Property of Polytopes: A Revision 3 Zaifu Ya ng y Octob er, 19 94 3
This r ese arch is part of the V F-p rogr am "Comp et ition and Co op er at ion". The pap er is a
re visionofap ap e runde r thes ame titlein 1994.
y
spa ce R , the question is to determinew hether P contains a n integ ral
p o intor not. Wepropose asimpliciala lg orithmto a nswerthis questio n
bas ed o n a sp ecic integ er labelin g rule and a speci c triang ula tio n
of R n
. Starting fro m an arbitrary integ ral point of R n
, the algo rithm
terminates within anite number of s tepsw itheith er an integ ralp o int
in P o rproving there is n o integral p o int in P. One pro minent feature
ofthealgo rithmisthatthe structureofthealgorithmisverys implea nd
it ca n be easily implemented o n a computer. Moreover, the algo rithm
is computationally very simple,exibleand stable.
Ke yw ords: Po lytope, integ ral p oint, simplicial method, integer linea r
1 Int roductio n
Given a polytop e P, fo r example, th e co nvex hull of n +1 anely indep endent
vectors of R n
, the questio n is to determin e whether P contains an integral p oint
or no t. We develo p a s implicia l a lg orith m to solve the problem. The algorithm
is ba sed on a sp ecic integ er la b eling rule and th e well- know n K
1
- trian gulation o f
R n
. The main feature of th ea lg orithmca nb e described as fo llows: The algorithm
subdivides R n
into n-dimensiona l simplices s uch that a ll integ ral p oints of R n
are
vertices of the tria ngulation, a nd then assigns an integer to each integ ral p oint
of R n
a cco rding to the labelin g rule. Sta rting from an a rbitra ry integral point,
the algo rithmg enerates a s equence ofa djacent simpliceso f varying d imens io n a nd
terminatesw itheither the YESor (exclusively) NO ans wer w ithin a nite numb er
of steps. In th e YES ca se, the algorithm n ds an integra l p o int in P. T he NO
ans wer s hows th at there is no integral point inP.
Our work was motivated by the wo rks of Sca rf [11 ] a nd of Dan g and va n
Maaren[1]. However,Scarf's alg orithm is based o nprimitive sets. Altho ughDang
andvanMaa ren'sa lg orith misa lsobas edo nsimplices,itdo esn otgu aranteeth atthe
p o lytopehas no integ ra lp oint if n ointegra l p ointca nbefo undby theira lg orith m.
Thealgo rithma nd thelab eling rule inthis pa p erareverydierent fromDan ga nd
van Ma aren theirs. We would also like to p o int o ut that o ur algorithmco uld date
back to the wo rk of va n der La an and Talma n [5 ], a ltho ugh th eir algo rithm was
introducedto compute a xed p o int of acontinuo us function.
The rema inder of the paper is su mmarized next. In Section 2 the lab eling
rule and basic theorems are intro d uced. Section 3 gives a full des criptio n o f the
algo rithmincasethep o lytopeisafull- dimensionalsimplexinsomestan dardfo rm.
InSection4wesha lldemonstra tehowtotransfo rmthepro blemofanarbitrary
full-dimensiona l simplex into the sta ndard fo rm. In Section 5 we ap ply the algorithm
p o lytopes . Co ncludingremarksa re fo und inSectio n7.
2 Int eg er labeling Rule
The p roblem in this sectio n is to tes t the integral prop erty of an n-d imen sio nal
simplex P given by P =fx2R n jAxbg; where a i > = (a i1 ;:::;a in
) is th e i- th row of the n+1 by n matrix A for i =1 , 111,
n+1, and b = (b 1 ;111;b n+1 ) > is a vector o f R n+ 1
. Withou t lo ss of g en era lity we
sha llas sumeth rougho utthepa p erthata
1
, 111, a
n +1
areinteg ralvecto rso f R n ,a nd b= (b 1 ;111;b n +1 ) > is a n integ ra l vecto r of R n+ 1
. No tice that s ince the simplex P
is full-dimensiona l, th e o rig in o f R n
is co nta in ed in the interio r of the convex hull
of the vecto rs a 1 , 111, a n+ 1 . As usua l, Z n
deno tes the set o f all integra l p o ints in
R n
. LetN deno te the set f1;:::;n+1ga nd N
0i
the setN without the indexi,fo r
i2N. Now we introduce the fo llowing labeling rule.
Lab eling Rule: Tox2Z n
the la b ell(x)=i is as sig ned if iis thesmallest index
forw hich a > i x0b i =ma xfa > j x0b j ja > j x0b j >0;j 2Ng: If a > i xb i
for a lli=1 ,...,n+1, then the lab ell (x)=0 is assigned to x.
Notice tha t if l (x) = 0, then P co nta ins at least one integra l point. Let T
b e the K
1
-triang ula tio n o f R n
to b e des crib ed in the next s ectio n. Th is simplicial
subdivisio nof R n
is suchthat the collection ofthe verticeso f simplicesin T is the
set of all integ ral points of R n
. We deno te a simplex with vertices x 1 ,...,x n+ 1 by (x 1 ;:::;x n+ 1
). Givenan n-dimensiona l s implex (x 1 ;:::;x n +1 ) in T, let L()=fl(x 1 );:::;l (x n +1 )g:
Ann-simplex iscalledacompletelylabeledsimplexifjL( )j=n+1 . Sp ecically,
ann-s implex iscalledacompletelyla b eledsimplexoftypeI ifL()=f0g S
N
0 i
foran ind ex i2N. Wh ereas an n- simplex iscalled aco mpletely labeledsimplex
oftype II if L( )=N. Obs ervetha t acompletely la b eledsimplex of typeI has a
vertex b eing an integralpoint inP.
Now we sta te o urba sic results.
Theorem 2.1 The Labe ling Ru le resul ts in at least on e compl ete ly l abe led
sim-ple x.
Proof: It can b e derived by inductio n. We o mitthe details. 2
Furthermo re,on eca n derive th e fo llowing s harper and more importa nt resu lts.
Theorem 2.2 If P doe s not contain any in tegral point , then t he Label ing Rule
resul ts in a uniqu e comple te ly l abe led simpl ex.
