Re：ゲーム理論入門 第13回 - 特性関数とコア -
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https://www.nicovideo.jp/watch/sm33711818 https://www.youtube.com/watch?v=3TE_tLKMS40&t=3s
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Re：ゲーム理論入門 第13回 - 特性関数とコア -
- 2. “ ” — —
- 8. v({1}) = 0 v({2}) = 0 v({3}) = 0
- 9. v({1,2,3}) = 100 v({1,2}) = 50 v({2,3}) = 80 v({1,3}) = 40 v({1}) = 0 v({2}) = 0 v({3}) = 0 23 − 1 = 7 v (N, v) v
- 11. S ⊃ T (N, v) v v v v v(S ∪ T) ≥ v(S) + v(T) v(S) ≥ v(T) v(N) > n ∑ i=1 v(i) v(N) = v(S) + v(NS) v({1,2,3}) = 100 v({1,2}) = 50 v({2,3}) = 80 v({1,3}) = 40 v({1}) = 0 v({2}) = 0 v({3}) = 0
- 12. v({1,2,3}) = 100 v({1,2}) = 50 v({2,3}) = 80 v({1,3}) = 40 v({1}) = 0 v({2}) = 0 v({3}) = 0 x = (x1, x2, x3) xi ≥ v(i) 3 ∑ i=1 xi = v({1,2,3})
- 13. v({1,2,3}) = 100 v({1,2}) = 50 v({2,3}) = 80 v({1,3}) = 40 v({1}) = 0 v({2}) = 0 v({3}) = 0 x = (x1, x2, x3) xi ≥ v(i) 3 ∑ i=1 xi = v({1,2,3}) v({1,2,3}) x x3 x1 x2
- 14. x x3 x1 x2 (N, v) x, y v(S) ≥ ∑ i∈S x(i) xi > yi, ∀i ∈ S x y (N, v)
- 15. x x3 x1 x2 (N, v) ∑ i∈S x(i) ≥ v(S), ∀S ⊂ N
- 16. v({1,2,3}) = 100 v({1,2}) = 50 v({2,3}) = 80 v({1,3}) = 40 v({1}) = 0 v({2}) = 0 v({3}) = 0 x = (x1, x2, x3) x1 + x2 ≥ 50 x2 + x3 ≥ 80 x1 + x3 ≥ 40 x1 ≥ 0 x2 ≥ 0 x3 ≥ 0 x1 + x2 + x3 = 100 x1 = 20 ⇒ 20 ≥ x1
- 17. v({1,2,3}) = 100 v({1,2}) = 50 v({2,3}) = 80 v({1,3}) = 40 v({1}) = 0 v({2}) = 0 v({3}) = 0 x = (x1, x2, x3) x1 + x2 ≥ 50 x2 + x3 ≥ 80 x1 + x3 ≥ 40 x1 ≥ 0 x2 ≥ 0 x3 ≥ 0 x1 + x2 + x3 = 100 x1 = 20 ⇒ 60 ≥ x2 x2 = 60
- 18. v({1,2,3}) = 100 v({1,2}) = 50 v({2,3}) = 80 v({1,3}) = 40 v({1}) = 0 v({2}) = 0 v({3}) = 0 x = (x1, x2, x3) x3 x1 + x2 ≥ 50 x2 + x3 ≥ 80 x1 + x3 ≥ 40 x1 ≥ 0 x2 ≥ 0 x3 ≥ 0 x1 + x2 + x3 = 100 x1 = 20 ⇒ 50 ≥ x3 x2 = 60 x3 = 50
- 19. v({1,2,3}) = 100 v({1,2}) = 50 v({2,3}) = 80 v({1,3}) = 40 v({1}) = 0 v({2}) = 0 v({3}) = 0 x = (x1, x2, x3) x3 x1 + x2 ≥ 50 x2 + x3 ≥ 80 x1 + x3 ≥ 40 x1 ≥ 0 x2 ≥ 0 x3 ≥ 0 x1 + x2 + x3 = 100 x1 = 20 x2 = 60 x3 = 50
- 20. x3 x1 = 20 x2 = 60 x3 = 50
- 21. v({1,2,3}) = 100 v({1,2}) = 50 v({2,3}) = 80 v({1,3}) = 80 v({1}) = 0 v({2}) = 0 v({3}) = 0 x3 x1 + x2 ≥ 50 x2 + x3 ≥ 80 x1 + x3 ≥ 80 x1 ≥ 0 x2 ≥ 0 x3 ≥ 0 x1 + x2 + x3 = 100 x1 = 20 x2 = 20 x3 = 50
- 22. (N, v) (N, v′) βi(i = 1,2,⋯, n) v′ = αv(S) + ∑ i∈S βi (N, v) v′(i) = 0, i = 1,⋯, n (N, v′) v′(N) = 1
- 23. (N, v) v({1,2}) + v({2,3}) + v({1,3}) ≤ 2v({1,2,3})
- 24. (N, v) min ∑ i∈N xi s . t . ∑ i∈S xi ≥ v(S), ∀S ⊊ N z * z* ≤ v(N)
- 27. See you next time