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 Show 2 previous comments 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 `JEET commented Dec 11, 2019 reply Follow Share Good answer. Post more answers now in your stronger subjects like TOC. 2 votes 2 votes neeraj_bhatt commented Dec 12, 2019 i edited by neeraj_bhatt Dec 12, 2019 reply Follow Share Akash Papnai thanks a lot man, but does this method works for implementing XOR gate using NAND gates only. I got the answer as 5, but the valid answer is 4. 0 votes 0 votes `JEET commented Dec 12, 2019 reply Follow Share Yeah, this doesn't work for that things. 0 votes 0 votes Akash Papnai commented Dec 12, 2019 reply Follow Share @`JEET and @neeraj2681 Implementing XOR using NAND or any other possible combination could be done using above method. I'm sure you are making a mistake while implementing it. This is how it's done: Also, note that I was able to combine because I know that A XOR A is 0 and B XOR B is also 0. 4 votes 4 votes `JEET commented Dec 12, 2019 reply Follow Share Yeah, that what exactly I did. But I still believe it can't be used at many other places, though I feel fundamental gates can be implemented. I mean most of the cases its applicable but it doesn't seem to be an universal approach for a tricky or maybe a little big cases. 0 votes 0 votes `JEET commented Dec 12, 2019 reply Follow Share @Akash Papnai You could combine them in the step $\mathbf 1$ itself. Like taking $\mathbf{B}$ at both the places and same for $\mathbf A$ as well. You just need to add Not which can be converted to AND later. Don't you think this can be done? 0 votes 0 votes neeraj_bhatt commented Dec 12, 2019 reply Follow Share @Akash Papnai The question might seem silly, but Is there an algorithm or something to check that the design can further be minimised after step 3rd or Is it just intuition? I mean how one can be sure to go ahead after step 3rd? Also, Is there a way to check that its final and no further steps needed now? 0 votes 0 votes Akash Papnai commented Dec 12, 2019 reply Follow Share @`JEET I have done a lot question on NAND NOR implementation and I haven't encountered such a case in which I follow the above steps and got drowned even in many complex questions. Also, what you said about combining them in step 1 could surely be done but for the sake of understandability, I did it at last. This was done to understand the general idea of the question but not just the answer to the above question. It's better to understand the answer intricately. 1 votes 1 votes `JEET commented Dec 12, 2019 reply Follow Share No actually there isn't any feasible algorithm. If I remember correctly, its time complexity is something exponential like. 0 votes 0 votes `JEET commented Dec 12, 2019 reply Follow Share @Akash Papnai Yeah, right! Let me see if I get any such question in the future, then will respond. 1 votes 1 votes 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.