Clea rly, the unique co mpletely labeleds implex must b e of typ e II. A proof of the
abovetheoremisdeferredtoS ectio n4 . T histheo remcanb eseenas ag en era liza tio n
ofthe fo llowing lemma (s ee va nder L aan [4 ]a nd Ta lman [1 4]).
Lemma 2.3 Choose an arbitrary poin t c 2 R n
. We assign x 2 Z n
with the label
l(x)=i if i is the small est in dex for whic h
x i 0c i =ma xfx j 0c j jx j 0c j >0; j 2N g: If x i c i
for all i=1, 111, n, w e assign x with the label l(x)=n+1 . Then there
ex istsa u niqu e comple tel y labe led simple x.
Theorem 2.4 If P has an integ ral point, the n the Label ing Rul e resul ts in at
Proof: T he rst pa rt ca n be derived by ind uction. The second part follows from
the same line o f the proof o f Theorem2:2. 2
We p oint o ut that all theorems a b ove will b e co nstructively demonstra ted by
the algo rithmto b e presented inthe next sectio n. Let usgiveso meexamples.
Example1 . We are g iven
P =fx2R 2 ja i > xb i ;i =1;2;3g where a 1 = (3 ;2 ) > , a 2 = (1 ;01 ) > and a 3 = (03;01) > , b 1 = 1, b 2 = 01 a nd b 3
=1. Thisexampleis shown inFigure1wh erethere arethreecompletely labeled
simplices. One o f themis of type II. The o ther two are of typ e I.
Example2 . We are g iven
P =fx2R 2 ja i > xb i ;i=1 ;2;3 g where a 1 =(2;01) > , a 2 = (3 ;1) > an d a 3 =(03 ;0) > , b 1 =1, b 2 = 2 a nd b 3 =01.
This example is illus trated inFig ure 2 where there is a un ique completely labeled
simplex oftyp e II.
3 The algorithm
In th is sectio n we sha ll dis cus s how to o p erate the a lg orith m to nd a co mp letely
la b eledsimplexw ithinanitenumb erofsteps . Inthe restofthesection wea ssume
that the simplex P ass o ciated with matrixA is g iven such that
a . a (n +1)j 0 forj =1, 111,n; b. a ii >0for i=1,111, n; c. a ij 0 and ja ij j<a ii for i6=j, i, j =1 , 111,n .
Such a fo rmulationo f the polyto p eP is referredto as the stand ard fo rm. Observe
thatthestanda rdformisra thersimila rtothewell- kn ownHermitenormalform(see
e.g., Section 6). In the next s ectio n we sh allshow th atany n-dimensiona l simplex
P ca nberestructuredinto thesta ndard form. We rstderivethe follow inglemma.
Lemma 3.1 Le t a simple x P be giv en in stan dard form. If P con tains tw o
integ ral points x 1
and x 2
, it also con tains t he in tegral poin t
x=(ma xfx 1 1 ;x 2 1 g;111;maxfx 1 n ;x 2 n g) > :
Proof: SinceP conta instwointegra lpointsx 1 andx 2 ,itimpliesthatAx 1 ba nd Ax 2 b. Notice tha t n X j=1 a (n+1)j x 1 j b n+ 1 ; and n X j=1 a (n+1)j x 2 j b n+ 1 : Sincea (n+1)j
0for j =1 , 111,n, itfollow s tha t
n X j=1 a (n+1)j maxfx 1 j ;x 2 j gb n+ 1 ; i.e., P n j=1 a (n+ 1)j x j b n +1
. Moreover,for h=1,2 , itho ldsthat
n X j=1 a ij x h j b i ;i=1;111;n: Sincea ij 0for j 6=i, it is ea sy to s ee tha t a ii x h i b i 0 P j6= i a ij x h j b i 0 P j6= i a ij maxfx 1 j ;x 2 j g = b i 0 P j6= i a ij x j
fori=1,...,n. It mea nstha t
a ii x i b i 0 X a ij x j
fori=1,..., n. Hence n X j=1 a ij x j b i
fori=1,...,n. In summary,we have Axb. 2
Now we introduce the K
1
-tria ngulation of R n
(s ee [4 , 14 , 15 ]) which underlies
the algo rithm. We dene aset ofn+1vectors of R n by q (i)=0e(i);i=1;:::;n and q (n+1)= n X i=1 e(i);
where e(i) denotes the i-th unit vector of R n
, i = 1,..., n. For a g iven integer t,
0 t n, a t-dimensiona l simplex or t- simplex, denoted by , is d ened as the
convex hull o f t+1 anely indep endent vectors x 1 , 111, x t+1 of Z n . We usua lly write = (x 1 ;111;x t+ 1 ) a nd callx 1 , 111, x t+ 1
the vertices of . A (t01)-simplex
b eing the co nvex hull of t vertices o f (x 1 ;111;x t+1 ) is said to b e a facet of . If x 1 2 Z n
and = ((1);111;(n)) is a p ermuta tio n of the elements o f the set
f1;2 ;:::;ng,then denote by (x 1
;) the n -simplexw ithverticesx 1 ,..., x n +1 where x i+1 =x i +e((i))fo rea chi=1,...,n. The K 1
-triang ula tio nofR n
isthe collectio n
ofa ll suchsimplices.
L et v be an integra l p oint o f R n
. The point v will b e th e starting p o int o f the
algo rithm. Dene for T being apro p er sub set of N the reg io ns A(T) by
A(T)=fx2R n jx=v+ X j2T j q (j); j 0 ;j 2T g:
No tice that the dimensio n of A(T) equa ls t with t = jTj. The K
1
-triang ula tio n
subdivides any s et A(T) into t-simplices (x 1 ; (T)) with vertices x 1 , 111, x t+1 , where x 1
is a vertex in A(T) , (T) = ((1);111;(t)) is a p ermuta tio n o f the
elements of the setT, an d x i+1
=x i
+q( (i)), i=1,111, t. Fo ra pro p er s ubset T
of carry a ll la b els o f the set T. Note tha t every vertex y as a zero-d imen sio nal
simplex fyg is fl (y )g- co mp lete in ca sel(y)6=0.
Now the algorithmgenera tes asequence of a djacentt-simp lices inA(T)having
T-co mplete common facets. Formally the steps of the algo rithm can be described
as follows . Step 0 . Set t = 0 , x 1 = v , T = ;, (T) = ;, = fx 1 g, x = x 1 , R i = 0, i = 1, ..., n+1,a nd b=1 .
