Tropical Rank and Beyond Alexander E. Guterman Moscow State - - PowerPoint PPT Presentation

Tropical Rank and Beyond

Alexander E. Guterman Moscow State University

Based on joint works with Marianne Akian, LeRoy B. Beasley, Stephane Gaubert, Yaroslav Shitov

• Definition. A semiring S consists of a set S and two

binary operations, addition and multiplication, such that:

• S is an Abelian monoid under addition (identity de-

noted by ε);

• S is a semigroup under multiplication (identity, if

any, denoted by e);

• multiplication is distributive over addition on both

sides;

• sε = εs = ε for all s ∈ S.

Examples:

• All rings are semirings
• Z+, R+, Q+

Are there semirings which are not “half of rings” ? Dedekind, 1894 Let R be a ring. Ideal (R) be the set of its two-sided ideals, together with R. (Ideal (R), +, ·) is a semiring: addition and multiplication of ideals are defined ele- mentwise {0R}, {R} are neutral elements Ideal (R) is not a ring since one can not subtract ideals

Now S is an antinegative semiring without zero divi- sors.

• Boolean algebras: algebras of subsets with respect

to ∪ and ∩ Binary boolean algebra:

0+0=0 0 · 0=0 1+0=1 1 · 0=0 0+1=1 0 · 1=0 1+1=1 1 · 1=1

• Max-algebras

• Definition. Max-algebra or tropical algebra:

Rmax := (R, ⊕, ⊗), a ⊕ b = max{a, b}, a ⊗ b = a + b, Rmax := Rmax ∪ {−∞} zero is −∞ and unit is 0. 2 ⊕ 3 = 3; 2 ⊗ 3 = 5 2⊗3 =?

• Definition. Max-algebra or tropical algebra:

Rmax := (R, ⊕, ⊗), a ⊕ b = max{a, b}, a ⊗ b = a + b, Rmax := Rmax ∪ {−∞} zero is −∞ and unit is 0. 2 ⊕ 3 = 3; 2 ⊗ 3 = 5 2⊗3 = 6

• Definition. Max-algebra or tropical algebra:

Rmax := (R, ⊕, ⊗), a ⊕ b = max{a, b}, a ⊗ b = a + b, Rmax := Rmax ∪ {−∞} zero is −∞ and unit is 0. 2 ⊕ 3 = 3; 2 ⊗ 3 = 5 2⊗3 = 6 √−1 =?

• Definition. Max-algebra or tropical algebra:

Rmax := (R, ⊕, ⊗), a ⊕ b = max{a, b}, a ⊗ b = a + b, Rmax := Rmax ∪ {−∞} zero is −∞ and unit is 0. 2 ⊕ 3 = 3; 2 ⊗ 3 = 5 2⊗3 = 6 √−1 = −0.5

Equations x ⊕ 5 = 5 ⇐ ⇒ max{5, x} = 5 ⇐ ⇒ x ∈ [−∞; 5] x ⊕ 3 = 5 ⇐ ⇒ max{3, x} = 5 ⇐ ⇒ x = 5 x ⊕ 5 = 3 ⇐ ⇒ max{5, x} = 3 ⇐ ⇒ x ∈ ∅

Matrices and vectors A = (aij), B = (bij) A ⊕ B = (aij ⊕ bij) = (max{aij, bij}) [A ⊗ B]ij =

• k

aik ⊗ bkj = max

k

{aik + bkj} α ⊗ A = (α ⊗ aij) = (α + aij)

Problem 1. Scheduling problem. Tropical approach Ports Airports

s

y1

s

y2

s

y3 x4

s

x3

s

x2

s

x1

s ❵❵❵❵❵❵❵❵❵❵❵❵❵❵ ③ ❜❜❜❜❜❜❜❜❜❜❜❜❜❜ ❝❝❝❝❝ ❝ ❝ ❝❝❝❝❝ ❝❝ ✏✏✏✏✏✏✏✏✏✏✏✏ ✶

a11 a12 a13 a31 max{x1 + a11, x2 + a21, . . . , xk + ak1} = y1

Scheduling problem. Tropical approach Ports Airports

s

y1

s

y2

s

y3 x4

s

x3

s

x2

s

x1

s ❵❵❵❵❵❵❵❵❵❵❵❵❵❵ ③ ❜❜❜❜❜❜❜❜❜❜❜❜❜❜ ❝❝❝❝❝ ❝ ❝ ❝❝❝❝❝ ❝❝ ✏✏✏✏✏✏✏✏✏✏✏✏ ✶

a11 a12 a13 a31

  

max{x1 + a11, x2 + a21, . . . , xk + ak1} = y1 max{x1 + a12, x2 + a22, . . . , xk + ak2} = y2

