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1

1 THE PRIMAL-DUAL ALGORITHM 1

1.1 Revision of Duality in Linear Programming 3

1.2 The Primal-Dual Algorithm 14

2 EFFICIENTLY SOLVABLE PROBLEMS 29

2.1 Max-Flow Problem in Networks 33

2.2 Application of the Primal-Dual Algorithm to the Transportation Problem 83

2.3 Introduction to the Theory of Matroids 123

3 HARD PROBLEMS 173

3.1 On the NP -completeness 173

3.2. Algorithms for NP -complete Problems 189

3

1 The primal-dual algorithm

1.1 Revision of Duality in Linear Programming

General LP:

f (x) = c1x1 + c2x2 + … + cnxn → min or max

subject to (s.t.)

ai1x1 + ai2x2 + … + ainxn {≤, ≥, =} bi (i = 1, ..., m)

x1, x2, … , xp ≥ 0

Example 1.1.1

3x1 – 2x2 → max

s.t.

2x1 + 5x2 ≥ 17

3x1 – 5x2 = –27

x1 – x2 ≤ 20

x2 ≥ 0

W.l.o.g. :

c1x1 + c2x2 + … + cnxn → min

ai1x1 + ai2x2 + … + ainxn = bi (i = 1, ..., m)

x1, x2, …, xn ≥ 0

5

f(x) = cTx → min

Ax = b

x ≥ 0

f (x) … objective function

Here

A = (aij) ∈ Rm×n

c = (c1, … , cn)T∈Rn

b = (b1, … , bm)T∈Rm

x = (x1, … , xn)T∈Rn

M = { x ∈Rn; Ax = b, x ≥ 0 } ……

…… set of feasible solutions

Mopt = { x ∈M; cTx ≤ cTz for all z ∈M } ……

…… set of optimal solutions

Theorem 1.1.1 (Fundamental Theorem of LP)

For every LP one of the following holds:

(i) LP is infeasible (i.e. M = Ø),

(ii) LP has an optimal solution (Mopt ≠ Ø),

(iii) )(min xfMx∈

= –∞ .

(SLP)

7

Definition of the dual problem:

f (x) = cTx → min g(π) = πTb → max

iTi bxa = i∈I πi ≥≤ 0

iTi bxa ≥ i∈ I πi ≥ 0

xj ≥ 0 j∈J πTAj ≤ cj

xj ≥≤ 0 j∈J πTAj = cj

( )JjAJjAIia

IiaA jjT

i

Ti ∈∈=

∈∈

= ,,,

,

Example 1.1.2:

f(x) = 3x1 – 4x2 → min g(π) = 5π1 + 7π2→ max

s.t. s.t.

2x1 – 7x2 ≥ 5 2π1 + 3π2 = 3

3x1 + x2 = 7 –7π1 + π2 ≤ –4

x2 ≥ 0 π1≥ 0

primal dual

9

Theorem 1.1.2 (On Symmetry)

Dual to the dual is the primal.

MP [optPM ] … set of feasible [optimal] solutions

to the primal LP

MD [optDM ] … set of feasible [optimal] solutions

to the dual LP

Theorem 1.1.3 (Weak Duality Theorem)

(∀x∈MP)(∀π∈MD) cTx ≥ πTb.

Corollary 1: If x∈MP, π∈MD, cTx = πTb then optD

optP , MMx ∈π∈ .

Corollary 2:

(a) If −∞=∈

xcT

Mx P

min then MD = Ø.

(b) If +∞=π∈π

bT

M D

max then MP = Ø.

Theorem 1.1.4 (Strong Duality Theorem) opt

MP ≠ Ø if and only if opt

MD ≠ Ø. If an optimal

solution exists then .maxminDP

bxcT

M

T

Mxπ=

∈π∈

11

Corollary: For every primal-dual pair exactly

one of the possibilities in the table below occurs:

optPM ≠ Ø min cTx = –∞ MP = Ø

optD M ≠ Ø

max πTb = +∞

MD = Ø

Important primal-dual pairs

f(x) = cTx → min g(π) = πTb → max

Ax ≥ b πTA ≤ cT

x ≥ 0 π ≥ 0

(CLP) “Symmetric pair”

f(x) = cTx → min g(π) = πTb → max

Ax = b πTA ≤ cT

x ≥ 0

(SLP)

13

cTx → min πTb → max





Important task: Find an optimal solution to the

dual only using the information about the

primal optimal solution.

Theorem 1.1.5 (Complementary Slackness)

Suppose x∈MP, π∈MD.

Then x∈ optMP and π∈opt

MD if and only if

(cj – πTAj )xj = 0 = πi ( Tia x – bi) for all i and j.

[End of sample. You can download more pages from CANVAS]

15

1.2 The Primal-Dual Algorithm


Ax = b πTA ≤ cT

x ≥ 0

(P) (D)

W.l.o.g. b ≥ 0.






By the Complementary Slackness:

Optimality conditions for x∈MP, π∈MD:

(cj – πTAj)xj = 0 = πi( Tia x – bi) for all i and j.

Equivalently:

(cj – πTAj)xj = 0 for all j ( 1.1)

17

An essential assumption:

Suppose at least one π ∈MD is known.

(Thus cj – πTAj ≥ 0 for all j) .

x = ? (cj – πTAj > 0 ⇒ xj = 0) for every j (*)

(then this x (and also π) would be optimal).

Figure 1.2.1

Start by any π∈MD (set π = 0 if c ≥ 0)

∴ πTAj ≤ cj for all j

J = { j; πTAj = cj }

Primal Dual Restricted primal (find x*

satisf. (*))

Dual of the restricted

primal

Adjustment to π

π

Stop success

fail

19

By the Complementary Slackness Theorem: If

we find x∈MP, xj = 0 for all j∉J then x is

optimal.

Therefore we need to find x such that

∑∈

=Jj

ijij bxa , i = 1, ... , m

( 1.2 )

xj ≥ 0 , j∈J

J …. the set of admissible columns (indices)

To search for such an x, we create a new LP,

called the restricted primal (RP):

ξ = ∑=

m

i

aix

1

→ min

∑∈

=+Jj

iaijij bxxa i = 1, ..., m

xj ≥ 0 j∈J

0≥aix i = 1, ..., m

aix (i = 1, ..., m) ……. artificial variables

Note: (1.2) has a solution ⇔ ξopt = 0

(RP)

21

If ξopt = 0 , then all aix = 0 ∴ x optimal

If ξopt > 0 then consider the dual of the

restricted primal (DRP), which is:

w = πTb → max

πTAj ≤ 0, j ∈J (DRP)

πi ≤ 1, i = 1, ..., m

π …. an optimal solution to (DRP)

Then by the SDT we have:

πTb = ξopt > 0 ( 1.3)

“Correction” of π, using π

π* = π + θπ, where θ = ?

