1 votes 1 votes To design n input NOR gate, the number of 2 input NOR gate =2n - 3. Somebody please verify and explain with example ( diagram). Digital Logic digital-logic self-doubt + – Overflow04 asked Sep 22, 2022 Overflow04 305 views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply ankitgupta.1729 commented Sep 23, 2022 reply Follow Share The result is based on a recurrence. First I am showing you the example for understanding and then prove it. Suppose, you have to design $3$-input NOR gate. So, $n=3.$ Suppose, inputs are $A,B,C$ and So, result would be: $\overline{A+B+C} = \overline{\overline{\overline{A+B}}+C}$ Here, $\overline{A+B}$ requires one 2-input NOR gate, Now, $\overline{\overline{A+B}}$ requires total two 2-input NOR gates (1 NOR gate from previous $\overline{A+B}$) and finally, $\overline{\overline{\overline{A+B}}+C}$ requires total threee 2-input NOR gates (2 NOR gates from previous $\overline{\overline{A+B}}$). So, you can verify $3$ is actually $2*3-3.$ Now, Based on this, you can prove the result as: Suppose, $ \overline{A_1+A_2+A_3+…+A_n} = \overline{\overline{\overline{A_1+A_2+A_3+...+A_{n-1}}}+A_n} $ Let, $T_n$ be the minimum number of two-inputs NOR gates. Then, you can write the recurrence as: $T_n = T_{n-1} + 1 + 1$ i.e. $T_n = T_{n-1} + 2$ with $T_2 = 1$ Hence, $T_n = 2n-3$ 1 votes 1 votes Overflow04 commented Sep 23, 2022 reply Follow Share Thank you But how you written(last two line) the equation after:- Tn = Tn-1 + 2 with T2= 1 Tn = 2n -3. 0 votes 0 votes ankitgupta.1729 commented Sep 23, 2022 reply Follow Share Let, $T_{n}$ be the number of two-input NOR gates for $\overline{A_1+A_2+A_3+...+A_{n-1}+A_n},$ So, $T_{n-1}$ be the number of two-input NOR NOR gates for $\overline{A_1+A_2+A_3+...+A_{n-1}}$ and So, $T_{n-1} +1 $ be the number of two-input NOR gates for $\overline{\overline{A_1+A_2+A_3+...+A_{n-1}}}$ and So, $T_{n-1} +1 +1 $ be the number of two-input NOR gates for $\overline{\overline{\overline{(A_1+A_2+A_3+...+A_{n-1})}} + A_n}$ Since, $\overline{\overline{\overline{(A_1+A_2+A_3+...+A_{n-1})}} + A_n}$ is same as $\overline{A_1+A_2+A_3+...+A_{n-1}+A_n},$ So, $T_n = T_{n-1} + 1 + 1$ When $n=2$ then you will get one two-input NOR for $\overline{A_1 + A_2}$ So, $T_2=1,$ You could also take, $T_3=3$ (This is your base condition) Hence, $T_n = T_{n-1} + 2$ with $T_2 = 1$ When you solve this recurrence, you will get, $T_n = 2n-3$ You can also understand all these things from the example which I have given and then generalize it. 1 votes 1 votes Please log in or register to add a comment.