Дискретка.Лекции, литература / Chast1_2012
.pdfКАЗАНСКИЙ (ПРИВОЛЖСКИЙ) ФЕДЕРАЛЬНЫЙ УНИВЕРСИТЕТ
Задачи по дискретной математике для контрольных и самостоятельных работ
Булевы функции
Издательство Казанского федерального университета
2012
ÓÄÊ 512.563.6
Печатается по решению методической комиссии Института ВМиИТ
Протокол 6 от 16 февраля 2012 г.
кафедры системного анализа и информационных технологий Казанского (Приволжского) федерального университета Протокол 6 от 15 февраля 2012 г.
Составители:
канд. физ.-мат. наук А.В. Васильев, докт. физ.-мат. наук, проф. Н.К. Замов, канд. физ.-мат. наук, доц. П.В. Пшеничный
Рецензент
канд. физ.-мат. наук, доц. Н.Н. Нурмеев
Задачи по дискретной математике для контрольных и самостоя-
тельных работ. Булевы функции: Учебный практикум / А.В. Васильев, Н.К. Замов, П.В. Пшеничный. 2-е изд., перераб. и доп. Казань: Издательство Казанского федерального университета, 2012. 57с.
Практикум предназначен для студентов, изучающих курс Дискретная математика , а также для преподавателей, ведущих практические занятия по данному курсу.
c Казанский федеральный университет, 2012
1.Выписать все пары соседних, противоположных и несравнимых наборов из множества A.
1.1.A = f(01101); (11100); (11000); (00111); (11111)g
1.2.A = f(01101); (01010); (10101); (01001); (11111)g
1.3.A = f(11110); (00001); (00110); (11111); (01111)g
1.4.A = f(11101); (00111); (01111); (01000); (10000)g
1.5.A = f(00001); (01000); (01001); (00111); (10111)g
1.6.A = f(01001); (11011); (01101); (10101); (00100)g
1.7.A = f(10011); (10101); (01100); (11010); (00100)g
1.8.A = f(00001); (10111); (10100); (11110); (01001)g
1.9.A = f(00101); (10101); (01011); (10010); (10100)g
1.10.A = f(11011); (10000); (11000); (00111); (10001)g
1.11.A = f(10111); (01001); (01101); (11100); (00011)g
1.12.A = f(01100); (10100); (11110); (10011); (00011)g
1.13.A = f(00000); (11011); (10111); (01010); (00100)g
1.14.A = f(01011); (00011); (11101); (11111); (00010)g
1.15.A = f(11000); (00111); (01111); (01100); (01011)g
1.16.A = f(00010); (11011); (01010); (00000); (11101)g
1.17.A = f(10110); (00001); (00011); (10111); (11110)g
1.18.A = f(01100); (10011); (11010); (01011); (01000)g
1.19.A = f(10011); (01100); (01001); (11101); (01011)g
1.20.A = f(00100); (01000); (10111); (11000); (11111)g
1.21.A = f(01011); (10100); (01111); (10110); (10000)g
1.22.A = f(10110); (11100); (01000); (00111); (10111)g
1.23.A = f(00111); (10011); (00100); (11000); (01100)g
1.24.A = f(00010); (00011); (10001); (11100); (10010)g
1.25.A = f(01010); (11110); (00000); (01111); (00001)g
1.26.A = f(11010); (10010); (10001); (01101); (11000)g
1.27.A = f(11100); (11001); (01011); (10100); (11110)g
1.28.A = f(11100); (00001); (10110); (00011); (10011)g
1.29.A = f(10010); (11000); (01000); (00101); (11010)g
1.30.A = f(10010); (00001); (11001); (01101); (01111)g
3
2.Выписать все пары соседних, противоположных и несравнимых наборов из множества A.
