whose capacity is c(S, T ) T ) = 4 + 4 + 2 + 3 = 13
ab
ac
ed
ef
as expected (edges labelled under their capacities).
5 (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 . ops
(b) If {an }n≥0 ←→
n≥0
ops
an xn = f ( f (x), then
∞
{an+k }n≥0 ←→
an+k xn
n=0
= ak + ak+1 x + ak+2 x2 + · · · ak xk + ak+1xk+1 + ak+2 xk+2 + · · · xk n k −1 ) n≥0 an x − (a0 + a1 x + · · · + ak−1 x = xk f ( f (x) − (a0 + a1 x + · · · + ak−1 xk−1 ) = xk =
as claimed. ops
ops
(c) If {an }n≥0 ←→ f ( f (x) and {bn }n≥0 ←→ g(x), then f ( f (x)g (x) =
an xn
n≥0
bn xn
n≥0
n
=
n≥0
k=0
n
ops
←→
an bn−k
k=0
ops
using {bn = 1} ←→
n≥0
xn
an bn−k
n≥0
xn = 1/(1 − x) = g(x) gives
n
{sn }n≥0 =
an bn−k
k=0
as claimed.
=
f ( f (x) 1−x
(d) Using Using ops
{n}n≥0 ←→
x ops x(x + 1) 2 and n { } ← → (1 − x)2 (1 − x)3
gives ops
3n + 1 } ←→ 2 {2n2 + 3n as required.
x(x + 1) x 1 1 + 3x 3x + 3 + = (1 − x)3 (1 − x)2 1 − x (1 − x)3
6 (e)
ops
(i) Let {cn } ←→ h(x), then using (b) and the recurrence relation gives h(x) − c0 − c1 x h(x) − c0 = 6 − 8h(x) x2 x substituting substituting in c0 = 1, c1 = 3 and rearranging gives h(x) =
1 − 3x 1 − 3x = 1 − 6x + 8x 8 x2 (1 − 2x)(1 − 4x)
by partial fractions 1 1 ops + h(x) = ←→ 2(1 − 2x) 2(1 − 4x)
1 n 1 n 2 + 4 2 2
n≥0
therefore cn = 2n−1 + 22n−1 = 2n−1 (1 + 2n ) as required. ops (ii) Let tn = kn=0 ck , then {tn }n≥0 ←→ h(x)/(1 − x) by (c), then
ops
{3tn + 2}n≥0 ←→ 3
h(x) 1 3h(x) + 2 +2 = 1−x 1−x 1−x
and ops
{8cn − cn+1 } ←→ 8h(x) −
h(x) − c0 (8x (8x − 1)h 1)h(x) + 1 = x x
therefore the equation 3tn + 2 = 8c 8cn − cn+1 holds if and only if the corresponding generating functions are equal 3h(x) + 2 (8x (8x − 1)h 1)h(x) + 1 = 1−x x solving for h(x), this has unique solution h(x) = which we know to be true.
1 − 3x 1 − 6x + 8x 8x2
7 (4) (a) The sieve formula is E (X ) = S (X − 1). See Theorem 4.1 in lecture notes for proof of n r et = (−1)r−t sr t r=t .
(b) In this contex contextt
• s0 = 55, • s1 = 95 = 25 + 23 + 24 + 23, • s2 = 67 = 8 + 14 + 14 + 9 + 11 + 11, • s3 = 18 = 3 + 5 + 7 + 3, • s4 = 1. so the number of elements of Ω contained in exactly one of A of A1 , . . . , A4 is by definition 4
e1 =
(−1)r−1
r =1
r sr = s1 − 2s2 + 3s 3s3 − 4s4 = 11 1
and the number of elements of Ω contained in at least one of A of A1 , . . . , A4 is |Ω| − e0 = 55 − (s0 − s1 + s2 − s3 + s4 ) = 45 (c) A seque sequence nce {an }n∞=0 has exponential generating function given by the formal power series n∞=0 an xn /n!. /n!. egf
(d) If {an }n≥0 ←→
n≥0
an xn /n! /n! = f ( f (x), then
xn−1 f (x) = nan = n ! n≥0
xn−1 an = ( n 1)! − n≥1
xn egf an+1 ←→ {an+1 }n≥0 n ! n≥0
and
{nan }n≥0
xn xn−1 =x = xf (x) nan nan ←→ n! n! n≥0 n≥0 egf
egf
(e) If {bn }n≥0 ←→ g (x), then
{nbn−1 }n≥0
xn nbn−1 = ←→ n ! n≥0 egf
xn bn−1 = ( n 1)! − n≥1
so using the recurrence relation and (d) gives g (x) = 2xg (x) + g (x) + 2xg 2 xg((x)
xn+1 bn = xg( xg (x) n ! n≥0
8 hence g (x) 1 + 2x 2x = and ln |g(x)| = g (x) 1 − 2x
1 + 2x 2x dx = −x − ln(1 − 2x) + c1 1 − 2x
for some constant c1 , hence −x−ln(1−2x)+c1
g (x) = e
c2 e−x = 1 − 2x
from b0 = g(0) we get c2 = 0, therefore g (x) =
1 ex (1 − 2x)
is the exponential generating function for {bn }n≥0 .
