Download - MAPAS DE KARNAUGHT
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Maurice Karnaugh
Ingeniero de Telecomunicaciones • AT&T Bell.
• 1953 Inventa el mapa-K o mapa de Karnaugh.
• Minimización de funciones por inspección visual.
Minimización de Funciones Booleanas
Mapas de Karnaugh
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Tabla o mapa de Karnaugh, Kmap
Procedimiento gráfico para la simplificación de funciones
algebraicas de un número de variables relativamente
pequeño
(en la práctica se puede utilizar para funciones de hasta seis variables).
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Tabla o mapa de Karnaugh
Un diagrama o
mapa de Karnaugh
es una tabla de
verdad dispuesta de
manera adecuada
para determinar por
inspección la
expresión mínima
de suma de
productos de una
función lógica.
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La factorización se efectúa cuando solo cambia una variable entre dos términos y esta variable se elimina
Con 2 variables A y B se pueden tener 4 Términos
Cada termino de dos variables tiene dos posibilidades de factorización
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Kmap para 2 variables
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Mapa de Karnaugh para dos variables
A’B’ AB’
A’B AB
m0 m2
m1 m3
0 2
1 3
0 1
0
1
AB
A
B
m A B S
0 0 0
1 0 1
2 1 0 AB’
3 1 1
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Kmap para 2 variables
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Kmap para 2 variables
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Como llenar el Kmap para 2 variables
1
0
1
1
F1 (A,B) = A’ B’ + A B’ + A B
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Como resolver Kmap para 2 variables
F1(A,B)= A
1
+ B’
0
F1 (A,B) = A’ B’ + A B’ + A B
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Kmap para 3 variables
Con 3 Variables se tienen 8 términos
y cada termino tiene 3 posibilidades
de factorización
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Kmap para 3 variables
Cada termino tiene 3 posibilidades de factorización
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Kmap para 3 variables
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Mapa de Karnaugh para 3 variables
A’B’C’ A’BC’ ABC’ AB’C’
A’B’C A’BC ABC AB’C
00 01 11 10
0
1
ABC
0 2 6 4
1 3 7 5
00 01 11 10
0
1
ABC
La idea con la codificación es poder usar el P9a. ab+ab’=a
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Mapa de Karnaugh para 3 variables
A’B’C’ A’BC’ ABC’ AB’C’
A’B’C A’BC ABC AB’C
00 01 11 10
0
1
ABC
0 2 6 4
1 3 7 5
00 01 11 10
0
1
ABC
La idea con la codificación es poder usar el P9a. ab+ab’=a
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Kmap para 3 variables
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Kmap para 3 variables
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F (A, B, C) = A´ B C’ + A B´ C´ + A B C’ + A B C
Kmap para 3 variables
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Kmap para 3 variables
F (A, B, C) = A´ B C’ + A B´ C´ + A B C’ + A B C
A´ B C’
1
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F (A, B, C) = A´ B C’ + A B´ C´ + A B C’ + A B C
A B´ C´
1 1
Kmap para 3 variables
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F (A, B, C) = A´ B C’ + A B´ C´ + A B C’ + A B C
A B C´
1 11
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F (A, B, C) = A´ B C’ + A B´ C´ + A B C’ + A B C
A B C
1 11
1
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F (A, B, C) = A´ B C’ + A B´ C´ + A B C’ + A B C
1 11
1
F (A, B, C) = B
11
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F (A, B, C) = A´ B C’ + A B´ C´ + A B C’ + A B C