(a) π*T b > πT b or, equivalently,

πT b + θπT b > πT b

∴ θ > 0

23

(b) To guarantee (dual) feasibility of π*:

π*TAj ≤ cj, j = 1, ..., n

or, equivalently,

πTAj + θπTAj ≤ cj, j = 1, ..., n .

We set:

>ππ

π−0j

T

jT

jT

jA

A

Acmin

def

= θ1 (1.4)

Clearly, θ1 > 0 and the new cost is

w* = πT b + θ1 bTπ > w = πT b

Next step: Redefine the set J and repeat the

procedure until either ξopt = 0 and optimality in

P is reached.

But what if πTAj ≤ 0 for all j? We show in

Theorem 1.2.1 that then P is infeasible.

Theorem 1.2.1

(Infeasibility in the primal-dual algorithm): If

(1) ξopt > 0 and

(2) πTAj ≤ 0 for all j ∉ J

then P is infeasible. Proof: →→→

25

The primal-dual algorithm

Primal Dual Restricted primal

π := π + θ1 π

π

Dual of restricted primal

J = {j; πTAj = cj}

ξopt = 0

stop

ξopt > 0

πTAj ≤ 0

for all j∉J

MP = Ø

∃ j∉J: πTAj > 0

27

The primal-dual algorithm – basic scheme

The primal-dual algorithm – detailed scheme

Primal Dual Restricted primal

Adjustment to π

π x

Given is P

cTx → min Ax = b

x ≥ 0

Take π feasible to D πTb → max πTA ≤ cT

Dual of restricted

primal

29

J := { j; πTAj = cj }

Solve RP

ξ = ∑=

m

i

aix

1

→ min

∑∈

=+Jj

iaijij bxxa i = 1, ..., m

xj ≥ 0 j∈ J 0≥aix i = 1, ..., m

ξopt = 0

∃ j∉J: πTAj > 0

θ1 :=

>ππ

π−0j

T

jT

jT

jA

A

Acmin

π := π + θ1π

Find π optimal to DRP

yes

yes

no

no

x is optimal to P

P is infeasible

31

Example:

6x1 + 16x2 – 15x3 → min

s.t.

2x1 – 4x2 + x3 = 1

x1 + 5x2 – 5x3 = 2

x1-3 ≥ 0

Solve by the PDA starting from π = (0, 3)T.

33

2 Efficiently solvable problems

2.1 Max-Flow Problem in Networks

2.1.1 Introduction

Example 2.1.1

Sunco Oil want to ship the maximum possible

amount of oil per hour via the pipeline from

node s to node t in Figure 2.1.1. On its way

from node s to node t, oil must pass through

some or all of stations a, b, and c. The various

arcs in Figure 2.1.1 represent pipes of different

diameters. The maximum number of barrels of

oil [millions of barrels] per hour that can be

pumped through each arc is indicated in

rectangles at each arc (“arc capacity”).

35

Figure 2.1.1

Maximum-flow problem (informally): to achieve

the maximum amount of flow from a starting

point (source s) to a terminal point (sink t).

s b

c

t

a

5

2

4 2

4

1

37

2.1.2 Directed graphs (digraphs) and flows

Digraph is D = (V, E) where

• V ≠ Ø is a finite set (set of nodes / vertices)

• E ⊆ V × V ={(u, v); u, v∈V} (set of arcs /edges)

Note that (u, v) ≠ (v, u).

(u, v) ∈E ….. u, v are adjacent

Usually: V = {v1, …, vm}, V = m

E = {e1, …, en}, E = n

A digraph (V, E) is called oriented if for every u,

v ∈ V we have: (u, v) ∈ E ⇒ (v, u) ∉ E.

Network is N = (V, E, s, t, b), where (V, E) is an

oriented digraph without loops, s∈V has no

incoming arcs, t ∈V has no outcoming arcs and

b: E → R+ ∪ {∞}.

b(i, j) or only bij ….. capacity of the arc (vi, vj).

39

If E = {e1, … , en}, then bk will stand for b(ek).

f: E → R+ is a flow in N if

(1) 0 ≤ f ≤ b (capacity constraints)

(2) ∑∑∈∈

=ExyyEyxy

xyfyxf),(:),(:

),(),( holds

for all x∈V, x ≠ s, x ≠ t (conservation law)

Convention: fij and fk

∑∑ =xek

xek

kk

ff enters leaves

for all x∈V, x ≠ s, x ≠ t.

∑=sek

k

fv leaves

….. the value of the flow f, notation

v(f) or just v.

Max-Flow Problem (MFP)

Given a network, find a flow (“maximal flow”)

in this network whose value is maximal.

41

Example 2.1.1 (cont.):

Conventions:

Free-standing numbers … flow values

Other numbers … capacities

s b

c

t

a

5

2

4 2

4

1

1

2 1 1

2 2

43

2.1.3 MFP is a linear program

Let D = (V, E) be a digraph without loops.

V = {v1, … ,vm}, E = {e1, … , en}.

AD = (aij) ∈ Rm×n is called the node-arc incidence

matrix of D if

1, if ej is leaving vi,

aij = -1, if ej is entering vi,

0, otherwise.

Example 2.1.2

s

b

t

a

e1

e2 e5

e3

e4

45

Structure of node-arc incidence matrices:

Each column contains exactly one 1 and one –1

since every arc is leaving one node and entering

one node.

Let N = (V, E, s, t, b) be a network and A be

the node-arc incidence matrix of N. Let f be a

flow in N.

Consider the sum jj

ij fa∑ for a fixed index i:

∑∑∑ −=ijij xe

j x e

jjj

ij fffaentering is leaving is

= 0,

if xi ∉{s, t};

vffaij x e

jjj

ij == ∑∑leaving is

, if xi = s ;

vffaij x e

jjj

ij ′=−= ∑∑entering is

, for some v′ ,

if xi = t.

∴ Af = (0, …, v, … , v′ , …, 0)T

Add up all equations:

vv ′+=0 ∴ vv −=′

47

Therefore the system of equations has the form:

Af + dv = 0,

where d =

−

⋮

⋮

⋮

1

1

Other conditions on f :

f ≤ b, f ≥ 0

Theorem 2.1.1 If A is a node-arc incidence

matrix of the network N = (V, E, s, t, b) then

f: E → R+ is a flow in N if and only if it

satisfies the conditions

Af + dv = 0

f ≤ b

f ≥ 0

and so the MFP in N is equivalent to the LP

… s

… t

49

v → max

s.t.

