How Mathematics Gives Us Superman X-ray Vision

Helen Moore, PhDBristol-Myers Squibb

January 28, 2017

Regular X-rays

• Picture of Anna Bertha Ludwig’s hand, taken by Wilhelm Röntgen, using x-rays, 1896

• Röntgen received the first Nobel Prize in Physics, 1901, for his discovery of x-rays

• We see shadows! Darkness shows density; cannot see inside the bones or tissue

Public domain image; https://commons.wikimedia.org/wiki/File:Wilhelm-Roentgen's-X-ray-photograph-of-his-wife's-hand.png 2

Superman X-ray Vision

Fair use File:Superboy98.jpgUploaded: 21 March 2008https://en.wikipedia.org/wiki/File:Superboy98.jpg#/media/File:Superboy98.jpg

Wired Magazine Online, March 22, 2016https://www.wired.com/2016/03/sure-superman-x-ray-vision-actually-work/

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Computerized Tomography (CT) Scan/Scanner

By U.S. Navy photo by Chief Warrant Officer 4 Seth Rossman. - This Image was released by the United States Navy with the ID 030819-N-9593R-125. Public Domain, https://commons.wikimedia.org/w/index.php?curid=8196091

By Yale Rosen from USA - Adenocarcinoma - CT scan, CC BY-SA 2.0, https://commons.wikimedia.org/w/index.php?curid=31127656

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How CT scans work

Distribution of log of attenuation, assuming density is uniform (so log of attenuation is proportional to width of object being scanned)

X-ray source

X-rays

Circular object being scanned

Projection screen

𝑦𝑦 = 2 1 − 𝑥𝑥2

Rotated

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Potential problem

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Solution: more angles

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Game of Nonograms

• Published in the Sunday Telegraph (UK) since 1990• Invented by Non Ishida (independently by Tetsuya Nishio) • Fill in the number of contiguous blocks according to the numbers

listed for each column and row2 2

9 9 2 2 4 44 . .6

2 2 . .2 2 . .

64 . .2 . . . .2 . . . .2 . . . . 8

Nonogram Handout

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Nonogram numbers can represent density

• Attenuation of photons is measured by the x-ray detector• The logarithm of attenuation is proportional to object density• If constant density, then log attenuation measures length of object• What if the density if not constant?

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Nonograms with any non-negative integer

• These represent different densities

A B 3

C D 1

2 2

A B 1

C D 1

1 1

A B 2

C D 0

1 1

A B 5

C D 6

8 3

2 1

0 1

1 1

0 0

1 2

1 0

1 0

0 1

0 1

1 0

? ?

? ?

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Solution: more angles

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Nonograms with any non-negative integer

• Three angles

A B 3

C D 1

2 2

A B 1

C D 1

1 1

A B 5

C D 6

8 3

2 1

0 1

0 1

1 0

4 1

4 2

10

1

13

0

16

4

13

12

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What if the grids are bigger?

• We need to use all possible angles for non-constant densities

A B C 3

D E F 3

G H I 3

3 3 3

2 0 1

0 1 2

1 2 0

0 2 1

2 1 0

1 0 2

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Handling different densities

• Approximate varying density using constants over small intervals• Total log attenuation is proportional to adding the density over each

interval (the Riemann sum)

Riemann sum = �𝑘𝑘=1

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𝑓𝑓 𝑥𝑥𝑘𝑘 � Δ𝑥𝑥

lim𝑛𝑛→∞

�𝑘𝑘=1

𝑛𝑛

𝑓𝑓 𝑥𝑥𝑛𝑛 � ∆𝑥𝑥 = �𝑎𝑎

𝑏𝑏𝑓𝑓 𝑥𝑥 𝑑𝑑𝑥𝑥

x3 x7x5 x9=ba=x1

𝑦𝑦 = 𝑓𝑓 𝑥𝑥 = density

𝑦𝑦

𝑥𝑥

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If we know the density line integrals for all lines through a slice, can we reconstruct the shapes in the slice?

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“All lines”• 𝑦𝑦 = 𝑚𝑚𝑥𝑥 + 𝑏𝑏 gives all lines in the plane except vertical lines• For a given line ℓ, take a vector 𝑛𝑛 perpendicular to it

• There exists an angle 𝜃𝜃 such that 𝑛𝑛 is parallel to the unit vector coming from the origin at angle 𝜃𝜃 from the 𝑥𝑥-axis

• The line coming from the origin at angle 𝜃𝜃 is perpendicular to l and intersects it at a point 𝓅𝓅 = (𝑡𝑡 cos 𝜃𝜃 , 𝑡𝑡 sin𝜃𝜃)

• Every line ℓ𝑡𝑡,𝜃𝜃 in the plane can be characterized by a 𝑡𝑡 and 𝜃𝜃: the line through 𝓅𝓅 = (𝑡𝑡 cos 𝜃𝜃 , 𝑡𝑡 sin𝜃𝜃)perpendicular to the unit vector (cos 𝜃𝜃 , sin𝜃𝜃)

