# A 4-bit combinatorial circuit that outputs 1 when integer greater than 7 and is odd number

I need to draw a combinatorial circuit that when an integer is greater than 7 and is an odd number, the output will be 1. There are 4 inputs representing 4 bits and 1 output wire. The output wire is only 1 if an integer is greater than 7 and is an odd number.

My attempt: If I segment the 4-bit input as r x y z where if the input was 8, the 4-bits would be 1 0 0 1. I know that r and z must have an AND gate but cannot figure out a way to get x y to work. I drew a truth table and found that

• 1 0 0 1
• 1 0 1 1
• 1 1 0 1
• 1 1 1 1

must all equal to 1 to satisfy the condition placed. Any help is appreciated.

• Hint: only one gate is required. – Yuval Filmus Jan 24 '18 at 19:09
• @YuvalFilmus My initial thought was that as well; however, how do I check for integers greater than 8? If I use only one AND gate on r and z, how do I check for 1 1 0 1 (11) or 1 1 1 1 (15)? I'm not allowed to ground a charge apparently. I know it would be a very easy solution to simply have the r z bits connected to an AND gate and x y being grounded since as long as r z bits output a 1, then the x y bits do not matter. – JJMin Jan 24 '18 at 19:16
• Combinatorial circuits don't care about electromagnetism. If your question is about electromagnetism then it doesn't belong in this site. – Yuval Filmus Jan 24 '18 at 19:24
• JJMin: You asked about greater than seven, not eight. But you should figure out why greater than eight makes no difference. – gnasher729 Aug 23 '18 at 8:50

Greater than $7$ along with being odd implies numbers $9, 11, 13, 15$ which you have already listed out. Going by the bit pattern generated, the resulting circuit consists of only $r . z$. You won't need to check for other even numbers $>=8$ since they have $z$ bit as $0$, and numbers smaller than $8$ will ahve $r$ bit as $0$ and will thus output $0$.