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APPROXIMATION ALGORITHMS FOR RECTANGLE

STABBING AND INTERVAL STABBING PROBLEMS

SOFIA KOVALEVA • FRITS CR. SPIEKSMA

OR 0433

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Approximation Algorithms for Rectangle

Stabbing and Interval Stab bing Pro blems*

Sofia Kovalevat and Frits C.R. Spieksma+

Abstract

In the weighted rectangle stabbing problem we are given a grid

in ]R2 consisting of columns and rows each having a positive integral

weight, and a set of closed axis-parallel rectangles each having a pos-

itive integral demand. The rectangles are placed arbitrarily in the

grid with the only assumption that each rectangle is intersected by

at least one column and at least one row. The objective is to find a

minimum-weight (multi)set of columns and rows of the grid so that

for each rectangle the total multiplicity of selected columns and rows

stabbing it is at least its demand. A special case of this problem

arises when each rectangle is intersected by exactly one row. We de-

scribe two algorithms, called STAB and ROUND, that are shown to

be constant-factor approximation algorithms for different variants of

this stabbing problem.

1 Introduction.

The weighted rectangle stabbing problem (WRSP) can be described as follows:

given is a grid in JR2 consisting of columns and rows each having a positive

integral weight, and a set of closed axis-parallel rectangles each having a

*This work grew out of the Ph.D. thesis [5]. This research was supported by EU-grant

APPOL, 1ST 2001-30027.

tCorresponding author, Department of Quantitative Economics, Maastricht Uni-

versity, P.O. Box 616, NL-6200 MD Maastricht,

sonja.kovaleva©ke.unimaas.nl.

+Department of Applied Economics, Katholieke Universiteit Leuven, Naamsestraat 69,

B-3000, Leuven, Belgium, e-mail: frits.spieksma©econ.kuleuven.ac.be.

The Netherlands, e-mail:

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positive integral demand. The rectangles are placed arbitrarily in the grid

with the only assumption that each rectangle is intersected by at least one

column and at least one row. The objective is to find a minimum-weight

(multi)set of columns and rows of the grid so that for each rectangle the total

multiplicity of selected columns and rows stabbing this rectangle equals at

least its demand. (A column or row is said to stab a rectangle if it intersects

it. )

A special case of the WRSP is the case where each rectangle is intersected

by exactly one row; we will refer to the resulting problem as the weighted

interval stabbing problem (WISP), or ISP in case of unit weights (see Figure 1

for an example of an instance of the ISP).

Figure 1: An instance of ISP with unit demands. The rectangles (or intervals

in this case) are in grey; the columns and row in black constitute a feasible

solution.

Motivation. Although at first sight the WRSP may seem rather specific,

it is not difficult to see that the following two problems can be reduced to

WRSP:

• Solving special integer programming problems: the following type of

integer linear programming problems can be reformulated as instances

of WRSP: minimize{ wxl (BIG)x ~ b, x E II,!}, where Band G are

both O)-matrices with consecutive 'l'-s in the rows (a so-called interval

matrix, see e.g. Schrijver [8]), b E I I ; ~ , w E Z ~ . Indeed, construct a grid

which has a column for each column in B and a row for each column

in G. For each row i of matrix BIG, draw a rectangle i such that it

intersects only the columns and rows of the grid corresponding to the

positions of '1 '-s in row i. Observe that this construction is possible

since Band G have consecutive 'l'-s in the rows. To complete the

construction, assign demand bi to each rectangle i and a corresponding

weight Wj to each column and row of the grid. Let the decision variables

x describe the multiplicities of the columns and rows of the grid. In this

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way we have obtained an instance of WRSP. In other words, integer

programming problems where the columns of the constraint matrix A

can be permuted such that A = (BIG) with Band G each being an

interval matrix, is a special case of WRSP .

