1. A Paper and Pencil Technique
for Variables More than 8
Prof. Sureshchander
19/10/2012
© Dr.SureshChander
1

2. 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.
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3. Of these Veitch and Karnaugh method, popularly known as K-maps, is very popular. It is
very powerful tool for functions up to 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 variables function. It may be noted that upper
bound of PIs for a function of n variables is (3n -2n). In this method, Q-M technique, a
trivial function with all TRUE minrterms will require determination of all (3n -2n) PIs to
conclude that f(x1, x2, ..., xn) = 1
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4. 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.
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5. 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 2r
minterms.
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© Dr.SureshChander

6. 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, m1
2, ..., m1
r.. Let these
terms be called m1 and Pi term as m0.
The m2, m3, ..., mr terms will be:
m2 = m1
i +m2
j –m0, i = 1, 2, ..., r-1
j = (i+1), ..., r; i ≠ j
m3 = m1
i +m1
j + m1
k – 2m0, i = 1, 2, ..., (r-2)
j = (i+1), ..., (r-1)
k = (j+1), ..., r; i ≠ j ≠ k
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7. In general,
mk, k = 2, 3, ..., r, terms are calculated as sum of k m1 terms
minus (k -1) times m0.
Definition 3: mk, k = 0, 1, 2, ..., r , terms are at a distance-k
from Pi minterm. mr will have only one minterm.
Definition 4: mk
j 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.
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8. 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 m1
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
0m
2
1m
1
1m0
1m
1
0m
3
0m
2
1m
1
1m
2
0m
4
0m
3
1m
2
1m
3
0m
3
0m
2
1m
1
1m
2
40 12
11
8
11
51 13
11
9
11
73 15
11
11
11
62 10
11
267 1125
251125
25
Fig. 1
AB
CD
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© Dr.SureshChander

9. 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 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 (3rd degree consensus terms with m0 are
minterms at distance-1 from m0)
m2 terms 8 + 11 – 9 = 10 (m2 terms are distance-2 from m0)
8 + 13 – 9 = 12
11 + 13 – 9 = 15
m3 term 8 + 11 + 13 – 9 – 9 = 14 (m3 term is at distance-3 from m0)
Since all m2 and m3 minterms (10, 12, 15, 14) are TRUE, minterm 9 has proper 3rd-
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).
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© Dr.SureshChander
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 A𝐵𝐶𝐷, 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.

11. 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.
--------------------------------------------------------
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
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© Dr.SureshChander

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© Dr.SureshChander
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
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.
------------------------------------------------------------------------------------------------------------------------
Pi (m0
), terms m1
terms m2 terms m3 terms minterns in sub-cube PI
------------------------------------------------------------------------------------------------------------------------
✓
0 1, 2, 4 6x
------------------------------------------------------------------------------------------------------------------------
*
2 0 (0, 2) 𝐴𝐵𝐷
------------------------------------------------------------------------------------------------------------------------
✓
4 0, 5, 12 8x
------------------------------------------------------------------------------------------------------------------------
*
9 1 (1, 9) 𝐵𝐶𝐷
------------------------------------------------------------------------------------------------------------------------
*
12 4 (4, 12) 𝐵𝐶𝐷
------------------------------------------------------------------------------------------------------------------------
*
7 15 (7, 15) 𝐵𝐶𝐷
------------------------------------------------------------------------------------------------------------------------
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.

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© Dr.SureshChander
Example 3: Simplify the switching function
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
, m2
, m3
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 m3
minterms are TRUE hence, a subcube
(0, 1, 4, 5, 8, 9, 12, 13) is formed. The PI is 𝐴𝐷. All the minterms covered by PI 𝐴𝐷 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 𝐵𝐶𝐸, 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 𝐴𝐷. It can be treated as
don’t care term, but still m2
is generated for a possible larger sub-cube. Resulting m2
minterm is 1, a minterm
already covered by PI 𝐴𝐷. 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.

