Why MSB's are discarded during 2's complement multiplication?

Question

I have a trouble with understanding why we discard MSB when we do multiplication of 2's complements numbers.

Let's take 101 (decimal: -3) and 011 (decimal: 3) for example. Algorithm goes like this: first we double length of our numbers, then we do usual multiplication as we did in school with decimals, then we take (doubled length)*2 = 6 least significant bits.

1. Double lengths:

   101 -> 111101
   011 -> 000011
   

2. Do multiplication: 111101 * 000011 = 10110111 (decimal: -73)

As we see, this result is wrong.

3. Take 6 least significant bits (drop 2 most significant bits). 10110111 -> 110111 (decimal: -9)

And so result became magically right. How can this be explained? I understand that MSB is kind of special and the same rules I've used in school cannot be 100% suitable for 2's complements, but while I fully understand school rules of multiplication, I can't wrap my head around the last step of 2's complement multiplication (first two steps I understand).

Answer 1

Your multiplication is simply incorrect. You did an unsigned multiplication. In other words this:

    111101  
    000011 x
    ------ 
    111101
   111101
  000000
 000000
000000     +
----------
0010110111

Or in decimal 61 x 3 = 183. But a signed multiplication requires extending the sign bit in the partial products as well. Like this:

    111101  
    000011 x
    ------ 
1111111101
111111101 
00000000
0000000
000000     +
----------
1111110111

So now you correctly compute -3 x 3 = -9. This distinction matters for processors as well, compare MUL vs IMUL on Intel processors.