• Stabbing geometric figures in the plane: given a set of arbitrary con-

nected closed geometric sets in the plane, use a minimum number of

straight lines of two given directions to stab each of these sets at least

once. Indeed, by introducing a new coordinate system specified by the

two directions, and by replacing each closed connected set by a closed

rectangle defined by the projections of the set to the new coordinate

axes, we obtain an instance of the problem of stabbing rectangles using

a minimum number of axis-parallel lines. More specifically, we define

a grid whose rows and columns are axes-parallel lines containing the

rectangles' edges. We can restrict attention to those lines since any

axis-parallel line stabbing some set of rectangles can be replaced by a

line stabbing this set and containing a rectangle's edge. Therefore, the

problem of stabbing the rectangles with axis-parallel lines reduces to

the problem of stabbing them with the rows and columns of the grid.

Literature. The WRSP and its special case WISP have received attention in

literature before. Motivated by an application in parallel processing, Gaur et

al. [2] present a 2-approximation algorithm for the WRSP with unit weights

and demands, which admits an easy generalization to arbitrary weights and

demands. Furthermore, Hassin and Megiddo [3] (mentioning military and

medical applications) study a number of special cases of the problem of stab-

bing geometric figures in JP2.2 by a minimum number of straight lines. In par-

ticular, they present a 2-approximation algorithm for the task of stabbing

connected figures of the same shape and size with horizontal and vertical

lines. Moreover, they study the case of stabbing horizontal line segments

of length K, whose endpoints have integral x-coordinates, with a minimum

number of horizontal and vertical lines, and give a 2 -

algorithm for this problem. In our setting this corresponds to the ISP with

unit demands, where each rectangle in the input is intersected by exactly K

columns.

Finally, concerning computational complexity, a special case of ISP where

each rectangle is stabbed by at most two columns, is shown to be APX-hard

in [7].

~ -approximation

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Our results. We present two approximation algorithms for different vari-

ants of WRSP (see e.g. Vazirani [9] for an overview on approximation algo-

rithms). First, we describe a q;l-approximation algorithm called ROUND

for the case where the demand of each rectangle is bounded from below by

an integer q. Observe that this provides a 2-approximation algorithm for the

WRSP described in the introduction, where q = 1. Thus, our algorithm is

an improvement upon the approximation ratio of the algorithm of Gaur et

al. [2] for instances with a lower bound on the rectangles' demands that is

larger than 1. Second, we present a ~ l - ( l ~ l / k ) k )

called STAB for ISPk, the variant of I P where each row intersects at most

k rectangles (e.g., the instance depicted in Figure 1 is an instance of ISP3).

Observe that STAB is a ~ - a p p r o x i m a t i o n algorithm for the case k = 2, and

that STAB is a e ~ l -approximation algorithm for the case where the num-

ber of rectangles sharing a row is unlimited (k = 00). Thus, STAB improves

upon the results described in Hassin and Megiddo [3] (for K 2: 3) and does

not impose any restrictions on the number of columns intersecting rectan-

gles. Third, we state here that STAB for the weighted case of ISP 00, i.e.,

the case where the columns and the rows of the grid have arbitrary positive

integral weights, is a e ~ l -approximation algorithm. For the proof of this

result, we refer to Kovaleva [5]. Our algorithms are based on rounding the

linear programming relaxation of an integer programming formulation in an

interesting way. We use the following property present in our formulation:

the variables can be partitioned into two sets such that when given the val-

ues of one set of variables, one can compute in polynomial time the optimal

values of the variables of the other set of variables, and vice versa. Next, we

consider different ways of rounding one set of variables, and compute each

time the values of the remaining variables, while keeping the best solution.

Summarizing our results:

approximation algorithm

• we generalize the results of Gaur et al. [2] to obtain a q+1 -approximation

algorithm called ROUND for the case where the demand of each rect-

angle is bounded from below by an integer q (Section 3),

q

• we describe an ( l - ( l ~ l / k ) k )

ISPk based on an original rounding idea (Section 4).

approximation algorithm called STAB for

We also show that there exist instances of the WRSP, ISP2 and ISP 00, for

which the ratio between the values of a natural ILP formulation and its LP-

relaxation is equal (or arbitrary close) to the obtained approximation ratios.

