minimization of switching functions

SIMPLIFICATION OF SWITCHING FUNCTIONS

A PAPER AND PENCIL TECHNIQUE

Prof. Sureshchander

Email: [email protected]

The rules of Boolean algebra are used for algebraic manipulation of switching functions. This process, however,

is quite involved, and finding the right line of attack requires considerable ingenuity, judgement, experience and

sometimes plain luck and yet there is no way of finding if a minimal or near minimal solution has been found.

Various non-algebraic methods of simplification are available. The notable among them are:

Map method of Veitch and Karnaugh

Tabular method of of Quine-McCluskey.

Svoboda’s method of grids

Sureshchander’s search technique. ("Minimization of Switching Functions - A Fast Technique", IEEE

Trans. on Computer, vol. C-24, pp. 753-756.). This may be called as Sureshchander’s Search

Technique or ST technique.

Of these Veitch and Karnaugh method, popularly known as K-maps, is very popular. It is very powerful tool for

functions upto 5-variables. The tabular method or Quine-McCluskey technique decomposes the problem into

two parts

Determination of prime-implicants.

Selection of prime implicants from a prime implicant chart for a minimal cover.

» This method is quite laborious and unsuitable for functions more than seven variables for paper and

pencil solutions. The method can be programmed but the number of prime implicants become very

large for values of n greater than 10. For example, the upper bound of PIs for a 10 variable function is

58024. It is more than three billions (3485735825) for a 20 variable function. It may be noted that

upper bound of PIs for a function of n variables is (3n -2

n). In this method, Q-M technique, a trivial

function with all TRUE minrterms will require determination of all (3n -2

n) PIs to conclude that f(x1, x2,

..., xn) = 1

Sureshchander’s Search Technique or ST technique:

The ST technique reduces the search space for generating minimal sum-of-products (or product-of-sums) form.

All EPIs and other PIs are generated with minimal effort without generating all the PIs as in Q-M technique.

The technique is programmable. Here paper and pencil procedure of ST technique is described.

Definition 1:

If there are r minterms that are at a distance-1 from a minterm Pi, Pi is said to have rth-degree consensus.

Definition 2:

If Pi and all the r minterms at a distance-1 from it are covered by a sub-cube, then Pi is said to have proper rth-

degree consensus.. Such a sub-cube will have 2r (TRUE) minterms including Pi and r minterms at a distance-1

from it, and will be an rth-order sub-cube.

In order to know that a Pi has proper consensus, it is necessary to know which (2r - (r + 1)) terms are

required to form an rth-order sub-cube and whether these terms are TRUE or not. This technique has three steps:

1. Generate (2r - (r + 1)) terms for a Pi term.

2. Check if these (2r - (r + 1)) terms are TRUE, for an SOP form.

3. If all the terms, (2r - (r + 1)), at step 2 above are TRUE, then PI is selected. This PI will have 2

r

minterms.

mailto:[email protected]

Selection of (2r-(r + 1)) terms:

The (2r-(r + 1)) terms required for proper rth-degree consensus are selected as follows:

There are r-minterms that are at distance-1 from Pi. The other minterms are given by following relations:

Let r-minterms, at distance-1 from Pi, be m1

1, m12, ..., m

1r.. Let these terms be called m

1 and Pi term as m

0.

The m2, m

3, ..., m

r terms will be:

m2 = m

1i +m

2j –m

0, i = 1, 2, ..., r-1

j = (i+1), ..., r; i ≠ j

m3 = m

1i +m

1j + m

1k – 2m

0, i = 1, 2, ..., (r-2)

j = (i+1), ..., (r-1)

k = (j+1), ..., r; i ≠ j ≠ k

In general,

mk, k = 2, 3, ..., r, terms are calculated as sum of k m

1 terms minus (k -1) times m

0.

Definition 3: mk, k = 0, 1, 2, ..., r , terms are at a distance-k from Pi minterm. m

r will have only one minterm.

Definition 4: mkj minterm is at a distance-k from a Pi minterm.