Step 1 . Calcula te l (x) a nd set L= l(x). If L=0, an integra l point is found a nd the
algo rithm termina tes . If L is no t a n element of T, go to Step 3. Otherwise
L=l(x s
) fo r exa ctly on evertex x s
6=xof .
Step 2 . If s = t+1 and R
(t)
= 0, go to Step 4. Oth erwise and R are ada pted
according to Table 1 by replacing x s
. Set b=b+1 . Return to Step 1 with x
equa l to the new vertex of .
Step 3 . If t =n , a co mpletely la b eleds implex of typ e II is found a nd the algorithm
terminates. Otherwis e, a (T S
fLg)- co mplete s implex is fou nd and T
be-co mes T S
fLg, (T) b ecomes ((1);:::;(t);L), beco mes (x 1
;(T)), a nd
t b eco mes t+1 . Set b=b+1 . Return to Step 1 with xequal to x t+1
.
Step 4 . Let,forsomek,k t,x k
bethevertex of withlabel(t). ThenT b ecomes
Tnf(t)g, (T)becomes( (1 );:::; (t0 1 )),beco mes (x 1
; (T)),tb ecomes
t01 ,a nd return to Step 2 with s=k and b=b+1.
In Ta ble 1 the vecto r E(i)denotes the i-th unit vecto rof R n+ 1
, i2N.
Table 1. Pivo t rules if the vertexx s
o f (x 1
x 1
b ecomes (T) b eco mes R beco mes
s=1 x 1 +q ( (1 )) ( (2);:::; (t);(1)) R+E((1)) 1<s<t+1 x 1 ( (1);:::; (s); (s01 );:::; (t)) R s=t+1 x 1 0q( (t)) ( (t);(1);:::;(t01 )) R0E( (t))
With out loss of g enera lity we may assume that the alg orithm is initia ted at a n
in feas ible integra l p o int v. No tice that every simplex (x 1
; (T)) generated by
the algo rithmlies in A(T) and is a t-simplex of the simplicial su b division o f A(T)
in duced by the K
1
- tria ngulation of R n
. Now in order to prove the convergence o f
the a lgo rithm, we need to borrow some no tio ns fro m gra ph theory. Firs t, let us
denea g raph co nsisting o f nodes and edg es,deno ted by 0. We s ay a s implex is
ano de if an d onlyif it s atises on eof the following conditio ns:
(1) =fvg;
(2) isa t-simplex inA(T) forso me propers ubset T of N w ith t=jTj1 a nd
at least o nefacet of is T-co mp lete.
We say two nodes
1
an d
2
in the gra ph 0 a re a djacent a nd therefo re con nected
by an edge if and o nly if b o th
1
and
2
are in A(T) fo r s ome pro p er s ubset T o f
N, and one o f the following ca sesoccurs :
(1)
1 and
2
are both t-s implices and share acommo n T- co mplete facet;
(2) either 1 isaT-completefa cetof 2 and 2 isat-simp lexo r 2 isaT-complete facet of 1 and 1 is at-simp lex.
Observethats incethea b overelationshipissymmetric,theedgesa ren otnecess arily
ordered. Finally, we dene the deg ree of a node in the grap h by the number o f
nodes b eing co nnected by an ed ge to , denoted by deg(). By ado pting the
Lemma 3.2 Let be a node in the g raph 0. The n
(1) deg( )=1 when =fvg;
(2) deg( ) is eit her zero or on e when is a compl ete ly l abel ed simplex ;
(3) deg( ) is eit her one or tw o in all ot he r cases.
L emma3.2impliesthatthe sequenceofa djacents implicesofva rying dimens io n
starting fro mthe 0-dimens ion al s implex fvg g enerated by the algo rithmmay lea d
to aco mpletely la b eled simplex, o r may terminate w ith an integral point in P, o r
may go to innity. Wewill prove that the latter case can b eexcluded.
AsnormweusetheEuclidea nnorminR n
. Wenowdeneano p enballo fra dius
centered at v by
B()=fx2R n
jjjx0v jj g:
We have the fo llowing lemma.
Lemma 3.3 For any prope r sub set T of N, the re is no T-comple te (t 0
1)-simpl ex in A(T)nB() provided that is c hosen to be a su cien tly l arge n umber.
Proof: It is astra ig htfo rwa rdcons equenceof the fa ct that w hen opera ting inR n
,
thea lg orithma lwaysmovesinto thedirectio ninwhichfo rs omej 2N,thefunctio n
a >
j x0b
j
is strictly decreas ing b ecause o f q (i) >
a
i
<0fo r alli2N. 2
Nowitiseasytoo btainthefo llowingresultbynoticingth atthenumb erofnodes
in th e g raph 0 is nite and the algo rithm ca n never return to a node previously
visited.
Lemma 3.4 Letan n- simpl ex P be giv en inst andardform. Then the al gorithm
will terminate w ith eithe r an integ ral poin t in P or a compl ete ly labe led simplex of
The g eometric co ntext o f the next theorem can b e easily understoo d in two
dimension.
Theorem 3.5 Ex cl usion Theore m
Le t a simple x P be g ive n in standard form w it h t he addit ional condition that
P j6=i ja ij j < a ii
hol ds for all i = 1 , 111, n. The n the Label ing Rule precl udes the
possibil ity of t he coex isten ce of a comple tel y labe led simpl ex of type I an d a
com-ple tel yl abe led simpl ex of type II. If P contains an int eg ral poin t, then the re ex ists
no compl ete ly l abeled simpl ex of type II.
Proof: We o nly n eed to consider the case in w hich P conta ins an integ ral point,
say x 0
, i.e., Ax 0
b. L et us suppose to the contrary that there is a co mp letely
la b eled simp lex o f typ e II, s ay (x 1 ; ) with vertices x 1 , 111, x n+1 , where =
( (1 );111; (n+1 )) is ap ermutation of the n+1 elments ofN,a nd
x i+1 =x i +q ( (i));i=1 ;111;n; x 1 =x n+ 1 +q ( (n+1)):
Nowitiseasytos eethatthereexis tno nnegativeinteg ersk 1 1 ,111,k 1 n+ 1 suchtha t x 1 =x 0 + X i2N k 1 i q(i); and min h 2N k 1 h =0 : Let l =ar gmin f 0 1 (h)jk 1 h =max j2N k 1 j g:
Thenthere exist no nnegativeinteg ersk i 1 ,111, k i n+ 1 such that x i =x 0 + X j2N k i j q (j)
fori=2, 111, n+1. No tice that k i l =ma x j2N k i j fora ny i2N.
