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Computer Architecture I: Digital Design Dr. Robert D. KentPowerPoint Presentation

Computer Architecture I: Digital Design Dr. Robert D. Kent

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Computer Architecture I: Digital Design Dr. Robert D. Kent

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Computer Architecture I: Digital Design Dr. Robert D. Kent

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Computer Architecture I: Digital Design

Dr. Robert D. Kent

Lecture 3

Simplification of Boolean Expressions

SimplificationofBoolean Expressions

A. Cost Models

B. Reduction Techniques -

i. Karnaugh Map Method

- Simplification of complex circuits, starting from their equivalent representation as complex Boolean expressions, requires an understanding of cost.
- We will see that the typical goal is to arrive at Boolean expressions that:
- are expressible in SOP (or POS) form
- involve minimal numbers of literals
- involve a minimal number of gate application levels

- To help us understand how to accomplish circuit minimization, we will study one powerful reduction technique:
- Karnaugh maps

- We focus on two types of cost model.
- Minimum time of completion of circuit logic
- both SOP and POS circuit representations are two-level designs with all the gates at each level performing in parallel with the same time characteristics.

- Minimum cost of circuit construction
- very often it is possible, starting with the SOP or POS forms, to rewrite a near-optimal cost form involving pure nand or nor gates only.

- This leads to four cost theorems.

- Before stating the theorems we need some definitions of terms:
- Literala variable or its complement (unique symbol)
- ImpliesF implies G, means that there does not exist a set of N input literals such that both F=1 and G=0; By contrast, if F=1 then G must be equal to 1.
- SubsumesA term T1subsumes a term T2if and only if (!)all the literals in T2 are also found in T1.If one term subsumes another in an expression, then the subsuming term can always be deleted with changing the function. In other words, if T1 subsumes T2 then T1 may be ignored because the T2 terms account for logic.

- … and some more definitions of terms:
- Implicanta product term is said to be an implicant of a complete function if the product term impliesthe function.
- Implicatea sum term is said to be an implicate of a complete function if the sum term impliesthe function.

- … and some more definitions of terms:
- Prime Implicantan implicant is a prime implicant if it does not subsume any other implicant with fewer literals.This means that if any literal is removed from the term, it no longer implies the function.
- Prime Implicatean implicate is a prime implicate if it does not subsume any other implicate with fewer literals.This means that if any literal is removed from the term, it no longer implies the function.

Note the symmetry between the definitions of implicant (product) and implicate (sum).

- Cost Theorem 1a.When the cost, assigned by some criterion, for a minimal Boolean formula is such that decreasing the number of literals in the disjunctive normal formula does not increase the cost of the formula, there is at least one minimal disjunctive normal formula that corresponds to a sum of prime implicants.
- Cost Theorem 2a. For any cost criterion such that the cost of a formula does not increase when a literal is removed, at least one minimal disjunctive normal formula describing a function is an irredundantdisjunctive normal formula (SOP).

- Cost Theorem 1b.When the cost, assigned by some criterion, for a minimal Boolean formula is such that decreasing the number of literals in the conjunctive normal formula does not increase the cost of the formula, there is at least one minimal conjunctive normal formula that corresponds to a product of prime implicates.
- Cost Theorem 2b. For any cost criterion such that the cost of a formula does not increase when a literal is removed, at least one minimal conjunctive normal formula describing a function is an irredundant conjunctive normal formula (POS).
- These theorems establish that implicates are dual to implicants.

- The algebraic techniques we have developed are too slow and uncertain to apply in cases where the numbers of variables are large.
- To meet the needs of modern circuit analysis and design several methods have been developed that
- are somewhat scalable, and
- permit limited degrees of automation (ie. Programmability).

- The use of mapping, or tableau-based, techniques were developed by Veitch and modified by Karnaugh.
- We will determine these techniques by studying examples in order to establish the rules for map manipulation.

