How many boolean functions of $F(x,y,z) = F(x',y,z') + F(x,y',z)$?
$begingroup$
How to find total number of boolean functions of the equation :
$F(x,y,z) = F(x',y,z') + F(x,y',z)$
Is there any procedure for this ?
functions discrete-mathematics boolean-algebra
asked Dec 9 '16 at 9:06
Jon GarrickJon Garrick
1,246828
$endgroup$
add a comment |
$begingroup$
How to find total number of boolean functions of the equation :
$F(x,y,z) = F(x',y,z') + F(x,y',z)$
Is there any procedure for this ?
functions discrete-mathematics boolean-algebra
asked Dec 9 '16 at 9:06
Jon GarrickJon Garrick
1,246828
$endgroup$
add a comment |
$begingroup$
How to find total number of boolean functions of the equation :
$F(x,y,z) = F(x',y,z') + F(x,y',z)$
Is there any procedure for this ?
functions discrete-mathematics boolean-algebra
asked Dec 9 '16 at 9:06
Jon GarrickJon Garrick
1,246828
$endgroup$
How to find total number of boolean functions of the equation :
$F(x,y,z) = F(x',y,z') + F(x,y',z)$
Is there any procedure for this ?
functions discrete-mathematics boolean-algebra
functions discrete-mathematics boolean-algebra
asked Dec 9 '16 at 9:06
Jon GarrickJon Garrick
1,246828
asked Dec 9 '16 at 9:06
Jon GarrickJon Garrick
1,246828
edited Dec 9 '16 at 10:35
Jon Garrick
asked Dec 9 '16 at 9:06
Jon GarrickJon Garrick
1,246828
asked Dec 9 '16 at 9:06
Jon GarrickJon Garrick
1,246828
asked Dec 9 '16 at 9:06
Jon GarrickJon Garrick
1,246828
1,246828
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
A function with three variables is described in a truth-table with eight (2*2*2) rows:
x y z F
0 0 0 f0
0 0 1 f1
0 1 0 f2
0 1 1 f3
1 0 0 f4
1 0 1 f5
1 1 0 f6
1 1 1 f7
As any value in the output column can be either 0 or 1, disregarding the equation constraint, the total number of different Boolean functions with three input variables is 2^(2*2*2) = 2^8 = 256
The given equation translates to a set of equations (+ = or):
f0 = f5 + f2 = f2 + f5
f1 = f4 + f3 = f3 + f4
f2 = f7 + f0 = f0 + f7
f3 = f6 + f1 = f1 + f6
f4 = f1 + f6 = f1 + f6
f5 = f0 + f7 = f0 + f7
f6 = f3 + f4 = f3 + f4
f7 = f2 + f5 = f2 + f5
Each of the eight equations has one other equation with the same right-hand-side. This leads to:
f0 = f7
f1 = f6
f2 = f5
f3 = f4
Inserted in the set of equations:
f0 = f2 = f5 = f7
f1 = f3 = f4 = f6
There are four solutions for (f0; f1): (0; 0), (0; 1), (1; 0) and (1; 1)
Translated back - as an example - to the original truth-table for (f0=1; f1=0):
x y z F x'y z' F1 x y' z F2 F=F1+F2
0 0 0 f0 1 1 0 1 1 0 1 0 1 1
0 0 1 f1 0 1 0 0 0 0 1 1 0 0
0 1 0 f2 1 1 1 1 1 0 0 0 1 1
0 1 1 f3 0 1 1 0 0 0 0 1 0 0
1 0 0 f4 0 0 0 1 0 1 1 0 0 0
1 0 1 f5 1 0 0 0 1 1 1 1 1 1
1 1 0 f6 0 0 1 1 0 1 0 0 0 0
1 1 1 f7 1 0 1 0 1 1 0 1 1 1
The total number of functions which fulfil the equation is four.
Two of these are trivial constant functions: always 0, or always 1
Alternative approach using the MiniZinc constraint solver:
% function F is determined by 2^n = 8 function values:
array[0..7] of var bool: f;
function var bool: F(bool: x, bool: y, bool: z) =
% bool variable are automatically converted (coerced) in 0 .. 1
f[4*x+2*y+z];
constraint forall(x,y,z in [false,true]) (
F(x, y, z) == (F(not x, y, not z) / F(x, not y, z))
);
MiniZinc comes up with four solutions as expected:
[false, false, false, false, false, false, false, false]
[false, true, false, true, true, false, true, false]
[true, false, true, false, false, true, false, true ]
[true, true, true, true, true, true, true, true ]
answered Dec 9 '16 at 12:20
Axel KemperAxel Kemper
3,33611418
$endgroup$
$begingroup$
Very Nice Explaination !!
