Proof. Let H be an r×n cycle-free parity-check matrix for , and let T = T(H) be the corresponding cycle-free Tanner graph that represents . The following procedure converts T into a graph , such that is the cut-set code of . We will describe this procedure assuming that T is a tree, in which case is connected. In case T is a forest consisting of trees, the same procedure should be carried out independently for each tree in T, and will have connected components.
Let
= {y1, y2,..., yr} be the set of check vertices in T,
and let
Xi denote the neighborhood of
yi
for
i = 1, 2,..., r. Further define
X*i = X1 X2 ... Xi.
Since T is a tree, it is always possible to enumerate
the check vertices in T in such a way that Xi intersects
X*i-1 in one and only one symbol vertex for all i.
Given such enumeration
y1, y2,..., yr,
we construct
iteratively, check-vertex by check-vertex. First, we represent
y1 and its neighborhood X1 by a cycle
1 consisting
of | X1| edges and | X1| vertices.
Now suppose that
X2 X1 = {x2}. Then we create
2 from
1
by appending | X2| - 1 edges -- one for each symbol vertex in
X2 except x2 -- and | X2| - 2 vertices, in such
a way that the edges corresponding to the symbol
vertices in X2 form a cycle in
2.
And so forth: if
Xi X*i-1 = {xi},
we create
i from
i-1
by appending | Xi| - 1 edges and | Xi| - 2 vertices,
in such a way that the edges corresponding to the symbols
in Xi form a new cycle.
It is easy to see that
= r will contain exactly n
edges and
n - (r-1) vertices. Furthermore, the code
generated by H is precisely the cycle code of . Since the
cut-set code of is the dual of its cycle code, our proof
is complete.
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For example, the cycle-free codes defined by the parity-check matrices in (3.16) and (3.19) are cut-set codes of the graphs depicted in Figure 3.7a and Figure 3.7b, respectively.
For cut-set codes, it is well-known [29,48] that 2nmd, where m is the number of vertices in the underlying graph . Indeed, this follows immediately from the fact that if the minimum distance of is d, then every vertex of must have degree at least d, otherwise the cut-set that isolates this vertex will have less than d edges. It is also well-known that , where () is the number of connected components in . Thus we obtain the following cut-set bound on the minimum distance of cycle-free codes
We have proved that every cycle-free binary linear code is a cut-set code. We now attempt to make this correspondence more precise by examining exactly which cut-set codes correspond to cycle-free binary linear codes. An answer to this question follows from a closer look at the construction in Theorem 3.11 and relies on star graphs, which are graphs where the removal of a vertex, which is connected to all the other vertices, results in a forest.
Proof. Consider a cycle-free Tanner graph T for a given cut-set code, the corresponding graph from the construction in Theorem 3.11, and its dual graph D. Absent the vertex corresponding to the outer region of , each path in D corresponds to a path through the Tanner graph T: vertices in D correspond to check nodes of T, and edges in D correspond to symbol nodes of T. Since a path in T cannot contain a cycle, the same is true for D absent the outer vertex. Thus, D is a star corresponding to T.
The other direction of the correspondence is similarly straightforward.
Given a cut-set code of a graph whose dual is a star, we may simply
reverse the algorithm in Theorem 3.11 to arrive at a
Tanner graph. This Tanner graph must be cycle free, or else
D has a cycle among vertices other than the outer vertex.
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