Approximation Algorithms

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Approximation Algorithms
EPIT 2007
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Independent Set
Instance G(V,E), k? N Question Is there an
independent set of size k in G (i.e. a subset V?
V such that no two vertices in V are joined by
an edge) ? This problem is NP-Complete Instance
G(V,E) Question Find an independent set of
maximal cardinality NP-hard
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Definition of Approximation Algorithms
Find a good solution in polynomial-time What is
a good solution? Let S be a solution given by a
heuristic Let S be an optimal solution The aim
S/S1
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Another problem vertex cover
Instance G(V,E), k ?N Question Find a set
cover of edges with minimal size (i.e. a subset
V ? V such that, for each edge u,v ? E, at
least one of u and v belongs to V).
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These two problems seem to be equivalent  
independent set vertex cover
and vertex cover
independent set Max independent set Min
vertex cover An existence of an independent set
of size k ltgt an
existence of the vertex cover of size n-k.
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In fact these two problems are very different in
term of approximability there exists an
algorithm for the vertex cover which gives a
solution S to a factor two of a optimal solution
S at most gt 2 ? S/S No constant ratio exists
for the independent set
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Definition
An algorithm A is ?(n)-approximated If Max
(S/S,S/S) ? ?(n) For all instances of size n
of the problem where S is the solution given by A
and S an optimal solution.
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FPTAS and PTAS
Q ? PTAS ??gt0, ?A?such that A? is
(1?)-approximated. Moreover, if the complexity
of A? is polynomial in 1/?, then Q ? FPTAS.
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Some problems of PTAS
Cover of points by minimum number of circles of
radius r Knapsack problem
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APX and no-APX
Q ? APX ?? ? R, ?A such that A is a
?-approximation algorithm Q ? no-APX ?? ? R,
there exists no A such that A is ?-approximation
algorithm
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Some problems APX and no-APX
Vertex cover of a graph is APX Travelling
salesman problem is APX Independent set of
maximum cardinality is no-APX Vertices coloring
is no-APX
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The world if N ? NP
No-APX
APX
PTAS
FPTAS
P
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An absolute approximation Algorithm
The edge coloring of a graph It exists a
4/3-approximated algorithm Vizings theorem
It is always possible to color the edges with
?(G)1 colors
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An absolute approximation Algorithm
This result is the best possible Indeed the
problem to know if a graph with ?(G)3 can be
colored with 3 colors is NP-complete. It is easy
to see that an algorithm ?-approximated with
?lt4/3 permits to solve this problem.
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An absolute approximation Algorithm
Indeed if the solution is clt4 then c 3, and
the graph can be colored with 3 colors. If the
solution is c?4, since c/clt4/3 we have cgt3 and
the graph cannot be colored with 3 colors.
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The problem of the k-center
Instance Let V be a set of n sites, D be a
matrix where dij is the distance between site i
and site j, and let k?N. Problem Find S ? V,
Sk which minimizes maxi?V(dist(i,S)) with
dist(i,S)minj?S(dij)
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Remark
There exists no ?-approximated algorithm with ?
lt2. Proof If such an algorithm existed, it
could be used to solve the problem of the
existence of a dominating set of size k.
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Some definitions
We call Gd(V,Ed) the graph where i,j? Ed if
and only if d ? dij . Let G(V,E), we call
G2(V,E) the graph such that i,j? E if and
only if either i,j? E or ?k such that i,k?
E and k,j? E.
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Here is a 2-approximated algorithm Sort the
dij d1 lt d2 lt lt dm , i 0
Repeat i i1 d di
Construct Gd2 Find
a maximal stable S of Gd2 until k ? S
Take S as solution
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• Remark
• Let i0 be the value of the last i in the loop.
  Then all the sites are at distance at most 2di0
  from S.
• The optimal solution is at least di0.
• Indeed Gd2 with ddi0-1, has a stable of
  cardinality at least k1.

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Conclusion Thus the algorithm is a
2-approximated. That is the best possible
result.
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Most of the scheduling problems are NP-Hard
Take the following simple problem Instance
Independent tasks (no precedence constraint),
each task ti has a processing time pi Question
Scheduling these tasks on two processors while
minimizing the length of the schedule
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This problem is NP-Hard (partition is a
particular case). But there exists a FPTAS
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A intermediate problem
Let be the sum problem a list E of n integers,
an integer t. The goal is to find a sublist E ?