      

max{x1 + a11, x2 + a21, . . . , xk + ak1} = y1 max{x1 + a12, x2 + a22, . . . , xk + ak2} = y2 max{x1 + a13, x2 + a23, . . . , xk + ak3} = y3

            

max{x1 + a11, x2 + a21, . . . , xk + ak1} = y1 max{x1 + a12, x2 + a22, . . . , xk + ak2} = y2 max{x1 + a13, x2 + a23, . . . , xk + ak3} = y3 max{x1 + a14, x2 + a24, . . . , xk + ak4} = y4

                        

max{x1 + a11, x2 + a21, . . . , xk + ak1} = y1 max{x1 + a12, x2 + a22, . . . , xk + ak2} = y2 max{x1 + a13, x2 + a23, . . . , xk + ak3} = y3 max{x1 + a14, x2 + a24, . . . , xk + ak4} = y4 · · · · · · · · · · · · max{x1 + a1s, x2 + a2s, . . . , xk + aks} = ys

                        

max{x1 + a11, x2 + a21, . . . , xk + ak1} = y1 max{x1 + a12, x2 + a22, . . . , xk + ak2} = y2 max{x1 + a13, x2 + a23, . . . , xk + ak3} = y3 max{x1 + a14, x2 + a24, . . . , xk + ak4} = y4 · · · · · · · · · · · · max{x1 + a1s, x2 + a2s, . . . , xk + aks} = ys

                        

max{x1 + a11, x2 + a21, . . . , xk + ak1} = y1 max{x1 + a12, x2 + a22, . . . , xk + ak2} = y2 max{x1 + a13, x2 + a23, . . . , xk + ak3} = y3 max{x1 + a14, x2 + a24, . . . , xk + ak4} = y4 · · · · · · · · · · · · max{x1 + a1s, x2 + a2s, . . . , xk + aks} = ys max{7, 2}+max{5, 8} = 15 = 12 = max{7+5, 2+8}

max ← → ⊕ + ← → ⊗

                        

x1 ⊗ a11 ⊕ x2 ⊗ a21 ⊕ . . . ⊕ xk ⊗ ak1 = y1 x1 ⊗ a12 ⊕ x2 ⊗ a22 ⊕ . . . ⊕ xk ⊗ ak2 = y2 x1 ⊗ a13 ⊕ x2 ⊗ a23 ⊕ . . . ⊕ xk ⊗ ak3 = y3 x1 ⊗ a14 ⊕ x2 ⊗ a24 ⊕ . . . ⊕ xk ⊗ ak4 = y4 · · · · · · · · · · · · x1 ⊗ a1s ⊕ x2 ⊗ a2s ⊕ . . . ⊕ xk ⊗ aks = ys A ⊗ x = y

We have a system of tropical equations: A ⊗ x = b How can we solve this?

We have a system of tropical equations: A ⊗ x = b How can we solve this? For our problem we need to find just one solution.

Identity matrix: I = diag (0, . . . , 0) =

      

−∞ . . . −∞ −∞ ... . . . . . . ... ... −∞ −∞ . . . −∞

      

I ⊗ A = A ⊗ I = A Definition Adjoint matrix to A is A• = (a•

ij), where

a•

ij = a⊗−1 ji

= −aji. Lemma A ⊗ A• ≥ I. Proof (A ⊗ A•)ij = max

k

{aik − akj} ≥ −∞. If i = j then (A ⊗ A•)ii = max

k

{aik − aki} ≥ 0, since if k = i then 0 is in the set.

Dual operations a ⊕′ b = min{a, b} a ⊗′ b = a + b = a ⊗ b For vectors and matrices: A ⊕′ B =

  aij ⊕′ bij   

A ⊗′ B =

  

• k

′aik ⊗′ bkj

   =   min

k (aik + bkj)

  

Properties are the same

Notation: x = A• ⊗′ b

• Theorem. x is a solution of A ⊗ x ≤ b iff x ≤ x.

Proof. A ⊗ x ≤ b ⇔ max

j

(aij + xj) ≤ bi ∀ i ⇔ aij + xj ≤ bi ∀ i, j ⇔ xj ≤ a⊗−1

ij

⊗ bi = −aij + bi ∀ i, j multiplicative inverses exist ⇔ xj ≤ min

i (−aij + bi) ∀ j

⇔ xj ≤ min

i (a• ji + bi) ∀ j def. of

a•

ji

⇔ x ≤ A• ⊗′ b = x

• Corollaries. Let x = A• ⊗′ b.
• 1. A ⊗ x ≤ b.
• 2. Any solution x of A ⊗ x = b satisfies x ≤ x.
• 3. A ⊗ x = b has a solution iff x is a solution.

