1. Approximation Algorithms
EPIT 2007
Approximation Algorithms

2. Independent Set Instance: G=(V,E), k N
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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

3. Definition of Approximation Algorithms
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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/S*≈1

4. Another problem: vertex cover
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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’).

5. These two problems seem to be equivalent
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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
<=>
an existence of the vertex cover of size n-k.

6. No constant ratio exists for the independent set
EPIT 2007
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 => 2  S/S*
No constant ratio exists for the independent set

7. Definition An algorithm A is (n)-approximated
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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.

8. >0, Asuch that A is (1+)-approximated.
EPIT 2007
FPTAS and PTAS
Q  PTAS:
>0, Asuch that A is (1+)-approximated.
Moreover, if the complexity of A is polynomial in 1/, then Q  FPTAS.

9. Cover of points by minimum number of circles of radius r
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Some problems of PTAS
Cover of points by minimum number of circles of radius r
Knapsack problem

10. EPIT 2007
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

11. Some problems APX and no-APX
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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

12. EPIT 2007
The world if N  NP
No-APX
APX
PTAS
FPTAS P

13. An absolute approximation Algorithm
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An absolute approximation Algorithm
The edge coloring of a graph:
It exists a 4/3-approximated algorithm
Vizing’s theorem :
It is always possible to color the edges with (G)+1 colors

14. An absolute approximation Algorithm
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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 <4/3 permits
to solve this problem.

15. An absolute approximation Algorithm
EPIT 2007
An absolute approximation Algorithm
Indeed if the solution is c<4 then c* ≤3,
and the graph can be colored with 3 colors.
If the solution is c4,
since c/c*<4/3 we have c*>3 and the graph cannot be colored with 3 colors.

16. The problem of the k-center
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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, |S|=k which minimizes maxiV(dist(i,S))
with dist(i,S)=minjS(dij)

17. Remark There exists no -approximated algorithm with  <2.
EPIT 2007
Remark
There exists no -approximated algorithm with  <2.
Proof : If such an algorithm existed,
it could be used to solve the problem
of the existence of a dominating set of size k.

18. EPIT 2007
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.

19. Here is a 2-approximated algorithm:
EPIT 2007
Here is a 2-approximated algorithm:
Sort the dij : d1 < d2 < … < dm ,
i :=0
Repeat
i := i+1 ; d := di
Construct Gd2
Find a maximal stable S of Gd2
until k  |S|
Take S as solution

20. The optimal solution is at least di0.
EPIT 2007
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 d=di0-1, has a stable of cardinality at least k+1.

21. Thus the algorithm is a 2-approximated.
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Conclusion
Thus the algorithm is a 2-approximated.
That is the best possible result.

22. Most of the scheduling problems are NP-Hard
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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

23. This problem is NP-Hard (partition is a particular case).
EPIT 2007
This problem is NP-Hard (partition is a particular case).
But there exists a FPTAS

24. A intermediate problem
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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)

25. A pseudo-polynomial-time algorithm
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A pseudo-polynomial-time algorithm
L0 = {0}
L1 = {0, e1 }
Li = Li-1  {x |y Li-1, x=y+ei, x t}
The solution is expressed as y*= max (Ln)
The complexity order of this algorithm is in
O (n * min(t,2n))

26. How can we reduce the size of the lists?
EPIT 2007
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

27. How do we reduce the size of the lists ?
EPIT 2007
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

28. How do we reduce the size of the lists ?
EPIT 2007
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}

29. How do we reduce the size of the lists ?
EPIT 2007
How do we reduce the size of the lists ?
We are given
the pseudo-polynomial algorithm,
and in each step we filter with =/n.
Let L’i be the intermediate lists.
Then the approximated solution is
y = max(L’n)

30. The sizes of the lists are small
EPIT 2007
The sizes of the lists are small
L’i = {0, z1 , z2 , … zm+1}
we have by construction :
zi/ zi-1 > 1/(1-) and t  zm+1 and z1 1
Thus
t  zm+1/z1 = i=2..m+1 zi/ zi-1  (1/(1- ) )m

31. The sizes of the lists are small
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The sizes of the lists are small
Thus t  (1/(1- ) )m
log (t)  - m * log (1- )
Finally m < n log(t)/ 
(because - log(1-x)>x)

32. The sizes of the lists are small
EPIT 2007
The sizes of the lists are small
1. m < n log(t)/ 
2. the size of the lists is in O(m)
The algorithm becomes polynomial.