E which the sum of the integers in E is maximum
but less than t (Find E? E, such that max(?
i?E i) ?t)
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A pseudo-polynomial-time algorithm
L0 0 L1 0, e1 Li Li-1 ? x ?y? Li-1,
xyei, x ?t The solution is expressed as y
max (Ln) The complexity order of this algorithm
is in O (n min(t,2n))
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How can we reduce the size of the lists?
The idea is to filter the list Example.
Suppose that the two integers 45 and 47 are in a
list Li. The sum which we would able to reach
from 45 will be closed to their from 47. Thus 45
will represent 47
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How do we reduce the size of the lists ?
Filtering (L, ?) consists in suppressing all the
integers y in L such that there exists an
integer z in L with y ? z ?(1- ?)y Example L
0, 1, 2, 4, 5, 11, 12, 20, 21, 23, 24, 25, 26
and ?0.2
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How do we reduce the size of the lists ?
Filtering (L, ?) consists in suppressing all the
integers y in L such that there exists an
integer z in L with y ? z ?(1- ?)y Example L
0, 1, 2, 4, 5, 11, 12, 20, 21, 23, 24, 25, 26
and ?0.2. Then Filtering(L,?) 0,1,2, 4, 11,
20, 26
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How do we reducethe size of the lists ?
We are given the pseudo-polynomial algorithm,
and in each step we filter with ??/n. Let Li
be the intermediate lists. Then the approximated
solution is y max(Ln)
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The sizes of the lists are small
Li 0, z1 , z2 , zm1 we have by
construction zi/ zi-1 gt 1/(1-?) and t ? zm1
and z1 ?1
Thus t ? zm1/z1 ?i2..m1 zi/ zi-1 ?
(1/(1- ?) )m
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The sizes of the lists are small
Thus t ? (1/(1- ?) )m log (t) ? - m
log (1- ?)
Finally m lt n log(t)/ ?
(because - log(1-x)gtx)
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The sizes of the lists are small
1. m lt n log(t)/ ? 2. the size of the lists is
in O(m)
The algorithm becomes polynomial.
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The error is small
We can prove by induction that ?y?Li ,
?z?Li such that y? z ?(1-?)i y
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The error is small
We can prove by induction that ?y?Li ,
?z?Li such that y? z ?(1-?)i y
This result is true for in and yy
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The error is small
We can prove by induction that ?y?Li ,
?z?Li such that y? z ?(1-?)i y
This result is true for in and yy
? z?Ln z ? (1-?)n y
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The error is small
? z?Ln z ? (1-?)n y and y ? z
Conclusion y / y ? (1-?/n)n ? 1-?
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Back to our problem
Let B be the sum of the processing time of
all the tasks. P1 P2
?
B/2
The solution S is B/2 ? If pi becomes ei and
tB/2 the solution S is B/2- ?
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Back to our problem
?
B/2
The solution S is B/2 ?, S is B/2? with ??? If
pi becomes ei and tB/2 the solution S is B/2-
? and the best solution S is B/2-?
It suffices to note that S/SltS/S to show that
our problem is FPTAS
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A non-APX problem
Instance Some tasks with the same processing
times and some incompatibility constraints.
Question Find a optimal schedule satisfing the
incompatibility constraints. Let G be the
conflict graph where vertices represent tasks a
scheduling vertex coloring. This problem is
not APX
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An APX problem without PTAS
Scheduling on a given number of processors
Instance Let G(V,A) be a d.a.g. with the
chronological constraints and let m be the number
of processors Question Minimize the length of
the schedule
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No ?-approximated algorithm if ?lt4/3
We are going to show that Even if the graph is
bipartite and the duration of all the tasks is
equal to 1, knowing whether the graph can be
scheduled in time 3 is an NP-complete problem The
proof of this result suffices to conclude
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The proof
The problem of well-balanced independent set in
bipartite graphs is NP-complete
Instance G(V1?V2,E) a bipartite graph
Question does there exist an independent set S
in G such that SV1 and V1V2 and V1?S
V2?S?
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W with Wn/2
X with Xn
Y with Yn
T with Tn/2
The edges of G
Complete Graph
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X with Xn
W with Wn/2
Y with Yn
T with Tn/2
Scheduling this DAG in three steps on n
processors is equivalent to showing the existence
of a well-balanced independent set in the graph G.