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This suggests that these approximation ratios are unlikely to be improved by

an LP-rounding algorithm based on the natural ILP formulation.

2 Preliminaries.

Let us formalize the definition of WRSP. Let the grid in the input consist of

t columns and m rows, numbered consecutively from left to right and from

bottom to top, with positive weight We (vr) attached to each column c (row

r). Further, we are given n rectangles such that rectangle i has demand

di E Z+ and is specified by leftmost column li' rightmost column ri, top row

ti and bottom row bi.

Let us give a natural ILP formulation of WRSP. In this paper we use notation

[a : b] for the set of integers {a, a+ 1, ... , b}. The decision variables Ye, Zr E Z+,

c E [1 : t], r E [1 : m], denote the multiplicities of column c and row r

respectively.

Minimize

subject to

2 : : ~ = 1 WeYe + 2 : : ~ = 1 VrZr

2::rE[kti] Zr + 2::eE[li:ri] Ye ~ di Vi E [1 : n]

Zr, Ye E Z ~

Vr,c.

(1)

(2)

(3)

The linear programming relaxation is obtained when replacing the integrality

constraints (3) by the nonnegativity constraints Zr, Ye ~

For an instance I of WRSP and a vector b E zn, we introduce two

auxiliary ILP problems:

0, Vr, c.

IPY(I, b):

Minimize

subject to

Minimize

subject to

2 : : ~ = 1 WeYe

2::eE[li:ri] Ye ~ bi

Ye E Z+,

2 : : ~ = 1 VrZr

2::rE [bi:ti] Zr ~ bi

Zr E Z+,

Vi E [1 : n]

Vc E [1 : t]

Vi E [1 : n]

Vc E [1 : m]

(4)

(5)

Lemma 2.1. For any b E zn, the LP-relaxation of each of the problems

Ipz (I, b) and IFY (I, b) is integral.

Proof. This follows from the unimodularity of the constraint matrix of (5)

which is implied by the "consecutive one's" -property (see e.g. Schrijver [8]).

o

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Corollary 2.2. The optimum value of IFY(I, b) (IPZ(I, b)) is smaller than

or equal to the value of any feasible solution to its LP-relaxation.

Corollary 2.3. The problem IFY(I, b) (IPZ(I, b)) can be solved in polynomial

time. Its optimal solution coincides with that of its LP-relaxation.

In fact, the special structure of Ipy (I, b) (IpZ (I, b)) allows us to solve it via

a minimum-cost flow algorithm: let JvICF(p, q) denote the time needed to

solve the minimum cost flow problem on a network with p nodes and q arcs.

Lemma 2.4. The problem IFY(I, b) (IPZ(I, b)) can be solved in time O(JvICF(t, n+

t)) (O(JvICF(m, n + m))).

Proof. Consider the LP-relaxation of formulation Ipy (I, b) and substitute

variables with new variables un, ... , Ut as Yc = Uc - Uc-I, Vc E [1 : t]. Then it

transforms into

Minimize

subject to

-WI Uo + (WI -

Uri -

Uc-I 2': 0,

W2)U2 + ... + (Wt-I -

Vi E [1 : n]

Ve E [1 : t].

Wt)Ut-1 + WtUt

UZi-1 2': bi ,

Uc -

(6)

Let us denote the vector of objective coefficients, the vector of right-hand

sides and the constraint matrix by w, band C respectively, and the vector of

variables by u. Then (4) can be represented as {minimize wul Cu 2': b}. Its

dual is: {maximize bxl CT x = W, x 2': O}. Observe that this is a minimum

cost flow formulation with flow conservation constraints CT x = w, since CT

has exactly one '1' and one '-1' in each column. Given an optimal solution

to the minimum cost flow problem, one can easily obtain the optimal dual

solution Uo, ... , Ut (see Ahuja et al. [1]), and thus optimal YI, ... , Yt as well. 0

3 The Algorithm ROUND.

Let WRSPq be the special case of the WRSP, where di 2': q, Vi E [1 : n]. In

Subsection 3.1 we describe an algorithm ROUND, and show that it achieves a

ratio of q+1. Subsection 3.2 shows that the integrality gap between a natural

q

integer programming formulation and its corresponding LP-relaxation equals

the same ratio.