Truth Table

- 1 variable mapx f(x)0 f(0)1 f(1)

Truth Table

- 1 variable mapx f(x)0 f(0)1 f(1)

Literal

Function Values

Binary Values

x0 1

f(0)

f(1)

Truth Table

- 1 variable mapx f(x)0 f(0)1 f(1)

Literal

Literal

Binary Values

Function Values

Function Values

Binary Values

x0 1

f(0)

f(1)

Truth Table

1x2 Karnaugh Map

- 1 variable mapx f(x)0 f(0)1 f(1)

Truth Table

- 2 variable mapx y f(x,y) 0 0 f(0,0)0 1 f(0,1)1 0 f(1,0)1 1 f(1,1)

Truth Table

y0 1

2x2 Karnaugh Map

f(0,0)

f(0,1)

0

x

1

f(1,0)

f(1,1)

- 2 variable mapx y f(x,y) 0 0 f(0,0)0 1 f(0,1)1 0 f(1,0)1 1 f(1,1)

yz00 01 11 10

f(000) f(001) f(011) f(010)

0

x

1

f(100) f(101) f(111) f(110)

2x4 Karnaugh Map

- 3 variable mapx y z f(x,y,z)0 0 0 f(000)0 0 1 f(001)0 1 0 f(010)0 1 1 f(011)1 0 0 f(100)1 0 1 f(101)1 1 0 f(110)1 1 1 f(111)

yz00 01 11 10

f(000) f(001) f(011) f(010)

0

x

1

f(100) f(101) f(111) f(110)

2x4 Karnaugh Map

Note the way that the column indices change by only 1 bit at a time from left to right.

- 3 variable mapx y z f(x)0 0 0 f(000)0 0 1 f(001)0 1 0 f(010)0 1 1 f(011)1 0 0 f(100)1 0 1 f(101)1 1 0 f(110)1 1 1 f(111)

y00 011110

f(000) f(001) f(011) f(010)

0

x

1

f(100) f(101) f(111) f(110)

2x4 Karnaugh Map

An alternative labelling scheme is based on which literal has value 1.

- 3 variable mapx y z f(x,y,z)0 0 0 f(000)0 0 1 f(001)0 1 0 f(010)0 1 1 f(011)1 0 0 f(100)1 0 1 f(101)1 1 0 f(110)1 1 1 f(111)

z

y00 011110

0 1 3 2

0

x

1

4 5 7 6

2x4 Karnaugh Map

One final alternative labelling scheme replaces the function value by the decimal minterm index value.

- 3 variable mapx y z f(x,y,z)0 0 0 f(000)0 0 1 f(001)0 1 0 f(010)0 1 1 f(011)1 0 0 f(100)1 0 1 f(101)1 1 0 f(110)1 1 1 f(111)

z

- 4 variable mapw x y z f(w,x,y,z)0 0 0 0 f(0000)0 0 0 1 f(0001)0 0 1 0 f(0010)0 0 1 1 f(0011)0 1 0 0 f(0100)0 1 0 1 f(0101)0 1 1 0 f(0110)0 1 1 1 f(0111)1 0 0 0 f(1000)1 0 0 1 f(1001)1 0 1 0 f(1010)1 0 1 1 f(1011)1 1 0 0 f(1100)1 1 0 1 f(1101)1 1 1 0 f(1110)1 1 1 1 f(1111)

- 4 variable mapw x y z f(w,x,y,z)0 0 0 0 f(0000)0 0 0 1 f(0001)0 0 1 0 f(0010)0 0 1 1 f(0011)0 1 0 0 f(0100)0 1 0 1 f(0101)0 1 1 0 f(0110)0 1 1 1 f(0111)1 0 0 0 f(1000)1 0 0 1 f(1001)1 0 1 0 f(1010)1 0 1 1 f(1011)1 1 0 0 f(1100)1 1 0 1 f(1101)1 1 1 0 f(1110)1 1 1 1 f(1111)

yz00 01 11 10

00

01

wx

11

10

f(0000) f(0001) f(0011) f(0010)

f(0100) f(0101) f(0111) f(0110)

f(1100) f(1101) f(1111) f(1110)

f(1000) f(1001) f(1011) f(1010)

Note the way that both the row and column indices change by only 1 bit at a time.

- 4 variable mapw x y z f(w,x,y,z)0 0 0 0 f(0000)0 0 0 1 f(0001)0 0 1 0 f(0010)0 0 1 1 f(0011)0 1 0 0 f(0100)0 1 0 1 f(0101)0 1 1 0 f(0110)0 1 1 1 f(0111)1 0 0 0 f(1000)1 0 0 1 f(1001)1 0 1 0 f(1010)1 0 1 1 f(1011)1 1 0 0 f(1100)1 1 0 1 f(1101)1 1 1 0 f(1110)1 1 1 1 f(1111)

yz00 01 11 10

00

01

wx

11

10

0 1 3 2

4 5 7 6

12 13 15 14

8 9 11 10

Note the way that both the row and column indices change by only 1 bit at a time.