2.1.A = f(10001); (10101); (01010); (11011); (00000); (01110); (11001); (00011)g
2.2.A = f(11101); (00010); (10111); (00110); (10001); (01111); (00001); (00101)g
2.3.A = f(11110); (00010); (00100); (10111); (00110); (01000); (01110); (10011)g
2.4.A = f(00010); (11010); (01001); (10000); (10010); (11110); (11011); (00001)g
2.5.A = f(11001); (10100); (10011); (10110); (01000); (11010); (00100); (10111)g
2.6.A = f(00100); (01111); (10101); (10111); (10010); (01101); (01110); (11001)g
2.7.A = f(01110); (10100); (00001); (11000); (00000); (10111); (10001); (11100)g
2.8.A = f(10100); (11100); (01011); (00001); (11000); (10110); (10011); (00000)g
2.9.A = f(11101); (01101); (00011); (11100); (11110); (01010); (01011); (00111)g
2.10.A = f(00110); (11100); (00111); (10000); (10111); (01000); (10101); (00100)g
2.11.A = f(10000); (01111); (10100); (10110); (01010); (10011); (11010); (10111)g
2.12.A = f(00101); (01100); (01111); (00011); (11111); (11100); (11010); (01110)g
2.13.A = f(10110); (11111); (01000); (01001); (00001); (11101); (11001); (00011)g
2.14.A = f(11111); (01100); (10110); (10011); (00100); (01000); (10101); (11010)g
2.15.A = f(00101); (01000); (11111); (11001); (01011); (10100); (01111); (11000)g
2.16.A = f(10011); (10010); (11011); (10000); (00000); (11001); (11111); (01011)g
2.17.A = f(11110); (11011); (10101); (01110); (10001); (01101); (11101); (00100)g
2.18.A = f(10111); (01110); (11000); (01010); (11100); (00000); (00111); (00101)g
2.19.A = f(00111); (11010); (00001); (11110); (11100); (11111); (00010); (00000)g
2.20.A = f(01011); (11100); (01000); (10101); (10111); (01101); (01111); (00111)g
2.21.A = f(01010); (01110); (10111); (10100); (11101); (10101); (00010); (11100)g
2.22.A = f(01001); (11010); (11100); (00010); (11101); (00001); (11110); (00111)g
2.23.A = f(01011); (11110); (01000); (01110); (10100); (00010); (01001); (00111)g
2.24.A = f(00000); (00111); (10111); (11110); (10001); (01110); (01101); (11100)g
2.25.A = f(01000); (10111); (00111); (00000); (11100); (11111); (00011); (11000)g
2.26.A = f(11000); (11010); (00010); (10010); (11101); (01101); (00101); (10110)g
2.27.A = f(00100); (01010); (10001); (01100); (00011); (10111); (11100); (01011)g
2.28.A = f(01000); (00011); (11110); (11101); (01111); (10000); (01001); (00001)g
2.29.A = f(10010); (01101); (11101); (11111); (01000); (01111); (01011); (01010)g
2.30.A = f(10011); (10000); (01011); (10110); (01000); (01001); (01101); (11100)g
4
3.Построить таблицу функции от 4 переменных, которая равна 1 на наборах вида ( 1; 2; 3; 4), ãäå
3.1.1 _ 2 = 3 4
3.2.1 _ 2 = 3 4
3.3.1 _ 2 = 3 4
3.4.1 _ 2 = 3 ! 4
3.5.1 _ 2 = 3 j 4
3.6.1 _ 2 = 3 # 4
3.7.1 2 = 3 4
3.8.1 2 = 3 4
3.9.1 2 = 3 ! 4
3.10.1 2 = 3 j 4
3.11.1 2 = 3 # 4
3.12.1 2 = 3 4
3.13.1 2 = 3 ! 4
3.14.1 2 = 3 j 4
3.15.1 2 = 3 # 4
3.16.1 2 = 3 ! 4
3.17.1 2 = 3 j 4
3.18.1 2 = 3 # 4
3.19.1 ! 2 = 3 j 4
3.20.1 ! 2 = 3 # 4
3.21.1 j 2 = 3 # 4
3.22.1 _ 2 > 3 4
3.23.1 2 > 3 j 4
3.24.1 2 > 3 # 4
3.25.1 j 2 > 3 ! 4
3.26.1 2 > 3 4
3.27.1 ! 2 3 4
3.28.1 2 3 _ 4
3.29.1 j 2 3 4
3.30.1 # 2 3 4
5
4. По таблице для функции f(x; y; z) построить таблицу для функции
f(z; x; y).
