‫מבנה מחשב – מבוא למחשבים ספרתיים‬
‫תרגול מספר ‪5‬‬
‫‪Encoders, Decoders & Shifters‬‬
EncoderF
• Specification:
input : x , weight ( x)  1
output : y, F
Functionality:
 y[n  1 : 0]   i  x[i]  1 , i  1,2,....
 y[n  1 : 0]   0  x[0]  1 or x  0
F 1  x  0
2n
2n
Proof of Correctness
• By induction.
• Base:
n 1
 y[0 : 0]  1  y[0]  1 
x[1]  1
 y[0 : 0]  0  x[0]  1 or x[1]  x[0]  0
F  NOR ( x[0], x[1])  0  x[1]  x[0]  0
Proof of Correctness
• Induction assumption:
EncoderF(n-1) is correct
• Induction step:
Prove that if the design is correct for k-1, it is
also correct for k.
Proof of Correctness
• Induction step:
• We need to examine the output closer
n 1
y[n  2 : 0]  OR ( EncoderF ( x[2  1 : 2 ]) ,
n
EncoderF ( x[2
n 1
 1 : 0])
Since weight(x)=1, one of the components must be all
zeros.
)
Induction step - continued
• There is at most one i for which x[i] = 1
• First case:
i  2 n 1
 y[n  2 : 0]   i
y[n  1]  NOT ( F ( x[2 n  1 : 2 n 1 ]))  NOT (1)  0
  y[n  1 : 0]   i
Induction step continued
• Second case: i  2 n 1
 y[n  2 : 0]   i  2
n 1
y[n  1]  NOT ( F ( x[2n  1 : 2n1 ] )  NOT (0)
  y[n  1 : 0]    y[n  2 : 0]  2
  y[n  1 : 0]   i
n 1
Induction Step continued
• Third case:
x0
2n
n 1
n 1
y[n  2 : 0]  OR (0 ,0 )  0
n 1
n 1
y[n  1]  NOT ( F ( x[2  1 : 2 ])  NOT (1)  0
n
Induction Step continued
• Calculating F
F ( x)  AND( F ( x[2 n  1 : 2 n 1 ]) ,
F ( x[2 n 1  1 : 0]) )  1
 x[2  1 : 2
n
 x0
2n
n 1
]0
2 n1
and x[2
n 1
 1 : 0]  0
2 n1
Induction Conclusion
• If EncoderF(k-1) is correct then
EncoderF(k) is correct.
• Since EncoderF(1) is correct (base), it
follows that EncoderF(n) is correct for all n.
Cost of EncoderF
C (1)  C ( NOR )
C (n)  2  C (n  1)  C ( AND)  C ( NOT )  (n  1)  C (OR )
 2  C (n  1)  O(n)
 O(2 n )
 O(n) in terms of input size
Delay of EncoderF
D(1)  D( NOR )
D(n)  D(n  1)  max{ D( AND), D( NOT ), D(OR )}
 D(n  1)  const
 O ( n)
 O(log n) in terms of input size
Cost of Decoder
• Each of the 2n output gates is fed by a
different non-trivial gate, therefore the cost
must be at least that much.
Cost of Encoder*
• There are 2n inputs which are fed into nontrivial gates. If the fan-in is bounded, then
there must be an order of that size of input
gates
• Another way to view it: each output bit has
a cone of exactly 2n-1. Why?
Delay of Decoder & Encoder*
Decoder - Using the theorem shown in class:
Delay  log c | cone( f ) |  O(log n)
Encoder - Using the same method:
Delay  log c | cone( f ) |  log c 2  O(n)
n
Barrel Shifter
Functionality Preservation
• If we swap the order of the blocks with
their inputs will the functionality be
preserved?
• If we swap the order of the blocks without
swapping their inputs will the functionality
be preserved?
Asymptotic Optimality of Delay
• The delay of the circuit is log(n)
• How can we prove optimality?
• We can view the circuit as having n+log(n)
inputs.
• Each output bit is in a certain situation
affected by every input bit. Therefore the
cone of every output bit is n+log(n)
• This results in a delay that is O(log(n)).