From Greek Mythology to Modern Manufacturing: The Procrustes

Problem

ByDr. Dan Curtis

Department of MathematicsCentral Washington University

Procrustes offers Theseus a bed for the night

Theseus gives Procrustes a dose of his own medicine.

q

x1

y1

x2

y2

p

X Y

The Alignment Problem

• We know the X-coordinates of the features and of p.

• We know the Y-coordinates of the features , but not the

Y-coordinates of q.

• When the part is assembled, these points will coincide in space, so the and give the coordinates of the same point in two different coordinate systems.

• What will be the Y-coordinates of q?

ix

iy

ix

iy

1x2x

3x

p

1y

3y

2y

X

Y

Map Registration Problem

• Coordinates of features known in X-coordinate system. Also, X-coordinates of feature p are known.

• Y-coordinates of same features, are known.

• What would the Y-coordinates of feature p be?

ix

iy

Common Thread:

1. Have two cartesian coordinate systems in space, X and Y.

2. Have points whose coordinates are known in both coordinate systems.

Common Thread:

1. Have two cartesian coordinate systems in space, X and Y.

2. Have points whose coordinates are known in both coordinate systems.

Find the transformation which maps the X-coordinates of a point to the Y-coordinates of the same point.

x y Qx + t

rotation matrix translation vector

The Orthogonal Procrustes Problem

Given: points and in space, i = 1, …, n

Find: optimal rotation Q and translation vector t

x = Qx + t

does the best possible job of mapping the points

to for i = 1,...,ni i

x y

“Best possible” means choose Q and t to minimize the following expression:

ix iy

1

2

i ii

n

Qx + t - y

1

2

i ii

n

Qx + t - y

The above expression can be written as:

n

1

T( ) ( ) i i i ii

Qx + t - y Qx + t - y

1

2

i ii

n

Qx + t - y

The above expression can be written as:

n

1

T( ) ( ) i i i ii

Qx + t - y Qx + t - y

or, multiplying it out, as

nT T T T T T

i i i i

T T

i i i ii 1

( 2 2 2 )

x x Q t x Q y t yt tx yy

We must minimize

nT T

i

T T T

1i i

T

ii

( 22 2 )

x Q t tyx Q t y t

We must minimize

nT T

i

T T T

1i i

T

ii

( 22 2 )

x Q t tyx Q t y t

So t must be chosen to minimize

nT T T T

i ii 1

(2 2 )

x Q t t t y t

,

We must minimize

nT T

i

T T T

1i i

T

ii

( 22 2 )

x Q t tyx Q t y t

So t must be chosen to minimize

nT T T T

i ii 1

(2 2 )

x Q t t t y t

or, equivalently,

nT T T T

i ii 1

(2 2 )

x Q t y t

Introduce centers of gravity

n n

i ii=1 i=1

1 1,

n n x x y y

Now minimize

T (2 2 ) t Qx t y

Introduce centers of gravity

n n

i ii=1 i=1

1 1,

n n x x y y

Now minimize

T (2 2 ) t Qx t y

This has the formT ( )t t b

where

2 2 b Qx y

We have the identity:2

2T 1 1( )

2 4 t t + b t + b b

Minimum is obtained when

1

2t b

We have the identity:2

2T 1 1( )

2 4 t t + b t + b b

Minimum is obtained when

1

2t b

Thus, take

t y Qxor

y Qx t

Original expression to be minimized was:

1

2

i ii

n

Qx + t - y

Original expression to be minimized was:

1

2

i ii

n

Qx + t - y

This now becomes:

i i1

2

i

n

Qx y

where

i i i i, . x x x y y y

This expression expands to

nT

n nT T

i i i ii 1 i

T

i ii 1 1

2

x Q yx x y y

This expression expands to

nT

n nT T

i i i ii 1 i

T

i ii 1 1

2

x Q yx x y y

Choose Q to maximize the expression

nT T

i ii 1 x Q y

This expression expands to

nT

n nT T

i i i ii 1 i

T

i ii 1 1

2

x Q yx x y y

Choose Q to maximize the expression

nT T

i ii 1 x Q y

Define the matrix A by

nT

i ii 1

A x y

For any two column vectors u and v, we have

T Ttr( )u v uv

For any two column vectors u and v, we have

T Ttr( )u v uv

n nT T T

i i i ii 1

nT T

i1

ii 1 i

= ( tr) )) (tr(

x Q y x y Qx Q y AQ

So,

For any two column vectors u and v, we have

T Ttr( )u v uv

New problem: Given a matrix A, find a

rotation matrix Q which maximizes tr(AQ).

n nT T T

i i i ii 1

nT T

i1

ii 1 i

= ( tr) )) (tr(

x Q y x y Qx Q y AQ

So,

The Singular Value Decomposition

1

2

n

0 0 0

0 0 0

0

TA U V

1 2 n 0

U and V are orthogonal matrices

(singular values)

Theorem 1: If A is an matrix and is the sum of the singular values of A, then

n×n AS

Atr(A) S

with equality if and only if A is symmetric and positivesemi-definite.

Theorem 1: If A is an matrix and is the sum of the singular values of A, then

n×n AS

Atr(A) S

with equality if and only if A is symmetric and positivesemi-definite.

Theorem 2: If A is an matrix, then there is an orthogonal matrix Q such that AQ is symmetric and positive semi-definite.If Y is any other orthogonal matrix, then

n×n

tr(AY) tr(AQ)with equality if and only if AY is symmetric and positivesemi-definite.

To find Q maximizing tr(AQ):

To find Q maximizing tr(AQ):

• Obtain SVD

TA = UΣV

To find Q maximizing tr(AQ):

• Obtain SVD

• Take

TA = UΣV

TQ = VU

To find Q maximizing tr(AQ):

• Obtain SVD

• Take

TA = UΣV

TQ = VU

Then: T T T( )( ) , AQ UΣV VU UΣU

which is symmetric and positive semi-definite.

Summary of Solution Steps

1. Find centers of gravity and .2. Form displacements

nT

i ii 1

A x y

yx

i i i i, . x x x y y y

3. Form the matrix 4. Obtain SVD

TA = UΣV

5. Take TQ = VU

6. Take t y Qx