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X with Xn
W with Wn/2
mn
T with Tn/2
Y with Yn
Scheduling in 3 steps no idle time
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X with Xn
W with Wn/2
mn
T with Tn/2
Y with Yn
Scheduling in 3 steps no idle time
Step 1 W and X1 ( X1? X, X1n/2) must be
executed.
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X with Xn
W with Wn/2
mn
T with Tn/2
Y with Yn
Scheduling in 3 steps no idle time
Step 1 W and X1 ( X1? X, X1n/2) must be
executed.
Step 3 T and Y1 ( Y1? Y, Y1n/2) must be
executed.
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X with Xn
W with Wn/2
mn
T with Tn/2
Y with Yn
Scheduling in 3 steps no idle time
Step 1 W and X1 ( X1? X, X1n/2) must be
executed.
Step 2 X2 and Y2 ( X2Y2n/2 ) must be
executed.
Step 3 T and Y1 ( Y1? Y, Y1n/2) must be
executed.
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X with Xn
W with Wn/2
mn
T with Tn/2
Y with Yn
Scheduling in 3 steps no idle time
Step 2 X2 and Y2 ( X2Y2n/2 ) must be
executed.
X2 ?Y2 is a well balanced independent set in the
graph G
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2-approximated list scheduling
Grahams Theorem all lists scheduling is a
2-approximation for this problem. Proof We can
construct by induction a path in the d.a.g. such
that at all the times, either a task in the path
is executing or all the processors are working.
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2-approximated list scheduling
construction of the path Take as t1 one of the
tasks finishing at the end of the scheduling, and
take as tk any predecessor of tk-1 which
finishing last among all its predecessors. The
construction is achieved when we arrive at a
source (task with no predecessor)
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2-approximated list scheduling
The sum of the times when all the processors are
working is smaller than the duration of the
optimal scheduling and the rest of the time is
smaller than the longer of the path and so the
duration of the optimal scheduling. This
concludes the proof .
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Some remarks
We have seen that the constraints of
incompatibility yield a very difficult problem
(non-APX) but the chronologic constraint yield an
easier problem (often APX) but even polynomial
if the number of processors is not bounded .
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Other constraints
Communication delays It is reasonable to assume
that two tasks executing on two different
processors need to comunicate and this takes a
certain delay. As a first approximation, we
suppose that this delay is always equal to 1 like
the duration of each task.
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Communication delay UET-UCT
When the number of processors is not bounded, the
problem become APX and the duplication techniques
permit to reduce sensitively the complexity
(Colin-Chretienne show that the problem remain
polynomial). Without duplication a lower bound
of 7/6 has been found (Hoogeveen, Lenstra,
Veltman).
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Communication delay UET - UCT
Munier and Co have given a 4/3-approximated
algorithm. Their idea is to solve a linear
problem which gives a solution smaller that the
solution of the original problem. A rounding
technique permit to give a realistic solution.
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Communication delay UET - UCT
cij
i
j
If di is the starting time of i dj ? di 1
cij where cij is equal to 0 only for one
successor task of ti. The problem then amounts to
giving each arc a value cij equal to 0 or 1, with
the constraints that, for each task at most one
incoming (resp. outcoming) arc has the value 0.

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Communication delay UET - UCT
cij
ti
tj
If every cij takes its value in 0,1 the
problem becomes easy the sum of the cij (resp.
cji ) for a fixed i must be greater than the
out-degree (resp. in-degree) minus 1. To find a
realistic solution the integer value of the arc
is 0 only if cij lt 1/2.
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Communication delay UET - UCT
cij
ti
tj
The rounding step distends the path by a factor
at most 4/3. Indeed suppose a path of length
(k1) with x arcs such that cij lt ½ (i.e the cost
of the path in the real solution is greater than
3/2k-1/2x1). After rounding, the cost is 2k-x1.
The ratio is always smaller of 4/3.
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Unbounded vs. bounded number of processors

Technique folding. (each step of the unbounded
scheduling where x tasks are executing is
transformed into ?x/m? steps for the scheduling
with m processors. Without communication a
polynomial problem becomes 2-approximated
algorithm With the uet-uct model a
?-approximated algorithm becomes a
(?1)-approximated algorithm.
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Other ratios have been used for approximation
problems, the differential ratio for
example. Other constraints and other models
there is no small communication time (thin
granularity i.e. more possibility of parallelism)