LT codes are practical fountain codes that are near-optimal erasure correcting codes.

Simply stated, for encoding a $n$-block message, each packet first chooses a degree $d\in\{1,\ldots,n\}$ according to a specific distribution, and then $d$ random blocks are xor-ed to create the packet's message.

The analysis shows that $O(n)$ packets that make to the receiver are enough for decoding, by allowing finding degree-one packets and xoring its content from all other packets that contain the same block (decreasing their degree by one).

What I haven't found in Luby's paper, or anywhere else, is the runtime complexity of the decoding. That is, what's the overall time spend on computing the original message.

A simple argument shows that $O(n^2)$ time is enough. Can we do better?


1 Answer 1


The approach you describe can be implemented to run in linear time.

Build a bipartite graph where left-vertices are blocks and right-vertices are packets, and with an edge from block $b$ to packet $p$ if $b$ is one of the blocks xor-ed to create $p$'s message. Store the bipartite graph in adjacency list format. Also keep track of the degree of each packet, so given a packet $p$ you can compute its degree in $O(1)$ time.

Then it's easy to implement the algorithm you describe in $O(n)$ time, using a worklist algorithm. The worklist is a list of all packets of degree one. In each iteration, you remove a degree-one packet $p$ from the worklist, then process it as follows. Find the corresponding block $b$, then find all other packets $p'$ containing the block. Xor $b$ into $p'$, and delete the edge $(b,p')$ (and update the degree of $p'$). This can be done in $O(1)$ time. If the new degree of $p'$ is 1, add $p'$ to the worklist. Repeat until the worklist is empty.

The running time is $O(1)$ per edge in the bipartite graph, i.e., $O(m)$ time where $m$ counts the number of edges in the graph (the sum of degrees of all of the packets).

  • $\begingroup$ Thanks for the answer. I agree that O(m) is achievable this way. The two questions I have are: (1) what is m, as a function of n, for the distribution (Soliton / robust Soliton) that gives O(n) packets overall? And (2) can we get a faster algorithm by getting more than O(n) packets? $\endgroup$
    – R B
    Jan 19, 2020 at 2:48
  • $\begingroup$ That is, the distribution presented in the paper aimed to minimize the number of packets, potentially without concerning the decode time. Is it optimal also when considering the decoding? $\endgroup$
    – R B
    Jan 19, 2020 at 2:53
  • $\begingroup$ @RB, I don't know; the distribution isn't specified in the question, so I'm not in a position to answer about the asymptotics of $m$ as a function of $n$. That sounds worth a separate question, perhaps in Math.SE. It appears that (1) basically boils down to what is the expectation of that distribution. $\endgroup$
    – D.W.
    Jan 19, 2020 at 3:44
  • $\begingroup$ (1) may boil to computing expectation, but perhaps we don't need to decode all edges to recover the message? That is, we need to get all blocks attached to a degree-1 packet at some point, but why do we have to remove all edges? Think of it this way -- say we decoded all blocks but 1. Now comes a new packet with high degree and without our missing block. Why should we bother with it? $\endgroup$
    – R B
    Jan 19, 2020 at 22:42
  • $\begingroup$ That is to say, I agree that your approach gives $O(m)$ time; I am just not convinced that this is the best we can hope for. $\endgroup$
    – R B
    Jan 19, 2020 at 22:46

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