__Problem :__

**Modular arithmetic and 2's complement representation**

Most computers choose a
particular word length (measured in bits) for representing integers and provide
hardware that performs various arithmetic operations on word-size operands. The
current generation of processors have word lengths of 32 bits; restricting the
size of the operands and the result to a single word means that the arithmetic
operations are actually performing arithmetic modulo 2

^{32}.
Almost all computers use
a 2's complement representation for integers since the 2's complement addition
operation is the same for both positive and negative numbers. In 2's complement
notation, one negates a number by forming the 1's complement (i.e., for each
bit, changing a 0 to 1 and vice versa) representation of the number and then
adding 1. By convention, we write 2's complement integers with the
most-significant bit (MSB) on the left and the least-significant bit (LSB) on
the right. Also by convention, if the MSB is 1, the number is negative;
otherwise it's non-negative.

A.
How many different values can be encoded in a 32-bit word?

Each bit can be either "0" or
"1", so there are 2

^{32}possible values which can be encoded in a 32-bit word.
B.
Please use a 32-bit 2's complement representation to answer the
following questions. What are the representations for

zero

the most positive integer that can be represented

the most negative integer that can be represented

the most positive integer that can be represented

the most negative integer that can be represented

What
are the decimal values for the most positive and most negative integers?

zero = 0000 0000 0000 0000 0000 0000 0000 0000

most positive integer = 0111 1111 1111 1111 1111 1111 1111 1111 = 2

most negative integer = 1000 0000 0000 0000 0000 0000 0000 0000 = -2

most positive integer = 0111 1111 1111 1111 1111 1111 1111 1111 = 2

^{31}-1most negative integer = 1000 0000 0000 0000 0000 0000 0000 0000 = -2

^{31}
C.
Since writing a string of 32 bits gets tedious, it's often
convenient to use hexadecimal notation where a single digit in the range 0-9 or
A-F is used to represent groups of 4 bits using the following encoding:

D.
hex bits hex bits hex bits hex bits

E.
0
0000 4 0100
8 1000 C
1100

F.
1
0001 5 0101
9 1001 D
1101

G.
2
0010 6 0110
A 1010 E
1110

H.
3
0011 7 0111
B 1011 F
1111

Give
the 8-digit hexadecimal equivalent of the following decimal and binary numbers:
37

_{10}, -32768_{10}, 11011110101011011011111011101111_{2}.
37 = 00000025

-32768 = FFFF8000

1101 1110 1010 1101 1011 1110 1110 1111

_{16}-32768 = FFFF8000

_{16}1101 1110 1010 1101 1011 1110 1110 1111

_{2}= DEADBEEF_{16}
D.
Calculate the following using 6-bit 2's complement arithmetic
(which is just a fancy way of saying to do ordinary addition in base 2 keeping
only 6 bits of your answer). Show your work using binary (base 2) notation.
Remember that subtraction can be performed by negating the second operand and
then adding it to the first operand.

E.
13 + 10

F.
15 - 18

G.
27 - 6

H.
-6 - 15

I.
21 + (-21)

J.
31 + 12

Explain
what happened in the last addition and in what sense your answer is
"right".

13 = 001101 15 = 001111 27 = 011011

+ 10 = 001010 -18 =
101110 - 6 = 111010

=========== =========== ===========

23 = 010111 -3 = 111101 21 = 010101

-6 = 111010 21 = 010101 31 = 011111

-15 = 110001 -21 = 101011 +12 = 001100

=========== ============ ===========

-21 = 101011 0 = 000000 -21 = 101011 (!)

In
the last addition, 31 + 12 = 43 exceeds the maximum representable positive
integer in 6-bit two's complement arithmetic (max int = 2

^{5}-1 = 31). The addition caused the most significant bit to become 1, resulting in an "overflow" where the sign of the result differs from the signs of the operands.
E.
At first blush "Complement and add 1" doesn't seem to an
obvious way to negate a two's complement number. By manipulating the expression
A+(-A)=0, show that "complement and add 1" does produce the correct
representation for the negative of a two's complement number. Hint: express 0
as (-1+1) and rearrange terms to get -A on one side and XXX+1 on the other and
then think about how the expression XXX is related to A using only logical
operations (AND, OR, NOT).

Start by expressing zero as (-1 + 1):

A + (-A) = -1 + 1

Rearranging terms we get:

(-A) = (-1 - A) + 1

The two's complement representation for -1 is all
ones, so looking at (-1 - A) bit-by-bit we see:

1 ...
1 1

- A

_{N-1}... A_{1}A_{0}
===============

~A

_{N-1}... ~A_{1}~A_{0}
where "~" is the bit-wise complement
operator. We've used regular subtraction rules (1 - 0 = 1, 1 - 1 = 0) and
noticed that 1 - A

_{i}= ~A_{i}. So, finally:
(-A) = ~A + 1

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