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I'm currently working on learning to minimize a circuit which has multiple outputs using K-maps. My universities script seems rather unhelpful to me which left me in a spot where I have a very rough understanding of the matter without actually being able to pull it through.

I currently am supposed to first minimize the functions on their own and then compare that to minimizing them together. Karnaugh-maps of Fa and Fb [1]

The resulting functions are:

$Fa = (bc)+(bd)+(\neg acd)$

$Fb = (b\neg c\neg d) + (b\neg cd) + (bc\neg d)$

So now I think I am supposed to minimize them together. My approach is to mark the 1s in the maps that are unique to one function red and group them first.
After that, I look at the ones that are shared. I can now group these the same way in both maps. K-maps of Fa and Fb with common terms in yellow[2] The groups bordered by yellow are the ones that are shared. So I now created the functions for these two.

$Fa = (bcd) + (\neg acd) + (b\neg cd) + (bc\neg d)$

$Fb = (b\neg c\neg d) + (b\neg cd) + (bc\neg d)$

Here I count that they use two unique AND gates for Fa, one unique AND gate in Fb, two unique AND gates across the two and then one OR gate for each of these two. So all in all, they use 7 gates. This is exactly the same as before I minimized them together, thought the assignment heavily implies that the amount of gates should go down.
I simply cannot figure out what I am doing wrong here.

Furthermore, they tell me to then do this without K-maps, which is a task that I don't even know how to begin working on.

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Good job on $f_a = \overline acd \lor bc \lor bd$.
With $f_b$, I'd denote it as $b\overline d\lor b\overline cd$ - you missed overlap with $\overline d$: $f_b = b\overline c \lor b\overline d$.
I'd tally that as and inputs: 11 complemented: 3 ands: 5 gates/ors: 2

In the combined implementation, groups are still allowed to overlap: both blue ones can be twice as large, use one input less.
One can share either $bc\overline d$:
$f_a = b\overline cd \lor \overline acd \lor bc$
$f_b = b\overline cd \lor b\overline d$
or $b\overline cd$:
$f_a = bc\overline d \lor \overline acd \lor bd$
$f_b = bc\overline d \lor b\overline c$
and inputs: 10 complemented: 3 ands: 4 gates/ors: 2

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  • $\begingroup$ Thank you very much! This exemplatory process really helped me understand what I was missing and where my problem was. Thanks for even showing those two possible solutions. Great answer! $\endgroup$ – Vladis Becker Nov 25 '19 at 13:13

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