Katholieke Universiteit Leuven

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• Approximation algorithms for
• rectangle stabbing and interval stabbing problems
• by
• Sofia Kovaleva ORTEC
• Frits Spieksma Katholieke Universiteit Leuven

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The problem

• Given a set of rectangles in the plane that
  satisfy the following two properties
• axis-parallel
• each corner-point has integral coordinates,
• FIND
• a minimum number of horizontal and vertical lines
  that stab each rectangle at least once.

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The problem

• Terminology a line stabs a rectangle when the
  line has a non-empty intersection with the
  rectangle
• Example of an instance

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The problem
A feasible solution
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The problem
Observation only a finite number of lines need
be considered.
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The problem
Observation only a finite number of lines need
be considered. Why?
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The problem
We refer to this problem as RS (Rectangle
Stabbing)
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Variants of problem RS

• Suppose that for each possible line a cost is
  known (WRS).
• Suppose that for each rectangle a requirement is
  known (RSR).
• Define the height of a rectangle as the
  difference between the y-coordinates of the
  upper-left and lower-left corner point.
• We refer to a unit-height rectangle as an
  interval.
• 3. The resulting problem is referred to as IS
  (Interval Stabbing)
• Define the level of an interval as the
  y-coordinate of its lower-left corner point.
• Observe that integrality of corner points implies
  that no horizontal line can stab two intervals
  that have different levels.

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2
1
4
3
5
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Putting RS into perspective (1)

• Notice that stabbing any set of connected figures
  in the plane, using lines of at most two
  different directions, yields problem RS.

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Putting RS into perspective (2)

• Notice that any integer programming instance of
  the form
• Min 1 x
• S.t.
• x (A B)T 1
• x ? 0,1,
• where A and B are each interval matrices (i.e.,
  a 0-1 matrix with the ones in each column
  appearing consecutively), can be transformed into
  an instance of RS.

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Putting RS into perspective (2)

• Example
• 1 1 0 1
• 0 1 1 1
• (y1 y2 y3 z1 z2 z3) 0 0 1 0 1
• 1 1 1 0
• 1 0 1 0
• 0 0 0 1

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Putting RS into perspective (2)

• Example
• 1 1 0 1
• 0 1 1 1
• (y1 y2 y3 z1 z2 z3) 0 0 1 0 1
• 1 1 1 0
• 1 0 1 0
• 0 0 0 1

z1
z2
z3
y3
y2
y1
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Putting RS into perspective (2)

• Example
• 1 1 0 1
• 0 1 1 1
• (y1 y2 y3 z1 z2 z3) 0 0 1 0 1
• 1 1 1 0
• 1 0 1 0
• 0 0 0 1

z1
z2
z3
y3
y2
y1
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Putting RS into perspective (2)

• Example
• 1 1 0 1
• 0 1 1 1
• (y1 y2 y3 z1 z2 z3) 0 0 1 0 1
• 1 1 1 0
• 1 0 1 0
• 0 0 0 1

z1
z2
z3
y3
y2
y1
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Putting RS into perspective (2)

• Example
• 1 1 0 1
• 0 1 1 1
• (y1 y2 y3 z1 z2 z3) 0 0 1 0 1
• 1 1 1 0
• 1 0 1 0
• 0 0 0 1

z1
z2
z3
y3
y2
y1
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Applications of RS and IS

• In parallel computing, the so-called rectilinear
  partitioning problem is of interest. RS is
  subproblem (Gaur, Ibaraki, Krishnamurty, 2002).
• In maintenance
• M1
• M2
• M3

time
Decision when to inspect (select a vertical
line) and/or for which machines to outsource
the inspection (select a horizontal line)
(Hassin and Megiddo, 1991).
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Known results

• ) There is a 2-approximation algorithm for RS
  (Gaur et al.).
• ) There is a 2 (1/K) approximation algorithm
  for IS (Hassin and Megiddo).
• ) IS is APX-hard, even if no more than two
  intervals are on the same level (Kovaleva, S.).

