Why do we take 4,5 and 5,13 instead of 4,6 and 13,15?

[tomekrx]

Homework Statement

Quine McCluskey (QM) Method

Relevant Equations

little problem

Hello,
can anyone tell me why do we take 4,5 and 5,13 instead of 4,6 and 13,15?Thats QM method

 

[jedishrfu]

There's a lot available online about this method including some videos:

https://www.google.com/search?newwi...59.269.0j2...0...2j1..gws-wiz...0.e7Nj18kpWXQ

 

[jedishrfu]

The wikipedia article summarizes it pretty well:

https://en.wikipedia.org/wiki/Quine–McCluskey_algorithm

Not really understanding how it works yet, could the choices be due to fewer bit variations?

basically 4,5 is $(0 1 0 -)$ and 5,13 is $(- 1 0 1)$ with the middle bits $(-10-)$ in common to both?

whereas 4,6 which is $(01-0)$ and 13,15 is $(11-1)$ only have the $(-1--)$ bit in common.

from the wikipedia article above:

Step 2: prime implicant chart[edit]
None of the terms can be combined any further than this, so at this point we construct an essential prime implicant table. Along the side goes the prime implicants that have just been generated, and along the top go the minterms specified earlier. The don't care terms are not placed on top—they are omitted from this section because they are not necessary inputs.

	4	8	10	11	12	15	⇒	A	B	C	D
m(4,12)*							⇒	—	1	0	0
m(8,9,10,11)							⇒	1	0	—	—
m(8,10,12,14)							⇒	1	—	—	0
m(10,11,14,15)*							⇒	1	—	1	—

To find the essential prime implicants, we run along the top row. We have to look for columns with only 1 "X". If a column has only 1 "X", this means that the minterm can only be covered by 1 prime implicant. This prime implicant is essential.

For example: in the first column, with minterm 4, there is only 1 "X". This means that m(4,12) is essential. So we place a star next to it. Minterm 15 also has only 1 "X", so m(10,11,14,15) is also essential. Now all columns with 1 "X" are covered.

The second prime implicant can be 'covered' by the third and fourth, and the third prime implicant can be 'covered' by the second and first, and neither is thus essential. If a prime implicant is essential then, as would be expected, it is necessary to include it in the minimized boolean equation. In some cases, the essential prime implicants do not cover all minterms, in which case additional procedures for chart reduction can be employed. The simplest "additional procedure" is trial and error, but a more systematic way is Petrick's method. In the current example, the essential prime implicants do not handle all of the minterms, so, in this case, one can combine the essential implicants with one of the two non-essential ones to yield one equation:

{\displaystyle f_{A,B,C,D}=BC'D'+AB'+AC\ }

[11]
or

{\displaystyle f_{A,B,C,D}=BC'D'+AD'+AC\ }

Both of those final equations are functionally equivalent to the original, verbose equation:

{\displaystyle f_{A,B,C,D}=A'BC'D'+AB'C'D'+AB'C'D+AB'CD'+AB'CD+ABC'D'+ABCD'+ABCD.\ }