i am asked to show the calculation of 11+6 using 5 bit two's complement. i find some rules confusing but have come up with the answer as shown below. please let me know if it is correct or what needs to be done if it is wrong.
two's complement of 11 is
01011
two's complement of 6 is
00110
now adding them using carriers if required:
01011 -----11 in binary
00110 -----6 in binary
10001 ---total
which in decimal is 17. is this the correct method of working out??? because my result in binary shows 10001. isn't 10001 supposed to mean -1 because the first bit in two's complement is a sign bit. please help me solve this if it is wrong. thanks for your help.
Let's explain how this works. You know that 1 is represent as 00001 in 5-bit binary representation. To get -1, a known method for (2's complement) is to :
Invert all bits, which means 11110.
Add 1 to the previous result which lead to 11111 equals to 31 in decimal base (unsigned).
Thus, 10001 is not equal to -1.
Now, let's take 11 (base 10) as an example. We know that 11 equals to 01011.
Invert all bits => 10100
add 1 => 10101 which equals to 21 in unsigned mode or -11 in signed mode.
You can then deduce that 10001 is -15 in signed mode or 17 in unsigned mode.
Also be careful, a signed 5-bit integer is bounded from -16 to 15. An unsigned 5-bit Integer is bounded from 0 to 31. In your case, the answer is 17 which means it's an unsigned integer or -15 as a signed integer.
Hope that helps you.
Related
Hi I had a Quick Question,
What does 81 represent in sign and magnitude based on 8 bits?
a. 129 b. -128 c. -127 d. -1 e. none of the above
My Logic is that it would be none of the above as 81 is a positive integer the signed magnitude representation would just be a bunch of 0's in front of it as the value is not negative, would I be correct in this assumption?
Yes none of the above is the answer.
In signed magnitude the most significant bit is used to represent the sign. The rest of the bits are used to represent the magnitude of the number. Here since 81 is positive the sign bit is 0. The magnitude bits are 1010001
81 is 01010001 when represented in 8 bits signed magnitude.
I was doing my exam prep and I have come across a problem that ive been having issues with mainly because of the lack of info provided. The question is
b.What integer does the 16 bit word F751 represent in the LC-3?
So do we convert the base 16 to base 10 or base 2, Im not really sure how to do this problem.
Take f751 and convert to binary
1111 0111 0101 0001
The most significant bit is 1 so we know the number is negative, so take the 2s complement
0000 1000 1010 1111
And Convert to decimal -2223
The High digit is greater or equal to 8 so the number is negative.
Take the complement to F (fifteen) of each digit: f751
f give 0
7 give 8
5 give A
1 give E
08AE is the 1 complement
08AF is the 2 complement which is in decimal -2223
This prevent to convert to binary
Let's say we want to represent a signed number with 5 bits where the first bit is used for the sign (+ or -) of the number. Then the zero can be represented by two bit representations (10000 and 00000).
How is this problem solved?
Okay. There are always two bit in binary 1 or 0
And then there could be any number of bits for example 1bit to 64bit
If the question is 5-bit string then it should be XXXXX where X can be any bit(1 or 0)
First bit(sign bit) we can have either +0 and -0. (Thanks #machinery)
So if it is positive, we put 0 at first position and if it is negative, we put 1 at first position.
Four Bits
Now, we got our first bit, we are left with another 4-bits 0XXXX or 1XXXX as the question asked for 0,
the rest bit will be zero.
therefore the answer is 00000 or 10000
Look how to convert decimal to binary and binary to decimal.
Bitwise unary complement operator (~) of 2 is -3. I read some where the value 2 in binary representation is 0010 and Bitwise unary complement operator changes bits from 0 to 1, or vice versa. So the value of ~2 is 1101. it means -3. But my confusion is why have they taken 2's binary representation as 0010. according to me int is 32bits. so why 2 cant be 00000000000000000000000000000010 and it's unary complement is 11111111111111111111111111111101? I know am wrong but why? please explain?
The answer to your question is "Two's complement was chosen over one's complement because of several convenient characteristics making arithmetic easier to implement in digital circuits".
I believe from the wording of your question that a bit of an illustration would help.
To fully appreciate this, you still need to read up on two's complement notation and arithmetic - both how they work and their history - but I will try to explain the basics here in a story-like fashion.
Let's say that we have 4 bits in which to represent a signed integer value.
Only 16 distinct values can be represented in 4 bits (16 distinct
different "patterns" can be made with 4-bits)... 0000, 0001, 0010,
0011, 0100, ... 1111 (try it, it's easier to see and develop the
pattern in a columnar format, which you'll see I've done below)
Decide which 16 values you want to be able to represent
It makes sense to say that 0000 stands for zero, 0001 for one, and so on for positives, but what about negatives?
Because zero has "taken one place", we can represent 15 other integers so it is immediately obvious that we cannot represent the same amount of positive and negative values.
We make a choice: our range will run from -8 to +7 (we might have said -9 to +6 or -7 to +8 etc but you'll see below how this choice pays off)
Now which bit-patterns should represent the negatives?
I'm sure you'll agree that it would be very nice if every number added to its additive inverse gave zero without us needing to resort to if-negative-then-else(if positive) logic. E.g. If +3 is represented 0011 and we do (binary) addition of 1101 we get the result (carry 1)0000. Ignore the carry and we've got zero. This makes the bit pattern 1101 obvious winner of the tag "-3".
You can play with the remaining values the same way and what you should arrive at is the following...
-8 1000
-7 1001
-6 1010
-5 1011
-4 1100
-3 1101
-2 1110
-1 1111
0 0000
+1 0001
+2 0010
+3 0011
+4 0100
+5 0101
+6 0110
+7 0111
With the following beautiful and convenient characteristics
"Natural counting bit patterns". Look down the column on the far right and you'll see 0 1 0 1..., then 0 0 1 1... in the next column, then 0 0 0 0 1... etc running perfectly in sequence into the positives
Incrementing "wraps around" (0,1,2,...7,-8,-7,-6,...-1,0,1,...etc and the same goes for decrementing)
Binary addition of additive inverses gives zero with no extra logic to deal with signs
All negative numbers have 1 for their first bit, zero and all positive numbers start with a 0. (The first bit is referred to as "the sign bit".)
Additive inverses can be obtained by the following rule/algorithm: "Invert all bits, increment, discard carry". Magic! Try 3 again:
3 : 0011
~ : 1100 (the NOT operator gives "one's complement")
+1: 1101 (the two's complement representation of -3)
~ : 0010
+1: 0011 (back to +3)
etc
This is two's complement notation
If you've understood this 4-bit story, you'll find that it can be easily extended to apply to 32 bit signed integers.
This from my Interview-MCQ module:
A computer represents information in groups of 64 bits. How many
different integers can be represented in BCD code?
The given answer is 1016, however no explanation is provided, I was just wondering if somebody could help me understand the answer.
BCD is binary coded decimal. In BCD, every 4 bits is used to represent a single digit from 0 to 9. So if you have 64 bits, that gives you 64/4 = 16 decimal digits, which means you can have 10^16 different integers.