COMPUTER PROGRAMMING
PRE-FASCICLE 3A
A DRAFT OF SECTION 7.2.1.3:
GENERATING ALL COMBINATIONS
DONALD E.KNUTH StanfordUniversity
ADDISON{WESLEY
6
77
urrent informationaboutthisbookandrelatedbooks.
Seealsohttp://www-s-faulty. sta nf or d. ed u/~ kn ut h/ sg b.h tm lforinformation
aboutTheStanfordGraphBase,inludingdownloadablesoftwarefordealingwiththe
graphsusedinmanyoftheexamplesinChapter7.
Seealsohttp://www-s-faulty. sta nf or d. ed u/~ kn ut h/ mm ixw ar e. ht ml fordown-
loadablesoftwaretosimulatetheMMIXomputer.
Copyright
2002byAddison{Wesley
Allrightsreserved.Nopartofthispubliationmaybereprodued,storedinaretrieval
system,ortransmitted,inanyform, orby anymeans,eletroni, mehanial,photo-
opying, reording, or otherwise, without the prior onsent of the publisher, exept
that the oÆial eletroni le may be used to print single opies for personal (not
ommerial)use.
Zerothprinting(revision11),31Marh2005
[TheArt of Combinations℄ has a relation
toalmost everyspeies of useful knowledge
that themind ofman an beemployed upon.
| JAMES BERNOULLI, Ars Conjetandi (1713)
This booklet ontains draft material that I'm irulating to experts in the
eld, in hopes that they an help remove its most egregious errors before too
many other people see it. I am also, however, posting it on the Internet for
ourageous and/or random readers who don't mind the risk of reading a few
pagesthathavenotyetreahedaverymaturestate. Beware: Thismaterialhas
notyetbeenproofreadasthoroughlyasthemanusriptsofVolumes1,2,and3
were at the time of theirrst printings. And those arefully-hekedvolumes,
alas,were subsequentlyfoundtoontainthousandsofmistakes.
Giventhisaveat, Ihopethatmyerrorsthistime willnotbesonumerous
and/orobtrusivethatyouwillbedisouragedfromreadingthematerialarefully.
Ididtrytomakeitbothinterestingandauthoritative,asfarasitgoes. Butthe
eldissovast,Iannothopetohavesurroundeditenoughtoorralitompletely.
ThereforeIbegyoutoletmeknowaboutanydeieniesyoudisover.
Toputthematerialinontext,thisisSetion7.2.1.3ofalong,longhapter
onombinatorialalgorithms. Chapter7willeventuallyllthreevolumes(namely
Volumes 4A, 4B, and 4C), assuming that I'm able to remain healthy. It will
begin with a short review of graph theory, with emphasis on some highlights
ofsigniantgraphsinTheStanfordGraphBase, from whihI willbedrawing
manyexamples. Then omesSetion 7.1, whih deals withthetopi of bitwise
manipulations. (Idraftedabout60pagesaboutthat subjetin1977,but those
pages need extensive revision; meanwhile I've deided to work for awhile on
the material that follows it, so that I an get a better feel for how muh to
ut.) Setion7.2isaboutgeneratingallpossibilities,anditbegins withSetion
7.2.1: Generating Basi Combinatorial Patterns|whih, in turn, begins with
Setion 7.2.1.1, \Generating all n-tuples,"and Setion 7.2.1.2, \Generatingall
permutations." (Readersof the present booklet should havealready looked at
those setions, drafts of whih are availableas Pre-Fasiles2A and 2B.) The
stage isnow setfor themain ontentsof thisbooklet, Setion7.2.1.3: \Gener-
atingallombinations." Thenwillome Setion7.2.1.4 (aboutpartitions), et.
Setion 7.2.2willdealwith baktrakingingeneral. And soit willgo on,ifall
goeswell;an outlineoftheentireChapter7 as urrently envisagedappearson
thetaopwebpagethatisitedonpageii.
Eventheapparentlylowlytopi ofombinationgeneration turnsouttobe
surprisingly rih, with ties to Setions 1.2.1, 1.2.4, 1.2.6, 2.3.2, 2.3.4.2, 3.4.2,
4.3.2, 4.6.1, 4.6.2, 5.1.2, 5.4.1, 5.4.2, 6.1, and 6.3 of the rst three volumes.
Istronglybelieveinbuildinguparmfoundation,soIhavedisussedthistopi
muh more thoroughlythanI will beableto dowith materialthat isneweror
lessbasi. To mysurprise,Iame upwith110 exerises,eventhough|believe
itornot|Ihadtoeliminatequiteabitoftheinterestingmaterialthatappears
inmyles.
Someofthethingspresentedarenew,tothebestofmyknowledge,although
Iwillnotbeatallsurprisedto learnthatmyownlittle\disoveries"havebeen
disoveredbefore. Pleaselook,forexample,attheexerisesthat I'velassedas
researhproblems(ratedwithdiÆultylevel46orhigher),namelyexerises53,
56, 67, and 83; I've also impliitly posed additional unsolved questions in the
answers to exerises 59, 63, 101,105, and 109. Are those problemsstill open?
Pleaseletmeknowifyouknowofasolutiontoanyoftheseintriguingquestions.
Andofourseifnosolutionisknowntodaybutyoudomakeprogressonanyof
theminthefuture,I hopeyou'll letmeknow.
I urgently need your help also with respet to some exerises that I made
upasIwaspreparingthismaterial. Iertainlydon'tliketogetreditforthings
thathavealreadybeenpublishedbyothers, andmostoftheseresultsarequite
natural \fruits" that were just waiting to be \pluked." Therefore please tell
meifyouknow whoI should haveredited,with respetto the ideas foundin
exerises9, 18, 19,20, 26, 27, 28, 30,31, 32, 33, 34, 35,36, 37, 41, 42, 43, 44,
45,48,51,59,62,63,64,65,66,69,79,82(b{f),85,86,87,93,and/or110.
Ishallhappilypayander'sfeeof$2.56foreaherrorinthisdraftwhenitis
rstreportedtome,whetherthaterrorbetypographial,tehnial,orhistorial.
ThesamerewardholdsforitemsthatIforgottoputintheindex. Andvaluable
suggestions forimprovementsto the text areworth 32/eah. (Furthermore, if
younda bettersolutiontoanexerise,I'llatuallyrewardyouwithimmortal
gloryinsteadofmeremoney,bypublishingyournameintheeventualbook: )
Cross referenes to yet-unwritten material sometimes appear as `00'; this
impossiblevalueisaplaeholderfortheatualnumberstobesuppliedlater.
Happyreading!
Stanford,California D.E.K.
13June 2002
7.2.1.3. Generatingallombinations. Combinatorialmathematisisoften
desribed as \the study of permutations, ombinations, et.," so we turn our
attentionnow to ombinations. A ombination of n things, takent ata time,
oftenalledsimplyat-ombinationofnthings,isawaytoseletasubsetofsizet
fromagivensetofsizen. WeknowfromEq.1.2.6{(2)thatthereareexatly n
t
ways to dothis; andwelearned inSetion 3.4.2 howto hoose t-ombinations
atrandom.
Seleting t of n objets is equivalent to hoosing the n t elements not
seleted. Wewillemphasizethissymmetrybyletting
n = s+ t (1)
throughoutourdisussion,andwewilloftenrefertoat-ombinationofnthings
as an \(s;t)-ombination." Thus, an (s;t)-ombination is a way to subdivide
s+t objetsintotwoolletionsof sizessandt.
If Iaskhowmanyombinations of 21anbetaken outof 25,
Ido ineetaskhowmanyombinations of 4maybetaken.