If a Pi term, in an n variables function, has nth degree consensuses, then all 2n minterms should be TRUE for a

proper nth-degree consensus. In this case, the function is reduced to f = 1.

The search technique will be illustrated with following examples.

Example 1: f = ∑ (4, 8,9,10, 11 12, 13, 14, 15)

Let us consider 12 as a Pi, Fig 1. Only minterms 4, 8, 13 and 14 are at distance-1 from minterm 12. Hence,

minterm 12 has 4th

-degree consensus.

Here m0 is 12, and m

1 terms are 4, 8, 13, 14.

m2 terms can be generated as:

4 + 8 – 12 = 0

4 + 13 – 12 = 5

4 + 14 – 12 = 6

8 + 13 – 12 = 9

8 + 14 – 12 = 10

13 + 14 – 12 = 15

the m3 terms are

4 + 8 + 13 – 2*12 = 1

4 + 8 + 14 – 2*12 = 2

4 + 13 + 14 – 2*12 = 7

8 + 13 + 14 – 2*12 = 11

and m4 is

4 + 8 + 13 +14 – 3*12 = 3

Obviously minterm 12 does not have proper 4th

-degree consensus as minterms 0, 5, 6, 9, 1, 2 and 3 are not

TRUE . We need not have generated all the (2r – (r + 1)) as minterm 0 is FALSE. The process of generation of

0 m

2

1 m1

1m0

1 m1

0 m

3

0 m2

1 m1

1 m2

0 m

4

0 m3

1 m2

1 m3

0 m

3

0 m2

1 m

1

1 m2

4

0

12

11

8

11

5

1

13

11

9

11

7

3

15

11

11

11

6

2

10

11

2

6

7

11

25

25

11

25

25

Fig. 1

AB

CD

further minterms (for proper consensus) is to be terminated when a (generated) minterm is not TRUE that is it is

FALSE.

In this example, minterm 9, has 3rd

-degree consensus with minterms 8, 11 and 13. For a proper consensus, the

following terms should be TRUE.

m0 term 9 (Pi)

m1 terms 8, 11, 13 (3

rd degree consensus terms with m

0 are minterms at

distance-1 from m0)

m2 terms: 8 + 11 – 9 = 10 (m

2 terms are distance-2 from m

0)

8 + 13 – 9 = 12

11 + 13 – 9 = 15

m3 term 8 + 11 + 13 – 9 – 9 = 14 (m

3 term is at distance-3 from m

0)

Since all m2 and m

3 minterms (10, 12, 15, 14) are TRUE, minterm 9 has proper 3

rd-degree consensus, i.e., there

exists a 3rd

order subcube (of 8 minterms). The sub-cube is (8, 9, 10, 11, 12, 13, 14, 15).

Definition 5:

The PI in product form is product of TRUE literals of the minterm having lowest decimal value and

FALSE (complimented) literals of the minterm having highest decimal value among the 2r minterms.

In present case minterm with lowest decimal value is 8 or A , i.e., A is TRUE, and minterm 15 (highest

decimal value) is ABCD that does not have any FALSE minterm. Hence the PI has only one literal namely A.

Don’t care conditions.

The don’t care minterms, in this method, are treated as if they have already been covered. In other words, the

don’t care terms are not used as Pi terms. However, don’t cares can be used for enlarging a sub-cube.

Now, paper and pencil procedure of ST technique is described with some examples.

Example 2: Simplify the switching function

f (A, B, C, D) = ∑ (0, 2, 4, 9, 12, 15 + ∑ (1, 5, 7, 10)

The minterms are grouped according to the number of 1’s in their binary representation, Fig.2.

A minterm in k-group is compared with the terms of both (k-1) and (k+1) groups to find

distance-1 terms from it. For example, minterm 7 is at a distance-1 from minterms 5 in

(k-1) and minterm 15 in (k+1) group. Note minterm 7 is not at a distance-1 from minterm

9 in (k-1)

A minterm in k-group is at a distance-1 from (k-1) group if

a) the minterm in (k-1) group has less numeric value

than the term in k-group.

b) their difference is a power of 2.

