quine - mccluskey · quine‐mccluskey algorithm(for e15) 1. starting with a list of cubes (groups)...

Quine ‐McCluskey

To Karnaugh and Beyond…

Minimization of a three variable map as seen on a cube

q1 q0

00 01 11 10q2

01m0

000

0m1001

0m3011

1m2010

11m4100

1m5101

0m7111

1m6110

Adapted from “Fundamentals of Digital Logic with Verilog Design” by Brown and Vranesic, McGraw‐Hill

2 1 0f q ,q ,q m(0,2,4,5,6) 0 2 1q q q

A four variable map as seen on a hypercube

Adapted from http://commons.wikimedia.org/wiki/File:Schlegel_wireframe_8‐cell.png

Karnaugh Map, with minterms and “Groups”Q1 Q0

00 01 11 10Q3 Q2

00m0

0000

G0

m1

0001

G1

m3

0011

G2

m2

0010

G1

01m4

0100

G1

m5

0101

G2

m7

0111

G3

m6

0110

G2

11m12

1100

G2

m13

1101

G3

m15

1111

G4

m14

1110

G3

10m8

1000

G1

m9

1001

G2

m11

1011

G3

m10

1010

G2

Every minterm in Group n (Gn) has n 1’s. So minterms in G2 have 2 ones.

Example 1, Generate Prime Implicants

Minimize f(d,c,b,a)=m(1,3,5,11,12,13,14,15)

0 1 1 0

0 1 0 0

1 1 1 1

0 0 1 0

Group minterm Implicant

List 1 (1 cube – a node) List 2 (2 cube – an edge) List 3 (3 cube – a face)

G0 ‐‐‐‐‐‐‐ G1,2 1,3 00x1 P1 G1,2,3 ‐‐‐‐‐‐

G1 1 0001 √ 1,5 0x01 P2 G2,3,4 12,13,14,15 11xx P6G2 3 0011 √ G2,3 3,11 x011 P3 12,14,13,15

5 0101 √ 5,13 x101 P412 1100 √ 12,13 110x √

G3 11 1011 √ 12,14 11x0 √13 1101 √ G3,4 11,15 1x11 P514 1110 √ 13,15 11x1 √

G4 15 1111 √ 14,15 111x √

Table to generate prime implicants

Note: Prime implicants can be identified on Karnaugh map for low number of inputs.

Example 1, Find Minimum Cover (1)0 1 1 0

0 1 0 0

1 1 1 1

0 0 1 0

PrimeImplicant m1 m3 m5 m11 m12* m13 m14* m15

P1 √ √P2 √ √P3 √ √P4 √ √P5 √ √P6 √ √ √ √

Any column with a single check mark is only covered by one prime implicant (i.e., IPE), so that implicant must be chosen and can be removed from the table,along with all the other minterms covered by that implicant.So we choose P6 as one of our implicants.We can then work with a simplified table.

minterms implicant

1,3 001x P1

1,5 0x01 P2

3,11 x011 P3

5,13 x101 P4

11,15 1x11 P5

12,13,14,15 11xx P6


0 1 0 0

1 1 1 1

0 0 1 0PrimeImplicant m1 m3 m5 m11

P1 √ √P2 √ √P3 √ √P4 √P5 √

• We say that row P2 dominates row P4 because P2 covers every minterm covered by P4, so we only need P2.

• So by including P2, we are sure that the resulting circuit will be no more “costly” than it would by including P4.

• So eliminate P4. This makes P2 essential (because it is the only implicant covering m5).• So we choose P2 and we can eliminate m1 and m5 from the table.• Note: book uses slightly different method (we will discuss later).• We can then work with a simplified table.


0 1 0 0

1 1 1 1

0 0 1 0PrimeImplicant m3 m11

P1 √P3 √ √P5 √

• In the same way P3 dominates P1 and P5, so we eliminate them.• So we choose P3.• We eliminate columns m3 and m11 and have an empty table, (because we pick P3) so

we can stop.

• The prime implicants we have chosen (P6=11xx, P2=0x01, and P3=x011) form our minimum cover.

• Note: there are no fundamental difficulties in extending this to more inputs (the bookkeeping just gets harder – but computers are good at that).

