MVI Function Review Input X is p -valued variable. Each Input can have Value in Set {0, 1, 2,..., p...
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Transcript of MVI Function Review Input X is p -valued variable. Each Input can have Value in Set {0, 1, 2,..., p...
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
1:{0,1,..., } {0,1, }iF p X
SOP Bit RepresentationSOP Bit Representation
X1 X2 X3
01 – 012 – 012311 – 100 – 100011 – 010 – 010111 – 001 – 001001 – 110 – 0001
c = X1S
1 X2S
2 . . . XnS
n , Si Pi
Cube in an n-dimensional hyper-cube
Restriction (Cofactor) Operation Restriction (Cofactor) Operation in MV cube calculusin MV cube calculus
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 = X1S
1 X2S
2 . . . XnS
n be a product. Then, the restriction F(|c) of F to c is obtained as follows:
(1) For each product term in F, make a logical product with c. Delete the zero terms.
(2) Let d = X1T
1 X2T
2 . . . XnT
n be a product obtained in (1). Replace d with X1(T
1
S1
) X2
(T2 S
2) . . . Xn
(Tn S
n)
Procedure for Finding F(|c)11 – 100 – 100011 – 010 – 010111 – 001 – 001001 – 110 – 0001
F = c = (01-101-1111)
Step 1: Bit-wise AND each product term in F with c01 – 100 – 100001 – 001 – 001001 – 100 – 0001
F c =
Step 2: Bit-wise OR each product term in F c with c
F(|c) =11 – 110 – 100011 – 011 – 001011 – 110 – 0001 Students: check this in
maps and equations
Cofactor ConceptBy 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 F =
x y z 01 – 01 – 1111 – 01 – 0101 – 11 – 01f (0,y,z)
c0= (10-11-11)
F c0 = [10- 01- 01]
F(|c0) = [11- 01- 01]
f (1,y,z)c1= (01-11-11)
F c1= F(|c1) =01 – 01 – 1101 – 01 – 0101 – 11 – 01
11 – 01 – 1111 – 01 – 0111 – 11 – 01
Students: check this in maps and equations
Tautology for MV
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
F1 =01 – 100 – 110011 – 111 – 0010
11 – 110 – 111011 – 110 – 000111 – 001 – 1111
F2 =
Z0 1 2 3
Example: No Yes
Y 0 1 2 0 1 2
X = 0
X = 1
Can confirm with K-Maps
Inclusion Relation for MV
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 – 100011 – 010 – 010111 – 001 – 001001 – 110 – 0001
F = Example:
c1= (01- 100 - 1001)
F(|c1) =11 – 111 – 111011 – 111 – 0111
F(|c1) 1, c1 F
c2= (11- 010 - 1101)
F(|c2) =11 – 111 – 011101 – 111 – 0011
F(|c2) 1, c2 F
Equivalence RelationLet
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
i = 1 j = 1
p q
F(|x) 1, F(|xy) 1, G(|xy) 1, and G(|y) 1, Thus F G
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
i = 1
k
i = 1
k
This was already illustrated graphically to check SAT, TAUTOLOGY and other similar
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 = XjS
A , c2 = XjS
B
SA SB = Pj and SA SB =
i = 1
k
Divide and Conquer MethodUsing 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 MethodSelection Method:1. Chose all the variables with the
maximum number of active columns 12. 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
Divide and Conquer MethodExample:
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 – 100011 – 010 – 010011 – 001 – 001001 – 110 – 0001
F =
c1= (11- 111 - 1100)
F(|c1) =11 – 100 – 101111 – 010 – 0111
F(|c2) =11 – 001 – 111001 – 110 – 1101
c2= (11- 111 - 0011)
We split with respect to some values, like cutting a KMAP in our previous examples
Complementation of SOPS
Let
F = ci 1 and ci cj = 0 (i j ).
Then, the complement of F is
F = ci F(|ci)
i = 1
k
i = 1
k
Algorithm for Complementation of SOPS
1. F consist of one product c
F = X1S
1 X2S
2 . . . XnS
n
Then
F = X1S
1 + X1S
1 X2S
2 + . . . + X1S
1 X2S
2 . . . Xn-1S
n-1 XnS
n
2. F consist of more than one product
Expand F into F = c1 F(|c1) + c2 F(|c2) , where
c1 = XjS
A , c2 = XjS
, SA SB = Pj and SA SB =
F = c1 F(|c1) + c2 F(|c2)
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 – 100011 – 010 – 010011 – 001 – 001101 – 110 – 0001
F =
c1= (11- 111 - 1100)
F1= F(|c1) =11 – 100 – 101111 – 010 – 0111
F2= F(|c2) =11 – 001 – 111101 – 110 – 1101
c2= (11- 111 - 0011)
Complementation of SOPS Example (Continued)
F1= F(|c1) =
11 – 111 – 1011 11 – 110 – 0111
2. Next, expand F1 variable X2 , F1 = c3 F1 (|c3) + c4 F1 (|c4)
c3= (11- 100 - 1111)
F3= F(|c3) = F4= F(|c4) =
c4= (11- 011 - 1111)
11 – 100 – 101111 – 010 – 0111
Complementation of SOPS Example (Continued)
F2= F(|c2) =
01 – 111 – 1101 11 – 111 – 1111
3. Next, expand F2 variable X2 , F2 = c5 F2 (|c5) + c6 F2 (|c6)
c5= (11- 110 - 1111)
F5= F(|c5) = F6= F(|c6) =
c6= (11- 001 - 1111)
11 – 001 – 111101 – 110 – 1101
Complementation of SOPS Example (Continued)
4. F3 through F6 are single products so we apply alg. Step 1.
11 – 111 – 0100F3=
10 – 111 – 111101 – 111 - 0010
F5=
11 – 001 – 111111 – 110 – 1000
F4=
F6= 0
11 – 111 – 1011F3=
01 – 111 – 1101F5=11 – 110 – 0111F4=
11 – 111 – 1111F6=
Complementation of SOPS Example (Completed)
5. Combining all the products gives:
F=
11 – 100 – 010011 – 001 – 110011 – 010 – 100010 – 110 – 001101 – 110 - 0010
F = c1F1 + c2F2 = c1(c3F3 + c4F4 ) + c2 (c5F5 + c6F6 )
= c1c3F3 + c1c4F4 + c2 c5F5 + c2 c6F6
Problems to remember and solveProblems to remember and solve
1. Multi-output Multi-valued prime implicants2. Covering for MV functions.3. Cofactors of MV functions.4. Visualization of MV functions5. Complementation of MV functions.6. Decision trees and Decision Diagrams for MV functions7. Cube Calculus operations and Algorithms for MV functions.