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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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abeer hafeez
abeer hafeez 2021-9-25
please comment if you know,
Erik Huuki
Erik Huuki 2021-9-25
编辑：Erik Huuki 2021-9-25
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So I believe you're asking whether the 2 expresssions are equivalent. Unfortunately your solution I believe is incorrect. Below is code that highlights one example of where your expression doesn't match.
x = 0;
y = 1;
z = 0;
if((~x*~y*~z + ~x*~y*~z + ~x*y*z + x*y*z)==(~x*z+~z))
fprintf("Equal for x'yz'\n")
else
fprintf("Unequal for x'yz'\n")
end
Unequal for x'yz'
Below is my work and how I got to it using a Kurnagh map.
abeer hafeez
abeer hafeez 2021-9-25
okah , great. but the problem is what are the groups are you taking , because i think the group will be from 2 and 3 , and will 1 and 8 make a group
?.
abeer hafeez
abeer hafeez 2021-9-25
i want like the solution of the above equation , to match it with my answer.

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回答（3 个）

Erik Huuki
Erik Huuki 2021-9-26
编辑：Erik Huuki 2021-9-26
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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
Not equivalent for x'yz'
Below is the kurnuagh map that arrives at the solution

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abeer hafeez
abeer hafeez 2021-9-26
okah, thanks very much , i even asked my computer sir , the asnwer was x' in the notes. why?.
Erik Huuki
Erik Huuki 2021-9-26
编辑：Erik Huuki 2021-9-26
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
Erik Huuki
Erik Huuki 2021-9-26
xor means “exclusive or” Ex. 0^0=0 0^1=1 1^0=1 1^1=0

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Walter Roberson
Walter Roberson 2021-9-26
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%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 )
ans = 8×6 table
x y z proposed_output answer_from_notes actual_output _____ _____ _____ _______________ _________________ _____________ false false false true true true true false false true false false false true false true true false true true false true false true false false true true true true true false true false false false false true true true true true true true true false false false
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.

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Walter Roberson
Walter Roberson 2021-9-26
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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.
[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)
ans = 8×6 table
x y z proposed_output answer_from_notes actual_output _____ _____ _____ _______________ _________________ _____________ false false false true true true true false false true false true false true false true true true true true false true false true false false true true true true true false true false false true false true true true true true true true true false false false

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abeer hafeez
abeer hafeez 2021-9-26

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this is the answer in notes ,
and this is my answer.

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Walter Roberson
Walter Roberson 2021-9-26
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[x, y] = ndgrid([false, true]);
x = x(:); y = y(:);
expression = (~x & ~y) | (x & y);
table(x, y, expression)
ans = 4×3 table
x y expression _____ _____ __________ false false true true false false false true false true true true
In order for your z'(y'x' + xy) ---> z' to be true, then the output expression above would have to be true in each case.
xy + x'y' is "not exclusive or"
nox = ~xor(x, y);
table(x, y, expression, nox)
ans = 4×4 table
x y expression nox _____ _____ __________ _____ false false true true true false false false false true false false true true true true
Walter Roberson
Walter Roberson 2021-9-26
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.
abeer hafeez
abeer hafeez 2021-9-26
no , this is the only questions , which have 3 variables , so 2 power 3 = 8.

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