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. 