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Our results

• Consider WRSR, where there is a requirement ri
  given for each rectangle i such that ri q
• Result There exists a (q1)/q-approximation
  algorithm for WRSR.
• Consider WIS
• Result There exists a e/(e-1)-approximation
  algorithm for WIS.
• Consider IS, and let k be the maximum number of
  intervals on the same level
• Result There exists an 1/(1-(1-1/k)k)-approximati
  on algorithm for IS.
• This yields a 4/3 approximation for
  the(difficult) case k2.
• Both for k2 and for k infinity, the
  integrality gap equals the corresponding ratios.

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A result for IS

• An IP-formulation for IS

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A result for IS

• An IP-formulation for IS

z1
z2
z3
y2
y1
y5
y4
y3
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A result for IS

• An IP-formulation for IS

z1
z2
z3
y2
y1
y5
y4
y3
Min y1 y2 y3 y4 y5 z1 z2
z3 s.t. y1 z1 1 y2
y3 z1 1 y1 y2
z2 1 y4
y5 z2 1 y1 y2 y3 y4
y5 z3 1 all variables in
0,1
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A result for IS

• An IP-formulation for IS

z1
z2
z3
y2
y1
y5
y4
y3
Min y1 y2 y3 y4 y5 z1 z2
z3 s.t. y1 z1 1 y2
y3 z1 1 y1 y2
z2 1 y4
y5 z2 1 y1 y2 y3 y4
y5 z3 1 all variables in
0,1
Observation If the z-values are given the
resulting problem becomes easy!
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An algorithm for IS called STAB

• Step 1 Solve the LP-relaxation of the given
  formulation

3/4
1/2
1/4
1/4
1/2

• Step 2 Construct 4 solutions (in general the
  number of z-variables 1) as follows

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STAB in action

• Step 2i) Set all z-variables to 0 solve the
  resulting instance. We find solution no. 0

1
1
1

• Resulting value 3

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STAB in action

• Step 2ii) Set the highest z-variable in the
  LP-relaxation to 1, and all others to 0 solve
  the resulting instance. We find solution no. 1

1
1
1

• Resulting value 3

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STAB in action

• Step 2iii) Set the highest two z-variables in the
  LP-relaxation to 1, and all others to 0 solve
  the resulting instance. We find solution no. 2

1
1
1

• Resulting value 3

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STAB in action

• Step 2iv) Set all z-variables in the
  LP-relaxation to 1 solve the resulting instance.
  We find solution no. 4

1
1
1

• Resulting value 3
• Step 3 output the best solution from Step 2.

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A Result

• Claim STAB is an e/(e-1) approximation
  algorithm for IS
• Argument
• Assume wlog that the zLP-variables are ordered,
  i.e.,
• 1 zLP1 zLPm
• Two inequalities imply the claim
• Value(ySTAB, zSTAB) minj (j
  value(yLP)/(1-zLPj1))
• and
• minj (j value(yLP)/(1-zLPj1)) e/(e-1)
  value(yLP, zLP)

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A Result (cntd)

• STAB ensures that
• value(ySTAB, zSTAB) minj (value(yj, zj)) ,
• where value(yj, zj) represents the value of the
  j-th candidate solution (j0,1,,m).
• Let value(yj, zj) value(yj) value(zj)
• Of course, for each j value(zj) j.
• Let us now argue that, for each j,
• value(yj) value(yLP)/(1-zLPj1).
• We do this by arguing that the RHS above is a
  feasible solution to the LP with given zj values
  (and recall that yj is optimal wrt zj ).

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A Result (cntd)

• So is it true that for each interval i
• value(yLP)/(1-zLPj1) 1 zj?(i) ?

i
S(i) the set of y-variables stabbing interval i
Or, equivalently Sj e S(i) yjLP /(1-zLPj1) 1
zj?(i)?
Case 1 zj?(i) 1
Case 2 zj?(i) 0. It follows that ?(i) j1,
or, by the ordering of the z-variables zLP?(i)
zLPj1. Thus Sj e S(i) yjLP /(1-zLPj1) (1
zj?(i))/(1-zLPj1) 1.
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A Result

• It follows that
• Value(ySTAB, zSTAB) minj (j
  value(yLP)/(1-zLPj1))
• Together with
• minj (j value(yLP)/(1-zLPj1)) e/(e-1)
  value(yLP, zLP)
• the claim follows.

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Worst-case instance for IS with k2
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Question

• Do we actually need to solve an LP?