Using 3 input LUT's to Implement a Boolean Function

[blue2004STi]

Homework Statement

Consider the function f(w1,w2,w3,w4) = SUM( m(0, 1, 3, 6, 8, 9, 14, 15)). Derive an implementation using the minimum possible number of three-input LUTs.

Homework Equations

Where SUM(m(0, 1, 3, 6, 8, 9, 14, 15)) is the sum of the min terms, a Sum of Products that from a K map is (~w1*~w2*~w3)+(~w1*~w2*w4)+(w1*w2*~w3)+(w1*~w2*w3)+(~w1*w2*w3*~w4) Where ~w is "not w".

The Attempt at a Solution

Thus far all I can do is get the minimum cost solution from the K map, which I gave above. I guess I'm just a little confused as where to start considering a 3 input LUT has 8 stored values and 3 different inputs and one output. Any help will be beneficial! Thanks,

Matt

 

[blue2004STi]

Noone have any ideas? I'm not looking for the answer just a push in the right direction. Thanks, Matt

 

[Nyasha]

Use shannon's expansion theorem