Tree-Tree Matrices and other Combinatorial Problems from Taxonomy

Abstract

Let A be a bipartite graph between two sets D and T. Then A defines by Hamming distance, metrics on both T and D. The question is studied which pairs of metric spaces can arise this way. If both spaces are trivial the matrix A comes from a Hadamard matrix or is a BIBD. The second question studied is in what ways A can be used to transfer (classification) information from one of the two sets to the other. These problems find their origin in mathematical taxonomy. Mathematics subject classification 1991: 05B20, 05B25, 05C05, 54E35, 62H30, 68T10 Key words & phrases: bipartite graph, Hamming distance, tree metric space, tree, mathematical taxonomy, design, BIBD, generalized projective space, Hausdorff distance, Urysohn distance, Lipshits distance, cocitation analysis, clustering, ultrametric, single link clustering, linked design, balanced design 1.

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Tree -Tree Matrices and other
Combinatorial Problems from Taxonomy
Michiel Hazewinkel
CWI
P.O. Box 94079
1090 GB Amsterdam
The Netherlands
mich@cwi.nl
Abstract.
Let A be a bipartite graph between two sets D and T. Then A defines, via Hamming distance,
metrics on both, T and D. The question is studied which pairs of metric spaces can arise this way.
If both spaces are trivial, the matrix A comes from a Hadamard matrix or is a BIBD. The second
question studied is how A can be used to transfer (classification) information from one of the two
sets to the other. These problems find their origin in mathematical taxonomy.
Mathematics subject classification 1991:05B20, 05B25, 05C05, 54E35, 62H30, 68T10
Key words & phrases: bipartite graph, Hamming distance, tree-like metric space, tree, mathematical
taxonomy, design, BIBD, generalized projective space, Hausdorff distance, Urysohn distance,
Lipshits distance, cocitation analysis, clustering, ultrametric, single link clustering, linked design,
balanced design, classification, transfer of classification information.
1. Introduction.
A great deal of the literature in mathematical taxonomy focusses on clustering; i.e.
summarizing the information present in a metric or dissimilarity on a set X by means of a
classification tree or something similar.
Here, we focus directly on the situation that one finds in the taxonomic problems of
scientific disciplines. Often, the data are in the form of a collection of documents and a collection
of key words and key phrases that is supposed to be sufficiently rich to descibe (up to a point)
the scientific field in question. Here, I am not concerned with how such a control list or thesaurus
is generated.
The data are thus in the form of a bipartite graph A (or, equivalently, a relation) between
two sets, a set D (of documents) and a set T (of terms). The bipartite graph A tells us which terms
occur in which documents.
These data can be used to define a metric space structure on both T and D by means of
Hamming distance —the distance between two terms is the number of documents in which one
term occurs and the other not. A first question that arises is what pairs of discrete metric spaces
can arise this way. For trivial metric space structures on both T and D it turns out that A must be
very regular (a Hadamard matrix, a Hadamard matrix minus one row or column, or a symmetric
BIBD. Section 2 below is devoted to some results in this direction.
It arises frequently in practice that on one of the spaces T or D there is available metric
information coming from other sources. For instance, in the case of a body of scientific literature,
co-citation analysis can be used to define ‘research clusters’ or ‘research fronts’ of strongly
linked clusters of documents. The question then arises how to transfer such information from one
of the sets, in this case D, to the other by means of the bipartite graph between them. This matter
is discussed in section three.

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Finally section 4 summarizes some recent ideas and results concerning metrics on the space
of all metrics on a given finite set. These things are fundamental for adressing the question of
finding, for instance, the best approximative ultrametric to a given metric or dissimilarity.
I thank one of the referees for some useful and informative comments.
2. The tree-tree problem.
2.1. Definiton of the problem.
As indicated above, we shall take as the basic available data a bipartite graph A between
terms and documents. Or, equivalently, A is a 0-1 matrix with the set of terms as column indices
and the set of documents as row indices. A 1 at spot (i,j) means that the term j occurs in the
document i. These data define two metric spaces as follows:
— the column space of A, cs AT A
( )( )
=
. As a set this is the set of terms. The distance
between two terms t t , ' is the Hamming distance between the corresponding columns, i.e. the
number of row indices with different entries at spots t and t'.
— the row space rs AD A
( )( )
=
of A. As a set this is the set of documents. The distance
between two documents d d , 'is the Hamming distance beteen the corresponding rows, i.e. the
number of column indices with different entries in rows d and d'.
This leads immediately to a number of natural basic questions, such as:
— Which metric spaces can arise as a T A
( ) or a D A
— To what extent is A determined by D A
( ) and T A
— Which pairs of metric spaces D, T can arise from a 0-1 matrix A?
In this paper I concentrate on the last question. Trees and classification schemes (which are
special kinds of trees) are ubiquitous in (mathematical taxonomy). Thus it is important and
natural to start with the question when both the column and row spaces of a 0-1 matrix are trees
or related to trees.
( )?
( )?
2.1.1. Definitions. A tree is an unoriented connected
graph such that there is a unique path between any two given
vertices. A leaf of a tree is a vertex with just one edge incident
with it. An edge weighted tree is a tree with each edge labelled
with a real number > 0. An example is shown on the right. The
distance between two vertices of an edge weighted tree is the
sum of the weights of the edges occurring in the unique path
between those vertices. This defines a metric on the set of
vertices (and on any subset, particularly the set of leafs). A
rooted tree is a tree with a special, selected vertex called the
root. A hierarchical tree is a rooted edge weighted tree such
that each leaf has the same distance to the root.
2
3
1
2
2
3
1
2
3
Figure 1