Fortherearejustasmanyways oftaking21asthere areofleaving4.
|AUGUSTUS DEMORGAN, AnEssayonProbabilities (1838)
There are two main waysto represent(s;t)-ombinations: We an listthe
elements
t :::
2
1
that havebeenseleted,orweanworkwithbinary strings
a
n 1 :::a
1 a
0
forwhih
a
n 1
++a
1 +a
0
= t: (2)
Thelatterrepresentationhass0sand t1s, orrespondingto elementsthat are
unseletedor seleted. Thelistrepresentation
t :::
2
1
tendsto workoutbest
ifwelettheelementsbemembersofthesetf0;1;:::;n 1gandifwelistthem
indereasing order:
n>
t
>>
2
>
1
0: (3)
Binarynotationonnetsthese two representationsniely,beausetheitemlist
t :::
2
1
orrespondsto thesum
2 t
++2 2
+2 1
= n 1
X
k=0 a
k 2
k
= (a
n 1 :::a
1 a
0 )
2
: (4)
Of ourse we ould also list the positions b
s :::b
2 b
1
of the 0s ina
n 1 :::a
1 a
0 ,
where
n>b
s
>>b
2
>b
1
0: (5)
Combinations are important not only beause subsets are omnipresent in
mathematisbutalsobeausetheyareequivalenttomanyotherongurations.
For example, every (s;t)-ombination orresponds to a ombination of s+1
thingstakentatatimewithrepetitionspermitted,alsoalledamultiombination,
namelya sequeneofintegersd
t :::d
2 d
1 with
sd
t
d
2 d
1
0: (6)
Onereasonisthatd
t :::d
2 d
1
solves(6)ifandonlyif
t :::
2
1
solves(3),where
t
=d
t
+t 1; :::;
2
=d
2
+1;
1
=d
1
(7)
(seeexerise 1.2.6{60). Andthereis anotherusefulwayto relateombinations
withrepetitiontoordinaryombinations,suggestedbySolomonGolomb[AMM
75(1968),530{531℄,namelytodene
e
j
=
j
; if
j s;
e
j s
; if
j
>s.
(8)
In this form the numbers e
t :::e
1
don't neessarily appear in desending or-
der, but the multiset fe
1
;e
2
;:::;e
t
g is equal to f
1
;
2
;:::;
t
g if and only if
fe
1
;e
2
;:::;e
t
g isaset. (See Table1andexerise 1.)
An(s;t)-ombinationisalsoequivalenttoaomposition ofn+1intot+1
parts,namelyanorderedsum
n+1 = p
t
++p
1 +p
0
; wherep
t
;:::;p
1
;p
0
1. (9)
Theonnetionwith(3)isnow
p
t
=n
t
; p
t 1
=
t
t 1
; :::; p
1
=
2
1
; p
0
=
1
+1: (10)
Equivalently,ifq
j
=p
j
1,wehave
s = q
t
++q
1 +q
0
; whereq
t
;:::;q
1
;q
0
0, (11)
aomposition ofsintot+1nonnegative parts,relatedto(6)bysetting
q
t
=s d
t
; q
t 1
=d
t d
t 1
; :::; q
1
=d
2 d
1
; q
0
=d
1
: (12)
Furthermore it is easy to see that an (s;t)-ombination is equivalent to a
path of length s+t from orner to orner of an st grid, beause suh a
path ontainss vertial steps and t horizontal steps. Thus, ombinations an
be studied in at least eight dierent guises. Table 1 illustrates all 6
3
= 20
possibilitiesintheases=t=3.
Theseousinsofombinationsmightseemratherbewilderingatrstglane,
but most of them an be understood diretly from the binary representation
a
n 1 :::a
1 a
0
. Consider,forexample,the\random"bitstring
a
23 :::a
1 a
0
= 011001001000011111101101; (13)
Table1
THE (3;3)-COMBINATIONS AND THEIR EQUIVALENTS
a5a4a3a2a1a0 b3b2b1 321 d3d2d1 e3e2e1 p3p2p1p0 q3q2q1q0 path
000111 543 210 000 210 4111 3000
001011 542 310 100 310 3211 2100
001101 541 320 110 320 3121 2010
001110 540 321 111 321 3112 2001
010011 532 410 200 010 2311 1200
010101 531 420 210 020 2221 1110
010110 530 421 211 121 2212 1101
011001 521 430 220 030 2131 1020
011010 520 431 221 131 2122 1011
011100 510 432 222 232 2113 1002
100011 432 510 300 110 1411 0300
100101 431 520 310 220 1321 0210
100110 430 521 311 221 1312 0201
101001 421 530 320 330 1231 0120
101010 420 531 321 331 1222 0111
101100 410 532 322 332 1213 0102
110001 321 540 330 000 1141 0030
110010 320 541 331 111 1132 0021
110100 310 542 332 222 1123 0012
111000 210 543 333 333 1114 0003
whih has s=11zeros and t=13 ones, henen=24. Thedual ombination
b
s :::b
1
liststhepositionsofthezeros,namely
2320191716141312114 1;
beause the leftmost position is n 1 and the rightmost is 0. The primal
ombination
t :::
1
liststhepositionsoftheones, namely
222118151098 76 53 20:
Theorrespondingmultiombinationd
t :::d
1
liststhenumberof0stotheright
ofeah 1:
10108 62 22 222 11 0:
Theompositionp
t :::p
0
liststhedistanesbetweenonseutive1s,ifweimagine
additional1s attheleftandtheright:
2 13 351 11 11 21 21:
Andthenonnegativeompositionq
t :::q
0
ountshowmany0sappearbetween
\feneposts"representedby1s:
10 22 400 00 01 01 0;
thuswehave
a
n 1 :::a
1 a
0
=0 qt
10 qt
1
1:::10 q1
10 q0
: (14)
ThepathsinTable1alsohaveasimpleinterpretation(seeexerise2).
Lexiographi generation. Table 1 shows ombinations a
n 1 :::a
1 a
0 and
t :::
1
inlexiographiorder, whih isalsothelexiographi orderof d
t :::d
1 .
Notiethatthedual ombinationsb
s :::b
1
andtheorrespondingompositions
p
t :::p
0 ,q
t :::q
0
thenappearinreverse lexiographiorder.
Lexiographiorder usually suggests the most onvenientway to generate
ombinatorial ongurations. Indeed, Algorithm 7.2.1.2L already solves the
problem for ombinations in the form a
n 1 :::a
1 a
0
, sine (s;t)-ombinations
inbitstringformarethesameaspermutationsofthemultisetfs0;t1g. That
general-purposealgorithmanbestreamlinedinobviouswayswhenitisapplied
to this speial ase. (See also exerise 7.1{00, whih presents a remarkable
sequeneofsevenbitwiseoperationsthatwill onvert anygivenbinarynumber
(a
n 1 :::a
1 a
0 )
2
to the lexiographially next t-ombination, assuming that n
doesnotexeedtheomputer'swordlength.)
Let'sfous,however,ongeneratingombinationsintheotherprinipalform
t :::
2
1
,whihismorediretlyrelevanttothewaysinwhihombinationsare
oftenneeded, and whih is more ompat than the bit strings when t is small
omparedton. Intherstplaeweshouldkeepinmindthatasimplesequene
ofnestedloopswill dothe job nielywhent is very small. For example,when
t=3thefollowinginstrutionssuÆe:
For
3
=2,3,:::,n 1(in thisorder)dothefollowing:
For
2
=1,2,:::,
3
1(inthisorder)dothefollowing:
For
1
=0,1,:::,
2
1 (inthisorder)dothefollowing:
Visit theombination
3
2
1 .
(15)
(Seetheanalogoussituationin7.2.1.1{(3).)
On the other hand when t is variable or not so small, we an generate
ombinations lexiographially by following the general reipe disussed after
Algorithm7.2.1.2L,namelytondtherightmostelement
j
thatanbeinreased
and then to set the subsequent elements
j 1 :::
1
to their smallest possible
values:
Algorithm L (Lexiographi ombinations). This algorithm generates all t-
ombinations
t :::
2
1
of the n numbers f0;1;:::;n 1g, given n t 0.
Additionalvariables
t+1 and
t+2
areused assentinels.
