permuted max-eigenvector problem is np-completeweb.mat.bham.ac.uk/p.butkovic/my papers/beamer...

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

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. . .

1CCCCCCCAI = diag(0, ..., 0)

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

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

a� a = a(a� b)k = ak � bk

(A� B)k 6= Ak � Bk

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

Transitive closures

∆(A) = I � Γ(A) = I � A� A2 � A3 � ... ... Kleene star

Ar ... greatest weights of paths of length r

λ(A) � 0 =) ∆(A) = I � A� A2 � ...� An�1

Transitive closures

λ(A) � 0 =) ∆(A) = I � A� A2 � ...� An�1

Transitive closures

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

Transitive closures

λ(A) � 0 =) ∆(A) = I � A� A2 � ...� An�1

Transitive closures

λ(A) � 0 =) ∆(A) = I � A� A2 � ...� An�1

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)! ...

(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)! ...

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)! ...

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)! ...

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

MMIPP: Steady state

x(r + 1) = λ x(r)

Sincex(r + 1) = A x(r) (r = 0, 1, . . .)

A x = λ x

A x = b

MMIPP: Steady state

x(r + 1) = λ x(r)

Sincex(r + 1) = A x(r) (r = 0, 1, . . .)

A x = λ x

A x = b

MMIPP: Steady state

x(r + 1) = λ x(r)

Sincex(r + 1) = A x(r) (r = 0, 1, . . .)

A x = λ x

A x = b

MMIPP: Steady state

x(r + 1) = λ x(r)

Sincex(r + 1) = A x(r) (r = 0, 1, . . .)

A x = λ x

A x = b

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)

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)

Permuted matrices and vectors

A = (aij ) 2 Rn�n

x = (x1, ..., xn)T 2 R

π, σ 2 Pn

de�ne

A(π, σ) =�aπ(i ),σ(j)

�x (π) =

�xπ(1), ..., xπ(n)

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?

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?

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.

A x = b (π)

has a solution?

A x (π) = x (π)?

Theorem

A x = b (π)

has a solution?

A x (π) = x (π)?

Theorem

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

Maximum cycle mean

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

Maximum cycle mean

k; i1, ..., ik 2 N

�A = (aij ) 2 R

Maximum cycle mean

k; i1, ..., ik 2 N

�A = (aij ) 2 R

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?

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?

BANDWIDTH

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

A x = b

x = (x1, ..., xn)T

A x = b

x = (x1, ..., xn)T

A x = b

x = (x1, ..., xn)T

A x = b

x = (x1, ..., xn)T

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

(a) x � x and(b) [

xj=x j

Mj = M

Corollary (1)

(I) S (A, b) 6= ∅

Sj2N Mj = M

(a) x � x and(b) [

xj=x j

Mj = M

Corollary (1)

(I) S (A, b) 6= ∅

(II) x 2 S (A, b)

(III)Sj2N Mj = M

(a) x � x and(b) [

xj=x j

Mj = M

Corollary (1)

(I) S (A, b) 6= ∅

Sj2N Mj = M

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 .

Corollary (2)

(a)Sj2N Mj = M and

Corollary (3)

Corollary (2)

(a)Sj2N Mj = M and

Corollary (3)

Corollary (2)

(a)Sj2N Mj = M and

Corollary (3)

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

Strong regularity

(9b 2 Rn) jS (A, b)j = 1

i2Nai ,π(i )

w(A,π)g

Strong regularity

(9b 2 Rn) jS (A, b)j = 1

i2Nai ,π(i )

w(A,π)g

Strong regularity

(9b 2 Rn) jS (A, b)j = 1

i2Nai ,π(i )

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?

Strong regularity

(9b 2 Rn) jS (A, b)j = 1

i2Nai ,π(i )

w(A,π)g

Strong regularity

(9b 2 Rn) jS (A, b)j = 1

i2Nai ,π(i )

w(A,π)g

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)

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¤

∆(A) = Γ(A) = An�1 = An = An+1 = ...

A normalised =)

�A3�� ...

� Im�An�1

�= Im (An) = Im

�An+1

�= ... = V (A)

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¤

∆(A) = Γ(A) = An�1 = An = An+1 = ...

A normalised =)

�A3�� ...

� Im�An�1

�= Im (An) = Im

�An+1

�= ... = V (A)

Strong regularity

λ(A) = 0 (A is de�nite) and

aii = 0 for all i 2 N (A is increasing, A � I )

∆(A) = Γ(A) = An�1 = An = An+1 = ...

A normalised =)

�A3�� ...

� Im�An�1

�= Im (An) = Im

�An+1

�= ... = V (A)

Strong regularity

∆(A) = Γ(A) = An�1 = An = An+1 = ...

A normalised =)

�A3�� ...

� Im�An�1

�= Im (An) = Im

�An+1

�= ... = V (A)

Strong regularity

∆(A) = Γ(A) = An�1 = An = An+1 = ...

