An algorithm to describe the solution set of any tropical linear system $A\odot x=B\odot x$

0.16 MB·English

by  E. Lorenzo

An algorithm to describe the solution set of any
A x = B x
tropical linear system ⊙ ⊙
E. Lorenzo∗ M.J.delaPuente†
1 Dpto. deMatema´ticaAplicada2 Dpto. deAlgebra
1
Facultad deMatema´ticasy Estad´ıstica FacultaddeMatema´ticas
0
2 UniversidadPolite´cnicadeCatalun˜a UniversidadComplutense

n [email protected] [email protected]
a
J
January 24,2011

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2

]
A
Abstract
R
. AMSclass.: 15–04;15A06;15A39.
h
Keywordsandphrases:tropicallinearsystem,algorithm.
t
a Analgorithmtogiveanexplicitdescriptionofallthesolutionstoanytrop-
m
icallinearsystemA⊙x=B⊙xispresented.Thegivensystemisconverted
[ intoafinite(rathersmall)numberpofpairs(S,T)ofclassicallinearsystems:

3 a system S of equations and a system T of inequalities. The notion, intro-
v ducedhere,thatmakespsmall,iscalledcompatibility.Theparticularfeature
3 ofbothS andT isthateachitem(equationorinequality)isbivariate,i.e.,it
9 involvesexactlytwo variables;onevariablewith coefficient1, andthe other
1
onewith−1. S issolvedbyGaussianelimination. Weexplainhowtosolve
5
. T byamethodsimilartoGaussianelimination.Toachievethis,weintroduce
7
the notion of sub–special matrix. The procedure applied to T is, therefore,
0
calledsub–specialization.
0
1
:
v 1 Introduction
i
X
r Consider the set R∪{−∞}, denoted T for short, endowed with tropical addition
a
⊕andtropical multiplication ⊙,wheretheseoperations aredefinedasfollows:
a⊕b = max{a,b}, a⊙b = a+b,
for a,b ∈ T. Here, −∞ is the neutral element for tropical addition and 0 is the
neutral element for tropical multiplication. Notice that a ⊕ a = a, for all a, i.e.,
∗PartiallysupportedbyaLaCaixagrantandaFPUgrant.
†PartiallysupportedbyUCMresearchgroup910444. Correspondingauthor.
1
tropical addition is idempotent. Notice also that a has no inverse with respect to
⊕. We will write ⊕ or max, (resp. ⊙ or +) at our convenience. In this paper we
will use the adjective classical as opposed to tropical. Most definitions in tropical
mathematicsjustmimictheclassicalones. Veryoften,workingwith(T,⊕,⊙)leads
toworkingwithmin,whichwillbedenoted ⊕′.
Given matrices A,B ∈ M (T), we want to describe all x ∈ Tn such that
m×n
A⊙x = B ⊙x. Thismeans
max{a +x : 1 ≤ j ≤n} = max{b +x : 1 ≤ j ≤ n}, i= 1,2,...,m.
ij j ij j
Ofcourse, x = −∞, all j = 1,2,...,n is a solution (the trivial solution). Using
j
the notions of winning pairs, compatibility and win sequence (introduced in this
paper; see. definitions 2, 7 and 11), the given problem is reduced to solving a
finite, rather small, number of pairs (S,T) of classical linear systems: a system S
ofequationsandasystemT ofinequalities,eachitem(equationorinequality)being
bivariate, i.e., itinvolves exactly twovariables, one withcoefficient 1, and another
with coefficient −1. Of course, S can be easily solved by Gaussian elimination.
On the other hand, the fact that T consists of bivariate inequalities allows us to