This implies that two rows, or columns, whose indices differ by only 1 bit value, are adjacent.

yz00 01 11 10

00

01

wx

11

10

0 1 3 2

4 5 7 6

12 13 15 14

8 9 11 10

Note the way that both the row and column indices change by only 1 bit at a time.

This implies that two rows, or columns, whose indices differ by only 1 bit value, are adjacent.

yz00 01 11 10

00

01

wx

11

10

0 1 3 2

4 5 7 6

12 13 15 14

8 9 11 10

WRAP-AROUND!

An alternative row/column labelling, highlighting the literal with value 1.

y00 01 11 10

00

01

x

11

10

0 1 3 2

4 5 7 6

12 13 15 14

w

8 9 11 10

z

x0 1

0

1

Place function values (from the defining function truth table) in the map positions.

- Case Study: 1 variable map
- Ex.

x0 1

0

1

- Case Study: 1 variable map

Circle all 1 entries that, taken together, form a subcube (i.e. rectangular shape formed from 1-boxes, without holes).

x0 1

0

1

- Case Study: 1 variable map

Circle all 1 entries that, taken together, form a subcube (i.e. rectangle).

DEFINITION:

When constructing SOP forms, a 2N -subcube is a rectangular region of a Karnaugh map consisting of 2Nadjacent cells, each containing the same value 1 (or 0 for POS forms), and where N must be an integer greater or equal to zero.

x0 1

0

1

- Case Study: 1 variable map

The entry 1 in the second column corresponds to the prime implicantx.

x0 1

0

1

- Case Study: 1 variable map

The entry 1 in the second column corresponds to the prime implicantx.

Recall that x is a prime implicant iff:

- x implies f(x) and,

- x does not subsume any other implicant: Since, removing x from itself leaves nothing, then x is clearly prime.

x0 1

0

1

- Case Study: 1 variable map

Thus, the minimal expression of the function is:

F(x) = x

x0 1

1

0

- Case Study: 1 variable map - Complementation

The entry 1 in the first column corresponds to the prime implicantx’.

x0 1

1

0

- Case Study: 1 variable map - Complementation

The entry 1 in the first column corresponds to the prime implicantx’.

Thus, the minimal expression of the function is:

F(x) = x’

x0 1

1

1

- Case Study: 1 variable map - 2nd variation

Circle all 1 entries that, taken together, form a subcube (i.e. rectangle).

x0 1

1

1

- Case Study: 1 variable map - 2nd variation

We note that both x and x’ terms are included in the rectangle. Their individual “product” contributions to the SOP expression must be summed (or’ed):

x + x’

But, this reduces to 1.

x0 1

1

1

- Case Study: 1 variable map - 2nd variation

Thus, the minimal expression of the function is:

F(x) = 1

x0 1

1

1

- Case Study: 1 variable map - 2nd variation

Thus, the minimal expression of the function is:

f = 1

In other words, when both 0 and 1 terms appear for a literal referenced within the cells of a subcube, that literal is removed from the implicant corresponding to that subcube, leaving a prime implicant.

yz00 01 11 10

0 1 1 0

0

x

1

1 0 1 0

2x4 Karnaugh Map

Place function values (from the defining function truth table) in the map positions.

- 3 variable map

yz00 01 11 10

0 1 1 0

0

x

1

1 0 1 0

- 3 variable map

- Circle all 1 entries that, taken together, form a subcube (i.e. rectangle).
- Start with the largest subcubes, then proceed to smaller subcubes
- Generally speaking, there will be more than one independent subcube, each reflecting a different prime implicant.

yz00 01 11 10

0 1 1 0

0

x

1

1 0 1 0

- 3 variable map

y’ z’

x

- For each subcube, write its algebraic expression using the variable name (eg. x) if the box occurs in the row/column for which the variable is 1 – otherwise, if the box occurs in the row/column for which the variable (eg. y,z) is 0, use the complemented variable (ie. y’, z’).