L1.[Initialize.℄ Set
j
j 1 for1jt;alsoset
t+1
nand
t+2 0.
L2.[Visit.℄ Visittheombination
t :::
2
1 .
L3.[Findj.℄ Setj 1. Then,while
j +1=
j+1 ,set
j
j 1andj j+1;
repeatuntil
j
+16=
j+1 .
L4.[Done?℄ Terminatethealgorithmifj>t.
L5.[Inrease
j .℄ Set
j
j
+1and returnto L2.
The running time of this algorithm is not diÆult to analyze. Step L3 sets
j
j 1 just after visiting a ombination forwhih
j+1
=
1
+j, and the
numberofsuh ombinationsisthenumberofsolutionsto theinequalities
n>
t
>>
j+1
j; (16)
but this formula is equivalent to a (t j)-ombination of the n j objets
fn 1;:::;jg,sotheassignment
j
j 1oursexatly n j
t j
times. Summing
for1jttellsusthat theloopinstepL3isperformed
n 1
t 1
+
n 2
t 2
++
n t
0
=
n 1
s
+
n 2
s
++
s
s
=
n
s+1
(17)
timesaltogether,oranaverageof
n
s+1 .
n
t
=
n!
(s+1)!(t 1)!
.
n!
s!t!
= t
s+1
(18)
times pervisit. This ratio is less than 1 when t s, so Algorithm L is quite
eÆientinsuhases.
But the quantity t=(s+1) an be embarrassingly large when t is near n
ands issmall. Indeed,AlgorithmL oasionally sets
j
j 1 needlessly, at
timeswhen
j
alreadyequals j 1. Furthersrutinyreveals that we neednot
alwayssearhfortheindexjthatisneededinstepsL4andL5,sinetheorret
value of j an often be predited from the ations just taken. For example,
afterwehaveinreased
4
andreset
3
2
1
totheirstartingvalues210,thenext
ombination willinevitably inrease
3
. These observations leadto a tuned-up
version ofthealgorithm:
AlgorithmT(Lexiographiombinations). ThisalgorithmislikeAlgorithmL,
butfaster. It alsoassumes,foronveniene,thatt<n.
T1.[Initialize.℄ Set
j
j 1 for1j t;thenset
t+1 n,
t+2
0,and
j t.
T2.[Visit.℄ (Atthispointj isthesmallestindex suhthat
j+1
>j.) Visitthe
ombination
t :::
2
1
. Then,ifj>0,set x j andgoto stepT6.
T3.[Easyase?℄ If
1
+1<
2 ,set
1
1
+1andreturntoT2. Otherwiseset
j 2.
T4.[Findj.℄ Set
j 1
j 2 andx
j
+1. Ifx=
j+1
, setj j+1 and
repeatthisstepuntilx6=
j+1 .
T5.[Done?℄ Terminatethealgorithmifj>t.
T6.[Inrease
j .℄ Set
j
x,j j 1,andreturntoT2.
Now j =0 instep T2 ifand onlyif
1
>0, so theassignments instepT4 are
neverredundant. Exerise6 arriesoutaompleteanalysis ofAlgorithmT.
Notie that the parameter n appears only in the initialization steps L1
and T1, notinthe prinipalparts ofAlgorithms L and T. Thus wean think
oftheproessas generatingtherst n
t
ombinationsofaninnite list,whih
dependsonlyon t. This simpliationarises beausethe listof t-ombinations
forn+1thingsbeginswiththelistfornthings,underouronventions;wehave
beenusinglexiographiorderonthedereasingsequenes
t :::
1
forthisvery
reason,insteadofworkingwiththeinreasingsequenes
1 :::
t .
DerrikLehmer notied another pleasant property ofAlgorithms L and T
[AppliedCombinatorialMathematis,editedbyE.F.Bekenbah(1964),27{30℄:
TheoremL. Theombination
t :::
2
1
isvisitedafter exatly
t
t
++
2
2
+
1
1
(19)
otherombinationshavebeenvisited.
Proof. There are k
k
ombinations 0
t :::
0
2
0
1 with
0
j
=
j
fort j >k and
0
k
<
k
,namely
t :::
k+1
followedbythek-ombinationsoff0;:::;
k 1g.
Whent=3,forexample,thenumbers
2
3
+ 1
2
+ 0
1
; 3
3
+ 1
2
+ 0
1
; 3
3
+ 2
2
+ 0
1
; :::; 5
3
+ 4
2
+ 3
1
that orrespondto the ombinations
3
2
1
inTable1 simply run through the
sequene 0, 1, 2, :::, 19. Theorem L gives us a nie way to understand the
ombinatorial number system of degree t, whih represents every nonnegative
integerN uniquelyintheform
N =
n
t
t
++
n
2
2
+
n
1
1
; n
t
>>n
2
>n
1
0: (20)
[SeeErnestoPasal,GiornalediMatematihe 25(1887),45{49.℄
Binomialtrees. ThefamilyoftreesT
n
dened by
T
0
= ; T
n
=
T
0
T
1
T
n 1 :::
0 1 n 1
forn>0; (21)
arises in several important ontexts and sheds further light on ombination
generation. Forexample,T
4 is
0
0 0
0 0
0 0
0 1
1 1
1 2
2 3
; (22)
and T
5
, rendered more artistially, appears as the frontispiee to Volume 1 of
thisseriesofbooks.
NotiethatT
n
islikeT
n 1
,exeptforanadditionalopyofT
n 1
;therefore
T
n has2
n
nodesaltogether. Furthermore, thenumberofnodesonlevelt isthe
binomialoeÆient n
t
;thisfataountsforthename\binomialtree." Indeed,
thesequene of labels enountered onthe path from theroot to eah node on
leveltdenesaombination
t :::
1
,andallombinationsourinlexiographi
orderfromlefttoright. Thus,AlgorithmsLandTanberegardedasproedures
totraversethenodesonleveltofthebinomialtreeT
n .
TheinnitebinomialtreeT
1
isobtainedbylettingn!1in(21). Theroot
ofthis treehas innitelymanybranhes, but every nodeexept fortheoverall
rootatlevel0istherootofanitebinomialsubtree. Allpossiblet-ombinations
appearinlexiographiorderonleveltofT
1 .
Let'sgetmorefamiliarwithbinomialtreesbyonsideringallpossibleways
to pak a ruksak. More preisely, suppose we have n items that take up
respetivelyw
n 1 ,:::,w
1 ,w
0
unitsofapaity,where
w
n 1
w
1 w
0
; (23)
wewantto generateallbinaryvetorsa
n 1 :::a
1 a
0
suh that
aw=a
n 1 w
n 1
++a
1 w
1 +a
0 w
0
N; (24)
where N is thetotal apaity of a ruksak. Equivalently, we want to nd all
subsetsC off0;1;:::;n 1gsuhthatw(C)= P
2C w
N;suhsubsetswill
bealledfeasible. We willwrite a feasible subsetas
1 :::
t
, where
1
>>
t
0, numbering the subsripts dierently from the onvention of (3) above
beauset isvariableinthisproblem.
Every feasible subsetorresponds to a node ofT
n
, and ourgoalis to visit
eahfeasiblenode. Clearlytheparentofeveryfeasiblenodeisfeasible,andsois
theleftsibling,ifany;thereforeasimpletreeexplorationproedureworkswell:
Algorithm F (Fillinga ruksak). Thisalgorithm generatesallfeasibleways
1 :::
t
tollaruksak,givenw
n 1 ,:::,w
1 ,w
0
,andN. WeletÆ
j
=w
j w
j 1
for1j<n.
F1.[Initialize.℄ Sett 0,
0
n,andr N.
F2.[Visit.℄ Visittheombination
1 :::
t
,whihusesN runitsofapaity.
F3.[Tryto add w
0 .℄ If
t
>0 and rw
0
, set t t+1,
t
0,r r w
0 ,
andreturntoF2.