$endgroup$
– Jon Garrick
Dec 9 '16 at 13:07
add a comment |
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$begingroup$
A function with three variables is described in a truth-table with eight (2*2*2) rows:
x y z F
0 0 0 f0
0 0 1 f1
0 1 0 f2
0 1 1 f3
1 0 0 f4
1 0 1 f5
1 1 0 f6
1 1 1 f7
As any value in the output column can be either 0 or 1, disregarding the equation constraint, the total number of different Boolean functions with three input variables is 2^(2*2*2) = 2^8 = 256
The given equation translates to a set of equations (+ = or):
f0 = f5 + f2 = f2 + f5
f1 = f4 + f3 = f3 + f4
f2 = f7 + f0 = f0 + f7
f3 = f6 + f1 = f1 + f6
f4 = f1 + f6 = f1 + f6
f5 = f0 + f7 = f0 + f7
f6 = f3 + f4 = f3 + f4
f7 = f2 + f5 = f2 + f5
Each of the eight equations has one other equation with the same right-hand-side. This leads to:
f0 = f7
f1 = f6
f2 = f5
f3 = f4
Inserted in the set of equations:
f0 = f2 = f5 = f7
f1 = f3 = f4 = f6
There are four solutions for (f0; f1): (0; 0), (0; 1), (1; 0) and (1; 1)
Translated back - as an example - to the original truth-table for (f0=1; f1=0):
x y z F x'y z' F1 x y' z F2 F=F1+F2
0 0 0 f0 1 1 0 1 1 0 1 0 1 1
0 0 1 f1 0 1 0 0 0 0 1 1 0 0
0 1 0 f2 1 1 1 1 1 0 0 0 1 1
0 1 1 f3 0 1 1 0 0 0 0 1 0 0
1 0 0 f4 0 0 0 1 0 1 1 0 0 0
1 0 1 f5 1 0 0 0 1 1 1 1 1 1
1 1 0 f6 0 0 1 1 0 1 0 0 0 0
1 1 1 f7 1 0 1 0 1 1 0 1 1 1
The total number of functions which fulfil the equation is four.
Two of these are trivial constant functions: always 0, or always 1
Alternative approach using the MiniZinc constraint solver:
% function F is determined by 2^n = 8 function values:
array[0..7] of var bool: f;
function var bool: F(bool: x, bool: y, bool: z) =
% bool variable are automatically converted (coerced) in 0 .. 1
f[4*x+2*y+z];
constraint forall(x,y,z in [false,true]) (
F(x, y, z) == (F(not x, y, not z) / F(x, not y, z))
);
MiniZinc comes up with four solutions as expected:
[false, false, false, false, false, false, false, false]
[false, true, false, true, true, false, true, false]
[true, false, true, false, false, true, false, true ]
[true, true, true, true, true, true, true, true ]
answered Dec 9 '16 at 12:20
Axel KemperAxel Kemper
3,33611418
$endgroup$
$begingroup$
Very Nice Explaination !!
