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In the KMAP shown below, I can think of three different ways to minimize it. How can we determine which is correct? I have listed the minimized equations and why I think each one is acceptable below.

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I believe this case is acceptable because it has the fewest number of groups, and the largest possible group.

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I believe this case is acceptable because it utilizes the overlap rule and doesn't violate any other rule.

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I believe this case is acceptable similar to Case 1.

They all are within the rules, so what makes one more correct than the other?

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    \$\begingroup\$ Only one of those hits all your 1's! \$\endgroup\$ – Scott Seidman Sep 18 '18 at 22:02
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    \$\begingroup\$ DeeTee, assuming your blank boxes mean "0" and not "X", you've done a good job of showing the two minimized groupings, in your first and third tables, that you need to OR together to create the right answer shown in your middle table. Only the middle result is correct, though. \$\endgroup\$ – jonk Sep 18 '18 at 22:24
  • \$\begingroup\$ Are you assuming any 1s not grouped are represented by a 3 input AND term? \$\endgroup\$ – crj11 Sep 18 '18 at 22:26
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Consider:

For Case 1:

A = 0

B = X

C = 1

Solution = 1

But AC = 0

Wrong

For Case 3:

A = X

B = 1

C = 1

Solution = 1

But B'C = 0

Wrong

If you are not using all your "true" solutions in the formulation then bizarre things start to happen :)

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    \$\begingroup\$ This answer is lacking in the structure department, I think. \$\endgroup\$ – Harry Svensson Sep 20 '18 at 0:39
  • \$\begingroup\$ @HarrySvensson true but it also seems like someones homework which is why I am trying to not give a complete answer. \$\endgroup\$ – EasyOhm Sep 20 '18 at 6:17

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