By Elias, Peter

Another concept of interpreting known as checklist interpreting, proposed independently via Elias

and Wozencraft within the overdue 50s, permits the decoder to output an inventory of all codewords that

dier from the got note in a undeniable variety of positions. even if restricted to

output a comparatively small variety of solutions, checklist interpreting allows restoration from mistakes well

beyond the d=2 barrier, and opens up the opportunity of significant error-correction from

large quantities of noise.

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**Additional resources for List decoding for noisy channels**

**Example text**

In applying such an algorithm, we ignore the transit times and the problem is thus a classical shortest path problem. For completeness, we describe the application of Dijkstra’s algorithm below and refer to it as SP. The algorithm maintain two sets S and S . The set S contains vertices for which the ﬁnal shortest path costs have been determined, while the set S contains vertices for which upper bounds on the ﬁnal shortest path costs are known. Initially, S contains only the source s, and the costs of the vertices in S are set to da (y, t).

If (x , g) ∈ A(T ), where A(T ) is the arc set of T , then we can restore the arc (f, x ) to a spanning tree TD of D with cost δE (D , ts ), where ts is the departure time at f in T . Combining TD and all other arcs in T except (f, x ), we can obtain a spanning tree of N , denoted as T . 3, we know that the cost to reach x from f with 37 Time-Varying Minimum Spanning Trees τ (f ) = ts is δ(x , ts , t). Then we have τ (x)−1 ζ(T ) = c(x, y, τ (x))+ (x,y)∈A(T \(f,x )) c(x, t)+δ(x , ts , t) x∈V (T \{x }) t=α(x) τ (x)−1 = c(x, y, τ (x)) + (x,y)∈A(T \(f,x )) c(x, t) + δE (D , ts ) x∈V (T \{x }) t=α(x) = ζ(T ) ≥ ζ(T ) The last inequality holds since T is the minimum spanning tree of N .

Since there is only one arc, N is a path and the claim holds obviously. Assume that when m < k, the claim is true. Now we consider the case with m = k. We examine the following cases: (i) N is a path. 1, the claim holds. (ii) N is a diamond. 2, the claim is also true. (iii) N is neither a path nor a diamond. Then we select a diamond D in N , and compute δI (D, ts , t) and δE (D , ts ) and change D to a path P (f, x , g) to obtain a new network N . Now, we prove ζ(T ) = ζ(T ), where T and T are the minimum spanning trees in N and N , respectively.

### List decoding for noisy channels by Elias, Peter

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