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Algebraic classification related to contrast optimization for MRI Bernard Bonnard 1 Jean-Charles Faugre 2 Alain Jacquemard 2 Mohab Safey El Din 2 Thibaut Verron 2 1 Institut de Mathmatiques de Bourgogne, Dijon, France UMR CNRS 5584 2


  1. Algebraic classification related to contrast optimization for MRI Bernard Bonnard 1 Jean-Charles Faugère 2 Alain Jacquemard 2 Mohab Safey El Din 2 Thibaut Verron 2 1 Institut de Mathématiques de Bourgogne, Dijon, France UMR CNRS 5584 2 Université Pierre et Marie Curie, Paris 6, France INRIA Paris-Rocquencourt, Équipe P OL S YS Laboratoire d’Informatique de Paris 6, UMR CNRS 7606 Journées Nationales de Calcul Formel, 05 novembre 2015

  2. Physical problem (N)MRI = (Nuclear) Magnetic Resonance Imagery 1. apply a magnetic field to a body 2. measure the radio waves emitted in reaction Optimize the contrast = be able to distinguish two biological matters from this measure Bio. matter 1 Bio. matter 2 Bad contrast Good contrast Known methods: ◮ inject contrast agents to the patient: potentially toxic ◮ make the field variable to exploit differences in relaxation times ⇒ requires finding optimal settings = (Images: Pr. Steffen Glaser, Tech. Univ. München)

  3. Numerical approach The Bloch equations Saturation method Find a path u x so that after some time T : � ˙ y i = − Γ i y i − u x z i ◮ matter 1 saturated: y 1 ( T ) = z 1 ( T ) = 0 = − γ i ( 1 − z i ) + u x y i ˙ z i ◮ matter 2 “maximized”: | ( y 2 ( T ) , z 2 ( T )) | maximal ( i = 1 , 2) Glaser’s team, 2012 : Control Theory method ◮ Numerical method to find a path towards a saturated system = solution u x ◮ Already used in some specific cases for the MRI, here applied in full generality

  4. Numerical approach... and computational problem The Bloch equations Saturation method Find a path u x so that after some time T : � ˙ y i = − Γ i y i − u x z i ◮ matter 1 saturated: y 1 ( T ) = z 1 ( T ) = 0 = − γ i ( 1 − z i ) + u x y i ˙ z i ◮ matter 2 “maximized”: | ( y 2 ( T ) , z 2 ( T )) | maximal ( i = 1 , 2) Glaser’s team, 2012 : Control Theory method ◮ Numerical method to find a path towards a saturated system = solution u x ◮ Already used in some specific cases for the MRI, here applied in full generality Problem: ◮ The complexity of the path u x is not bounded Goal: ◮ Classify singular trajectories for the control ◮ Obtain control policies for the contrast problem This classification problem can be modelled with polynomials!

  5. System − z 1 − 1 ( 2 γ 1 − 2 Γ 1 ) y 1   − Γ 1 y 1 − Γ 1 + ( γ 1 − Γ 1 ) z 1 2 Γ 1 − γ 1 − ( 2 γ 1 − 2 Γ 1 ) z 1 − γ 1 z 1 y 1 ( γ 1 − Γ 1 ) y 1   ◮ M := − z 2 − 1 ( 2 γ 2 − 2 Γ 2 ) y 2   − Γ 2 y 2 − Γ 2 + ( γ 2 − Γ 2 ) z 2   2 Γ 2 − γ 2 − ( 2 γ 2 − 2 Γ 2 ) z 2 − γ 2 z 2 y 2 ( γ 2 − Γ 2 ) y 2 ◮ D := det ( M ) Problem Find all zeroes of D which are singular in ( y 1 , y 2 , z 1 , z 2 ) Equivalent formulation Find the zeroes of � � D , ∂ D , ∂ D , ∂ D , ∂ D ∂ y 1 ∂ y 2 ∂ z 1 ∂ z 2 ◮ Method described in [Bonnard et al. 2013] ◮ Proof of concept: they used this method to solve the problem for the 4 experimental settings serving as examples to the saturation method ◮ Question : solutions in full generality?

  6. Method Obvious method? Compute a Gröbner basis of this system ◮ Works in theory: method used in [Bonnard et al. 2013] ◮ Impracticable in full generality This work Decomposition into simpler problems ◮ easy simplifications (e.g. γ 1 = 1) ◮ specific physical cases: e.g. matter 1 is water ⇐ ⇒ Γ 1 = γ 1 ◮ specific structure of the system ◮ systematic study of factorizations What is “simpler”? ◮ More constraints: study I + � f � ⇐ ⇒ study V ( I ) ∩ V ( f ) ◮ Less components: study I + � U f − 1 � ⇐ ⇒ study V ( I ) � V ( f )

  7. Method Obvious method? Compute a Gröbner basis of this system ◮ Works in theory: method used in [Bonnard et al. 2013] ◮ Impracticable in full generality This work Decomposition into simpler problems ◮ easy simplifications (e.g. γ 1 = 1) ◮ specific physical cases: e.g. matter 1 is water ⇐ ⇒ Γ 1 = γ 1 ◮ specific structure of the system ◮ systematic study of factorizations What is “simpler”? ◮ More constraints: study I + � f � ⇐ ⇒ study V ( I ) ∩ V ( f ) ◮ Less components: study I + � U f − 1 � ⇐ ⇒ study V ( I ) � V ( f )