A minterm in k-group is at a distance-1 from (k+1) group if

a) the minterm in (k+1) group has more numeric

value than the term in k-group.

b) their difference is a power of 2.

The search chart for this example is shown in Fig. 3.

The don’t cares are underlined in figures 2 and 3.

The minterms selected in a PI (prime implicant) are

✓ marked in Fig.3. Double ✓ mark indicates that

--------------------------------------------------------

No. of 1’s minterms

--------------------------------------------------------

0 0✓✓

----------------------- ---------------------------------

1 1, 2✓, 4✓✓

--------------------------------------------------------

2 5, 9✓, 10, 12✓

--------------------------------------------------------

3 7

--------------------------------------------------------

4 15✓

--------------------------------------------------------

Fig. 2 k - chart

Note: minterms 0 and 4 were not selected as PI

------------------------------------------------------------------------------------------------------------------------

Pi (m0), terms m1 terms m2 terms m3 terms minterns in sub-cube PI

------------------------------------------------------------------------------------------------------------------------ ✓0 1, 2, 4 6x

------------------------------------------------------------------------------------------------------------------------ *2 0 (0, 2)

------------------------------------------------------------------------------------------------------------------------ ✓4 0, 5, 12 8x

------------------------------------------------------------------------------------------------------------------------ *9 1 (1, 9)

------------------------------------------------------------------------------------------------------------------------ *12 4 (4, 12)

------------------------------------------------------------------------------------------------------------------------ *7 15 (7, 15)

------------------------------------------------------------------------------------------------------------------------

Fig. 3 Search Chart with sub-cubes and PIs

Pi terms are ✓marked once they are included in a subsequent selected PI. The ✓ and * Pi terms are not

considered in subsequent iterations.

a particular minterm was not selected on the first encounter.

In this example, there are 4 PIs, all are 1-order sub-cube. It may be observed that minterm 9 is not at

distance-1 from any non-don’t care term, minterm 1 (a don’t care term) is used to enlarge the subcube.


f (A, B, C, D, E) = ∑ (0, 1, 3, 4, 5, 7, 8, 9,10, 11,12, 13, 21,24, 25,26, 28, 29)

The minterms of the function are grouped according to the number of 1s in a minterm, Fig. 4a. The search

chart with m1, m

2, m

3 minterms, and their corresponding subcube and PI are shown in Fig. 5. It may be

observed that procedure described is a paper-pencil technique. It can be programmed easily.

minterm 0 has proper 3rd

degree consensus, All the generated m2 and m

3 minterms are TRUE hence, a subcube

(0, 1, 4, 5, 8, 9, 12, 13) is formed. The PI is . All the minterms covered by PI are ✓ marked, Fig. 4a.

Next non ✓ marked minterm is 3, it is now chosen as next Pi. It does not have proper consensus of 2nd

-degree

as minterm 15 is FALSE.

We continue to look for next non ✓ marked minterm that is 10. It results in a subcube (8, 10, 24, 26). The

minterms in sub-cube (8, 10, 24, 26), PI , are ✓ marked, Fig. 4b. Next Pi is minterm 7. It has 2nd

-degree

consensus with mniterms 3 and 5. Note that minterm 5 has already been included in PI . It can be treated as

don’t care term, but still m2 is generated for a possible larger sub-cube. Resulting m

2 minterm is 1, a minterm

already covered by PI . The PI thus formed is (1, 3, 5, 7), only minterms 3 and 7 are the non-covered

minterms. So ✓ mark minterm 3 and 7, Fig. 4c. Likewise, remaining terms (11), (21, 29), (25, 28) are ✓

marked after PIs with 11, 21 and 25 respectively are formed, Fig. 4d.

