Axial Tomography by
filtered least squares
Kenneth C. Holmes
Max Planck Institute
for Medical Research, 69120 Heidelberg, Germany
e-mail – holmes@mpimf-heidelberg.mpg.de
and
Rasmus
R. Schroeder
CellNetworks, BioQuant, University of Heidelberg
Im Neuenheimer Feld 267, 69120 Heidelberg, Germany
e-mail
– rasmus.schroeder@bioquant.uni-heidelberg.de
Introduction
We address the problem of reconstructing
the 2-dimensional density of a plane object from a number of projections of the
density in the plane of the object. The one-dimensional projections are
generated by rotating the object around an axis at right angles to the plane
and projecting down a constant direction in the plane of the object. Thus the
direction of projection is defined to be at right angles to the rotation axis.
i.e. the geometry of the single axis projection. Klug and DeRosier [1] first
solved this problem by showing how the two-dimensional Fourier transform of the
required density could be built up from transforms of the projected density.
This is the basis of the method of reciprocal-space analysis and Fourier
synthesis of the density. Another frequently used method makes use of back
projection, where each projection is used to generate a set of lines along the
direction of projection, each line weighted by the projected density. The sum
over all such lines for each point in the object gives an estimate of the
reconstructed density. Back projection yields an image in which the high
frequency components are attenuated. A way to compensate for this is to compute
the Fourier transform of the back projection (with its origin on the axis of
rotation) and to sharpen the Fourier transform of the density computed by back
projection by a weighting function that is proportional to the distance from
the origin in Fourier space (Gilbert [2]). Thus the center of the transformed
back projection receives zero weight, and at some resolution there is a sharp
cut off.
Crowther DeRosier and Klug [3] examined
the possibility of using a real space method of reconstructing the required
density based on filtered least squares. One defines the sought-after density
on a grid and various projections of this grid are available as observations.
This problem can then be solved by least squares. Moreover, an algebraic
formulation allows an analysis of the information content of the available
projections. Typically there will be too few projection equations to determine
the problem at the resolution of the grid. The rank of the normal matrix will
be less than the order and therefore the normal matrix has no unique inverse.
The problem actually goes deeper: the
rotation axis introduces a singularity into the matrix that cannot be removed
by adding more views. This singularity has bedeviled all algebraic formulations
of the axial tomography problem. However, as was suggested by Crowther DeRosier
and Klug (loc. cit.) one
can get round these problems by diagonalizing the normal matrix and by
inverting the normal matrix in the subspace of significant eigenvalues. This
method of inverting ill-defined matrices was described by Diamond [4]. The
truncation of the inversion limits the resolution in the final density.
However, this filtered least squares was never used as a method of
reconstruction because at the time the computing load would have been
prohibitive. However, computer power in no longer limiting and the method is
now applicable. In the following we show how the density of the object may be
recovered from the projected density using the method of least squares. The
method extracts the maximum possible amount of information about the reconstructed
density in an unbiased way. Moreover, the crude shape of the object can be used
to define the shape of the domain in which the solution is sought. We refer to
this as a mask. The use of a mask (which is rather like solvent flattening on
crystallography) yields more accurate reconstructions. It is particularly
valuable for asymmetric objects. Furthermore, it can significantly reduce the
order and rank of the normal matrix, which is computationally very
advantageous.
Finally, it turns out that the right hand
side of the normal equations is the
back projection. Thus the theory demonstrates that an analytical procedure
exists for extracting the three dimensional density from the back projection.
Indeed, the correct filter
for turning the back projection into the reconstructed object is in fact the
inverted normal matrix!
The method we have developed has close
parallels with the weighted back projection. However, in place of the
sine-cosine transform we operate on the back projection with an orthogonal
matrix of eigenvectors derived from the normal matrix. The nature of the
eigenvectors (and the derived eigenfunctions) depends on the shape of the mask
and the number and spacing of the projections. For a round mask they look like
cylinder functions. This operation is analogous to a Fourier transformation. In
the transformed space the coefficients are divided by the appropriate calculated eigenvalue:
the eigenvalues are large for the low orders and fall off monotonically. Thus
the transformed density is subjected to a sharpening analogous to r weighting
(Gilbert [2]). Somewhere, because the eigenvalues get very small, we have to
truncate. The point depends on the noise in the data. Thus we see that noise
limits resolution. Finally, we operate on the sharpened transform with the
inverse orthogonal matrix of eigenvectors, which is analogous to a back Fourier
transform. Compared with the weighted back projection method the least-squares
method uses custom-made orthogonal functions for computing the transform of the
back projection. These functions are determined by and are characteristic of
the geometry of the problem. Moreover, the weighting function does not go to
zero at the origin of the transform and is also appropriate for the geometry.
Thus the method deals quantitatively with the relationship between low and high
order terms. The method yields the best possible solution for the given
geometry without any ad hoc assumptions
and moreover, provides a theoretical basis for analyzing the weighted back
projection method.
Setting up the problem
We
represent the unknown density in the selected plane as values on a lattice
within some chosen radius (Fig 1a). The use of a circular boundary is
arbitrary. However, round is the best choice in the absence of prior
information about the object. The lattice points are tagged with serial numbers
so that they can be written down as a long list of numbers (a vector x). For example, in the first observation
(rotation zero) the values at the points 2-8 add up to the projected density B0. The points 9-19 sum to the projected
density C0 etc. We can write the projector for each
point A0, B0 etc.
as a vector h, which
spans all points. In the present case by inspection the projector for A0
has components
h1=1, h2=0,
h3=0, h4=0
etc.
and for B0
h1
= 0 h5
= 1 h9 = 0
h2 = 1 h6
= 1 h10
= 0
h3 = 1 h7
= 1 h11
= 0
h4
= 1 h8
= 1 h12
= 0 etc.
Similar results are obtained by inspection for each of the
other points C,D,E,F etc.

