‫ﻧﻅﺭﻳﺔ ﺍﻟﺗﻌﻘﻳﺩ‬
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4 ‫ﺍﻟﺟﻠﺳﺔ‬
DTime(T(n)) is a set of all computable function (algorithm) in at
most c.T(n) ,where c > 0.
DTime(T(n2)) is a set of all computable function (algorithm) in at
most O(n2).
If T1(n)=O(T2(n)) ; T2(n) > T1(n)
⟹ DTime(T1(n)) ⊆ DTime(T2(n))
⟹ DTime(n) ⊆ DTime(n2)
Proved that DTime(n) ⊆ DTime(n2) ?
Proof : n2= O(n)..\\ just this!!
Or: for DTime(n) there is a Turing machine that computed a
function in (c.n) steps.
DTime(n2) : the same but in (c.n2 ) steps. Now if we use c.n
steps to compute the function, we still have n steps for nothing
(because it's c.n2); ⟹ any machine could compute (c.n2) it can
do it for (c.n).
:‫ﺻﻑ ﺍﻟﺗﻌﻘﻳﺩ‬
!‫؟‬.‫ ﺃﻭ ﻻ ﺗﻧﺗﻣﻲ‬p ‫ ﻫﻝ ﺗﻧﺗﻣﻲ ﺍﻟﻣﺳﺄﻟﺔ ﺇﻟﻰ‬: ‫ﺑﺎﻟﻧﺳﺑﺔ ﻟﻠﺳﺅﺍﻝ‬
‫ﺑﺈﻣﻛﺎﻧﻧﺎ ﺃﻥ ﻧﻘﺩﻡ ﺧﻭﺍﺭﺯﻣﻳﺔ ﻟﺣﺳﺎﺏ ﺍﻟﺗﺎﺑﻊ ﺑﺯﻣﻥ ﻓﻌﺎﻝ ﺃﻭ ﺍﻻﺧﺗﺻﺎﺭ ﺇﻟﻰ ﻣﺳﺄﻟﺔ‬
.‫ﻣﻌﺭﻭﻓﺔ‬
:SAT ‫ﻣﺳﺄﻟﺔ ﺍﻟـ‬
: ‫ﻟﺩﻳﻧﺎ ﺑﺷﻛﻝ ﻋﺎﻡ ﻣﺟﻣﻭﻋﺔ ﻣﺗﺣﻭﻻﺕ ﻣﻧﻁﻘﻳﺔ‬
X1,X2…………..Xn
:‫ ﻣﻥ ﺍﺣﺩ ﺍﻷﺷﻛﺎﻝ ﺍﻟﺗﺎﻟﻳﺔ‬f ‫ﻭﺗﺎﺑﻊ‬
F= X1∨ X2 ∧ X1 ∨ X3……
‫ﻧﻅﺭﻳﺔ ﺍﻟﺗﻌﻘﻳﺩ‬
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‫؟‬..true ‫ ﻳﻛﻭﻥ‬f ‫ ﻫﻝ ﻳﻭﺟﺩ ﻗﻳﻡ ﻟﻠﻣﺗﺣﻭﻻﺕ ﺑﺣﻳﺙ‬:‫ﺍﻟﻣﺳﺄﻟﺔ‬
.SAT in general ‫ﺇﻧﻬﺎ ﻣﺳﺄﻟﺔ‬
:‫ ﻳﺄﺧﺫ ﺍﻟﺷﻛﻝ‬CNF SAT ‫ﻫﻧﺎﻙ ﻧﻭﻉ ﻣﻧﻬﺎ‬
F= C1 ∧ C2 ∧ C3…..
C : clause , consists of literal ex: Ci = (l1 ∨ l2 ∨…..) , and li is a Xi or
Xi .
Note that between clauses there's (∧) , and between literals
we had (∨) .
:‫ﻣﺛﺎﻝ‬
(X3 ∨ X5) ∧ (X3 ∨ X1 ∨ X4) ∧ (X2 ∨ X3 ∨ X5)
m : num of clauses = 3
n : num of variables = 5
X1 = 1, X4 = 1 ,X2 = 1, X3 =0, X5=0
2 SAT means 2 CNF SAT ; in each clause we have 2 literals at
most.
It is ok if we have one attachment , we put it with itself by or
(∨) .
