Interval packing problem
Multicommodity demand flow
in a line
Jian Li
Sep. 2005
Interval packing problem
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Input:
1) A set I of n intervals, I1, I2, ..., In, with each Ii = [li, ri] and
having a THICKNESS (or HEIGHT) hi, such that li and ri
are all integers in {1,2, ..., n} and hi is an integer > 0
2) n-1 CAPACITY constraints ci, with each ci located at the
point pi = (i + 1/2, 0) on the x-axis, for every i = 1,2, ..., n-1,
such that each ci is an integer > 0. We may call each pi a
thickness constrained point
Output:
The MAXIMUM NUMBER of intervals in I such that at each
thickness constrained point pi = (i + 1/2, 0), the sum of
thickness values of all output intervals that contain pi is
less then ci.
In
fact, it is a special
case of the
multicommodity
demand flow problem
Multicommodity demand flow
problem
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T=(V,E,u) be a capacitated network, where each
edge capacity ue is an integer.
A number of demand vertices pairs (si,ti) with
demand value di and profit wi
GOAL: simultaneously route the demand flow
without violating the capacity constraints and
maximize the profit.
NOTE: di must be fully satisfied to obtain a profit
wi
Interval packing problem can be
modeled as multicommodity flow
problem in a line.
LP formulation
max i wixi
i:e2 path(i)dixi · ue
xi=0,1
Some special case
All capacities are one.
 All demand are one.
 The problem is exactly to compute a
independent set in a interval graph, which is
in P.
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Algorithm:
 Sort intervals according their left endpoints.
 Use the reverse order to greedy construct
the independent set.
(reverse perfect elimination order)
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Some special case.
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the problem can be formulated as the following
min-cost flow problems.
There are n vertice, namely v1,v2,…..,vn, and
arcs from vi to v(i+1) with capacity ui and cost
zero and arc from vn to v1 with capacity +infinity
and cost zero.
For every interval Ii, there is a arc from v(ri) to
v(li) with capacity di and cost -1/di(we call it
back-arc).
Capacity=ui, cost=-1/di
Some special case
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However, the min cost circulation may not give the
correct answer, because in the original problem
only in case that the back-arc are
saturated(flow=capacity) can incur a cost 1. So we
can treat it as the relax of the original problem. In
fact it solve the relaxed LP of the original problem.
Moreover, if all di is one, then this special case can
be solved optimally in polytime.
Some special case
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if the number of intervals passing pi is
bounded( less than a constant C), then we
can solve the problem optimally by
standard dynamic programming technique.
Small and Large demand
Bottleneck capacity b(f): the smallest
capacity edge on Path(f)
 -large demand: d(f)¸  b(f)
 Otherwise ,-small
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Bounding large demand
Lemma:
 If dmax· umin (a wild assumption)in a feasible
solution, the number of -large demands
that cross any edge is at most 2(1/2)
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So, we can use DP to solve the case with
only large task optimally.
Bounding large demand
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Proof of the lemma:
Fix a feasible solution S, consider an edge e.
Let Se be the set of all -large demand cross e.
Se=Sl[ Sr, where Sl is the set where its bottleneck
is to the left of e(including e), and Sr otherwise.
Sr
Sl
e
Bounding large demand
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Proof Cont.
We can show Sl<=1/2 (similar for Sr)
Let A be the set of bottleneck edge for Sl. Let e’ be
the rightmost one in A. Suppose e’ is the
bottleneck edge for demand Ij, thus dj> c(e’). But
all demands in Sl pass through e’(since e’ is
rightmost one). But each demand Ik is -large. dk¸
 bk¸  cmin.
cminSl· c(e’), thus
Sl· |c(e’)/cmin| · |di/2 cmin| · 1/2
\QED
Suppose we have a -approximation algorithm for
the case where only small demands exist.
 Then we choose the larger one from the optimal
solution for large demand and approximation
solution for small demand
 We get 2 approximation for original problem
OPT<=OPTlarge+OPTsmall<=SOLlarge+ SOLsmall
<=2max{SOLlarge+ SOLsmall}
<= 2 max{SOLlarge+ SOLsmall}
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Dealing with Small demands
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Randomized rounding:
G.Calineascu et IPCO2002 for uniform
capacity. (1+e)-approx.
A.Chakrabarti et APPROX2002 for general
case. O(1)-approx.
Packing integer program
C.Chekuri et ICALP2003 for general case.
(1+e)-approx.