Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer...
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Permuted max-eigenvector problem isNP-complete
P.ButkoviµcUniversity of Birmingham
http://web.mat.bham.ac.uk/P.Butkovic/
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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De�nitions and basic properties
a� b = max(a, b)a b = a+ b
a, b 2 R := R[ f�∞g
Properties (ε = �∞, a�1 = �a):
a� b = b� a(a� b)� c = a� (b� c)
a� ε = a = ε� a
a b = b a(a b) c = a (b c)
a ε = ε = ε aa 0 = a = 0 a
a a�1 = 0 = a�1 a
(a� b) c = a c � b ca� b = a or b
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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De�nitions and basic properties
Extension to matrices and vectors:
A� B = (aij � bij )A B =
�∑�k aik bkj
�α A = (α aij )
diag(d1, ..., dn) =
0BBBBBBB@
d1. . . ε
. . .
ε. . .
dn
1CCCCCCCAI = diag(0, ..., 0)
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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De�nitions and basic properties
A� B = B � A(A� B)� C = A� (B � C )
A ε = ε = ε A
[not A B = B A](A B) C = A (B C )
A I = A = I A
(A� B) C = A C � B CA (B � C ) = A B � A C
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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A piece of magic ...
Invertibility of � �! Idempotency of �
a� a = a
(a� b)k = ak � bk
(A� B)k 6= Ak � Bk
A � B =) A C � B C for any compatible A,B,C
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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A piece of magic ...
Invertibility of � �! Idempotency of �
a� a = a(a� b)k = ak � bk
(A� B)k 6= Ak � Bk
A � B =) A C � B C for any compatible A,B,C
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 7: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/7.jpg)
A piece of magic ...
Invertibility of � �! Idempotency of �
a� a = a(a� b)k = ak � bk
(A� B)k 6= Ak � Bk
A � B =) A C � B C for any compatible A,B,C
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 8: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/8.jpg)
A piece of magic ...
Invertibility of � �! Idempotency of �
a� a = a(a� b)k = ak � bk
(A� B)k 6= Ak � Bk
A � B =) A C � B C for any compatible A,B,C
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 9: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/9.jpg)
Transitive closures
For A = (aij ) 2 Rn�n :
Γ(A) = A� A2 � A3 � ... ... metric matrix
∆(A) = I � Γ(A) = I � A� A2 � A3 � ... ... Kleene starAr ... greatest weights of paths of length r
λ(A) � 0 =) Ar � A� A2 � ...� An for every r � 1λ(A) � 0 =) Γ(A) = A� A2 � ...� An
λ(A) � 0 =) ∆(A) = I � A� A2 � ...� An�1
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Transitive closures
For A = (aij ) 2 Rn�n :
Γ(A) = A� A2 � A3 � ... ... metric matrix
∆(A) = I � Γ(A) = I � A� A2 � A3 � ... ... Kleene star
Ar ... greatest weights of paths of length r
λ(A) � 0 =) Ar � A� A2 � ...� An for every r � 1λ(A) � 0 =) Γ(A) = A� A2 � ...� An
λ(A) � 0 =) ∆(A) = I � A� A2 � ...� An�1
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Transitive closures
For A = (aij ) 2 Rn�n :
Γ(A) = A� A2 � A3 � ... ... metric matrix
∆(A) = I � Γ(A) = I � A� A2 � A3 � ... ... Kleene starAr ... greatest weights of paths of length r
λ(A) � 0 =) Ar � A� A2 � ...� An for every r � 1λ(A) � 0 =) Γ(A) = A� A2 � ...� An
λ(A) � 0 =) ∆(A) = I � A� A2 � ...� An�1
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Transitive closures
For A = (aij ) 2 Rn�n :
Γ(A) = A� A2 � A3 � ... ... metric matrix
∆(A) = I � Γ(A) = I � A� A2 � A3 � ... ... Kleene starAr ... greatest weights of paths of length r
λ(A) � 0 =) Ar � A� A2 � ...� An for every r � 1
λ(A) � 0 =) Γ(A) = A� A2 � ...� An
λ(A) � 0 =) ∆(A) = I � A� A2 � ...� An�1
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Transitive closures
For A = (aij ) 2 Rn�n :
Γ(A) = A� A2 � A3 � ... ... metric matrix
∆(A) = I � Γ(A) = I � A� A2 � A3 � ... ... Kleene starAr ... greatest weights of paths of length r
λ(A) � 0 =) Ar � A� A2 � ...� An for every r � 1λ(A) � 0 =) Γ(A) = A� A2 � ...� An
λ(A) � 0 =) ∆(A) = I � A� A2 � ...� An�1
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Transitive closures
For A = (aij ) 2 Rn�n :
Γ(A) = A� A2 � A3 � ... ... metric matrix
∆(A) = I � Γ(A) = I � A� A2 � A3 � ... ... Kleene starAr ... greatest weights of paths of length r
λ(A) � 0 =) Ar � A� A2 � ...� An for every r � 1λ(A) � 0 =) Γ(A) = A� A2 � ...� An
λ(A) � 0 =) ∆(A) = I � A� A2 � ...� An�1
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Multi-machine interactive production process (MMIPP)
(R.A.Cuninghame-Green 1962)
Machines M1, ...,Mn work interactively and in stages
xi (r) . . . starting time of the r th stage on machine Mi
(i = 1, . . . , n; r = 0, 1, ...)aij . . . time Mj needs to prepare the component for Mi
xi (r + 1) = max(x1(r) + ai1, . . . , xn(r) + ain)(i = 1, . . . , n; r = 0, 1, ...)xi (r + 1) = ∑�
k aik xk (r) (i = 1, . . . , n; r = 0, 1, ...)x(r + 1) = A x(r) (r = 0, 1, . . .)A : x(0)! x(1)! x(2)! ...
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Multi-machine interactive production process (MMIPP)
(R.A.Cuninghame-Green 1962)
Machines M1, ...,Mn work interactively and in stages
xi (r) . . . starting time of the r th stage on machine Mi
(i = 1, . . . , n; r = 0, 1, ...)
aij . . . time Mj needs to prepare the component for Mi
xi (r + 1) = max(x1(r) + ai1, . . . , xn(r) + ain)(i = 1, . . . , n; r = 0, 1, ...)xi (r + 1) = ∑�
k aik xk (r) (i = 1, . . . , n; r = 0, 1, ...)x(r + 1) = A x(r) (r = 0, 1, . . .)A : x(0)! x(1)! x(2)! ...