The fo llowin gcas es needto be addressed:
(1). If 1l n, we have tha t for any i2N, itho lds
a > l x i 0b l = a > l x 0 0b l + P n+ 1 h= 1 k i h a > l q (h) 0k i l a ll 0 P n h= 1;h 6=l k i h a lh +k i n+1 P n h =1 a lh 0k i l a ll 0 P n h= 1;h 6=l k i l a lh +k i n+ 1 P n h=1 a lh (k i n +1 0k i l ) P n h= 1 a lh 0: It impliestha t l (x i )6=l fori=1, 111, n+1.
(2). If l =n+1,wehave tha t fora ny i2N, it holds
a > n+ 1 x i 0b n+ 1 = a > n+1 x 0 0b n +1 + P n+ 1 h= 1 k i h a > n+ 1 q (h) P n+ 1 h= 1 k i h a > n +1 q (h) 0 P n h= 1 k i h a (n+ 1)h +k i n+ 1 P n h=1 a (n+ 1)h 0 P n h= 1 k i n+ 1 a (n+ 1)h +k i n+ 1 P n h=1 a (n+ 1)h (k i n +1 0k i n +1 ) P n h= 1 a (n+ 1)h = 0: It impliestha t l (x i
)6=n+1 for a lli=1 ,111, n+1 . We conclud efro m (1) a nd
(2) w ith acontradictio n. Weare do ne. 2
Moreover, itis ea sy to derive the follow ing lemma .
Lemma 3.6 Let a simpl ex P be give n in standard form with the additional
condition that P j6=i ja ij j < a ii
holds for all i =1, 111, n. If P con tains an in tegral
Proof: C onsiderthe simplest ca sein whichP contains a singleintegra l point, say
w . The g en era l ca se can b e sh own in a s imilar way. Choo se w to be the sta rting
p o int since it is allowed. Take an arbitrary ind ex k fro m N a nd set l (w ) = k
articially. Then the algo rithm w ill terminate with a completely la b eled simplex,
say
k
, withina nitenumbero f steps. Ifw isa vertex of
k
,th enrestore the true
la b el of w, i.e., l(w) = 0 . Hence
k
is of typ e I. Otherwis e, it mu st be of typ e
II. However, th is can no t happen a ccording to Theorem 3.5. Hence rep eat the
procedure overa ll indiceso f N. Wecompletethe proof. 2
It is n ot clear w hether the Exclusion T heo rem a lso holds fo r the pro blems in
stan dard form without the additio nal co ndition . In o rder to provide a complete
ans wer intha t case, let usdene for k2N a s ubset C
k o f Z n by C k =fx2Z n ja j > x>b j fo r j 2N 0k g:
Now we establis h the follow ing procedure.
Step (1) Set k=1 .
Step (2) Chooseanystartingp o intv k
2C
k
a ndimp lementthea lg orith m. Ifaninteg ral
p o intin P is fo und, then sto p. Otherwis e, k beco mesk+1.
Step (3) If k =n+2 , then stop. Otherwise, go to S tep (2 ).
We still have to discuss how to o btain astarting p oint v k 2C k for each k2N. For eachk 2N, we dene q(k)=0q(k):
We ado ptthe follow ing lab eling rule.
Modied La b eling Rule: To x2Z n
,weass ig nx with the lab el
l (x)=minfj 2N 0 k ja > x0b j =minfa > x0b h ;h2N 0k ja > x0b h 0gg:
If a > h x0b h >0 fora ll h2N 0k ,then the la b el l(x)=0is a ssignedto x.
Nowweca napplythe algorithmbyu sing
l (:)a nd q (:) insteadofl(:)and q (:). It
is easy to verifytha t the alg orithm will n da n integra l p o int in C
k
w ithin a n ite
numb ero f steps. Wea re ledto the following res ult.
Theorem 3.7 Let a simple x P be g ive n in st an dard form. The procedure te
r-minate switheithe ran int egral point in P oracomple tel ylabe led simple xof type II
whic hshow sthatthereisnoin tegral poin tinP, w ithinan itenumberof iterat ion s.
Aproofo fthe abovetheoremwillb eg iveninthe nexts ection. Letusillus tratethe
algo rithmby some exa mp les.
Example3 . The polyto p e isg iven by
P =fx2R 2 ja > i xb i ;i=1;2;3 g; where a 1 = (2 ;01 ) > , a 2 = (01 ;3) > , a nd a 3 = (01;01) > , b 1 = 1, b 2 = 01, and b 3
= 1 . T he pa ths genera ted by the algo rithm lead fro m v 1 = (4;04 ) > a nd v 2 = (4 ;4) > to th e integra l point (0;01) >
in P, resp ectively, a nd are shown in
Fig ure 3.
Example4 . The polyto p e isg iven by
P =fx2R 2 ja > i xb i ;i=1;2;3 g; where a 1 = (5;01) > , a 2 = (0 ;1) > , and a 3 = (03;0) > , b 1 = 1, b 2 = 2, and b 3 =
01. The path generated by the a lg orithm lea ds from v = (4 ;04 ) >
to the u nique
completely lab eled simplex of typeII and is demo nstrated in Figure4.
4 Refo rmulation
Inorderto co nformtoth es tanda rdform,letuscomeb ack to theoriginalproblem.
We are g iven a nn- dimensio nal simplex
wherea > i =(a i1 ;:::;a in
) is th ei-th rowo f A for i2N, and b=(b
1 ;:::;b n +1 ) > .
Alineartra nsformationU iscalledaunimo dulartransfo rma tionifU isbijective
on Z n
. U is unimodula r if and only if the entries of U are integ ral a nd the
deter-mina nt of U is equal to 1 o r 01. Let
A b e th e product AU. If U is unimodular,
it is rea dily seen that g iven an integ ral p o int x a nd y =Ux, Ay b if a nd o nly if
Axb.