F4.[Try to inrease
t
.℄ Terminate ift = 0. Otherwise, if
t 1
>
t
+1 and
rÆ
t +1
,set
t
t
+1,r r Æ
t
,andreturntoF2.
F5.[Remove
t
.℄ Setr r+w
t
,t t 1,andreturntoF4.
NotiethatthealgorithmimpliitlyvisitsnodesofT
n
inpreorder,skippingover
unfeasiblesubtrees. Anelement >0 is plaedinthe ruksak, ifit ts, just
aftertheproedurehasexploredallpossibilitiesusingelement 1initsplae.
Therunningtimeisproportionaltothenumberoffeasibleombinationsvisited
(seeexerise20).
Inidentally,thelassial\knapsakproblem"ofoperationsresearh isdif-
ferent: ItasksforafeasiblesubsetC suhthat v(C)= P
2C
v()ismaximum,
whereeahitemhasbeenassignedavaluev(). AlgorithmFisnotapartiu-
larlygoodwaytosolvethatproblem,beauseitoftenonsidersasesthatould
beruledout. Forexample,ifCandC 0
aresubsetsoff1;:::;n 1gwithw(C)
w(C 0
)N w
0
andv(C)v(C 0
),AlgorithmFwillexaminebothC[f0gand
C 0
[f0g,butthelattersubsetwillneverimprovethemaximum. Wewillonsider
methodsforthelassialknapsakproblem later;AlgorithmFisintendedonly
forsituationswhenall ofthefeasiblepossibilitiesarepotentiallyrelevant.
Grayodesforombinations. Insteadofmerelygeneratingallombinations,
weoftenprefertovisittheminsuh awaythateahoneisobtainedbymaking
onlya smallhangeto itspredeessor.
For example, we an ask for what Nijenhuis and
Wilfhavealleda\revolvingdooralgorithm": Imagine
tworoomsthatontainrespetivelysandtpeople,with
a revolving door between them. Whenever a person
goesintotheoppositeroom,somebodyelseomesout. Canwedeviseasequene
ofmovessothat eah(s;t)-ombinationoursexatlyone?
The answeris yes, and infat a huge number of suh patterns exist. For
example, it turns out that if we examineall n-bit strings a
n 1 :::a
1 a
0 in the
well-known order of Gray binary ode (Setion 7.2.1.1), but selet only those
thathaveexatlys0sandt1s,theresultingstringsformarevolving-doorode.
Here'stheproof: Graybinaryodeisdenedbythereurrene
n
=0
n 1 ,
1 R
n 1
of7.2.1.1{(5),soits(s;t) subsequenesatisesthereurrene
st
= 0
(s 1)t
; 1 R
s(t 1)
(25)
when st > 0. We also have
s0
= 0 s
and
0t
= 1 t
. Therefore it is lear by
indutionthat
st
begins with 0 s
1 t
and ends with 10 s
1 t 1
when st > 0. The
transition at the omma in (25) is from the last element of 0
(s 1)t to the
lastelementof 1
s(t 1)
,namelyfrom 010 s 1
1 t 1
=010 s 1
11 t 2
to 110 s
1 t 2
=
110 s 1
01 t 2
when t 2, and this satises the revolving-door onstraint. The
aset=1 alsoheksout. Forexample,
33
isgivenbytheolumnsof
000111 011010 110001 101010
001101 011100 110010 101100
001110 010101 110100 100101
001011 010110 111000 100110
011001 010011 101001 100011
(26)
and
23
an be found in the rst two olumns of this array. One more turn
ofthe door takesthe last elementinto the rst. [These properties of
st were
disovered by D. T. Tang and C. N. Liu, IEEE Trans. C-22 (1973), 176{180;
alooplessimplementationwaspresentedbyJ.R.Bitner,G.Ehrlih, andE.M.
Reingold,CACM 19(1976),517{521.℄
Whenweonvertthebitstringsa
5 a
4 a
3 a
2 a
1 a
0
in(26)totheorresponding
index-listforms
3
2
1
,astrikingpatternbeomesevident:
210 431 540 531
320 432 541 532
321 420 542 520
310 421 543 521
430 410 530 510
(27)
Therstomponents
3
ourininreasingorder;butforeahxedvalueof
3 ,
thevalues of
2
our indereasing order. Andfor xed
3
2
, thevalues of
1
are againinreasing. The same is true ingeneral: Allombinations
t :::
2
1
appearinlexiographiorderof
(
t
;
t 1
;
t 2
; :::; ( 1) t 1
1
) (28)
intherevolving-doorGrayode
st
. Thispropertyfollowsbyindution,beause
(25)beomes
st
=
(s 1)t
; (s+t 1) R
s(t 1)
(29)
forst>0 whenweuseindex-list notationinsteadofbitstringnotation. Conse-
quentlythesequeneanbegeneratedeÆientlybythefollowingalgorithmdue
toW.H.Payne[seeACMTrans.Math. Software 5(1979),163{172℄:
Algorithm R (Revolving-door ombinations). This algorithm generatesall t-
ombinations
t :::
2
1
of f0;1;:::;n 1g in lexiographi order of the alter-
nating sequene (28), assuming that n > t > 1. Step R3 has two variants,
dependingonwhether tisevenor odd.
R1.[Initialize.℄ Set
j
j 1 fortj1,and
t+1 n.
R2.[Visit.℄ Visittheombination
t :::
2
1 .
R3.[Easyase?℄ Ift is odd: If
1
+1<
2
, inrease
1
by1 and returntoR2,
otherwiseset j 2and goto R4. If tis even: If
1
>0,derease
1 by 1
and returnto R2,otherwisesetj 2 andgotoR5.
R4.[Trytoderease
j
.℄ (Atthispoint
j
=
j 1
+1.) If
j
j,set
j
j 1 ,
j 1
j 2,andreturnto R2. Otherwiseinreasej by1.
R5.[Try to inrease
j
.℄ (At this point
j 1
= j 2.) If
j
+1 <
j+1 , set
j 1
j ,
j
j
+1,andreturn toR2. Otherwiseinreasej by 1,and
goto R4ifjt.
Exerises21{25 explorefurther propertiesof this interesting sequene. Oneof
themisanieompaniontoTheoremL:Theombination
t
t 1 :::
2
1
isvisited
byAlgorithmR afterexatly
N =
t +1
t
t 1 +1
t 1
++( 1) t
2 +1
2
( 1) t
1 +1
1
[todd℄ (30)
otherombinationshavebeenvisited. WemayallthistherepresentationofN
inthe\alternatingombinatorialnumbersystem"ofdegreet;oneonsequene,
for example, is that every positive integer has a unique representation of the
formN = a
3
b
2
+
1
witha>b>>0. AlgorithmRtellsushowtoadd1
toN inthissystem.
Although the strings of (26) and (27)are notin lexiographi order, they
are examplesof a more generalonept alledgenlex order, a name oinedby
TimothyWalsh. Asequeneofstrings
1 ,:::,
N
issaidto beingenlexorder
when all strings with a ommon prex our onseutively. For example, all
3-ombinationsthat beginwith53appeartogetherin(27).
Genlex ordermeansthatthestringsan bearranged ina triestruture, as
inFig.31ofSetion6.3,but withthehildrenofeahnodeorderedarbitrarily.
Whenatrieistraversedinanyordersuhthateahnodeisvisitedjustbeforeor
justafteritsdesendants,allnodeswithaommonprex|thatis,allnodesof
asubtrie|appearonseutively. Thispriniplemakesgenlexorderonvenient,
beauseitorrespondstoreursivegenerationshemes. Manyofthealgorithms
wehaveseenforgeneratingn-tupleshavethereforeproduedtheirresultsinsome
version of genlex order; similarly, the method of \plain hanges" (Algorithm
7.2.1.2P) visits permutations ina genlex order of the orresponding inversion
tables.