Ports Airports

s

y1

s

y2

s

y3 x4

s

x3

s

x2

s

x1

s ❵❵❵❵❵❵❵❵❵❵❵❵❵❵ ③ ❜❜❜❜❜❜❜❜❜❜❜❜❜❜ ❝ ❝❝❝❝ ❝ ❝❝ ❝❝❝❝ ❝❝ ✏✏✏✏✏✏✏✏✏✏✏✏ ✶

a11 a12 a13 a31

• Example. A =

   

2 1 3 3 4 3 1 1 0 2 −∞ 1

    ,

b =

   

6 5 4

    .

A• =

      

−2 −4 −1 −3 −2 −3 −1 ∞ −3 −1 −1

      

x = A• ⊗′ y =

      

−2 −4 −1 −3 −2 −3 −1 ∞ −3 −1 −1

      

⊗′

   

6 5 4

    =       

1 2 3 3

      

Problem 2. Train time schedule

✈ ✈ ★ ✧ ✥ ✦ ★ ✧ ✥ ✦

S1 S2 a11 = 2 a22 = 3 a21 = 3 a12 = 5

• 1. Frequency must be maximal possible.
• 2. Frequency must be constant on all roads (regular

time schedule). 3. Passengers must have a possibility to change a train at the station.

• 4. Trains must stay at stations the minimal possible

intervals.

A = (aij) =

2 5

3 3

• . Thus

  

x1(k + 1) ≥ max{x1(k) + 2, x2(k) + 5} x2(k + 1) ≥ max{x1(k) + 3, x2(k) + 3} By optimality

  

x1(k + 1) = max{x1(k) + 2, x2(k) + 5} x2(k + 1) = max{x1(k) + 3, x2(k) + 3} If x(0) =

x1(0)

x2(0)

• is given then the rest is uniquely

defined!

  

x1(k + 1) = max{x1(k) + 2, x2(k) + 5} x2(k + 1) = max{x1(k) + 3, x2(k) + 3} x1(0) = x2(0) = 0 ⇒ x(k):

  0  ,   5

3

 ,   8

8

 ,   13

11

 ,   16

16

  , . . .

x1(0) = 1, x2(0) = 0 ⇒ x(k):

  1  ,   5

4

 ,   9

8

 ,   13

12

 ,   17

16

  , . . .

The second schedule is more regular, isn’t it ?

Is it possible to construct a better schedule? Cycle S1S2 takes 8 hours. Hence, an interval between trains can not be less than 8/2 =4

Is it possible to construct a better schedule? Cycle S1S2 takes 8 hours. Hence, an interval between trains can not be less than 8/2 =4 So, this is the optimal schedule!

  

x1(k + 1) = max{x1(k) + 2, x2(k) + 5} x2(k + 1) = max{x1(k) + 3, x2(k) + 3} , A =

2 5

3 3

• max −

→ ⊕ : x(k + 1) = A ⊗ x(k) + − → ⊗ x(1) = A ⊗ x(0) x(2) = A ⊗ x(1) = A ⊗ (A ⊗ x(0)) = A⊗2 ⊗ x(0) Similarly, x(k) = A⊗k ⊗ x(0).

x1(0) = x2(0) = 0 ⇒ x(k):

• ,
• 5

3

• ,
• 8

8

• ,
• 13

11

• ,
• 16

16

• , . . .

x1(0) = 1, x2(0) = 0 ⇒ x(k):

• 1
• ,
• 5

4

• ,
• 9

8

• ,
• 13

12

• ,
• 17

16

• , . . .

A ∈ Mn. Let A ⊗ v = λ ⊗ v, v ≡ −∞. λ ⊗ v := (λ ⊗ vi)i=1,...,n = (λ + vi)i=1,...,n v is an eigenvector, λ is an eigenvalue If x(0) is an eigenvector with an eigenvalue λ, then x(k) = λ⊗k ⊗ x(0)(= kλ + x(0)). Initial conditions – eigenvectors ⇒ regular timetable

Graphs and matrices Directed graph G is (V, E), V – vertices, E ⊆ V × V – edges G is weighted if w : D → R is defined Path from i to j is p = ((ik, jk) ∈ D(A), k = 1, m) if i = i1, jk = ik+1, jm = j Length |p|l = m, weight |p|w =

m

• k=1

aik+1ik.

A circuit is a closed path: i = j. It is elementary if ik = il ∀ k, l.

• Theorem. A is indecomposable ⇒ Eigenvalue

λ = max

γ

|γ|w |γ|l where γ – elem. circuit in G(A).

• Theorem. A is indecomposable

⇒ λ = min

i=1,...,n

 

max

k=1,...,n−1 (A⊗n)i,j−(A⊗k)i,j n−k

  ∀ j

A+=

∞

• k=1

A⊗k Theorem. A is indecomposable ⇒ Eigenvector is any i’th column of A+

λ , where vertex i lies in the el-

ementary circuit with the maximal average weight, Aλ = A − λJ, λ is the eigenvalue.

