Simplification of two lane Karnaugh Map Q: simplify the logical function to the simplest and or, where F = ∑ (M5. M6. M7. M8. M9), and the given constraint condition is M10 + M11 + M12 + M13 + M14 + M15 = 0 Simplify the logical function to the simplest or and F = ∏（M1.M2.M3.M6.M7.M9.M11.M13）

Simplification of two lane Karnaugh Map Q: simplify the logical function to the simplest and or, where F = ∑ (M5. M6. M7. M8. M9), and the given constraint condition is M10 + M11 + M12 + M13 + M14 + M15 = 0 Simplify the logical function to the simplest or and F = ∏（M1.M2.M3.M6.M7.M9.M11.M13）

The thick line is the Karnaugh map of the first question, the red one is the answer, the rest of the thin line is the Karnaugh map of the second question, the purple one. The rest is the four variable Karnaugh map

The cycle method of Karnaugh map. Which of the two cycle methods is right? What is the principle of Karnaugh map cycle
 

Both are correct
The above one is the simplest, but there is competition and risk (some schools don't talk about it)
The following is the simplest form that there is no competitive risk, that is, there is one more redundant item to avoid competitive risk
The principle of Karnaugh circle is: circle "1" with n power of 2 (2,4,8.)
The top, bottom, left and right of Karnaugh circle are adjacent phases

Why can the same grid be surrounded by different circles in digital Karnaugh map?

Each square is the minimum term, which appears n times in the and or expression, and will not affect the value of the whole expression
For example: y = a + a'B = a + a'B + a'B

Karnaugh map simplification
Question: F = ABC + abd + CD + BD
How to make the final CD and BD in the form of ABCD?

First, write the original formula in the form of minimum term
F=ABC(D+D')+ABD(C+C')+CD(AB+A'B+AB'+A'B')+BD(AC+A'C+AC'+A'C')
=ABCD+ABCD'+ABC'D+A'BCD+AB'CD+A'B'CD+A'BC'D
Then draw Karnaugh map and merge it

F=ABCD+ABD+BCD+ABC+BD+BC
The D of ABCD has a horizontal line, the D of BCD has a horizontal line, and the C of BC has a horizontal line
It is the problem of simplifying logic function of digital circuit

(1) F = ABC (dnon) + abd + BC (dnon) + ABC + BD + B (cnon)
ABC (d-non) can be absorbed by ABC, and abd can be absorbed by BD, so f = BC (d-non) + ABC + BD + B (c-non)
(2) Let B be f = B [C (dnon) + AC + D + (cnon)]
Where D + (cnon) = [cnon]
So f = B {C (D non) + [C (D non)] non + AC}
(3) So f = B (1 + AC) = B
The final result is f = B

Boolean algebra reduction AB + abc'd + ABDE '+ a'bc'e + a'b'c'e

AB+ABC'D+ABDE'+A'BC'E+A'B'C'E
= AB+A'C'E

Proof of Boolean Algebra B * C + D +! D (! B +! C) (d * a + b) = B + D

When d = 1, the left and right sides are obviously 1
When d = 0, the right side is B
On the left is BC + (! B +! C) B = (c +! B +! C) B = 1 * b = B, which is also B
So the equation holds

Use the basic formula of logic algebra to simplify the following logic functions
(1) Y = (a non) B + a (b non) + A
(2) Y = a (b not) (C not) + ABC + a (b not) C + AB (C not) + (a not) B

The process of the second question is shown in figure 840y3 = AB + AB + C + bdy4 = BCD + ABC + ABC + bcdy5 = BD + AC

Simple calculation 20

What is the sum of "and" or "not" in Boolean algebra?
For example: what are the results of 34 and 55?

Logic algebra or Boolean algebra. Although it also uses letters to represent variables like ordinary algebra, the values of variables are only "1" and "0". The so-called logic "1" and logic "0" represent two opposite logical states