Задачи по дискретной математике для контрольных и самостоятельных работ. Булевы функции. Васильев А.В - 4 стр.

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2. По функциям f и g, заданным векторно, построить векторное представ-
ление функции h.
2.1. f = (0001), g = (1101), h(x
1
, x
2
, x
3
) = f(x
2
, g(x
1
, x
3
)) ∨ g(x
2
, x
3
)
2.2. f = (0111), g = (1101), h(x
1
, x
2
, x
3
) = f(g(x
1
, x
2
), x
1
) & g(x
1
, x
3
)
2.3. f = (1001), g = (1101), h(x
1
, x
2
, x
3
) = f(x
1
, x
2
) ⊕ g(f(x
1
, x
1
), x
3
)
2.4. f = (0110), g = (1101), h(x
1
, x
2
, x
3
) = f(x
1
, x
3
) ∼ g(x
2
, f(x
1
, x
1
))
2.5. f = (1110), g = (1101), h(x
1
, x
2
, x
3
) = f(x
1
, f(x
1
, x
1
)) → g(x
2
, x
3
)
2.6. f = (1000), g = (1101), h(x
1
, x
2
, x
3
) = f(f(x
2
, x
2
), x
1
) ↓ g(x
2
, x
3
)
2.7. f = (1011), g = (1101), h(x
1
, x
2
, x
3
) = f(x
1
, x
3
) | g(g(x
2
, x
3
), x
3
)
2.8. f = (0010), g = (1101), h(x
1
, x
2
, x
3
) = f(x
3
, x
1
) ∨ g(x
2
, g(x
3
, x
3
))
2.9. f = (0100), g = (1101), h(x
1
, x
2
, x
3
) = f(x
3
, g(x
2
, x
1
)) & g(x
2
, x
2
)
2.10. f = (0001), g = (0100), h(x
1
, x
2
, x
3
) = f(g(x
2
, x
3
), x
1
) ⊕ g(x
1
, x
2
)
2.11. f = (0111), g = (0100), h(x
1
, x
2
, x
3
) = f(x
3
, x
1
) → g(f(x
2
, x
3
), x
1
)
2.12. f = (1001), g = (0100), h(x
1
, x
2
, x
3
) = f(x
2
, f(x
3
, x
2
)) ↓ g(x
1
, x
3
)
2.13. f = (0110), g = (0100), h(x
1
, x
2
, x
3
) = f(f(x
1
, x
3
), x
3
) | g(x
3
, x
2
)
2.14. f = (1000), g = (0100), h(x
1
, x
2
, x
3
) = f(x
2
, x
1
) ∨ g(g(x
1
, x
1
), x
3
)
2.15. f = (0010), g = (0100), h(x
1
, x
2
, x
3
) = f(x
3
, x
3
) & g(x
3
, g(x
1
, x
2
))
2.16. f = (1110), g = (0100), h(x
1
, x
2
, x
3
) = f(x
1
, g(x
3
, x
2
)) ⊕ g(x
3
, x
1
)
2.17. f = (0111), g = (1110), h(x
1
, x
2
, x
3
) = f(g(x
3
, x
1
), x
2
) ∼ g(x
2
, x
1
)
2.18. f = (1001), g = (1110), h(x
1
, x
2
, x
3
) = f(x
3
, f(x
2
, x
3
)) → g(x
3
, x
1
)
2.19. f = (0110), g = (1110), h(x
1
, x
2
, x
3
) = f(x
2
, x
3
) ↓ g(f(x
3
, x
2
), x
1
)
2.20. f = (1000), g = (1110), h(x
1
, x
2
, x
3
) = f(x
3
, x
1
) | g(g(x
3
, x
3
), x
2
)
2.21. f = (0001), g = (1110), h(x
1
, x
2
, x
3
) = f(x
1
, g(x
1
, x
1
)) ∨ g(x
2
, x
3
)
2.22. f = (0010), g = (1110), h(x
1
, x
2
, x
3
) = f(f(x
2
, x
1
), x
3
) & g(x
1
, x
1
)
2.23. f = (0001), g = (0110), h(x
1
, x
2
, x
3
) = f(x
1
, x
2
) ⊕ g(x
3
, g(x
3
, x
3
))
2.24. f = (0111), g = (0110), h(x
1
, x
2
, x
3
) = f(x
2
, g(x
2
, x
3
)) ∼ g(x
1
, x
2
)
2.25. f = (1110), g = (0110), h(x
1
, x
2
, x
3
) = f(g(x
1
, x
2
), x
3
) → g(x
3
, x
2
)
2.26. f = (1000), g = (0110), h(x
1
, x
2
, x
3
) = f(x
1
, f(x
3
, x
1
)) ↓ g(x
1
, x
2
)
2.27. f = (0010), g = (0110), h(x
1
, x
2
, x
3
) = f(x
3
, x
1
) | g(f(x
1
, x
3
), x
2
)
2.28. f = (0001), g = (1001), h(x
1
, x
2
, x
3
) = f(f(x
3
, x
2
), x
1
) ∨ g(x
2
, x
2
)
2.29. f = (0111), g = (1001), h(x
1
, x
2
, x
3
) = f(x
2
, x
3
) ⊕ g(x
1
, g(x
1
, x
1
))
2.30. f = (1000), g = (1001), h(x
1
, x
2
, x
3
) = f(x
3
, g(x
3
, x
1
)) ∼ g(x
2
, x
3
)
4
2. По функциям f и g, заданным векторно, построить векторное представ-
   ление функции h.