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Multi-machine interactive production process (MMIPP)
(R.A.Cuninghame-Green 1962)
Machines M1, ...,Mn work interactively and in stages
xi (r) . . . starting time of the r th stage on machine Mi
(i = 1, . . . , n; r = 0, 1, ...)aij . . . time Mj needs to prepare the component for Mi
xi (r + 1) = max(x1(r) + ai1, . . . , xn(r) + ain)(i = 1, . . . , n; r = 0, 1, ...)xi (r + 1) = ∑�
k aik xk (r) (i = 1, . . . , n; r = 0, 1, ...)x(r + 1) = A x(r) (r = 0, 1, . . .)A : x(0)! x(1)! x(2)! ...
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Multi-machine interactive production process (MMIPP)
(R.A.Cuninghame-Green 1962)
Machines M1, ...,Mn work interactively and in stages
xi (r) . . . starting time of the r th stage on machine Mi
(i = 1, . . . , n; r = 0, 1, ...)aij . . . time Mj needs to prepare the component for Mi
xi (r + 1) = max(x1(r) + ai1, . . . , xn(r) + ain)(i = 1, . . . , n; r = 0, 1, ...)
xi (r + 1) = ∑�k aik xk (r) (i = 1, . . . , n; r = 0, 1, ...)
x(r + 1) = A x(r) (r = 0, 1, . . .)A : x(0)! x(1)! x(2)! ...
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Multi-machine interactive production process (MMIPP)
(R.A.Cuninghame-Green 1962)
Machines M1, ...,Mn work interactively and in stages
xi (r) . . . starting time of the r th stage on machine Mi
(i = 1, . . . , n; r = 0, 1, ...)aij . . . time Mj needs to prepare the component for Mi
xi (r + 1) = max(x1(r) + ai1, . . . , xn(r) + ain)(i = 1, . . . , n; r = 0, 1, ...)xi (r + 1) = ∑�
k aik xk (r) (i = 1, . . . , n; r = 0, 1, ...)
x(r + 1) = A x(r) (r = 0, 1, . . .)A : x(0)! x(1)! x(2)! ...
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 20: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/20.jpg)
Multi-machine interactive production process (MMIPP)
(R.A.Cuninghame-Green 1962)
Machines M1, ...,Mn work interactively and in stages
xi (r) . . . starting time of the r th stage on machine Mi
(i = 1, . . . , n; r = 0, 1, ...)aij . . . time Mj needs to prepare the component for Mi
xi (r + 1) = max(x1(r) + ai1, . . . , xn(r) + ain)(i = 1, . . . , n; r = 0, 1, ...)xi (r + 1) = ∑�
k aik xk (r) (i = 1, . . . , n; r = 0, 1, ...)x(r + 1) = A x(r) (r = 0, 1, . . .)
A : x(0)! x(1)! x(2)! ...
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Multi-machine interactive production process (MMIPP)
(R.A.Cuninghame-Green 1962)
Machines M1, ...,Mn work interactively and in stages
xi (r) . . . starting time of the r th stage on machine Mi
(i = 1, . . . , n; r = 0, 1, ...)aij . . . time Mj needs to prepare the component for Mi
xi (r + 1) = max(x1(r) + ai1, . . . , xn(r) + ain)(i = 1, . . . , n; r = 0, 1, ...)xi (r + 1) = ∑�
k aik xk (r) (i = 1, . . . , n; r = 0, 1, ...)x(r + 1) = A x(r) (r = 0, 1, . . .)A : x(0)! x(1)! x(2)! ...
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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MMIPP: Steady state
The system is in a steady state if it is moving forward inregular steps
Equivalently, if there is a λ such that
x(r + 1) = λ x(r)
Sincex(r + 1) = A x(r) (r = 0, 1, . . .)
x (0) should satisfy
A x = λ x
One-o¤ process (b is the vector of completion times):
A x = b
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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MMIPP: Steady state
The system is in a steady state if it is moving forward inregular steps
Equivalently, if there is a λ such that
x(r + 1) = λ x(r)
Sincex(r + 1) = A x(r) (r = 0, 1, . . .)
x (0) should satisfy
A x = λ x
One-o¤ process (b is the vector of completion times):
A x = b
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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MMIPP: Steady state
The system is in a steady state if it is moving forward inregular steps
Equivalently, if there is a λ such that
x(r + 1) = λ x(r)
Sincex(r + 1) = A x(r) (r = 0, 1, . . .)
x (0) should satisfy
A x = λ x
One-o¤ process (b is the vector of completion times):
A x = b
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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MMIPP: Steady state
The system is in a steady state if it is moving forward inregular steps
Equivalently, if there is a λ such that
x(r + 1) = λ x(r)
Sincex(r + 1) = A x(r) (r = 0, 1, . . .)
x (0) should satisfy
A x = λ x
One-o¤ process (b is the vector of completion times):
A x = b
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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MMIPP: Steady state
The system is in a steady state if it is moving forward inregular steps
Equivalently, if there is a λ such that
x(r + 1) = λ x(r)
Sincex(r + 1) = A x(r) (r = 0, 1, . . .)
x (0) should satisfy
A x = λ x
One-o¤ process (b is the vector of completion times):
A x = b
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Two basic problems
Problem (LINEAR SYSTEM [LS])
Given A 2 Rm�n
and b 2 Rm�nd all x 2 R
nsatisfying
A x = b
Problem (EIGENVECTOR [EV])
Given A 2 Rn�n
�nd all x 2 Rn, x 6= (ε, ..., ε)T (eigenvectors)
such thatA x = λ x
for some λ 2 R (eigenvalue)
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Two basic problems
Problem (LINEAR SYSTEM [LS])
Given A 2 Rm�n
and b 2 Rm�nd all x 2 R
nsatisfying
A x = b
Problem (EIGENVECTOR [EV])
Given A 2 Rn�n
�nd all x 2 Rn, x 6= (ε, ..., ε)T (eigenvectors)
such thatA x = λ x
for some λ 2 R (eigenvalue)
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Permuted matrices and vectors
For
A = (aij ) 2 Rn�n
x = (x1, ..., xn)T 2 R
n
π, σ 2 Pn
de�ne
A(π, σ) =�aπ(i ),σ(j)
�x (π) =
�xπ(1), ..., xπ(n)
�T
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Permuted basic problems
Problem (PERMUTED LINEAR SYSTEM [PLS])
Given A 2 Rm�n
and b 2 Rm, is there a π 2 Pm such that
A x = b (π)
has a solution?
Problem (PERMUTED EIGENVECTOR [PEV])
Given A 2 Rn�n
and x 2 Rn, x 6= (ε, ..., ε)T , is there a π 2 Pn
such thatA x (π) = λ x (π)
for some λ 2 R?
