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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 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:
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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
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
Page 16
Fix now some j E [l
1, \fT E [l + 1 : m]):
m l, and construct y'j as follows (recall that z7 <
1
Ip
( l  z ~ ) Ye ,
yJ
if c E Sj+l
if C E Sj+2 \Sj+l
if C E Sm \(Sj+l U ... U Smd
otherwise.
(17)
Let us now establish feasibility of y'j with respect to the LPrelaxation of
IPY(I, b). For any interval i we show that the following inequality holds:
"" y'j > 1  zj .
~
e 
eE[li:ri]
Pi
(18)
If Pi < j + 1, where Pi is the row number of interval i, then Z ~ i = 1, and the
inequality holds automatically. Consider the case Pi ~
C E SPi' either c E Spi\(Sj+l U ... U Spid, or c E (Sj+l U ... U Spi1). The
first case implies that y1 = yJ /(1 
to yJ multiplied by some coefficient at least as large as 1/ (1 z ~ ) (see (17)).
Then, since [li : Til E SPi' we have y2 ~ yJ /(1 
U sing this and remembering that (yip, Zip) satisfies z ~ + LeE [Ii :ri] yJ ~ 1, we
have:
"" Ye ' j
1 ""
~
> (IziP) ~
eE[kri]
p, eE[kri]
Thus we have shown that inequality (18) holds for any i E [1 : n], and
therefore y'j is feasible to the LPrelaxation of IPY(b).
implies that
value(yj) ::; value(y,j).
j + 1. Clearly, for any
z ~ ) ; the second means that y2 is equal
z ~ ) for any c E [li : Til·
I
y; >
1 Zip
_,P,..::..i = 1
1 z ~
.
Now Lemma 2.1
(19)
( I") (I value(yIP))
[l
1
In what follows we show that value y J
By construction of y'j, using notation Y(S) = LeEs yJ :
::; k . qj z P,
k
' \f j E : m .
15
Page 17
Observe that for the Y Uterms the following equality holds:
(21 )
lVIoreover, using the definition of Sr, our assumption (12), and the fact that
there are at most k intervals per row, we have for each r = j + 1, ... , Tn:
(22)
Consider now the following optimization problem:
max
(\PYJ+1 + ~
lZj+l
YJ+1 + ... + Ym + ~ ~ m + 1 Yr ::; value(ip)
o ::; Yr ::; k(l  z7),
o ::; Yr ::; k,
YJ+2 + ... + ~Ym + ~ ~ m + 1 Yr)
lj+l,lj+2,'"
subject to
lZj+2 lzm
(23)
(24)
(25)
Vr = j + 1, ... , Tn
Vr = Tn + 1, ... , 00
Due to (21) and (22) the following solution is feasible to it:
Yr = Y(Sr \(Sj+1USj+2U",USrl)) for each r = j+1, ... , Tn, and ~ : m + l Yr =
Y((l : t)\(Sj+1USj+2U ... USm)) (distributed arbitrary among the components
of the sum). Therefore the optimum value of this optimization problem is
an upper bound on the righthand side of (20).
How does the optimum solution to this optimization problem look like?
It is easy to see that to achieve the optimum value one should assign the
maximum possible value (according to the constraints) to the variables with
the largest objective coefficient. Since the objective coefficients are non
increasing, we increase the values of the variables YJ+l, YJ+2, ... in this order
until the limits are met. We obtain the following optimal solution:
for some number q E lR+, which due to (23) has to satisfy:
16
Page 18
where we put: z ~ P = 0 for any r > m. Notice that q = qj(ZIP, valu:(ylP)) (see
(14)), and the optimum value of the problem (23)(25), which bounds the
righthand side of (20) from above, is k·qj(ZIP, valu:(ylP)). This proves Lemma
4.2. D
D
Remark 4.1. With small adjustments, STAB and Theorem 4.1 can be gener
alized to the case of arbitrary rectangle demands. The main idea is then to
sort the rows according to the fractional parts of the zvalues of the optimal
LP solution in case they are greater than 1.
Remark 4.2. It is proven in [5] that STAB also is a e/(e I)approximation
algorithm for the weighted case of ISP 00, i.e., the case where the columns
and rows of the grid may have arbitrary integer positive weights. The proof
given in this paper does not extend automatically to this case.