4.1.f(x; y; z) = (01000111)
4.2.f(x; y; z) = (00101011)
4.3.f(x; y; z) = (11001101)
4.4.f(x; y; z) = (00000100)
4.5.f(x; y; z) = (11101100)
4.6.f(x; y; z) = (11010010)
4.7.f(x; y; z) = (11001000)
4.8.f(x; y; z) = (01111001)
4.9.f(x; y; z) = (11011110)
4.10.f(x; y; z) = (01111100)
4.11.f(x; y; z) = (00001011)
4.12.f(x; y; z) = (01111111)
4.13.f(x; y; z) = (01100111)
4.14.f(x; y; z) = (11011101)
4.15.f(x; y; z) = (10001001)
4.16.f(x; y; z) = (00000111)
4.17.f(x; y; z) = (10001110)
4.18.f(x; y; z) = (10010101)
4.19.f(x; y; z) = (10000110)
4.20.f(x; y; z) = (00000010)
4.21.f(x; y; z) = (00001111)
4.22.f(x; y; z) = (11010111)
4.23.f(x; y; z) = (11010000)
4.24.f(x; y; z) = (11101010)
4.25.f(x; y; z) = (11011011)
4.26.f(x; y; z) = (01110101)
4.27.f(x; y; z) = (11110010)
4.28.f(x; y; z) = (01010000)
4.29.f(x; y; z) = (10000010)
4.30.f(x; y; z) = (11010101)
6
5.По функциям f è g, заданным векторно, построить векторное представление функции h.
5.1.f = (0001); g = (1101); h(x1; x2; x3) = f(x2; g(x1; x3)) _ g(x2; x3)
5.2.f = (0111); g = (1101); h(x1; x2; x3) = f(g(x1; x2); x1) & g(x1; x3)
5.3.f = (1001); g = (1101); h(x1; x2; x3) = f(x1; x2) g(f(x1; x1); x3)
5.4.f = (0110); g = (1101); h(x1; x2; x3) = f(x1; x3) g(x2; f(x1; x1))
5.5.f = (1110); g = (1101); h(x1; x2; x3) = f(x1; f(x1; x1)) ! g(x2; x3)
5.6.f = (1000); g = (1101); h(x1; x2; x3) = f(f(x2; x2); x1) # g(x2; x3)
5.7.f = (1011); g = (1101); h(x1; x2; x3) = f(x1; x3) j g(g(x2; x3); x3)
5.8.f = (0010); g = (1101); h(x1; x2; x3) = f(x3; x1) _ g(x2; g(x3; x3))
5.9.f = (0100); g = (1101); h(x1; x2; x3) = f(x3; g(x2; x1)) & g(x2; x2)
5.10.f = (0001); g = (0100); h(x1; x2; x3) = f(g(x2; x3); x1) g(x1; x2)
5.11.f = (0111); g = (0100); h(x1; x2; x3) = f(x3; x1) ! g(f(x2; x3); x1)
5.12.f = (1001); g = (0100); h(x1; x2; x3) = f(x2; f(x3; x2)) # g(x1; x3)
5.13.f = (0110); g = (0100); h(x1; x2; x3) = f(f(x1; x3); x3) j g(x3; x2)
5.14.f = (1000); g = (0100); h(x1; x2; x3) = f(x2; x1) _ g(g(x1; x1); x3)
5.15.f = (0010); g = (0100); h(x1; x2; x3) = f(x3; x3) & g(x3; g(x1; x2))