One may won der how to g et a unimodular ma trix. Clea rly, the identity I
n is
unimodular. In order to obta in ano n-trivialunimodularmatrix, however,weneed
to study so me other examples a nd their e ect when they postmultiply a matrix A
and premultiplya vector x:
(i) Intercha nge: U isequalto I
n
except thatthe k-thcolumnofU ise(l )a nd the
l- th column o f U ise(k). This tra nsformation switches co lumnsk a nd l o f A
and switches the k- th and l- th comp on entso f x.
(ii) Revers alo f sign: U is equal to I
n
except that the k-th column of U is equal
to 0e(k). U chang es the sign of the entries of th e k-th column of A a nd o f
the k-th component ofx.
(iii) Add ition: U isequa ltoI
n
excepttha tthe(k;l )-thentryofU,k6=l,isequalto
one. Thistra nsformationrepla cesth el -thcolumnofAbythesumofcolumns
k an d l and replaces the k-th comp o nent of x by the sum of comp o nents k
and l.
The following result ca n be found inNewma n[9].
Theorem 4.1 Eve ry u nimodul ar matrix can be ex pressed as a nite product of
unimodular matrice s of t ype (i), (ii) and (iii).
Thenextba sicres ultsaystha teverysimplexcanbebro ughtintothe stand ardform
Theorem 4.2 Tran sformation Theorem
For any g ive n n -dimension al simpl ex
P =fx2R n
jAxbg;
there e xists a unimodular matrix U suc h that
A=AU has the stan dard form.
Proof: Weshallrst provebyinductio ntha t thereisaunimodular ma trix V such
that AV = 2 6 6 6 6 6 6 6 6 6 6 6 6 4 + 0 111 0 0 0 + 111 0 0 . . . . . . . . . . . . 0 0 111 0 + 0 0 111 0 0 3 7 7 7 7 7 7 7 7 7 7 7 7 5
where"+" standsfor ap o sitiveentry,and "0"fo r azero o r neg ative entry. The
above res ult is basically equivalent to a n exampleg ivenby White[1 6] on pag e 5 1.
For completeness, we shall give a proof of it. L et us cons ider the last row of A.
By app lying revers als of sig n, we may as sume that each entry o f th e last row is
less tha n or equa l to zero. Ta ke a ny two negative entries a
(n+ 1)k a nd a (n+ 1)l such that a (n+ 1)k a (n+ 1)l
. Replace co lu mn l by co lumnl minuscolumn k. Rep ea t this
simple tra nsformation. We ca n red uce all entries of the las t row bu t o ne to zero.
Bya ninterch angewemayas sumethatthis elementisthela stentryofthelas trow.
So a fter a nite number o f steps we end w ith the last row having th e sign pattern
(0;0;111;0;0). Notice that the s ubma trix obta ined by deleting the last row a nd
the las t co lumnof A must a lso have the p rop ertytha t the zero rowvectorin R n 01
isinthe interio rofthe convexhullo f allrowso f thissubmatrix. Thenby inductio n
2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4 + 0 111 0 ? 0 + 111 0 ? . . . . . . . . . . . . 0 0 111 + ? 0 0 111 0 ? 0 0 111 0 0 3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5 :
Thesub ma trixofA,whichiso btainedbydeletingthe la stcolumnandthelas trow,
is a pro du ctive Leontief ma trix. Thus there exists a positive integ ral combina tio n
of co lumns one thro ugh n01 for which th e la st element is zero, the n-th element
is strictly nega tive an d th e oth er elements are strictly p os itive. We can therefore
trans fo rmAto thema trixB =AV bysubtractingalarg ep o sitiveinteg ralmultiple
ofthis co mbination fro mth elas t colu mn.
Nextwesha llgiveaproceduretotra nsformthematrixB =(b
ij
)toth es tanda rd
form. The proced ure is des crib ed a sfollows:
Step (a) If there are ind ices i a nd j (i 6= j) for s ome 1 i;j n w ith b
ii jb
ij j, we
ca n nd a positive integer c a nd a n integ er d 2 f0;1;:::;b
ii 01g such tha t jb ij j=cb ii
+d,wherecisthe lowerinteg erpartof jb
ij j
b
ii
,andthena ddcmultiple
of column i to co lumn j.
Step (b) Repeat Step(a )until thereareno indicesiand j (i6=j)fo r 1i;j n with
b
ii jb
ij j.
It iso bvious that the above opera tion isa unimodular trans formatio n. We still
have to demons tra te th at the pro cedureis feas ible. Reca ll that the origin of R n
is
in the interior ofthe convexhull ofthe vecto rsa
1 ,...,a
n +1
. It impliesthat there are
n+1strictlyp os itiveconvex combination coecients
1 ,..., n +1 such that n+ 1 X i a i =0 : (4 :1 )
We sha ll showthat b ii jb ij jimplies jb ji j<b jj
. System (4:1) a ls oimplies that
n +1 X i=1 i b ih =0 ;h=1 ;111;n:
It is ea sily derived that
i b ii j jb ji j and i jb ij j j b jj :
No tice that a t leas t one of the a b ove inequa lities ho lds with strict inequa lity, say,
i b ii > j jb ij
j. Mo reover, it h old s tha t
i b ii i jb ij
j. All o f this together implies
that j jb ji j< i b ii i jb ij j j b jj : It follows that jb ji j<b jj :
We a re now rea dy to prove that the new generated column j, den oted by
( b 1j ;111; b (n+ 1)j ) >
,preserves the sa me sign pattern a sbefore. Notetha t
b hj =b hj +cb hi ;h=1;111;n+1:
It is rea dily seen that
b ij =0d0 and j b ij j<b ii b hj 0 fo rall h, h6=i;j: Observe tha t i b ii i jb ij j = i (cb ii +d) j b jj = j ( b jj +cjb ji j):
It follows that j b jj i cb ii + i d0 j cjb ji j = c( i b ii 0 j jb ji j)+ i d > i d 0: Hence we have b jj >0 :
On the o ther hand, it is not d icult to see that the procedure will terminate
withinanitenumb erofiteratio nssinceea chentryb
ij
isnite. Hen cetheprocedure
produces the matrix
A = AU in stan dard form where U is a unimodular matrix.
We completethe proof. 2
We rema rk that the o rigin o f R n
is still contained in the interior of the convex
hull o f the vecto rs a
1 ,..., a
n+1
. Moreover, the co rresponding convex combina tio n
co ecientsremainuncha nged. Itisalsoea sytoshowth atthevolumeo fth esimplex
does n ot chan ge u nder unimo du lar tra nsformation. For n = 2, we ca n co nstruct
suchunimo d ula rmatrices by ad apting Sca rf's methodin [1 0]. The procedure now
can be applied.