4.2 Tightness.
In this subsection we demonstrate that the ratio between the optimum val
ues of ISPk and the LPrelaxation of its ILP formulation (8)(10) can be
arbitrarily close to the bounds achieved by STAB in case k = 2 and k = ()()
(which is respectively 4/3 and e/(e  1)).
For the case k = 2 this is shown by the instance of ISP2 depicted in
Figure 5 (recall that all the column and row demands and rectangle weights
are unit). Here the optimal value of the problem is 2, since at least two
elements (columns or rows) are needed to stab the 3 rectangles, whereas the
optimal fractional solution has the value of 3/2 .
• ~">'''~
J I
C",;
'H''''T'JfC
!
Figure 5: An instance of ISP2 and an optimal fractional solution.
In the remainder of the section we consider the problem ISP 00, or simply ISP,
without any limitation on the number of rectangles sharing a row. We will
exhibit a family of instances {Im}mEN of ISP, such that the ratio between the
17
Page 19
optimal values of integral and fractional solutions for Im tends to e/ (e  1)
as m increases.
Theorem 4.4. The integrality gap of (8)(1 0) is arbitrarily close to e ~ l '
Proof. For each mEN we construct an instance Im as follows. Let the
grid have m rows and t = m! columns. Let the rows be numbered conseCll
tively and let each row j intersect exactly j rectangles of the instance. Let
rectangles intersected by row j be numbered j1, ... , jj. All these rectangles
are disjoint and each intersects exactly m! columns (see Figure 6). So, for
a rectangle ji we have that its row number Pji is r, and its leftmost and
rightmost columns are lji = j! (i  1) + 1 and rji = j! i. The total number of
rectangles in the instance is then n = 1 + 2 + ... + m.
J
Figure 6: Instance I 4.
We claim that the following solution (y, z) is optimal to the LPrelaxation of
(8)(10) for Im:
{ 0,
\lj = 1, ... , P
Zj =
Yc = ..;, \Ie = 1, ... , m!,
m.
1  Pjj, \lj = P + 1, ... , m
where P = P( m) is the number satisfying:
1
+
m
1 1 1
(26)
1
++ ... +<1 and
m
ml
1 1
P+l
l+···++P>l.
P+l
m
18
Page 20
It is easy to verify that the value of this solution equals:
mIl
'""' Yc + L Zr = m  P( +  + ... + ).
~
c=l r=l
t
1
P+1 P+2
m
First, we show that (y, z) is a feasible solution to the LPrelaxation. Take
any rectangle ji and show that the constraint Zpji + 2:cE[lji,rjil Yc ~ 1 is
satisfied. Notice that the zvalues of our solution can be also expressed as:
Zj = max(l  y, 0), Vj = I, .... ,m. Substituting these values, and rewriting
the lefthand side of constraints (9) gives:
PPm! P
max(l :,0) + '""' 
= max(l :,0) + . 
Ji
~
m!
CE[ljJjJ
P P
Ji
P
Ji
= max(l :,0) +:.
Ji Ji m!
Clearly, the last expression is at least equal to I, which proves feasibility of
the solution (y, z).
Now, let us prove optimality of (y, z). For this goal we present a feasible
dual solution to the LPrelaxation of (8)(10) which has the same value as
(y, z). Below we present the dual problem, where a variable Xi corresponds
to rectangle i:
. 2 : ~ = 1 Xi
2:i:Pi=j Xi ::; 1
2:i:CE[li,ril Xi ::; 1 'lie = I, .. , t
o ::; Xi ::; 1
Maximize
subject to
Vj = I, .. , m
Vi = I, .. , n
Consider now the following dual solution:
{
l/j,
~  ( r k + m ~ l + ... + P ~ l ) '
Xji=
if j = P + 1, ... , m,
if j = P,
otherwise.
(27)
(28)
In words, we assign 1/j to the dual variables corresponding to rectangles on
row j = P+1, ... ,m, l  ( r k + m ~ l + ... + P ~ l ) to the variables corresponding
to rectangles on row j = P, and 0 to the other variables.
It is easy to verify that this solution is feasible to problem (27) and has value
nIl
'""' Xi = m  P(  +  + .. , + ),
~
m
i=l
1
m1 P+1
(29)
19
Page 21
which is equal to the value of solution (y, z) (26). This proves optimality of
the latter to the LPrelaxation of (8) (10). We denote the optimal value of
this LPrelaxation by LP(Im) (see (29)).
