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1.  $\sum m(2,3,6,7)$
2. $\sum m(0,1,2,3)$
3. $\sum m(0,1,4,5)$
4. $\sum m(4,5,6,7)$
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Option (c) ? Please tell me it is (c)
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Yah C
AIR 1 incoming
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I am unable to understand
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Haha - one can hope.

Were you able to solve it ?
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Nope

I got stuck in the decoder part
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Is the last bubble just adjacent to f a NOT?
+1
I hope not. That's a very odd place to insert a negation. Even if it is, just complement the final output and see if it matches any of the options. If not, ignore that particular bubble.

+1 vote
Stage 1 is a MUX - finding the output should be straightforward

$Y = \overline{S_1}.\overline{S_0}.1 + \overline{S_1}.{S_0}.0 + {S_1}.\overline{S_0}.0 + {S_1}.{S_0}.1$
substitute for $S_1, S_0$

$Y = \overline{A}\overline{C} + {A}{C}$

Move on to the decoder now. We're only bothered with the outputs $O_0$ and $O_2$, lets not bother with any others:

$O_0 = \overline{B}\overline{Y}$, think about why this is. The decoder output $O_0$ is selected when input is $00$ (converts encoded binary to decoded decimal)

$O_2 = \overline{B}{Y}$, Output $O_2$ selected when input is $10$

And the outputs are bubbled (negated) before fanning into the NAND:

$f(A,B,C) = \overline{\overline{O_0}.\overline{O_2}} = O_0 + O_2$

Now just substitute for Y in the expressions for $O_0 + O_2$ we obtained earlier, and you will get the required minterms of $\Sigma(0, 1, 4, 5)$
by (1.9k points)
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+1
This is a case where going the SOP method will take time (4 variables here, not a good approach). So try manipulating the expression.

$(\overline{A}+\overline{B})(C + D) = (\overline{A}+\overline{B}).C + (\overline{A}+\overline{B}).D$

$=\overline{AB}.C + \overline{AB}.D$

double complement

$=\overline{\overline{\overline{AB}.C + \overline{AB}.D}}$

$=\overline{\overline{\overline{AB}.C}.\overline{\overline{AB}.D}}$

Count the bars - $\overline{AB}$ can be calculated once and reused. So I'm guessing answer is 4..
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Beautiful.. :)
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Haha thanks. The solution I typed out seems elegant. The working is anything but. As soon as I decided to expand the given expression - things fell in place. I assume this is the kind of problem I'll skip and tackle at the end if I have enough time. If it clicks - its a very short solution. Else, like Vimal mentioned, its NP-C.
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Though NP-C it amazes me how a human mind can apply various procedures best suited according to the given problem of that context to reach a solution quite quickly, where as any well defined algorithm fails to do so in general. There is lack of flexibility in algorithms, where a human mind has it in plenty.
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Which is precisely why all those RSA crypto questions in ME where they ask you find $e$ or $d$ for some insanely large $n$ are to be considered bad. The basis of RSA is the discrete logarithm problem, which itself is NP-I. But alas.