Our systems: Problem 1 A ⊗ x = b and Problem 2 A ⊗ x = λ ⊗ x General system: A ⊗ x ⊕ c = B ⊗ x ⊕ d (A ⊗ x ⊕ c ≤ B ⊗ x ⊕ d)

To deal with linear systems we would like to develop tropical linear algebra:

To deal with linear systems we would like to develop tropical linear algebra:

• 1. LINEAR INDEPENDENCE.

To deal with linear systems we would like to develop tropical linear algebra:

• 1. LINEAR INDEPENDENCE.
• 2. RANK.

To deal with linear systems we would like to develop tropical linear algebra:

• 1. LINEAR INDEPENDENCE.
• 2. RANK.
• 3. DETERMINANT.

Linear independence over semirings Definition A system of elements,

m1, . . . , mk

in a semimodule M over a semiring S is linearly depen- dent in the Gondran-Minoux sense if there exist two subsets I, J ⊆ K := {1, . . . , k}, I ∩ J = ∅, I ∪ J = K and scalars α1, . . . , αk ∈ S, = (0, . . . , 0), such that

• i∈I

αimi =

• j∈J

αjmj

• Theorem. [Gondran, Minoux] Any n + 1 vectors of

the size n are linearly dependent

• Theorem. [Gondran, Minoux] Any n + 1 vectors of

the size n are linearly dependent Minus: often there is no linear independent generating sets in a semimodule

Example [Butkoviˇ c, Cuninghame-Green]

v1 =

   

1 −1

    , v2 =    

2 −2

    , v3 =    

3 −3

    , v4 =    

4 −4

   

are linearly dependent over Rmax since max

          

1 − 1 0 − 1 −1 − 1

    ,    

3 + 0 0 + 0 −3 + 0

          

= max

          

2 + 0 0 + 0 −2 + 0

    ,    

4 − 1 0 − 1 −4 − 1

          

V = v1, v2, v3, v4 contains no l.in. generating set. No 3 vectors generate V ∀ 4 vectors from V are linearly dependent.

Are there better variants of Linear dependence? Definition A subset P of elements in a semimodule M is called weakly linearly dependent if there is an element in P that can be expressed as a linear com- bination of other elements of P

Definition A subset P of elements in a semimodule M is called weakly linearly dependent if there is an element in P that can be expressed as a linear com- bination of other elements of P Plus: Any f.g. module has a finite weakly linearly independent generating set.

Definition A subset P of elements in a semimodule M is called weakly linearly dependent if there is an element in P that can be expressed as a linear com- bination of other elements of P Plus: Any f.g. module has a finite weakly linearly independent generating set. {x1, . . . , xk} are weakly linearly dependent. ⇒ xi =

• j=i

λjxj ⇒ {xj|j = i} generates {x1, . . . , xk}. Etc

Minus: There exist infinite systems of weakly linearly independent 3-vectors.

Minus: There exist infinite systems of weakly linearly independent 3-vectors. [Butkoviˇ c, Cuninghame-Green] Vectors

   

xi −xi

    ∈ R3

max, i = 1, 2, . . . , m are

weakly linearly independent for any m and for different xi.

Minus: There exist infinite systems of weakly linearly independent 3-vectors. [Butkoviˇ c, Cuninghame-Green] The vectors

   

xi −xi

    ∈ R3

max, i = 1, 2, . . . , m are weakly linearly in-

dependent for any m and for different xi. Idea: due to max any linear combination disturbs ei- ther 0 in the middle or symmetry of x and −x.

Minus: There exist infinite systems of weakly linearly independent 3-vectors. [Butkoviˇ c, Cuninghame-Green] The vectors

   

xi −xi

    ∈ R3

max, i = 1, 2, . . . , m are weakly linearly in-

dependent for any m and for different xi. Idea: due to max any linear combination disturbs ei- ther 0 in the middle or symmetry of x and −x. Any 4 of these vectors are Gondran-Minoux linearly dependent!

weak Gondran-Minoux linear = ⇒ linear dependence dependence ⇐ = X :

Definition [Izhakian] A system of elements,

m1, . . . , mk, mi = [m1

i , . . . , mn i ]t, i = 1, . . . , k, in a semimodule M is

strongly linearly dependent if there exist two series of subsets Il, Jl ⊆ K := {1, . . . , k}, Il ∩ Jl = ∅, Il ∪ Jl = K, l = 1, . . . , n, and α1, . . . , αk ∈ S, = (0, . . . , 0):

• i∈Il

αiml

i =

• j∈Jl

αjml

j

Gondran-Minoux strong linear = ⇒ linear dependence dependence ⇐ = X :

   

−1

    ,    

−1

    ,    

−1

    ∈ R3

max

are strongly linearly dependent (coefficients 0, 0, 0, note that in max-algebra 0 is not a neutral element by addition, −∞ is), but linearly independent by Gondran- Minoux.

