43 votes 43 votes What is the minimum number of gates required to implement the Boolean function $\text{(AB+C)}$ if we have to use only $2\text{-input NOR}$ gates? $2$ $3$ $4$ $5$ Digital Logic gatecse-2009 digital-logic min-no-gates normal + – Kathleen asked Sep 22, 2014 edited Jun 19, 2018 by Pooja Khatri Kathleen 19.9k views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply abhiarns commented May 9, 2020 reply Follow Share Is there any rule to solve "minimum number of gates" question? I am able to solve it but not optimally. Please tell any method or way of thinking that you all implement. 1 votes 1 votes mohan123 commented Jun 13, 2020 reply Follow Share @abhiarns SOP : 2 - level AND-OR then NAND gate POS : 2 level OR -AND then NOR gate now coming to questions AB+C = (A+C)(B+C) ( first "or" then "and" that means NOR gate ) number of NOR Gate : 3 11 votes 11 votes abhiarns commented Jun 14, 2020 reply Follow Share @mohan123 Thanks 1 votes 1 votes dutta18 commented Jan 13, 2021 reply Follow Share Can you just tell me how did you convert AB+C to (A+B)(B+C) ? Is it SOP to POS ? 0 votes 0 votes mohan123 commented Feb 7, 2021 reply Follow Share firstly u write the wrong AB+C to (A+B)(B+C) correct AB+C to (A+C)(B+C) Distributive Law states that the multiplication of two variables and adding the result with a variable will result in the same value as the multiplication of addition of the variable with individual variables. For example: A + BC = (A + B) (A + C) 0 votes 0 votes Please log in or register to add a comment.
Best answer 90 votes 90 votes Given boolean function is $f = AB + C$ $\quad = (A+C) . (B +C)$ $\quad =((A+C)' +(B+C)')'$ Therefore, $3$ NOR gates required . Correct Answer: $B$ Mithlesh Upadhyay answered Jun 4, 2015 edited May 2, 2021 by gatecse Mithlesh Upadhyay comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Rajesh Panwar commented Dec 15, 2018 reply Follow Share beautiful explanation. 1 votes 1 votes realdonaldtrump commented Feb 5, 2021 reply Follow Share @Barney Because OR distributed over AND. 0 votes 0 votes neel19 commented Aug 23, 2021 reply Follow Share The given function is in SOP form. You can find the POS form using K-map. 0 votes 0 votes Please log in or register to add a comment.
58 votes 58 votes Here is a simple approach to these kinds of questions. This can be applied to any questions and you'll get min number of gates always. Akash Papnai answered Dec 5, 2019 Akash Papnai comment Share Follow See all 13 Comments See all 13 13 Comments reply Show 10 previous comments neeraj_bhatt commented Dec 12, 2019 reply Follow Share @Akash Papnai and @`JEET thanks for the explanations! 0 votes 0 votes yash22899 commented Sep 21, 2020 reply Follow Share Thanks for the method. Will definitely help many including me 0 votes 0 votes Abhrajyoti00 commented Jan 5, 2023 reply Follow Share Very nice method. Thanks @Akash Papnai 0 votes 0 votes Please log in or register to add a comment.
7 votes 7 votes Answer should be 3 nor gates...A.B + C can be written as (A+C).(B+C) which is a OR-AND realization and can be made with 3 NOR gates as OR- AND is similar to a NOR-NOR realization... Nitish Shrivastava 1 answered Jun 29, 2016 edited Jun 29, 2016 by Nitish Shrivastava 1 Nitish Shrivastava 1 comment Share Follow See all 0 reply Please log in or register to add a comment.
6 votes 6 votes Almost same solution, Prateek Raghuvanshi answered May 20, 2018 Prateek Raghuvanshi comment Share Follow See all 0 reply Please log in or register to add a comment.