Af + dv = 0

f ≤ b

f ≥ 0

Note: (Here m is the number of nodes and n is

the number of arcs)

♦ (MFP) has n + 1 variables f1, …, fn, v and

m + n non-trivial constraints (m equations

and n inequalities).

♦ v can be assumed to be non-negative without

loss of generality.

♦ f = 0 (with v = 0) is always a feasible solution,

and thus it is a flow in any network.

Corollary: A maximal flow exists in every

network with finite capacities of arcs leaving the

source (or entering the sink) – this will be

assumed from now on.

Proof: → → →

(MFP)

51

Example 2.1.3:

s b

c

t

a

1

2

2 2

3

1

53

2.1.4 The Ford and Fulkerson algorithm for

solving the MFP

Example 2.1.3 (contd.):

s b

c

t

a

1

2

2 2

3

1

1

1 0 1

1 1

55

Observation:

An “augmenting path” has one or both of the

two types of arcs:

♦ forward arcs whose direction is the same as

that of the path and the flow is strictly less

than the capacity of arc;

♦ backward arcs whose direction is opposite and

the flow is positive.

s b

c

t

a

1

2

2 2

3

1

0

2 2 0

2 2

57

Definition of an augmenting path

Given a network N = (V, E, s, t, b), and a flow

f, an augmenting path p is a path from s to t

in the undirected graph resulting from (V, E) by

ignoring arc directions, with the following

properties:

♦ For every arc (i, j) that is traversed by p in

the forward direction (“forward arc”), we

have f(i, j) < b(i, j). That is, forward arcs are

unsaturated.

♦ For every arc (i, j) that is traversed by p in

the reverse direction (“backward arc”), we

have f(i, j) > 0. That is backward arcs are

non-empty.

59

Using an augmenting path

How to find a flow of a bigger value (if an

augmenting path is known):

♦ increase the flow value on every forward arc

but still satisfying the capacity constraint and

♦ decrease the flow value on every backward arc

but keeping it ≥ 0.

The maximum amount of flow augmentation

possible along p is

b(i, j) – f(i, j) along a forward arc

δ = pji on ),(

min

f(j, i) along a backward arc

61

How to find an augmenting path

A systematic way of finding an augmenting

path:

Start by f = 0 (or other flow).

Propagate labels from node s until we reach

node t or get stuck. Each node x will have

assigned to it a two-part label:

label(x) := (L1[x], L2[x])

Here

L1[x] ….. from where x was labelled

L2[x] ….. the amount of extra flow that can be

“brought” to x from s.

Labelling outward from a node x: “scanning x”

label(s) := (∅, +∞)

63

Scanning from a node x:

For every unlabelled node y adjacent to x

distinguish two cases:

Case 1: (x, y) ∈ E

Label y if f (x, y) < b(x, y)

L1[y] := x

L2[y] := min {L2[x], b(x, y) – f(x, y)}

y

x

(L1[x], L2[x] )

(x, min{L2[x], b(x,y) – f(x,y)})

f(x,y) < b(x,y)

65

Case 2: (y, x) ∈ E

Label y if f(y, x) > 0

L1[y] := –x

L2[y] := min {L2[x], f(y, x)}

y

x

(L1[x], L2[x] )

(–x, min{L2[x], f(y, x)})

f (y, x) > 0

67

LIST … the set of nodes that are labelled but not

scanned yet

The labelling algorithm:

(1) LIST := { s }.

(2) Repeat until LIST = Ø:

Select a node x from LIST,

remove x from LIST,

scan x, and

add all nodes labelled from x to LIST.

Termination of the labelling algorithm:

♦ either t gets labelled: reconstruct an

augmenting path backwards from t using

L1’s and augment the flow along the

augmenting path by δ = L2(t), that is,

fj → fj’ =

δ−δ+

arcbackward

arcforward

,f

,f

j

j

♦ or LIST becomes empty: flow is optimal (to be

proved later on).

69

When we achieve optimality after an application

of FFA to MFP, we say we are at a

nonbreakthrough.

Example 2.1.4: Apply FFA to solve MFP in the

network below starting from the given flow:

s b

c

t

a

2

2

3 2

3

2

0

2

2

2

2

0

71

FORD AND FULKERSON ALGORITHM

(FFA)

Input: A network N = (V, E, s, t, b)

Output: A maximum flow f in N

f: = 0 (or any other flow if known)

again: delete all existing labels (if any)

LIST := {s}, label(s) := (Ø, +∞)

while LIST ≠ Ø do

begin

take any x∈LIST;

remove x from LIST;

scan x;

extend LIST by all nodes scanned from x

if t is labelled then

begin

augment flow f along augmenting path

by δ = L2(t);

go to again

end

end

73

♦ If FFA terminates, does it find a maximal flow

(correctness)?

♦ Does FFA terminate?

♦ Correctness of FFA

Let N = (V, E, s, t, b) be a network.

An s-t cut in N is (W, W ), where

V = W ∪W ,

W ∩W = Ø,

s ∈W, t ∈W .

The capacity of the cut (W, W ) is

c(W, W ) = ∑{b(x, y); x ∈W, y ∈W , (x, y) ∈E}.

W

s

W

t

75

Example (2.1(c) in problem sheets):

For each cut ),( WW , find c(W, W ),

∑∈∈ WyWx

yxf,

),( and ∑∈∈ WyWx

yxf,

),( .

Lemma: For any cut (W, W ) and flow f the

following relation holds:

v(f) = ∑∑∈

∈∈∈

∈∈−

EyxWyWx

EyxWyWx

yxfyxf

),(,

),(,

),(),( .

Proof: →→→

3 8

6 2

s

a

b

t 6 3

2

5

0 3 5

2

77

Theorem 2.1.2 (Min-Cut, Max-Flow)

(a) The relation v(f ) ≤ c(W, W ) holds for

every flow f and every cut (W, W ).

(b) The value of a maximal flow is equal to the

capacity of a minimal cut.

Proof : →→→

W*

s

*W

t

Labelled nodes Unlabelled nodes

79

Theorem 2.1.3 (Ford-Fulkerson)

If FFA terminates, it does so at an optimal flow.

Proof : →→→

Example 2.1.5:

s b

c

t

a

1

2

3 2

3

1

1

2 1 1

2 2

(Ø,+∞)

(s,1)

(c,1)

W

81

♦ Termination of FFA

Theorem 2.1.4: If the capacities in MFP are

integers then FFA will terminate after a finite

number of steps if it starts with an integral flow.

Proof: → → →

Corollary: If the capacities in MFP are rational

numbers then the MFP is solvable using the

FFA in a finite number of steps.

Proof: → → →

Note (following from the construction of flows in

FFA): If the capacities in (MFP) are integers,

FFA starts from an integral flow then it finds an

integral max flow.

