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for columns two through five, but with the better estimate of hubbiness given by the fifth
column.
The limit of this process may not be obvious, but it can be computed by a simple pro-
gram. The limits are:
This result makes sense. First, we notice that the hubbiness of
E
is surely 0, since it leads
nowhere. The hubbiness of
C
depends only on the authority of
E
and vice versa, so it should
not surprise us that both are 0.
A
is the greatest hub, since it links to the three biggest au-
thorities,
B
,
C
, and
D
. Also,
B
and
C
are the greatest authorities, since they are linked to by
the two biggest hubs,
A
and
D
.
For Web-sized graphs, the only way of computing the solution to the hubsand-authorities
equations is iteratively. However, for this tiny example, we can compute the solution by
solving equations. We shall use the equations
h
=
λµLL
T
h
. First,
LL
T
is
Let
ν
= 1/(λ
µ
) and let the components of
h
for nodes
A
through
E
be
a
through
e
, respect-
ively. Then the equations for
h
can be written
νa
= 3
a
+
b
+ 2
d
νb
=
a
+ 2
b
νc
=
c
νd
= 2
a
+ 2
d
νe
= 0
The equation for
b
tells us
b
=
a
/(
ν
− 2) and the equation for
d
tells us
d
= 2
a
/(
ν
− 2). If we
substitute these expressions for
b
and
d
in the equation for
a
, we get
νa
=
a
(3 + 5/(
ν
− 2)).
From this equation, since
a
is a factor of both sides, we are left with a quadratic equation
for
ν
which simplifies to
ν
2
− 5
ν
+ 1 = 0. The positive root is Now that we know
ν
is neither 0 or 1, the equations for
c
and
e
tell us immediately that
c
=
e
= 0.
Finally, if we recognize that
a
is the largest component of
h
and set
a
= 1, we get
b
=
0
.
3583 and
d
= 0
.
7165. Along with
c
=
e
= 0, these values give us the limiting value of
h
.
The value of
a
can be computed from
h
by multiplying by
L
T
and scaling.
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