Fig. 1
In
the second measurement (plane rotated for example by 360/13) the situation is
not quite so straightforward (fig 1b). Again the integration is carried out by
sampling the density at equal intervals along the projecting lines but the
integration points (shown as small circles) no longer coincide with the lattice
points. We can no longer write down the projector by inspection.
An interpolation scheme
is needed. We have used Gaussian interpolation. Each lattice point in the plane
is convoluted with a gauss function of suitable width (Fig 2).

Fig. 2
As before, integration is carried out at
equal distances along projector lines but at each integration point the
contributions of all nearby lattice points are taken into account (weighted
according to the value of the appropriate Gaussian at the integration point).
Thus a number of lattice points will contribute to each integration point. By
extending this method to all projected points for each rotational position we
finally obtain observational equations relating all the unknown density points
with the known projections, which we can write as
H x = b
(1)
where
b is the list of all
observables i.e. all the projected densities A0,B0, ... A1,B1... A2,B2....A12,
B12 in order and x is a vector of the unknown weights of the
Gaussians sitting on each of the defined lattice points which we wish to
determine. H is a projector matrix derived by assembling all the available
vectors h arranged in
rows. Each row is as long as the list of unknown density points and consists
mainly of zero's. The number of rows is equal to the number of observations, so
that H is very wide and not very high! We wish to determine x.
The
Solution
The
classical solution to the equations
H x= b+
ε (1a)
which minimizes the
sum of squares of errors eTe
(least squares solution) is
HTH x = HTb (2)
HTH is the normal or Hessian matrix, which
from its definition is square symmetric. If HTH is non-singular it can be inverted to
give the solution
x
= (HTH)-1 HT b (3)
However, in our case b
is shorter than x (we do not have enough observations to
determine the density x).
Because of the form of H, the vector HTb is longer than b.
Normally, in least squares there is a reduction of dimensionality on
pre-multiplying by H. However, in the present case since the number of unknown
exceeds the number of observations the dimensionality of the normal matrix is
actually increased. Equation 2 is
ill determined and cannot be solved by normal methods.
We
proceed by expressing HTH
in diagonal form. Since HTH
is square and positive semi-definite one can find a transformation
HTH = VTΛV (4)
where
Λ is a diagonal matrix of the eigenvalues λi. (the eigenvalues of HTH). V is a matrix of which the columns
are the orthonormal eigenvectors of HTH. Premutiplication by V rotates into a
representation where the eigenvectors become the principle axes of the problem.
Substituting (4) in (3) we have
VTΛVx = HTb (5)
or
ΛVx= VHTb (6)
We are now in a representation where the
principle axes of the Hessian or normal matrix are the base and all equations
have been separated. Now we may solve (6) element by element by dividing by λi
(Vx)i = 1/λi
(V HTb)i (7)
We can simply stop solving for elements
of Vx when λ gets
small compared with the errors in the observations. Hence we determine some
(but not all) values of Vx. The
rest we set to zero. We call this truncated vector (Vx)
Rotating (transforming) back gives
x
= VT
(Vx) (8)
that
is the required solution.
We actually start
out with the back projection
HTb, the right hand side of equation 2, is a vector derived from
the observations that is central feature of the least squares method.
Noteworthy, in the present context, HTb assumes a special significance; it is in fact the back
projection! This comes about because H is a projector matrix.
The Eigenfunctions
Each of the eigenvectors of HTH (one of the the columns of V) may be
mapped onto the lattice since each component of the vector is associated with a
lattice point. When we do this we generate two-dimensional functions in the
circular space we have chosen for the solution. These functions are (by
definition) orthonormal. Because we have chosen a circular boundary, they look
rather like cylinder functions. In fact they have a symmetry characteristic of
the geometry of the problem. Some examples from a case considered below are
shown in Fig.3