P6 P2 P3

f(d,c,b,a) m 1,3,5,11,12,13,14,15

d c d b a c b a

P6

P2

P3

Example 2, QM and don’t‐caresGenerate Prime Implicants (include don’t cares)Minimize f(d,c,b,a)=m(0,4,5,7,8,10,14,15) + D(1,2,12)

Group Implicant


G0 0 0000 √ G0,1 0,1 000x √ G0,1,2 0,1,4,5 0x0x P4

G1 1 0001 √ 0,2 00x0 √ 0,4,1,5

2 0010 √ 0,4 0x00 √ 0,2,8,10 x0x0 P5

4 0100 √ 0,8 x000 √ 0,8,2,10

8 1000 √ G1,2 1,5 0x01 √ 0,4,8,12 xx00 P6

G2 5 0101 √ 2,10 x010 √ 0,8,4,12

10 1010 √ 4,5 010x √ G1,2,3 8,10,12,14 1xx0 P7

12 1100 √ 4,12 x100 √ 8,12,10,14

G3 7 0111 √ 8,10 10x0 √

14 1110 √ 8,12 1x00 √

G4 15 1111 √ G2,3 5,7 01x1 P1

10,14 1x10 √

12,14 11x0 √

G3,4 7,15 x111 P2

14,15 111x P3

1 X 0 X

1 1 1 0

X 0 1 1

1 0 0 1

Example 2, Find minimum Cover (1)(exclude don’t care minterms)

1 X 0 X

1 1 1 0

X 0 1 1

1 0 0 1

PrimeImplicant m0 m4 m5 m7 m8 m10 m14 m15

P1 √ √P2 √ √P3 √ √P4 √ √ √P5 √ √ √P6 √ √ √P7 √ √ √

• No essential prime implicants.• No row domination.• Column m0 dominates column m4. So we eliminate column m0

(when we cover m4 we also cover m0, so m0 is not needed).• In the same way column m8 dominates column m10, so remove m8.• Now we can work with a simplified table.

minterms Implicant

5,7 01x1 P1

7,15 x111 P2

14,15 111x P3

0,1,4,5 0x0x P4

0,2,8,10 x0x0 P5

0,4,8,12 xx00 P6

8,10,12,14 1xx0 P7

Example 2, Find minimum Cover (2)1 X 0 X

1 1 1 0

X 0 1 1

1 0 0 1Prime

Implicant m4 m5 m7 m10 m14 m15

P1 √ √P2 √ √P3 √ √P4 √ √P5 √P6 √P7 √ √

• No essential prime implicants.• Row P4 dominated P6, so eliminate P6.• This makes P4 essential.• So we choose P4 to be in our cover, and we can eliminate columns m4 and m5.• Row P7 dominate P5, so eliminate P5.• This makes P7 essential (for m10).• Choose P7 and eliminate columns m10 and m14.• This lets us work with a simplified table (we eliminate chosen implicants (P4 and P7)

from simplified table).

Example 2, Find minimum Cover (3)1 X 0 X

1 1 1 0

X 0 1 1

1 0 0 1PrimeImplicant m7 m15

P1 √P2 √ √P3 √

• No essential prime implicants.• Row P2 dominates P1 and P3, so choose P2.• This lets us eliminate row P1 and P3 and we have an empty table

(since we have picked P2).

• The solution includes P2=x111, P4=0x0x, P7=1xx0.

P2 P4 P7

f(d,c,b,a) m 0,4,5,7,8,10,14,15 D 1,2,12

c b a d b d a

P2

P4

P7

Example 3, Minimize f(d,c,b,a)=m(5,6,10,15)+D(2,4,7,11,13,14)

Group Implicant


G0 ‐‐‐‐ G1,2 2,6 0x10 √ G1,2,3 2,6,10,14 xx10 P1

G1 2 0010 √ 2,10 x010 √ 2,10,6,14

4 0100 √ 4,5 010x √ 4,5,6,7 011x P2

G2 5 0101 √ 4,6 01x0 √ 4,6,5,7

6 0110 √ G2,3 5,7 01x1 √ G2,3,4 5,7,13,15 x1x1 P3

10 1010 √ 5,13 x101 √ 5,13,7,15

G3 7 0111 √ 6,7 011x √ 6,7,14,15 011x P4

11 1011 √ 6,14 x110 √ 6,14,7,15

13 1101 √ 10,11 101x √ 10,11,14,15 1x1x P5

14 1110 √ 10,14 1x10 √ 10,14,11,15

G4 15 1111 √ G3,4 7,15 x111 √

11,15 1x11 √

13,15 11x1 √

14,15 111x √

0 X 0 X

X 1 X 1

0 X 1 X

0 0 X 1

Example 3, Find minimum Cover (1)(exclude don’t care minterms)

PrimeImplicant m5 m6 m10 m15

P1 √ √P2 √ √P3 √ √P4 √ √P5 √ √

• No essential prime implicants.• No row domination.• No column domination• Use branching – try including implicants and solving.