suggest a Gaussian–like procedure to solve system T. More precisely, by certain
row operations (see p. 14), we do not triangulate a coefficient matrix for T, but
nearlyso. Whatwedoistotransformsuchamatrixintotwomatrices,oneofwhich
is a sub–special matrix (see definition 15). From these two matrices the solution
settotheT isdirectly read. Wecallthisprocedure sub–specialization, (seeremark
18). Notice that we need not use the simplex algorithm or other well–known ones
tosolveT.
Compatibility of winning pairs (see definition 7) turns out a very handy nec-
essary condition to work with. Indeed, two given winning pairs yield some bi-
variate linear inequalities. Imagine that two of these are x − x + a ≤ 0 and
1 2
−x +x −b ≤ 0, forsomea,b ∈ T. Nowthecompatibility ofthe winning pairs
1 2
guaranteesthattheseinequalitiesholdsimultaneously,i.e.,theycanbeconcatenated
into x +a ≤ x ≤ x +b, so that a ≤ b. Therefore, by requiring compatibility
1 2 1
of winning pairs, all we are doing is ruling out, at the very beginning, wasting out
time with systems of inequalities which, for sure, the only solution is trivial. The
problemA⊙x = B⊙xhasbeenaddressed before. Indeed,in[4],itisprovedthat
thesolution setcanbefinitelygenerated. In[5],astrongly polynomial algorithm is
foundwhicheitherfindsasolutionortellsusthatnosolutionexists. In[1]sec. 3.5,
allthesolutions arecomputed byatechnique calledsymmetrization andresolution
ofbalances. In[4],generatorsforthesolutionsetarecomputed. Theideaoffinding
(aminimal family of)generators ispursued in[12,14,15]forthe equivalent prob-
lemA⊙x ≤ B⊙x. Aniterativemethodinpresented in[7]foranotherequivalent
problem, namely A⊙ x = B ⊙ y, where x and y are unknown. Also, there is a
techniquein[16]tosolvetheproblemA⊙x⊕a= B⊙x⊕b,relyingonarecursive
2
formulation of the closure operator (also called Kleene star operator) on matrices.
In [9] ch. 4, the closely related problem A⊙ x ⊕ b = x (similar to the classical
Jacobiiterativemethod)issolvedusingKleenestars.
WepresentthesolutionsettoA⊙x = B⊙xasafiniteunionofsets,whichare
(attending to their presentation) obviously convex. This, ofcourse, agrees withthe
cellular decomposition in[8]. Convexity issues arealsostudied in[13]. Therecent
paper [14]addresses theproblem A⊙x ≤ B ⊙x. Thedrawback ofthealgorithm
presented in [14] is, in our opinion, that this algorithm calls for the calculation of
certainKleenestars,andthisisnotaneasytask.
Wedo not describe the solution set by generators. Wedo not use Kleene stars.
Instead, our method provides adescription ofeach convex piece ofthesolution set
by parameters and bivariate linear inequality relations among them, in a reduced
form.
Letm,n,s ∈ Nbegiven. Thefollowingarekindredproblemsintropicallinear
algebra:
• P1: A⊙x = 0,
• P2: A⊙x = b,
• P3: A⊙x ≤ b,
• P4: A⊙x = B⊙x,
• P5: A⊙x ≤ B⊙x,
• P6: C ⊙x = D⊙y,
• P7: A⊙x⊕a = B⊙x⊕b.
HerethedataarematricesA,B ∈ M (T),C ∈ M (T),D ∈ M (T)and
m×n s×n s×m
vectorsa,boverT,andthejthproblemiscomputingallvectorsx ∈ Tn,y ∈ Tm,