In general (for real capacities):

• FFA may not terminate.

• FFA may converge to a flow value strictly less

than optimal!

Modifications of FFA which finitely terminate

after a polynomial number of steps (Karzanov).

83

(ai) Ai

Bj (bj)

cij

xij = ?

2.2 Application of the Primal-Dual Algorithm to

the Transportation Problem

Transport of a single type of product that can be

arbitrarily divided (oil, flour, coal or electricity).

Given are:

• Producers A1, …, Am with capacities a1, …, am.

• Customers B1, …, Bn with demands b1, …, bn.

• Unit transportation costs cij (from Ai to Bj) for

every i = 1, …, m; j = 1, …, n.

Wanted:

• The amount of product to be transported

from Ai to Bj for every i = 1, …, m and

j = 1, …, n [transportation plan] which

minimises the total cost of the transport and

meets the capacity and demands constraints.

85

(ai) Ai

Bj (bj)

cij

xij = ?

Transportation table (TT)

Demands C

ap

aci

ties

Unit costs and/or amount of transported products

bj

…..

ai ………….. xij or cij

Cell (i, j) ↔ Arc AiBj Row i ↔ Producer Ai

Column j ↔ Customer Bj

cell (i, j)

87

Example 2.2.1:

Powerco has 3 electric power plants that supply power for 4 cities. The table below shows capacities of the plants in 106 kWh, peak power demands of the cities and the costs [£1,000] of sending 106 kWh from plants to cities. Find the distribution of the electricity minimising the cost and meeting each city’s peak power demand.

Demands→ 45 20 30 30 Capacities

↓

35 8 6 10 9

50 9 12 13 7

40 14 9 16 5

GREEDY:

45 20 30 30

35

50

40

Cost:

89

TP as an optimisation problem:

∑∑= =

→m

i

n

jijij xc

1 1

min

subject to

∑=

=n

jiij ax

1

(i = 1, ..., m) (A)

∑=

=m

ijij bx

1

(j = 1, ..., n) (B)

xij ≥ 0 (i = 1, ..., m, j = 1, ..., n) (C)

W.l.o.g.: ai > 0, bj > 0 for all i, j.

∑∑= =

m

i

n

jijij xc

1 1

…… objective function.

X = (xij) satisfying (A), (B) and (C) is a feasible

solution.

A feasible solution minimising the objective

function is called an optimal solution.

91

Theorem 2.2.1: TP has an optimal solution if it

has a feasible solution.

Theorem 2.2.2: TP has a feasible solution if and

only if ∑ ∑= =

=m

i

n

jji ba

1 1

.

∑ ∑= =

=m

i

n

jji ba

1 1

… TP is balanced

If ∑ ∑= =

<m

i

n

jji ba

1 1

then a dummy producer Am+1

is introduced with the capacity

am+1 = ∑ ∑= =

−n

j

m

iij ab

1 1

and costs

cm+1,j = 0 ( j = 1, ..., n).

∴ We may assume w.l.o.g. : TP is balanced (and

therefore has a feasible and optimal solution).

Now apply the primal-dual algorithm to TP.

93

∑∑= =

→m

i

n

jijij xc

1 1

min

subject to

∑=

=n

jiij ax

1

(i = 1, ..., m)

∑=

=m

ijij bx

1

( j = 1, ..., n)

xij ≥ 0 (i = 1, ..., m, j = 1, ..., n)

is in standard form - consider it as P in the

primal-dual algorithm.

Construction of the dual.

Two types of equations → use two types of

variables: αi’s and βj’s.

w = ∑ ∑= =

β+αm

i

n

jjjii ba

1 1

→ max

s.t.

αi + βj ≤ cij for all i = 1, ..., m and j = 1, ..., n.

(P)

(D)

95

A feasible solution to the dual:

αi = 0 for all i = 1, ..., m;

βj = min{cij ; i = 1, ..., m}, for j = 1, …, n.

I = {1, ..., m}, J = {1, ..., n}

Convention: “node Ai” → “node i”, similarly

for Bj.

i

j αi + βj ≤ cij

97

Construction of RP.

IJ = {(i, j) ∈ I × J; αi + βj = cij}.

Thus, RP is

ξ = ∑+

=

nm

i

aix

1

→ min

∑∈

=+IJjij

iaiij axx

),(:

i = 1, ..., m

∑∈

+ =+IJjii

ja

jmij bxx),(:

j = 1, ..., n

xij ≥ 0 (i, j) ∈ IJ

0≥aix i = 1, ..., m + n

How to solve (RP): Transform to an equivalent

(RP’) by eliminating all aix :

Add up all equations in RP:

ξ = ∑ ∑ ∑= = ∈

−+m

i

n

j IJjiijji xba

1 1 ),(

2 ( 2.1)

(RP)

99

∴ RP can be written equivalently as:

g(x) = ∑∈IJji

ijx),(

→ max

∑∈

≤IJjij

iij ax),(:

i = 1, ..., m

∑∈

≤IJjii

jij bx),(:

j = 1, ..., n

xij ≥ 0 (i, j) ∈ IJ

RP’ ≡ MFP of Fig. 2.2.1, where:

AiBj is an arc ⇔ (i, j) ∈ IJ

(RP’)

⋮ ⋮

⋮ ⋮

s t

Admissible arcs

N

A1

A2

Ai

Am

B1

B2

Bj

Bn Figure 2.2.1

a1

a2

ai

am

b1

b2

bj

bn

+∞ +∞

+∞

+∞

+∞

+∞

101

Lemma 1:

(a) If f is a flow in N and xij = f(i, j), (i, j) ∈ IJ

then x = (xij) ∈MRP’ and g(x) = v(f).

(b) If x = (xij) ∈MRP’ ,

f(i, j) = xij ((i, j) ∈ IJ),

f(s, i) = ∑∈IJjij

ijx),(:

(i = 1,…, m),

f(j, t)= ∑∈IJjii

ijx),(:

(j = 1, …, n),

then f is a flow in N and v(f) = g(x).

Proof: → → →

Lemma 1 shows that RP for TP is MFP of

Fig.2.2.1.

In PDA we need next to check whether ξopt > 0

or ξopt = 0.

103

How to check ξopt = 0:

(2.1) was: ξ = ∑ ∑ ∑= = ∈

−+m

i

n

j IJjiijji xba

1 1 ),(

2

Hence by Lemma 1:

ξ = ∑ ∑= =

−+m

i

n

jji fvba

1 1

)(2 = ∑=

−m

ii fva

1

)(22

ξ = 0 ⇔ v(f) = ∑=

m

iia

1

(recall that v(f) = ∑=

m

i

isf1

),( and f(s, i) ≤ ai, ∀i)

or, equivalently:

ξ = 0 ⇔ f(s, i) = ai for all i

We can now apply FFA to this MFP.