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1. solve the LP-relaxation of (1)-(3) for I and obtain its optimal solution

(yIP, ZIp)

2. solve Ipy (I, a), where ai = l q;l LCE[li:r;j y ~ j, for all i E [1 : n];

obtain y;

3. solve Ipz (I, b), where b i = l q;l LCE[bi:ti] z ~ j, for all i E [1 : n];

obtain z;

4. return (y, z)

Figure 2: Algorithm ROUND.

3.1 An approximation result.

Figure 2 describes algorithm ROUND applied to an instance I of WRSPq.

Theorem 3.1. Algorithm ROUND is a q+l-approximation algorithm for

the problem WRSPq.

q

Proof. Let I be an instance of WRSPq. First, we show that the solution

(y, z) returned by ROUND is feasible for I, i.e., satisfies constraints (2) and

(3). Obviously, vectors y and z are integral, hence, constraint (3) is satisfied.

For each i E [1 : n], consider the left-hand side of constraint (2) for (y, z).

By construction, y and z are feasible to IPY(I, a) and IPZ(I, b) respectively.

U sing this, and the way vectors a and b were constructed, obtain:

'"' -

D

rE[bi:ti]

'"' -- l q + 1 '"' IPj

q

cE[li:ri]

l q + 1 '"' IPj

Yc + -- D

q

Zr + D

Yc 2: ai + bi - -- D

zr .

cE[li:ri] rE[bi:ti]

(7)

Since l a j + lp j 2: l a + p j - 1, for any positive a and p, and since ZIp and

yIp satisfy constraint (2), the right-hand side of (7) is at least equal to:

where the last inequality holds because di 2: q, \:Ii E [1 : n] for WRSPq.

Thus, inequality (2) holds for (y, z) and hence (y, z) constitutes a feasible

solution to instance I of WRSPq.

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The approximation ratio of ROUND is obtained from the ratio between

the value of the returned solution (y, z) and the value of the optimal fractional

solution (yIP, ZIp) (and of course by observing that the value of the latter

solution does not exceed the optimum value of \VRSPq).

We claim that L ~ = l wcYc :s; q;l L ~ = l wcyt Observe that q;lylP is a feasi-

ble solution to the LP-relaxation of Ipy (I, a) with ai = l q;l LCE[li:ri] y;r J, Vi E

[1 : n].

Indeed, constraints (4) are clearly satisfied: LCE[li:r;j q;ly;r 2:

l q;l LCE[li:ri] y;r J, Vi E [1 : n]. Thus, vectors y and q;lylp are respectively

an optimal solution to Ipy (I, a) and a feasible solution to its LP-relaxation.

Now Corollary 2.2 implies the claim.

Similarly, L ~ = l vrzr :s; q;l L ~ = l vrz;? This proves the ratio of q;l be-

tween the value of solution (y, z) and solution (yIp, ZIP).

Observe that in case of unit weights and unit demands, ROUND boils

down to the algorithm described in Gaur et al. [2].

0

3.2 Tightness.

In this section we provide instances showing that the ratio between the the

optimal value ofWRSPq and the LP-relaxation of its natural ILP formulation

(1)-(3) can be arbitrary close to q+l, the approximation factor of algorithm

q

ROUND. We will refer to this ratio as the integrality gap of (1)-(3).

Theorem 3.2. For each q E N, the integrality gap of (1)-(3) is arbitrarily

close to q+l.

q

Proof. For any q E N and any integral n 2: 2, we construct an instance I ~ of

WRSPq, such that the ratio between the optimal values of the formulation

(1 )-(3) and its LP-relaxation for the instance I ~ tends to q+1 as n increases.

