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Implement f(a, b, c, d) = Σ m(3, 4, 5, 6, 7, 11, 15) as a 2-level gate circuit
(a) Using OR gates and NOR gates.
(b) Using NOR gates only.

I have found that F=ab+d using Karnaugh map. I have also been able to construct the logic expression using OR gates and NOR gates as in part a of the question.

Can you please provide me some idea on how to construct the logic expression ab+d using NOR gates only?

Please do not say this is a homework help question and close this thread. I'm not asking to do the entire homework, I have only asked for some idea on how to do part b.

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    $\begingroup$ Try to build NOT, AND, and OR gates only with NOR gates as components first. After you solved that task, the rest of the question becomes much more simple. $\endgroup$ – DCTLib Sep 19 '15 at 9:28
  • $\begingroup$ @DCTLib I'm afraid that in this scenario that advice is entirely useless, since the asker has to construct a two-level circuit. $\endgroup$ – orlp Sep 19 '15 at 14:14
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    $\begingroup$ The fact that you solved part (a) does not mean it's OK to copy-paste a dump of part (b) and ask us to solve it for you. What have you tried, for part (b)? Where did you get stuck? We do not want to just do your exercise for you; we want you to gain understanding. However, as it is we do not know what your underlying problem is, so we can not begin to help. See here for a relevant discussion. $\endgroup$ – D.W. Sep 20 '15 at 4:20
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    $\begingroup$ 1. Please be constructive. Rude language is not appropriate here. 2. I didn't say it was homework (and I frankly don't care whether it is homework or not). My comment applies regardless, whether or not this is not homework. $\endgroup$ – D.W. Sep 21 '15 at 15:08
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Hint: you know it's a two-level circuit using only NOR gates. There is only one such circuit:

  OUT
   |
  NOR
  / \
 |   | 
NOR NOR
/ \ / \
? ? ? ?

Can you now figure out where to place a, b and d to correctly construct the circuit? Since you have to repeat (at least) one input, does looking at symmetry help?

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  • $\begingroup$ Without myself understanding this question entirely, I think I can say that you have 3!*(4 chooses 3) = 6 * 4 = 24 possibilities for 3 inputs into Orlp's diagram, I hope this is correct and of some help. I'm not encouraging using a search here :) $\endgroup$ – alan2here Sep 20 '15 at 13:01
  • $\begingroup$ One of the inputs of the NOR at the top could be not a NOR, but just one of the original input values. Obviously that means the whole circuit has only three inputs, so it can't possibly handle functions that are affected by all four inputs. $\endgroup$ – gnasher729 Feb 18 '17 at 23:44

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