1 11
1
11
0
F (A, B, C) = B C’ +
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F (A, B, C) = A´ B C’ + A B´ C´ + A B C’ + A B C
1 11
1
F (A, B, C) = B C’ +
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F (A, B, C) = A´ B C’ + A B´ C´ + A B C’ + A B C
1 11
1
F (A, B, C) = B C’ +
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Kmap para 3 variables
F (A, B, C) = A´ B C’ + A B´ C´ + A B C’ + A B C
1 11
1
F (A, B, C) = B C’ +
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Kmap para 3 variables
F (A, B, C) = A´ B C’ + A B´ C´ + A B C’ + A B C
1 11
1
F (A, B, C) = B C’ +
1 1
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Kmap para 3 variables
F (A, B, C) = A´ B C’ + A B´ C´ + A B C’ + A B C
1 11
1
F (A, B, C) = B C’ + A
1 1
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Kmap para 3 variables
F (A, B, C) = A´ B C’ + A B´ C´ + A B C’ + A B C
1 11
1
F (A, B, C) = B C’ + A
0
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Kmap para 3 variables
F (A, B, C) = A´ B C’ + A B´ C´ + A B C’ + A B C
1 11
1
F (A, B, C) = B C’ + A C’
0
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Kmap para 3 variables
F (A, B, C) = A´ B C’ + A B´ C´ + A B C’ + A B C
1 11
1
F (A, B, C) = B C’ + A C´ +
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Kmap para 3 variables
F (A, B, C) = A´ B C’ + A B´ C´ + A B C’ + A B C
1 11
1
F (A, B, C) = B C’ + A C´ +
11
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Kmap para 3 variables
F (A, B, C) = A´ B C’ + A B´ C´ + A B C’ + A B C
1 11
1
F (A, B, C) = B C’ + A C´ + A B
11
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Kmap para 4 variables
Con 4 Variables se tienen 16 términos
y cada termino tiene 4 posibilidades
de factorización
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Kmap para 4 variables
Cada termino tiene 4 posibilidades de factorización
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Cada termino tiene 4 posibilidades de factorización
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K map para 4 variables
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AB00 01 11 1010
K map para 4 variables
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Kmap para 4 variables
AB
CD
00 01 11
00
01
11
1010
10
10
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Kmap para 4 variables
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Mapa de Karnaugh para 4 variables
A’B’C’D’ A’BC’D’ ABC’D’ AB’C’D’
A’B’C’D A’BC’D ABC’D AB’C’D
A’B’CD A’BCD ABCD AB’CD
A’B’CD’ A’BCD’ ABCD’ AB’CD’
00 01 11 10
00
01
11
10
ABCD
0 4 12 8
1 5 13 9
3 7 15 11
2 6 14 10
00 01 11 10
00
01
11
10
ABCD
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Kmap para 4 variables
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Kmap para 5 variables
Con 5 Variables se tienen 32 términos
y cada termino tiene 5 posibilidades
de factorización
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Kmap para 5variables
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Kmap para 5variables
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Kmap para 5 variables
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Kmap para 6 variables
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Reglas para el uso del Kmap
1.- Formar el menor numero de grupos2.- Cada grupo lo mas grande posible3.- Todos los unos deberán de ser agrupados4.- Un solo uno puede formar un grupo5.- Casillas de un grupo pueden formar parte de otro grupo
Grupo = Unos adyacentes enlazados (paralelogramos) en una cantidad igual a una potencia entera de dos, eje. (1, 2, 4, 8,
…).