33. The error is small We can prove by induction that :
EPIT 2007
The error is small
We can prove by induction that :
yLi , zL’i such that y z (1-)i y

34. The error is small We can prove by induction that :
EPIT 2007
The error is small
We can prove by induction that :
yLi , zL’i such that y z (1-)i y
This result is true for i=n and y=y*

35. The error is small We can prove by induction that :
EPIT 2007
The error is small
We can prove by induction that :
yLi , zL’i such that y z (1-)i y
This result is true for i=n and y=y*
 zL’n z  (1-)n y*

36. The error is small  zL’n z  (1-)n y* and y  z Conclusion:
EPIT 2007
The error is small
 zL’n z  (1-)n y* and y  z
Conclusion:
y / y*  (1-/n)n  1-

37. Let B be the sum of the processing time of all the tasks.
EPIT 2007
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 t=B/2 : the solution S’ is B/2- 

38. EPIT 2007
Back to our problem 
B/2
The solution S is B/2+ , S* is B/2+ with 
If pi becomes ei and t=B/2 : the solution S’ is B/2-  and the best solution S’* is B/2-
It suffices to note that S/S*<S’*/S’ to show that our problem is FPTAS

39. EPIT 2007
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

40. An APX problem without PTAS
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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

41. No -approximated algorithm if <4/3
EPIT 2007
No -approximated algorithm if <4/3
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

42. EPIT 2007
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 |S|=|V1| and |V1|=|V2| and |V1S|= |V2S|?

43. W with |W|=n/2 X with |X|=n Y with |Y|=n T with |T|=n/2 The edges of G
EPIT 2007
W with |W|=n/2
X with |X|=n
Y with |Y|=n
T with |T|=n/2
The edges of G
Complete Graph

44. EPIT 2007
X with |X|=n
W with |W|=n/2
Y with |Y|=n
T with |T|=n/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.

45. Scheduling in 3 steps : no idle time
EPIT 2007
X with |X|=n
W with |W|=n/2
m=n
Y with |Y|=n
T with |T|=n/2
Scheduling in 3 steps : no idle time

46. Step 1 : W and X1 ( X1 X, |X1|=n/2) must be executed.
EPIT 2007
X with |X|=n
W with |W|=n/2
m=n
Y with |Y|=n
T with |T|=n/2
Scheduling in 3 steps : no idle time
Step 1 : W and X1 ( X1 X, |X1|=n/2) must be executed.

47. Step 1 : W and X1 ( X1 X, |X1|=n/2) must be executed.
EPIT 2007
X with |X|=n
W with |W|=n/2
m=n
Y with |Y|=n
T with |T|=n/2
Scheduling in 3 steps : no idle time
Step 1 : W and X1 ( X1 X, |X1|=n/2) must be executed.
Step 3 : T and Y1 ( Y1 Y, |Y1|=n/2) must be executed.

48. Step 1 : W and X1 ( X1 X, |X1|=n/2) must be executed.
EPIT 2007
X with |X|=n
W with |W|=n/2
m=n
Y with |Y|=n
T with |T|=n/2
Scheduling in 3 steps : no idle time
Step 1 : W and X1 ( X1 X, |X1|=n/2) must be executed.
Step 2 : X2 and Y2 ( |X2|=|Y2|=n/2 ) must be executed.
Step 3 : T and Y1 ( Y1 Y, |Y1|=n/2) must be executed.

49. Step 2 : X2 and Y2 ( |X2|=|Y2|=n/2 ) must be executed.
EPIT 2007
X with |X|=n
W with |W|=n/2
m=n
Y with |Y|=n
T with |T|=n/2
Scheduling in 3 steps : no idle time
Step 2 : X2 and Y2 ( |X2|=|Y2|=n/2 ) must be executed.
X2 Y2 is a well balanced independent set in the graph G

50. 2-approximated list scheduling
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2-approximated list scheduling
Graham’s 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.

51. 2-approximated list scheduling
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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)

52. 2-approximated list scheduling
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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 .

53. EPIT 2007
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 .

54. Other constraints Communication delays
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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.

55. Communication delay : UET-UCT
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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).

56. Communication delay : UET - UCT
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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.

57. Communication delay: UET - UCT
EPIT 2007
Communication delay: UET - UCT
cij i j
If di is the starting time of i: dj  di 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.

58. Communication delay : UET - UCT
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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 < 1/2.

59. Communication delay : UET - UCT
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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 (k+1) with x arcs such that cij < ½ (i.e the cost of the path in the real solution is greater than 3/2k-1/2x+1). After rounding, the cost is 2k-x+1. The ratio is always smaller of 4/3.

60. Unbounded vs. bounded number of processors
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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.

61. EPIT 2007
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)