0 ⊗ m = 0 + m = m. Thus we have: x1 =

   

−1

    ,

x2 =

   

−1

    ,

x3 =

   

−1

   

Consider I1 = {1, 2}, J1 = {3}. Then x1

1 ⊕ x1 2 = max{−1, 0} = 0 = x1 3.

Similarly, for I2 = I3 = {1}, J2 = J3 = {2, 3} x2

2 ⊕ x2 3 = max{−1, 0} = 0 = x2 1 and

x3

2 ⊕ x3 3 = max{−1, 0} = 0 = x3 1.

Definition A row rank of A ∈ Mm,n(S), r(A), is the minimal cardinality of weakly l.in. generating set of the linear span of the rows of A.

It is useless to consider analogs of this function for other types of l.in. since they are either coincide or do not exist

Example Y =

  

1 0 0 0 1 0 0 0 1 1 1 0 1 0 1

   ,

X =

 

0 1 0 0 0 1 1 1 0 1 0 1

  .

Hence r(Y ) = 3<4 = r(X).

Example Y =

  

1 0 0 0 1 0 0 0 1 1 1 0 1 0 1

   ,

X =

 

0 1 0 0 0 1 1 1 0 1 0 1

  =   

e2 e3 e1+e2 e1+e3

  .

Hence r(Y ) = 3<4 = r(X).

Example X =

 

0 1 0 0 0 1 1 1 0 1 0 1

  .

Then c(X) = 3=4 = r(X).

Definition A ∈ Mm,n(S) is of maximal row rank k, mrGM(A) = k, mrS(A) = k, or mrw(A) = k, if A contains k l.in. rows and any k + 1 rows are l.d. for any one type of l.d. Lemma 1 r(A) ≤ mrw(A) Example A =

  3 −

√ 7 √ 7 − 2

  ∈ M2,1(Z[

√ 7]+). mrw(A) = 2: 3 − √ 7 = α( √ 7 − 2) & α(3 − √ 7) = √ 7 − 2 in Z[ √ 7]+ r(A) = 1: 1 = (3 − √ 7) + ( √ 7 − 2) generates the row space of A.

Definition A matrix A ∈ Mmn(S) is of factor rank k, f(A) = k if k is the smallest with ∃ B ∈ Mmk(S), C ∈ Mkn(S) s.t. A = BC; f(A) = 0 iff A = 0 Factor rank over R+ differs from usual rank over R: A =

 

0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0

  ⇒fR(A) = 3 but fR+(A) = 4.

Dn =

         

−1 . . . 0 −1 . . . 0 −1 . . . . . . . . . . . . ... . . . . . . −1

         

∈ Mn(Rmax) Theorem [Develin, Santos, Sturmfels] f(Dn) is the smallest integer r: n ≤ C⌊r

2⌋

r

.

• Ex. f(D6) = 4, f(D36) = 8

• Lemma. mrS(A) ≤ mrGM(A) ≤ f(A) ≤ r(A) ≤ mrw(A)

Examples Dn =

         

−1 . . . 0 −1 . . . 0 −1 . . . . . . . . . . . . ... . . . . . . −1

         

∈ Mn(Rmax) 2 = mrS(D3)<mrGM(D3) = 3; 3 = mrGM(D4)<f(D4) = 4

A sum of any 2 columns with 0 coeffs is (0, . . . , 0), ⇒ any 4 columns are l.d.

A =

      

1 0 −1 2 0 −2 3 0 −3 4 0 −4

      

∈ M3,4(Rmax) 3 = f(A)<r(A) = 4

Rank via determinant?

Rank via determinant? There is no classical definition of the determinant

There is no semiring definition of the determinant. Instead it is usual to consider the following invariant Definition A bi-determinant of A = [aij] ∈ Mn(S) is a pair (A+, A−), where A+ =

• σ∈An

(a1σ(1) · . . . · anσ(n)) A− =

• σ∈Sn\An

(a1σ(1) · . . . · anσ(n)) Sn is the permutation group of order n, An ⊂ Sn is the subgroup of the even permutations.

Some Properties Let A = [aij].

• 1. bid A = bid (At).
• 2. bid

         

a11 . . . a1n . . . . . . . . . αai1 . . . αain . . . . . . . . . an1 . . . ann

         

= αbid A.

• 3. bid

   

a11 . . . αa1j . . . a1n . . . . . . . . . . . . . . . an1 . . . αajn . . . ann

    = αbid A.

• 4. A is invertible. Then |A|+ = |A|−.

The converse does not hold: A =

  1 2

3 4

  ∈ Mn(Q+, max, ·)

Then A+ = 4 = 6 = A− but ∃ B : AB = I.

