asalam.A.

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.

[abeer hafeez]

please comment if you know,

[Erik Huuki]

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]

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]

i want like the solution of the above equation , to match it with my answer.

Answer 1

[Erik Huuki]

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

[abeer hafeez]

okah, thanks very much , i even asked my computer sir , the asnwer was x' in the notes. why?.

[Erik Huuki]

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]

xor means “exclusive or” Ex. 0^0=0 0^1=1 1^0=1 1^1=0

Answer 2

[Walter Roberson]

%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.

[Walter Roberson]

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    

Answer 3

[abeer hafeez]

this is the answer in notes ,

and this is my answer.

[Walter Roberson]

[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]

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]

no , this is the only questions , which have 3 variables , so 2 power 3 = 8.