"mitteesitavaks" - 1 õppematerjal

Mis on Diskreetne Matemaatika

leidmist mingi avaldis 0 ⊕ 1 [ ⊕  [ ⊕ . . . ⊕ 3 [3   # c0 ⊕ c1 ⋅ 0 ⊕ c2 ⋅ 0 = 1(00)               
      c0 ⊕ c1 ⋅ 0 ⊕ c2 ⋅ 1 = f(01) 1 ( [1 [2  [n  mitteesitavaks. c0 ⊕ c1 ⋅ 1 ⊕ c2 ⋅ 0 = f(10 Selliselt saadud avaldis 0 ⊕ 1 x1 ⊕ 2 x2 ⊕ . . . ⊕ n xn   2-muutuja funktsiooni f  [ [ )                      &           . .