Note: minterm 11 has 0th

–order consensus as minterms 3, 9 and 10 are treated as don’t cares having been

included earlier in sub-cubes (0, 1, 4, 5, 8, 9, 12, 13), (1, 3, 5, 7) and (8, 10, 24, 26).

The process is continued till all the minterms are covered. In case some minterms remain uncovered (non ✓

marked terms). The process is reiterated from the top. A cyclic condition may exist in certain cases. The cycle is

broken by reducing the rth-degree consensus of the first (or any) minterm in the cycle to (r-1)-degree consensus

by removing one of the distance-1 minterm of the selected Pi.

--------------------------------------------------------


--------------------------------------------------------

0 0✓

--------------------------------------------------------

1 1✓, 4✓, 8✓

--------------------------------------------------------

2 3, 5✓, 9✓, 10, 12✓, 24

--------------------------------------------------------

3 7, 11, 13✓, 21, 25, 26, 28

--------------------------------------------------------

4 29

--------------------------------------------------------

Fig 4a k chart

(0, 1, 4, 5, 8, 9, 12, 13) are ✓ marked after proper

3rd –degree proper consensus for minterm 0 is

established.

3r

------------------------------------------------------------------------------------------------------------------------


------------------------------------------------------------------------------------------------------------------------

*0 1, 4, 8 5, 9, 12 13 (0, 1, 4 ,5, 8, 9, 12, 13)

------------------------------------------------------------------------------------------------------------------------ ✓3 1, 7, 11 5, 9, 15x

------------------------------------------------------------------------------------------------------------------------

*10 8. 26 24 (8, 10, 24, 26)

------------------------------------------------------------------------------------------------------------------------ *7 3, 5 1 (1, 3, 5, 7)

------------------------------------------------------------------------------------------------------------------------ *11 3, 9, 10 1, 2x (1, 3, 9, 11)

------------------------------------------------------------------------------------------------------------------------ *21 5, 29 13 (5, 13, 21, 29)

------------------------------------------------------------------------------------------------------------------------

*25 9, 24, 29 8, 13, 28 12 (8, 9, 12, 13, 24, 25, 28, 29) B

------------------------------------------------------------------------------------------------------------------------


--------------------------------------------------------


--------------------------------------------------------

0 0✓

--------------------------------------------------------

1 1✓, 4✓, 8✓

--------------------------------------------------------

2 3, 5✓, 9✓, 10✓, 12✓, 24✓

--------------------------------------------------------

3 7, 11, 13✓, 21, 25, 26✓, 28

--------------------------------------------------------

4 29

--------------------------------------------------------

Fig 4b k chart

(10, 24, 26) are ✓ marked after proper

2nd –degree proper consensus for minterm 10 is

established. 8 has already been ✓ marked

--------------------------------------------------------


--------------------------------------------------------

0 0✓

--------------------------------------------------------

1 1✓, 4✓, 8✓

--------------------------------------------------------

2 3✓, 5✓, 9✓, 10✓, 12✓, 24✓

--------------------------------------------------------

3 7✓, 11, 13✓, 21, 25, 26✓, 28

--------------------------------------------------------

4 29

--------------------------------------------------------

Fig 4c k chart

(3, 7) are ✓ marked after proper 2nd –degree

proper consensus for minterm 7 is established.

1 and 5 have already been ✓ marked.

--------------------------------------------------------


--------------------------------------------------------

0 0✓

--------------------------------------------------------

1 1✓, 4✓, 8✓

--------------------------------------------------------

2 3✓, 5✓, 9✓, 10✓, 12✓, 24✓

--------------------------------------------------------

3 7✓, 11✓, 13✓, 21✓, 25✓, 26✓, 28✓

--------------------------------------------------------

4 29✓

--------------------------------------------------------

Fig. 4d k Chart

Likewise, remaining terms (11), (21, 29),

(25, 28) are ✓ marked after PIs with 11, 21 and

25 respectively are formed.