$endgroup$
– Jon Garrick
Dec 9 '16 at 13:07
add a comment |
$begingroup$
A function with three variables is described in a truth-table with eight (2*2*2) rows:
x y z F
0 0 0 f0
0 0 1 f1
0 1 0 f2
0 1 1 f3
1 0 0 f4
1 0 1 f5
1 1 0 f6
1 1 1 f7
As any value in the output column can be either 0 or 1, disregarding the equation constraint, the total number of different Boolean functions with three input variables is 2^(2*2*2) = 2^8 = 256
The given equation translates to a set of equations (+ = or):
f0 = f5 + f2 = f2 + f5
f1 = f4 + f3 = f3 + f4
f2 = f7 + f0 = f0 + f7
f3 = f6 + f1 = f1 + f6
f4 = f1 + f6 = f1 + f6
f5 = f0 + f7 = f0 + f7
f6 = f3 + f4 = f3 + f4
f7 = f2 + f5 = f2 + f5
Each of the eight equations has one other equation with the same right-hand-side. This leads to:
f0 = f7
f1 = f6
f2 = f5
f3 = f4
Inserted in the set of equations:
f0 = f2 = f5 = f7
f1 = f3 = f4 = f6
There are four solutions for (f0; f1): (0; 0), (0; 1), (1; 0) and (1; 1)
Translated back - as an example - to the original truth-table for (f0=1; f1=0):
x y z F x'y z' F1 x y' z F2 F=F1+F2
0 0 0 f0 1 1 0 1 1 0 1 0 1 1
0 0 1 f1 0 1 0 0 0 0 1 1 0 0
0 1 0 f2 1 1 1 1 1 0 0 0 1 1
0 1 1 f3 0 1 1 0 0 0 0 1 0 0
1 0 0 f4 0 0 0 1 0 1 1 0 0 0
1 0 1 f5 1 0 0 0 1 1 1 1 1 1
1 1 0 f6 0 0 1 1 0 1 0 0 0 0
1 1 1 f7 1 0 1 0 1 1 0 1 1 1
The total number of functions which fulfil the equation is four.
Two of these are trivial constant functions: always 0, or always 1
Alternative approach using the MiniZinc constraint solver:
% function F is determined by 2^n = 8 function values:
array[0..7] of var bool: f;
function var bool: F(bool: x, bool: y, bool: z) =
% bool variable are automatically converted (coerced) in 0 .. 1
f[4*x+2*y+z];
constraint forall(x,y,z in [false,true]) (
F(x, y, z) == (F(not x, y, not z) / F(x, not y, z))
);
MiniZinc comes up with four solutions as expected:
[false, false, false, false, false, false, false, false]
[false, true, false, true, true, false, true, false]
[true, false, true, false, false, true, false, true ]
[true, true, true, true, true, true, true, true ]
answered Dec 9 '16 at 12:20
Axel KemperAxel Kemper
3,33611418
$endgroup$
$begingroup$
Very Nice Explaination !!
$endgroup$
– Jon Garrick
Dec 9 '16 at 13:07
add a comment |
$begingroup$
A function with three variables is described in a truth-table with eight (2*2*2) rows:
x y z F
0 0 0 f0
0 0 1 f1
0 1 0 f2
0 1 1 f3
1 0 0 f4
1 0 1 f5
1 1 0 f6
1 1 1 f7
As any value in the output column can be either 0 or 1, disregarding the equation constraint, the total number of different Boolean functions with three input variables is 2^(2*2*2) = 2^8 = 256
The given equation translates to a set of equations (+ = or):
f0 = f5 + f2 = f2 + f5
f1 = f4 + f3 = f3 + f4
f2 = f7 + f0 = f0 + f7
f3 = f6 + f1 = f1 + f6
f4 = f1 + f6 = f1 + f6
f5 = f0 + f7 = f0 + f7
f6 = f3 + f4 = f3 + f4
f7 = f2 + f5 = f2 + f5
Each of the eight equations has one other equation with the same right-hand-side. This leads to:
f0 = f7
f1 = f6
f2 = f5
f3 = f4
Inserted in the set of equations:
f0 = f2 = f5 = f7
f1 = f3 = f4 = f6
There are four solutions for (f0; f1): (0; 0), (0; 1), (1; 0) and (1; 1)
Translated back - as an example - to the original truth-table for (f0=1; f1=0):
x y z F x'y z' F1 x y' z F2 F=F1+F2
0 0 0 f0 1 1 0 1 1 0 1 0 1 1
0 0 1 f1 0 1 0 0 0 0 1 1 0 0
0 1 0 f2 1 1 1 1 1 0 0 0 1 1
0 1 1 f3 0 1 1 0 0 0 0 1 0 0
1 0 0 f4 0 0 0 1 0 1 1 0 0 0
1 0 1 f5 1 0 0 0 1 1 1 1 1 1
1 1 0 f6 0 0 1 1 0 1 0 0 0 0
1 1 1 f7 1 0 1 0 1 1 0 1 1 1
The total number of functions which fulfil the equation is four.
