Digital Design - decoders and multiplexer help

AI Thread Summary
The discussion focuses on implementing a function f(x3, x2, x1, x0) that determines the parity of the number of 1's in a four-bit input, producing outputs y1 and y0. Participants explore using four 2-to-4 decoders and a 4-to-1 multiplexer for implementation. A truth table is created to clarify the outputs for various input combinations, confirming that for the input (0, 0, 0, 0), both outputs y1 and y0 should indicate even and odd counts, respectively. The conversation highlights the confusion around the outputs for zero inputs, emphasizing that zero is considered even. The thread ultimately seeks clarity on the next steps for solving the problem.
Saterial
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Homework Statement


Let f(x3; x2; x1; x0) = (y1; y0) such that y1 = 1 if the number of 1's in x3x2x1x0 is even, 0 otherwise, and, output y0 = 1 if the number of 1's in x3x2x1x0 is odd, 0 otherwise.

1. Implement f with four 2-to-4 decoders.
2. Implement f with a 4-to-1 multiplexer with selection inputs x0; x1 (in this order)

Homework Equations


The Attempt at a Solution



How do I even start solving this? This was a bonus as it was not really taught to me so I have to learn myself.

(IN THE CASE OF 0, 0, 0, 0, both y1 and y0 are 0 right? because there are no 1's at all)

I begin by making a truth table of:
Code:
x3 x2 x1 x0   y1 y0
 0  0  0  0    0  0
 0  0  0  1    0  1
 0  0  1  0    0  1
 0  0  1  1    1  0
 0  1  0  0    0  1
 0  1  0  1    1  0
 0  1  1  0    1  0
 0  1  1  1    0  1
 1  0  0  0    0  1
 1  0  0  1    1  0
 1  0  1  0    1  0
 1  0  1  1    0  1
 1  1  0  0    1  0
 1  1  0  1    0  1
 1  1  1  0    0  1
 1  1  1  1    1  0

Where do I go from here?
 
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Saterial said:
(IN THE CASE OF 0, 0, 0, 0, both y1 and y0 are 0 right? because there are no 1's at all)

Zero is an even number.

If x3..x0 | y1,y0 = 0,0,0,0 | 0,0 then y1 would indicate: "number of ones is not even" y0 would indicate "number of ones is not odd" --a contradiction.
 
So that would mean the output of y1 and y0 in my case is 1, 0 for 0, 0 ,0 0 in my case because the number of 1's is even and not odd?
 
Yes. And it should make your job somewhat easier as well.
 

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