yz00 01 11 10

0 1 1 0

0

x

1

1 0 1 0

- 3 variable map

y’ z’

Collecting literals into a product gives:

xy’z’

x

1-subcubes are expressed using all variable symbols

yz00 01 11 10

0 1 1 0

0

x

1

1 0 1 0

- 3 variable map

z

y+y’=1

x’

Circle all 1 entries that, taken together, form a subcube (i.e. rectangle). Generally speaking, there will be more than one independent subcube, each reflecting a different prime implicant.

yz00 01 11 10

0 1 1 0

0

x

1

1 0 1 0

- 3 variable map

z

y+y’=1

Collecting literals into a product gives:

x’z

x’

Circle all 1 entries that, taken together, form a subcube (i.e. rectangle). Generally speaking, there will be more than one independent subcube, each reflecting a different prime implicant.

yz00 01 11 10

0 1 1 0

0

x

1

1 0 1 0

- 3 variable map

y z

x+x’ = 1

Circle all 1 entries that, taken together, form a subcube (i.e. rectangle). Generally speaking, there will be more than one independent subcube, each reflecting a different prime implicant.

yz00 01 11 10

0 1 1 0

0

x

1

1 0 1 0

- 3 variable map

y z

x+x’ = 1

Collecting literals into a product gives:

yz

yz00 01 11 10

0 1 1 0

0

x

1

1 0 1 0

- 3 variable map

Each subcube (rectangle) corresponds to a prime implicant term.

Gathering all terms in SOP form,

f = xy’z’ + x’z + yz

yz00 01 11 10

0 1 1 0

0

x

1

1 0 1 0

- 3 variable map

f = xy’z’ + x’z + yz

yz00 01 11 10

0 1 1 0

0

x

1

1 0 1 0

- 3 variable map

f = xy’z’ + x’z + yz

Each subcube contains at least one 1-cell that can ONLY be included within that subcube. Such 1-cells are called essential 1-cells and their corresponding prime implicant is called an essential prime implicant.

yz00 01 11 10

0 1 1 0

0

x

1

1 0 1 0

- 3 variable map

f = xy’z’ + x’z + yz

xy’z’ ( xy’z’ )

Essential 1-cell (essential prime implicant).

Each subcube contains at least one 1-cell that can ONLY be included within that subcube. Such 1-cells are called essential 1-cells and their corresponding prime implicant is called an essential prime implicant.

yz00 01 11 10

0 1 1 0

0

x

1

1 0 1 0

- 3 variable map

f = xy’z’ + x’z + yz

x’y’z ( x’z )

Essential 1-cell (essential prime implicant).

Each subcube contains at least one 1-cell that can ONLY be included within that subcube. Such 1-cells are called essential 1-cells and their corresponding prime implicant is called an essential prime implicant.

yz00 01 11 10

0 1 1 0

0

x

1

1 0 1 0

- 3 variable map

f = xy’z’ + x’z + yz

xyz ( yz )

Essential 1-cell (essential prime implicant).

yz00 01 11 10

1 1 1 1

0

x

1

1 0 0 1

- 3 variable map - 2nd variation

yz00 01 11 10

1 1 1 1

0

x

1

1 0 0 1

- 3 variable map - 2nd variation

Circle all 1 entries that, taken together, form a subcube (i.e. rectangle) of the largest size, but containing 2N squares (for all possible N).

yz00 01 11 10

1 1 1 1

0

x

1

1 0 0 1

- 3 variable map - 2nd variation

Circle all 1 entries that, taken together, form a subcube (i.e. rectangle) of the largest size, but containing 2N squares (for all possible N).

yz00 01 11 10

1 1 1 1

0

x

1

1 0 0 1

- 3 variable map - 2nd variation

The “wrap-around” technique.

Circle all 1 entries that, taken together, form a subcube (i.e. rectangle) of the largest size, but containing 2N squares (for all possible N).

yz00 01 11 10

1 1 1 1

0

x

1

1 0 0 1

- 3 variable map - 2nd variation

x’[yz + y’z + yz’ + y’z’]

= x’[(y+y’)z + (y+y’)z’]

= x’[z+z’]

= x’

x’y’z’+x’yz’+xy’z’+xyz’

= x’(y’+y)z’ +x(y’+y)z’

= (x’+x)(y’+y)z’

= z’

yz00 01 11 10

1 1 1 1

0

x

1

1 0 0 1

- 3 variable map - 2nd variation

Collecting literals into a product gives:

x’ + z’

Place function values in the map positions.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

1 1 1 0

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

1 1 1 0

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

There are no subcubes of sizes:

24 =16 or 23 = 8.