1. If f(s, i) = ai for all i then ξ = 0

∴ x = (xij) is optimal and conversely;

xij = else,0

),(if),,( IJjijif ∈

2. If f(s, i) < ai for some i. Need optimal

solution, say βα, , of the dual to the

restricted primal.

105

Finding βα, :

At nonbreakthrough, we denote

I* = { i ∈ I; node i is labelled },

J* = { j ∈ J; node j is labelled }.

Lemma 2: At nonbreakthrough the implication

i ∈I* ⇒ j ∈J*

holds for every (i, j) ∈IJ.

Proof: →→→

107

Theorem 2.2.3: At nonbreakthrough, an optimal

solution to DRP is given by

1=αi i ∈I*,

1−=αi i ∉I*,

1−=β j j ∈J*,

1=β j j ∉J*.

Proof: In two steps:

(a) We show that ( )βα, is a feasible solution to the DRP.

(b) We show that the objective function value at

( )βα, is also attained in the RP.

The statement will then follow from the duality

theory.

→→→

109

( )βα, ……. optimal solution to DRP Case 1: 0≤β+α ji for all (i, j) ∉ IJ – this

means that (P) is infeasible – impossible in TP.

Case 2: 0>β+α ji for at least one (i, j) ∉ IJ –

in this case θ1 and an update of π are

computed as follows:

θ1 = =

>β+αβ+α

β−α−0min ji

ji

jiijc

=

∉∈β−α−

**,2

min JjIic jiij

( 2.2)

111

The update of π:

αi + θ1, if i∈I*

αi := αi + θ1 iα = ( 2.3)

αi – θ1, if i∉I*

βj – θ1, if j∈J*

βj := βj + θ1 jβ = ( 2.4)

βj + θ1, if j∉J*

113

Algorithm ALPHABETA

Transportation problem with inputs cij, ai, bj

αi := 0 for i = 1, ..., m; βj := min{cij; i = 1, ..., m} for j = 1, ..., n.

IJ := {(i, j)∈I × J; αi + βj = cij}

Using FFA solve MFP of Figure 2.2.1 with the set of admissible arcs IJ.

If (∀i) f(s, i) = ai x is optimal

Let I* and J* be the sets of labelled nodes

at nonbreakthrough. Find θ1 and update (α, β) using (2.2), (2.3) and (2.4).

yes

no

115

Extended transportation table:

Demands C

ap

aci

ties

Unit costs and/or amount of transported goods

Du

al

va

ria

ble

s α

i

Dual variables βj

bj

ai cij or xij αi

βj

cell (i, j)

117

Example 2.2.2:

Powerco has 3 electric power plants that supply power for 4 cities. The table below shows capacities of the plants in 106 kWh, peak power demands of the cities and the costs [£1,000] of sending 106 kWh from plants to cities. Find the distribution of the electricity minimising the cost and meeting each city’s peak power demand.

Demands→ 45 20 30 30 Capacities

↓

35 8 6 10 9

50 9 12 13 7

40 14 9 16 5

Initialization:

α = (0, 0, 0)T; β = (8, 6, 10, 5)T.

45 20 30 30

35 8 6 10 9

50 9 12 13 7

40 14 9 16 5

119

Conventions: +∞ omitted; no flow value means flow is 0.

Iteration 1:

45 20 30 30

35 8 6 10 9 0

50 9 12 13 7 0

40 14 9 16 5 0

8 6 10 5

35

40

45

30

20

s t 30

50

121

Iteration 2:

45 20 30 30

35 8 6 10 9 – ½

50 9 12 13 7 ½

40 14 9 16 5 ½

8.5 6.5 10.5 4.5

35

40

45

30

20

s t 30

50

123

45 20 30 30

35 8 6 10 9 – 1.5

50 9 12 13 7 1.5

40 14 9 16 5 1.5

7.5 7.5 11.5 3.5

Optimal transportation plan:

45 20 30 30

35

50

40

Total cost:

35

40

45

30

20

s t 30

50

125

2.3 Introduction to the Theory of Matroids

2.3.1 Undirected graphs

G = (V, E) is called an undirected graph (shortly

graph) if

• V ≠ Ø is a finite set (set of nodes/vertices)

• E ⊆ {{u, v}; u, v∈V, u ≠ v} (set of arcs/edges)

We denote {u, v} by uv. Clearly, uv = vu.

Usually: V = {v1, … ,vm}, V = m

E = {e1, … , en}, E = n

Let G′ = (V′, E′) and G = (V, E) be graphs.

G′ is called a subgraph of G (notation G′ ⊆ G)

if V′ ⊆ V and E′ ⊆ E. G′ is called proper if

G′ ⊆ G but G′ ≠ G.

If E = {{u, v}; u, v ∈V, u ≠ v} then G is called

complete, notation Km:

In every graph: 0 ≤ n ≤ 2

1m(m – 1)

K1 K2 K3

127

V = V(G), E = E(G)

p = (v0, v1, … , vk) is a walk (in G) if

• v0, v1, … , vk ∈ V

• vi–1vi ∈E (i = 1, 2, …, k)

If, moreover

• vi ≠ vj (i ≠ j; i, j = 0, 1, …, k)

then p is called a path.

p is a “u-v path” if v0 = u and vk = v;

p “passes” through v0, v1, … , vk;

p “contains” v0, v1, … , vk;

v is “reachable” from u if there is a u-v path

c = (v0, v1, … , vk) is a closed walk (in G) if

c is a walk and v0 = vk.

If, moreover, (v0, v1, … , vk–1) is a path in G

and k ≥ 3 then c is called a cycle.

Hamilton cycle of G … cycle that contains all

nodes of G

deg(v) …the number of arcs incident with v

129

A graph G = (V, E) is called

• acyclic (or forest) if there is no cycle in G;

• connected if there is a u-v path in G for all

u, v ∈ V(G);

• a tree if G is both acyclic and connected.

acyclic but not connected connected but not acyclic tree (m = 9, n = 8)

131

Theorem 2.3.1 If G is a tree then n = m – 1.

Proof: By induction on m.

Lemma (obvious): G acyclic ⇒ every subgraph

of G is acyclic.

Graph H is called a (connected) component of

the graph G if

• H ⊆ G

• H is connected

• H ⊆ H′, H′⊆ G, H′ connected ⇒ H = H′

p(G) ….. number of components of the graph G

is not is not p(G) = 2

is a component

Every component of a forest is a tree.

G

133

Theorem 2.3.2

If G is acyclic then n = m – p(G).

