Algorithmic Coalitional Game Theory Lecture 12: Anytime Coalition - - PowerPoint PPT Presentation
Algorithmic Coalitional Game Theory Lecture 12: Anytime Coalition - - PowerPoint PPT Presentation
Algorithmic Coalitional Game Theory Lecture 12: Anytime Coalition Structure Generation Oskar Skibski University of Warsaw 19.05.2020 Coalition Structure Generation Coalition Structure Generation Find a partition of players = { ! ,
Coalition Structure Generation
In other words: which coalition structure will form? 2
Oskar Skibski (UW) Algorithmic Coalitional Game Theory
Find a partition of players π = {π!, β¦ , π"} such that the sum
- f values of coalitions, i.e. π€ π! + β― + π€(π"), is maximized.
Coalition Structure Generation
Coalition Structure Generation
3
Oskar Skibski (UW) Algorithmic Coalitional Game Theory
The exact algorithms requires a lot of timeβ¦ Let πβ = arg max
$βπ¬ ' π€(π).
Can we find a subset π β π¬ π s.t. πΎ β₯
( $β )*+ ( $ βΆ$βπ for
every game (π, π€)? We define:
bound π = min πΎ β β βΆ β !,# πΎ β₯ π€ πβ max π€ π βΆ π β π
Can we search through only a subset of coalition structures and be guaranteed to find a solution that is within a certain bound from the optimum?
?
Coalition Structure Generation
4
Oskar Skibski (UW) Algorithmic Coalitional Game Theory
π¬
.(π)
π¬/(π) π¬0(π) π¬
!(π)
1|2|3|4 12|3|4 13|2|4 14|2|3 1|23|4 1|24|3 1|2|34 124|3 1234 14|23 13|24 123|4 12|34 1|234 134|2
Worst-case guarantee
Proof: On the blackboard. 5
Oskar Skibski (UW) Algorithmic Coalitional Game Theory
For π = π¬
! π βͺ π¬0 π we have bound π = π, π =
212!, and π is the minimal set with bound smaller than β. Minimal set with a bound [Sandholm et al. 1999]
Worst-case guarantee
Oskar Skibski (UW) Algorithmic Coalitional Game Theory
6 Sketch of proof: Fix π3 = arg max
$βπ π€(π) and πβ = arg max 4β' π€(π).
- bound π β€ π: We know π€ π3 β₯ π€ πβ . Hence,
π€ πβ β€ |πβ| β π€ πβ β€ |πβ| β π€ π3 β€ π β π€ π3 .
- bound π β₯ π: Assume π€ π = 1 if π = 1 and π€ π =
0, oth. Then: π€ π3 = 1 = !
1 π€
1 , β¦ , π = !
1 π€(πβ).
- Clearly, π =
π β π βΆ 1 β π = 212!.
- In every β¬ with bound β¬ β€ β we have at least 1
partition for every coalition with player 1 (they do overlap, so cannot be in the same partition), so β¬ β₯ 212!.
Worst-case guarantee
7
Oskar Skibski (UW) Algorithmic Coalitional Game Theory
What subset of coalition structures should we search next?
?
Worst-case guarantee
8
Oskar Skibski (UW) Algorithmic Coalitional Game Theory
What subset of coalition structures should we search next?
?
For π = π¬
! π βͺ π¬0 π βͺ π¬ 1 π and π > 3 we have
bound π = βπ/2β. Improving the bound [Sandholm et al. 1999] Proof: On the blackboard.
Worst-case guarantee
Oskar Skibski (UW) Algorithmic Coalitional Game Theory
9 Sketch of proof: Assume πβ contains π singletons { π! , β¦ , π" }. If π = 0, then πβ β€
1 0 and π€ πβ β€ 1 0 β π€ π3 .
Assume π > 0. We know that π€ π! , β¦ , π" β€ π€ 1 , β¦ , π β€ π€ π3 . Also, πβ\{ π! , β¦ , π" } β€
12"
β€
12!
. Hence, π€ πβ β€
12!
+ 1 π€ π3 =
1 0 β π€ π3 .
Worst-case guarantee
Oskar Skibski (UW) Algorithmic Coalitional Game Theory
10 Sketch of proof (continued): Assume π€ π = 1 if π = 2 and 1 β π, π€ {1} = 1 and π€ π = 0, oth. Then:
- π€ π3 = 1 (since π > 3)
- and π€
1 , 2 , {3,4} β¦ , {π β 1, π} = 1
0 if π is even,
- and π€
1 , {2,3} β¦ , {π β 1, π} = 16!
