By Martin Rubey
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Consequently, hz; C~ i is equal to zero as well, which implies that C~ is in the orthogonal complement of Ker B. It remains to show, that dim Ker B = q p + c and dim(Ker B) = p c: Let x be any vector with p components, and for any edge e, let h(e) be its head and t(e) its tail in the (arbitrary) orientation of G. Then we have Bt x(e) = x(h(e)) x(t(e)): Hence, x is in the kernel of Bt , if and only if x is constant on each component of G, which implies that dim Ker Bt = c. Bt is a function de ned on the vertices of G, therefore we have dim Im Bt = p dim Ker Bt = p c: By the `row rank=column rank' theorem, dim Im B = dim Im Bt = p c.
Let V be a set containing the vertices of G H1 ; H2 ; : : : ; Hp] except of rx. Produce the corresponding tree T as follows: WHILE not all of the words in w are empty Let ui be the vertex with the smallest label in V which has outdegree zero in Fu and does neither occur in wv , nor in wu . Remove ui from V . If ui has a predecessor in Fu , then let vj be this vertex and remove (vj; ui) from Fu . If ui does not have a predecessor in Fu but wu 6= ;, then let vj be the rst letter in wu . Otherwise let vj be the rst letter in wv , where v is the predecessor of u in F .
Given a linear order on the vertices of G and H1 , H2 , : : : , Hp , in what follows we will use the lexicographic order on the vertices of the lexicographic product G H1 ; H2 ; : : : ; Hp], id est: ui < vj :, u < v or u = v and i < j: in a forest a leaf is not a root and has degree one (outdegree zero) in digraphs, an arc (v; u) starts in v and terminates in u 3. CODES 43 The encoding we will present enables us to encode the spanning trees of any (di)graph of the form G H1 ; H2 ; : : : ; Hp]. 4 below.