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 X{j} where j is the logic value
Empty Literal: X{}
Full Literal has Values S={0, 1, 2, …, p-1}
X{0,1,…,p-1} Equivalent to Don’t Care

2. SOP Bit Representation
X X X3
01 – 012 – 0123
11 – 100 – 1000
11 – 010 – 0101
11 – 001 – 0010
01 – 110 – 0001
c = X1S1 X2S 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 X2S 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 X2T 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
F =
c = ( )
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 Shannon’s 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= ( )
c1= ( )
01 – 01 – 11
01 – 01 – 01
01 – 11 – 01
11 – 01 – 11
11 – 01 – 01
11 – 11 – 01
F c0 = [ ]
F(|c0) = [ ]
F(|c1) =
F c1=

6. 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 Y 1 2
Can confirm
with K-Maps
X = 0
X = 1

7. 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.
Example:
11 – 100 – 1000
11 – 010 – 0101
11 – 001 – 0010
01 – 110 – 0001
F =
c1= ( )
c2= ( )
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

8. 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

9. 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

10. 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

11. Divide and Conquer Method
Using Divide and Conquer we use the recursive application of the restriction operation to attempt to get columns of all 0’s or 1’s (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 0’s in the array is maximum
3. For all variables in step 2, find a column that has the maximum number of 0’s and from among them choose the one with the minimum number of 0’s

12. 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}
X X X3
11 – 100 – 1000
11 – 010 – 0100
11 – 001 – 0010
01 – 110 – 0001
F =
c1= ( )
c2= ( )
11 – 100 – 1011
11 – 010 – 0111
11 – 001 – 1110
01 – 110 – 1101
F(|c2) =
F(|c1) =

13. 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

14. Algorithm for Complementation of SOPS
F consist of one product c
F = X1S1 X2S XnSn
Then
F = X1S1 + X1S1  X2S X1S1  X2S 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)

15. Complementation of SOPS Example
1. Expand F w.r.t. X3 and let SA = {0,1} and SB = {2,3}
X X X3
11 – 100 – 1000
11 – 010 – 0100
11 – 001 – 0011
01 – 110 – 0001
F =
c1= ( )
c2= ( )
11 – 100 – 1011
11 – 010 – 0111
11 – 001 – 1111
01 – 110 – 1101
F2= F(|c2) =
F1= F(|c1) =

16. 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= ( )
c4= ( )
11 – 110 – 0111
11 – 111 – 1011
F4= F(|c4) =
F3= F(|c3) =

17. 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= ( )
c6= ( )
11 – 111 – 1111
01 – 111 – 1101
F6= F(|c6) =
F5= F(|c5) =

18. Complementation of SOPS Example (Continued)
11 – 111 – 1011
F3=
11 – 110 – 0111
F4=
01 – 111 – 1101
F5=
11 – 111 – 1111
F6=
4. F3 through F6 are single products so we apply
alg. Step 1.
11 – 111 – 0100
F3=
11 – 001 – 1111
11 – 110 – 1000
F4=
10 – 111 – 1111
01 –
F5=
F6= 0

19. 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 – F=