Denote by 0 PT(I) the optimum value of ISP for I. Let us show by induction
on Tn that OPT(Im) = Tn. It is clear that OPT(Id = 1, since exactly one
element (column or row) is needed to stab the only rectangle of the instance
I 1. Assume now that OPT(Im1) = m  1 is proven. Consider an instance
Im. It is easy to see that OPT(Im) ::; Tn (assign, for example, multiplicity
1 to each row). We claim that OPT(Im) 2: m. Indeed, take an optimal
solution to ISP for I m , (y, z), and consider the following two cases. First,
assume Zm = O. This implies that we have to stab all the Tn nonoverlapping
rectangles on row Tn with columns of the grid, which would require exactly
Tn columns and would imply OPT(Im) 2: Tn.
Second, assume Zm = 1. In this case all rectangles intersecting row Tn are
stabbed by the row. Observe that the rectangles that are not yet stabbed
can be considered separately and constitute instance I m 1. By the induction
hypothesis we know that one has to select at least m  1 elements to stab
all the rectangles in I m 1, and therefore OPT(Im) 2: 1 + Tn  1 = Tn. So,
OPT(Im) has to be equal to Tn.
We use Lemma 6.3 given in the appendix to prove that the ratio
Tn
T n  P ( ~ + m ~ l + ... + P ~ l )
approaches e ~ l when Tn increases. This establishes our tightness result.
0
As mentioned in the introduction, Theorems 4.1 and 4.4 imply that it is
unlikely that a better ratio for ISP!Xl can be achieved using formulation (8)
(10). Another example of a formulation with an integrality gap that equals
e ~ l is described in Hoogeveen et al. [4].
5
Conclusion
We discussed two variants of the weighted rectangle stabbing problem. For
one problem, called WRSPq, an approximation algorithm ROUND is pro
posed that achieves a ratio of q ~ l . For the other problem, called ISPk, an ap
proximation algorithm STAB is proposed that achieves a ratio of l  ( l ~ l / k ) k .
20
Page 22
Each of these algorithms is based on rounding the LPrelaxation of a straight
forward integer programming formulation. ROUND is a generalization of an
algorithm proposed in Gaur et al. [2], and it is shown that the ratio proved
equals the integrality gap. STAB considers different ways of rounding the
LPrelaxation, and outputs the best solution found in this way; again, it is
shown that the ratio proved equals the integrality gap when k = 2 and when
k = 00.
6 Appendix.
We restate Lemma 4.3, before giving its proof.
Lemma 4.3 Given are real numbers 1 2:: 6.1 2:: 6.2 2:: ... 2:: 6.m 2:: 0, a positive
real number Y, an integer p 2:: 2 and an integer 0 ::; l < m. The following
holds:
(30)
where qi = qi(6., Yip) for each i E [0: m] is uniquely defined by the equality:
where we put 6.i = 0, if i > m.
Proof. It is enough to prove this lemma for l = O. The case of other l < m
can be reduced to the case of l = 0 by changing the index to j = i  land
observing that qj+1 (6., a) = qi (6. I, Yip), where vector 6.1 is obtained by
deleting the first l elements from vector 6.. So we will prove that
The proof consists of two lemmas. In Lemma 6.1 we show that the left
hand side of (30) is upper bounded by the following supremum:
sup
G(J(·))
(32)
f(·) E H
21
Page 23
where
G(f(·)) =
min (f(x) + k(f(x + Yip)  f(x))),
x E lR+
(33)
and the class of functions H is defined as follows:
H = {f() . lR
7 lR I f(·) is continuous, increasing, concave,}
+
(34)
. +
f ( O ) = o , f ( x ) : : ; X + L ~ = l ~ r
.
In Lemma 6.2 we show, that this supremum is upper bounded by the right
hand side of (30), which proves the lemma.
Lemma 6.1.
mm (i + p. q i ( ~ ' Yip)) ::;
i=O, ... ,m
sup
f(·) E H
G(f()),
where G(f(·)) and H are defined in (33) and (34).
Proof. To establish this, it is sufficient to exhibit a particular function
j() E H, such that:
G(}(·)) = . min (i + p. q i ( ~ ' YIp)).
