Multi-Valued Input Two-Valued Output Functions
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Transcript of Multi-Valued Input Two-Valued Output Functions
Multi-Valued InputTwo-Valued Output
Functions
Multi-Valued Input Slide 2
Example
• Automobile features
0 1 2 3
X1 Trans Man Auto
X2 Doors 2 3 4
X3 Colour Silver Red Black Blue
Multi-Valued Input Slide 3
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
Multi-Valued Input Slide 4
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 X1
S1 X2S2… Xn
Sn 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.
Multi-Valued Input Slide 5
Bit Representation
• Example: Bit representation for the previous example:
x1 x2 x3
01-012-012311-100-100011-010-010111-001-001001-110-0001
Multi-Valued Input Slide 6
Restriction
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Multi-Valued Input Slide 7
Example
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Multi-Valued Input Slide 8
Restriction
• Theorem c F = c F(|c)• The restriction is also called cofactor• What is the relation Shannon’s expansion?
Multi-Valued Input Slide 9
Example
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Multi-Valued Input Slide 10
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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Multi-Valued Input Slide 11
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.
Multi-Valued Input Slide 12
Example
• When c2 = (11-010-1101)?
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Multi-Valued Input Slide 13
Equivalence
• The following theorem shows that the determination of the equivalence of two SOPs is transformed to the tautology problem:
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Multi-Valued Input Slide 14
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Divide and Conquer Method
(9.1)
Multi-Valued Input Slide 15
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.
Multi-Valued Input Slide 16
Divide and Conquer Methods
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Multi-Valued Input Slide 17
Selection Method for Variables
• Chose variables that have the maximum number of active columns
• Among those, chose variables where the total sum of 0’s in the array is maximum
• Example
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Multi-Valued Input Slide 18
Complementation of SOPs
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Multi-Valued Input Slide 19
Complementation of SOPs
• When n = 10, this method is about 100 times faster than using De Morgan’s theorem
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Multi-Valued Input Slide 20
Example
• Find the complement for F
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Multi-Valued Input Slide 21
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.
Multi-Valued Input Slide 22
Example
• Consider the F. Is F weakly unate?
1111-1110-11101111-1101-11010110-0110-11010101-0111-1101
Multi-Valued Input Slide 23
Theorems
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Multi-Valued Input Slide 24
Tautology Decision
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Multi-Valued Input Slide 25
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(|X=j) ≤ F(|X=k), 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(|X=j) is contained by all the product terms of the SOP F(|X=k). 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.
Multi-Valued Input Slide 26
Generation of Prime Implicants
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Multi-Valued Input Slide 27
Generation of Prime Implicants
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Multi-Valued Input Slide 28
The Sharp Operation
• Sharp operation (#) and disjoined sharp operation ( # ) compute FG.
• Example. Let a = (11-11-11) and b = (01-01-01). Find a#b and a # b
• Example. Let B = {b1,b2}, where b1 = (11-11-11) and b2 = (10-10-10). Let C={c1,c2,c3}, where c1 = (10-11-11), c2 = (11-10-11), and c3 = (11-11-10). Find a#C and a # C