Delay complexity of Boolean functions

[peripatein]

Homework Statement

I don't really understand how/why every Boolean function of n variables may be implemented with a delay complexity of O(n).
Could someone please try and explain?

Homework Equations

The Attempt at a Solution

I was trying to show it using minterms (SOP). There is a maximum of 2^n minterms but how do I derive the delay? There is also a maximum of n-1 OR gates between these minterms.

 

[KataKoniK]

The delay complexity of a Boolean function refers to the number of logic gates and their connections needed to implement the function. In the case of a function with n variables, there are 2^n possible input combinations, each of which can be represented by a minterm.

To implement a Boolean function with minterms, we can use an AND gate for each minterm and then connect all the AND gates to a single OR gate. This means that for a function with n input variables, we will need a maximum of 2^n AND gates and 1 OR gate.

Now, to determine the delay complexity, we need to consider the worst-case scenario. This would be when all 2^n minterms are combined in a way that results in the longest path from an input to the output (i.e. the longest delay). In this case, the maximum delay would be n-1, as there are n-1 OR gates between the input and the output.

Therefore, the delay complexity for a Boolean function with n variables is O(n), as the maximum number of logic gates and their connections needed is n-1. This means that the implementation of any Boolean function with n variables can be achieved with a delay complexity of O(n).