Therevolving-doormethodofAlgorithmRisagenlexroutinethathanges
onlyoneelementoftheombinationateahstep. Butitisn'ttotallysatisfatory,
beauseitfrequentlymusthangetwo oftheindies
j
simultaneously,inorder
topreservetheondition
t
>>
2
>
1
. Forexample,AlgorithmRhanges
210into320,and(27)inludes ninesuh\rossing"moves.
Thesoure ofthis defetan betraedto ourproof that (25)satises the
revolving-door property: We observed that the string010 s 1
11 t 2
is followed
by 110 s 1
01 t 2
when t 2. Hene the reursive onstrution
st
involves
transitionsof theform 110 a
0 $010 a
1, whena substring like11000 is hanged
to01001orvie versa;thetwo1s rosseahother.
AGraypathforombinationsis saidtobehomogeneous ifithanges only
one of the indies
j
at eah step. A homogeneous sheme is haraterized
in bitstring form by having only transitions of the forms 10 a
$ 0 a
1 within
strings,for a 1, when wepass from one string
tothenext. Withahomogeneousshemewean,
for example, play all t-note hords on an n-note
keyboardbymovingonlyone ngerata time.
A slight modiation of (25) yields a genlex
sheme for (s;t)-ombinations that is pleasantly
homogeneous. The basi idea is to onstrut a
sequenethat begins with0 s
1 t
and endswith 1 t
0 s
, andthefollowingreursion
suggestsitselfalmostimmediately: LetK
s0
=0 s
,K
0t
=1 t
,K
s( 1)
=;,and
K
st
= 0K
(s 1)t
; 10K R
(s 1)(t 1)
; 11K
s(t 2)
forst>0: (31)
At the ommas of this sequene we have 01 t
0 s 1
followed by 101 t 1
0 s 1
, and
10 s
1 t 1
followed by 110 s
1 t 2
; both of these transitions are homogeneous, al-
thoughtheseondonerequiresthe1tojumparosss0s. TheombinationsK
33
fors=t=3are
000111 010101 101100 100011
001011 010011 101001 110001
001101 011001 101010 110010
001110 011010 100110 110100
010110 011100 100101 111000
(32)
inbitstringform,andtheorresponding\ngerpatterns"are
210 420 532 510
310 410 530 540
320 430 531 541
321 431 521 542
421 432 520 543.
(33)
Whenahomogeneousshemeforordinaryombinations
t :::
1
isonverted
to the orresponding sheme (6) for ombinations with repetitions d
t :::d
1 , it
retains thepropertythat only one oftheindies d
j
hangesat eah step. And
whenit isonvertedto theorresponding shemes(9)or (11)forompositions
p
t :::p
0 or q
t :::q
0
,onlytwo(adjaent)partshange when
j
hanges.
Near-perfet shemes. But wean do even better! All(s;t)-ombinations
an be generated by a sequene of strongly homogeneous transitions that are
either01$10or001$100. Inotherwords,weaninsistthateahstepauses
a single index
j
to hange by at most 2. Let's all suh generation shemes
near-perfet.
Imposing suh strong onditions atually makes it fairly easy to disover
near-perfet shemes, beause omparativelyfew hoies areavailable. Indeed,
ifwerestritourselvestogenlexmethodsthat arenear-perfetonn-bitstrings,
T.A. Jenkynsand D.MCarthy observed that allsuh methods an be easily
haraterized[ArsCombinatoria40(1995),153{159℄:
Theorem N. Ifst>0,thereareexatly2s near-perfetwaystolistall(s;t)-
ombinations ina genlex order. Infat, when 1 a s, there is exatly one
suh listing, N
sta
, that begins with 1 t
0 s
and ends with 0 a
1 t
0 s a
; the other s
possibilitiesarethereverselists,N R
sta .
Proof. Theresultertainlyholdswhens=t=1;otherwiseweuseindutionon
s+t. ThelistingN
sta
,ifitexists,musthavetheform1X
s(t 1) ,0Y
(s 1)t
forsome
near-perfet genlex listings X
s(t 1) and Y
(s 1)t
. Ift =1, X
s(t 1)
is thesingle
string0 s
;heneY
(s 1)t
mustbeN
(s 1)1(a 1)
ifa>1,andit mustbeN R
(s 1)11
ifa=1. Ontheotherhandift>1,thenear-perfetonditionimpliesthatthe
laststringofX
s(t 1)
annotbeginwith 1;heneX
s(t 1)
=N
s(t 1)b
forsomeb.
Ifa>1, Y
(s 1)t
mustbeN
(s 1)t(a 1)
,henebmustbe1;similarly,bmustbe1
ifs =1. Otherwise we have a =1 <s, and this fores Y
(s 1)t
=N R
(s 1)t for
some. Thetransitionfrom10 b
1 t 1
0 s b
to0 +1
1 t
0 s 1
isnear-perfetonlyif
=1 andb=2.
TheproofofTheoremNyieldsthefollowingreursiveformulaswhenst>0:
N
sta
= 8
>
<
>
: 1N
s(t 1)1
; 0N
(s 1)t(a 1)
; if1<as;
1N
s(t 1)2
; 0N R
(s 1)t1
; if1=a<s;
1N
1(t 1)1
; 01 t
; if1=a=s.
(34)
Also,ofourse,N
s0a
=0 s
.
LetussetA
st
=N
st1 andB
st
=N
st2
. Thesenear-perfetlistings,disovered
byPhillipJ.Chasein1976,havetheneteetofshiftinga leftmostblokof1s
totherightbyoneor twopositions,respetively,andtheysatisfythefollowing
mutualreursions:
A
st
=1B
s(t 1)
; 0A R
(s 1)t
; B
st
=1A
s(t 1)
; 0A
(s 1)t
: (35)
\To takeonestepforward,taketwo steps forward, thenone stepbakward; to
taketwo steps forward, takeone stepforward,thenanother." These equations
Table2
CHASE'S SEQUENCES FOR (3;3)-COMBINATIONS
A
33
= b
C R
33
543 531 321 420
541 530 320 421
540 510 310 431
542 520 210 430
532 521 410 432
B
33
=C
33
543 520 432 410
542 510 430 210
540 530 431 310
541 531 421 320
521 532 420 321
holdforallinteger valuesofsandt,ifwedeneA
st andB
st
to be;whensor
tis negative,exeptthat A
00
=B
00
= (theemptystring). ThusA
st
atually
takes min(s;1) forward steps, and B
st
atually takes min(s;2). For example,
Table2 showstherelevantlistings fors=t =3,using anequivalentindex-list
form
3
2
1
insteadofthebitstringsa
5 a
4 a
3 a
2 a
1 a
0 .
Chasenotiedthat aomputerimplementationofthesesequenesbeomes
simplerifwedene
C
st
=
A
st
; ifs+tisodd;
B
st
; ifs+tiseven;
b
C
st
=
A R
st
; ifs+tiseven;
B R
st
; ifs+tisodd.
(36)
[SeeCongressus Numerantium69(1989),215{242.℄ Thenwehave
C
st
= (
1C
s(t 1)
; 0 b
C
(s 1)t
; ifs+tisodd;
1C
s(t 1)
; 0C
(s 1)t
; ifs+tiseven;
(37)
b
C
st
= (
0C
(s 1)t
; 1 b
C
s(t 1)
; ifs+tiseven;
0 b
C
(s 1)t
; 1 b
C
s(t 1)
; ifs+tisodd.
(38)
Whenbit a
j
is ready to hange, wean tell where we are in the reursion by
testingwhetherj isevenorodd.