  8. Method Obvious method? Compute a Gröbner basis of this system ◮ Works in theory: method used in [Bonnard et al. 2013] ◮ Impracticable in full generality This work Decomposition into simpler problems ◮ easy simplifications (e.g. γ 1 = 1) ◮ specific physical cases: e.g. matter 1 is water ⇐ ⇒ Γ 1 = γ 1 ◮ specific structure of the system ◮ systematic study of factorizations Typical example If I contains f · g , we can decompose the system into: ◮ either f = 0 → add f to the system ◮ or f � = 0 and g = 0 → add U f − 1 and g to the system

  9. First decomposition: rank of the matrix We split the problem depending on the rank of the matrix: Base problem rank ( M ) < 3 rank ( M ) = 3 Why the rank? Because of... Theorem Consider M = ( P i , j ( X )) 1 ≤ i , j ≤ n Then generically: det ( M ) singular = ⇒ rank ( M ) < n − 1 For our specific matrix, we do not know if the theorem applies. ⇒ We need to consider both branches. =

  10. First decomposition: rank of the matrix We split the problem depending on the rank of the matrix: Base problem rank ( M ) < 3 rank ( M ) = 3 Why the rank? Because of... Theorem Consider M = ( P i , j ( X )) 1 ≤ i , j ≤ n Then generically: det ( M ) singular = ⇒ rank ( M ) < n − 1 For our specific matrix, we do not know if the theorem applies. ⇒ We need to consider both branches. =

  11. The case rank ( M ) < 3: classification 4 � � D , ∂ D , ∂ D , ∂ D , ∂ D a d (Γ 1 , Γ 2 , γ 2 ) y 2 d , 3-minors of M contains P = � 2 ∂ y 1 ∂ y 2 ∂ z 1 ∂ z 2 d = 0 We classify depending on the number of roots of P in Y 2 = y 2 2 : ◮ First bound: degree of P ◮ Need to handle multiple roots (for example using discriminants) . . . rank ( M ) < 3, Y 2 = y 2 2 . . No solution in Y 2 1 solution in Y 2 . . . . . . a 1 = · · · = a 4 = 0 a 2 = · · · = a 4 = 0 a 3 = a 4 = 0 . . . . a 0 � = 0 a 1 � = 0 a 2 � = 0 disc ( P ) = 0

  12. First decomposition: rank of the matrix We split the problem depending on the rank of the matrix: Base problem rank ( M ) < 3 rank ( M ) = 3 Why the rank? Because of... Theorem Consider M = ( P i , j ( X )) 1 ≤ i , j ≤ n Then generically: det ( M ) singular = ⇒ rank ( M ) < n − 1 For our specific matrix, the theorem does not apply. ⇒ We need to consider both branches. =

  13. The case rank ( M ) = n − 1: incidence varieties Theorem Consider M = ( P i , j ( X )) 1 ≤ i , j ≤ n and let I be the incidence variety defined by 0  λ 1      . .  · . . M  =  .   .      0 λ n − 1 If ( x ) is a point of V ( det ( M )) , then: ◮ there exists a non-zero vector Λ = ( λ i ) such that ( x , Λ) ∈ I

  14. The case rank ( M ) = n − 1: incidence varieties Theorem Consider M = ( P i , j ( X )) 1 ≤ i , j ≤ n and let I be the incidence variety defined by 0  λ 1      . .  · . . M  =  .   .      0 λ n − 1 If ( x ) is a singular point of V ( det ( M )) such that M ( x ) has rank n − 1, then: ◮ there exists a non-zero vector Λ = ( λ i ) such that ( x , Λ) ∈ I , and ◮ Λ is unique up to scalar multiplication, and ◮ ( x , Λ) is a singular point of I w.r.t. X

  15. The case rank ( M ) = n − 1: incidence varieties Theorem Consider M = ( P i , j ( X )) 1 ≤ i , j ≤ n and let I be the incidence variety defined by 0  λ 1      . .  · . . M  =  .   .      0 λ n − 1 If ( x ) is a singular point of V ( det ( M )) such that M ( x ) has rank n − 1, then: ◮ there exists a non-zero vector Λ = ( λ i ) such that ( x , Λ) ∈ I , and ◮ Λ is unique up to scalar multiplication, and ◮ ( x , Λ) is a singular point of I w.r.t. X � � ∂ D , M · Λ , ∂ M · Λ , M k � = 0 , λ i = 1 D , ∂ y i , z i ∂ y i , z i 1 ≤ k ≤ 16 1 ≤ i ≤ 4 Non-zero 3 × 3 minor Non-zero coordinate Incidence variety (16 choices) (4 choices)

  16. Overview of the classification Base problem rank ( M ) < 3 rank ( M ) = 3 Y 2 ← y 2 16 branches according to 2 the nonzero minor 6 branches according to 4 branches according to the number of roots of P the non-zero coordinate in Λ Branches according to More branches from factorizations the degree of P and the roots multiplicity More branches from factorizations

  17. Overview of the classification Suggest ways to branch earlier Base problem rank ( M ) < 3 rank ( M ) = 3 Y 2 ← y 2 16 branches according to 2 the nonzero minor 6 branches according to 4 branches according to the number of roots of P the non-zero coordinate in Λ Branches according to More branches from factorizations the degree of P and the roots multiplicity More branches from factorizations Recurring factors or with physical interpretation

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