17 votes 17 votes How many minimum number of two input AND gates and two input OR gates are required to realize $Y = BD+CE+AB$ ? $2, 2$ $4, 2$ $3, 2$ $2, 3$ Digital Logic tbb-digital-logic-2 + – Bikram asked Nov 26, 2016 edited Aug 19, 2019 by Counsellor Bikram 1.1k views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Sheshang commented Dec 15, 2016 reply Follow Share upvote for question. really a nice one. 1 votes 1 votes Rishabh Gupta 2 commented Oct 10, 2017 reply Follow Share I think the question should mention "minimum" number of gates. 2 votes 2 votes Shaik Masthan commented Jan 5, 2019 reply Follow Share updated ! 0 votes 0 votes Please log in or register to add a comment.
Best answer 5 votes 5 votes Given : Y = BD+CE+AB here 3 and gate and 2 or gate [ we can minimize 1 AND gate] Y = B(D + A) + CE Now we see that 2 AND Gate i.e.1st B & (D+A), 2nd C & E.. Prashant. answered Dec 1, 2016 selected Aug 29, 2017 by Bikram Prashant. comment Share Follow See 1 comment See all 1 1 comment reply Wanted commented Jan 5, 2017 reply Follow Share twisted answer 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes BD+AB+CE = B(D+A)+CE 1 or gate for X= D+A 1 and gate for Y=B.(D+A) 1 and gate for Z=CE 1 or gate for Y and Z hence 2 AND gate and 2 OR gate I_am_winner answered Jul 6, 2018 I_am_winner comment Share Follow See all 0 reply Please log in or register to add a comment.