• For uniqueness, let 0 ≤ 𝜃𝜃 < 𝜋𝜋

𝑦𝑦

𝑥𝑥ℓ

𝑛𝑛

𝜃𝜃

𝓅𝓅 = (𝑡𝑡 cos𝜃𝜃, 𝑡𝑡 sin𝜃𝜃)

(cos𝜃𝜃, sin𝜃𝜃)

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Line parametrization

• Each line ℓ𝑡𝑡,𝜃𝜃 can be described as the line through 𝓅𝓅=(𝑡𝑡 cos𝜃𝜃, 𝑡𝑡 sin𝜃𝜃) perpendicular to the unit vector (cos𝜃𝜃, sin𝜃𝜃)

• The vector (−sin𝜃𝜃 , cos𝜃𝜃) is perpendicular to (cos𝜃𝜃, sin𝜃𝜃) since the dot product of the two vectors is zero

• So ℓ𝑡𝑡,𝜃𝜃 can be parametrized as 𝑡𝑡 cos𝜃𝜃 , 𝑡𝑡 sin𝜃𝜃 + 𝑠𝑠(− sin𝜃𝜃 , cos𝜃𝜃)

𝑦𝑦

𝑥𝑥ℓ𝑡𝑡,𝜃𝜃

𝜃𝜃

𝓅𝓅 = (𝑡𝑡 cos𝜃𝜃, 𝑡𝑡 sin𝜃𝜃)

(cos𝜃𝜃, sin𝜃𝜃)

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Radon transform and back transform

• Define the Radon transform (1917) of a function 𝑓𝑓 as• ℛ𝑓𝑓 𝑡𝑡,𝜃𝜃 = ∫−∞

∞ 𝑓𝑓(𝑡𝑡 cos 𝜃𝜃 − 𝑠𝑠 sin𝜃𝜃, 𝑡𝑡 sin𝜃𝜃 + 𝑠𝑠 cos 𝜃𝜃)𝑑𝑑𝑠𝑠• If 𝑓𝑓 is the density function, ℛ𝑓𝑓 gives the line integral for the density function

• Define the back transform of a function ℎ as• ℬℎ 𝑥𝑥,𝑦𝑦 = 1

𝜋𝜋 ∫0𝜋𝜋 ℎ(𝑥𝑥 cos 𝜃𝜃 + 𝑦𝑦 sin𝜃𝜃,𝜃𝜃)𝑑𝑑𝜃𝜃

• First, note that 𝑥𝑥,𝑦𝑦 ⋅ cos 𝜃𝜃 , sin𝜃𝜃 = 𝑥𝑥 cos 𝜃𝜃 + 𝑦𝑦 sin𝜃𝜃• So 𝑥𝑥 cos 𝜃𝜃 + 𝑦𝑦 sin𝜃𝜃 is the length of the projection of (𝑥𝑥,𝑦𝑦) on the line

determined by 𝜃𝜃• If ℎ = ℛ𝑓𝑓, then ℬℎ gives the average density function values of all lines that

pass through (𝑥𝑥,𝑦𝑦)

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Does back transform give the reconstruction?

• The back transform calculates the average density at (𝑥𝑥,𝑦𝑦)• This gives a smoothed-out version of the original density at every

point• How do we know it is smoothed out? • Because lines with constant density would give same average

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Fourier Transform and Inverse

• Fourier transform: ℱ𝑓𝑓 𝜔𝜔 = ∫−∞∞ 𝑓𝑓(𝑥𝑥) � 𝑒𝑒−𝑖𝑖𝜔𝜔𝑥𝑥𝑑𝑑𝑥𝑥

• This captures the part of 𝑓𝑓 that is at frequency 𝜔𝜔2𝜋𝜋

• Inverse Fourier transform: ℱ−1𝑔𝑔 𝑥𝑥 = ∫−∞∞ 𝑔𝑔(𝜔𝜔) � 𝑒𝑒𝑖𝑖𝜔𝜔𝑥𝑥𝑑𝑑𝜔𝜔

• CT scan image is given by :

𝑓𝑓 𝑥𝑥,𝑦𝑦 =12ℬ ℱ−1 𝑆𝑆 ℱ ℛ𝑓𝑓 𝑆𝑆,𝜃𝜃 (𝑥𝑥,𝑦𝑦)

• Inserting the Fourier transform allows us to “pull out” the density at each location (corresponding to a frequency) on a given line; applying the inverse later allows us to put it back

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Notes

• We assumed rays leave the source parallel, but they actually spread like a fan – this does not change the results, since rotating the source still gives us all lines through the object of interest

• We assumed the rays continue straight into the object, or reflect back; but they might diffract and pass through at an angle

• In practice, we only measure a finite number of directions, and not each angle

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References

• The Mathematics of Medical Imaging: A Beginner’s Guide, Timothy G. Feeman, Springer, New York, NY, 2010.

• Introduction to the Mathematics of Medical Imaging, Second Edition, Charles L. Epstein, SIAM, Philadelphia, PA, 2008.

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