0 oth.
Anytime CSG
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Oskar Skibski (UW) Algorithmic Coalitional Game Theory
How to search through the coalition structures to improve the guarantee over time?
?
Anytime CSG
12
Oskar Skibski (UW) Algorithmic Coalitional Game Theory
1|2|3|4 12|3|4 13|2|4 14|2|3 1|23|4 1|24|3 1|2|34 124|3 1234 14|23 13|24 123|4 12|34 1|234 134|2
π! π0 π/ π. π7
Anytime CSG
13
Oskar Skibski (UW) Algorithmic Coalitional Game Theory
How to search through the coalition structures to improve the guarantee over time?
?
Divide the search space into subsets: π¬ π = π! βͺ π0 βͺ β― βͺ π", such that: bound π! β₯ bound π! βͺ π0 β₯ β― β― β₯ bound π! βͺ β― βͺ π" = 1. Anytime Coalition Structure Generation
Anytime CSG
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Oskar Skibski (UW) Algorithmic Coalitional Game Theory
- 1. Search π¬
! π .
- 2. Search π¬0 π
- 3. Search π¬
1 π .
- 4. Search π¬12! π .
- 5. β¦
- 6. Search π¬/ π .
Anytime CSG-99 [Sandholm et al. 1999] π! = π¬
! π , π0 = π¬0 π , and
π" = π¬16/2" π for 2 < π β€ π.
Anytime CSG
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Oskar Skibski (UW) Algorithmic Coalitional Game Theory
π¬!(π)
[1,1,1,1,1,1,1,1] [2,1,1,1,1,1,1] [3,1,1,1,1,1] [2,2,1,1,1,1] [4,1,1,1,1] [3,2,1,1,1] [2,2,2,1,1] [5,1,1,1] [4,2,1,1] [3,3,1,1] [3,2,2,1] [2,2,2,2] [6,1,1] [5,2,1] [4,3,1] [4,2,2] [3,3,2] [7,1] [6,2] [8]
π¬"(π) π¬#(π) π¬$(π) π¬
%(π)
π¬&(π) π¬'(π) π¬
((π)
[5,3] [4,4]
π! π0 π/ π. π7 π; π< π=
Anytime CSG
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Oskar Skibski (UW) Algorithmic Coalitional Game Theory
Define β π = (π β π)/2 + 2. After searching π¬> π for π > 3, the bound is β π/β(π). Specifically, for π = π¬
! π βͺ π¬0 π βͺ π¬ 1 π βͺ β― βͺ π¬> π :
bound π = 4 π/β(π) ππ π β‘ β1 πππ β π , π β‘ π πππ 2 , π/β(π) ππ’βππ π₯ππ‘π.
Bounds for Anytime CSG-99 [Sandholm et al. 1999] Proof: On the blackboard.
Anytime CSG
Oskar Skibski (UW) Algorithmic Coalitional Game Theory
17 Sketch of proof: β π = (π β π)/2 + 2 = π β π β 2 /2 + 1 is a number such that partition of the form:
- [β π , β π β 1, 1, β¦ , 1] if 2 β€ π β π (case A) or
- β π , β π β 2, 1, β¦ , 1 if 2|π β π (case B)
appears in level π ([β¦] contains the list of sizes of coalitions). After searching level π we know that disjoint coalitions of sizes π and π such that π + π β€ π β π β 2 appeared in one partition.
Anytime CSG
Oskar Skibski (UW) Algorithmic Coalitional Game Theory
18 Sketch of proof (continued): Lower bound for the bound: Let π = π mod β π = π β π/β(π) β β π . Consider partition πβ of the form [β π , β(π), β¦ , β π , π ] and game π€ π = 1 if π = β(π). We have: π€ πβ = π/β(π) and π€ πβ² = 1. Thus, bound π β₯ π/β(π) . If we have case B (i.e., 2|π β π) and π = β π β 1, then by adding π€ π = 1 for one specific coalition of size π we get: bound π β₯ βπ/β(π)β + 1 = βπ/β(π)β
Anytime CSG
Oskar Skibski (UW) Algorithmic Coalitional Game Theory
19 Sketch of proof (continued): Upper bound for the bound: To see that bound π β€ π/β(π) consider the game (π, π€) such that π€ πβ /π€(π3) is the highest. We can assume that π€ π = 0 for every π β πβ. Also, we can assume that π3 β© πβ = 1: if π, π3 β π3 β© πβ, then replacing π, π3 with π βͺ π3 and defining game analogously would result in the same value π€ πβ /π€(π3).