Proof: Consider the components of G:

1 2 3 p m1–n1=1 m2–n2=1 m3–n3=1 ……… mp–np=1

Let G be a connected graph. Then a subgraph

T of the graph G is called a spanning tree of G

if

• T is a tree

• V(T) = V(G) (“T is spanning G”)

(∴ Every spanning tree of a graph with m

nodes has …….(?) arcs.)

135

F … maximal acyclic subgraph (MAS) of G, if F

is an acyclic subgraph of G, and F is not a

proper subgraph of any acyclic subgraph of G.

Theorem 2.3.3 Let G be a connected graph with

m nodes. Then every MAS F of G is a spanning

tree of G and hence has m – 1 arcs.

Proof: →→→

Graph F is a spanning forest of the graph G if

V(F) = V(G) (let us denote this number by m)

and each component of F is a spanning tree of

the corresponding component of G.

Hence p(G) = p(F) = p and so by Theorem 2.3.2

|E(F)| = m – p.

Corollary of Theorem 2.3.3: Let G be a graph

with m nodes and p components. Then every

maximal acyclic subgraph of G is a spanning

forest of G and therefore has m – p arcs.

137

2.3.2 The greedy algorithm

GREEDY: Repeatedly select a most attractive

feasible element until there are no feasible

elements any more.

The assignment problem (AP): Given an m × m

matrix A, find m entries in A, no two from the

same row or column so that their sum is

minimal.

If

=

53

21A then GREEDY selects 1+ 5 but

2 + 3 is minimal. Thus GREEDY is not a

correct solution method for AP.

The shortest path problem (SPP):

GREEDY selects 2 + 3 but 4 is minimal.

Thus GREEDY is not a correct solution method

for SPP.

s

a

t

2 3

4

139

The Minimal Spanning Tree Problem (MST)

Given a connected graph G = (V, E) and a

function w: E → R, find a spanning tree

T = (V, E*) of G such that

∑∈

=*

)(*)(Ee

ewEw

is minimal.

MST is solvable by GREEDY - this will follow

from the conversion of MST to MWF:

The Maximum Weight Forest Problem (MWF)

Given a graph G = (V, E) and a function

w: E → R+, find a forest F = (V, E*) of G such

that

∑∈

=*

)(*)(Ee

ewEw

is maximal (note that w(∅) = 0).

For the conversion see Problem Sheet 4.

Is MWF solvable by GREEDY? See matroids

(below).

141

2.3.3 Independence systems and matroids

Aim: To characterise combinatorial

optimisation problems that are solvable by

GREEDY.

S = (X,F ) is an independence system (IS) if

(1) X is a finite set (called the ground set), and

(2) F is a collection of subsets of X closed

under inclusion, that is A∈F , A′ ⊆ A ⇒ A′∈F

A∈F …… A is called an independent set

143

Four examples of independence systems:

Let G = (V, E) be a graph.

• S1 = (E, A) is an independence system where

A = { A ⊆ E; (V, A) acyclic}.

Proof : Immediate by “Obvious Lemma”.

Consider

Then

V = {a, b, c}, E = {ab, bc, ac},

A = {Ø,{ab},{bc},{ac},{ab, bc},{ab, ac},{bc, ac}}=

= 2E – {E}.

(E, A) is an instance of S1

a

b c

145

A bit more of graph theory:

(will be needed for S2)


M ⊆ E is called a matching if e ∩ f = Ø for all

e, f ∈M, e ≠ f.

Thus |M| ≤ 2

1 m for every matching M.

Ø is always a matching.

{e} is always a matching.

If |M| = 2

1m then M is called perfect.

147

If M is a matching in G and M is not a

proper subset of any other matching in G then

M is called maximal (“it cannot be extended to a

bigger one”).

If M is a matching in G of biggest size then M

is called maximum.

(In the figures below M is formed by bold arcs.)

The difference between the size of a maximum

matching and 2

1 m can be arbitrarily big.

G Not a

matching

Matching but not maximal

Maximal but not maximum

Maximum and

perfect

149


• S2 = (E, M) is an independence system where

M = {M ⊆ E; M is a matching}

Proof : →→→

Consider

V = {a, b, c, d}, E = {ab, bd, dc, ca},

M = {Ø, {ab}, {bd}, {dc}, {ac}, {ab, cd}, {ac, bd}}

(E, M) is an instance of S2.

Note:

For the graph above set

F = {Ø, {ab}, {ac}, {ab, ac}}.

(E, F ) is an independence system (different

from S1 and S2).

Proof: →→→

a

b d

c

151

• S3 = (E, I) is an independence system where

E = {(i, j); i, j = 1, …, m} (the set of all

positions in an m × m matrix) and I∈ I if

and only if I is a set of positions such that no

two are from the same row or column.

Proof: →→→

Consider

A =

2221

1211

aa

aa

E = {(1, 1), (1, 2), (2, 1), (2, 2)}

I = {Ø,

{(1, 1)}, {(1, 2)}, {(2, 1)}, {(2, 2)},

{(1, 1), (2, 2)}, {(1, 2), (2, 1)}}

(E, I) is an instance of S3.

153

• S4 = (C, L) is an independence system where

C = {1, …, n} (the set of column indices of an

m × n matrix) and

L = {J ⊆ C; columns with indices from J are

linearly independent}.

Proof: →→→

Consider

=

00110

02101A

C = {1, 2, 3, 4, 5}

L = {Ø, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {2, 3},

{2, 4}, {3, 4}}

(C, L) is an instance of S4.

155

Upper bounds for the sizes of independent sets:

S1: |A| ≤ m – p

S2: |M| ≤ 2

1 m

S3: |I| ≤ m

S4: |J| ≤ rank of A ≤ min(m, n)

“Independent” means:

S1: acyclic (graph)

S2: matching (in a graph)

S3: independent (positions in a matrix)

S4: linearly independent (vectors)

157

Let S = (E, F ) be an independence system. The

combinatorial optimisation problem (COP)

associated with S, notation (S, w), is:

Given a weight function w: E → R+, find an

independent set (called an optimal solution) that

has the biggest total weight, i.e. find X*∈F such

that

w(X*) = )(max XwX F∈

where w(X) = ∑∈Xe

ew )( for X ⊆ E.

COP associated with S1:

Given a graph G = (V, E) and w: E → R+, find

an acyclic subgraph of G of maximal total

weight [= Maximum Weight Forest Problem].

COP associated with S3:

Given an m × m matrix A = (aij) ≥ 0, find m

entries in A, no two either from the same row or

column, of maximal total value [= Assignment

Problem].

159

THE GREEDY ALGORITHM (GREEDY)

Input: Independence system S = (E, F ),

w: E → R+.