such that Pj holds. By deciding the jth problem we mean either finding one
solutionordeclaringthattheproblemhasno(non–trivial)solution. Somereferences
since1984are[1,4,6,7,12,15,16]. Earlierbooksandpaperscanbefoundthere.
Ofcourse, x = −∞, y = −∞are solutions to P3,P4,P5 and P6. These are
thetrivialsolutions.
Only if the vector b is real, problem P2 reduces to P1. More generally, one
must realize that, contrary to classical linear algebra, problems P7 and P4 do not
reduce to problem P2 or P1, because there are no inverses for tropical addition,
so there is no tropical analogue for the matrix −B. Nevertheless, there are well–
knownconnections amongtheseproblems,i.e.,beingabletosolvesomeofthemis
equivalent tobeingabletosolvesomeother.
Weneedsomenotations:
3
• Forc,d ∈ T,c⊕′dmeansmin{c,d}andc⊙′dmeansc+d.
• Forc,d ∈ Tn,c⊙′dT meansmin{c +d ,c +d ,...,c +d }.
1 1 2 2 n n
• IfA = (a )∈ M (R)thenA∗ = (−a )istheconjugate matrix.
ij m×n ji
Therelationship amongtheseproblemsisasfollows:
• DecidingP3ispossible, ifAisreal.
Indeed, x# = A∗ ⊙′ b is a solution (called principal solution) and x ≤ x#
if and only if x is a solution; see [6], p. 31; in [1] this process is called
residuation.
• DecidingP3helpswithdeciding P2,ifAisreal.
Indeed, P2 might be incompatible but, if it has a solution, then x# is the
greatest one;see[6],p.31.
• DecidingP6impliesdeciding P2.
GivenAandb, wedecide A⊙x = I ⊙y. Foreach pair ofsolutions x,y, if
any,wesety = b,ifpossible.
• DecidingP4isequivalent todecidingP6.
Supposexisasolution toP4andwriteA⊙x = y. Concatenating matrices,
write C = A ∈ M (T), D = I ∈ M (T), where I is the
B 2m×n I 2m×n
tropical idenhtityimatrix, sothatC ⊙x =hDi⊙y. Therefore, ifwecandecide
P6,thenwecandecideP4.
Suppose now x,y are solutions to P6 and write z = x , A = [C,−∞],
y
B = [−∞,D]sothatA⊙z = B ⊙z. Therefore, ifwehcaindecide P4,then
wecandecideP6.
• P4andP5areequivalent.
A⊙x = B⊙xisequivalent toA⊙x ≤ B⊙xandA⊙x ≥ B⊙x. Onthe
otherhand,A⊙x ≤ B ⊙xisequivalent to(A⊕B)⊙x = B ⊙x.
• DecidingP4impliesdeciding P7.
WeintroduceanewscalarvariablezandwriteA⊙x⊕a⊙z = B⊙x⊕b⊙z.
Concatenating matrices, write t = x , C = [A,a], D = [B,b] so that
z
C ⊙t = D⊙t. AftersolvingP4,sett = z = 0.
n+1
(cid:2) (cid:3)
4
2 The problem
GivenmatricesA,B ∈ M (T),wewanttodescribeallnon–trivialx ∈ Tnsuch
m×n
that
A⊙x = B ⊙x. (1)
Notations:
• A = (a ),B = (b ),with
ij ij
a ifa ≥ b , b ifa ≤ b ,
ij ij ij ij ij ij
a = b =
ij ij
(−∞ otherwise, (−∞ otherwise.
• M =A⊕B = A⊕B = (m )isthemaximummatrix.
ij
Let
A⊙x = B⊙x = M ⊙x. (2)
Notice that (1) is equivalent to (2) and (2) is simpler than (1) because it involves
fewerrealcoefficients. Thus, wewillassume thatA = AandB = B,(assumption
1)inthefollowing.
Morenotations:
• [n]= {1,...,n},forn∈ N.
• Foranyc ∈ T,x = c ∈ Tn meansx = c,forallj ∈[n].
j
• For convenience, we extend T to T, by adding +∞. Eventually we will
remove+∞.
• Ifk ∈ [m]andj,l ∈ [n],then
m −m , ifm 6= −∞,
kj kl kl
dif(M;j,l) = +∞ ifm 6= −∞andm = −∞,
k  kj kl
undetermined, otherwise.