Indeed, the sequene C
st
an be generated by a surprisingly simple algo-
rithm,basedongeneralideasthat applytoany genlexsheme. Letussaythat
bita
j
isativeinagenlexalgorithmifitissupposedtohangebeforeanythingto
itsleftisaltered. (Thenodeforanativebitintheorrespondingtrieisnotthe
rightmosthildofits parent.) Supposewehaveanauxiliarytablew
n :::w
1 w
0 ,
where w
j
= 1 if and only if either a
j
is ative or j < r, where r is the least
subsriptsuhthata
r 6=a
0
;wealsoletw
n
=1. Thenthefollowingmethodwill
ndthesuessorofa
n 1 :::a
1 a
0 :
Set j r. Ifw
j
=0,set w
j
1,j j+1, andrepeat until
w
j
=1. Terminateifj =n; otherwiseset w
j
0. Changea
j
to1 a
j
,and makeanyotherhangestoa
j 1 :::a
0
andrthat
applyto thepartiulargenlexshemebeingused.
(39)
Thebeautyofthisapproah omesfromthefatthattheloopisguaranteedto
beeÆient: Wean prove that theoperationj j+1 will beperformedless
thanonepergenerationstep,ontheaverage (seeexerise36).
Byanalyzingthetransitionsthat ourwhen bitshange in(37)and(38),
weanreadilyeshouttheremainingdetails:
Algorithm C (Chase'ssequene). Thisalgorithmvisitsall(s;t)-ombinations
a
n 1 :::a
1 a
0
,wheren=s+t,inthenear-perfetorderofChase'ssequeneC
st .
C1.[Initialize.℄ Seta
j
0 for0 j <s, a
j
1 forsj <n, andw
j 1
for0jn. Ifs>0,set r s;otherwisesetr t.
C2.[Visit.℄ Visit theombinationa
n 1 :::a
1 a
0 .
C3.[Find j and branh.℄ Set j r. If w
j
= 0, set w
j
1, j j+1, and
repeatuntil w
j
=1. Terminateifj =n;otherwisesetw
j
0 andmakea
four-way branh: Go to C4 if j is odd anda
j
6=0, to C5 ifj is even and
a
j
6=0,toC6ifj isevenanda
j
=0,toC7ifj isoddanda
j
=0.
C4.[Move right one.℄ Set a
j 1
1, a
j
0. If r = j > 1, set r j 1;
otherwiseifr=j 1setr j. Returnto C2.
C5.[Moverighttwo.℄ If a
j 2
6=0,goto C4. Otherwiseset a
j 2 1,a
j 0.
Ifr=j,setr max(j 2;1);otherwiseifr=j 2,setr j 1. Return
toC2.
C6.[Moveleftone.℄ Seta
j 1,a
j 1
0. Ifr=j>1,setr j 1;otherwise
ifr=j 1 setr j. ReturntoC2.
C7.[Movelefttwo.℄ Ifa
j 1
6=0,gotoC6. Otherwiseseta
j 1,a
j 2
0. If
r=j 2,setr j;otherwiseifr=j 1,setr j 2. ReturntoC2.
*Analysis of Chase's sequene. The magialproperties of AlgorithmC ry
outforfurther exploration,and a loserlook turns outto bequite instrutive.
Givena bitstringa
n 1 :::a
1 a
0
,letusdenea
n
=1,u
n
=nmod2,and
u
j
=(1 u
j+1 )a
j+1
; v
j
=(u
j
+j)mod2; w
j
=(v
j +a
j
)mod2; (40)
forn>j0. For example,wemighthaven=26and
a
25 :::a
1 a
0
=11001001000011111101101010;
u
25 :::u
1 u
0
=10100100100001010100100101;
v
25 :::v
1 v
0
=00001110001011111110001111;
w
25 :::w
1 w
0
=11000111001000000011100101:
(41)
Withthese denitions wean prove by indutionthat v
j
=0 ifandonlyif bit
a
j
isbeing\ontrolled"byCratherthanby b
C inthereursions(37){(38) that
generatea
n 1 :::a
1 a
0
,exeptwhena
j
ispart ofthenalrun of0sor1s atthe
rightend. Thereforew
j
agreeswiththevalueomputedbyAlgorithmCatthe
momentwhena
n 1 :::a
1 a
0
isvisited,forrj<n. Theseformulasanbeused
todetermineexatlywhereagivenombinationappearsinChase'ssequene(see
exerise39).
If we want to work with the index-list form
t :::
2
1
instead of the bit
strings a
n 1 :::a
1 a
0
, it is onvenient to hange the notation slightly, writing
C
t
(n)forC
st and
b
C
t (n)for
b
C
st
whens+t=n. Then C
0 (n)=
b
C
0
(n)=,and
thereursionsfort0 taketheform
C
t+1
(n+1)= (
nC
t (n);
b
C
t+1
(n); ifniseven;
nC
t (n); C
t+1
(n); ifnisodd;
(42)
b
C
t+1
(n+1)= (
C
t+1 (n); n
b
C
t
(n); ifnisodd;
b
C
t+1 (n); n
b
C
t
(n); ifniseven.
(43)
Thesenewequationsan beexpandedtotellus,forexample,that
C
t+1
(9)=8C
t (8); 6C
t (6); 4C
t
(4); :::; 3 b
C
t (3); 5
b
C
t (5); 7
b
C
t (7);
C
t+1
(8)=7C
t (7); 6C
t (6); 4C
t
(4); :::; 3 b
C
t (3); 5
b
C
t (5);
b
C
t+1
(9)= 6C
t (6); 4C
t
(4); :::; 3 b
C
t (3); 5
b
C
t (5); 7
b
C
t (7); 8
b
C
t (8);
b
C
t+1
(8)= 6C
t (6); 4C
t
(4); :::; 3 b
C
t (3); 5
b
C
t (5); 7
b
C
t (7);
(44)
notiethatthesamepatternpredominatesinallfoursequenes. Themeaningof
\:::"inthemiddledependsonthevalueoft: WesimplyomitalltermsnC
t (n)
andn b
C
t
(n)wheren<t.
Exeptfor edgeeets at the very beginningor end, allof theexpansions
in(44)arebasedontheinniteprogression
:::; 10; 8; 6; 4; 2; 0; 1; 3; 5; 7; 9; :::; (45)
whihisanaturalwaytoarrangethenonnegativeintegersintoadoublyinnite
sequene. If we omit all terms of (45) that are < t, given any integer t 0,
theremainingtermsretain thepropertythat adjaentelementsdierbyeither
1 or 2. Rihard Stanleyhas suggested the nameendo-order for this sequene,
beause we an rememberit by thinking\even numbers dereasing, odd:::".
(NotiethatifweretainonlythetermslessthanN andomplementwithrespet
toN, endo-orderbeomesorgan-pipeorder;see exerise6.1{18.)
Weouldprogramthereursionsof(42)and(43)diretly,butitisinterest-
ingto unwindthem using(44),thusobtaininganiterativealgorithmanalogous
toAlgorithmC.TheresultneedsonlyO(t)memoryloations,anditisespeially
eÆientwhentisrelativelysmallomparedton. Exerise45ontainsthedetails.
*Near-perfet multiset permutations. Chase's sequeneslead ina natural
way to an algorithm that will generate permutations of any desired multiset
fs
0 0;s
1
1;:::;s
d
dgina near-perfetmanner,meaningthat
i) everytransition iseithera
j+1 a
j
$a
j a
j+1 or a
j+1 a
j a
j 1
$a
j 1 a
j a
j+1
;
ii) transitionsoftheseondkindhavea
j
=min(a
j 1
;a
j+1 ).
AlgorithmCtellsushowto dothiswhend=1,andweanextendittolarger
valuesofdbythefollowingreursiveonstrution[CACM 13(1970),368{369,
376℄: Suppose
0
;
1
; :::;
N 1
isanynear-perfetlistingofthepermutationsoffs
1
1;:::;s
d
dg. ThenAlgo-
rithmC,withs=s
0
andt=s
1
++s
d
,tellsushowtogeneratea listing
j
=
j 0
s
; :::; 0 a
j 0
s a
(46)
inwhihalltransitionsare0x$x0or00x$x00;thenalentryhasa=1or2
leadingzeros,dependingonsandt. Thereforealltransitionsofthesequene
0
; R
1
;
2
; :::; (
N 1 or
R
N 1
) (47)
arenear-perfet;andthislistlearlyontainsallthepermutations.