The process of forming PIs is stopped after all

terms in the chart have been ✓ marked


f (a, b, c, d, e, f) = ∑ (1, 2, 3, 4, 5, 8, 9, 10, 17, 20, 24, 25, 27, 32, 33, 34, 36, 37, 40, 41, 42, 43, 44. 45, 46, 47, 48, 56, 59, 62)

The minterms of the function are grouped according to the number of 1s in a minterm, Fig. 6a . The search

chart with m1, m

2, m

3 minterms, and their corresponding subcube and PI are shown in Fig. 7. It may be

observed that it is a paper-pencil technique.

Using search technique, seven PIs were chosen in the first iteration, four in the second and final iteration.

--------------------------------------------------------


--------------------------------------------------------

1 1, 2, 4, 8, 32

--------------------------------------------------------

2 3, 5, 9, 10, 17, 20

24, 33, 34, 36, 40, 48

--------------------------------------------------------

3 25, 37, 41, 42, 44, 56

--------------------------------------------------------

4 27, 43, 45, 46

--------------------------------------------------------

5 47, 59, 62

--------------------------------------------------------

Fig. 6a k Ckart

3r

-------------------------------------------------------------------------------------------------------------------------------------------


-------------------------------------------------------------------------------------------------------------------------------------------

✓1 3, 5, 9, 17, 7x

-------------------------------------------------------------------------------------------------------------------------------------------

✓✓2 3, 10, 34 11x

-------------------------------------------------------------------------------------------------------------------------------------------

✓4 5, 20, 36 21x, 37

-------------------------------------------------------------------------------------------------------------------------------------------

✓8 9, 10, 24 11x,

------------------------------------------------------------------------------------------------------------------------------------------

✓32 33, 34, 36, 40 35x

-------------------------------------------------------------------------------------------------------------------------------------------

**3 1, 2 0x (2, 3)

-------------------------------------------------------------------------------------------------------------------------------------------

**5 1, 4, 37 0x, 33 (1, 5, 33, 37)

-------------------------------------------------------------------------------------------------------------------------------------------

✓9 1, 8, 25, 41 0x

-------------------------------------------------------------------------------------------------------------------------------------------

**10 2, 8, 42 0x, 34 (2, 10, 34, 42)

-------------------------------------------------------------------------------------------------------------------------------------------

*17 1, 25 9 (1, 9, 17, 25)

-------------------------------------------------------------------------------------------------------------------------------------------

*20 4 (4, 20)

-------------------------------------------------------------------------------------------------------------------------------------------

*24 8, 25, 56 40, 9, 57x (8, 24, 40, 56)

-------------------------------------------------------------------------------------------------------------------------------------------

✓✓33 1, 32, 37, 41 0x, 36, 40, 45 44

-------------------------------------------------------------------------------------------------------------------------------------------

✓✓34 2, 32, 42 0x , 10

-------------------------------------------------------------------------------------------------------------------------------------------

**36 4, 32, 37, 44 0x. 33, 40, 45 41 (32, 33, 36, 37, 40, 41, 44, 45)

-------------------------------------------------------------------------------------------------------------------------------------------

*48 32, 56 40 (32, 40, 48, 56)

-------------------------------------------------------------------------------------------------------------------------------------------

✓✓37 5, 33, 36, 45 1, 4, 13x

-------------------------------------------------------------------------------------------------------------------------------------------

✓41 9, 33, 40, 43, 45 35, 37, 47 39x

-------------------------------------------------------------------------------------------------------------------------------------------

✓42 10, 34, 40, 43, 46 2,11x

-------------------------------------------------------------------------------------------------------------------------------------------

✓44 36, 40, 45, 46 32, 37, 38x

-------------------------------------------------------------------------------------------------------------------------------------------

*27 25, 59 57x (27, 59)

-------------------------------------------------------------------------------------------------------------------------------------------

*43 41, 42, 47, 59 40, 45, 46, 57x 44 (40, 41, 42, 43, 44, 45, 46, 47)

-------------------------------------------------------------------------------------------------------------------------------------------

*62 46 (46, 62) --

-------------------------------------------------------------------------------------------------------------------------------------------