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Permuted basic problems
Problem (PERMUTED LINEAR SYSTEM [PLS])
Given A 2 Rm�n
and b 2 Rm, is there a π 2 Pm such that
A x = b (π)
has a solution?
Problem (PERMUTED EIGENVECTOR [PEV])
Given A 2 Rn�n
and x 2 Rn, x 6= (ε, ..., ε)T , is there a π 2 Pn
such thatA x (π) = λ x (π)
for some λ 2 R?
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Permuted basic problems - integer versions
Problem (INTEGER PERMUTED LINEAR SYSTEM [IPLS])
Given A 2 Zm�n and b 2 Zm , is there a π 2 Pm such that
A x = b (π)
has a solution?
Problem (INTEGER PERMUTED EIGENVECTOR [IPEV])
Given A 2 Zn�n and x 2 Zn, is there a π 2 Pn such that
A x (π) = x (π)?
Theorem
Both IPEV and IPLS are NP-complete.
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Permuted basic problems - integer versions
Problem (INTEGER PERMUTED LINEAR SYSTEM [IPLS])
Given A 2 Zm�n and b 2 Zm , is there a π 2 Pm such that
A x = b (π)
has a solution?
Problem (INTEGER PERMUTED EIGENVECTOR [IPEV])
Given A 2 Zn�n and x 2 Zn, is there a π 2 Pn such that
A x (π) = x (π)?
Theorem
Both IPEV and IPLS are NP-complete.
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Permuted basic problems - integer versions
Problem (INTEGER PERMUTED LINEAR SYSTEM [IPLS])
Given A 2 Zm�n and b 2 Zm , is there a π 2 Pm such that
A x = b (π)
has a solution?
Problem (INTEGER PERMUTED EIGENVECTOR [IPEV])
Given A 2 Zn�n and x 2 Zn, is there a π 2 Pn such that
A x (π) = x (π)?
Theorem
Both IPEV and IPLS are NP-complete.
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Maximum cycle mean
The maximum cycle mean of A :
λ(A) = max�ai1 i2 + ai2 i3 + ...+ aik i1
k; i1, ..., ik 2 N
�
A = (aij ) 2 Rn�n �! DA = (N, f(i , j); aij > �∞g, (aij ))
... associated digraph (N = f1, ..., ng)A is irreducible i¤ DA strongly connected
(Cuninghame-Green, 1962) A is irreducible =) λ(A) is theunique eigenvalue of A and all eigenvectors are �nite
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Maximum cycle mean
The maximum cycle mean of A :
λ(A) = max�ai1 i2 + ai2 i3 + ...+ aik i1
k; i1, ..., ik 2 N
�A = (aij ) 2 R
n�n �! DA = (N, f(i , j); aij > �∞g, (aij ))
... associated digraph (N = f1, ..., ng)
A is irreducible i¤ DA strongly connected
(Cuninghame-Green, 1962) A is irreducible =) λ(A) is theunique eigenvalue of A and all eigenvectors are �nite
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Maximum cycle mean
The maximum cycle mean of A :
λ(A) = max�ai1 i2 + ai2 i3 + ...+ aik i1
k; i1, ..., ik 2 N
�A = (aij ) 2 R
n�n �! DA = (N, f(i , j); aij > �∞g, (aij ))
... associated digraph (N = f1, ..., ng)A is irreducible i¤ DA strongly connected
(Cuninghame-Green, 1962) A is irreducible =) λ(A) is theunique eigenvalue of A and all eigenvectors are �nite
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Maximum cycle mean
The maximum cycle mean of A :
λ(A) = max�ai1 i2 + ai2 i3 + ...+ aik i1
k; i1, ..., ik 2 N
�A = (aij ) 2 R
n�n �! DA = (N, f(i , j); aij > �∞g, (aij ))
... associated digraph (N = f1, ..., ng)A is irreducible i¤ DA strongly connected
(Cuninghame-Green, 1962) A is irreducible =) λ(A) is theunique eigenvalue of A and all eigenvectors are �nite
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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BANDWIDTH
Problem (BANDWIDTH)
Given an undirected graph G = (N,E ) and a positive integerK � n, is there a π 2 Pn such that jπ(u)� π(v)j � K for alluv 2 E?
Equivalently:
Problem (BANDWIDTH - MATRIX VERSION)
Given an n� n symmetric 0� 1 matrix M = (mij ) with zerodiagonal, and a positive integer K � n, is there a π 2 Pn suchthat ji � j j � K whenever mπ(i ),π(j) = 1?
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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BANDWIDTH
Problem (BANDWIDTH)
Given an undirected graph G = (N,E ) and a positive integerK � n, is there a π 2 Pn such that jπ(u)� π(v)j � K for alluv 2 E?
Equivalently:
Problem (BANDWIDTH - MATRIX VERSION)
Given an n� n symmetric 0� 1 matrix M = (mij ) with zerodiagonal, and a positive integer K � n, is there a π 2 Pn suchthat ji � j j � K whenever mπ(i ),π(j) = 1?
