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  #11  
Old 10-13-2011, 03:31 AM
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Papabravo Papabravo is offline
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The end-around-carry means that you take the carry bit and add it to the result of the addition. In 4-bit words
Code:
   0 1 1 1  -->  7
+  1 1 1 0  --> -1
-----------------------
 1 0 1 0 1       6
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So here we have a 4-bit result of 0101 == 5 and a carry
Code:
   0 1 0 1  --> Result of 5
+  0 0 0 1  --> Carry from the original addition
-------------
   0 1 1 0  --> True result of 6
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That was an end-around-carry
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  #12  
Old 10-18-2011, 12:32 PM
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Thank you very much, GeoRacer, MrChips, Papabravo.

I'm sorry to say this but I'm still a bit confused. But the good thing is I can understand the topic in my book. What I'm inquiring about here is for my personal learning. But as advised by GeoRacer I won't pursue the topic further after finishing this discussion. Please be patient! Thanks.

In binary system 2's complement is the maximum complement one can find. I mean to say you can't think of finding 3's complement in base 2. As I have been told 2's complement doesn't give create any problem as compared to 1's complement which has two representation for 0. I'm just curious to know if this also happens in other bases such as base 10 that using, say, 4's complement gives problems such as two representation for 0.

Quote:
Originally Posted by Georacer View Post
Lemme take a shot at it: 0029+5532=5561. Obviously there is an overflow and thus the result must be negative. We will complement the result to find the results absolute value: |5555-5561|=|-6|=6
Bingo! (Didn't think that would work so well)
For example, in 2's complement in the binary, MSB is a sign bit. We can know by looking at the binary if it's a negative one or a positive (assuming we know that we are using complementary method). Here, in the decimal, you have overflow condition to infer the negativity of the number. Is this so?

Do we discard carry in case of decimal too if there is one?

Quote:
Originally Posted by Georacer View Post
Lastly, your book reference wasn't at fault. It's just that it only applies to the <base>'s complement of a number in the <base> numbering system. Check these two examples:
a)9-3
10's complement of 3 is 6 (by his method) which makes it
9+6=15
We disregard 1 as instructed by step 3 and the answer is 5.


b)5-7
10's complement of 7 is 2.
5+2=7
There is no carry thus the answer is
-complement(7)=-2
If I was using doing this: 9-3. I would do it as follows: 3's complement is 7 because 3+7=10 (discard the "1"), now 9+7=16. Obviously, my method doesn't work here. Where am I going wrong? Please help me.

Quote:
Originally Posted by MrChips View Post
Here we go again.
If you want to use a single digit of base 10, the numerals are 0,1,2,3,4,5,6,7,8,9.
But you are mixing up unsigned with signed numbers. If you want signed numbers then the conversion is:
0 0
1 1
2 2
3 3
4 4
5 -5
6 -4
7 -3
8 -2
9 -1

Therefore the value 9 does not exist. 9 is really -1. So your result is -1, the correct answer.
So, we have two side by side sets. MrChips, I have a big confusion here. So, please don't mind my question. Suppose we want to use double digits of base 10, then I think we have numbers from 0 - 99.

Suppose I want to do (-4 x 2 = 8) using complementary method. What would be -4's complement? 96 or 6? I'm saying this because (4+96=100) and (4+6=10).
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  #13  
Old 10-18-2011, 12:58 PM
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If you want signed numbers, you will have to either divide your value range in half or double the number of states.
So if we want 0 to 99 in both positive and negative:

0 0
1 1
:
:
99 99
100 -100
101 -99
:
:
191 -9
192 -8
193 -7
194 -6
195 -5
196 -4
197 -3
198 -2
199 -1

So -4 x 2 is 196 x 2 = 392 = -8
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  #14  
Old 10-18-2011, 01:57 PM
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@PG1995

You quoted it yourself:
Quote:
To find the complement of a n-digit long number: (base)^n - 1 - number. This is how subtraction is done using complementary method: 1. Find the complement of the subtrahend. 2. Add the complement to minuend. 3. If there is a carry of 1, add it to obtain the result; if there is no carry, re-complement the sum and attach a negative sign to obtain the result.
That said, the complement of 3 is 10-1-3=6 and 9-3->9+6->15->1+5->6

What you did wrong is the calculation of the complement. I know that 3+6 isn't equal to 10, but your source's method works, so I 'll stick to it.
As I said, I don't master the art of enumeration methods and numerical analysis.
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