unicity in discrete tomography - lorentz center tijdeman.pdf · 1. introduction discrete tomography...

Unicity in discrete tomography

Rob Tijdeman, Leiden, 12 February 2015 2013

1. Introduction Discrete tomography if

few materials which have lattice structure few grey values (few pixel values) Then 5 – 40 pictures may suffice

Advantages:

less damage of material smaller time range

For unicity: phenomena are clearer

Mathematical model

Simple model (binary case)

Horizontal line sums

Vertical line sums

Diagonal sums

Anti-diagonal sums

Questions 1.  Is the solution unique?

If not: 2. What can be said about other solutions? 3. Can we guess what was the original

solution?

Equal line sums

0 1 0 0

0 0 0 1

1 0 0 0

0 0 1 0

0 0 1 0

1 0 0 0

0 0 0 1

0 1 0 0

Subtract:

‘Switching component’ 0 1 -1 0

-1 0 0 1

1 0 0 -1

0 -1 1 0

Line sums are zero in the four directions. Any two tables with this difference have equal line sums.

Remarks

1.  Any two solutions have the same line sums, iff the difference has zero line sums.

2.  In general there are many more unknowns than line sums (linear equations). Therefore, in principle, there are many solutions.

3.  However, since we have only few grey values, the possibilities are restricted.

Unique solution

6

6

5

4

3

2

1

6 5 4 3 2 1

2. Algebraic structure Joint work with Lajos Hajdu (Debrecen, HU) J. reine angew. Math. 534 (2001), 119-128. Working over 0,1 is very difficult, but

working over the integers is not so hard. In this section we work with integer pixel

values.

Generating polynomial

3x2 - 2y + 7xy + 4y2 + 5xy2 + 6x2y2 – y3 + xy3

-1 1 0 0 4 5 6 0 -2 7 0 0 0 0 3 0

Directions

If vector between two consecutive pixels on a line is (a,b), then we write xayb-1 if b > 0, xa-1 if b=0, xa-y|b| if b < 0. (We assume a≥0.)

Thus: Row sum has (1,0) and therefore x-1, column sum has (0,1) and therefore y-1, diagonal sum has (1,1) and therefore xy-1, anti-diagonal sum has (1,-1)

and therefore x-y. Combination has (x-1)(y-1)(xy-1)((x-y).

Minimal switching component

(x-1)(y-1)(x-y)(xy-1) =

x3y2 - x3y + x2 - x2y3 + xy3 – x + y - y2.

Corresponding rectangle: Theory:This is the smallest configuration.

0 -1 1 0

1 0 0 -1

-1 0 0 1

0 1 -1 0

Properties

Size of minimal switching configuration is degree of x (sum over all ai) by degree of y (sum over all |bi|). Two tables with equal line sums cannot be very close to each other. All switching polynomials have a corresponding polynomial which is divisible by the minimal polynomial.

Example of switching component

-1 1

1 -1

-1 1-1 3 -3 1

1 -3 -1+1 -1 3

-1 3 1 -3

1 -1 -3 3

Directions (1,0), (0,1), (1,1), (1,-1).

Another example Consider directions (1,0), (2,1), (1,-3). Corresponding polynomial: (x-1)(x2y-1)(x-y3) = x4y -x3y –x2 +x -x3y4 +x2y4 +xy3 –y3. Minimal switching polynomial:

0 0 1 -1 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 1 -1 0 1 -1 0 0

3. Geometric structure Suppose m by n block with integers. Then representation by vector of length mn

and entries in fixed order. If L line sums are given, then L linear

conditions on these entries. Hence solutions are on a linear manifold.

Line sums will be linear dependent, e.g.

sum of column sums = sum of row sums.

Solutions are on a hypersphere

We know the sums of the entries of a solution, as it is the sum of all the line sums in one direction, e.g. the sums of all row sums.

Suppose we have only entries 0 and 1. (‘binary’) Since 02=0 and 12=1, we then also know the sum

of the squares of the entries, and therefore the length R of the solution vector.

Hence all 0-1 solutions are on a hypersphere with radius R around the origin.

Theorem of Pythagoras Suppose 0-1 solutions. We know: -  all solutions are in some linear manifold - all solutions are on a hypersphere of radius R

around the origin. We can compute the orthogonal projection x* from

the origin on the linear manifold. By Pythagoras we can then compute the distance

from x* to the solution vector.

In a picture

-  Sol

All binary solutions are on the intersection of the sphere around the origin and on the linear manifold. Thus they are on the ‘circle’. We know the length of x and the length of x*, and therefore the length of x-x*.

Distances between solutions

≥: Difference between two solutions is multiple of minimal switching component.

≤: Distance between two solutions is at

most 2 √(|x|2 - |x*|2) where |x| and |x*| can be computed independent of solutions.

|x| is fixed. After adding a new direction the minimal switching component is larger

and upper bound probably smaller.

Conclusions

Instead of asking for binary solutions, one can more generally ask for the integer solutions with smallest vector length.

The Pythagorean argument gives an upper

bound on the distance between two tables with equal line sums.

The greater the distance between the origin

and its orthogonal projection, the better the approximation to the solutions.

Literature See: www.math.unideb.hu/~hajdul/ www.math.leidenuniv.nl/~tijdeman/ up to 2010, on arXiv from 2010 on.