Output: X* ∈F for which w(X*) is maximal.

(1) X* := Ø

(2) (“Find the most attractive element in E”)

Find e ∈ E such that w(e) = )z(wmaxEz∈

(3) E := E – {e}

(4) (“Check feasibility”)

If X* ∪ {e} ∈F then X* := X* ∪ {e}

(5) If E = Ø then stop else go to (2)

161

Theorem 2.3.4 Let S = (E, F ) be an

independence system. Then the following

statements are equivalent:

(A) For every w: E → R+ GREEDY correctly

solves (S, w).

(B) [Exchange Proposition (EP)]

For all F, F′∈F, if |F′| = |F| + 1 then there

is e ∈ F′ – F such that F ∪ {e} ∈ F.

(C) [Rank Proposition (RP)]

If A ⊆ E and F, F′ are maximal

independent subsets of A then |F| = |F′|.

An independence system satisfying one (and

therefore all) of the conditions (A), (B), (C) in

Theorem 2.4.4 is called a matroid.

163

Let S = (E, F ) be a matroid. The (common) size

of maximal independent subsets of a set A ⊆ E

is called the rank of A and any maximal

independent subset of A is called a basis of A.

A minimal dependent subset of E (that is a

dependent set whose every proper subset is

independent) is called a circuit.

• S1 = (E, A) is a matroid:

Use RP: Let A ⊆ E be fixed and take F ⊆ A.

F is an independent subset of A if and only if

(V, F) is an acyclic subgraph of GA = (V, A).

∴F is a maximal independent subset of A if

and only if (V, F) is a maximal acyclic

subgraph of GA = (V, A).

∴ |F| = m – p(GA) (Cor. to Theorem 2.3.3)

∴ S1 is a matroid (notation Gr(G)).

Hence: GREEDY correctly solves every instance

of MWF and, in particular, the MST.

165

• S2 = (E, M ) is not a matroid:

A = {ac, ab, bd}

{ac,bd},{ab} – maximal independent subsets of A

∴ GREEDY does not always solve the maximal

weighted matching problem correctly.

• S3 = (E, I ) is not a matroid:

2221

1211

aa

aa

A = { (1, 1), (1, 2), (2, 1)}

{(1, 2), (2, 1)}, {(1, 1)} – maximal independent

subsets of A

∴ GREEDY does not always solve AP correctly.

• S4 is a matroid.

Proof: →→→

a

b d

c

167

K = is called a field if the following

axioms hold:

a + b = b + a a.b = b.a

a + (b + c) = (a + b) + c a.(b.c) = (a.b).c

a .(b + c) = a.b + a.c

(∃0∈K) a + 0 = a (∃1∈K) a.1 = a

(∀a∈K)(∃–a∈K) a + (–a) = 0

(∀a∈K, a ≠ 0)(∃a-1∈K) a.a-1 = 1

Vectors, matrices, linear (in)dependence.

col (A) ….. the set of column indices of A

Let A be an m × n matrix over K and

LA = { X ⊆ col(A); columns of A with indices

from X are linearly independent}.

Then (col(A), LA) is a matroid. Notation M(A).

S4 is M(A) for K = R.

169

Independence systems S = (E, F ) and

S′ = (E′, F ′ ) are called isomorphic (notation

S ≈ S’) if a bijection φ: E → E′ exists such that

F ∈F ⇔ φ(F) ∈F ′

where φ(F) = {φ(x); x ∈F }.

A matroid S is called matric if S ≈ M(A) for

some field K = and for some matrix

A over K.

A matroid S is called graphic if S ≈ S1 for some

graph G = (V, E). Notation Gr(G).

Theorem 2.3.5 Every graphic matroid is matric.

Notes:

• Not every matric matroid is graphic.

• Not every matroid is matric.

171

Graphic

matroids

Matric matroids

Matroids

Independence systems

GREEDY works

GREEDY does not work

173

3 Hard problems

3.1 On the NP -completeness

3.1.1 Why “efficient” means “polynomial”

Computer speed: 10 GHz (1010 ops/sec)

Running times of algorithms

Size of input

No of ops

20 60 … 100 1000

L 0.002µs 0.006µs … 0.01µs 0.1µs

L2 0.04µs 0.36µs … 1µs 0.1ms

L4 16µs 1.296ms … 10ms 100s

2L 0.1ms 3yrs … 1012yrs .

L! 7.7yrs . … . .

L … the size of input

f(L) … number of operations required for

instances of size L in the worst case

175

Today (within a time limit): size 100

Computer speed increase: 1000-times

Within the same time limit:

No of ops Size

L 100,000

L2 3,162

L4 562

2L 109

L! 101

Informally: A method for solving a problem Q

is called polynomial if f(L) ≤ p(L) for some

polynomial p.

177

3.1.2 Examples of hard problems

• HAM - Existence of a Hamiltonian cycle:

A cycle in a graph G = (V, E) is called

Hamiltonian (or a tour) if it contains all nodes of

G and G is called Hamiltonian if it contains a

Hamiltonian cycle.

HAM: Given a graph, is it Hamiltonian?

No polynomial method is known.

Similarly for digraphs.

• TSP – The Travelling Salesman Problem

Given a complete arc-weighted graph/

digraph, find a tour of minimal total length.


Hamiltonian graph

Hamiltonian graph

Non-Hamiltonian graph

179

• INTEGER LP (ILP)

Given an integer matrix A, integer vectors b, c,

find an integer vector x which solves the

optimisation problem

cTx → min

Ax = b

x ≥ 0


• (INTEGER) KNAPSACK

ILP is hard even if it contains only one equation

and only feasibility is wanted:

Given integers a1, …, an and K, are there

integers x1, …, xn ≥ 0 such that

Kxan

jjj =∑

=1 ?

An important variant: 0-1 KNAPSACK:

Given integers a1, …, an and K, is there a

subset S of {1, …, n} such that KaSj

j =∑∈

?

No polynomial method is known for either.

181

3.1.3 NP -complete problems

Class P : The class of all polynomially solvable

problems.

Class NP : The class of problems where an

object can be verified to be a solution in

polynomial time. This is a very wide class of CO

problems containing all “reasonable” problems

(including those for which a polynomial solution

method is not known and those which are

polynomially solvable).

ThusP ⊆NP.

Conjecture: P ≠ NP

183

Q ∈NP is called NP – complete if it satisfies the

following:

If Q ∈ P then P = NP .

Class NPC : The class of NP – complete

problems. Thus, if Q ∈NPC then Q is very

unlikely to be polynomially solvable.

