VKontakte P-35 Задача 23 as of 29.05.17

System above is equvalent this one :-

(┐x1 v ┐x2)=1
(┐x2 v ┐x3)=1
(┐x3 v ┐x4)=1
(┐x4 v ┐x5)=1
(┐x5 v ┐x6)=1
(┐x6 v ┐x7)=1
(┐x7 v ┐x8)=1
(┐y7 v y8 )=1

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For each j=1,..,7 . We have
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¬xj  v  ¬ x(j+1)) = 1
 ¬ (xj ^ x(j+1))    =1
(xj ^ x(j+1)) = 0 

Follow known schema ( see it-n.ru/attachment.aspx?id=150390 )
for system

(┐x1 v ┐x2)=1
(┐x2 v ┐x3)=1
(┐x3 v ┐x4)=1
(┐x4 v ┐x5)=1
(┐x5 v ┐x6)=1
(┐x6 v ┐x7)=1
(┐x7 v ┐x8)=1


Let К(n) to be a  number of bit's chains of length "n", which
don't have two ones "1" coming one by one

K1(n) - end up with 1
К0(n) - end up with 0

Then we get :-

K1(n+1)= K0(n)
K0(n+1)= K1(n) + K0(n) = K(n)
K(n+1) = K1(n+1) + K0(n+1) = K0(n) + K0(n+1) = K0(n) + K(n)

Due to :-
K0(n)=K(n-1)

We are getting
K(n+1) = K(n) + K(n-1)
Proof commmited.


Answer is  eight's Fibonacci number , which is 55 (solutions )


Thus final result would be 55*3 =165