Anytime CSG
Oskar Skibski (UW) Algorithmic Coalitional Game Theory
20 Sketch of proof (continued): Now, since π3 β© πβ = 1 it means that in πβ there are no two coalitions that appeared in the same partition considered so far. Hence, we get the limit on the number of such coalitions. In case A or case B where π β β π β 1, we get maximum βπ/β π β coalitions for: β π , β π , β¦ , β π , β π + π . In case B with π β β π β 1, we get maximum βπ/β π β coalitions for: β π , β π , β¦ , β π , β π β 1 This implies our thesis.
Anytime CSG
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Oskar Skibski (UW) Algorithmic Coalitional Game Theory
- 1. Search π¬
! π .
- 2. Search π¬0 π .
- 3. Search π¬
1 π .
- 4. Search π¬?120 π .
- 5. β¦
- 6. Search π¬?0 π .
Anytime CSG-04 [Dang & Jennings 2004] Define: π¬?@ π = π β π¬ π βΆ π > 2, max
4β$ π β₯ π .
Since many of this steps will not improve the bound, the authors considered π¬?β1(@2!)/@β π for π from β(π + 1)/4β down to 2 and then search the remaining coalition structures.
Anytime CSG
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Oskar Skibski (UW) Algorithmic Coalitional Game Theory
π¬!(π)
[1,1,1,1,1,1,1,1] [2,1,1,1,1,1,1] [3,1,1,1,1,1] [2,2,1,1,1,1] [4,1,1,1,1] [3,2,1,1,1] [2,2,2,1,1] [5,1,1,1] [4,2,1,1] [3,3,1,1] [3,2,2,1] [2,2,2,2] [6,1,1] [5,2,1] [4,3,1] [4,2,2] [3,3,2] [8]
π¬"(π) π¬#(π) π¬$(π) π¬
%(π)
π¬&(π) π¬'(π) π¬
((π)
[7,1] [6,2] [5,3] [4,4]
π! π0 π/ π= π< π; π7 π.
Anytime CSG
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Oskar Skibski (UW) Algorithmic Coalitional Game Theory
For π = π¬
! π βͺ π¬0 π βͺ π¬ 1 π βͺ π¬?β1(@2!)/@β π
we have: bound π β€ 2π β 1. Bounds for Anytime CSG-04 [Dang & Jennings 2004] Proof: On the blackboard.
Anytime CSG
Oskar Skibski (UW) Algorithmic Coalitional Game Theory
24 Sketch of proof: Assume πβ = π!, π0, β¦ , π" s.t. π! β₯ π0 β₯ β― β₯ π" . As before, we have π€ π! + β― + π€ π@2! β€ π β 1 β π€(π3). Now, consider π@, π0@, β¦ , πβ"/@β@. These coalitions combined have at most π/π players. So, they do not contain at least π β π/π = π(π β 1)/π. Hence, they appeared already in one partition and we have: π€ π@ + β― + π€ πβ"/@β@ β€ π€(π3). With the same analysis for groups (π@6!, π0@6!, β¦ , π "/@ @6!), (π@60, π0@60, β¦ , π "/@ @60) and so on, we get that: π€ πβ β€ 2π β 1 π€ π3 .
Anytime CSG
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Oskar Skibski (UW) Algorithmic Coalitional Game Theory
Conclusions
We consider the Anytime Coalition Structure Generation problem.
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Oskar Skibski (UW) Algorithmic Coalitional Game Theory
References
- [Dang & Jennings 2004] V.D. Dang & N.R. Jennings.
Generating coalition structures with finite bound from the optimal
- guarantees. Proceedings of the 3rd International Conference on
Autonomous Agents and Multiagent Systems (AAMAS), 564-571, 2004.
- [Rahwan et al. 2011] Rahwan, T.; Michalak, T. P. & Jennings, N. R.
Minimum search to establish worst-case guarantees in coalition structure generation. Proceedings of the 22nd International Joint Conference on Artificial Intelligence (IJCAI), 338-343, 2011.
- [Sandholm et al. 1999] T. Sandholm, K. Larson, M. Andersson,
- O. Shehory, F. TohmΓ©. Coalition structure generation with worst case
- guarantees. Artificial Intelligence 111, 209-238, 1999.
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Oskar Skibski (UW) Algorithmic Coalitional Game Theory