Using 3-8 Demux and two logic gates, make a circuit that received 4 bits and returns 1 if the number of 1's is odd and else returns 0.

My try:

Using truth table, we get that the desired function is: $$f=a'b'c'd+a'b'cd'+a'bc'd'+a'bcd+ab'c'd'+ab'cd+abc'd+abcd'$$

Please any help or direction.


The idea of this question is that you have to think beyond the truth table or K-map. I suggest to consider that a 3-8 demux has thee inputs that get decoded to eight outputs. Use those as three of the four input bits and realize that you can OR together the decode outputs that represent inputs that have an odd combination of '1's for the three input bits.

From there you have to figure out how to add the 4th input bit to achieve the net desired logic function. From the problem statement it appears that you can select any type of logic gate with appropriate number of inputs.

  • \$\begingroup\$ Good answer because you pointed him in the right direction but did not give it all away, forcing Um to puzzle over it. \$\endgroup\$ – richard1941 Nov 17 '18 at 18:26

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