permuted max-eigenvector problem is np-completeweb.mat.bham.ac.uk/p.butkovic/my papers/beamer...
Post on 15-Jul-2020
1 Views
Preview:
TRANSCRIPT
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
BANDWIDTH
K
0
0
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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)
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)
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)
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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)
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)
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)
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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)
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)
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)
The simple image set
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
The simple image set
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
The simple image set
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
The simple image set
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
The simple image set
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
The simple image set
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
The simple image set
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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)
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)
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)
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)
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
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
THANK YOU
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
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)
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)
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)
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)
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)
ReferencesHistorical remarks
R.A.Cuninghame-Green 1960N.N.Vorobyov 1967, Litvinov, Maslov, Kolokoltsov, Sobolevski...M.Gondran, M.Minoux 1975K.Zimmermann 1972P.Butkoviµc 1977R.E.Burkard, U.Zimmermann 1981H.SchneiderG.Cohen, D.Dubois, J.-P.Quadrat, M.Viot 1983K.Cechlárová, P.Szabó, J.Plávka 1985G.-J.Olsder, C.Roos 1988, B.Heidergott 2000R.D.Nussbaum 1991M.Akian 1998, S.Gaubert 1992, C.Walsh 2001R.B.Bapat, D.Stanford, P.van den Driessche 1993M.Gavalec 1995Tropical algebra from 1995: B.Sturmfels, M.Develin et al
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
F.L.Baccelli, G. Cohen, G.-J. Olsder and J.-P. Quadrat,Synchronization and Linearity, Chichester, New York: J.Wileyand Sons, 1992.
R.E. Burkard and P. Butkovic: Finding all essential terms of acharacteristic maxpolynomial, Discrete Applied Mathematics130 (2003) 367-380.
P.Butkoviµc, Strong regularity of matrices �a survey of results,Discrete Appl. Math. 48, 45�68 (1994).
P.Butkoviµc, Regularity of matrices in min-algebra and itstime-complexity, Discrete Appl. Math. 57, 121�132 (1995).
P.Butkoviµc, Simple image set of (max, +) linear mappings,Discrete Appl. Math. 105, 73�86 (2000).
P.Butkovic, Max-algebra: the linear algebra of combinatorics?Lin.Alg. and Appl. 367 (2003) 313-335.
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
P.Butkovic, H.Schneider and S.Sergeev: Generators, Extremalsand Bases of Max Cones, Linear Algebra and Its Applications421 (2007) 394-406.
P. Butkovic, S.Gaubert and R.A. Cuninghame-Green:Reducible spectral theory with applications to the robustnessof matrices in max-algebra, The University of Birmingham,preprint 2007/16.
G.Cohen, D.Dubois, J.-P.Quadrat, M.Viot, ALinear-System-Theoretic View of Discrete-Event Processes andIts Use for Performance Evaluation in Manufacturing, IEEETransactions on Automatic Control, Vol. AC-30, No3, 1985.
G.Cohen, S.Gaubert and J.-P.Quadrat, Duality and separationtheorems in idempotent semimodules. Tenth Conference of theInternational Linear Algebra Society. Linear Algebra and ItsApplications 379 (2004), 395�422.
R.A.Cuninghame-Green, Minimax Algebra, Lecture Notes inEconomics and Math. Systems 166, Berlin: Springer, 1979.
R.A. Cuninghame-Green and P. Butkoviµc, The equation Ax =By over (max, +), Theoretical Computer Science 293 (2003)3-12.
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
R.A.Cuninghame-Green, Minimax algebra and applications, inAdvances in Imaging and Electron Physics, Vol. 90, pp. 1�121,Academic Press, New York, 1995.
S.Gaubert and R.Katz, The Minkowski Theorem for max-plusconvex sets, Linear Algebra and Its Applications421(2007)356-369.
B.Heidergott, G.J.Olsder and J. van der Woude (2005), MaxPlus at Work: Modeling and Analysis of SynchronizedSystems, A Course on Max-Plus Algebra, PUP.
G. J.Olsder and C. Roos, Cramér and Cayley-Hamilton in themax algebra. Linear Algebra and Its Applications 101 (1988)87�108.
M.Plus, Linear systems in (max,+) algebra, in: Proceedings of
29th Conference on Decision and Control Honolulu, 1990.
B. Sturmfels, F. Santos and M. Develin, On the tropical rankof a matrix, in Discrete and Computational Geometry, (eds.J.E. Goodman, J. Pach and E. Welzl), Mathematical SciencesResearch Institute Publications, Volume 52, CambridgeUniversity Press, 2005, pp. 213-242.
P.Butkoviµc University of Birmingham Permuted eigenvector (Manchester 20 May 2008)
top related