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Ejemplos del Kmap
m X Y F0 0 0 11 0 1 12 1 0 03 1 1 1
F 0
1F (X, Y)= X’ + Y
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ejemplos del Kmap
F2(X, Y, Z) =Σm(1, 2, 5, 7)
1
1 1 1
00
0
0
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F2(X, Y, Z) =Σm(1, 2, 5, 7)
1
1 1 1
00
0
0
F2(X, Y, Z) = X Z
1 1
1
+
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+ Y’
F2(X, Y, Z) =Σm(1, 2, 5, 7)
1
1 1 1
00
0
0
F2(X, Y, Z) = X Z Z
00
1
+
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Y’
F2(X, Y, Z) =Σm(1, 2, 5, 7)
1
1 1 1
00
0
0
F2(X, Y, Z) = X Z Z + X’ Y Z’
01
0
+
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FX(A, B, C, D) =Σm(3, 4, 5, 7, 9, 13, 14, 15)
1
1
1
1
11
1
1
000
0
0
0 0 0
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FX(A, B, C, D) =Σm(3, 4, 5, 7, 9, 13, 14, 15)
1
1
1
1
11
1
1
000
0
0
0 0 0
FX(A, B, C, D) = A’ B
01
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FX(A, B, C, D) =Σm(3, 4, 5, 7, 9, 13, 14, 15)
1
1
1
1
11
1
1
000
0
0
0 0 0
FX(A, B, C, D) = A’ B
0
0
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FX(A, B, C, D) =Σm(3, 4, 5, 7, 9, 13, 14, 15)
1
1
1
1
11
1
1
000
0
0
0 0 0
FX(A, B, C, D) = A’ B C’
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FX(A, B, C, D) =Σm(3, 4, 5, 7, 9, 13, 14, 15)
1
1
1
1
11
1
1
000
0
0
0 0 0
FX(A, B, C, D) = A’ B C’ + A
01
C’D
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FX(A, B, C, D) =Σm(3, 4, 5, 7, 9, 13, 14, 15)
1
1
1
1
11
1
1
000
0
0
0 0 0
FX(A, B, C, D) = A’ B C’ + A
01
C’D
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FX(A, B, C, D) =Σm(3, 4, 5, 7, 9, 13, 14, 15)
1
1
1
1
11
1
1
000
0
0
0 0 0
FX(A, B, C, D) = A’ B C’ + A C’D+
0
A’
0
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FX(A, B, C, D) =Σm(3, 4, 5, 7, 9, 13, 14, 15)
1
1
1
1
11
1
1
000
0
0
0 0 0
FX(A, B, C, D) = A’ B C’ + A C’D+ A’
11
CD
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FX(A, B, C, D) =Σm(3, 4, 5, 7, 9, 13, 14, 15)
1
1
1
1
11
1
1
000
0
0
0 0 0
FX(A, B, C, D) = A’ B C’ + A C’D+ A’CD+
11
AB
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FX(A, B, C, D) =Σm(3, 4, 5, 7, 9, 13, 14, 15)
1
1
1
1
11
1
1
000
0
0
0 0 0
FX(A, B, C, D) = A’ B C’ + A C’D+ A’CD+AB
1
C
1
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1
1
1
1
11
1
1
000
0
0
0 0 0
FX(A, B, C, D) = A’ B C’ + A C’D+ A’CD+ABC
1.- Formar el menor número de grupos
2.- Cada grupo lo más grande posible
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0
0
0
0
0
FY(A, B, C, D) = Π m (1, 3, 8, 9, 11, 12, 14)
0
0
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0
1
1
1
01
0
1
001
0
0
1 1 1
FY(A, B, C, D) = Π m (1, 3, 8, 9, 11, 12, 14)
1.- Formar el menor número de grupos
2.- Cada grupo lo más grande posible
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0
1
1
1
01
0
1
001
0
0
1 1 1
FY(A, B, C, D) = Π m (1, 3, 8, 9, 11, 12, 14)
FY(A, B, C, D) =
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0
1
1
1
01
0
1
001
0
0
1 1 1
FY(A, B, C, D) = Π m (1, 3, 8, 9, 11, 12, 14)
FY(A, B, C, D) =
00
B’
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0
1
1
1
01
0
1
001
0
0
1 1 1
FY(A, B, C, D) = Π m (1, 3, 8, 9, 11, 12, 14)
FY(A, B, C, D) = B’
10
CD’
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0
1
1
1
01
0
1
001
0
0
1 1 1
FY(A, B, C, D) = Π m (1, 3, 8, 9, 11, 12, 14)
FY(A, B, C, D) = B’ C D’ +
1 1
B
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0
1
1
1
01
0
1
001
0
0
1 1 1
FY(A, B, C, D) = Π m (1, 3, 8, 9, 11, 12, 14)
FY(A, B, C, D) = B’ C D’ + B
1
1
D
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0
1
1
1
01
0
1
001
0
0
1 1 1
FY(A, B, C, D) = Π m (1, 3, 8, 9, 11, 12, 14)
FY(A, B, C, D) = B’ C D’ + B D +
0 0
A’
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0
1
1
1
01
0
1
001
0
0
1 1 1
FY(A, B, C, D) = Π m (1, 3, 8, 9, 11, 12, 14)
FY(A, B, C, D) = B’ C D’ + B D + A’
0
0
D’
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0
1
1
1
01
0
1
001
0
0
1 1 1
FY(A, B, C, D) = Π m (1, 3, 8, 9, 11, 12, 14)
FY(A, B, C, D) = B’ C D’ + B D + A’ D’
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F3(A, B, C, D) =Σm(0,2,5,6,7,8,12,14)
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F3(A, B, C, D) =Σm(0,2,5,6,7,8,12,14)
F3= A'B'D' + A C'D' + A'B D + B C D‘F3= B'C'D' + A'C D' + A'B D + A B D'
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F4(A, B, C) =Πm(2, 7)
0
0
1 1 1
1 1 1
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Reglas para el uso del Kmap1.- Formar el menor numero de grupos.2.- Cada grupo lo mas grande posible.3.- Todos los unos deberán de ser agrupados.4.- Un solo uno puede formar un grupo.5.- Casillas de un grupo pueden formar parte de otro grupo.