The construction is as follows. Denote: c =

consist of qn/c columns and qn/c rows. The instance includes all different

rectangles intersecting exactly q/ c columns and rows in total (here 'different'

implies intersecting different subsets of columns and rows). The weight of

each column and row is unit, and the demand of each rectangle is equal to q.

q

2 ( n - l ) ~ ( q + 1 ) 2 ' and let the grid

Claim 1. The optimum value of the LP-relaxation of formulation (1)-(3) for

I ~ is less than or equal to 2qn.

To show this, we introduce a feasible solution to the LP relaxation for I ~ of

the value 2qn: assign multiplicity c to each column and row of the grid. It

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is feasible, since each rectangle in I ~ is then stabbed by a total multiplicity

of q. The value of this solution is 2qn, hence the claim follows.

Claim 2. The optimum value of WRSPq for I ~ is greater than 2(n-1)(q+ 1).

Suppose the opposite is true, that is, suppose that there exists a feasible

(integral) solution to WRSPq for I%, (y, z), with value 2(n - 1)(q + 1). Sup-

pose that this solution assigns total multiplicity C to the columns and to-

tal multiplicity R to the rows of the grid ( L ~ ! ~ C yc = C, L ~ ~ t Zr = R,

C + R = 2(n -l)(q + 1)). Denote by x the maximum number of consecutive

columns having total multiplicity less than or equal to q - 1.

Claim 2.1. x 2: C / ~ ~ l - 1.

Indeed, by definition of x, the total multiplicity of any x + 1 consecutive

columns is at least q. Therefore, the total multiplicity C of the columns has

to be at least l : ~ ~ J . q. This inequality, i.e., C 2: l : ~ ~ J . q, implies the claim.

Claim 2.2. x < q/c - 1.

Suppose that this does not hold. Since the number of rows in the grid is

obviously larger than their total multiplicity R (otherwise the value of our

solution would be much larger than 2 (n - 1) (q + 1)), there exists a row with

multiplicity O. By construction our instance contains a rectangle intersecting

this row and q/c - 1 of the x consecutive columns. Obviously, the total

multiplicity of the columns and rows stabbing this rectangle is at most q -1,

which is a contradiction with the assumption that solution (y, z) is feasible.

Claim 2.3. C > (n-l)q. This follows from 0 ~ ~ 1 < q/c, which in turn follows

from Claims 2.1 and 2.2.

We continue with the proof of Claim 2. Consider all the rectangles in I ~

intersecting exactly those x consecutive columns. Each of them intersects

q/c - x rows. It receives multiplicity of at most q - 1 from the columns, and

therefore needs to receive multiplicity of at least 1 from its rows. The fact

that our instance contains all the possible different rectangles implies that

each set of q/ c - x consecutive rows has to have the total multiplicity of at

least 1. By a similar argument as in the proof of Claim 2.1, and by Claim

2.1 itself, the total multiplicity of the rows has to satisfy:

nq/c nq/c

nq/E:

c/q+1 +

n

R2: l /c-xJ 2: l /

q

1J = ll __ n_+fJ >

C/q+l

q c -

q

>

n

-1=

n(C+q) C

C + q - nq + c( q+q)

-1.

1 -

C / ~ + l + ~

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lVIultiplying both sides by the denominator, which is positive due to Claim

2.3, obtain: R(C + q - nq + s(Cq+q)) > n(C + q) - (C + q - nq + c(cq+q)).

Interchanging the role of columns and rows, and using a similar argument,

we have: C(R+q - nq+ s(R+q)) > n(R+q) - (R+q - nq+ e:(R+q)). Summing

q

up these inequalities, and collecting the coefficients of the terms C Rand

(C + R), we arrive at:

q

2CR(1 + c/q) > (C + R)( -q + nq - c + n - 1 - c/q) + 4nq - 2q - 2c =

= (n - l)(q + l)(C + R) - (c + c/q)(C + R) + 4nq - 2q - 2c.