Two of these are trivial constant functions: always 0, or always 1
Alternative approach using the MiniZinc constraint solver:
% function F is determined by 2^n = 8 function values:
array[0..7] of var bool: f;
function var bool: F(bool: x, bool: y, bool: z) =
% bool variable are automatically converted (coerced) in 0 .. 1
f[4*x+2*y+z];
constraint forall(x,y,z in [false,true]) (
F(x, y, z) == (F(not x, y, not z) / F(x, not y, z))
);
MiniZinc comes up with four solutions as expected:
[false, false, false, false, false, false, false, false]
[false, true, false, true, true, false, true, false]
[true, false, true, false, false, true, false, true ]
[true, true, true, true, true, true, true, true ]
answered Dec 9 '16 at 12:20
Axel KemperAxel Kemper
3,33611418
$endgroup$
A function with three variables is described in a truth-table with eight (2*2*2) rows:
x y z F
0 0 0 f0
0 0 1 f1
0 1 0 f2
0 1 1 f3
1 0 0 f4
1 0 1 f5
1 1 0 f6
1 1 1 f7
As any value in the output column can be either 0 or 1, disregarding the equation constraint, the total number of different Boolean functions with three input variables is 2^(2*2*2) = 2^8 = 256
The given equation translates to a set of equations (+ = or):
f0 = f5 + f2 = f2 + f5
f1 = f4 + f3 = f3 + f4
f2 = f7 + f0 = f0 + f7
f3 = f6 + f1 = f1 + f6
f4 = f1 + f6 = f1 + f6
f5 = f0 + f7 = f0 + f7
f6 = f3 + f4 = f3 + f4
f7 = f2 + f5 = f2 + f5
Each of the eight equations has one other equation with the same right-hand-side. This leads to:
f0 = f7
f1 = f6
f2 = f5
f3 = f4
Inserted in the set of equations:
f0 = f2 = f5 = f7
f1 = f3 = f4 = f6
There are four solutions for (f0; f1): (0; 0), (0; 1), (1; 0) and (1; 1)
Translated back - as an example - to the original truth-table for (f0=1; f1=0):
x y z F x'y z' F1 x y' z F2 F=F1+F2
0 0 0 f0 1 1 0 1 1 0 1 0 1 1
0 0 1 f1 0 1 0 0 0 0 1 1 0 0
0 1 0 f2 1 1 1 1 1 0 0 0 1 1
0 1 1 f3 0 1 1 0 0 0 0 1 0 0
1 0 0 f4 0 0 0 1 0 1 1 0 0 0
1 0 1 f5 1 0 0 0 1 1 1 1 1 1
1 1 0 f6 0 0 1 1 0 1 0 0 0 0
1 1 1 f7 1 0 1 0 1 1 0 1 1 1
The total number of functions which fulfil the equation is four.
Two of these are trivial constant functions: always 0, or always 1
Alternative approach using the MiniZinc constraint solver:
% function F is determined by 2^n = 8 function values:
array[0..7] of var bool: f;
function var bool: F(bool: x, bool: y, bool: z) =
% bool variable are automatically converted (coerced) in 0 .. 1
f[4*x+2*y+z];
constraint forall(x,y,z in [false,true]) (
F(x, y, z) == (F(not x, y, not z) / F(x, not y, z))
);
MiniZinc comes up with four solutions as expected:
[false, false, false, false, false, false, false, false]
[false, true, false, true, true, false, true, false]
[true, false, true, false, false, true, false, true ]
[true, true, true, true, true, true, true, true ]
answered Dec 9 '16 at 12:20
Axel KemperAxel Kemper
3,33611418
edited Dec 26 '18 at 12:02
answered Dec 9 '16 at 12:20
Axel KemperAxel Kemper
3,33611418
answered Dec 9 '16 at 12:20
Axel KemperAxel Kemper
3,33611418
answered Dec 9 '16 at 12:20
Axel KemperAxel Kemper
3,33611418
3,33611418
$begingroup$
Very Nice Explaination !!
$endgroup$
– Jon Garrick
Dec 9 '16 at 13:07
add a comment |
$begingroup$
Very Nice Explaination !!
$endgroup$
– Jon Garrick
Dec 9 '16 at 13:07
$begingroup$
Very Nice Explaination !!
$endgroup$
– Jon Garrick
Dec 9 '16 at 13:07
$begingroup$
Very Nice Explaination !!
$endgroup$
– Jon Garrick
Dec 9 '16 at 13:07
add a comment |
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