Definition A ∈ Mn(S) is semiinvertible if ∃ A1, A2 ∈ Mn(S) such that

  

I + AA1 = AA2 I + A1A = A2A 5. S is a semifield. Then A+ = A− = ⇒ A is semiinvertible. The converse does not hold: A =

  1 2

2 4

  ∈ Mn(R+)

Then A is semiinvertible with A1 = A2 = I, but A+ = A− = 4.

• 6. Multiplicativity:

AB+ = A+B+ + A−B− + r AB− = A+B− + A−B+ + r for some r ∈ S.

What is an analog of classical “determinantal” definition of rank ? Definition A determinantal rank rk det(A) is the biggest k such that there exists a k × k-submatrix B of A with B+ = B−

Another way ! Definition A permanent of A = [aij] ∈ Mn(S) is de- fined by per (A) =

• σ∈Sn

(a1σ(1) · . . . · anσ(n)) if S = Rmax then it is max

σ∈Sn{a1σ(1) + . . . + anσ(n)}

where Sn is the permutation group on the set {1, . . . , n}

Definition A matrix A ∈ Mn(Rmax) is said to be trop- ically singular if the maximum is achieved at least twice. Definition A tropical rank trop(A) is the biggest k such that A has a tropically non-singular k×k-submatrix.

What is about general semirings ? Definition A matrix A ∈ Mn(S) is said to be tropically singular if ∃ T ⊆ Sn such that

• σ∈T

(a1σ(1) · . . . · anσ(n)) =

• σ∈Sn\T

(a1σ(1) · . . . · anσ(n)). Definition A tropical rank trop(A) is the biggest k such that A has a tropically non-singular k×k-submatrix.

More differences in other rank functions over Rmax A =

         

−∞ −∞ −∞ −∞ −∞ −∞ −∞ −∞ −∞ −∞ −∞ −∞ −∞ −∞

         

∈

      

M5×6(B)

• r

M5×6(Rmax) .

• Theorem. [Ya. Shitov] mrGM(A) = 5, mcGM(A) =

rk det(A) = 4, trop(A) = 3. A is the minimal example distinguishing mrGM and mcGM, mrGM and rk det.

Factor rank ? Tropical rank ? Dn =

         

−1 . . . 0 −1 . . . 0 −1 . . . . . . . . . . . . ... . . . . . . −1

         

∈ Mn(Rmax) Then (1) rk det(D3) = 3 > 2 = trop(D3)

Max is achieved twice but both times on even substitutions

(2) f(D4) = 4 > 3 = rk det(D4)

mrw(A) mcw(A) r(A) c(A)

❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ✟✟✟✟✟✟✟ ✟ ✟✟✟✟✟✟✟ ✟ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍

f(A) mrGM(A) mcGM(A)

✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ❍❍❍❍❍❍❍ ❍

rk det(A)

❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ✟✟✟✟✟✟✟ ✟

mrS(A) = trop (A) = mcS(A)

• Theorem. [Izhakian, Rowen]. S = Rmax then

mrS(A) = mcS(A) = trop(A) [Akian, Gaubert, Guterman]: Another proof based on the game theory approach.

Mean payoff games: G = (V, E) – directed bipartite graph, aij, bkl – weights

• f arcs.

Max and Min move a pawn. Payments correspond to moves.

Player Max — maximizer Player Min — minimizer States=vertices: I∪J, disjoint, I = {1, . . . , m}, J = {1, . . . , n} Steps: j•

aij

− → i — Min plays and receives aij from Max k

bkl

− → •l — Max plays and Min pays bkl to Max

Example. 1 2 3

②1 ②2 P P P P P P P P P P P P P P P ✐ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✮ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✮

8 2 1 −3 5 −9

❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳ ③

−12 A =

   

2 −∞ 8 −∞ −∞

   

B =

   

1 −∞ −3 −12 −9 5

   

Min starts at 1. If 1•

2

− → 1 then cycle with value −2 + 1 = −1. But if 1•

8

− → 2 then Max can do 2

−12

− → •2 and Min have to do 2• − →

• 3. Then cycle with value

0 + 5 = 5. If Min starts at 2, she has no choice. 1• is winning state for Min, 2• is not

Natural Assumptions: ∀ j ∈ J, ∃ i ∈ I such that aij = −∞. ∀ i ∈ I, ∃ j ∈ J such that bij = −∞.

• Theorem. [AGG] Let A, B ∈ Mm,n(Rmax). ∃ solution
• f Problem Is a tropical cone A ⊗ x ≤ B ⊗ x non-

trivial? ⇐ ⇒ ∃ winning for Max initial state in mean pay-off game with matrices A and B.

• Corollary. Let A = (aij) ∈ Mm,n(Rmax), m ≥ n. Then

columns of A are strongly independent iff A contains a tropically non-singular n × n-submatrics.

How big can be the difference between rank functions?

Method of tropical matrix patterns Definition Tropical pattern of A = (aij) ∈ Mn m(Rmax) is P(A) = (bij) ∈ Mn m(B) defined by buv =

  

1 if auv = maxn

i=1{aiv} > −∞,

if either auv = −∞ or auv < maxn

i=1{aiv}.