2=O, ... ,m
(35)
Then, the supremum of G (f (. )) over all the possible f (.) E H is clearly larger
or equal to G(}()).
Before we describe the function j (.), let us define an auxiliary function F(·) :
lR+ 7 lR+ as follows:
lqJ
F(q) :L(1 ~ r ) + (q  lqJ) (1 
r=l
where we set ~ r = 0, Vr 2: m + l.
Observe that F(·) is
 continuous,
 increasing, since ~ r < 1, and therefore (1 
 convex, since the coefficients ~ r are nonincreasing with increasing r, and
therefore the coefficients (1 
~ l q J + d ,
~ r ) > 0, Vr = 1, ... ,00,
(36)
~ r ) are nondecreasing with increasing r,
22
Page 24
 F(O) = 0,
 F(q) :2: (q 
~ ~ = 1 6 r ) ' Vq E ffi.+) since F(q) can be also represented as:
l qJ
F(q) = q  (2: 6 r + (q  lqJ )6lqJ+l),
r=l
and obviously ( ~ ~ ~ 1 6 r + (q  lqJ)6lqJ+d :::; ~ ~ = 1 6 T l Vq E ffi.+,
 F(q) is linear on each of the intervals [i, i + 1], i = 0, ... , m  1, and on
[m, +(0).
We are now ready to present } (.) : ffi.+ + ffi.+. We define:
(since F (.) is increasing, F1 (.) exists.)
We claim that }() E H. Indeed, }U has the following properties:
 } ( .) : ffi.+ + ffi.+ since F ( .) : ffi.+ + ffi.+;
 } (.) is continuous, increasing, concave, since F (.) is continuous, increasing,
convex;
 }(O) = 0, since F(O) = 0;
 }(x) :::; x + ~ ~ = l 6 r , Vx E ffi.+. This can be obtained from F(q) :2: (q 
~ ~ = 1 6 r ) ' Vq E ffi.+, using: F(q) = x, q = }(x).
This proves that }U E H.
To prove the lemma it remains to show that
G (} ( . )) = . min (i + pqi ( 6, Yip)) .
~ = O , ... ,m
Comparing the definition of qi(6, Yip) (see 31) and F(·) (36), observe that
for each i E [0 : m] qi satisfies:
F(i + qi)  F(i) = Yip·
(37)
Thus, qi = Fl(F(i) + Yip)  i. Setting: Xi = F(i), Vi = 0, ... , m, we find
that i = F1(Xi) and qi = F1(Xi + Yip))  F1(Xi). Replacing F1U by
}(.), we obtain:
qi = }(Xi + Yip))  }(Xi), Vi = 0, ... , m.
23
Page 25
Using this together with i = FI(Xi) = j(Xi), we can rewrite:
. min (i + pQi(6, Yip)) =
2=O, ... ,m
Xi = jl(i)
min (j(Xi) + p(}(Xi + Yip)  j(Xi))) (38)
, ... ,
i = ° rn
N ow we need to show that the latter expression is equal to:
c(}()) = min (j(x) + p(}(x + Yip)  j(x)))
X E IR+
(39)
We do this by showing that the function j(x) + p(}(x + Yip)  j(x)) is
continuous and concave in each of the intervals [Xi, Xi+l], 'Vi = 0, ... , m  1,
and is increasing in [xm' +00). Therefore the minimum can be achieved only
at one of the endpoints xo, Xl, ... , Xm.
Indeed, consider function j(x) +p(j(x + Ylp)j(x)) in [Xi, Xi+l] for some
i E [0: m1]. It can also be written as pj(x + Yip)  (p l)j(x). We know
that j(x+ Yip) is concave on [Xi, Xi+l], since it is concave everywhere in IR+.
Furthermore, j(x) is linear on each [Xi,Xi+I],i E [0 : ml], since F(·) is
linear on [i, i + 1], i E [0 : m1]. Obviously, a concave function minus a
linear function is again concave.
Now we show that pj(x + Yip)  (p  l)j(x) is increasing in [xm, +00).