5.16.f = (1110); g = (0100); h(x1; x2; x3) = f(x1; g(x3; x2)) g(x3; x1)
5.17.f = (0111); g = (1110); h(x1; x2; x3) = f(g(x3; x1); x2) g(x2; x1)
5.18.f = (1001); g = (1110); h(x1; x2; x3) = f(x3; f(x2; x3)) ! g(x3; x1)
5.19.f = (0110); g = (1110); h(x1; x2; x3) = f(x2; x3) # g(f(x3; x2); x1)
5.20.f = (1000); g = (1110); h(x1; x2; x3) = f(x3; x1) j g(g(x3; x3); x2)
5.21.f = (0001); g = (1110); h(x1; x2; x3) = f(x1; g(x1; x1)) _ g(x2; x3)
5.22.f = (0010); g = (1110); h(x1; x2; x3) = f(f(x2; x1); x3) & g(x1; x1)
5.23.f = (0001); g = (0110); h(x1; x2; x3) = f(x1; x2) g(x3; g(x3; x3))
5.24.f = (0111); g = (0110); h(x1; x2; x3) = f(x2; g(x2; x3)) g(x1; x2)
5.25.f = (1110); g = (0110); h(x1; x2; x3) = f(g(x1; x2); x3) ! g(x3; x2)
5.26.f = (1000); g = (0110); h(x1; x2; x3) = f(x1; f(x3; x1)) # g(x1; x2)
5.27.f = (0010); g = (0110); h(x1; x2; x3) = f(x3; x1) j g(f(x1; x3); x2)
5.28.f = (0001); g = (1001); h(x1; x2; x3) = f(f(x3; x2); x1) _ g(x2; x2)
5.29.f = (0111); g = (1001); h(x1; x2; x3) = f(x2; x3) g(x1; g(x1; x1))
5.30.f = (1000); g = (1001); h(x1; x2; x3) = f(x3; g(x3; x1)) g(x2; x3)
7
6.Построить таблицу функции, заданной формулой.
6.1.((x1 j x2) j x3) ! x1
6.2.(x1 j x2) _ x2x3
6.3.((x1 x2) x2) j x3
6.4.x1 _ (x2x3 x1)
6.5.x1 # (x2 # (x2 j x3))
6.6.((x1 x2) j x1)x3
6.7.(x1 x2) _ (x3 # x2)
6.8.((x1 _ x2) x2) x3
6.9.x1 _ (x2 # x1x3)
6.10.(x1 _ x2) (x3 # x1)
6.11.(x1 x2) # (x3 _ x1)
6.12.(x1 j x2x3)x1
6.13.(x1 ! x2) # (x1 x3)
6.14.x1(x1 # x2)x3
6.15.x1 ((x2 j x3) x1)
6.16.x1 j (x2x3 x3)
6.17.x1 # ((x2 x3) # x1)
6.18.(x1 (x2 # x3)) x1
6.19.x1 x2x3 x3
6.20.x1 _ x2(x1 x3)
6.21.(x1 # x2)(x2 ! x3)
6.22.x1 j (x2x1 x3)
6.23.(x1 # x2)(x2 # x3)
6.24.x1 j ((x2 x3) x3)
6.25.(x1 j x2) (x2 # x3)
6.26.x1 ! ((x1 _ x2) ! x3)
6.27.(x1 _ (x1 # x2)) j x3
6.28.(x1 j x2) j (x3 ! x2)
6.29.(x1 # (x2 x3)) _ x2
6.30.((x1 j x2) x3)x1
8
7.Построить дèаграмму, характеризующую строение формулы над множеством = f ; &; _; ; ; !; j; #g. Указание: в случае ассоциативных операций применять группировку влево.
