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It depends on the precise nature of the coder and there are many possibilities. Here are three:

A: f(x) = 2x

B: f(x) = 4*(x - 1)

C: f(x) = 2*(x2 - 3x + 4)

The simplest of these is f(x) = 2x.

In that case the decoder is y = ln(x)/ln(2) where ln is the natural logarithm. Logs to base 10, or any other base can also be used and, if the chosen base is 2, then the decoder becomes:

y = log2(x)

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Q: How do you design an decoder 2 to 4 and 3 to 8?
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