Theundetermined casewillneverappearinthefollowing.
• Ω = {j ∈ [n] :x = −∞}.
j
We will deal with bivariate equalities and inequalities, linear on the x ’s with
j
coefficients in T. Tautological linear equalities or inequalities will be always re-
moved (assumption 2). A non–tautological linear equality not involving ±∞ will
be reduced to an equivalent equation of the form a = 0, for some a, by the usual
algebraic rules. Of course, −a = 0 is also possible. A non–tautological linear in-
equality not involving ±∞ will be reduced to an equivalent inequality of the form
5
a ≤ 0, for some a. These will be called normal forms. Notice that normal forms
haverealcoefficients.
What will we do with certain normal forms φ, if we know that j ∈ Ω, i.e.,
x = −∞? Wewillremoveandenlargeasfollows:
j
• Ifφisx −x +a ≤ 0,forsomek 6= j anda ∈R,thenremoveφ.
j k
• If φ is x −x +a ≤ 0, for some k 6= j and a ∈ R, then remove φ and set
k j
x = −∞,i.e.,enlargeΩwithk.
k
Remark1. 1. x = −∞satisfies (1). Thisisthetrivialsolution.
2. If row(A,i) > row(B,i) or row(A,i) < row(B,i) for some i ∈ [m], then
x = −∞istheonlysolution to(1).
3. If row(A,i) = row(B,i) for some i ∈ [m], then these two rows can be
removed,sothatmcanbedecreased tom−1.
4. If col(A,j) = col(B,j) = −∞ for some j ∈ [n], then no restriction is
imposed on x . Then these two columns and x can be removed, so that n
j j
decreases ton−1.
Wewillassumethatrow(A,i) 6= row(B,i)(assumption3),row(A,i) ≮ row(B,i)
(assumption 4) and row(A,i) ≯ row(B,i) (assumption 5), for all i ∈ [m], and
col(A,j) = col(B,j) = −∞(assumption 6),fornoj ∈[n],inthefollowing.
Thesetsinthenextdefinition aredenoted I,J,K,L in[4].
Definition2. Foreachi∈ [m],let
1. WA(i) = {j : a > b },WB(i)= {j :a < b }.
ij ij ij ij
2. E(i) = {j : a = b 6= −∞},F(i) = {j : a = b =−∞}.
ij ij ij ij
3. win(i) = (WA(i)×WB(i))∪(E(i)×E(i)) ⊂ [n]×[n]. Eachelementof
win(i)iscalledawinningpair.
Foreachi ∈ [m],WA(i)∪WB(i)∪E(i)∪F(i) = [n]isadisjoint union. By
assumption3,E(i)∪F(i) 6= [n]. Byassumptions4and5,WA(i) 6=[n] 6= WB(i).
Moreover,∩m F(h) = ∅,byassumption 6.
h=1
Example3. Given
3 7 −1 −∞ −∞ −∞ −∞ 8
A= 6 7 −∞ −∞ , B = −∞ −∞ 5 1 ,
   
