Approximation of Rectangle Stabbing and Interval Stabbing Problems.
ABSTRACT The weighted rectangle multistabbing problem (WRMS) can be described as follows: given is a grid in IR2\mathop{I\!\!R}^2consisting of columns and rows each having a positive integral weight, and a set of closed axisparallel rectangles each having
a 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 it is at least
its demand. (A column or row is said to stab a rectangle if it intersects it.)

Chapter: Rounding to an Integral Program
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ABSTRACT: We present a general framework for approximating several NPhard problems that have two underlying properties in common. First, the problems we consider can be formulated as integer covering programs, possibly with additional side constraints. Second, the number of covering options is restricted in some sense, although this property may be well hidden. Our method is a natural extension of the threshold rounding technique.05/2005: pages 4454;  SourceAvailable from: etd.uwaterloo.ca[Show abstract] [Hide abstract]
ABSTRACT: Abstract Piercing problems arise often in facility location, which is a wellstudied area of computational geometry. The general form of the piercing problem discussed in this dissertation asks for the minimum number of facilities for a set of given rectangular demand regions such that each region has at least one facility located within it. It has been shown that even if all regions are uniform sized squares, the problem is NPhard. Therefore we concentrate on approximation algorithms for the problem. As the known approximation ratio for arbitrarily sized rectangles is poor, we restrict our efiort to designing approximation algorithms for unitheight rectangles. Our †approximation scheme requires n,(1=†  SourceAvailable from: Retsef LeviACM Transactions on Algorithms 01/2008; 4. · 0.54 Impact Factor
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APPROXIMATION ALGORITHMS FOR RECTANGLE
STABBING AND INTERVAL STABBING PROBLEMS
SOFIA KOVALEVA • FRITS CR. SPIEKSMA
OR 0433
Page 2
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 axisparallel 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
minimumweight (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 constantfactor 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 axisparallel rectangles each having a
*This work grew out of the Ph.D. thesis [5]. This research was supported by EUgrant
APPOL, 1ST 200130027.
tCorresponding author, Department of Quantitative Economics, Maastricht Uni
versity, P.O. Box 616, NL6200 MD Maastricht,
sonja.kovaleva©ke.unimaas.nl.
+Department of Applied Economics, Katholieke Universiteit Leuven, Naamsestraat 69,
B3000, Leuven, Belgium, email: frits.spieksma©econ.kuleuven.ac.be.
The Netherlands, email:
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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 minimumweight
(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 socalled 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 axisparallel lines. More specifically, we define
a grid whose rows and columns are axesparallel lines containing the
rectangles' edges. We can restrict attention to those lines since any
axisparallel 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 axisparallel 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 2approximation 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 2approximation 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 xcoordinates, 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 APXhard
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;lapproximation 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 2approximation 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 LProunding 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 LPrelaxation 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 LPrelaxation.
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 LPrelaxation.
In fact, the special structure of Ipy (I, b) (IpZ (I, b)) allows us to solve it via
a minimumcost 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 LPrelaxation of formulation Ipy (I, b) and substitute
variables with new variables un, ... , Ut as Yc = Uc  UcI, Vc E [1 : t]. Then it
transforms into
Minimize
subject to
WI Uo + (WI 
Uri 
UcI 2': 0,
W2)U2 + ... + (WtI 
Vi E [1 : n]
Ve E [1 : t].
Wt)Ut1 + WtUt
UZi1 2': bi ,
Uc 
(6)
Let us denote the vector of objective coefficients, the vector of righthand
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 LPrelaxation equals
the same ratio.
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1. solve the LPrelaxation 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+lapproximation 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 lefthand 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 righthand 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 LPrelaxation 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 LPrelaxation.
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 LPrelaxation 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 LPrelaxation 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 LPrelaxation 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(n1)(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 > (nl)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 /cxJ 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(n1?(q+1)2 c(1+1/q).2(n1)(q+1)+4nq2q2c.
Obviously, the value of the term CR, given that C+R = 2(n1)(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(nl);(q+l)2):
1
 > 
q
1 1
+ 4nq  2q  ;:::;;;::
(n1)2(q+1)2· (n1)(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 LPrelaxation for I ~ is at
least equal to 2(n21)(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 LPrelaxation 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 worstcase 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 LPrelaxation 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 (ll/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
yIpvalues) 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 zIPvalues, and we
denote by I (l ~ 0) the number of zIpvalues 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 zIpvalues 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:
13
Page 15
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
righthand 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 LPrelaxation 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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 Available from Frits C. R. Spieksma · May 29, 2014
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