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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BANDWIDTH
K
0
0
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Solvability of linear systems
Let A 2 Rm�n and b 2 Rm , M = f1, ...,mg ,N = f1, ..., ng
A x = b
S (A, b) = fx 2 Rn;A x = bg
x j = �maxi(aij � bi ), j 2 N
x = (x1, ..., xn)T
Mj = fk 2 M; akj � bk = maxi(aij � bi )g, j 2 N
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Solvability of linear systems
Let A 2 Rm�n and b 2 Rm , M = f1, ...,mg ,N = f1, ..., ng
A x = b
S (A, b) = fx 2 Rn;A x = bg
x j = �maxi(aij � bi ), j 2 N
x = (x1, ..., xn)T
Mj = fk 2 M; akj � bk = maxi(aij � bi )g, j 2 N
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Solvability of linear systems
Let A 2 Rm�n and b 2 Rm , M = f1, ...,mg ,N = f1, ..., ng
A x = b
S (A, b) = fx 2 Rn;A x = bg
x j = �maxi(aij � bi ), j 2 N
x = (x1, ..., xn)T
Mj = fk 2 M; akj � bk = maxi(aij � bi )g, j 2 N
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Solvability of linear systems
Let A 2 Rm�n and b 2 Rm , M = f1, ...,mg ,N = f1, ..., ng
A x = b
S (A, b) = fx 2 Rn;A x = bg
x j = �maxi(aij � bi ), j 2 N
x = (x1, ..., xn)T
Mj = fk 2 M; akj � bk = maxi(aij � bi )g, j 2 N
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Solvability of linear systems
Let A 2 Rm�n and b 2 Rm , M = f1, ...,mg ,N = f1, ..., ng
A x = b
S (A, b) = fx 2 Rn;A x = bg
x j = �maxi(aij � bi ), j 2 N
x = (x1, ..., xn)T
Mj = fk 2 M; akj � bk = maxi(aij � bi )g, j 2 N
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Solvability of max-linear systems
Theorem (R.A.Cuninghame-Green, 1960)
Let A 2 Rm�n, b 2 Rm and x 2 Rn. Then x 2 S (A, b) if andonly if
(a) x � x and(b) [
xj=x j
Mj = M
Corollary (1)
The following three statements are equivalent:
(I) S (A, b) 6= ∅
(II) x 2 S (A, b)(III)
Sj2N Mj = M
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Solvability of max-linear systems
Theorem (R.A.Cuninghame-Green, 1960)
Let A 2 Rm�n, b 2 Rm and x 2 Rn. Then x 2 S (A, b) if andonly if
(a) x � x and(b) [
xj=x j
Mj = M
Corollary (1)
The following three statements are equivalent:
(I) S (A, b) 6= ∅
(II) x 2 S (A, b)(III)
Sj2N Mj = M
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Solvability of max-linear systems
Theorem (R.A.Cuninghame-Green, 1960)
Let A 2 Rm�n, b 2 Rm and x 2 Rn. Then x 2 S (A, b) if andonly if
(a) x � x and(b) [
xj=x j
Mj = M
Corollary (1)
The following three statements are equivalent:
(I) S (A, b) 6= ∅
(II) x 2 S (A, b)
(III)Sj2N Mj = M
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Solvability of max-linear systems
Theorem (R.A.Cuninghame-Green, 1960)
Let A 2 Rm�n, b 2 Rm and x 2 Rn. Then x 2 S (A, b) if andonly if
(a) x � x and(b) [
xj=x j
Mj = M
Corollary (1)
The following three statements are equivalent:
(I) S (A, b) 6= ∅
(II) x 2 S (A, b)(III)
Sj2N Mj = M
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Unique solution to a max-linear system
Corollary (2)
S (A, b) = fxg if and only if
(a)Sj2N Mj = M and
(b)Sj2N 0 Mj 6= M for any N 0 � N,N 0 6= N.
Corollary (3)
If m = n then S (A, b) = fxg if and only if there is a π 2 Pn suchthat Mπ(j) = fjg for all j 2 N. Equivalently
ai ,π(j) � bi < aj ,π(j) � bj
for all i , j 2 N, i 6= j .
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Unique solution to a max-linear system
Corollary (2)
S (A, b) = fxg if and only if
(a)Sj2N Mj = M and
(b)Sj2N 0 Mj 6= M for any N 0 � N,N 0 6= N.
Corollary (3)
If m = n then S (A, b) = fxg if and only if there is a π 2 Pn suchthat Mπ(j) = fjg for all j 2 N. Equivalently
ai ,π(j) � bi < aj ,π(j) � bj
for all i , j 2 N, i 6= j .
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Unique solution to a max-linear system
Corollary (2)
S (A, b) = fxg if and only if
(a)Sj2N Mj = M and
(b)Sj2N 0 Mj 6= M for any N 0 � N,N 0 6= N.
Corollary (3)
If m = n then S (A, b) = fxg if and only if there is a π 2 Pn suchthat Mπ(j) = fjg for all j 2 N. Equivalently
ai ,π(j) � bi < aj ,π(j) � bj
for all i , j 2 N, i 6= j .
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Unique solution to a max-linear system
Corollary (2)
S (A, b) = fxg if and only if
(a)Sj2N Mj = M and
(b)Sj2N 0 Mj 6= M for any N 0 � N,N 0 6= N.
Corollary (3)
If m = n then S (A, b) = fxg if and only if there is a π 2 Pn suchthat Mπ(j) = fjg for all j 2 N. Equivalently
ai ,π(j) � bi < aj ,π(j) � bj
for all i , j 2 N, i 6= j .
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Strong regularity
A = (aij ) 2 Rn�n is strongly regular i¤
(9b 2 Rn) jS (A, b)j = 1
Linear assignment problem for A :Find a π 2 Pn maximising w(A,π) = ∑
i2Nai ,π(i )
ap(A) = fσ 2 Pn;w(A, σ) = maxπ2Pn
w(A,π)g
(PB + Hevery, 1985) A = (aij ) 2 Rn�n is strongly regular ifand only if jap(A)j = 1What are those b if A is strongly regular?
A 2 Rn�n �! SA = fb 2 Rn;A x = b has a uniquesolutiong
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Strong regularity
A = (aij ) 2 Rn�n is strongly regular i¤
(9b 2 Rn) jS (A, b)j = 1
Linear assignment problem for A :Find a π 2 Pn maximising w(A,π) = ∑
i2Nai ,π(i )
ap(A) = fσ 2 Pn;w(A, σ) = maxπ2Pn
w(A,π)g
(PB + Hevery, 1985) A = (aij ) 2 Rn�n is strongly regular ifand only if jap(A)j = 1What are those b if A is strongly regular?
A 2 Rn�n �! SA = fb 2 Rn;A x = b has a uniquesolutiong
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Strong regularity
A = (aij ) 2 Rn�n is strongly regular i¤
(9b 2 Rn) jS (A, b)j = 1
Linear assignment problem for A :Find a π 2 Pn maximising w(A,π) = ∑
i2Nai ,π(i )
ap(A) = fσ 2 Pn;w(A, σ) = maxπ2Pn
w(A,π)g
(PB + Hevery, 1985) A = (aij ) 2 Rn�n is strongly regular ifand only if jap(A)j = 1What are those b if A is strongly regular?
A 2 Rn�n �! SA = fb 2 Rn;A x = b has a uniquesolutiong
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Strong regularity
A = (aij ) 2 Rn�n is strongly regular i¤
(9b 2 Rn) jS (A, b)j = 1
Linear assignment problem for A :Find a π 2 Pn maximising w(A,π) = ∑
i2Nai ,π(i )
ap(A) = fσ 2 Pn;w(A, σ) = maxπ2Pn
w(A,π)g
(PB + Hevery, 1985) A = (aij ) 2 Rn�n is strongly regular ifand only if jap(A)j = 1
What are those b if A is strongly regular?
A 2 Rn�n �! SA = fb 2 Rn;A x = b has a uniquesolutiong
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Strong regularity
A = (aij ) 2 Rn�n is strongly regular i¤
(9b 2 Rn) jS (A, b)j = 1
Linear assignment problem for A :Find a π 2 Pn maximising w(A,π) = ∑
i2Nai ,π(i )
ap(A) = fσ 2 Pn;w(A, σ) = maxπ2Pn
w(A,π)g
(PB + Hevery, 1985) A = (aij ) 2 Rn�n is strongly regular ifand only if jap(A)j = 1What are those b if A is strongly regular?
