14 votes 14 votes A variable that takes thirteen possible values can be communicated using? Thirteen bits. Three bits. $\log_{2}13$ bits. Four bits. None of the above. Digital Logic tifr2011 number-representation + – makhdoom ghaya asked Oct 19, 2015 edited Feb 21, 2018 by go_editor makhdoom ghaya 2.0k views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply HeadShot commented Dec 4, 2018 reply Follow Share @Pragy Agarwal Sir, Isnt it the same case where you answered that question as none of the above because sufficient data isnt given. Here also, its not case of addressing but in communication we need to send actual data and size of variable is not given then how can we state that 14 bits would be sufficient ? 0 votes 0 votes Pragy Agarwal commented Dec 4, 2018 reply Follow Share You have to send 1 value $x$, but that $x$ can be one of 13 values that you know. It doesn't matter what the values are, since you can just say that you're sending 1st one from those 13 values, or 7th one from those 13 values. So, you're not sending the value itself, but the index of that value. The index $i$ can be between anything between $c \leq i \leq c + 12$. Then, in order to send that index, you will need to send $k$ bytes, where $k = \lceil \log_2 (c+12)\rceil$. Since we would like to send the minimum number of bits possible, we would use $c=0$ (that is, start indexing from 0). We thus need 4 bits. 4 votes 4 votes HeadShot commented Dec 5, 2018 reply Follow Share Okay. So only index is being communicated. thanx @Pragy Agarwal sir. 0 votes 0 votes shashankrustagi commented Feb 6, 2021 reply Follow Share option c is incorrect because it is not ceil correct answer is D this is a trap examiner wanted us to think like that 1 votes 1 votes Please log in or register to add a comment.
Best answer 15 votes 15 votes As there are only $13$ possible values a variable can take, we need to use $\lceil 13 \rceil = 4-bits.$ PS: As variable can take only 13 values, we don't need to worry what those values are. Answer : Option D Akash Kanase answered Nov 5, 2015 selected Nov 5, 2015 by Arjun Akash Kanase comment Share Follow See all 2 Comments See all 2 2 Comments reply pritishc commented Dec 5, 2019 reply Follow Share This is $\left \lceil log_213 \right \rceil$ isn't it? 1 votes 1 votes arpit_18 commented Dec 26, 2020 reply Follow Share Yes, this is $\left \lceil \log 13 \right \rceil$ =4. But in options, only $ \log 13$ is given. 0 votes 0 votes Please log in or register to add a comment.
–2 votes –2 votes I think Option e because what kind of value(range) its not mentioned so we cant decide the bits for the variable. Umang Raman answered Oct 19, 2015 Umang Raman comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Akash Kanase commented Nov 5, 2015 reply Follow Share Yes. Range is not required here. There are only 13 possible values. 0 votes 0 votes Prateek kumar commented Sep 18, 2016 reply Follow Share @Arjun sir,can't we think in "K-MAP" way as variable can take 13 possible way so to cover all these possible ways we must need 4 variables or 4 bits "Kmap" 0 votes 0 votes rajinder singh commented Aug 24, 2018 reply Follow Share But question asking about no of bits required to communicate which will be 4bit because 13 value only possible through 4 bit.so option d is correct 0 votes 0 votes Please log in or register to add a comment.