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Thank you Klaus, If i follow correct your example, then for a 64:1 Mux then we need 16 pieces of 6 inputs LUTs (4:1 Mux) for the first stage, then another 4 piece, and in the final stage one more. This give us a total of 21 LUTs and a total of 6 address lines (2^6) = 64
I want to find out how many LUTs I need to build a 32 input MUX.
I know that a 6 input LUT for example can be used to map a 4:1 MUX. Is there a way to calculate how many LUT we need for N inputs MUXs? Also is there any common LUT size used in FPGAs (4LUT, 6LUT or something else?)
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