Examples. A1 =

  1 2

2 4

  ∈ M2(Rmax) → P(A1) =   0 0

1 1

  ∈

M2(B)

Examples. A1 =

  1 2

2 4

  ∈ M2(Rmax) → P(A1) =   0 0

1 1

  ∈

M2(B) A2 =

  2 3

2 4

  ∈ M2(Rmax) → P(A2) =   1 0

1 1

  ∈

M2(B)

Examples. A1 =

  1 2

2 4

  ∈ M2(Rmax) → P(A1) =   0 0

1 1

  ∈

M2(B) A2 =

  2 3

2 4

  ∈ M2(Rmax) → P(A2) =   1 0

1 1

  ∈

M2(B) A3 =

  1 2

3 4

  ∈ M2(Rmax) → P(A3) =   0 0

1 1

  ∈

M2(B)

Examples. A1 =

  1 2

2 4

  ∈ M2(Rmax) → P(A1) =   0 0

1 1

  ∈

M2(B) A2 =

  2 3

2 4

  ∈ M2(Rmax) → P(A2) =   1 0

1 1

  ∈

M2(B) A3 =

  1 2

3 4

  ∈ M2(Rmax) → P(A3) =   0 0

1 1

  ∈

M2(B) A4 =

  3 4

3 4

  ∈ M2(Rmax) → P(A4) =   1 1

1 1

  ∈

M2(B)

Gondran-Minoux linear dependence of rows: A1 =

  1 2

2 4

  l.in. → P(A1) =   0 0

1 1

 

l.d. A2 =

  1+1 2+1

2 4

  l.in. → P(A2) =   1 0

1 1

 

l.in. A3 =

  1 2

3 4

  l.d. → P(A3) =   0 0

1 1

 

l.d. A4 =

  1+2 2+2

3 4

  l.d. → P(A4) =   1 1

1 1

 

l.d.

• Theorem. Let A =

a1·

···

an·

• ∈ Mnm(Rmax).

The rows of A are GM-independent over Rmax iff ∃ λ1, . . . , λn ∈ Rmax: the rows of P

λ1⊗a1·

··· λn⊗an·

• are GM-independent over B.

Easy part: Lemma. Let A= (aij) ∈ Mnm(Rmax), P(A) = (wij) ∈ Mnm(B). If rows of A are GM-dependent then rows P(A) are GM-dependent.

• Proof. Assume ∃ I, J ⊂ {1, . . . , n}, I ∩ J = ∅, I ∪ J =

{1, . . . , n}, (λ1, . . . , λn) = (−∞, . . . , −∞): max

i∈I {λiaik} = max j∈J {λjajk} for all k.

Set µt = 1 if λt = maxn

u=1{λu} and µt = 0 if λt <

maxn

u=1{λu}.

Then maxi∈I{µiwik} = maxj∈J{µjwjk} for all k.

Hard part:

• Theorem. Let A = (aij) ∈ Mnm(Rmax), aij = −∞ for

all i, j. Assume the rows of A are GM-independent. Then ∃ A′ ∈ Mnm(Rmax) obtained from A by the mul- tiplication of rows with certain positive numbers s.t. the rows of P(A′) are GM-independent.

• Theorem. Let A ∈ Mnm(Rmax), P(A) ∈ Mnm(B).

Then trop(A) trop(P(A)).

• Theorem. Let A ∈ Mnm(Rmax), P(A) ∈ Mnm(B).

Then trop(A) trop(P(A)). Examples.

• 1. trop(In) = trop(P(In)) = trop(In(B)) = n.

• Theorem. Let A ∈ Mnm(Rmax), P(A) ∈ Mnm(B).

Then trop(A) trop(P(A)). Examples.

• 1. trop(In) = trop(P(In)) = trop(In(B)) = n.
• 2. P(Dn) = P

         

−1 . . . 0 −1 . . . 0 −1 . . . . . . . . . . . . ... . . . . . . −1

         

=

         

0 1 1 . . . 1 1 0 1 . . . 1 1 1 0 . . . 1 . . . . . . . . . ... . . . 1 1 1 . . . 0

         

. Thus trop(Dn) = trop(P(Dn)) = 2.

• Theorem. Let A ∈ Mnm(Rmax), P(A) ∈ Mnm(B).

Then trop(A) trop(P(A)).

• Ex. 1. trop(In) = trop(P(In)) = trop(In(B)) = n.
• 2. P(Dn) = P

         

−1 . . . 0 −1 . . . 0 −1 . . . . . . . . . . . . ... . . . . . . −1

         

=

         

0 1 1 . . . 1 1 0 1 . . . 1 1 1 0 . . . 1 . . . . . . . . . ... . . . 1 1 1 . . . 0

         

. Thus trop(Dn) = trop(P(Dn)) = 2. 3. trop(A1) = trop

  1 2

2 4

  = 2 since 5 = 4, but

trop(P(A1)) = trop

  0 0

1 1

  = 1.