Since j (x) = FI ( .) is increasing and linear in [xm' +00 ), the growth rate of
j(x) is the same as the growth rate of j(x + Yip) in [xm' +00), and thus the
growth rate of pj(x+ Yip)  (p1)j(x) is negative. We have proved that the
minimum in (39) is always achieved at one of the points xo, Xl, ... , Xm , and
therefore (39) is equal to (38). This completes the proof of Lemma 6.1. D
Lemma 6.2.
where
1
sup C(j(·)) < 1 _ (1 _ 1/k)k C,
f(·)EH
m
r=l
C (j ()) =
min (j (x) + k (j ( X + Yip)  f (x) ) )
xEIR+
and set of functions H (via notation C) is
H = {f(') . IR
> IR I f() is continuous, increasing, concave, }
+ f(O) = 0, f(x) :::; X + C  Y
. +
.
24
Page 26
Proof. We will prove several claims and subclaims.
Claim 1.
sup G(f(·)) =
f(')EH
where for each 9 E ffi.+ function fg () is defined as follows:
 fg(j . Yip) = g(l  (1  1/p)j), Vj E ° U N,
 fg (x) is continuous in [0, +(0) and linear in each [(j 1)· Yip, j . Yip], j E N.
Notice that f9 (.) is completely defined by the above characterization. An
example of this function is shown in Figure 7.
sup
g,
g:f9()EH
. ..
Figure 7: Illustration of function fg (x) for k = 5, 9 = 1.
To prove this claim it is enough to show that for any f (.) E H there exists a
function f9(.) E H, with j) ~ 0, such that
G(f(·)) = G(f9()) = j).
To show that, we prove 2 subsidiary claims.
Claim 1.1. For any 9 ~ 0:
G(fg(·)) =
mm
x E ffi.+
(P(x) + p(P(x+Ylp)P(x))) = g.
Indeed, by construction fg(x) is linear in each of the intervals [(j 1)· Yip, j.
Yip], j E N. This implies that function (fg(x) + p(fg(x+Ylp)fg(x)))
is linear in each of these intervals as well. Therefore the minimum over
all x ~ ° is achieved in one of the endpoints 0, Yip, 2Ylp, .... Consider
(fg(x) + p(fg(x+Ylp)fg(x))) at the point x = jYlp, for some j EN U 0:
P(j . Yip) + p(P((j + 1) . Yip)  P(j . YIp))·
25
Page 27
U sing the definition of f9 (.) we can rewrite it as follows:
g(l  (1  1/p)j) + p(g(l  (1  1/p)j+1)  g(l  (1  1/p)j))
With simple computations one can verify that the last expression is equal to
g. This proves Claim 1.1.
Claim 1.2. For any f(·) E H it holds that f9(.) E H, where § = G(J(·)).
Clearly, f9(X) is concave. To prove that f9(X) ::; x + C  Y, \Ix E ffi.+, it
is sufficient to show that f9(X) ::; f(x), since fe) E H means e.g. f(x) ::;
x + C  Y, \I x E ffi.+.
So, let us establish that j9(x) ::; f(x), \Ix E ffi.+. Recall that f9(X) is linear
in each of the intervals [(j  1) . Ylp,j' Yip], j EN, and f(x) is concave in
ffi.+. Then it is sufficient to show that:
f9(X) ::; f(x), \Ix = j . Yip, j E 0 u N.
We use mathematical induction on j. For j = 0, f9(0) = f(O) = 0 and the
inequality trivially holds. Suppose, for j  1 we have proven: f9 ((j  1) .
Yip) ::; f((j 1) . Yip), and let us show that f9(j. Yip) ::; f(j . YIp)·
Observe, that f9 (.) can be represented in a recursive way as follows:
f9 (j . Yip) = § I p + f9 ( (j  1) . Yip) (1  1 I p ) . ( 40)
Since § = G(J(·)) we know:
§ ::; f ((j 1) . Yip) + p(J (j . Yip)  f ((j 1) . YIp))·
Rearranging the expression, we obtain:
f(j . Yip) ~ §Ip + f((j 1) . Yip) (1  lip)·
By the induction hypothesis and (40) we can bound the righthand side as:
§Ip+ f((j1)· Yip) (ll/p) ~ §Ip+ f9((j1). Yip) (ll/p) = f9(j. YIp).
This proves Claim 1.2.
These 2 claims imply that for any f() E H, there exists f9() E H, with
§ ~
G(J(·)) = G(J9()) = §.
0, such that
26
Page 28
This implies Claim 1.
