Minimal boolean representation of an AND via two-input NOR
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Let $x_1,x_2,x_3,x_4$ be boolean variables (i.e $x_i in {0,1}$)
Consider
$f(x_1,x_2,x_3,x_4) = x_1 wedge x_2 wedge x_3 wedge x_4 $
I want to write $f$ in terms of two-input NOR gates. I.e, $mbox{NR}(x,y)= neg ( x vee y ) $.
What is the minimum number of NOR gates to write it?
I found that it can be done with 9. Can it be done with less.
boolean-algebra
asked Nov 20 at 18:09
M.A
1599
add a comment |
Let $x_1,x_2,x_3,x_4$ be boolean variables (i.e $x_i in {0,1}$)
Consider
$f(x_1,x_2,x_3,x_4) = x_1 wedge x_2 wedge x_3 wedge x_4 $
I want to write $f$ in terms of two-input NOR gates. I.e, $mbox{NR}(x,y)= neg ( x vee y ) $.
What is the minimum number of NOR gates to write it?
I found that it can be done with 9. Can it be done with less.
boolean-algebra
asked Nov 20 at 18:09
M.A
1599
It seems to me that $9$ is the minimum, since one two-input AND needs three two-input NORs.
– user376343
Nov 23 at 21:25
add a comment |
Let $x_1,x_2,x_3,x_4$ be boolean variables (i.e $x_i in {0,1}$)
Consider
$f(x_1,x_2,x_3,x_4) = x_1 wedge x_2 wedge x_3 wedge x_4 $
I want to write $f$ in terms of two-input NOR gates. I.e, $mbox{NR}(x,y)= neg ( x vee y ) $.
What is the minimum number of NOR gates to write it?
I found that it can be done with 9. Can it be done with less.
boolean-algebra
asked Nov 20 at 18:09
M.A
1599
Let $x_1,x_2,x_3,x_4$ be boolean variables (i.e $x_i in {0,1}$)
Consider
$f(x_1,x_2,x_3,x_4) = x_1 wedge x_2 wedge x_3 wedge x_4 $
I want to write $f$ in terms of two-input NOR gates. I.e, $mbox{NR}(x,y)= neg ( x vee y ) $.
What is the minimum number of NOR gates to write it?
I found that it can be done with 9. Can it be done with less.
boolean-algebra
boolean-algebra
asked Nov 20 at 18:09
M.A
1599
asked Nov 20 at 18:09
M.A
1599
asked Nov 20 at 18:09
M.A
1599
asked Nov 20 at 18:09
M.A
1599
asked Nov 20 at 18:09
M.A
1599
1599
It seems to me that $9$ is the minimum, since one two-input AND needs three two-input NORs.
– user376343
Nov 23 at 21:25
add a comment |
It seems to me that $9$ is the minimum, since one two-input AND needs three two-input NORs.
– user376343
Nov 23 at 21:25
It seems to me that $9$ is the minimum, since one two-input AND needs three two-input NORs.
– user376343
Nov 23 at 21:25
It seems to me that $9$ is the minimum, since one two-input AND needs three two-input NORs.
– user376343
Nov 23 at 21:25
add a comment |
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It seems to me that $9$ is the minimum, since one two-input AND needs three two-input NORs.
– user376343
Nov 23 at 21:25