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Figure 1 is not a hierarchical tree but figures 2
and 3 are. In these figures and those below an
unlabelled edge is supposed to have weight 1.
A hierarchical tree defines an ultrametric on its
set of leaves. And inversely, [6, 11], every
finite ultrametric space arises that way. By
inserting, if necessary, extra vertices of
valency two (as was done in figure 3), each
ultrametric space arises as the space of leafs of
some ‘hierarchically organized’ tree like the
one in figure 3 in which, for each vertex, all
the edges pointing towards the leafs have the same weight.
It is rather easy to see that each edge weighted tree with integer
weights can be realized as a T A
( ) (or a D A
( ) and D A
( ), are required to be trees or tree-like (definition below). This
appears to be quite difficult to realize. In particular, it seems difficult to realize a pair of spaces
that are not (nearly) isomorphic. This is, roughly, what I like to call the tree-tree problem. To
make the problem more precise, let us define:
2.1.2. Definition. A finite metric space ( , )
X m is tree-like if it is isometric to a subspace of
the vertex metric space defined by an edge weighted tree.
2.1.3. Tree-tree problem. Which pairs of tree-like spaces can be realized by a 0-1 matrix.
I view these 0-1 matrices as some sort of generalized hierarchical block designs. The
reason for that is theorem 2.2.5 below.
Related to the tree-tree problem is the problem of finding a good characterization of those
matrices for which both the column metric space and the row metric space are tree like.
Of course tree-like metric spaces are characterized by the so-called four-point condition:
2.1.4. Four-point condition. A finite metric space ( , )
not necessarily distinct four points a a b bX
1212
,, ,
∈
m a am b bm a b
( , )( , ) max{ ( , )
121211
+≤+
This gives a necessary and sufficient condition for a 0-1 matrix A to yield a pair of tree-like
spaces. But certainly a very inelegant and unsatisfying one.
2.1.5. Ultrametric tree-tree problem. Which pairs of ultrametric spaces can be realized
by a 0-1 matrix?
2.1.6. Complete tree-tree problem. Which (complete) pairs of edge weighted trees can be
realized by a 0-1 matrix?
( )). Things are rather
different if both, T A
X m is tree-like if and only if for all
m a b
( ,
m a b
), ( ,
m a b
( , )}
2
)
22121
+
(2.1.4.1)
2.2. Trivial tree - trivial tree matrices and BIBD’s.
Let us start with some very simple examples where the column and row metric spaces are
‘trivial’ in the sense of the definition below.
2.2.1. Definition. A trivial discrete metric space ( , )
a positive number a such that
m x yaxyX
( , ) =≠
for all in
(and of course m x xxX
( , ) =∈
0 for all ).
2.2.2. Example. Hadamard matrices. A Hadamard matrix is an n
entries 1 1,− such that
HHnI
n
=
.
It follows that also HH nI
n
=
(and that n is even, n
X m is a metric space such that there is
n
× matrix H with
T
T
k
= 2 ). It is immediate from these two
Figure 3
2
Figure 2