L etu s give s ome examples.
Example5 . We are g iven
A= 2 6 6 6 6 6 4 0 01 1 1 01 0 3 7 7 7 7 7 5 : Then U = 2 6 4 1 1 01 0 3 7 5
suchthat A= 2 6 6 6 6 6 4 1 0 0 1 01 01 3 7 7 7 7 7 5 :
Example6 . We are g iven
A= 2 6 6 6 6 6 4 3 2 1 01 03 01 3 7 7 7 7 7 5 : Then U = 2 6 4 0 1 1 02 3 7 5 suchthat A= 2 6 6 6 6 6 4 2 01 01 3 01 01 3 7 7 7 7 7 5 :
Observe tha t this example is ta ken fro m Exa mp les 1 a nd 3. The matrix A co
rre-sponds to Example 1 a nd th e matrix
A to Example 3 . See Figure 3 fo r E xa mple
3 w here there are three co mpletely la b eled s implices of typ e I after having b een
trans fo rmed. In this case there is n o co mpletely labeled simplex of type II. The
readersho uld co mpa re Figu re 1 with Figure3.
Example7 . We are g iven
A= 2 6 6 6 6 6 4 2 01 3 1 03 0 3 7 7 7 7 7 5 :
Then U = 2 6 4 1 0 03 1 3 7 5 suchthat A= 2 6 6 6 6 6 4 5 01 0 1 03 0 3 7 7 7 7 7 5 : Let b 1 = 1 , b 2 = 2 a nd b 3 = 01. Then A and A corresp o nd to E xa mple 2 a nd
Example4 , respectively. SeeFig ure 2a nd Figure 4.
Example8 . We are g iven
P =fx2R 2 ja i > xb i ;i =1;2;3g where a 1 = (3 ;01 ) > , a 2 = (03;2) > and a 3 = (01;01) > , b 1 = 2, b 2 = 01 a nd b 3
= 0. See Fig ure 5 where there are three completely labeled simplices. One o f
themis of type II. T heother twoare o f typeI. We use the fo llowing u nimodula r
trans fo rma tio n
U = 2 6 4 1 0 1 1 3 7 5 suchthat A= 2 6 6 6 6 6 4 2 01 01 2 02 01 3 7 7 7 7 7 5 :
Nowitisea sy tocheck tha t the resulting p olytop egeneratesno completelylabeled
simplex o f type II. In fact the La b eling Rule results in three completely labeled
Example9 . We are g iven A= 2 6 6 6 6 6 6 6 6 4 01 01 01 3 1 1 1 3 1 1 1 3 3 7 7 7 7 7 7 7 7 5 : Then U = 2 6 6 6 6 6 4 0 1 0 0 0 1 01 01 01 3 7 7 7 7 7 5 suchthat A= 2 6 6 6 6 6 6 6 6 4 1 0 0 01 2 0 01 0 2 03 02 02 3 7 7 7 7 7 7 7 7 5 :
Nowlet usg ive ap ro o fof T heo rem2:2 . We are g ivenapolyto p e w hich has the
stan dard form. Since P co nta ins n o integral p o int, the a lgo rithmterminates with
a co mpletely la b eled simplex o f type II, s ay,
1 (x
1
; ), within a nite numb er o f
steps. Supp o seto the co ntrarythat thereis anoth er co mpletely la b eled simplex o f
tpyeII,say,
2 (y
1
;). W itho utloss o fg enerality wemayass umethat is equalto
(1;:::;n),l (x i )=ifor any i2N, a ndx i =y i
foralli2N except forso me indexk,
1k n+1 . It implies that l (x i
)=l (y i
)=i fo r all i2N. We have to consider
the follow ing ca ses:
(1). If k =1,then y 1 =x n+ 1 +q (1). Sincel(x n +1 )=n+1,wehave a > n+ 1 x n +1 0b n+1 >a > i x n+1 0b i ;i=1;111;n :
Further, we have a > n+ 1 y 1 0b n +1 = a > n +1 (x n+1 +q(1))0b n+1 a > n +1 x n+ 1 0b n+ 1 > a > 1 x n+1 0b 1 > a > 1 (x n+1 +q(1))0b 1 = a > 1 y 1 0b 1 :
It means that l(y 1 )6=1 . It is acontradictio n. (2). If 1 <k <n+1, then y k =x k 01 +q (k). Since l(x k 0 1 ) =k01, it implies that a > k 0 1 x k01 0b k01 a > k x k 0 1 0b k : Hence we have a > k 0 1 y k 0b k 0 1 = a > k 0 1 (x k 0 1 +q (k))0b k01 = a > k 0 1 x k 0 1 0b k 0 1 +a > k 0 1 q (k) a > k x k 0 1 0b k +a > k q (k) a > k (x k 0 1 +q (k))0b k = a > k y k 0b k :
It means that l(y k
)6=k. It is ag ain a co ntra diction.