Using our assumption C + R = 2(n - l)(q + 1), we rewrite as follows:

2CR(1+c/q) > 2(n-1?(q+1)2- c(1+1/q).2(n-1)(q+1)+4nq-2q-2c.

Obviously, the value of the term CR, given that C+R = 2(n-1)(q+1), can

not exceed (n - 1)2(q + 1)2. Then, if solution (y, z) is feasible, the following

should be satisfied (recall that c = 2(n-l);(q+l)2):

1

- > -

q

1 1

+ 4nq - 2q - -;----:-::-;---;---;::

(n-1)2(q+1)2· (n-1)(q+1)

This inequality does not hold for any q ~ 1, n ~ 2. Thus, solution (y, z) can

not be feasible. This proves Claim 2.

From Claims 1 and 2 it follows that the ratio between the optimum value of

formulation (1)-(3) and the optimum value of its LP-relaxation for I ~ is at

least equal to 2(n-21)(q+l), which tends to q+l as n increases.

nq

Remark 3.1. Notice that this example shows that the results in Gaur et al. [2]

are tight as well.

D

q

As mentioned in the introduction, Theorems 3.1 and 3.2 imply that it is

unlikely that a better ratio for WRSPq can be achieved using formulation

(1)-(3).

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4 Algorithm STAB.

Recall that the interval stabbing problem WISPk refers to the restriction of

vVRSP where each rectangle in the input is intersected by exactly one row

and each row intersects at most k rectangles. Moreover, we assume in this

section that all the weights and demands are unit: We = Vr = di = 1, Yc E

[1 : t], r E [1 : m] and i E [1 : n], i.e., we concentrate on ISPk with unit

demands. In Subsection 4.1 we describe an algorithm STAB, and show that

it achieves a ratio of 1- l ~ l / k ) k . Subsection 4.2 shows that the integrality

gap between the values of a natural integer programming formulation and its

corresponding LP-relaxation equals the same approximation ratio for k = 2

and k = 00, namely ~ and e ~ l respectively. An alternative algorithm for

the case k = 2 yielding the same worst-case ratio (i.e., ~ ) is described in

Kovaleva and Spieksma [6].

4.1 An approximation result.

In this subsection we describe an algorithm STAB for ISPk and show that it is

a l - ( l ~ l / k ) k

approximation algorithm. Let us first adapt the ILP formulation

(1)-(3) to ISPk with unit demands:

Minimize

subject to

L ~ = l Ye + L ~ = l Zr

ZPi + LeE[li:ril Ye ~ 1 Yi E [1 : n]

ZTl Ye E Z+

Yr, c.

(8)

(9)

(10)

Informally, algorithm STAB can be described as follows: solve the LP-

relaxation of (8)-(10), and denote the solution found by (yIp, ZIp). Assume,

without loss of generality, that the rows are sorted as Z{ ~ zi ~ ... ~ z ~ .

At each iteration j (j = 0, ... , m) we solve the problem (8)-(10) with a fixed

vector z, the first j elements of which are set to 1, and the others to o. As

shown in Section 2, this can be done in polynomial time using a minimum

cost flow algorithm. Finally, we take the best of the resulting m + 1 solutions.

A formal description of STAB is shown in Figure 3.

We use notation: value(y, z) = L ~ = l Ye + L ~ = l Zr, value(y) =

value(z) -

L ~ = l Ye, and

L ~ = l Zr·

Theorem 4.1. Algorithm STAB is a l - ( l ~ l / k ) k - approximation algorithm for

ISPk ·

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1. solve the LP-relaxation of (8)-(10), and obtain its optimal solution

(yIp, ZIp);

2. reindex the rows of the grid so that z ~ P ~ zi ~ ... ~ z ~ ;

3. V+- 00;

4. for j = 0 to m

for i = 1 to j Zi +- 1,

for i = j + 1 to m

solve IPY(I, b), where bi = 1 -

if value(y, z) < V then V +- value(y, z), y* +- y, z* +- z;

Zi +- O.

ZPi' Vi E [1 : n], and obtain y;

5. return (y*,z*).

