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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