A 2 Rn�n �! SA = fb 2 Rn;A x = b has a uniquesolutiong
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Strong regularity
A = (aij ) 2 Rn�n is strongly regular i¤
(9b 2 Rn) jS (A, b)j = 1
Linear assignment problem for A :Find a π 2 Pn maximising w(A,π) = ∑
i2Nai ,π(i )
ap(A) = fσ 2 Pn;w(A, σ) = maxπ2Pn
w(A,π)g
(PB + Hevery, 1985) A = (aij ) 2 Rn�n is strongly regular ifand only if jap(A)j = 1What are those b if A is strongly regular?
A 2 Rn�n �! SA = fb 2 Rn;A x = b has a uniquesolutiong
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Strong regularity
SA = fb 2 Rn;A x = b has a unique solutiong
SA ... the simple image set (of the mapping x 7�! A x)A = (aij ) 2 Rn�n is normalised i¤
λ(A) = 0 (A is de�nite) andaii = 0 for all i 2 N (A is increasing, A � I )
A normalised =) ∆(A) = Γ(A) = A� A2 � ...� An�1 andI � A � A2 � ..., hence
∆(A) = Γ(A) = An�1 = An = An+1 = ...
A normalised =)
Im (A) � Im�A2�� Im
�A3�� ...
� Im�An�1
�= Im (An) = Im
�An+1
�= ... = V (A)
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Strong regularity
SA = fb 2 Rn;A x = b has a unique solutiongSA ... the simple image set (of the mapping x 7�! A x)
A = (aij ) 2 Rn�n is normalised i¤
λ(A) = 0 (A is de�nite) andaii = 0 for all i 2 N (A is increasing, A � I )
A normalised =) ∆(A) = Γ(A) = A� A2 � ...� An�1 andI � A � A2 � ..., hence
∆(A) = Γ(A) = An�1 = An = An+1 = ...
A normalised =)
Im (A) � Im�A2�� Im
�A3�� ...
� Im�An�1
�= Im (An) = Im
�An+1
�= ... = V (A)
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Strong regularity
SA = fb 2 Rn;A x = b has a unique solutiongSA ... the simple image set (of the mapping x 7�! A x)A = (aij ) 2 Rn�n is normalised i¤
λ(A) = 0 (A is de�nite) andaii = 0 for all i 2 N (A is increasing, A � I )
A normalised =) ∆(A) = Γ(A) = A� A2 � ...� An�1 andI � A � A2 � ..., hence
∆(A) = Γ(A) = An�1 = An = An+1 = ...
A normalised =)
Im (A) � Im�A2�� Im
�A3�� ...
� Im�An�1
�= Im (An) = Im
�An+1
�= ... = V (A)
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Strong regularity
SA = fb 2 Rn;A x = b has a unique solutiongSA ... the simple image set (of the mapping x 7�! A x)A = (aij ) 2 Rn�n is normalised i¤
λ(A) = 0 (A is de�nite) and
aii = 0 for all i 2 N (A is increasing, A � I )
A normalised =) ∆(A) = Γ(A) = A� A2 � ...� An�1 andI � A � A2 � ..., hence
∆(A) = Γ(A) = An�1 = An = An+1 = ...
A normalised =)
Im (A) � Im�A2�� Im
�A3�� ...
� Im�An�1
�= Im (An) = Im
�An+1
�= ... = V (A)
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Strong regularity
SA = fb 2 Rn;A x = b has a unique solutiongSA ... the simple image set (of the mapping x 7�! A x)A = (aij ) 2 Rn�n is normalised i¤
λ(A) = 0 (A is de�nite) andaii = 0 for all i 2 N (A is increasing, A � I )
A normalised =) ∆(A) = Γ(A) = A� A2 � ...� An�1 andI � A � A2 � ..., hence
∆(A) = Γ(A) = An�1 = An = An+1 = ...
A normalised =)
Im (A) � Im�A2�� Im
�A3�� ...
� Im�An�1
�= Im (An) = Im
�An+1
�= ... = V (A)
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Strong regularity
SA = fb 2 Rn;A x = b has a unique solutiongSA ... the simple image set (of the mapping x 7�! A x)A = (aij ) 2 Rn�n is normalised i¤
λ(A) = 0 (A is de�nite) andaii = 0 for all i 2 N (A is increasing, A � I )
A normalised =) ∆(A) = Γ(A) = A� A2 � ...� An�1 andI � A � A2 � ..., hence
∆(A) = Γ(A) = An�1 = An = An+1 = ...
A normalised =)
Im (A) � Im�A2�� Im
�A3�� ...
� Im�An�1
�= Im (An) = Im
�An+1
�= ... = V (A)
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Strong regularity
SA = fb 2 Rn;A x = b has a unique solutiongSA ... the simple image set (of the mapping x 7�! A x)A = (aij ) 2 Rn�n is normalised i¤
λ(A) = 0 (A is de�nite) andaii = 0 for all i 2 N (A is increasing, A � I )
A normalised =) ∆(A) = Γ(A) = A� A2 � ...� An�1 andI � A � A2 � ..., hence
∆(A) = Γ(A) = An�1 = An = An+1 = ...
A normalised =)
Im (A) � Im�A2�� Im
�A3�� ...
� Im�An�1
�= Im (An) = Im
�An+1
�= ... = V (A)
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The simple image set
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The simple image set
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The simple image set
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The simple image set
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The simple image set
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The simple image set
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The simple image set
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The simple image set
Theorem (PB, 1999)
If A 2 Rn�n is normalised and strongly regular (that is SA 6= ∅)then
V (A) = cl(SA)
Corollary
If A 2 Rn�n is normalised and strongly regular then
1 A b = b for every b 2 SA2 For every b 2 V (A) there is a sequence fb(k )g∞
k=0 � SA such thatb(k ) �! b
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The simple image set
Theorem (PB, 1999)
If A 2 Rn�n is normalised and strongly regular (that is SA 6= ∅)then
V (A) = cl(SA)
Corollary
If A 2 Rn�n is normalised and strongly regular then
1 A b = b for every b 2 SA2 For every b 2 V (A) there is a sequence fb(k )g∞
k=0 � SA such thatb(k ) �! b
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The simple image set
Theorem (PB, 1999)
If A 2 Rn�n is normalised and strongly regular (that is SA 6= ∅)then
V (A) = cl(SA)
Corollary
If A 2 Rn�n is normalised and strongly regular then
1 A b = b for every b 2 SA
2 For every b 2 V (A) there is a sequence fb(k )g∞k=0 � SA such that
b(k ) �! b
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The simple image set
Theorem (PB, 1999)
If A 2 Rn�n is normalised and strongly regular (that is SA 6= ∅)then
V (A) = cl(SA)
Corollary
If A 2 Rn�n is normalised and strongly regular then
1 A b = b for every b 2 SA2 For every b 2 V (A) there is a sequence fb(k )g∞
k=0 � SA such thatb(k ) �! b
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Normalised and strongly regular matrices
b(k ) 2 SA means���S �A, b(k )���� = 1
If m = n :
jS (A, b)j = 1() (9π 2 Pn)Mπ(j) = fjg for all j 2 N
Equivalently
ai ,π(j) � b(k )i < aj ,π(j) � b
(k )j
for all i , j 2 N, i 6= j .
