General Computer Science 320201 GenCS I & II Lecture ... - Kwarc

So, the conditional sum adder has a smaller depth than the carry chain adder. This smaller
depth is paid with higher cost.
There is another adder that combines the small cost of the carry chain adder with the low
depth of the conditional sum adder. This carry lookahead adder CLA n has a cost C(CLA n ) ∈
O(n) and a depth of dp(CLA n ) ∈ O(log 2(n)).
c○: Michael Kohlhase 250
Instead of perfecting the n-bit adder further (and there are lots of designs and optimizations out
there, since this has high commercial relevance), we will extend the range of arithmetic operations.
The next thing we come to is subtraction.
3.2.2 Arithmetics for Two’s Complement Numbers
This of course presents us with a problem directly: the n-bit binary natural numbers, we have
used for representing numbers are closed under addition, but not under subtraction: If we have
two n-bit binary numbers B(n), and B(m), then B(n + m) is an n + 1-bit binary natural number.
If we count the most significant bit separately as the carry bit, then we have a n-bit result. For
subtraction this is not the case: B(n − m) is only a n-bit binary natural number, if m ≥ n
(whatever we do with the carry). So we have to think about representing negative binary natural
numbers first. It turns out that the solution using sign bits that immediately comes to mind is
not the best one.
Negative Numbers and Subtraction
Note: So far we have completely ignored the existence of negative numbers.
Problem: Subtraction is a partial operation without them.
Question: Can we extend the binary number systems for negative numbers?
Simple Solution: Use a sign bit. (additional leading bit that indicates whether the number is positive)
Definition 426 ((n + 1)-bit signed binary number system)
〈〈an, . . . , a0〉〉 −
:=
〈〈an−1, . . . , a0〉〉
−〈〈an−1, . . . , a0〉〉
if an = 0
if an = 1
Note: We need to fix string length to identify the sign bit. (leading zeroes)
Example 427 In the 8-bit signed binary number system
10011001 represents -25 ((〈〈10011001〉〉 − ) = −(2 4 + 2 3 + 2 0 ))
00101100 corresponds to a positive number: 44
c○: Michael Kohlhase 251
Here we did the naive solution, just as in the decimal system, we just added a sign bit, which
specifies the polarity of the number representation. The first consequence of this that we have to
keep in mind is that we have to fix the width of the representation: Unlike the representation for
binary natural numbers which can be arbitrarily extended to the left, we have to know which bit
is the sign bit. This is not a big problem in the world of combinational circuits, since we have a
fixed width of input/output edges anyway.
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