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© Dr.SureshChander
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.
------------------------------------------------------------------------------------------------------------------------
Pi (m0
), terms m1
terms m2
terms m3
terms minterns in sub-cube PI
------------------------------------------------------------------------------------------------------------------------
*0 1, 4, 8 5, 9, 12 13 (0, 1, 4 ,5, 8, 9, 12, 13) 𝐴𝐷
------------------------------------------------------------------------------------------------------------------------
✓
3 1, 7, 11 5, 9, 15x
------------------------------------------------------------------------------------------------------------------------
*10 8. 26 24 (8, 10, 24, 26) 𝐵𝐶𝐸
------------------------------------------------------------------------------------------------------------------------
*
7 3, 5 1 (1, 3, 5, 7) 𝐴𝐵𝐸
------------------------------------------------------------------------------------------------------------------------
*
11 3, 9, 10 1, 2x
(1, 3, 9, 11) 𝐴𝐶𝐸
------------------------------------------------------------------------------------------------------------------------
*
21 5, 29 13 (5, 13, 21, 29) 𝐶𝐷𝐸
------------------------------------------------------------------------------------------------------------------------
*25 9, 24, 29 8, 13, 28 12 (8, 9, 12, 13, 24, 25, 28, 29) B 𝐷
------------------------------------------------------------------------------------------------------------------------
Fig. 5 Search Chart with sub-cubes and PIs
--------------------------------------------------------
No. of 1’s minterms
--------------------------------------------------------
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
--------------------------------------------------------
No. of 1’s minterms
--------------------------------------------------------
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
--------------------------------------------------------
No. of 1’s minterms
--------------------------------------------------------
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
--------------------------------------------------------
No. of 1’s minterms
--------------------------------------------------------
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.

15. Example 4: Simplify the switching function
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
, m2
, m3
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.
--------------------------------------------------------
No. of 1’s minterms
--------------------------------------------------------
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
--------------------------------------------------------
No. of 1’s minterms
--------------------------------------------------------
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.
--------------------------------------------------------
No. of 1’s minterms
--------------------------------------------------------
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.
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16. 19/10/2012
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© Dr.SureshChander
No explanation is given for charts
in this example as the procedure
has been explained earlier and
process is self explanatory.
.
Sureshchander’s
Search Technique or
ST technique

17. 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 m1
term, let it be term 8. Now Pi 0 has 1st
-order consensus.
The terms in sub-cube (0, 1), prime implicant 𝐴𝐵𝐶, so formed are treated as don’t cares for subsequent Pi terms.
The remaining PIs, 𝐴𝐵𝐷, 𝐴𝐵𝐷, 𝐴𝐵𝐶, as outlined earlier.
--------------------------------------------------------
No. of 1’s minterms
--------------------------------------------------------
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
------------------------------------------------------------------------------------------------------------------------
Pi (m0
), terms m1
terms m2 terms m3 terms minterns in sub-cube PI
------------------------------------------------------------------------------------------------------------------------
**0 1, 8 9x
(0, 1) 𝐴𝐵𝐶
------------------------------------------------------------------------------------------------------------------------
✓
1 0,5 4x (0, 2)
------------------------------------------------------------------------------------------------------------------------
**
8 0, 10 2x (8, 10) 𝐴𝐵𝐷
------------------------------------------------------------------------------------------------------------------------
**5 1, 7 3x
(5, 7) 𝐴𝐵𝐷
------------------------------------------------------------------------------------------------------------------------
✓
10 8, 14 12x (8, 10)
------------------------------------------------------------------------------------------------------------------------
✓
7 5, 15 13x
------------------------------------------------------------------------------------------------------------------------
**14 10,15 11x (14, 15) 𝐴𝐵𝐶
------------------------------------------------------------------------------------------------------------------------
✓
15 7, 14 6x (14, 15) -
-----------------------------------------------------------------------------------------------------------------------
Fig. 9 Search Chart with sub-cubes and PIs
Cyclic condition. Here double asterisk mark is plced against all the selected PIs as they all have
been selected in second iteration..
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© Dr.SureshChander

18. 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.
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© Dr.SureshChander

19. Thank you for viewing.
For comments or questions please contact
Prof. Sureshchander at
suresh.chander@gmail.com
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© Dr.SureshChander