Minimum number of 2-input NOR gates required to implement the given expression

To find the minimum number of 2-input NOR gates required to implement the given expression, we need to analyze the expression and simplify it using Boolean algebra. We can break down the expression into smaller parts to identify the common terms that can be shared among the different sub-expressions.

Given Expression:
Y = ACE + AFC + ADE + ADF + BCE + BDF + BCF + BDE

Step 1: Simplifying the Expression
To simplify the given expression, we can use Boolean algebra rules such as the distributive law and De Morgan's theorem. By applying these rules, we can identify the common terms and reduce the number of gates required.

Step 2: Identifying the Common Terms
After simplifying the given expression, we can identify the common terms that can be shared among the sub-expressions, which helps in reducing the number of gates required. The common terms are:

- A
- C
- D
- E
- F
- B

Step 3: Implementing the Reduced Expression
Using the identified common terms, we can implement the reduced expression by sharing the common terms among the different sub-expressions. By doing this, we can eliminate the redundant gates and reduce the overall number of gates required.

Step 4: Counting the Number of Gates
After implementing the reduced expression, we count the number of 2-input NOR gates required. In this case, the minimum number of 2-input NOR gates required is 6.

Conclusion
By simplifying the given expression and identifying the common terms, we can implement the expression using a minimum number of 2-input NOR gates. In this case, the minimum number of gates required is 6.