0 X 0 X

X 1 X 1

0 X 1 X

0 0 X 1

Example 3, Find minimum Cover (2)

PrimeImplicant m5 m15

P2 √P3 √ √P4 √P5 √

0 X 0 X

X 1 X 1

0 X 1 X

0 0 X 1

P1

P3


P1 √ √P2 √ √P3 √ √P4 √ √P5 √ √


P2 √P3 √ √P4 √P5 √

• Try including P1 (eliminate columns m6 and m10)• P3 covers both m5 and m15, so choose it.• Cover includes P1 and P3

P1

P3


• Next, we try including P2 (eliminate columns m5and m6)

• We can form a cover of the simplified table with just P5.

• Cover include P2 and P5

0 X 0 X

X 1 X 1

0 X 1 X

0 0 X 1

P5


P1 √ √P2 √ √P3 √ √P4 √ √P5 √ √

P2


P1 √P3 √P4 √P5 √ √


• We don’t need to try with P3 because it is part of a previous solution with only P1.

• Try including P4 (eliminate columns m6 and m15)• Cover include P4 and two others (P1 and P2;

P1 and P3; P5 and P2; P5 and P3) for a total of 3 implicants, so we reject the solution (it is more complex than others)

• We don’t need to try P5 because it was included in a previous solution with only P2.

• So… use one of the previous solutions (either use P1 and P3, or P2 and P5).

0 X 0 X

X 1 X 1

0 X 1 X

0 0 X 1


P1 √ √P2 √ √P3 √ √P4 √ √P5 √ √

P4


P1 √P2 √P3 √P5 √

The AlgorithmQuine‐McCluskey algorithm (for E15)1. Starting with a list of cubes (groups) that represent the minterms where f = 1 or a don't‐care

condition, generate the prime implicants by successive pairwise comparisons of the cubes.2. Derive a cover table which indicates the minterms where f = 1 that are covered by each prime

implicant.3. Include the essential prime implicants (if any) in the final cover and reduce the table by removing both

these prime implicants and the covered minterms.4. Use the concept of row and column dominance to reduce the cover table further. 5. Repeat steps 3 and 4 until the cover table is either empty or no further reduction of the table is

possible.6. If the reduced cover table is not empty, then use the branching approach to determine the remaining

prime implicants that should be included in a minimum cost cover.

Quine‐McCluskey algorithm (from text)The algorithms are the same except for step 4.4. Use the concept of row and column dominance to reduce the cover table further. A dominated row is

removed only if the cost of its prime implicant is greater than or equal to the cost of the dominating row's prime implicant.

The two methods result in the same number of gates. However, since the text method may end up using a prime implicant with lower cost (fewer inputs) the resulting gates used may have fewer inputs. However, it complicates the process enough (step 6, branching, is often required) so we won’t use it. Depending on the technology used, a gate with more inputs may not be more costly than one with fewer inputs.

An example follows.

Explore difference in methods

f m 5,13,15 D ︵0,1,4,14 ︶

Group Implicant


G0 0 0000 G0,1 0,1 000x √ G0,1,2 0,1,4,5 0x0x P4

G1 1 0001 √ 0,4 0x00 √ 0,4,1,5

4 0100 √ G1,2 1,5 0x01 √

G2 5 0101 √ 4,5 010x √

G3 13 1101 √ G2,3 5,13 X101 P1

14 1110 √ G3,4 13,15 11x1 P2

G4 15 1111 √ 14,15 111x P3

P1

P2 P3

P4X X 0 0

X 1 0 0

0 1 1 X

0 0 0 0

Find Cover

f a,b,c,d m 5,13,15 D ︵0,1,4,14 ︶

m5 m13 m15

P1 = x101 √ √P2 = 11x1 √ √P3 = 111x √P4 = 0x0x √

• By our rules P1 dominates P4 so we would pick P1 and P2 (P2 dominates P3)• This results in

• By book rules you would keep P1 and P4 (because P4 is lower cost (fewer inputs)) and use branching to find cover. The final result would be P4 and P2.• This results in

•We will use simpler rules. This results in the same number of gates (but some gates may have more inputs).

f c b a d c a

f d b d c a

P1

P2 P3

P4X X 0 0

X 1 0 0

0 1 1 X

0 0 0 0

quine - mccluskey · quine‐mccluskey algorithm(for e15) 1. starting with a list of cubes (groups)...

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