1. No category

‫ﻧﻅﺭﻳﺔ ﺍﻟﺗﻌﻘﻳﺩ‬
‫ﻣﺣﺎﺿﺭﺓ ‪9‬‬
‫ﺍﻟﻌﻼﻗﺔ ‪ ≤p‬ﻫﻲ ﻋﻼﻗﺔ ﻣﺗﻌﺩﻳﺔ ‪:‬‬
‫‪1/4/2014‬‬
‫‪A ≤p B , B ≤p C ⟹ A ≤p C‬‬
‫ﺍﻟﺑﺭﻫﺎﻥ‪ :‬ﻣﻥ ﺍﻟﻔﺭﺽ‪ :‬ﻳﻭﺟﺩ ﺗﺎﺑﻌﺎﻥ *∑ ⟶ *∑ ‪ ، f ,g :‬ﻗﺎﺑﻠﻳﻥ ﻟﻠﺣﺳﺎﺏ ﺑﺯﻣﻥ ﻛﺛﻳﺭﺓ‬
‫ﺍﻟﺣﺩﻭﺩ ‪ ، nkg , nkf‬ﺑﺣﻳﺙ ﺃﻥ ‪:‬‬
‫‪f(x) ∈ B ⟺ x ∈ A‬‬
‫ﻧﺄﺧﺫ ﺍﻟﺗﺎﺑﻊ‪x ∈ A ⟺ g(f(x)) ∈ C ، g(f(x)) :‬‬
‫‪g(x)∈ C ⟺ x ∈ B‬‬
‫‪x ∈ A ⟺ f(x) ∈ B ⟺ g(f(x)) ∈ C‬‬
‫ﺯﻣﻥ )‪. (nkf)kg = g(f‬‬
‫ﻣﻼﺣﻅﺔ ‪ :‬ﻳﻧﻁﺑﻕ ﻫﺫﺍ ﺍﻟﻛﻼﻡ ﻋﻠﻰ ‪.≤m‬‬
‫ﻣﻼﺣﻅﺔ ‪DTime (nk) ⊆ NTime (nk) ⊆ DTime(2n)k :2‬‬
‫ﺗﻌﺭﻳﻑ‪ :‬ﻧﻘﻭﻝ ﻋﻥ ﻣﺟﻣﻭﻋﺔ ﺃﻧﻬﺎ ‪NP_hard‬‬
‫ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻛﺎﻥ‪:‬‬
‫‪) ⋀B∈NP B ≤p A‬ﺃﻱ ﻣﻥ ﺃﺟﻝ ﻛﻝ ‪(B∈NP‬‬
‫ﻭﻧﻘﻭﻝ ﻋﻥ ﻣﺟﻣﻭﻋﺔ ﻣﻥ ‪ NP_hard‬ﺃﻧﻬﺎ ‪ NP_complete‬ﺇﺫﺍ ﻛﺎﻧﺕ ﻣﻥ ‪NP‬‬
‫ﺗﻌﺭﻳﻑ‪ :‬ﻧﻘﻭﻝ ﻋﻥ ﻣﺟﻣﻭﻋﺔ ‪ A‬ﺃﻧﻬﺎ ‪ NP_complete‬ﺇﺫﺍ ﻛﺎﻧﺕ‪:‬‬
‫‪A ∈ NP (۱‬‬
‫‪B ≤p A (۲‬‬
‫‪⋀B∈NP‬‬
‫ﻧﻅﺭﻳﺔ ‪:1971 – cook‬‬
‫‪SAT is NP_complete‬‬
‫ﻧﻅﺭﻳﺔ ‪ :‬ﺇﺫﺍ ﻛﺎﻧﺕ ‪ ، A ∈ NP_complete‬ﻭ ‪ ، A ≤p B‬ﻭ ‪B ∈ NP‬‬
‫ﻋﻧﺩﺋ ٍﺫ ﺗﻛﻭﻥ ‪B is NP_complete‬‬
‫ﺍﻟﺑﺭﻫﺎﻥ‪ :‬ﺑﻣﺎ ﺃﻥ ‪: A ∈ NP_complete‬‬
‫‪۱‬‬
‫ﻧﻅﺭﻳﺔ ﺍﻟﺗﻌﻘﻳﺩ‬
‫ﻣﺣﺎﺿﺭﺓ ‪9‬‬
‫‪1/4/2014‬‬
‫‪; C ≤p A‬‬
‫ﻭﺑﻣﺎ ﺃﻥ ‪A ≤p B :‬‬
‫‪⋁fc ; ∑∗ ⟶ ∑∗ ; x ∈ C ⟺ f(x) ∈ A‬‬
‫;‬
‫‪⋀C∈NP‬‬
‫‪⋀C∈NP‬‬
‫‪⋁f ; ∑∗ ⟶ ∑∗ ; x ∈ A ⟺ f(x) ∈ B‬‬
‫ﻣﻧﻪ ﻳﻧﺗﺞ ‪:‬‬
‫‪⋀C∈NP ; ⋁g=f(fc ) ; x ∈ c ⟺ g(x) ∈ B‬‬