1 1 0 0

1 1 1 0

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

Subcubes of size:

22 =4.

1 1 1 0

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

Subcubes of size:

22 =4.

1 1 1 0

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

Subcubes of size:

21 =2.

1 1 1 0

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

Subcubes of size:

21 =2.

1 1 1 0

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

No more subcubes!

1 1 1 0

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

1 1 1 0

Now to identify the prime implicants:

y’z’ + xy’ + x’z’ + wxz + wx’y

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

1 1 1 0

Now to identify the prime implicants:

xy’+ x’z’ + wxz + wx’y

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

1 1 1 0

Now to identify the prime implicants:

xy’ + x’z’+ wxz + wx’y

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

1 1 1 0

Now to identify the prime implicants:

xy’ + x’z’ + wxz+ wx’y

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

1 1 1 0

Now to identify the prime implicants:

xy’ + x’z’ + wxz + wx’y

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

1 1 1 0

Now to identify the prime implicants:

f(w,x,y,z) = xy’ + x’z’ + wxz + wx’y

1 0 1 1

- Using the procedure of identifying the largest possible subcubes in the Karnaugh map first, then dealing with smaller sized subcubes, we arrive at minimal representations of SOP (POS) expressions for the function.
- These expressions may not be unique, however.

Previously, we had presented this example.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

1 1 1 0

1 0 1 1

f(w,x,y,z) = xy’ + x’z’ + wxz + wx’y

Consider these subcubes ...

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

1 1 1 0

1 0 1 1

f(w,x,y,z) = xy’ + x’z’ + wxz + wx’y

Now, choose the other, alternative subcube containing these 1-cells …

…thereby, minimizing the number of subcubes, hence the number of product terms in the function.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

1 1 1 0

1 0 1 1

f(w,x,y,z) = xy’ + x’z’ + wyz+ wx’y

Both solutions are equivalent, but the second one is minimal.

- Case Study: 4 variable map

f(w,x,y,z) = xy’ + x’z’ + wxz + wx’y

f(w,x,y,z) = xy’ + x’z’ + wyz

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

1 1 1 0

1 0 1 1

- For a function of N variables (literals) draw the Karnaugh map consisting of 2N cells
- 1x2, 2x2, 2x4, 4x4, and so on
- we have not considered the cases of N=5, or 6 (See Textbook).

- Identify all rectangular subcubes of decreasing subcube size: 2L cells [L=N,N-1,...,1,0], using wraparound if possible.
- Always seek to minimize the degree of subcube overlap
- look for essential 1-cells to identify essential prime implicants and minimize the number of subcubes

- For each subcube, assign a prime implicant algebraic expression term by removing from the full implicant term all literals for which both 0 and 1 row or column labels appear, that is, keep only those literals that do not change.

- The Karnaugh mapping technique can also be applied to prime implicate (POS) forms.

- The Karnaugh mapping technique can also be applied to prime implicate (POS) forms.
- Instead of grouping 1-cells, we group 0-cells.

- The Karnaugh mapping technique can also be applied to prime implicate (POS) forms.
- Instead of grouping 1-cells, we group 0-cells.
- For each subcube of 0-cells we assign a sum expression, removing all literals whose row/column indices are both 0 and 1, and listing the literal itself for index label 0, or its complement for index label 1.
- Review this point with respect to algebraic representations
- Note the duality relationships

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

1 1 1 0

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

No subcubes of sizes 16, 8 or 4.

1 1 1 0

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

Subcubes of size 2.

1 1 1 0

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

1 1 0 0

1 1 1 0

Note the essential 0-cells that denote essential prime implicates.

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

This subcube contains 0-cells that may be associated with other subcube choices, hence they are not essential 0-cells.

This choice minimizes the number of subcubes.

1 1 0 0

1 1 1 0

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

Note there are no smaller subcubes.

The subcube choices reflect a minimal POS expression.

1 1 0 0

1 1 1 0

1 0 1 1

Identify 2N subcubes by decreasing N=4,3,2,1,0; using wraparound.