Grupo = Unos adyacentes enlazados (paralelogramos) en una cantidad igual a una potencia entera de dos ejemplo (1, 2, 4,
8,…).
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F5(X, Y, Z, W) =Σm(0,2,7,8,10,12,13,14)F6(A, B, C, D) =Πm(0,15)F7(A, B, C, D) =Πm(9, 11,15)F8(X, Y, Z, W) =Σm(0,2,3,5,6,7,8,10,11,14,15)F9 ( A,B,C,D )= Πm ( 2, 5, 7, 13, 15)F10 ( X,Y,Z,W )= Σm ( 5, 13, 15)F11 (X, Y, Z, W ) = X Y’ + X Y W’ + X’ Y’ W + X’ Y’ Z’ W’ F12 ( X,Y,Z,W )= Σm ( 4,7,9,10,12,13,14,15) F13 ( X,Y,Z,W )= Σm ( 1, 3, 6, 7, 9, 11, 12) F14 (A,B,C,D) = Σm ( 3,5,6,7, 9,10,11,12,13,14)F15 (A,B,C,D) =(B’+C+D)(B’+C’+D)(A’+B’+C’+D’)(A’+B +C+D’) F16 (A,B,C,D) = Σm ( 0, 2, 4, 5, 6, 7, 8, 9, 10, 13, 15)F17 (A,B,C,D) = Σm ( 0, 1, 2, 3, 5, 8, 9, 10, 13, 14, 15)
La mejor forma de Huir de un problema es resolverlo.
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F5(X, Y, Z, W) =Σm(0,2,7,8,10,12,13,14)
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F6(A, B, C, D) =Πm(5,15)
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F6(A, B, C, D) =Πm(5,15)
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F7(A, B, C, D) =Πm(9, 11,15)
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F7(A, B, C, D) =Πm(5, 7,15)Agrupando ceros POS
F7(A, B, C, D)=(B'+C'+D')(A+B'+D') (POS)
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+Y’ W’+Z
F8(X, Y, Z, W) =Σm(0,2,3,5,6,7,8,10,11,14,15)
1
1
1
1
0
1
1
0
1
1
0
0
1
1
0
1
F8(X, Y, Z, W)=X’YW
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F8(X, Y, Z, W) =Σm(0,2,3,5,6,7,8,10,11,14,15)
1
1
110
1101
1
00
110
1
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F8(X, Y, Z, W) =Σm(0,2,3,5,6,7,8,10,11,14,15)
1
1
110
1101
1
00
110
1
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F9 (A,B,C,D )= Πm ( 2, 5, 7, 13, 15)
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F9 (A,B,C,D )= Πm ( 2, 5, 7, 13, 15)
0
1
001
1111
1
01
111
0
F9 = B D' + B'D + A D' + C'D' F9 = B D' + B'D + A D' + B'C' F9 = B D' + B'D + A B' + C'D' F9 = B D' + B'D + A B' + B'C' ***********************************F9 = (B'+ D') (A + B + C'+ D )
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F9 (A,B,C,D )= Πm ( 2, 5, 7, 13, 15)
0
1
001
1111
1
01
111
0
F9 = B D' + B'D + A D' + C'D' F9 = B D' + B'D + A D' + B'C' F9 = B D' + B'D + A B' + C'D' F9 = B D' + B'D + A B' + B'C' ***********************************F9 = (B'+ D') (A + B + C'+ D )