(3). If k =n+1 ,then y n+ 1 =x 1 0q (n). Noticetha t a > n x 1 0b n a > n+ 1 x 1 0b n +1 : We have a > n y n +1 0b n = a > n (x 1 0q(n ))0b n a > n +1 x 1 0b n+ 1 0a > n q(n) > a > n +1 x 1 0b n+ 1 0a > n+ 1 q (n) = a > n +1 (x 1 0q(n))0b n +1 = a > y n +1 0b n+1 :
It means that l(y n+ 1
)6=n+1 . It isalso a contradiction . 2
We still haveto prove Theorem3 .7 . It suces to co nsider the case in w hichP
has an integ ral p oint x 0
, i.e., Ax 0
b. Following the sa me lineo f the p roo f of the
above theorem, we ca n s how tha t there is a t most one completely labeled simplex
of typ e II in this cas e. It isclea rweon ly need to demo nstra te that th e procedure
will produce at least twodierent completely la b eled s implices . Letus su pp os e to
the contrary that the pro cedure only generates a co mpletely la b eled s implex, say,
(x 1
;111;x t+ 1
), oftyp eII. It is ea sy to seetha t for each k 2N, the starting p oint
v k 2C k can be expressed a s v k =x 0 0 X h 2N 0k k h q(h) where k h
a repositiveinteg ersforallh 2N
0 k
. NoticethatinStep(2)the procedure
opera tes by on ly using q(h) fo r h 2 N
0k
fo r ea ch k 2 N. Hence, s ta rting from
v k 2C k ,th evertices x k j
g en era tedby the proced ure ca n b ew ritten as
x k j =x 0 + X h2N 0k kj h q (h) where kj h a re integers for a ll h 2 N 0k
. Since sta rting from the n+1 sta rting
p o ints v 1
, 111, v n +1
, the p ro cedure g enerates a unique completely lab eled simplex
(x 1
;111;x n+1
) o f type II,it implies that fo r each k 2N,
x k = x 0 + P h2N01 1j1 h q (h) = x 0 + P h2N 02 2j 2 h q (h) . . . = x 0 + P h2N 0(n+1) n +1j n+1 h q (h) (4 .2 ) where k j k h
a re integ ers for a llh 2N
0 k
. It isno t dicult to derivethat
k j k h = lj l h
fora ll k, l 2N. Nowit immediatelyfo llows from(4.2) tha t k j k h =0
for all h and k. Hence x k
=x 0
and l(x k
) =0. It is a co ntra diction. We complete
the proof. 2
Theorem2:4 is astra ig htfo rward result of the ab ove res ults .
5 Extension to general n- dimensio nal polytopes
Let M denote the set o f integers f1;111;mg. T he problem is to test the integ ral
property of a g eneral n- dimensio nal polyto p e P g iven by
P =fx2R n jAxb g; where a i > = (a i1 ;:::;a in
) is the i- th row o f the m by n ma trix A for i = 1, ..., m,
and b = (b 1 ;111;b m ) > is a vecto r o f R m
. It is clea r tha t m n+1 . As u sual we
may assu me tha t a
1 ,..., a
m
are integ ral vectors of R n , and b = (b 1 ;111;b m ) > is a n integra l vector of R m
. Fin ally we ass ume that none of the con straints a >
i
x b
i ,
i2M,isred undant,an dthatthere is asu bset J,w ithcardinalityn+1,ofM such
that fx2R n ja > i xb i ;i2Jg
isan n-dimensiona ls implex. InthesequelwetakeJ =N fors implicityofno tation.
Co mpa redw iththe la b eling ruleinSection2 ,wehavethefo llowinggeneralized
la b eling rule.
Gen eralized Labeling Rule: Ifthere isa nindexi 2M for whicha > i x0b i >0, weass ign x2Z n
with the label
l (x)=minfj 2Nja > j x0b j =ma xfa > h x0b h g g:
If a > h xb h fo r all h2 M, then l(x)=0.
As befo re,wehave the fo llowin gresults.
Theorem 5.1 If P doe s not cont ain an y integ ral point , the n the Ge ne ralized
Label ing Rul e re sults in a un ique complet ely l abel ed simpl ex. Moreove r, t he un ique
compl et ely l abel ed simplex must be of type II.
Theorem 5.2 If P contain s an in tegral poin t, then t he Ge neral ized Label ing
Rul e resul ts in at l east t wo compl ete ly l abel ed simplices. Moreove r, t here e xists at
most one comple tel y labe led simple x of type II.
Correspondingly, we ca n reformula te any n-dimens io nal p o lytope P into the
stan-dardfo rm,meaningtha tthe rst n+1 row softhema trixAsa tisfytheco nditions
(a), (b) a nd (c) as dened in Section 3 . Observe that the s tandard fo rm does no t
impose any conditio n on the cons traint vectors a
i
fo r i 2 MnN. Th is indica tes
thatifwetra nsformthepro blemsinto the s tanda rd fo rm, weon ly needto fo cuso n
the rst n+1 cons traint vecto rs an d then p o stmultip ly the rema ining co nstraint
vectors by the resulting unimodular matrixU.
Nowweca ndirectly a pply the procedure to polytop es. L et usgive s ome exa
m-ples. Example1 0. We a regiven P =fx2R 2 ja > i xb i ;i=1 ;111;5g where a 1 = (2;01) > , a 2 = (01;3) > , a 3 = (01;02) > , a 4 = (2;1 ) > , and a 5 = (02;0) > , b 1 = 1 , b 2 = 3, b 3 = 2, b 4 = 2, a nd b 5
= 3. T his example is shown in
Fig ure 7w herethere are three co mp letely lab eled simplices o f typ e I.
Example1 1. We a regiven P =fx2R 2 ja > i x b i ;i=1 ;111;5 g
where a 1 = (3;02) > , a 2 = (01;4) > , a 3 = (05;08) > , a 4 = (3;2 ) > , and a 5 = (0;04) > , b 1 =2 , b 2 = 3,b 3 =04 , b 4 = 4,and b 5
=01 . This example is illus trated
in Fig ure 8wh ere there is a u nique completely labeled simplexof type II.
Example1 2. We a regiven P =fx2R 2 ja > i x b i ;i=1 ;111;5 g where a 1 = (3;02) > , a 2 = (01;2) > , a 3 = (01;02) > , a 4 = (0;5 ) > , and a 5 = (0;05) > , b 1 =6 , b 2 =2 , b 3 = 0, b 4 = 4, and b 5 =01 . Th is example is depicted in
Fig ure 9w herethere is a uniquecompletely labeled simplex of typ e II.
Example1 3. We a regiven P =fx2R 2 ja i > xb i ;i=1 ;2;3;4g where a 1 = (2;01) > , a 2 = (01;2) > a 3 = (0 ;02 ) > , a nd a 4 = (010 ;0) > , b 1 = 4, b 2 =02, b 3 =3, and b 4
=1 1. This example is shownin Fig ure 10 w here there are
threeco mpletelyla b eledsimplices. On eofthemisof typeII. The o thertwoare o f
type I. Th ep ro cedureis shown in Figure1 0wh erev 1 2C 1 , v 2 2C 2 ,and v 3 2C 3 .
Theprocedu releads to the integ ralp o int(2 ;0 ) >
. Usingthe s ameexample,we a lso
illustrateinFigure1 1howto nd astartingpointv k 2C k . Takefo rin stancek =3 and v=(02 ;03 ) >
. T healgorithmnds an integ ralp o int inC
3
, namely,(3 ;1) >
.
