Minimization of AND-OR-EXOR Three Level
Networks with AND gate Sharing
Hasnain Heickal (SH-223)
Overview
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Introduction
AND-OR-EXOR networks
Objective
Preliminary Definitions
Properties of EX-SOPs
Minimization of EX-SOPs
Idea of Minimization
Summary
Reference
Introduction
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Logic networks are usually designed using AND
and OR gates (SOP).
AND-EXOR networks (EX-SOP) are
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More compact.
Easily testable.
Fault tolerant
AND-OR-EXOR Networks
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A two input EXOR gate is used.
AND gates can be shared or not shared.
If not shared an EX-SOP for a function F can be
written as F = Fa xor Fb
If shared the EX-SOP
can be written as
F = (Fa + Fs) xor
(Fb + Fs)
Objective
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Designing a AND-OR-EXOR three level network.
Minimizing the number of products.
We will discuss an exact algorithm for
minimization.
Preliminary Definitions
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τ(F)
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Number of products in an expression F.
τ(ABC + A’BC + AC) = 3
τ(SOP:f)
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Number of product in a minimum SOP for f.
τ(SOP : (ABC + A’BC + AC)) = 2 because it can be
minimized as BC + AC.
Preliminary Definitions
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τ(EX-SOPNS:f)
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τ(EX-SOPPS:f)
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Number of products in a minimum EX-SOP for f
with no product sharing.
Number of products in a minimum EX-SOP for f
with product sharing.
A logic function f can represented as
f = (fa + g) xor (fb + g)……………………(1)
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τ(EX-SOPPS:f) = min{τ(SOP:g) + τ(SOP:fa) + τ(SOP:fb)}
τ(EX-SOPNS:f) = min{τ(SOP:fa) + τ(SOP:fb)} while
considering g = 0
Properties of EX-SOPs
On the Karnaugh map of a function, a cell that
contains a 1(one) is called a 1-cell and a cell that
contains a 0(zero) is called 0-cell.
 Property 1:
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In a K-map for an EX-SOP, any 1-cell must be
covered by the loop(s) for exactly one SOP.
If a 0-cell is covered, then it must be covered by at
least one loop from both SOPs.
Definition 6:
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Let g(x) and h(x) be n variable functions. B = {0,1}, if
for every a ε Bn g(a)=1 satisfies h(a)=1 then g  h
Minimization of EX-SOPs
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Let g represent the shared products of an EXSOP of function f. The number of different
products in a minimum EX-SOP for f with
product sharing is denoted by τ(EX-SOPPS:f:g).
To compute τ(EX-SOPPS : f : g) using the Eq 1, g
is fixed and we choose fa and fb such that Eq 1
satisfies. Thus we have
τ(EX-SOPPS:f:g) = τ(SOP:g) + min{ τ(SOP:fa) +
τ(SOP:fb) }
Minimization of EX-SOPs
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Lemma 2:
 ( EX  SOPPS : f : g )  ( SOP :g ) 
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min
h g
 ( EX  SOPNS : f  h)
The proof of the lemma is out of scope.
The proof can be found on the paper [1].
Idea of Minimization
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The idea is for 5 of less number of variables.
We will try for all possible g and minimize the
following Eq for all possible g.
 ( EX  SOPPS : f : g )  ( SOP :g ) 
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We need to use K-map.
min
h g
 ( EX  SOPNS : f  h)
Example
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Let us consider g = A’C’D.
Possible values of h are
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A’BC’D
A’B’C’D
A’C’D
We have to find h that makes
 ( EX  SOPNS : f  h) minimum.
C’D’ C’D
A
A’B’
CD
1
A’B
1
g
1
AB
1
1
1
AB’
CD’
1
Example
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Lets first try with h = A’BC’D
So K-map for f v h will be
C’D’ C’D
A
CD
A’B’
1
A’B
1
1
1
AB
1
1
1
AB’
CD’
1
Example
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Rules for EX-SOPNS
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From the K-map we can see
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Loop 1-cell entries odd number
of times.
Loop 0-cell entries even number
of times.
fa = B
fb = A’CD’
 ( EX  SOPNS : f  h) = 2
τ(SOP:g) = 1
Τ(EX-SOPPS:f:g) = 3
We need to do this for every h.
C’D’ C’D
A
CD
A’B’
CD’
1
A’B
1
1
1
AB
1
1
1
AB’
fa
fb
1
Choosing g
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We can choose g using the following lemma :
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To obtain minimum EX-SOP of f it is sufficient to
consider only the prime implicants of f’ as shared
product of candidate.
The proof of this lemma can also be found in the
paper [1].
To find the prime implicants of f’ we can also use Kmap.
Drawbacks
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Choosing g is very time consuming.
We can use “Lookup Tables” to optimize it.
Overall an NP equivalent problem.
Summary
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We have seen the algorithm for minimizing
AND-OR-EXOR three level networks.
We have seen the algorithm for 5 or less
variables.
There exists algorithm for more variables.
References
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D. Debnath and T. Sasao, “Minimization of
AND-OR-EXOR three level networks with AND
gate sharing.”
Thank You