Предмет: Информатика, автор: arefnelly

Большая просьба! Помочь
Сколько различных решений имеет система уравнений?
(x1 → x2) / (x2 → x3) / (x3 → x4) / (x4 → x5) = 1
(у1 → у2) / (у2 → у3) / (у3 → у4) / (у4 → у5) = 1
(z1 → z2) / (z2 → z3) / (z3 → z4) / (z4 → z5) = 1
х1 / у1 / z1 = 1
где x1,x2,…,x5, у1,у2,…,у5, z1,z2,…,z5 – логические переменные?
В ответе не нужно перечислять все различные наборы значений переменных, при которых выполнено данное равенство.
В качестве ответа нужно указать количество таких наборов.

Ответы

Автор ответа: ludmilagena
0
из последнего уравнения: х1 + у1 +z1 =1  следует,
что  не может быть одновременно    x1=0 y1=0 z1=0

рассмотрим первое уравнение,  это логическое умножение, каждый сомножитель должен быть равен 1, такое будет, если:
(не х1 +  х2)    *   (не х2 +х3) *(не х3 +х4)  *        ( не х4 +х5) = 1
  1            1                  1      1          1        1                     1         1

   0            0                  0      0          1        1 (2реш)        1       1  (3 решения)
   0           1                   0      1          0         1                    0       1
                                     1       1         1         0                    0       0
Итак, при х1=1 - одно решение, при х1=0   - 5 решений
Для второго ( с Y) и третьего ( сZ ) ур-я - аналогично по 5 решений.
Возможные варианты сочетаний значений х1, y1,z1   :
  х1   y1  z1
 0     0     0    - (четвёртое уравнение исключает это сочетание)
 0     0     1         5*5*1 =25 решений
0      1    0         5*1*5 =25 реш
  0     1    1          5*1*1 =5
1     0      0            1*5*5=25
1     0      1           5 реш
1     1     0            5
1     1      1            1
                               5+5+5+25+25+25+1 = 91  <------- ответ
Если ответ не верный, напишите. Буду думать.

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