Examples of NP – complete problems:

HAM, TSP, ILP, KNAPSACK, 0-1 KNAPSACK

Examples of problems in P : LP, MFP, SPP, AP

NPC

P

NP

185

Two special cases of the TSP:

TSP is metric (notation ∆TSP) if the direct-

distances matrix A = (aij) satisfies the triangle

inequality, that is aik + akj ≥ aij for all i, j, k.

Metric:

Not metric:

3

2

1 4

2

3

a b

c d

3

2

1 4

5

7

a b

c d

187

“Distance” may have various meanings:

geographical distance, time, cost…

TSP is Euclidean if the direct-distances matrix

A = (aij) is the matrix of Euclidean distances

between points in a plane with the Cartesian

coordinate system.

aij = d(zi, zj) for all i, j

where for a = (x1, y1), b = (x2, y2)

d(a, b) = ( ) ( )221

2

21yyxx −+−

(Euclidean distance)

x

y z1

z2

z4 z3

189

Hence every Euclidean TSP is ∆TSP.

Further NP – complete problems:

∆TSP

Euclidean TSP

Remark: The hardness of two similar problems

may be very different. Pairs of closely related

problems of profoundly different hardness:

In P In NPC

LP

AP

AP

MINCUT

SEPP (≥ 0)

ILP

3AP

TSP

MAXCUT

LEPP (≥ 0)

Our final aim: To design approximate

algorithms for the ∆TSP.

191

3.2. Algorithms for NP -complete Problems

3.2.1 Approximation algorithms - introduction

In order to measure the quality of an algorithm

we define the relative error of an algorithm:

Q … a combinatorial optimisation problem

(minimisation)

inst(Q) … the set of all instances of Q

c > 0 … cost function on the set of feasible

solutions of an instance

A … an algorithm: I∈inst(Q) → f(I) (a

feasible solution)

c(f(I)) … the cost of f(I)

c*(I) … optimal cost value for I∈inst(Q)

re(A , I) )I(*c

)I(*c))I(f(cdef −= … the relative error

of A with respect to instance I. Let ε ≥ 0. Then

an algorithm A for solving Q is called ε-

approximate if re(A , I) ≤ ε for all instances I.

193

3.2.2 Multigraphs and Eulerian spanning graphs

Multigraph is a diagram that can be obtained

from an undirected graph by adding a

nonnegative number of parallel edges.

A walk in a multigraph is a sequence of nodes

and arcs: (v1, e1, v2, e2, …, vk) where each ei is

an arc with endpoints vi and vi+1 (i = 1, …, k-1).

If v1 = vk then the walk is called closed.

Eulerian walk in a multigraph is a walk

(1) which is closed,

(2) in which every node appears at least once

and

(3) in which every arc appears exactly once.

195

(v1, f8, v4) … not a walk

w = (v1, f1, v2, f2, v1, f3, v2, f4, v1, f6, v4, f8, v2, f7, v3, f5, v1)

w … Eulerian walk (EW)

A multigraph G is called Eulerian if there is an

Eulerian walk in G.

v1 v2

v3 v4

f1 f2

f3

f4 f6 f7 f5 f8

walk but not closed

closed walk but not Eulerian

197

Theorem 3.2.1 (Euler): A multigraph G = (V, E)

is Eulerian ⇔

(1) G is connected and

(2) The degree of every node is even.

Sketch of proof →→→

Note: If G is Eulerian then an Eulerian walk in

G can easily be found.

Our aim is to find a good approximation of a

solution to TSP, that is a tour whose length is

close to the minimal length.

But how to construct a tour from an Eulerian

walk?

We will see two ways of doing this, leading to

two algorithms: TREE and Christofides’.

Note that a TSP for can be given in two

equivalent ways:

By a symmetric matrix A = (aij) ∈ Rn×n or by a

complete arc-weighted graph (Kn, (aij)).

A … direct-distances matrix

199

A common tool for both algorithms is the

concept of an Eulerian spanning graph:

Let A = (aij) be a symmetric n × n matrix with

zero diagonal. An Eulerian spanning graph of A

(ESG) is any Eulerian multigraph G = (V, E)

with V = {1, …, n}. The cost of G is

c(G) = ∑∈Eji

ija),(

.

Hence c(G) = the cost of any Eulerian walk in G.

The two algorithms differ by the construction of

the ESG. But once we have an ESG we construct

a tour, called embedded tour, from any Eulerian

walk w in the ESG in a unique way:

Start from the first node in w and then

repeatedly take the first node in w that has not

been taken yet. At the end we add the first node.

(See the example on the next page)

201

w = (v1, e1, v2, e2, v1, e7, v4, e6, v3, e3, v2, e4, v3, e5, v1)

Embedded tour τ =

c(τ) =

v1 v2

v3 v4

e2

e7 e5 e4

e3

e6

e1

203

3.2.3 Approximation algorithms for ∆TSP

Convention: We will draw only arcs of Kn that

are needed.

Theorem 3.2.2 Let (Kn, (aij)) be an instance of

∆TSP for a symmetric matrix A = (aij). If

G = (V, E) is an Eulerian spanning graph for

A = (aij) then every tour in Kn embedded in an

Eulerian walk in G has length not exceeding the

cost of G.

Proof: →→→

204

e1

205

TREE ALGORITHM

Input: ∆TSP given by an n × n symmetric

direct-distances matrix A = (aij) ≥ 0

Output: A tour in Kn

(1) Find an MST T in (Kn, (aij)) (note: we may

use GREEDY!)

(2) Create a multigraph G = (V, E) by using

two copies of each edge of T.

(3) Find an Eulerian walk w in G and a tour

embedded in w.

Note: G is an ESG since it is connected and

every node has an even degree.

Theorem 3.2.3 TREE ALGORITHM is a 1-

approximate algorithm for ∆TSP.

Proof: →→→

207

Lemma (to the next theorem):

The number of nodes of odd degree in any

graph is even.

Proof: →→→

CHRISTOFIDES’ ALGORITHM (CA)

Input/Output: As TREE ALGORITHM

(1) Find an MST T in (Kn, (aij)).

(2) Take the nodes of odd degrees in T and

find a perfect matching M of minimal total

weight in the complete graph induced by

these nodes. Let G be the multigraph with

arcs from T and in M (note: G is an ESG

by the Lemma!).

(3) Find an Eulerian walk w of G and a tour

embedded in w.

Theorem 3.2.4 CA is a 2

1- approximate

algorithm for ∆TSP.

Remark: Average error = 19.5%.

209

Proof of Theorem: →→→

Illustration to the proof:

i8

i7

i6

i5

i4

i3

i2 i1

τ*

1 the primal-dual algorithm 1 2 efficiently ...web.mat.bham.ac.uk/p.butkovic/3m02b/3m02b transp...

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