Forexample,thepermutationsoff0;0;0;1;1;2ggeneratedinthis wayare
211000,210100,210001,210010,200110,200101,200011,201001,201010,201100,
021100,021001,021010,020110,020101,020011,000211,002011,002101,002110,
001120,001102,001012,000112,010012,010102,010120,011020,011002,011200,
101200,101020,101002,100012,100102,100120,110020,110002,110200,112000,
121000,120100,120001,120010,100210,100201,100021,102001,102010,102100,
012100,012001,012010,010210,010201,010021,000121,001021,001201,001210.
*Perfet shemes. Why shouldwesettlefora near-perfetgeneratorlikeC
st ,
insteadofinsistingthatalltransitionshavethesimplestpossibleform01$10?
One reason is that perfet shemes don't always exist. For example, we
observed in7.2.1.2{(2)that thereis noway togenerate allsixpermutations of
f1;1;2;2gwithadjaentinterhanges;thusthereisnoperfetshemefor(2;2)-
ombinations. Infat,ourhanesofahievingperfetionareonlyabout1in4:
Theorem P. Thegenerationofall(s;t)-ombinationsa
s+t 1 :::a
1 a
0
byadja-
entinterhanges 01$10ispossibleifandonlyifs1or t1 orstisodd.
Proof. Consider all permutations of the multiset fs0;t1g. We learned in
exerise5.1.2{16 thatthenumberm
k
ofsuhpermutations havingkinversions
istheoeÆientofz k
inthez-nomialoeÆient
s+t
t
z
= s+t
Y
k=s+1
(1+z++z k 1
) .
t
Y
k=1
(1+z++z k 1
): (48)
Everyadjaentinterhangehangesthenumberofinversionsby1,soaperfet
generationshemeispossibleonlyifapproximatelyhalfofallthepermutations
have an odd number of inversions. More preisely, the value of s+t
t
1
=
m
0 m
1 +m
2
mustbe0 or1. Butexerise49showsthat
s+t
t
1
=
b(s+t)=2
bt=2
[stiseven℄; (49)
andthisquantityexeeds1unlesss1 ort1 orstisodd.
Conversely, perfet shemes are easy with s 1 or t 1, and they turn
out to be possible also whenever st is odd. The rst nontrivial ase ours
for s = t = 3, when there are four essentially dierent solutions; the most
symmetrialofthese is
210 310 410 510 520 521 531 532 432 431
421 321 320 420 430 530 540 541 542 543 (50)
(see exerise 51). Several authors have onstruted Hamiltonian paths in the
relevant graph for arbitrary odd numbers s and t; for example, the method
of Eades, Hikey, and Read [JACM 31 (1984), 19{29℄ makes an interesting
exeriseinprogrammingwithreursiveoroutines. Unfortunately,however,none
ofthe known onstrutionsare suÆientlysimple to desribeina short spae,
ortoimplementwithreasonableeÆieny. Perfetombinationgeneratorshave
thereforenotyetprovedto beofpratialimportane.
Insummary,then, wehaveseenthat thestudyof(s;t)-ombinationsleads
to many fasinating patterns, some of whih are of great pratialimportane
andsomeofwhiharemerelyelegantand/orbeautiful. Figure26illustratesthe
prinipaloptionsthatareavailableintheases=t=5,when 10
5
=252ombi-
nationsarise. Lexiographiorder(AlgorithmL),therevolving-doorGrayode
(AlgorithmR),thehomogeneous shemeK
55
of(31), andChase'snear-perfet
sheme(AlgorithmC)areshowninparts(a),(b),(),and(d)oftheillustration.
Part(e) showsthenear-perfetshemethatisas losetoperfetionaspossible
whilestillbeingingenlexorderofthearray(seeexerise34),whilepart(f)is
theperfet shemeof Eades, Hikey, and Read. Finally, Figs. 26(g) and 26(h)
arelistingsthatproeedbyrotatinga
j a
j 1 :::a
0 a
j 1 :::a
0 a
j
orbyswapping
a
j
$a
0
,akinto Algorithms7.2.1.2C and7.2.1.2E(seeexerises55and56).
*Combinations of a multiset. Ifmultisets an havepermutations, theyan
haveombinationstoo.Forexample,onsiderthemultisetfb;b;b;b;g;g;g;r;r;r;
w;wg,representingasakthatontainsfourblueballsandthreethataregreen,
threered, two white. There are 37waysto hooseveballs from this sak; in
lexiographiorder(butdesendingineah ombination)theyare
gbbbb;ggbbb;gggbb;rbbbb;rgbbb;rggbb;rgggb;rrbbb;rrgbb;rrggb;
rrggg;rrrbb;rrrgb;rrrgg;wbbbb;wgbbb;wggbb;wgggb;wrbbb;wrgbb;
wrggb;wrggg;wrrbb;wrrgb;wrrgg;wrrrb;wrrrg;wwbbb;wwgbb;wwggb;
wwggg;wwrbb;wwrgb;wwrgg;wwrrb;wwrrg;wwrrr: (51)
This fatmight seem frivolous and/or esoteri, yetwewill see inTheorem W
belowthatthelexiographigenerationofmultisetombinationsyieldsoptimal
solutionstosigniantombinatorialproblems.
James Bernoulliobserved inhis ArsConjetandi (1713),119{123, that we
an enumerate suh ombinations by looking at the oeÆient of z 5
in the
produt(1+z+z 2
)(1+z+z 2
+z 3
) 2
(1+z+z 2
+z 3
+z 4
). Indeed,hisobservation
iseasytounderstand,beausewegetallpossibleseletionsfromthesakifwe
multiplyoutthepolynomials
(1+w+ww)(1+r+rr+rrr)(1+g+gg+ggg)(1+b+bb+bbb+bbbb):
Multisetombinationsarealsoequivalenttoboundedompositions, namely
toompositionsinwhihtheindividualpartsarebounded. Forexample,the37
multiombinationslistedin(51)orrespondto37solutionsof
5=r
3 +r
2 +r
1 +r
0
; 0r
3
2; 0r
2
;r
1
3; 0r
0 4;
namely5=0+0+1+4=0+0+2+3=0+0+3+2=0+1+0+4= =2+3+0+0.
Fig.26. Examples
of(5;5)-ombinations:
a)lexiographi;
b)revolving-door;
)homogeneous;
d)near-perfet;
e)nearer-perfet;
f)perfet;
g)suÆx-rotated;
h)right-swapped.
(a) (b) () (d) (e) (f) (g) (h)
Boundedompositions,inturn,arespeialasesofontingenytables,whih
areof greatimportane instatistis. Andallof these ombinatorial ongura-
tionsan be generated with Gray-like odes as well as in lexiographi order.
Exerises60{63 exploresomeofthebasiideas involved.
*Shadows.Setsofombinationsappearfrequentlyinmathematis. Forexample,
asetof2-ombinations(namelyasetofpairs)isessentiallyagraph,andasetof
t-ombinationsforgeneralt isalleda uniform hypergraph. If thevertiesofa
onvexpolyhedronareperturbedslightly,sothatnothreeareollinear,nofour
liein a plane, and ingeneralno t+1 liein a (t 1)-dimensional hyperplane,
theresulting(t 1)-dimensionalfaesare\simplexes"whosevertieshavegreat
signianein omputerappliations. Researhers havelearned that suh sets
ofombinationshaveimportantpropertiesrelatedtolexiographigeneration.
If is any t-ombination
t :::
2
1
, its shadow is the set of all its
(t 1)-element subsets
t 1 :::
2
1
, :::,
t :::
3
1 ,
t :::
3
2
. For example,
5310 = f310;510;530;531g. We an also represent a t-ombination as a bit
stringa
n 1 :::a
1 a
0
,inwhihaseisthesetofallstringsobtainedbyhang-
inga 1 to a 0: 101011=f001011;100011;101001;101010g. If A is any set of
t-ombinations,wedeneitsshadow
A =
S
fj2Ag (52)
to be the set of all (t 1)-ombinations in the shadows of its members. For
example,5310=f10;30;31;50;51;53g.