1 0 1 −∞ 1 0 1 2
   
6
weget
3 7 −1 8
M = 6 7 5 1 .
 
1 0 1 2
 
Then WA(1) = {1,2,3}, WB(1) = {4}, WA(2) = {1,2}, WB(2) = {3,4},
WB(3) = {4} and E(3) = {1,2,3}. Therefore, win(1) = {(1,4),(2,4),(3,4)},
win(2) = {(1,3),(1,4),(2,3),(2,4)}, win(3) = {(1,1),(2,2),(3,3)}.
Remark4. Supposethatwin(i) = ∅. Thisisequivalenttoeitherrow(A,i) = −∞
or row(B,i) = −∞, but not both, by assumption 3. Say row(A,i) = −∞ and
b 6= −∞forsomej ∈ [n]. IfA⊙x = B⊙x = y,theny = −∞andx = −∞.
ij i j
Withthisinformation inmind,wecanremovethei–throws,sothatmdecreases to
m−1. Therefore,wewillassumethatrow(A,i) 6= −∞androw(B,i) 6= −∞,for
alli∈ [m](assumption 7). Thenwin(i) 6= ∅,foralli∈ [m].
Non–trivial solutions to (1) arise from winning pairs. Let us see how. Recall
thatM = A⊕B = (m ) ∈ M (T). M mightnotbereal(seeexample9).
ij m×n
Definition 5. Consider i ∈ [m] and I ∈ win(i). Let x ∈ Tn, y ∈ Tm be any
vectorssatisfying A⊙x = B ⊙x = y (inparticular,
row(A,i)⊙x = row(B,i)⊙x = row(M,i)⊙x = y ).
i
Wesaythat thesolution xto(1)arisesfromI if
m +x ≤ y , (3)
ij j i
forallj ∈ [n]\F(i),withequalityforallj ∈ |I|.
Noticethatj ∈ F(i)ifandonlyifm = −∞and,insuchacase,theinequality
ij
(3)istautological. Otherwise,m isreal.
ij
Suppose that x arises from I and write |I| = {i ,i }. Then, equality for all
1 2
j ∈ |I|means
m +x = y = m +x ,
ii1 i1 i ii2 i2
whenceweobtainonebivariate linearequation
x = dif(M;i ,i ) +x . (4)
i2 1 2 i i1
Notice that dif(M;i ,i ) is real. In addition, (3) amounts to, at most, 2n − 2
1 2 i
bivariatelinearinequalitiesinthex ’s. Noticethat(4)istautological,wheni = i .
j 1 2
Example 3. (Continued) Take i = 1, I = (1,4) ∈ win(1) and suppose that a
solution xto(1)arisesfromI. Thismeans
3+x = 8+x
1 4
7
(onebivariate equation) and
7+x ≤ y , −1+x ≤ y ,
2 1 3 1
where 3+x = y . Replacing y by its value, we obtain two additional bivariate
1 1 1
inequalities. Altogether, (3)becomes, innormalform,
x −x −5 = 0, (5)
1 4
−x +x +4 ≤ 0, (6)
1 2
−x +x −4 ≤ 0. (7)
1 3
Remark 6. Suppose i,k ∈ [m], i < k, I ∈ win(i),K ∈ win(k). Assume that the
solution xarisesfromI andfromK. Thenforalli∈ |I|andk ∈ |K|,wehave
m +x = y , (8)
ii i i
m +x ≤ y , (9)
ik k i
m +x = y , (10)
kk k k
m +x ≤ y . (11)
ki i k
Addingup,
m +m +x +x ≤ m +m +x +x = y +y ,
ik ki i k ii kk i k i k
whence
m +m ≤ m +m . (12)
ik ki ii kk
Inotherwords,thevalueofthe2×2tropicalminorofM,denoted M(i,k;i,k),
m m
ii ik = max{m +m , m +m }, (13)
ii kk ik ki
m m
(cid:12) ki kk (cid:12)trop
(cid:12) (cid:12)
(cid:12) (cid:12)
isattained atth(cid:12)emaindiagon(cid:12)al. OnemorewaytowritethisoverTis
dif(M;i,k) ≤ dif(M;i,k) . (14)
k i
Thisremarkleadstothefollowingkeydefinition.
Definition 7. Consider i,k ∈ [m], i < k, I ∈ win(i),K ∈ win(k). We say that
K is compatible with I if the value of (13) is attained at the main diagonal, for all
i ∈ |I| and all k ∈ |K|. Equivalently, if (14) holds in T, for all i ∈ |I| and all
k ∈ |K|.
8
Example3. (Continued)Inourrunningexample,takei= 1,k = 2,I = (1,4)and
K = (1,3). ThenK iscompatible withI,sinceeachtropical minor
3 3 3 −1 8 3 8 −1
= 9, = 8, = 14, = 13
6 6 6 5 1 6 1 5
(cid:12) (cid:12)trop (cid:12) (cid:12)trop (cid:12) (cid:12)trop (cid:12) (cid:12)trop
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