Theorem. Let A ∈ Mnm(Rmax). Then trop(A)

• GMr(A).
• 1. Let the rows of A ∈ Mnm(B) be GM-independent.

Then trop(A) √n.

• 2. Let A ∈ Mnm(B). Then trop(A)
• GMr(A).

• Theorem. Let A ∈ Mn(Rmax). Then

trop(A) ≥ rk det(A) + 2 3 .

• Theorem. Let A ∈ Mn(Rmax). Then

trop(A) ≥ rk det(A) + 2 3 . Idea of the proof

• 1. Reduction to the nonsingular case: rk det(A) = n,

i.e. A+ = A−. 2. Reduction to boolean case (patterns). So, we prove Let A ∈ Mn(B) be such that A+ = A−. Then trop(A) ≥ n+2

3 .

• 3. A+, A− ∈ {0, 1}, A+ = A− ⇒ can not be

both 0 ⇒ A+ = 1 ∨ A− = 1. WLOG a11 = . . . = ann = 1. 4. Assume, G(A) has (j1, . . . , j2k) – elem. cycle of even length. Then aj1j2 = . . . = aj2k−1,j2k = aj2k,j1 = 1 ⇒ aj1j2 · . . . · aj2k−1,j2k · aj2k,j1 ·

• j /

∈{j1,...,j2k}

ajj = 1 Corresponding σ = (j1, j2, . . . , j2k) ∈ Sn is odd ⇒ A− = 1 ⇒ A+ = A− ⇒ G does not have an elementary cycle of even length.

• 5. A directed graph G is called strongly connected if

each its vertex can be reached from any other vertex. A directed graph G is called strongly k-connected if any graph G′ which is obtained from G by the removal

• f less than k vertices is strongly connected.
• Theorem. [Thomassen, 92] Each strongly 3-connected

directed graph contains an elementary cycle of even length.

• 6. Thus G is not strongly 3-connected

⇒ A contains 0 submatrix p × q, p + q = n − 2.

• 7. Lemma. Let A ∈ Mn(B), and p1, . . . , pt ∈ {1, . . . , n}

be different. Assume, ∀ q1, . . . , qt ∈ {1, . . . , n}, we have A[p1, . . . , pt|q1, . . . , qt]+ = A[p1, . . . , pt|q1, . . . , qt]−. Then A+ = A−.

• 8. Thus A contains a submatrix perm. equivalent to

       

A1 = A[ρ1, . . . , ρg|γ′

1, . . . , γ′ g]

O A[ρ′

1, . . . , ρ′ h|γ′ 1, . . . , γ′ g]

A2 = A[ρ′

1, . . . , ρ′ h|γ1, . . . , γh]

       

with d-nonsingular diagonal blocks, i.e., A1+ = A1− and A2+ = A2−.

• 9. Induction by n concludes the proof:

trop(A) ≥ trop(A1) + trop(A2) ≥ g+2

3

+ h+2

3

= = g+h+2+2

3

≥ n−2+2+2

3

≥ n+2

3

• Corollary. Let A ∈ Mn(Rmax). Then trop(A) ≥ rk det(A)+2

3

.

How big can be the difference between rank functions?

• Theorem. [Guterman, Shitov]

a. trop(A) ≥

• mrGM(A)
• b. rk det(A) ≥
• mrGM(A)

c. trop(A) ≥

rk det(A)+2 3

Factor rank: However, f(A), rw(A), sr(A) can be arbitrary large under fixed trop(A), mrGM(A), or rk det(A).

• Theorem. n ≥ 3. Let Dn = (dij)

dij =

  

if i = j −1 if i = j Then trop(Dn) = 2, rk det(Dn) = mrGM(Dn) = 3 ∀ n, but [Develin, Santos, Sturmfels] f(Dn) → ∞ if n → ∞.

✟✟✟✟✟✟✟✟✟✟✟✟ ✟ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍

Arct (A) mrw(A) mcw(A) r(A) c(A) t(A)

❍ ❍ ❍ ❍ ❍ ❍ ❍ ✟✟✟✟✟✟ ✟ ✟✟✟✟✟✟ ✟ ❍ ❍ ❍ ❍ ❍ ❍ ❍

f(A) mrGM(A) mcGM(A)

✟ ✟ ✟ ✟ ✟ ✟ ✟ ❍❍❍❍❍❍ ❍

rk det(A)

❍ ❍ ❍ ❍ ❍ ❍ ❍ ✟✟✟✟✟✟ ✟

mrS(A) = trop (A) = mcS(A)

What is about Arctic rank ?

Thank you !