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properties that for each two rows there are precisely k entries that are equal and k entries that are
unequal. And similarly for the columns. Let A be the matrix obtained from H by replacing each
−1with 0. Then both the column and the row space of A are the trival metric space of n
points with distance k.
2.2.3. Example. Hadamard matrices with one row or column deleted. Now let H be a
Hadamard matrix for which one row or column consists entirely of +1’s or entirely of -1’s.
Delete that row or column. Again replace -1 with 0 everywhere. The result is an 0-1 matrix with
trivial column and trivial row space of sizes n and n −1 and distance n/2.
Not every Hadamard matrix has such a column or row. However if D is diagonal with each
diagonal element equal to 1 or -1, and if H is an Hadamard matrix, then so are HD and DH. So it
is easy to modify a Hadamard matrix so as to obtain one with such a column or row.
2.2.4. Example. Symmetric BIBD’s. A balanced incomplete block design (BIBD) is a
zero-one matrix A such that each row has the same number , r, of 1’s, each column has the same
number, s, of 1’s, and further, for each pair of column indices i
which have a 1 at both locations i and j. This last condition is the same as saying that each two
different columns have λ common 1’s.
A BIBD is symmetric if A is square. It then follows that r
rows also have λ common 1’s, see e.g. [3].
It follows immediately that the row space and the column space of a symmetric BIBD are
trivial metric spaces with n points and distance 2(
2.2.5. Theorem. Let A be an mn
× zero-one matrix such that both the column space and
the row space are trivial. Then A is one of the examples 2.2.2 - 2.2.4. I.e. A ‘is’ a Hadamard
matrix, a Hadamard matrix with one constant row or column deleted, or it is a symmetric BIBD.
Let B be the matrix obtained from A by replacing each 0 with -1. Then the trivial column
and row space condition on A translates for B in the statement that the rows of B form a system
of m length n vectors all of whom make the same angle with one another and the columns form a
system of n vectors of length m that also all make the same angle with one another.
2.2.6. Proof of theorem 2.2.5. Let B be the m
0 with -1. Let d be the distance between each two distinct rows of B (or A) and e the distance
between each two distinct columns. Then
npp
pn
p
ppn


L
k
= 2
j
≠ there are precisely λ rows
s
= and that each two distinct
)
r − λ .
n
× matrix obtained from A by replacing each

BB
T=








L
OM
MO O
, pnd
=− 2(2.2.6.1)

B B
m
q
q
m
q
q
mqq
qme
T
=










=−
L
OM
M O O
L
,2
(2.2.6.2)
Interchanging rows and columns if necessary we can assume m
mm
×
matrix BBT is nonsingular except when p
happen because d > 0. The second case can happen. Then, because m
p = −1. Now, add one column of 1’s (or -1’s) to B to obtain an m
B is a Hadamard matrix. So in this case, we are dealing with an instance of the examples 2.2.3.
Continuing, we can assume that BBT is nonsingular and hence that
nm
=

n
−
≥ . By the lemma below, the
mp
) 1 . The first case can not
n
≥ , n
=
m
×
matrix B. It follows that
n
= or n
= −
(
m
−1, and
(2.2.6.3)

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Let c c
(2.2.6.1) with B on the right to obtain
cn
12
, ,,
L
be the column sums of B, and let r rrn
12
, ,,
L
be the row sums of B. Multiply

BB Bn p B
)
p
cc
cc
T
n
n
=−+








(
1
1
L
MM
L
(2.2.6.4)
and, using nm
=
, multiply (2.2.6.2) on the left with B to obtain
r
=−+

L
Subtracting (2.2.6.5) from (2.2.6.4), we see that the matrix (
≤ 2. If nm
=≥ 3 this is only possible if p
Now, there are two cases.
Case 1, pq
== 0. Then, B is a Hadamard matrix by (2.2.6.1).
Case 2, pq
=≠ 0. Then, it follows from (2.2.6.4) and (2.2.6.5) that
ccrr
nn
11
=====
LL
so that A is a symmetric BIBD with r
λ =+−
( )/
ncd
1
2.
This proves the theorem for n, m ≥3; it is trivial to deal with the remaining cases.
2.2.7. Lemma. The determinant of the m
npp
pn
p
ppn
L


Proof. Straightforward.
Using similar but more complicated arguments one can show that if A is an m
matrix such that each two distinct rows have exactly µ ones in common and each two distinct
columns have exactly λ ones in common, then A is a symmetric BIBD. Interpreting the column
indices of A as points and the row indices of A as lines, this gives, [8]:
2.2.8. Theorem. Let X be a finite set (of points), with a system of subsets called lines. Let
there be n points and m lines. Suppose that lines distinguish points (i.e. no two distinct points
have the same set of lines through them) and points distinguish lines, and that:
(i) Each two distinct lines meet in µ points
(ii) Through each pair of distinct points there pass λ lines.
Then nm
=
and λµ=
, each line has r points and through each point there pass r lines
(where r rn
()()
−=−
11
λ
).
This is a special case of a more general result of Röhmel, [16]; see also [3], p. 102ff.

BB Bnq B
)
q
r
rr
T
nn







(
1
M
1
M
L
(2.2.6.5)
)
q
= , because B is invertible.
p B
−
d
is equal to a matrix of rank
q
= and hence e
nc
=+
()/
1
2 entries 1 in each column and row and
m
×
matrix in (2.2.6.1) is equal to

det()(( ) )
p
1
npnm
m
L
OM
MO O








=−+−
−1
n
× zero-one
2.3. More Examples.
Using the various symmetric BIBD’s as main building blocks, a variety of examples of
tree-tree matrices can be constructed. Here is a small selection. In the pictures below (and above)
the black nodes in a tree make up the tree like space that is being realized.