If A is normalised and strongly regular then π = id , hence
aij � b(k )i < �b(k )j for every i , j 2 N, i 6= j and k = 0, 1, ....
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Normalised and strongly regular matrices
b(k ) 2 SA means���S �A, b(k )���� = 1
If m = n :
jS (A, b)j = 1() (9π 2 Pn)Mπ(j) = fjg for all j 2 N
Equivalently
ai ,π(j) � b(k )i < aj ,π(j) � b
(k )j
for all i , j 2 N, i 6= j .
If A is normalised and strongly regular then π = id , hence
aij � b(k )i < �b(k )j for every i , j 2 N, i 6= j and k = 0, 1, ....
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Normalised and strongly regular matrices
b(k ) 2 SA means���S �A, b(k )���� = 1
If m = n :
jS (A, b)j = 1() (9π 2 Pn)Mπ(j) = fjg for all j 2 N
Equivalently
ai ,π(j) � b(k )i < aj ,π(j) � b
(k )j
for all i , j 2 N, i 6= j .
If A is normalised and strongly regular then π = id , hence
aij � b(k )i < �b(k )j for every i , j 2 N, i 6= j and k = 0, 1, ....
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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Normalised and strongly regular matrices
b(k ) 2 SA means���S �A, b(k )���� = 1
If m = n :
jS (A, b)j = 1() (9π 2 Pn)Mπ(j) = fjg for all j 2 N
Equivalently
ai ,π(j) � b(k )i < aj ,π(j) � b
(k )j
for all i , j 2 N, i 6= j .
If A is normalised and strongly regular then π = id , hence
aij � b(k )i < �b(k )j for every i , j 2 N, i 6= j and k = 0, 1, ....
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Normalised and strongly regular matrices
If A is normalised and strongly regular then
aij � b(k )i < �b(k )j for every i , j 2 N, i 6= j and k = 0, 1, ....
Let A = (aij ) 2 Rn�n be normalised, strongly regular andb 2 Rn. Then b 2 V (A) if and only if
aij � bi � �bj for every i , j 2 N
Let A = (aij ) 2 Zn�n be normalised, strongly regular, b 2 Zn
and π 2 Pn. Then b(π) 2 V (A) if and only if
aπ(i ),π(j) � bi � bj for every i , j 2 N
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Normalised and strongly regular matrices
If A is normalised and strongly regular then
aij � b(k )i < �b(k )j for every i , j 2 N, i 6= j and k = 0, 1, ....
Let A = (aij ) 2 Rn�n be normalised, strongly regular andb 2 Rn. Then b 2 V (A) if and only if
aij � bi � �bj for every i , j 2 N
Let A = (aij ) 2 Zn�n be normalised, strongly regular, b 2 Zn
and π 2 Pn. Then b(π) 2 V (A) if and only if
aπ(i ),π(j) � bi � bj for every i , j 2 N
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Normalised and strongly regular matrices
If A is normalised and strongly regular then
aij � b(k )i < �b(k )j for every i , j 2 N, i 6= j and k = 0, 1, ....
Let A = (aij ) 2 Rn�n be normalised, strongly regular andb 2 Rn. Then b 2 V (A) if and only if
aij � bi � �bj for every i , j 2 N
Let A = (aij ) 2 Zn�n be normalised, strongly regular, b 2 Zn
and π 2 Pn. Then b(π) 2 V (A) if and only if
aπ(i ),π(j) � bi � bj for every i , j 2 N
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The NP-completeness result
TheoremIPEV is NP-complete for the class of normalised, strongly regularmatrices.
Proof
M = (mij ), 0 < K � n ... an instance of BANDWIDTH.
Let A = (aij ) 2 Zn�n:
aij =
8<:�K if mij = 1�n if mij = 0, i 6= j0 if i = j
A is normalised, strongly regular
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The NP-completeness result
TheoremIPEV is NP-complete for the class of normalised, strongly regularmatrices.
Proof
M = (mij ), 0 < K � n ... an instance of BANDWIDTH.Let A = (aij ) 2 Zn�n:
aij =
8<:�K if mij = 1�n if mij = 0, i 6= j0 if i = j
A is normalised, strongly regular
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The NP-completeness result
TheoremIPEV is NP-complete for the class of normalised, strongly regularmatrices.