- Case Study: 4 variable map

yz00 01 11 10

00

01

wx

11

10

1 0 0 1

Note there are no smaller subcubes.

The subcube choices reflect a minimal POS expression.

1 1 0 0

1 1 1 0

1 0 1 1

f(w,x,y,z) = (x+y+z’)(x’+y’+z)(w+y’+z’)

- Case Study: 4 variable map
- SOP: f(w,x,y,z) = xy’ + x’z’ + wxz
- the 7 literals provide input to the and gates (operators)
- the 3 terms provide input to the or gates
- there are 7+3=10 gate inputs for this expression (circuit)

- Case Study: 4 variable map
- SOP: f(w,x,y,z) = xy’ + x’z’ + wxz
- the 7 literals provide input to the and gates (operators)
- the 3 terms provide input to the or gates
- there are 7+3=10 gate inputs for this expression (circuit)

- POS: f(w,x,y,z) = (x+y+z’)(x’+y’+z)(w+y’+z’)
- the 9 literals provide input to the or gates
- the 3 terms provide input to the and gates
- there are 9+3=12 gate inputs for this circuit

- Case Study: 4 variable map
- SOP: f(w,x,y,z) = xy’ + x’z’ + wxz
- the 7 literals provide input to the and gates (operators)
- the 3 terms provide input to the or gates
- there are 7+3=10 gate inputs for this expression (circuit)

- POS: f(w,x,y,z) = (x+y+z’)(x’+y’+z)(w+y’+z’)
- the 9 literals provide input to the or gates
- the 3 terms provide input to the and gates
- there are 9+3=12 gate inputs for this circuit

- CONCLUSION: In this case, the SOP form is a minimal cost circuit expression.

- Sometimes the logic for a circuit function is specified incompletely
- Some possible inputs may not have a corresponding “fixed” output value

- What do we do when the client does not care about these special cases

Dealing with apathetic circuit specification

The case of Don’t Care Conditions

- When an incompletely specified function is given, for which there exist “don’t care conditions”, we indicate those terms by a hyphen (or other dc token).

- When an incompletely specified function is given, for which there exist “don’t care conditions”, we indicate those terms by a hyphen (or other dc token).

yz00 01 11 10

00

01

wx

11

10

1 0 - 1

0 1 - 0

1 1 1 0

- 0 0 1

- When an incompletely specified function is given, for which there exist “don’t care conditions”, we indicate those terms by a hyphen (or other dc token).
- We may treat those termsas if they are 1’s OR 0’s,whichever is more usefulin leading to a minimalform of expression.

yz00 01 11 10

00

01

wx

11

10

1 0 - 1

?

0 1 - 0

?

1 1 1 0

- 0 0 1

?

- We may treat those termsas if they are 1’s OR 0’s,whichever is more usefulin leading to a minimalform of expression.

yz00 01 11 10

00

01

wx

11

10

1 0 - 1

0 1 - 0

1

1 1 1 0

- 0 0 1

1

- We may treat those termsas if they are 1’s OR 0’s,whichever is more usefulin leading to a minimalform of expression.
- Ignore all unused dc conditions.

yz00 01 11 10

00

01

wx

11

10

1 0 - 1

0 1 - 0

1

1 1 1 0

- 0 0 1

1

PRIME IMPLICATES

yz00 01 11 10

00

01

wx

11

10

1 0 - 1

0

0 1 - 0

1 1 1 0

- 0 0 1

PRIME IMPLICATES

yz00 01 11 10

00

01

wx

11

10

1 0 - 1

0

0 1 - 0

1 1 1 0

- 0 0 1

- Ignore all unused dc conditions.

PRIME IMPLICATES

yz00 01 11 10

00

01

wx

11

10

1 0 - 1

0

0 1 - 0

1 1 1 0

- 0 0 1

- We have studied and developed several techniques for simplifying Boolean expressions.
- These are based on the axioms, definitions and theorems of the Boolean Algebra, applied through the Boolean Calculus.
- Powerful tabular techniques have been developed for rapid reduction to some minimal cost forms using
- Karnaugh maps

- An even more powerful technique has been developed by Quine and McCluskey (and Petrick). Time prevents covering this topic but it is described in advanced books on computer engineering and logic design.