A normalised =)

�A3�� ...

� Im�An�1

�= Im (An) = Im

�An+1

�= ... = V (A)

Strong regularity

∆(A) = Γ(A) = An�1 = An = An+1 = ...

A normalised =)

�A3�� ...

� Im�An�1

�= Im (An) = Im

�An+1

�= ... = V (A)

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

Theorem (PB, 1999)

V (A) = cl(SA)

Corollary

Theorem (PB, 1999)

V (A) = cl(SA)

Corollary

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

Theorem (PB, 1999)

V (A) = cl(SA)

Corollary

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

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, ....

b(k ) 2 SA means��S �A, b(k )�� = 1

If m = n :

Equivalently

ai ,π(j) � b(k )i < aj ,π(j) � b

b(k ) 2 SA means��S �A, b(k )�� = 1

If m = n :

Equivalently

ai ,π(j) � b(k )i < aj ,π(j) � b

b(k ) 2 SA means��S �A, b(k )�� = 1

If m = n :

Equivalently

ai ,π(j) � b(k )i < aj ,π(j) � b

If A is normalised and strongly regular then

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

The NP-completeness result

TheoremIPEV is NP-complete for the class of normalised, strongly regularmatrices.

M = (mij ), 0 < K � n ... an instance of BANDWIDTH.

Let A = (aij ) 2 Zn�n:

8<:�K if mij = 1�n if mij = 0, i 6= j0 if i = j

A is normalised, strongly regular

M = (mij ), 0 < K � n ... an instance of BANDWIDTH.Let A = (aij ) 2 Zn�n:

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

Set b = (1, ..., n)T

() 9π 2 Pn:

()�K � i � j if mπ(i )π(j) = 1

()K � ji � j j if mπ(i )π(j) = 1

Set b = (1, ..., n)T

() 9π 2 Pn:

()�K � i � j if mπ(i )π(j) = 1

()K � ji � j j if mπ(i )π(j) = 1

Set b = (1, ..., n)T

() 9π 2 Pn:

()�K � i � j if mπ(i )π(j) = 1

()K � ji � j j if mπ(i )π(j) = 1

Set b = (1, ..., n)T

() 9π 2 Pn:

()�K � i � j if mπ(i )π(j) = 1

()K � ji � j j if mπ(i )π(j) = 1

Set b = (1, ..., n)T

() 9π 2 Pn:

()�K � i � j if mπ(i )π(j) = 1

()K � ji � j j if mπ(i )π(j) = 1

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

More on regularity

∑�j2U αj vj = ∑�

j2V αj vj

Strongly LD i¤

∑�j=1,...,n vj xj = v

SLI =) GMLI =) WLI

More on regularity

∑�j2U αj vj = ∑�

j2V αj vj

Strongly LD i¤

∑�j=1,...,n vj xj = v

SLI =) GMLI =) WLI

More on regularity

∑�j2U αj vj = ∑�

j2V αj vj

Strongly LD i¤

∑�j=1,...,n vj xj = v

SLI =) GMLI =) WLI

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

More on regularity

A = (aij ) 2 Rn�n,A = (A1, ...,An)A is called:

∑�j2U αj Aj = ∑�

j2V αj Aj

∑�j=1,...,n Aj xj = b

More on regularity

∑�j2U αj Aj = ∑�

j2V αj Aj

∑�j=1,...,n Aj xj = b

More on regularity

∑�j2U αj Aj = ∑�

j2V αj Aj

∑�j=1,...,n Aj xj = b

More on regularity

∑�j2U αj Aj = ∑�

j2V αj Aj

∑�j=1,...,n Aj xj = b

More on regularity

∑�j2U αj Aj = ∑�

j2V αj Aj

∑�j=1,...,n Aj xj = b

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)

More on regularity

SR|{z}O (n3)

=) WR|{z}O (n3)

More on regularity

SR|{z}O (n3)

=) WR|{z}O (n3)

More on regularity

SR|{z}O (n3)

=) WR|{z}O (n3)

THANK YOU

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

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

0BBBB@�2 2 2�5 �3 �2

ε ε 3�3 �3 21 4 ε

1CCCCA0@ x1x2x3

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

0BBBB@�2 2 2�5 �3 �2

ε ε 3�3 �3 21 4 ε

1CCCCA0@ x1x2x3

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

0BBBB@�2 2 2�5 �3 �2

ε ε 3�3 �3 21 4 ε

1CCCCA0@ x1x2x3

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

0BBBB@�2 2 2�5 �3 �2

ε ε 3�3 �3 21 4 ε

1CCCCA0@ x1x2x3

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

ReferencesHistorical remarks

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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.

permuted max-eigenvector problem is np-completeweb.mat.bham.ac.uk/p.butkovic/my papers/beamer...

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