Proof
M = (mij ), 0 < K � n ... an instance of BANDWIDTH.Let A = (aij ) 2 Zn�n:
aij =
8<:�K if mij = 1�n if mij = 0, i 6= j0 if i = j
A is normalised, strongly regular
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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The NP-completeness result
Set b = (1, ..., n)T
The answer to IPEV for A and b is "yes"
() 9π 2 Pn:
aπ(i ),π(j) � i � j for all i , j 2 N
()�K � i � j if mπ(i )π(j) = 1
()K � ji � j j if mπ(i )π(j) = 1
() "Yes" to BANDWIDTH
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 90: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/90.jpg)
The NP-completeness result
Set b = (1, ..., n)T
The answer to IPEV for A and b is "yes"
() 9π 2 Pn:
aπ(i ),π(j) � i � j for all i , j 2 N
()�K � i � j if mπ(i )π(j) = 1
()K � ji � j j if mπ(i )π(j) = 1
() "Yes" to BANDWIDTH
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 91: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/91.jpg)
The NP-completeness result
Set b = (1, ..., n)T
The answer to IPEV for A and b is "yes"
() 9π 2 Pn:
aπ(i ),π(j) � i � j for all i , j 2 N
()�K � i � j if mπ(i )π(j) = 1
()K � ji � j j if mπ(i )π(j) = 1
() "Yes" to BANDWIDTH
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 92: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/92.jpg)
The NP-completeness result
Set b = (1, ..., n)T
The answer to IPEV for A and b is "yes"
() 9π 2 Pn:
aπ(i ),π(j) � i � j for all i , j 2 N
()�K � i � j if mπ(i )π(j) = 1
()K � ji � j j if mπ(i )π(j) = 1
() "Yes" to BANDWIDTH
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 93: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/93.jpg)
The NP-completeness result
Set b = (1, ..., n)T
The answer to IPEV for A and b is "yes"
() 9π 2 Pn:
aπ(i ),π(j) � i � j for all i , j 2 N
()�K � i � j if mπ(i )π(j) = 1
()K � ji � j j if mπ(i )π(j) = 1
() "Yes" to BANDWIDTH
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 94: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/94.jpg)
The NP-completeness result
Set b = (1, ..., n)T
The answer to IPEV for A and b is "yes"
() 9π 2 Pn:
aπ(i ),π(j) � i � j for all i , j 2 N
()�K � i � j if mπ(i )π(j) = 1
()K � ji � j j if mπ(i )π(j) = 1
() "Yes" to BANDWIDTH
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 95: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/95.jpg)
More on regularity
v1, ..., vn 2 Rmare called
WLD i¤ for some k and αj 2 R
vk = ∑�j 6=k αj vj
Gondran-Minoux LD i¤ for someU,V � f1, ..., ng ,U \ V = ∅,U,V 6= ∅ and αj 2 R
∑�j2U αj vj = ∑�
j2V αj vj
Strongly LD i¤
∑�j=1,...,n vj xj = v
does not have a unique solution for any v 2 Rm
SLI =) GMLI =) WLI
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 96: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/96.jpg)
More on regularity
v1, ..., vn 2 Rmare called
WLD i¤ for some k and αj 2 R
vk = ∑�j 6=k αj vj
Gondran-Minoux LD i¤ for someU,V � f1, ..., ng ,U \ V = ∅,U,V 6= ∅ and αj 2 R
∑�j2U αj vj = ∑�
j2V αj vj
Strongly LD i¤
∑�j=1,...,n vj xj = v
does not have a unique solution for any v 2 Rm
SLI =) GMLI =) WLI
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 97: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/97.jpg)
More on regularity
v1, ..., vn 2 Rmare called
WLD i¤ for some k and αj 2 R
vk = ∑�j 6=k αj vj
Gondran-Minoux LD i¤ for someU,V � f1, ..., ng ,U \ V = ∅,U,V 6= ∅ and αj 2 R
∑�j2U αj vj = ∑�
j2V αj vj
Strongly LD i¤
∑�j=1,...,n vj xj = v
does not have a unique solution for any v 2 Rm
SLI =) GMLI =) WLI
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 98: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/98.jpg)
More on regularity
v1, ..., vn 2 Rmare called
WLD i¤ for some k and αj 2 R
vk = ∑�j 6=k αj vj
Gondran-Minoux LD i¤ for someU,V � f1, ..., ng ,U \ V = ∅,U,V 6= ∅ and αj 2 R
∑�j2U αj vj = ∑�
j2V αj vj
Strongly LD i¤
∑�j=1,...,n vj xj = v
does not have a unique solution for any v 2 Rm
SLI =) GMLI =) WLI
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 99: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/99.jpg)
More on regularity
A = (aij ) 2 Rn�n,A = (A1, ...,An)
A is called:
Weakly regular i¤ the following is not true for any k andαj 2 R :
Ak = ∑�j 6=k αj Aj
Gondran-Minoux regular i¤ the following is not true for anyU,V � f1, ..., ng , U \ V = ∅, U,V 6= ∅ and αj 2 R :
∑�j2U αj Aj = ∑�
j2V αj Aj
Strongly regular i¤
∑�j=1,...,n Aj xj = b
has a unique solution for some b 2 Rn
Gondran-Minoux regularity =) Weak regularity
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 100: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/100.jpg)
More on regularity
A = (aij ) 2 Rn�n,A = (A1, ...,An)A is called:
Weakly regular i¤ the following is not true for any k andαj 2 R :
Ak = ∑�j 6=k αj Aj
Gondran-Minoux regular i¤ the following is not true for anyU,V � f1, ..., ng , U \ V = ∅, U,V 6= ∅ and αj 2 R :
∑�j2U αj Aj = ∑�
j2V αj Aj
Strongly regular i¤
∑�j=1,...,n Aj xj = b
has a unique solution for some b 2 Rn
Gondran-Minoux regularity =) Weak regularity
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 101: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/101.jpg)
More on regularity
A = (aij ) 2 Rn�n,A = (A1, ...,An)A is called:
Weakly regular i¤ the following is not true for any k andαj 2 R :
Ak = ∑�j 6=k αj Aj
Gondran-Minoux regular i¤ the following is not true for anyU,V � f1, ..., ng , U \ V = ∅, U,V 6= ∅ and αj 2 R :
∑�j2U αj Aj = ∑�
j2V αj Aj
Strongly regular i¤
∑�j=1,...,n Aj xj = b
has a unique solution for some b 2 Rn
Gondran-Minoux regularity =) Weak regularity
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 102: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/102.jpg)
More on regularity
A = (aij ) 2 Rn�n,A = (A1, ...,An)A is called:
Weakly regular i¤ the following is not true for any k andαj 2 R :
Ak = ∑�j 6=k αj Aj
Gondran-Minoux regular i¤ the following is not true for anyU,V � f1, ..., ng , U \ V = ∅, U,V 6= ∅ and αj 2 R :
∑�j2U αj Aj = ∑�
j2V αj Aj
Strongly regular i¤
∑�j=1,...,n Aj xj = b
has a unique solution for some b 2 Rn
Gondran-Minoux regularity =) Weak regularity
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 103: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/103.jpg)
More on regularity
A = (aij ) 2 Rn�n,A = (A1, ...,An)A is called:
Weakly regular i¤ the following is not true for any k andαj 2 R :
Ak = ∑�j 6=k αj Aj
Gondran-Minoux regular i¤ the following is not true for anyU,V � f1, ..., ng , U \ V = ∅, U,V 6= ∅ and αj 2 R :
∑�j2U αj Aj = ∑�
j2V αj Aj
Strongly regular i¤
∑�j=1,...,n Aj xj = b
has a unique solution for some b 2 Rn
Gondran-Minoux regularity =) Weak regularity
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 104: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/104.jpg)
More on regularity
A = (aij ) 2 Rn�n,A = (A1, ...,An)A is called:
Weakly regular i¤ the following is not true for any k andαj 2 R :
Ak = ∑�j 6=k αj Aj
Gondran-Minoux regular i¤ the following is not true for anyU,V � f1, ..., ng , U \ V = ∅, U,V 6= ∅ and αj 2 R :
∑�j2U αj Aj = ∑�
j2V αj Aj
Strongly regular i¤
∑�j=1,...,n Aj xj = b
has a unique solution for some b 2 Rn
Gondran-Minoux regularity =) Weak regularity
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 105: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/105.jpg)
More on regularity
Theorem (Gondran-Minoux 1977)
A = (aij ) 2 Rn�n is regular if and only if all permutations inap(A) are of the same parity.
