P1 = ( 1,6,11,12; 2,3,4,5,7,8,9,10 ) P5 = ( 2,3,5,6,7,10,12; 1,4,8,9,11 )
Zadanie 1
Korzystając z metody ESPRESSO zminimalizować funkcję:
F: 0000 R: 0001
0100 0010
0110 0011
1010 0101
1011
Zadanie 2
Dla funkcji F zdefiniowanej poniżej wyznaczyć wszystkie realizacje minimalno-argumentowe:
P1 = ( 1,6,11,12; 2,3,4,5,7,8,9,10 ) P5 = ( 2,3,5,6,7,10,12; 1,4,8,9,11 )
P2 = ( 1,11,12; 2,3,4,5,6,7,8,9,10 ) P6 = ( 1,3,5,7,8,10,11,12; 2,4,6,9 )
P3 = ( 2,5,7,10; 1,3,4,6,8,9,11,12 ) P7 = ( 1,2,4,6,7,9,11,12; 3,5,8,10 )
P4 = ( 2,4,7,8,9,10; 1,3,5,6,11,12 ) P8 = ( 1,4,6,8,10; 2,3,5,7,9,11,12 )
PF = (1,2,3,5,6,8,9,11,12; 4,7,10 )
k |
x1 |
x2 |
x3 |
x4 |
x5 |
x6 |
x7 |
x8 |
F |
1 |
0 |
0 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
2 |
1 |
1 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
3 |
1 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
4 |
1 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
5 |
1 |
1 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
6 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
7 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
8 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
9 |
1 |
1 |
1 |
0 |
1 |
1 |
0 |
1 |
0 |
10 |
1 |
1 |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
11 |
0 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
0 |
12 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
1 |
0 |
Zadanie 3
Dla funkcji F zdefiniowanej poniżej znaleźć dekompozycję dla U = { x1, x3 } oraz
V = { x2, x4, x5 }. W rozwiązaniu podać tablice prawdy funkcji G i H.
k |
x1 |
x2 |
x3 |
x4 |
x5 |
y1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
2 |
0 |
0 |
0 |
1 |
1 |
0 |
3 |
0 |
1 |
0 |
1 |
0 |
1 |
4 |
0 |
1 |
1 |
1 |
1 |
0 |
5 |
0 |
1 |
1 |
0 |
1 |
0 |
6 |
0 |
1 |
0 |
0 |
0 |
0 |
7 |
1 |
1 |
0 |
1 |
0 |
0 |
8 |
1 |
0 |
0 |
1 |
1 |
1 |
9 |
1 |
0 |
0 |
1 |
0 |
0 |
10 |
1 |
0 |
1 |
1 |
1 |
0 |