The amount of ROM needed to implement a 4-bit multiplier?

Question Detail: 

For a 4-bit multiplier there are $2^4 \cdot 2^4 = 2^8$ combinations.

The output of 4-bit multiplication is 8 bits, so the amount of ROM needed is $2^8 \cdot 8 = 2048$ bits.

Why is that? Why does the ROM need all the combinations embedded into it?

What will be the case with RAM?

Asked By : Ravi Teja

Best Answer from StackOverflow

Question Source : http://cs.stackexchange.com/questions/9471

Answered By : Reza

As shown in the figure bellow (general architecture of ROM):

to address 8-bit numbers correctly your ROM size should be: $2^8 \times 8$ bits.