Take the 2-minute tour ×

The question I am seeking to solve is to find the decimal values in the rage of x to y that can be represented using a 32bit excess-65540 system.

I know that excess-N means that all values are represented in a positive way such that if you have an excess-65540 system then decimal value of 0 would equal -65540

I am confused as to

  1. How to find the range of a specific excess-n notation
  2. How to actually find a value of an excess-n notation and determine if it is positive or negative.

My guess for 1 would be a range of -65540 to 65539 for decimals but I am unsure if that is right.
2 I am a little lost on.

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

Hint: Invent your own smaller sample problems until you get the idea.

For example, if the question asked about 8-bit excess-17, could you answer it?

Hint 2: There are 2^32 32-bit values, and each of them corresponds to a different excess-65540 number.

share|improve this answer
Oh, well done Nemo :) your hints are much better than mine. –  sarnold Jun 11 '11 at 22:40
@sarnold: Thanks. I used to teach CS classes in grad school :-) –  Nemo Jun 11 '11 at 22:41
@sarnold: In fact, I think I said too much. :-) –  Nemo Jun 11 '11 at 22:46
@Nemo, I am still very confused do you think you could help me walk through your example of 8 bit excess-17? –  drainojunkie Jun 11 '11 at 22:49
Well, a base-8 number has a range of 0 (all bits zero) to 255 (all bits one). In excess-17, 0 represents -17. 1 represents -16. 2 represents -15. And so on... So fill that in all the way to 255 (which represents... what?) and you will have your range. –  Nemo Jun 11 '11 at 23:29
show 11 more comments

Your Answer