Corollary: Strong regularity =) Gondran-Minoux regularity
(PB 1992) The problem "Given A, are all permutations inap(A) of the same parity?" is equivalent to the Even CycleProblem in digraphs
SR|{z}O (n3)
=) GMR| {z }� Even Cycle
=) WR|{z}O (n3)
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 106: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/106.jpg)
More on regularity
Theorem (Gondran-Minoux 1977)
A = (aij ) 2 Rn�n is regular if and only if all permutations inap(A) are of the same parity.
Corollary: Strong regularity =) Gondran-Minoux regularity
(PB 1992) The problem "Given A, are all permutations inap(A) of the same parity?" is equivalent to the Even CycleProblem in digraphs
SR|{z}O (n3)
=) GMR| {z }� Even Cycle
=) WR|{z}O (n3)
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 107: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/107.jpg)
More on regularity
Theorem (Gondran-Minoux 1977)
A = (aij ) 2 Rn�n is regular if and only if all permutations inap(A) are of the same parity.
Corollary: Strong regularity =) Gondran-Minoux regularity
(PB 1992) The problem "Given A, are all permutations inap(A) of the same parity?" is equivalent to the Even CycleProblem in digraphs
SR|{z}O (n3)
=) GMR| {z }� Even Cycle
=) WR|{z}O (n3)
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 108: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/108.jpg)
More on regularity
Theorem (Gondran-Minoux 1977)
A = (aij ) 2 Rn�n is regular if and only if all permutations inap(A) are of the same parity.
Corollary: Strong regularity =) Gondran-Minoux regularity
(PB 1992) The problem "Given A, are all permutations inap(A) of the same parity?" is equivalent to the Even CycleProblem in digraphs
SR|{z}O (n3)
=) GMR| {z }� Even Cycle
=) WR|{z}O (n3)
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 109: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/109.jpg)
THANK YOU
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 110: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/110.jpg)
WHAT IS MAX-ALGEBRAKey players: The principal solution
0BBBB@�2 2 2�5 �3 �2
ε ε 3�3 �3 21 4 ε
1CCCCA0@ x1x2x3
1A =
0BBBB@3
�2105
1CCCCA
�aij b�1i
�=
0BBBB@�5 �1 �1�3 �1 0
ε ε 2�3 �3 2�4 �1 ε
1CCCCAM1 = f2, 4g ,M2 = f1, 2, 5g ,M3 = f3, 4gx = (3, 1,�2)T is a solution since Sj=1,2,3Mj = MM2 [M3 = M hence
S(A, b) =n(x1, x2, x3)
T 2 R3; x1 � 3, x2 = 1, x3 = �2
o.
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 111: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/111.jpg)
WHAT IS MAX-ALGEBRAKey players: The principal solution
0BBBB@�2 2 2�5 �3 �2
ε ε 3�3 �3 21 4 ε
1CCCCA0@ x1x2x3
1A =
0BBBB@3
�2105
1CCCCA�aij b�1i
�=
0BBBB@�5 �1 �1�3 �1 0
ε ε 2�3 �3 2�4 �1 ε
1CCCCA
M1 = f2, 4g ,M2 = f1, 2, 5g ,M3 = f3, 4gx = (3, 1,�2)T is a solution since Sj=1,2,3Mj = MM2 [M3 = M hence
S(A, b) =n(x1, x2, x3)
T 2 R3; x1 � 3, x2 = 1, x3 = �2
o.
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 112: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/112.jpg)
WHAT IS MAX-ALGEBRAKey players: The principal solution
0BBBB@�2 2 2�5 �3 �2
ε ε 3�3 �3 21 4 ε
1CCCCA0@ x1x2x3
1A =
0BBBB@3
�2105
1CCCCA�aij b�1i
�=
0BBBB@�5 �1 �1�3 �1 0
ε ε 2�3 �3 2�4 �1 ε
1CCCCAM1 = f2, 4g ,M2 = f1, 2, 5g ,M3 = f3, 4g
x = (3, 1,�2)T is a solution since Sj=1,2,3Mj = MM2 [M3 = M hence
S(A, b) =n(x1, x2, x3)
T 2 R3; x1 � 3, x2 = 1, x3 = �2
o.
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 113: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/113.jpg)
WHAT IS MAX-ALGEBRAKey players: The principal solution
0BBBB@�2 2 2�5 �3 �2
ε ε 3�3 �3 21 4 ε
1CCCCA0@ x1x2x3
1A =
0BBBB@3
�2105
1CCCCA�aij b�1i
�=
0BBBB@�5 �1 �1�3 �1 0
ε ε 2�3 �3 2�4 �1 ε
1CCCCAM1 = f2, 4g ,M2 = f1, 2, 5g ,M3 = f3, 4gx = (3, 1,�2)T is a solution since Sj=1,2,3Mj = M
M2 [M3 = M hence
S(A, b) =n(x1, x2, x3)
T 2 R3; x1 � 3, x2 = 1, x3 = �2
o.
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
![Page 114: Permuted max-eigenvector problem is NP-completeweb.mat.bham.ac.uk/P.Butkovic/My papers/Beamer Manchester... · 2008-05-01 · P.Butkoviµc University of Birmingham Permuted eigenvector](https://reader033.fdocument.pub/reader033/viewer/2022050113/5f4a44d63e59663569640a64/html5/thumbnails/114.jpg)
WHAT IS MAX-ALGEBRAKey players: The principal solution
0BBBB@�2 2 2�5 �3 �2
ε ε 3�3 �3 21 4 ε
1CCCCA0@ x1x2x3
1A =
0BBBB@3
�2105
1CCCCA�aij b�1i
�=
0BBBB@�5 �1 �1�3 �1 0
ε ε 2�3 �3 2�4 �1 ε
1CCCCAM1 = f2, 4g ,M2 = f1, 2, 5g ,M3 = f3, 4gx = (3, 1,�2)T is a solution since Sj=1,2,3Mj = MM2 [M3 = M hence
S(A, b) =n(x1, x2, x3)
T 2 R3; x1 � 3, x2 = 1, x3 = �2
o.
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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ReferencesHistorical remarks
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