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Table 10.5
Applying g
sa
¼
x
1
x
2
to multivibrator states
Truth table
x
1
x
2
g
as
Prepared list
Tagged list
Assumed list structure
0
0
0
a
1
b
1
¼ 1
1
¼a
1
b
1
0
1
1
a
1
b
2
¼ 1
1
¼a
1
b
2
1
0
1
a
2
b
1
¼ 1
1
¼a
2
b
1
1
1
0
a
2
b
2
¼ 1
1
¼a
2
b
2
identifying an unknown g
sa
is illustrated for the case n
2 in Table
10.5
. The two
multivibrators are represented by [a
1
a
2
]
0
,[b
1
b
2
]
0
. The two multivibrators are
prepared to have equal probability and zero phase as suggested by the Prepared
List in the table. This list is ordered like a binary count merely for convenience:
a
1
b
1
, a
1
b
2
, a
2
b
1
, a
2
b
2
. This form is inspired by the terms of a direct product
a
¼
b
.
Probability normalization factors have been ignored in the tables to simplify the
notation.
The end result of applying the multivibrator function g
sa
is selected phase
reversals in a
1
b
1
, a
1
b
2
, a
2
b
1
, a
2
b
2
, as suggested by the Tagged List. The locations
of the negative signs correspond to the 1s in the truth table under g
as
. This is what
the function does; it selectively tags the combinations a
1
b
1
, a
1
b
2
, a
2
b
1
, a
2
b
2
with
negative signs. This in turn results in certain sign changes within the qubits [a
1
a
2
]
0
,
[b
1
b
2
]
0
. A solving procedure is a simple way to determine what these sign changes
are.
First equate the Assumed List entries to the Tagged List entries, for
example a
1
b
1
¼
1, a
1
b
2
¼
1, a
2
b
1
¼
1, a
2
b
2
¼
1. Begin by assuming
that a
1
¼
1. Solve the equations to obtain a
1
¼
b
1
¼
1, a
2
¼
b
2
¼
1. Then
[a
1
a
2
]
0
,[b
1
b
2
]
0
¼
1]
0
;thisinturncanbetransformedto[01]
0
,
[0 1]
0
using h-transforms, which upon sampling can be observed as
11
,and
will be
11
with certainty each time it is observed. This classifies the function,
since the math is such that only symmetric and antisymmetric functions with
truth tables “0110” or “1001” would be observed to be
11
.
A function with the truth table “1 0 0 1” is also symmetric (about its center).
Prepared multivibrators can be transformed as in the above table to become [
1]
0
,[1
[1
11]
0
,
[0 1]
0
,[01]
0
but this is still observed to be
11
.
A negative sign associated with
11
would not be seen by sampling, because
sampling does not respond to phase.
Functionssuchas“0110”and“1001”aretermedcomplementary functions.
Complementary multivibrator functions will give the same sampled output. Thus
a sampled result of
11
classifiesagivenfunctiontobe“0110”oritscomplement
“1001”withintheclassofsymmetricandantisymmetricfunctions.
As another example of this, consider Table
10.6
, which is a function with a truth
table “0 0 1 1” and so is antisymmetric; it transforms [1 1]
0
,[11]
0
1]
0
. This, in turn, transforms to
[1
1]
0
,
[1 1]
0
. Then it transforms to
10
. This ultimately classifies the function as being
either “0 0 1 1” or “1 1 0 0.”
into [1
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