TIFR CSE 2011 | Part A | Question: 16

Question

A variable that takes thirteen possible values can be communicated using?

• A. Thirteen bits.
• B. Three bits.
• C. $\log_{2}13$ bits.
• D. Four bits.
• E. None of the above.

Comments

@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 ? 

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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.

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Okay. So only index is being communicated.

thanx @Pragy Agarwal sir.

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option c is incorrect because it is not ceil

correct answer is D

this is a trap

examiner wanted us to think like that

Answer 1

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

Comments

This is $\left \lceil log_213 \right \rceil$ isn't it?

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Yes, this is $\left \lceil \log 13 \right \rceil$ =4. But in options, only $ \log 13$  is given.

Answer 2

I think Option e
because what kind of value(range) its not mentioned so we cant decide the bits for the variable.

Comments

Yes. Range is not required here. There are only 13 possible values.

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@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"

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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