Claim 2.
sup
g: fgUEH
1
9 <
 1  (1  l/p)p
C.
Indeed, fg() E H implies: fg(x) ~
for x = Y. From this, using the definition of f9U, obtain:
J9(Y) = J9(p * Yip)  g(l (1  l/p)P) ~ Y + C  Y = C,
x+C  Y, for all x E 1R.+, and in particular,
from the last inequality:
1
9 ~ (1  (1  l/p)p) C,
which proves Claim 2 and establishes Lemma 6.2.
Lemma 6.3. Let P(m) E N be defined as follows:
1
 +
m
1 1
+ ... + ()
m1
< 1 and
Pm+1
1
+
m
1 1 1
+ ... + ()
m1
+>1.
P(m)Pm +1
Then,
. m
hm
m>oo m  P m m + ml + ... + P(m)+l)
( ) ( 1 1 1
e1
e
o
o
( 41)
( 42)
Proof. Let us first find limm>00 P(m)/m. Observe that the following in
equalities hold:
1
 +
m
1 1 1
m+l 1
P(m)+l x
m + 1
(
P m) + 1
+ ... + () >  dx = In ,
m  1
P m + 1 
1
 +
m
111m
+ ... + () <
m  1
1
 dx = In ()
P(m)l X
m
,
P m 
Pm  1
(the equalities follow from J: l/x dx = Inb/a.) Then (41) and (42) imply
m+1
1 2': In P() ,
m +1
m
1 ~ In P(m) _ 1
27
Page 29
From this we have:
m+1
 1 S P(m) S  + 1.
e
m
e
Dividing by m:
1 + 11m
  1 m <  <  + 1 m.
e

I
P(m)
m
1
I
 e
Now we see, that limm+oo P(m)/m = lie.
Let us now find l i m m    + o o ( ~ + m ~ l + ... + P ( ~ ) + l ) ' From (41) and (42) we
have:
1 1 1
1<++ ... +
P(m)  m m  1
1
<1
P(m) + 1 
Since we already know that limm+oo P(m) = 00, we have:
111
lim ( +
m+oo m
+ ... + P () ) = 1.
'IT/,  1 m + 1
Now consider:
m
1
m  P ( m ) ( ~ + m ~ l + ... + P ( ~ ) + l )
1
P(m) ( 1
 :;:;:;: ~ + m1 + ... + P(m)+l
1 1 ) .
Using limm+oo P ~ ) = lie and limm+oo ( ~ + m ~ l + ... + P ( ~ ) + 1 ) = 1 we
have:
. 1
11m
m+oo 1 _ P(m) ( ~ + _1_ +
m
+
1 )
m
m1 ... P(m)+l
1
e
e 1'
1  lie
which establishes the lemma.
o
References
[1] Ahuja, R.K., T.L. Magnanti, and J.B. Orlin [1993], Network Flows: The
ory, Algorithms, and Applications, PrenticeHall.
28
Page 30
[2] Gaur, D.R, T. Ibaraki, and R Krishnamurti [2002]' Constant ratio ap
proximation algorithms for the rectangle stabbing problem and the recti
linear partitioning problem, Journal of Algorithms 43, 138152.
[3] Hassin, R, and N. Megiddo [1991]' Approximation algorithm for hitting
objects with straight lines, Discrete Applied Mathematics, 30, 2942.
[4] Hoogeveen H., M. Skutella, and G.J. Woeginger [2003], Preemptive
scheduling with rejection, Mathematical Programming 94, 361374.
[5] Kovaleva, S. [2003], Approximation of Geometric Set Packing and Hit
ting Set Problems, Ph.D. thesis of Maastricht University, Maastricht, The
Netherlands.
[6] Kovaleva, S., and F.C.R Spieksma (2001), Approximation of a geometric
set covering problem, in: Proceedings of the 12th Annual International
Symposium on Algorithms and Computation (ISAAC'01), Lecture Notes
in Computer Science 2223, 49350l.
[7] Kovaleva, S., and F.C.R Spieksma [2002]' Primaldual approximation
algorithms for a packingcovering pair of problems, RAIROOperations
Research 36, 5372.
[8] Schrijver, A. [1986], Theory of Linear and Integer Programming, Wiley.
[9] Vazirani, V.V. [2003], Approximation Algorithms, Springer.
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