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This is my homework: "Given the function \$f(x_1, x_2,x_3) = \sum (2,3,4,6,7)\$ Show how it can be realized using two 2-input LUTs. Give the truth table implemented in each LUT."

At the beginning it seems very simple, but this is a beast! I don't want you to solve it, but just show me a way to come up with a general formula to solve these kinds of LUT optimization problems.

I only know that an n-input LUT can be used to implement a \$ 2^n\$ input function, so in this case we have 3 inputs so we need a 3-input LUT. Meanwhile if we are forced to use 2-input LUT, then I can see that we can divide the function output values into two LUTs with \$x_1, x_2\$ as their inputs, and then put a MUX with \$x_3\$ as its selector to produce the function \$f\$. But in this problem we are not allowed to use the MUX, but only two 2-input LUTs. How can it be?!

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    \$\begingroup\$ Have you tried using Karnaugh diagrams? This might simplify your function quite radically. After that you could use de Morgan's law and other laws to get a function you want \$\endgroup\$ Commented Aug 28, 2016 at 14:08
  • \$\begingroup\$ This problem is not about function optimization. It is about implementation of a function using LUTs with constraints. \$\endgroup\$ Commented Aug 28, 2016 at 14:11
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    \$\begingroup\$ ... which is the same problem, @Ehsan. \$\endgroup\$ Commented Aug 28, 2016 at 14:16
  • \$\begingroup\$ For example, 2,3, 6 and 7 look like could be grouped and written by two or even one variable which would make everything so much easier, wouldn't it? \$\endgroup\$ Commented Aug 28, 2016 at 14:24
  • \$\begingroup\$ No. In LUT implementation, optimization do not have an effect on the number of LUTs being used. There is no need for Boolean optimization here. \$\endgroup\$ Commented Aug 28, 2016 at 14:28

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