The minimal sum-of-products is:

+ + + + + + + +

--------------------------------------------------------


--------------------------------------------------------

1 1✓, 2, 4✓, 8✓, 32✓

--------------------------------------------------------

2 3, 5, 9✓, 10, 17✓, 20✓

24✓, 33, 34, 36, 40✓, 48✓

--------------------------------------------------------

3 25✓, 37, 41✓, 42✓, 44✓, 56✓

--------------------------------------------------------

4 27✓, 43✓, 45✓, 46✓

--------------------------------------------------------

5 47✓, 59✓, 62✓

--------------------------------------------------------

Fig. 6b k-chart

Seven PIs were selected in first iteration, the

minterms covered by them are ✓ marked.

3r

--------------------------------------------------------


--------------------------------------------------------

1 1✓, 2✓✓, 4✓, 8✓, 32✓

--------------------------------------------------------

2 3✓✓, 5✓✓, 9✓, 10✓✓, 17✓, 20✓

24✓, 33✓✓, 34✓✓, 36✓✓, 40✓, 48✓

--------------------------------------------------------

3 25✓, 37✓✓, 41✓, 42✓, 44✓, 56✓

--------------------------------------------------------

4 27✓, 43✓, 45✓, 46✓

--------------------------------------------------------

5 47✓, 59✓, 62✓

--------------------------------------------------------

Fig. 6c k-chart

Four PIs were selected in second iteration, the

minterms covered by them are ✓✓ marked.

No explanation is given for above charts as the procedure has been explained earlier and process is self

explanatory.

There may be instances when a switching function may not have an EPI. Consider, for example, the function

Example 5: f (A, B, C, D) = ∑ (0, 1, 5, 7, 8, 10, 14, 15)

The search chart, Fig. 9, is cyclic as no PI can be selected in first iteration. The cycle is broken by reducing the

consensus of a Pi term by one. Any Pi can be chosen. Let us consider Pi 0. Pi 0 has 2nd

order consensus. It has

to be reduced to 1st-order consensus by dropping one m

1 term, let it be term 8. Now Pi 0 has 1

st-order consensus.

The terms in sub-cube (0, 1), prime implicant so formed are treated as don’t cares for subsequent Pi terms.

The remaining PIs, , as outlined earlier.

A technique for minimization of switching functions has been described in this presentation that can be easily

programmed and is equally suitable for paper and pencil as well.

The technique does not generate all the prime implicants but looks for existence of a prime implicant thus

reducing the search space. It allows simplification of switching functions having variables greater than 8 as well.

***

--------------------------------------------------------


--------------------------------------------------------

0 0✓✓

----------------------- ---------------------------------

1 1, 8✓✓

--------------------------------------------------------

2 5 ✓✓, 10,

--------------------------------------------------------

3 7, 14✓✓

--------------------------------------------------------

4 15

--------------------------------------------------------

Fig. 8 k - chart

Note: minterms 0 and 4 were not selected as PI

------------------------------------------------------------------------------------------------------------------------


------------------------------------------------------------------------------------------------------------------------

**0 1, 8 9x (0, 1)

------------------------------------------------------------------------------------------------------------------------ ✓1 0,5 4x (0, 2)

------------------------------------------------------------------------------------------------------------------------ **8 0, 10 2x (8, 10)

------------------------------------------------------------------------------------------------------------------------

**5 1, 7 3x (5, 7)

------------------------------------------------------------------------------------------------------------------------ ✓10 8, 14 12x (8, 10)

------------------------------------------------------------------------------------------------------------------------ ✓7 5, 15 13x

------------------------------------------------------------------------------------------------------------------------

**14 10,15 11x (14, 15)

------------------------------------------------------------------------------------------------------------------------ ✓15 7, 14 6x (14, 15) -

-----------------------------------------------------------------------------------------------------------------------


Cyclic condition. Here double asterisk mark is placed against all the selected PIs as they all have

been selected in second iteration..

minimization of switching functions

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