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F10 ( X,Y,Z,W )= Σm ( 4,7,9,10,12,13,14,15)
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F11 (X, Y, Z, W ) = X Y’ + X Y W’ + X’ Y’ W + X’ Y’ Z’ W’
F11X, Y
Z, W
X Y’1
1
1
1
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F11 (X, Y, Z, W ) = X Y’ + X Y W’ + X’ Y’ W + X’ Y’ Z’ W’
F11X, Y
Z, W
1
1
1
1
X Y W’1
1
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F11 (X, Y, Z, W ) = X Y’ + X Y W’ + X’ Y’ W + X’ Y’ Z’ W’
F11X, Y
Z, W
1
1
1
1
X’ Y’ W1
1
1
1
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F11 (X, Y, Z, W ) = X Y’ + X Y W’ + X’ Y’ W + X’ Y’ Z’ W’
F11X, Y
Z, W
1
1
1
1
X’ Y’ Z’ W’
1
1
1
1
1
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F11 (X, Y, Z, W ) = X Y’ + X Y W’ + X’ Y’ W + X’ Y’ Z’ W’
F11X, Y
Z, W
1
1
1
11
1
1
1
1
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F11 (X, Y, Z, W ) = X Y’ + X Y W’ + X’ Y’ W + X’ Y’ Z’ W’
F11X, Y
Z, W
1
1
1
11
1
1
1
1 0
0
0
0
0
0
0
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F11 (X, Y, Z, W ) = X Y’ + X Y W’ + X’ Y’ W + X’ Y’ Z’ W’
F11X, Y
Z, W
1
1
1
1
1
1
1
1
1
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F12 ( X,Y,Z,W )= Σm ( 1, 3, 6, 7, 9, 11, 12)
F12X,Y
Z,W1
1
1
1
1
1
1
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F12 ( X,Y,Z,W )= Σm ( 1, 3, 6, 7, 9, 11, 12)
F12X,Y
Z,W1
1
1
1
1
1
10 0
0
0
0
0
0
0 0
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F13 (A,B,C,D) = Σm (3,5,6,7, 9,10,11,12,13,14)
F13
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M A B C D P0 0 0 0 0 11 0 0 0 1 12 0 0 1 0 13 0 0 1 1 04 0 1 0 0 15 0 1 0 1 06 0 1 1 0 07 0 1 1 1 08 1 0 0 0 19 1 0 0 1 010 1 0 1 0 011 1 0 1 1 012 1 1 0 0 013 1 1 0 1 014 1 1 1 0 015 1 1 1 1 0
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M A B C D S0 0 0 0 0 11 0 0 0 1 12 0 0 1 0 13 0 0 1 1 14 0 1 0 0 15 0 1 0 1 06 0 1 1 0 07 0 1 1 1 08 1 0 0 0 19 1 0 0 1 110 1 0 1 0 011 1 0 1 1 012 1 1 0 0 113 1 1 0 1 014 1 1 1 0 015 1 1 1 1 0
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F14 ( A, B , C ,D)= Σm(4, 8, 9, 10, 11, 12, 14, 15)
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F15 (A,B,C,D) =(B’+C+D)(B’+C’+D)(A’+B’+C’+D’)(A’+B +C+D’)
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F15 (A,B,C,D) =(B’+C+D)(B’+C’+D)(A’+B’+C’+D’)(A’+B +C+D’)
0
0
0
0
0
0
1
1
1
1
1
1
1
1 1
1