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A comment on A maximally-dense encoding for n-choose-k (2013). Parent comment
So, the result is in T. You subtract n, leaving 4, and use whatever bijection you had for T to decode that into an element of T.
If, on the other hand, you had n-4, you know it must decode to an element of S, and directly use its decoder to find that element.
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So, the result is in T. You subtract n, leaving 4, and use whatever bijection you had for T to decode that into an element of T.
If, on the other hand, you had n-4, you know it must decode to an element of S, and directly use its decoder to find that element.