MVI Function Review

1
MVI Function Review

• Input X is p-valued variable. Each Input can have
  Value in Set 0, 1, 2, ..., pi-1
• literal over X corresponds to subset of values
  of S ? 0, 1, ... , p-1 denoted by XS or Xj
  where j is the logic value
• Empty Literal X?
• Full Literal has Values S0, 1, 2, , p-1
• X0,1,,p-1 Equivalent to Dont Care

2
SOP Bit Representation

• X1 X2 X3
• 01 012 0123
• 11 100 1000
• 11 010 0101
• 11 001 0010
• 01 110 0001

c X1S1 X2S2 . . . XnSn , Si ?
Pi Cube in an n-dimensional hyper-cube
3
Restriction (Cofactor) Operation

• Restriction of Two-Valued Output Function F
  obtained by restricting Domain to D, denoted by
  F(D)
• For SOP, the restriction is defined as follows
• Let F be a SOP, and c X1S1 X2S2 . . .
  XnSn be a product. Then, the restriction F(c)
  of F to c is obtained as follows
• For each product term in F, make a logical
  product with c. Delete the zero terms.
• Let d X1T1 X2T2 . . . XnTn be a product
  obtained in (1). Replace d with X1(T1? S1) X2(T2?
  S2) . . . Xn(Tn? Sn)

4
Procedure for Finding F(c)

• 11 100 1000
• 11 010 0101
• 11 001 0010
• 01 110 0001

c (01-101-1111)
Step 1 Bit-wise AND each product term in F with
c
01 100 1000 01 001 0010 01 100 0001
F? c
Step 2 Bit-wise OR each product term in F ? c
with c
11 110 1000 11 011 0010 11 110 0001
F(c)
5
Cofactor Concept
By Shannons Expansion f (x1, x2 , . . , xn )
x1 f (0, x2 , . . , xn ) x1 f (1, x2 , . . ,
xn )
F(c0)
F(c1)
Where c0 and c1 are cubes with x1 0 and x1 1,
respectively. Example f xy yz zx
x y z 01 01 11 11 01 01 01
11 01
F
f (0,y,z)
f (1,y,z)
c0 (10-11-11)
c1 (01-11-11)
11 01 11 11 01 01 11 11 01
01 01 11 01 01 01 01 11 01
F(c1)
F? c0 10- 01- 01 F(c0) 11- 01- 01
F? c1
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Tautology
When the logical expression F is equal to
logical 1 for all the input combinations, F is a
tautology. Tautology Decision Problem -
determining if logical expression is or is not a
tautology
Example No
Yes
11 110 1110 11 110 0001 11 001 1111
01 100 1100 11 111 0010
F1
F2
Z 0 1 2 3
Y 0 1 2 0 1 2
Can confirm with K-Maps
X 0 X 1
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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 F contains a product
c then c is an implicant of F.
11 100 1000 11 010 0101 11 001
0010 01 110 0001
Example
F
c1 (01- 100 - 1001)
c2 (11- 010 - 1101)
11 111 1110 11 111 0111
11 111 0111 01 111 0011
F(c2)
F(c1)
F(c1) ? 1, c1 ? F
F(c2) ? 1, c2 ? F
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Equivalence Relation
Let F ? fj and G ? gj then F ? G ?
F(gj) ?1 (j 1, . . , q) and G(fj) ?1 (i 1,
. . , p) Example F xy y and G x xy
p
q
i 1
j 1
F(x) ? 1, F(xy) ? 1, G(xy) ? 1, and G(y) ? 1,
Thus F ? G
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Divide and Conquer Method
Let F be a SOP and ci (i 1, 2,. . , k) be the
cubes satisfying the following conditions
F ? ci ? 1 and ci ? cj 0 (i ? j
). Then, can partition SOP into k SOPs
F ? ci ? F(ci) Operations can be done on
each F(ci) independently and then combined to
get result on F
k
i 1
k
i 1
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Divide and Conquer Method
Let t(F) be the number of products in an SOP F.
We can use Divide and Conquer Theorem to minimize
? t(F(ci) ) and thus the number of
products. Partition Example k 2, c1 XjSA
, c2 XjSB SA ? SB Pj and SA ? SB
?
k
i 1
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Divide and Conquer Method
Using Divide and Conquer we use the recursive
application of the restriction operation to
attempt to get columns of all 0s or 1s (they
can be ignored). A column with both 0 and 1 is
active. Selection Method 1. Chose all the
variables with the maximum number of active
columns 1 2. Among the variables chosen in step
1, choose variables where the total sum of 0s in
the array is maximum 3. For all variables in
step 2, find a column that has the maximum number
of 0s and from among them choose the one with
the minimum number of 0s
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Divide and Conquer Method
Example X2 and X3 have the largest
number of active columns. Choose X3 and let SA
0,1 and SB 2,3

• X1 X2 X3
• 11 100 1000
• 11 010 0100
• 11 001 0010
• 01 110 0001

F
c1 (11- 111 - 1100)
c2 (11- 111 - 0011)
11 100 1011 11 010 0111
11 001 1110 01 110 1101
F(c2)
F(c1)
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Complementation of SOPS

Let F ? ci ? 1 and ci ? cj 0 (i ? j
). Then, the complement of F is F ? ci
? F(ci)
k
i 1
k
i 1
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Algorithm for Complementation of SOPS

• F consist of one product c
• F X1S1 X2S2 . . . XnSn
• Then
• F X1S1 X1S1 ? X2S2 . . . X1S1 ? X2S2 .
  . . Xn-1Sn-1 ? XnSn
• F consist of more than one product
• Expand F into F c1 ? F(c1) c2 ? F(c2) ,
  where
• c1 XjSA , c2 XjS , SA ? SB Pj and SA
  ? SB ?
• F c1 ? F(c1) c2 ? F(c2)

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Complementation of SOPS Example

1. Expand F w.r.t. X3 and let SA 0,1
and SB 2,3

• X1 X2 X3
• 11 100 1000
• 11 010 0100
• 11 001 0011
• 01 110 0001

F
c1 (11- 111 - 1100)
c2 (11- 111 - 0011)
11 100 1011 11 010 0111
11 001 1111 01 110 1101
F2 F(c2)
F1 F(c1)
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Complementation of SOPS Example (Continued)


11 100 1011 11 010 0111
F1 F(c1)
2. Next, expand F1 variable X2 , F1 c3 ? F1
(c3) c4 ? F1 (c4)
c3 (11- 100 - 1111)
c4 (11- 011 - 1111)
11 110 0111
11 111 1011
F4 F(c4)
F3 F(c3)
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Complementation of SOPS Example (Continued)


11 001 1111 01 110 1101
F2 F(c2)
3. Next, expand F2 variable X2 , F2 c5 ? F2
(c5) c6 ? F2 (c6)
c5 (11- 110 - 1111)
c6 (11- 001 - 1111)
11 111 1111
01 111 1101
F6 F(c6)
F5 F(c5)
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Complementation of SOPS Example (Continued)


4. F3 through F6 are single products so we apply
alg. Step 1.
11 001 1111 11 110 1000
F4
10 111 1111 01 111 - 0010
F5
F6 0
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Complementation of SOPS Example (Completed)

5. Combining all the products gives
F c1F1 c2F2 c1(c3F3 c4F4 ) c2 (c5F5
c6F6 ) c1c3F3 c1c4F4 c2 c5F5 c2
c6F6
11 100 0100 11 001 1100 11 010
1000 10 110 0011 01 110 - 0010
F