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Here is the expression I am trying to represent in the Karnaugh Map:

enter image description here

This is what I have done: Can somebody confirm if this is right? I have done the truth table right, however I am having doubts on whether the map has been done correctly.

enter image description here

Any help would be appreciated

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    \$\begingroup\$ Seems correct. \$\overline{(xy)}\$ simplifies to \$\overline{x}+\overline{y}\$. Then withinin the left part you get a \$\overline{x}+x\$ which is always 1, so only the right part is of interest. The right part is exactly the same as your K-map. \$\endgroup\$
    – jippie
    Commented Feb 28, 2015 at 23:07
  • \$\begingroup\$ @jippie, you should post your comment as an answer so that this question does not remain unanswered. \$\endgroup\$
    – JRN
    Commented Mar 1, 2015 at 0:25
  • \$\begingroup\$ @jippie - I understood what you mean, but do i have to change anything on the picture above or is that right? \$\endgroup\$
    – Mathematica
    Commented Mar 1, 2015 at 9:40
  • \$\begingroup\$ Also asked at math.stackexchange.com/q/1169384/18398 \$\endgroup\$
    – JRN
    Commented Mar 1, 2015 at 9:56
  • \$\begingroup\$ @Mathematica not in my opinion. \$\endgroup\$
    – jippie
    Commented Mar 1, 2015 at 11:40

1 Answer 1

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Your K-Map is correct.

R = (X + X' + Y' + Z)(XY + X'Y'Z+YZ')

R = (1 + Y' + Z)(XY + X'Y'Z+YZ')

R = (1)(XY + X'Y'Z+YZ')

R = XY + X'Y'Z+YZ'

R = XYZ' + XYZ + X'Y'Z + XYZ' + X'YZ'

R = XYZ + X'Y'Z + XYZ' + X'YZ'

respective terms of last equation will be logic one in K-Map

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