Figure 3: Algorithm STAB

Proof. Consider an instance I of ISP k, and let (yIp, ZIp) and (y*, z*) be re-

spectively an optimal LP solution and the solution returned by the algorithm

for I. We prove the theorem by establishing that

1

value(y*,z*) ~ 1- (l-l/k)k value(ylp, ZIp).

(11)

It is enough to prove the result for instances satisfying the following as-

sumption: we assume that the optimal LP solution satisfies constraints (9)

at equality, i.e.

zh + L Yd = 1, Vi E [1 : n].

cE(li:ri)

(12)

Indeed, if (12) does not hold for some intervals i, we change the instance by

shortening the appropriate intervals (and perhaps splitting the columns with

yIp-values) so that the assumption becomes true (see Figure 4). It is easy

to check that the optimal LP solution remains the same (up to the splitted

columns). Since in the new instance the intervals become shorter, algorithm

STAB returns a solution with a value equal to or larger than the value of

the solution returned for the initial instance. Then inequality (11) proven

for the new instance implies this inequality for the initial instance as well.

We order the rows of the grid in order of nonincreasing zIP-values, and we

denote by I (l ~ 0) the number of zIp-values equal to 1. Then: zi =

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Figure 4: Example of an initial instance (left) and a new instance satisfying

the assumption (right).

z!p = 1,1 > Z ! ~ l 2': ... 2': z ~ 2': O. We assume that value(ylp) is positive

(otherwise all the zIp-values have to be equal to 1 and the theorem obviously

holds).

By construction:

value(y*, z*) = min value(yj, zj) ::; min value(yj, zj),

jE[O:m]

(13)

jE[I:m]

where (yj, zj) is the jth solution generated in Step 4 of STAB.

Let us proceed by defining for each j E [0 : m], a number qj E ffi. that

depends on some given .6. E [0, l]m and given (3 > 0 as follows:

where we put .6.j = 0 if j > m.

Observe that qj for each j E [0 : m] is uniquely defined by this equality; we

denote the solution of (14) by qj(.6., (3).

We will prove the following lemma:

Lemma 4.2.

Then, assuming that Lemma 4.2 holds, it follows from (13) that:

value(ylp)

k

value(y*, z*) ::; min (j + k . qj (ZIp,

JE[I:m]

)).

(15)

The Theorem follows now from the following lemma, the proof of which can

be found in the appendix:

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Lemma 4.3. Given are real numbers 1 2: 61 2: 62 2: ... 2: 6 m 2: 0, a

positive real number Y, an integer p 2: 2, and an integer l 2: 0. Then the

following holds:

1

min (i + p. Qi(.6., Yip)) ::;

iE[I:m]

(

I ) (Y + "" .6.r ) + l,

1 - 1 p P

1 -

~

r=l+l

m

(16)

By applying this Lemma with p = k, .6. = ZIp, and Y = value(ylp), the

right-hand side of (15) can be bounded by:

and since z ~ P

equal to:

= z!P = 1, the right hand side of this last expression is

1

l

(

IP IP)

(

Ik)k va ue y ,z .

1- 1- 1

The theorem is then proven.

To complete the proof of the theorem, we now proceed with the

Proof of Lemma 4.2. Consider (yj, zj), for some j E [l : m], let us find an

upper bound for value (yj , zj). By construction:

- zj = 1 Vr < J'

T , -,

-z? = 0, Vr 2: j + 1,

- yj is an optimal solution to Ipy (I, b), where bi = 1 - Zii' Vi E [1 : n].

Obviously, value(zj)

solution y'j, which is feasible to the LP-relaxation of Ipy (I, b). Then, Lemma

2.1 implies that value(yj) ::; value(y'j).

= j. In order to bound value(yj) we introduce a

First, let us define subsets 81,82, ... , 8m , where 8r C [1 : t], Vr = 1, ... , Tn,

(i.e., each subset consists of a set of columns of the grid) in the following

way:

i:Pi=r

Thus, 8r is the set of columns stabbing intervals in row r.

14

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