Multi-Valued Input Two-Valued Output Functions

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Multi-Valued InputTwo-Valued OutputFunctions
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Example

• Automobile features

0 1 2 3
X1 Trans Man Auto
X2 Doors 2 3 4
X3 Colour Silver Red Black Blue
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Example

• Function Table

X1 X2 X3 F
0 0 0 1
0 0 1 0
0 0 2 0
0 0 3 0
0 1 0 0
0 1 1 1
0 1 2 0
0 1 3 1
X1 X2 X3 F
0 2 0 0
0 2 1 0
0 2 2 1
0 2 3 0
1 0 0 1
1 0 1 0
1 0 2 0
1 0 3 1
X1 X2 X3 F
1 1 0 0
1 1 1 1
1 1 2 0
1 1 3 1
1 2 0 0
1 2 1 0
1 2 2 1
1 2 3 0
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Definition

• A mapping F P1 ? P2 ? ??????Pn???? is a
  multi-valued input two-valued output function, Pi
  0,1, pi-1, and B 0,1. Let X be a
  variable that takes one value in P 0,1, ,
  p-1. Let S ? P. Then XS is a literal of X. When
  X ? S, XS 1, and when X ? S, XS 0. Let Si ?
  Pi, then X1S1 X2S2 XnSn is a logical product.
• We can also define minterm and SOP
• Any binary function can be represented this way
• Problem Find the SOP for the function given in
  the previous slide.

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Bit Representation

• Example Bit representation for the previous
  example
• x1 x2 x3
• 01-012-012311-100-100011-010-010111-001-00100
  1-110-0001

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Restriction
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Example
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Restriction

• Theorem c F c F(c)
• The restriction is also called cofactor
• What is the relation Shannons expansion?

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Example
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Tautology

• When the logical expression F is equal to 1 for
  all the input combinations, F is a tautology. The
  problem of determining whether a given logical
  expression is a tautology or not is the tautology
  decision problem.
• Example. Which are tautologies?

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Inclusion Relation

• Let F and G be logic functions. For all the
  minterms c such that F(c) 1, if G(c) 1, then
  F G, and G contains F. If a logic function F
  contains a product c, then c is an implicant of
  F.
• Let c be a logical product and F be a logical
  expression. Then c F ? F(c) ?1.

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Example

• When c2 (11-010-1101)?

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Equivalence

• The following theorem shows that the
  determination of the equivalence of two SOPs is
  transformed to the tautology problem

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Divide and Conquer Method
(9.1)
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Divide and Conquer Method

• By using the previous theorem, a given SOP is
  partitioned into k SOPs. In performing some
  operation on F, first expand F into the form
  (9.1). Then, for each F(ci), apply the operation
  independently. Finally, by combining the results
  appropriately, we have the results for the
  operation on F. Since, the same operation can be
  applied to F(ci) as to F, this method can be
  computed by a recursive program.
• Definition. Let t(F) be the number of products in
  an SOP F.

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Divide and Conquer Methods
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Selection Method for Variables

• Chose variables that have the maximum number of
  active columns
• Among those, chose variables where the total sum
  of 0s in the array is maximum
• Example

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Complementation of SOPs
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Complementation of SOPs

• When n 10, this method is about 100 times
  faster than using De Morgans theorem

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Example

• Find the complement for F

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Tautology Decision

• When there is a variable Xi and at least one
  constant a ? Pi satisfying F(Xi a) F(X1, ,
  Xi, , Xn), the function F is weakly unate with
  respect to the variable Xi.
• In an array F, consider the sub-array consisting
  of cubes that depend on Xi. In the variable Xi in
  this array, if all the values in a column are 0,
  then the SOP F is weakly unate with respect to
  the variable Xi.

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Example

• Consider the F. Is F weakly unate?
• 1111-1110-11101111-1101-11010110-0110-11010101
  -0111-1101

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Theorems
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Tautology Decision
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Generation of Prime Implicants

• Let X be a variable that takes a value in P 0,
  1, , p-1. If there is a total order (?) on the
  values of variable X in function F, such that j ?
  k (j,k ? P) implies F(Xj) F(Xk), then the
  function is strongly unate with respect to X. If
  F is strongly unate with respect to all
  variables, then the function F is strongly unate.
• Assume that F is an SOP. If there is a total
  order (?) among the values of a variable X, and
  if j ? k, then each product term of the SOP
  F(Xj) is contained by all the product terms of
  the SOP F(Xk). In this case the SOP F is
  strongly unate with respect to X.
• Lemma. If F is strongly unate with respect to
  Xi, then F is weakly unate with respect to Xi.

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Generation of Prime Implicants
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Generation of Prime Implicants
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The Sharp Operation

• Sharp operation () and disjoined sharp operation
  ( ) compute F??G.
• Example. Let a (11-11-11) and b (01-01-01).
  Find ab and a b
• Example. Let B b1,b2, where b1 (11-11-11)
  and b2 (10-10-10). Let Cc1,c2,c3, where c1
  (10-11-11), c2 (11-10-11), and c3 (11-11-10).
  Find aC and a C