We conclud ewiththe following observation.
Theorem 5.3 Giv en a polytope P in standard form, the procedure te rminate s
eithe r with an in tegral poin t in P or a complet el y label ed simple x of type II w hic h
prove s there is no in tegral poin t in P, within a nite nu mbe r of ste ps.
6 Extension to lower-dimensional polyto pes
In the previous section we as sumed th at the p o lytope P in R n
is n-d imen sio nal
and tha t n one o f th e constra ints a >
x b
i
lower-dimensiona l no nempty p o lytope in R n
we may a ssume under the same co nditions
that P can beexpressed as
P =fx2R n ja > i x b i ;i=1;111;m; and c > i x=d i ;i=1 ;111;m 1 g; whered imP =n0m 1 , for s ome m 1 , 0 m 1
n. As us ual we as sume tha t a
1 ,..., a m ,c 1 ,..., c m
1 are integralvectors of R n ,and b 1 ,111, b m , d 1 ,...,d m
1 a re integ ers . Let
A denote an m2n ma trix who se rows are a > 1 , 111, a > m , a nd let b= (b 1 ;111;b m ) > . Moreover, let C b e a n m 1
2n ma trix wh ose rows are c > 1 , 111, c > m 1 , and let d = (d 1 ;111;d m 1 ) > . We de ne aset Qby Q=fx2Z n jCx=dg:
It is clear tha t if Q is empty, then P has no integ ralp o int.
Now we ca n tra nsfo rm the polyto p e into a full-dimens ion al polyto p e in R n 0m
1
byred ucingthenumberofvaria blesinp o lyno mialtime. Todoso ,we rstintroduce
matrices inHermiten ormal form.
De nition 6.1 An m 1
2m 1
n onsing ularin tege rmatrix H is call ed in He rmite
normal form if it satise s the followin g conditions:
(1) H is low ertriangul ar and h
ij =0 for i<j; (2) h ii >0for i=1 , 111, m 1 ; (3) h ij 0 an d jh ij j<h ii for i>j.
The fo llowin gtworesults can b e fou ndin [8 ,13].
Theorem 6.2 Give n an m 1
2n matrix C with rank(C) =m 1
, t he re e xists an
n2n u nimodul armat rix U such t hat the follow ing holds:
(i) CU =(H;0 ) w he re H is an m 1
2m 1
matrix in H ermit e n ormal form and 0
(ii) H 0 1
C is an in teger matrix.
Apolynomia l-timea lg orith mforn ding U andH ca nb ealsofo undin[8 ,13 ]. Now
letH a ndU =(U
1 ;U
2
)b eas in T heo rem6.2, with U
1 an n2m 1 matrixand U 2 a n n2(n0m 1 )matrix. Theorem 6.3
(i) Q isn on empty if and on ly if H 01
d2Z m
1
.
(ii) If Q is n onempty, e ve ry point x of Q can be writ ten as
x=U 1 H 0 1 d+U 2 z ; for some z 2Z n0 m 1 :
When Q isno nempty, we have
P 0 =fy2R n j AUyb; and CUy=dg =fy2R n j AUyb; and [H;0 ]y =dg =fy2R n j y=((H 01 d) > ;z > ) > ; and AU((H 01 d) > ;z > ) > b;z 2R n0 m 1 g =fy2R n j y=((H 01 d) > ;z > ) > ; and Az b; z2R n0 m 1 g: Let P =fz 2R n 0m 1 j Az bg: Do ingsolea dstoan(n0m 1 )- dimensionalpolytop e P inR n0m 1
. Nowtheremaining
discu ssions are the same a sin th eprevious sectio n. Letus demonstra te this by a n
exa mple. Weare g iven apolyto p e
P =fx 2R 3 ja > i xb i ;i=1;2;3 ; and c > 1 x=d 1 g; where a 1 = (01;0;0 ) > , a 2 = (0;01;0) > , a 3 =(0 ;0;01) > , c 1 =(4 ;12;2 ) > , b 1 =0, b 2 =0,b 3 =1 , a nd d 1 =2. Thenwe have
U = 2 6 6 6 6 6 4 1 3 07 0 01 2 01 0 2 3 7 7 7 7 7 5 ; H =[2]; and CU =[2 0 0]: Moreover,wedecomp o se U a s U =(U 1 ;U 2 ) where U 1 = 2 6 6 6 6 6 4 1 0 01 3 7 7 7 7 7 5 and U 2 = 2 6 6 6 6 6 4 3 07 01 2 0 2 3 7 7 7 7 7 5 : No tice that y 1 =1. We g et a2 -dimensiona l polytop e P =fz2R 2 j Az bg where A= 2 6 6 6 6 6 4 03 7 1 02 0 02 3 7 7 7 7 7 5 ; and b=(1;0 ;0 ) > .
7 Concluding rema rks
Some preliminary numerica l resu lts indica te th at the algo rithm works remarkably
well. A large number of large-sca le in stances ( more than 100 ;000 variables ) has
alsobeencarriedo ut. Theinsta ncescanbeeas ilycreatedacco rding tothes tanda rd
form. Weshallrep ortnumerica lresu ltsandthecomplexitya nalys isofthealgorithm
in an ext pa p er.
Herewedo no tdiscussa ny indextheo ry. Infactthere ca nb ebuilt up an index
theory for the algorithm. The interested read er is referred to van der Laa n [4] a nd
Scarf[11 ]forins ig htfuldiscuss io ns. Fina llywecon jecturetha tthenumb ertwob o th
in Theorem2.4 a nd inTheorem5.2 ca n b ereplaced by th enumb er n+1 .
Acknowledgement
I a m extremely g rateful to Gerard va nder La an and pa rticula rly Dolf Talma n
who se in sig htful comments, s ugg estions an d discussions signica ntly improved the
paper. IwouldalsoliketothankCu rtisEaveswithw homIhaveha dveryinteres ting
conversations o n the genera l to pic of this paper wh en I was vis iting him. I a m,
however, so lely res p on sible for any rema ining erro rs. Th is res ea rch is part o f the
VF- progra m"Co mpetitio n a nd Coopera tio n".
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Delft, 1 993 , submitted to Mat hematics of Operations Researc h.
[2] B.C.Eaves ,"Homotopiesforcomputa tio nof xedp o ints" ,Mathe mat ical
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