 2.1. f = (0001), g = (1101), h(x1 , x2 , x3 ) = f (x2 , g(x1 , x3 )) ∨ g(x2 , x3 )
 2.2. f = (0111), g = (1101), h(x1 , x2 , x3 ) = f (g(x1 , x2 ), x1 ) & g(x1 , x3 )
 2.3. f = (1001), g = (1101), h(x1 , x2 , x3 ) = f (x1 , x2 ) ⊕ g(f (x1 , x1 ), x3 )
 2.4. f = (0110), g = (1101), h(x1 , x2 , x3 ) = f (x1 , x3 ) ∼ g(x2 , f (x1 , x1 ))
 2.5. f = (1110), g = (1101), h(x1 , x2 , x3 ) = f (x1 , f (x1 , x1 )) → g(x2 , x3 )
 2.6. f = (1000), g = (1101), h(x1 , x2 , x3 ) = f (f (x2 , x2 ), x1 ) ↓ g(x2 , x3 )
 2.7. f = (1011), g = (1101), h(x1 , x2 , x3 ) = f (x1 , x3 ) | g(g(x2 , x3 ), x3 )
 2.8. f = (0010), g = (1101), h(x1 , x2 , x3 ) = f (x3 , x1 ) ∨ g(x2 , g(x3 , x3 ))
 2.9. f = (0100), g = (1101), h(x1 , x2 , x3 ) = f (x3 , g(x2 , x1 )) & g(x2 , x2 )
2.10. f = (0001), g = (0100), h(x1 , x2 , x3 ) = f (g(x2 , x3 ), x1 ) ⊕ g(x1 , x2 )
2.11. f = (0111), g = (0100), h(x1 , x2 , x3 ) = f (x3 , x1 ) → g(f (x2 , x3 ), x1 )
2.12. f = (1001), g = (0100), h(x1 , x2 , x3 ) = f (x2 , f (x3 , x2 )) ↓ g(x1 , x3 )
2.13. f = (0110), g = (0100), h(x1 , x2 , x3 ) = f (f (x1 , x3 ), x3 ) | g(x3 , x2 )
2.14. f = (1000), g = (0100), h(x1 , x2 , x3 ) = f (x2 , x1 ) ∨ g(g(x1 , x1 ), x3 )
2.15. f = (0010), g = (0100), h(x1 , x2 , x3 ) = f (x3 , x3 ) & g(x3 , g(x1 , x2 ))
2.16. f = (1110), g = (0100), h(x1 , x2 , x3 ) = f (x1 , g(x3 , x2 )) ⊕ g(x3 , x1 )
2.17. f = (0111), g = (1110), h(x1 , x2 , x3 ) = f (g(x3 , x1 ), x2 ) ∼ g(x2 , x1 )
2.18. f = (1001), g = (1110), h(x1 , x2 , x3 ) = f (x3 , f (x2 , x3 )) → g(x3 , x1 )
2.19. f = (0110), g = (1110), h(x1 , x2 , x3 ) = f (x2 , x3 ) ↓ g(f (x3 , x2 ), x1 )
2.20. f = (1000), g = (1110), h(x1 , x2 , x3 ) = f (x3 , x1 ) | g(g(x3 , x3 ), x2 )
2.21. f = (0001), g = (1110), h(x1 , x2 , x3 ) = f (x1 , g(x1 , x1 )) ∨ g(x2 , x3 )
2.22. f = (0010), g = (1110), h(x1 , x2 , x3 ) = f (f (x2 , x1 ), x3 ) & g(x1 , x1 )
2.23. f = (0001), g = (0110), h(x1 , x2 , x3 ) = f (x1 , x2 ) ⊕ g(x3 , g(x3 , x3 ))
2.24. f = (0111), g = (0110), h(x1 , x2 , x3 ) = f (x2 , g(x2 , x3 )) ∼ g(x1 , x2 )
2.25. f = (1110), g = (0110), h(x1 , x2 , x3 ) = f (g(x1 , x2 ), x3 ) → g(x3 , x2 )
2.26. f = (1000), g = (0110), h(x1 , x2 , x3 ) = f (x1 , f (x3 , x1 )) ↓ g(x1 , x2 )
2.27. f = (0010), g = (0110), h(x1 , x2 , x3 ) = f (x3 , x1 ) | g(f (x1 , x3 ), x2 )
2.28. f = (0001), g = (1001), h(x1 , x2 , x3 ) = f (f (x3 , x2 ), x1 ) ∨ g(x2 , x2 )
2.29. f = (0111), g = (1001), h(x1 , x2 , x3 ) = f (x2 , x3 ) ⊕ g(x1 , g(x1 , x1 ))
2.30. f = (1000), g = (1001), h(x1 , x2 , x3 ) = f (x3 , g(x3 , x1 )) ∼ g(x2 , x3 )


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