:P ‫ ﻫﻲ ﻣﺳﺄﻟﺔ‬2 SAT ‫ﻟﺑﺭﻫﺎﻥ ﺃﻥ ﻣﺳﺄﻟﺔ‬
:‫ ﻭﺳﻳﻛﻭﻥ ﻟﺩﻳﻧﺎ ﺍﻟﺣﺎﻻﺕ ﺍﻟﺗﺎﻟﻳﺔ‬:‫ﻧﻌﻁﻲ ﻷﺣﺩ ﺍﻟﻣﺗﺣﻭﻻﺕ ﻗﻳﻣﺔ‬
Xi not set && Xj not set
Xi =1
&& Xj not set
Xi =0
&& Xj not set
Xi =0
&& Xj =0
2
‫ﻧﻅﺭﻳﺔ ﺍﻟﺗﻌﻘﻳﺩ‬
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Xi =1
&& Xj=0
4 ‫ﺍﻟﺟﻠﺳﺔ‬
5
.‫ ﻷﻥ ﺍﻟﺣﺎﻟﺔ ﺗﻧﺎﻅﺭﻳﺔ ﺑﻳﻧﻬﻣﺎ‬Xi ، Xj ‫ﻟﻳﺱ ﻫﻧﺎﻙ ﻓﺭﻕ ﻓﻲ ﺣﺎﻝ ﻋﻛﺳﻧﺎ ﺍﻟﺣﺎﻟﺔ ﺑﻳﻥ‬
‫ ﻧﺣﺻﻝ ﻋﻠﻰ‬5 ، 2 ‫ ﺇﺫﺍ ﺣﺫﻓﻧﺎ ﻣﻥ ﺍﻟﺣﺎﻻﺕ ﺍﻟﺳﺎﺑﻘﺔ‬. ‫ ﻳﻣﺛﻝ ﻣﺎ ﺳﺑﻕ‬f ‫ﺍﻵﻥ ﺍﻟﺗﺎﺑﻊ‬
.true ‫ ﺳﻳﻛﻭﻥ ﺣﻛﻣﺎ‬f ‫ ﻫﺫﺍ ﻳﻌﻧﻲ ﺃﻥ‬true ‫ ﻣﺣﻘﻕ ﺃﻭ‬f` ‫ ﻭﺍﻓﺗﺭﺿﻧﺎ ﺃﻥ‬f` ‫ﺍﻟﺗﺎﺑﻊ‬
.NOT SAT f ‫ ﻓﻠﻳﺱ ﺑﺎﻟﺿﺭﻭﺭﺓ ﺃﻥ ﻳﻛﻭﻥ‬NOT SAT ‫ ﺃﻭ‬false f` ‫ﺃﻣﺎ ﺇﺫﺍ ﻛﺎﻥ‬
X1 =1 ‫ ﻟﻧﻔﺭﺽ‬: ‫ﻋﻠﻰ ﺍﻟﻣﺛﺎﻝ ﺍﻟﺳﺎﺑﻕ‬
C1 → 2 X
C2 → 1
C3 → 2 X
F` = ( X2 ∨ X3)
Again X2 =1 ⇒ f`` = (X2 ∨ X3 ) : X2 has 1 before and now X3 : if
X3 = 0 that's ok and f`` =1 , else if it's 1 then f`` = 0 and we will
move to 4th state. ⟹ X1 =1, X2 =1, X3 =0 .
Another example:
f= (X1 ∨ X2) ∧ (X2 ∨ X3) ∧ (X1 ∨ X2)
X1 =0 : C1 → 3 , C2 → 1 , C3 → 2
We avoid the 3rd clause.
F``= (X1 ∨ X2) ∧ (X2 ∨ X3)
Now X2 =0 : C1 → 5 , C2 → 3
We avoid the 1st clause.
F``` = (X2 ∨ X3)
X3 =1 ⇒ f``` = 1 ⇒ F is SAT ,
X3 = 0 ⇒ f``` = 0 we move to 4th state .
‫ﻧﻅﺭﻳﺔ ﺍﻟﺗﻌﻘﻳﺩ‬
‫ﺍﻟﺟﻠﺳﺔ ‪4‬‬
‫‪30/3/2014‬‬
‫ﻣﻣﺎ ﺳﺑﻕ ﻧﺟﺩ ﺃﻥ ﻣﺳﺄﻟﺔ ‪ SAT‬ﻫﻲ ﻣﺳﺄﻟﺔ ‪.P‬‬
‫ﺗﻌﻘﻳﺩ ﺍﻟﺧﻭﺍﺭﺯﻣﻳﺔ‪:‬‬
‫ﺃﺳﻭﺃ ﺣﺎﻟﺔ ‪ :‬ﺍﻟﺗﺟﺭﺑﺔ ‪2n‬‬
‫ﻓﻲ ﺃﻭﻝ ﻣﺭﺓ ﻳﺣﺫﻑ ‪ clause‬ﻭﺍﺣﺩ ﻭﻛﺫﻟﻙ ﻓﻲ ﺍﻟﻣﺭﺓ ﺍﻟﺛﺎﻧﻳﺔ ﻭﺍﺣﺩ ﻭﻫﻛﺫﺍ‪.....‬‬
‫)‪m(m+1‬‬
‫‪2‬‬
‫= ‪m + m-1 +m-2 +…..‬‬
‫‪O(nm2) ∈ P.‬‬
‫‪2 coloring graph:‬‬
‫)‪G(V ,E‬‬
‫ﻣﻥ ﺃﺟﻝ ﻛﻝ ﻋﻘﺩﺗﻳﻥ ﻧﺿﻳﻑ ﺍﻟﺗﺎﺑﻊ ‪:‬‬
‫ﺃﺳﻭﺩ‬
‫)‪For each c(Vi ,Vj) ∈ E; c(Vi) ≠ c(Vj‬‬
‫ﺃﺳﻭﺩ‬
‫)‪F= (Vi ∨ Vj) ∧ (Vi ∨ Vj‬‬
‫ﺃﺳﻭﺩ‬
‫ﺃﺣﻣﺭ‬
‫ﺑﻣﻌﻧﻰ ﺃﻧﻪ ﻻ ﻳﻣﻛﻥ ﺃﻥ ﻳﻛﻭﻥ ﻓﻲ ﻁﺭﻓﻲ ﺍﻟﻭﺻﻠﺔ ﺍﻟﻠﻭﻥ ﻧﻔﺳﻪ‪.‬‬
‫‪Complexity : O(|E|) ..// length of the graph.‬‬