Minimal boolean representation of an AND via two-input NOR












0














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.










share|cite|improve this question






















  • 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
















0














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.










share|cite|improve this question






















  • 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














0












0








0







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.










share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • 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















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