7.1.((x2 # x3) ! x1) j ((x3x1) x1x3)
7.2.(((x1x2) _ x1) # (x2 ! x3)) (x3x2)
7.3.(x3x1) (x3 _ x2) (x3 j (x3 # x1))
7.4.(x1 _ x2) (((x1x2) _ x2) j (x2 x1))
7.5.((x2x3) x3) ! ((x2 x1) ! x1 _ x2)
7.6.((x1 (x3 # x1)) # (x1 x3))(x1 j x2)
7.7.((x3 x2 x3) j (x1 x2)) j (x3 # x1)
7.8.x2x1(((x3 x2) j x2) # (x3 # x1))
7.9.x2(x3 j x2)(x1 (x2 x3) x2)
7.10.(((x3 x2) ! x1) ((x3 j x2) j x3)) _ x2
7.11.x3 x1 ((x2 x3)(x3 (x2 x3)))
7.12.(x3x1 _ x2) (((x2 x1) # x3) x3)
7.13.(x2 (x2 j (x1 _ x3))) j ((x2 # x1) x1)
7.14.(x3 ! x1) ((x1 # x2) j ((x1 ! x3) j x3))
7.15.(x2x3x2) ((x2 x3) (x1 ! x3))
7.16.((x1 (x2 x3)) j x1 x2) (x1 _ x2)
7.17.(x1 x2 (x2 (x2x1))) j (x2 j x3)
7.18.(((x2 _ x3)x3) _ x2) ! (x1 _ (x2 x3))
7.19.((x3 _ x1)(x1 j (x2 x3))) (x3 j x1)
7.20.(((x3 x2) # x3)x2) (x1 # (x2 # x3))
7.21.((x2 # x3) # x1) j ((x2 j x1) # (x2 x3))
7.22.(x2 x3 x1) ! (x1 _ x3) j (x1 x2)
7.23.(((x3 j x1)x1) ! (x3 # x1)) (x3 # x2)
7.24.((x3 x1) # x3) j ((x3 ! (x3 x2)) j x3)
7.25.x1 j (((x2 x3) (x1 # x2)) (x3 # x1))
7.26.(x1((x2 ! x1) x3)) # (x1 # (x3 j x1))
7.27.x2 j x1 # ((x2 x3) _ ((x3x1) ! x1))
7.28.(x1 ! x3)x2 ! (x2 ! (x3 ! (x3 j x2)))
7.29.((x2 x3) _ x1) (x1 ! ((x3 x1)x2))
7.30.((x3 # (x3 x2)) ! (x1 _ x3)) # (x2 x1)
9
8. По диаграмме восстановить формулу.
|
|
|
|
|
|
|
|
|
_ |
|
|
! |
|
8.1. |
j |
|
x3 |
& |
x2 |
8.5. |
|
x3 |
x1 |
! |
x1 |
|
|
|
|
x2 |
x3 |
|
||
|
|
|
|
|
|
|
|
! |
|
|
! |
|
|
8.2. |
x2 x1 |
x3 |
! |
8.6. |
||
|
|
j |
|
|
|
|
|
|
x2 |
|
x1 |
x1 |
x3 |
|
|
|
|
|
|
|
|
_ |
|
|
|
|
|
8.3. |
x2 |
|
|
# |
x2 |
8.7. |
|
x2 |
x3 |
j |
|
x2 |
|
|
|
x3 |
x2 |
|
||
|
|
|
|
j |
|
|
|
|
& |
|
|
|
|
8.4. |
|
|
x2 |
x3 |
# |
8.8. |
|
& |
x2 |
|
|
x3 |
x1 |
|
|
|
|
|||
|
x2 x3 |
|
|
|
|
|
|
|
|
|
|
! |
|
|
|
|
|
x2 x3 |
& |
|
|
_ |
|
x2 |
x3 |
|
x3 |
|
|
|
|
|
|
x1 |
x3 |
|
|
|
|
# |
|
|
! |
|
|
|
! |
! |
|
|
x3 |
x2 |
|
x2 |
_ x3 |
|
x1 |
|
|
x1 |
x3 |
|
! |
|
|
|
|
|
|
||
|
j |
|
|
# |
|
_ |
x1 |
|
x3 |
||
_ |
x2 |
x1 |
x3 |
|
|
x2 |
x1 |
|
|
! |
|
|
|
|
|
|
|
|
! |
|
|
|
|
& |
|
|
# x3 |
x2 |
|
x3 |
x2 x2 |
|
& |
|
|
|
|
|
|||
|
|
x3 |
x2 |
|
10