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Question

What will be the minimum size of ROM which maintains truth table of square of 3 bit numbers(in bits) ?

Answer 1

Truth Table will be as:

Input (Address lines)			Output (Data)
I0	I1	I2	B0	B1	B2	B3	B4	B5
									0	0	0	0	0	0	0	0	0
0	0	1	0	0	0	0	0	1
0	1	0	0	0	0	1	0	0
0	1	1	0	0	1	0	0	1
1	0	0	0	1	0	0	0	0
1	0	1	0	1	1	0	0	1
1	1	0	1	0	0	1	0	0
1	1	1	1	1	0	0	0	1

example for 111 (7), square will 110001 (49)

so it will have 3 address line  mean 8 address [000 to 111] and 6 bits [B0 to B5] at each address = 8 X 6 = 48 bits 

but we do not need to store B4 and B5 [as B4 = 0 and B5 = I2 ]

so ROM will be of size 8 x 4 = 32bits.

Comments

cannot understand why B5 is not required. could u plz elaborate?

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As we provide I0, I1 and I2 in address line to select address then I2 can be pass to B5 output line. 

so There is no need to store B5 in ROM as in 

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@Praveen Saini I think in total we need 8*7 = 56 bits. Because its ROM so every combination will be stored. So from above table and your logic we have 8 rows × 7 columns = 56 bits.

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why 7, 8*6 = 48 rt?

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@Arjun Sir what i thought is that ROM cant be written again and again. So we have to take care of all possibilities. Now since we have to store a truth table therefore we have to consider all possibilities. Earlier i was also straightforward thinking that 48 bits is the answer but after looking Praveen Saini logic, i thought he is right. I mean the table above given by Parveen Saini tells that select lines I0 I1 I2 will make 8 combinations i.e.8 rows. Now for each row we have to have 9 column's( I0 I1 I2 B0 B1 B2 B3 B4 B5). But now according to Praveen Saini logic, we dont need column B4 and B5 since they can taken from I0 and I2 . Therefore in storing whole truth table we are left with 8 rows and 9-2=7 columns. Hence 8×7=56 bits.

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Your logic does not seem correct. ROM is a fixed memory- Read Only. So, as long as we know its purpose we don't need to store redundant data in it. More importantly isn't this portion out of syllabus?

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@Arjun Then what is the correct logic for this?

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I go with this answer. There is no doubt answer is <= 48 bits. I have a doubt if we can skip the 2 columns as this is happening due to the property of the squares- ROM designer may not know this. But question asks "minimum size" so 32 seems the best answer. Ideally you can see similar questions from Digital text, but AFAIK this is not in syllabus.

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@Arjun yes even i too agree 32 bits seems best answer but i could not understand Praveen Saini when he says 8*4=32. What is 8 here ( rows i suppose) and what is 4 then? Can you make this clear?.

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We need 6 bits- maximum square is 49. But two of those can be skipped - the last two. So, for 8 numbers we have 8 * 4.

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@Arjun That means we are not storing I0 I1 I2. We are  just focussing on B0 B1 B2 B3. I0 I1 I2 will be taken care by some decoder logic.

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We store those 8 rows in ROM. Now to select any one of them we need 3 address lines- which are I0, I1 and I2- they need not be stored.

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Thanks got it.

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@arjun sir , why did u say that 0, I1 and I2- they need not be stored ?
I didn't Understand why are you skipping this ?
Please make it simple

Answer 2

no is 3 bit long.. square of 3 bit no is 6 bit long..
total 8 nos (0 to 7) which truth table has to store..
total size is 8*6=48 bit