at(cid:12)tains its(cid:12)value at t(cid:12)he main d(cid:12)iagonal. T(cid:12)he ineq(cid:12)ualities (14(cid:12)) are 0 ≤(cid:12) 0, 1 ≤ 4,
−5 ≤ 5 and −4 ≤ 9 in this case. However, K = (1,4) is not compatible with I,
sincethetropical minor
3 8
= 14
6 1
(cid:12) (cid:12)trop
(cid:12) (cid:12)
doesnotattainitsvalueatthem(cid:12)aindiag(cid:12)onal.
(cid:12) (cid:12)
Remark 8. Suppose i,k ∈ [m], i < k, I ∈ win(i),K ∈ win(k), K compatible
with I. If i ∈ |I| and k ∈ |K| are fixed, then dif(M,i,k) is decreasing on the
subscripts, bycompatibility. Therefore
[dif(M;i,k) ,dif(M;i,k) ] (15)
k i
is a non–empty closed interval, denoted int(M;i,k;i,k). It may degenerate to
[−∞,a],[a,+∞],[−∞,+∞]or[a,a]forsomereala.
Example9. Given
1 −∞ −∞ −∞ 1 −∞
A = , B = ,
a a 0 b b 0
21 22 21 22
(cid:20) (cid:21) (cid:20) (cid:21)
weget
1 1 −∞
M = ,
m m 0
21 22
(cid:20) (cid:21)
somea ,a ,b ,b ,m andm ∈ T. Takei = 1, I = (1,2) ∈ win(1), k = 2
21 22 21 22 21 22
and K = (3,3) ∈ win(2). Then K is compatible with I, for any m ,m since
21 22
m = −∞. Thentheinterval(15)equals[m ,+∞],fori= 1andk = 3.
13 21
Definition 10. An interval relation is an expression x ∈ [a,b]+x , where a ≤ b
i k
in T and i,k ∈ [n]. Equivalently, an interval relation is a pair of concatenated
bivariate linearinequalities x +a ≤ x ≤ x +b.
k i k
Expressions (8)–(11) imply that the following, at most, four interval relations
mustbetrue:
x ∈ int(M;i,k;i,k)+x , i∈ |I|, k ∈|K|. (16)
k i
Noticethat(16)istautological wheni = k.
9
Example3. (Continued)Inourrunningexample,takei = 1,k = 2andI = (1,4).
The winning pair K = (1,3) is compatible with I and this gives rise to the non–
emptyclosedintervals
int(M;1,2;1,1) = [0,0], int(M;1,2;1,3) = [1,4],
int(M;1,2;4,1) = [−5,5], int(M;1,2;4,3) = [−4,9]
andtheintervalrelations
x ∈ [1,4]+x , x ∈[−5,5]+x , x ∈ [−4,9]+x .
3 1 1 4 3 4
Equivalently, wecanwrite
x +1≤ x ≤x +4,
1 3 1
x −5≤ x ≤x +5,
4 1 4
x −4≤ x ≤x +9.
4 3 4
Where does, say, the interval relation x ∈ [−4,9]+x come from? Weknow
3 4
thatI = (1,4) ∈ win(1)translate into(5)-(7)and,similarly, K = (1,3) ∈ win(2)
translate into
x −x +1 = 0, (17)
1 3
−x +x +1 ≤ 0, (18)
1 2
−x +x −5 ≤ 0. (19)
1 4
Thesesixexpressions implytheconcatenated inequalities −4+x ≤ x ≤ 9+x .
4 3 4
8 −1
ThisispossiblebycompatibilityofKwithI,sincethetropicalminor =
1 5
(cid:12) (cid:12)trop
13attainsitsvalueatthemaindiagonal. (cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
Four tropical minors of the maximum matrix M must be checked out, in order
to decide compatibility of K with I. We can forget repeated minors, keeping just
oneofthem. Anyminorwithrepeatedcolumnswillbecalledtrivial. Trivialminors
will be disregarded; they will play no role. A minor is tropically singular if it
attains its value at both diagonals; otherwise the minor is tropically regular. Of
course, M(i,k;i,i) is tropically singular. Now, if i 6= k, the minor M(i,k;i,k)
is tropically singular if and only if the interval relation (16) reduces to the non–
tautological bivariateequation
x = dif(M;i,k) +x (= dif(M;i,k) +x ). (20)
k k i i i
Summingup,if1 ≤ i< k ≤ m,thentheconditions |I| = {i ,i },I ∈ win(i),
1 2
|K| = {k ,k }, K ∈ win(k) and K compatible with I provide, at most, two
1 2
10