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The size bound follows immediately. The depth bound requires a more careful analysis.
ShowninFig. 2.21 (a) is a full adder together with notation showing the amount by
which the length of a path from one input to another is increased in passing through it
when the full-adder circuit used is that shown in Fig. 2.14 and described by Equation 2.6 .
From this it follows that
D Ω c ( l + 1 )
j + 1 =max D Ω c ( l + 1 )
+ 2, D Ω s ( l j + 3
j
D Ω s ( l + 1 )
=max D Ω c ( l + 1 )
+ 1, D Ω s ( l j + 2
j
j
1, where s ( l )
l−
1 = c ( l )
l and 0
j
l
for 2
1 . It can be shown by induction that
D Ω c ( k )
= 2 ( k + j )
1, and D Ω s ( k )
= 2 ( k + j )
l−
j
k
j
k
3, 1
2, 0
2,
j
j
k . (See Problem 2.16 .) Thus, D Ω f ( n )
count = D Ω c ( k )
1 =( 4 k
5 ) .
both for 2
k
We now use this bound to derive upper bounds on the size and depth of symmetric func-
tions in the class S n , m .
THEOREM 2.11.1 Every symmetric function f ( n ) : B
n
m can be realized with the following
→B
where φ ( k )= 5 ( 2 k
circuit size and depth over the basis Ω=
{∧
,
,
⊕}
k
1 ) :
C Ω f ( n )
2 ) 2 ( n + 1 )
m
( n + 1 ) / 2
+ φ ( k )+ 2 ( n + 1 )+( 2
log 2 ( n + 1 )
D Ω f ( n )
log 2 ( n + 1 )
+
log 2
log 2 ( n + 1 )
5
4
for k =
log 2 ( n + 1 )
even.
Proof Lemma 2.11.1 establishes bounds on the size and depth of the function f ( n )
for
count
n = 2 k
and fill out the 2 k
1. For other values of n ,let k =
log 2 ( n + 1 )
n
1
variables with 0's.
The elementary symmetric functions are obtained by applying the value of f ( n )
count as
argument to the decoder function. A circui t for this function has been constructed that has
size 2 ( n + 1 )+( 2
2 ) 2 ( n + 1 ) and depth
log 2 ( n + 1 )
log 2
log 2 ( n + 1 )
+ 1.
(See Lemma 2.5.4 .Weusethefactthat2 log 2 m
2 m .) Thus, all elementary symmetric
functions on n variables can be realized with the following circuit size and depth:
C Ω e ( n )
0
2 ) 2 ( n + 1 )
, e ( n )
1
, ... , e ( n )
φ ( k )+ 2 ( n + 1 )+( 2
log 2 ( n + 1 )
n
D Ω e ( n )
0
, e ( n )
1
, ... , e ( n )
n
4 k
5 +
log 2
log 2 ( n + 1 )
+ 1
The expansion of Equation ( 2.15 ) can be used to realize an arbitrary Boolean symmetric
function. Clearly, at most n OR gates and depth
suffice to realize each one of m
arbitrary Boolean symmetric functions. (Since the v t are fixed, no AND s are needed.) This
number of OR scanbereducedto ( n
log 2 n
1 ) / 2 as follows: if
( n + 1 ) / 2
or more elementary
functions are needed, use the complementary set (of at most
( n + 1 ) / 2
functions) and
take the complement of the result. Thus, no more than
( n + 1 ) / 2
1 OR s are needed per
symmetric function (plus possibly one NOT ), and depth at most
log 2
(( n + 1 ) / 2 )
+ 1
log 2 ( n + 1 )
.
 
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