asalam.A.
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i have a question regarding computer ( karnaugh map) , i certainly wish that a computer teacher explains me a question , x'y'z' + x'y'z + x'yz + xyz' . this is the equation, the apostrophe represents inverse, my answer is coming x'z+z' , is it right or wrong?. i want the answer till tom , like within 10 hours.
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Antworten (3)
Erik Huuki
am 26 Sep. 2021
Bearbeitet: Erik Huuki
am 26 Sep. 2021
The above answer you provided was incorrect. Below shows a case to prove it.
x = 0;
y = 1;
z = 0;
if (~x*~y*~z+~x*~y*z+~x*y*z+x*y*~z==~x*z+~z)
fprintf("Equivalent for x'yz'\n");
else
fprintf("Not equivalent for x'yz'\n")
end
Below is the kurnuagh map that arrives at the solution
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/749829/image.jpeg)
3 Kommentare
Erik Huuki
am 26 Sep. 2021
Bearbeitet: Erik Huuki
am 26 Sep. 2021
The only answer I can come up with is that your mistaking xor with or. Symbolically they look similar but have very different meanings. So an alternative answer could be x^yz’. where ^ is the operator for xor
Walter Roberson
am 26 Sep. 2021
%x'y'z' + x'y'z + x'yz + xyz'
[x, y, z] = ndgrid([false true]);
x = x(:); y = y(:); z = z(:);
actual_output = (~x & ~y & ~z) | (~x & ~y & z) | (~x & y & z) | (x & y & ~z);
proposed_output = (~x & z) | (~z);
answer_from_notes = (~x);
table(x, y, z, proposed_output, answer_from_notes, actual_output )
You can see from this that both the proposed expression you made, and the answer from the notes, must be incorrect. The proposed output differs from the actual output in the second and third lines.
1 Kommentar
Walter Roberson
am 26 Sep. 2021
When I look carefully at the way the question is drawn, it looks like possibly instead of
(that you implemented) that the term is
. and likewise in the second term. The notation needs to be examined carefully.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/750029/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/750034/image.png)
[x, y, z] = ndgrid([false true]);
x = x(:); y = y(:); z = z(:);
actual_output = (~(x & y) & ~z) | (~(x & y) & z) | (~x & y & z) | (x & y & ~z);
proposed_output = (~x & z) | (~z);
answer_from_notes = (~x);
table(x, y, z, proposed_output, answer_from_notes, actual_output)
abeer hafeez
am 26 Sep. 2021
3 Kommentare
Walter Roberson
am 26 Sep. 2021
The only thing I can think of for why they mark z in the diagram the way they do, is if there is another equation you have not shown us that defines z in terms of x and y. In such a case there would only be 4 states instead of 8.
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