‫ﻭﻟﺩﻳﻧﺎ ﺑﺎﻟﻔﺭﺽ ‪. B is NP_complete ⇐ B ∈ NP‬‬
‫ﻧﻅﺭﻳﺔ ‪:‬‬
‫‪α1‬‬
‫‪α2‬‬
‫‪α3‬‬
‫‪3_SAT = {C1 ∧ C2 ∧…. Cn} ; Ci = (X i1‬‬
‫‪∨ X i2‬‬
‫‪∨ X i3‬‬
‫)‬
‫ﺃﻱ ‪ 3_SAT‬ﻫﻲ ﻣﺟﻣﻭﻋﺔ ﺍﻟﺗﻌﺎﺑﻳﺭ ﺫﺍﺕ ﺍﻟﺷﻛﻝ ﺍﻟﺳﺎﺑﻕ ‪ ،‬ﻭﻗﺎﺑﻠﺔ ﻟﻠﺗﺣﻘﻳﻕ‪.‬‬
‫ﻣﺛﻼً‪� 5 ∨ X1 ) :‬‬
‫‪(X1 ∨ �X 2 ) ∧ X3 ∧ (X2 ∨ X‬‬
‫ﺍﻟﺑﺭﻫﺎﻥ‪ :‬ﻳﻛﻔﻲ ﺃﻥ ﻧﺑﺭﻫﻥ ﺃﻥ ‪SAT ≤p 3SAT‬‬
‫ﻷﻥ ‪3SAT ∈ NP :‬‬
‫ﻟﻛﻲ ﻧﺑﺭﻫﻥ ﻋﻥ ﻣﺳﺄﻟﺔ ﻣﺎ ﺃﻧﻬﺎ ‪ ، NP‬ﻓﻧﻌﻠﻡ ﺃﻥ ﻣﺳﺎﺋﻝ ‪ NP‬ﻛﻠﻬﺎ ﺗﻌﺗﻣﺩ ﻋﻠﻰ ﺧﻳﺎﺭﺍﺕ‬
‫ﻛﺛﻳﺭﺓ ‪ ،‬ﺇﺫﺍ ﻛﺎﻥ ﺯﻣﻥ ﺍﺧﺗﺑﺎﺭ ﺍﻟﺧﻳﺎﺭ ﺍﻟﻭﺍﺣﺩ ﻫﻭ ‪ P‬ﻓﺈﻥ ﺍﻟﻣﺳﺄﻟﺔ ﺗﻛﻭﻥ ‪.NP‬‬
‫‪ (۱‬ﻣﻥ ﺃﺟﻝ‬
‫‪Ci = Xi ، |Ci|=1‬‬
‫‪�i1 ∨ Yi2 ) ∧ (X i ∨ Yi1 ∨ Y‬‬
‫∧ ) ‪�i2‬‬
‫‪f(Ci) = (X i ∨ Yi1 ∨ Yi2 ) ∧ (X i ∨ Y‬‬
‫‪�i1 ∨ Y‬‬
‫) ‪�i2‬‬
‫‪(X i ∨ Y‬‬
‫‪Ci = (Xi ∨ Yj) ، |Ci|=2 (۲‬‬
‫� ‪�i1‬‬
‫‪f(Ci) = �X j ∨ X k ∨ Yi1 � ∧ �X j ∨ X k ∨ Y‬‬
‫‪ ، |Ci|=3 (۳‬ﻳﺑﻘﻰ ﺍﻟﺗﻌﺑﻳﺭ ﻛﻣﺎ ﻫﻭ‪.‬‬
‫‪Ci = (u1 ∨ u2 ……. ∨ uk) ، |Ci|>3 (٤‬‬
‫‪۲‬‬
‫ﻧﻅﺭﻳﺔ ﺍﻟﺗﻌﻘﻳﺩ‬
‫ﻣﺣﺎﺿﺭﺓ ‪9‬‬
‫‪1/4/2014‬‬
‫∨ ‪f(Ci) = (u1 ∨ u2 ∨ yi1 ) ∧ (y� i1 ∨ u3 ∨ Yi2 ) … … ∧ (y� ik−3‬‬
‫) ‪uk−1 ∨ uk‬‬
‫ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﺗﻌﺑﻳﺭ ﻣﺣﻘﻕ ﻣﻥ ﺃﺟﻝ ﺃﺣﺩ ﺍﻟـ )‪ (u‬ﻓﻠﻥ ﺗﺅﺛﺭ ﻗﻳﻡ ‪ y‬ﻣﻬﻣﺎ ﻛﺎﻧﺕ‪.‬‬
‫ﻭﺇﺫﺍ ﻛﺎﻥ ﻏﻳﺭ ﻣﺣﻘﻕ ﻓﻠﻥ ﺗﺅﺛﺭ ﻗﻳﻡ ‪.y‬‬
‫ﻭﻟﻥ ﻳﻛﻭﻥ ﻣﺣﻘﻕ ﺇﻻ ﺇﺫﺍ ﺗﺣﻘﻘﺕ ‪.Ci‬‬
‫__________________‬
‫__________‬
‫___‬
‫‪۳‬‬