Fig 3
A transformation
and back transformation yields the answer
The significance of
equations 6, 7 and is as follows: equation 6 is a transformation of the back
projection using a set of orthonormal functions that are the eigenfunctions of
the outer product of the projector matrix with itself. The result of this
tranformation is a list of numbers (with associated errors). To proceed we divide
each of these numbers by the appropriate eigenfunction arranged from large to
small in order to determine how much of each eigenfunction contribute to the
final answer (7). At some point the eigenfunction becomes as small as the
errors and we have to stop, otherwise we amplify up noise – there is no point in going on
– a truncation. This yields a list of numbers telling us how much of each
eigenfunction to put into the answer. The result is calculated by a back
transformation (8). The effect of truncation is to limit the resolution in the
answer. It transpires that the eigenfunctions with the longest spatial
frequencies have the largest eigenvalues so that the resolution goes up with
more eigenvalues. Since the base we have used is orthonormal and final we would
not get another answer if we considered a larger subspace – just more noise.
A Donkey

Fig 4
To test the theory, we
took a donkey as a trial object. This object is demanding in a number of
respects. It has no symmetry, It has sharp graduations of density. It has long
legs! For first trial the projected density was calculated every 3 between 0 and 180 (the symmetry of the
projection precludes the addition of any new information in the second half
circle). Projections at 0 and 30 are shown in Fig 4. The 61 projections were
used to reconstruct the donkey using the theory given above. The diameter of
the donkey was about 150 pixels, so we set up a circle of radius 75 pixels.
This produced 17645 lattice points (.752
= 17671). Thus the order of the normal matrix is 17645.
Calculation shows that the effective rank is 5000-6000 (the cut off is not
sharp). Using a Fortran program running on a Silicon Graphics Origin 3000 and
four processors it took about 1/2h. to set up the normal matrix and another 3
h. to extract the 4000 most significant eigenvectors using the Lapack routine
SSYEVX. It is noteworthy that this considerable computing load has only to be
undertaken for a new geometry. If the geometry remains unaltered (same number
of projections in the same angular range) then a reconstruction can be
calculated from stored matrices in a fraction of a second. In Fig 5 we show the
results of this reconstruction. Shown are; the original object; the raw back
projection, the r-weighted back-projection, the reconstructed image using 2000
largest eigenfunctions; the reconstructed image using 4000 eigenfunctions.

Fig 5
The
use of a mask
We
are not obliged to use a round area to define the boundaries of the lattice. If
we have some knowledge of the shape of the object to be reconstructed this can
be used as a mask to define the area within which we seek the density of the
object. The use of a mask will reduce the size of the normal matrix, which has
two desirable effects. The calculation of the diagonalized matrix is much
faster; and the resolution obtainable from the given data is improved since we
are seeking to solve for a smaller number of variables. The use of a mask,
which fits the object, is rather like the use of solvent flattening in protein
crystallography to improve the resolution of electron density maps. It the
following example the projections have been made every 6, giving 30
independent projections. Fig.6 shows the back projection and r-weighted back
projection. Fig 7, shows the effects of the mask.
This method has
been successfully used to reconstruct actin decorated with myosin cross-bridges
at 13 resolution [5].

Fig 6 Back projection and r-weighted back
projection for 30 projections at 6 intervals
Fig 7. (see next page)
Projections at 6 intervals. The effects on resolution of a mask compared with
a circle for a given number of eigenfunction. The eigenvalues fall off more
quickly for the mask than for the circle. Thus the circle eigenvalues are down
to 1% of the origin value at 1300 eigenvalues and 0.1% at 2300 eigenvalues. The
corresponding numbers for the mask are 1000 and 1800.

Fig. 7
References
2. Gilbert,
P. Ph.D. thesis 1972 University of Cambridge
5. Holmes, K. C., Angert, I., Kull, F.
J., Jahn, W. and Schroder, R. R. 2003. Electron cryo-microscopy shows how
strong binding of myosin to actin releases nucleotide. Nature 425: p. 423-7.