These denitions apply also to ombinations with repetitions, namely to
multiombinations: 5330= f330;530;533g and 5330 =f30;33;50;53g. In
general, when A is a set of t-element multisets, A is a set of (t 1)-element
multisets. Notie,however,thatAneverhasrepeatedelementsitself.
Theupper shadow
withrespettoa universeU isdenedsimilarly,but
itgoesfrom t-ombinationsto(t+1)-ombinations:
= fU j2g; for2U; (53)
A = S
f
j2Ag; forAU: (54)
If,for example,U =f0;1;2;3;4;5;6g, wehave
5310=f53210;54310;65310g;
ontheotherhand,ifU =f10;11;:::;16g,wehave
5310=f53100;53110;
53210;53310;54310;55310;65310g.
Thefollowingfundamentaltheorems,whihhavemanyappliationsinvar-
ious branhes of mathematis and omputer siene, tell us how small a set's
shadowsan be:
TheoremK. IfAisasetofNt-ombinationsontainedinU =f0;1;:::;n 1g,
then
jAjjP
Nt
j and j
Ajj
Q
Nnt
j; (55)
whereP
Nt
denotestherstN ombinationsgenerated byAlgorithmL,namely
the N lexiographially smallest ombinations
t :::
2
1
that satisfy (3), and
Q
Nnt
denotestheN lexiographiallylargest.
Theorem M. IfAisasetofN t-multiombinationsontainedinthemultiset
U =f10;11;:::;1sg,then
jAjj b
P
Nt
j and j
Ajj
b
Q
Nst
j; (56)
where b
P
Nt
denotestheNlexiographiallysmallestmultiombinationsd
t :::d
2 d
1
thatsatisfy(6),and b
Q
Nst
denotes theN lexiographiallylargest.
Both ofthese theorems areonsequenes of a strongerresult thatweshall
provelater. TheoremKisgenerallyalledtheKruskal{Katonatheorem,beause
it was disovered by J. B.Kruskal[Math. OptimizationTehniques,edited by
R.Bellman(1963),251{278℄ andredisoveredbyG.Katona[TheoryofGraphs,
Tihany 1966, edited by Erd}os and Katona (Aademi Press, 1968), 187{207℄;
M.P.Shutzenberger hadpreviouslystateditinaless-well-knownpubliation,
with inomplete proof [RLE Quarterly Progress Report 55 (1959), 117{118℄.
Theorem M goes bak to F. S. Maaulay, many years earlier [Pro. London
Math.So.(2)26 (1927),531{555℄.
Beforeproving(55)and(56),let'stakealoserlookatwhatthoseformulas
mean. Weknow from TheoremL that therstN of allt-ombinationsvisited
byAlgorithmL arethosethatpreeden
t :::n
2 n
1 ,where
N =
n
t
t
++
n
2
2
+
n
1
1
; n
t
>>n
2
>n
1 0
isthedegree-tombinatorialrepresentationofN. Sometimesthisrepresentation
hasfewerthantnonzeroterms,beausen
j
anbeequaltoj 1;let'ssuppress
thezeros,andwrite
N =
n
t
t
+
n
t 1
t 1
++
n
v
v
; n
t
>n
t 1
>>n
v
v1: (57)
Nowtherst n
t
t
ombinations
t :::
1
arethet-ombinationsoff0;:::;n
t 1g;
thenext nt
1
t 1
arethoseinwhih
t
=n
t and
t 1 :::
1
isa(t 1)-ombination
off0;:::;n
t 1
1g;andsoon. Forexample,ift=5andN = 9
5
+ 7
4
+ 4
3
,the
rstN ombinationsare
P
N5
=f43210;:::;87654g[f93210;:::;96543g[f97210;:::;97321g: (58)
Theshadowofthis setP
N5
is,fortunately,easytounderstand: Itis
P
N5
=f3210;:::;8765g[f9210;:::;9654g[f9710;:::;9732g; (59)
namelytherst 9
4
+ 7
3
+ 4
2
ombinationsinlexiographiorderwhent=4.
Inotherwords,ifwedeneKruskal'sfuntion
t
bytheformula
t N =
n
t
t 1
+
n
t 1
t 2
++
n
v
v 1
(60)
whenN hastheuniquerepresentation(57),wehave
P
Nt
= P
(tN)(t 1)
: (61)
Theorem K tells us, for example, that a graph with a million edges an
ontainat most
1414
3
+
1009
2
= 470;700; 300
triangles,that is, at most 470,700,300sets ofvertiesfu;v;wg with u v
w u.Thereasonisthat1000000= 1414
2
+ 1009
1
byexerise17,andtheedges
P
(1000000)2
dosupport 1414
3
+ 1009
2
triangles;butifthereweremore,thegraph
would neessarilyhave atleast
3
470700301= 1414
2
+ 1009
1
+ 1
0
=1000001
edgesintheirshadow.
Kruskaldened theompanionfuntion
t N =
n
t
t+1
+
n
t 1
t
++
n
v
v+1
(62)
to deal with questions suh as this. The and funtions are related by an
interestinglawprovedinexerise 72:
M +N =
s+t
t
implies
s M+
t N =
s+t
t+1
; ifst>0. (63)
Turning toTheoremM,thesizesof b
P
Nt and
b
Q
Nst
turn outtobe
j b
P
Nt j =
t
N and j
b
Q
Nst
j = N+
s
N (64)
(seeexerise 81),wherethefuntion
t
satises
t N =
n
t 1
t 1
+
n
t 1 1
t 2
++
n
v 1
v 1
(65)
whenN hastheombinatorialrepresentation (57).
Table 3 shows how these funtions
t N,
t
N, and
t
N behave for small
values of t and N. When t and N are large, they an be well approximated
intermsof a remarkablefuntion (x) introdued by TeijiTakagi in1903; see
Fig.27andexerises82{85.
TheoremsKandMareorollariesofamuhmoregeneraltheoremofdisrete
geometry,disoveredbyDa-LunWangandPingWang[SIAMJ.AppliedMath.
33 (1977), 55{59℄, whih we shall now proeed to investigate. Consider the
disreten-dimensional torus T(m
1
;:::;m
n
)whoseelementsareinteger vetors
x=(x
1
;:::;x
n
)with0 x
1
<m
1
,:::, 0x
n
<m
n
. Wedene thesumand
diereneoftwo suhvetorsxandy asinEqs.4.3.2{(2) and4.3.2{(3):
x+y = (x
1 +y
1
)modm
1
;:::;(x
n +y
n
)modm
n
; (66)
x y = (x
1 y
1
)modm
1
;:::;(x
n y
n
)modm
n
: (67)
Wealsodene theso-alledrossorder onsuh vetors bysayingthat xy if
andonlyif
x<y or (x=y andxylexiographially); (68)
here,as usual,(x
1
;:::;x
n )=x
1
++x
n
. For example,whenm
1
=m
2
=2
andm
3
=3,the12vetorsx
1 x
2 x
3
inrossorder are
000; 100; 010; 001; 110; 101; 011; 002; 111; 102; 012; 112; (69)
Table3
EXAMPLESOF THE KRUSKAL{MACAULAYFUNCTIONS, , AND
N= 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
1
N= 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2N= 0 2 3 3 4 4 4 5 5 5 5 6 6 6 6 6 7 7 7 7 7
3
N= 0 3 5 6 6 8 9 9 10 10 10 12 13 13 14 14 14 15 15 15 15
4N= 0 4 7 9 10 10 13 15 16 16 18 19 19 20 20 20 23 25 26 26 28
5
N= 0 5 9 12 14 15 15 19 22 24 25 25 28 30 31 31 33 34 34 35 35
1N= 0 0 1 3 6 10 15 21 28 36 45 55 66 78 91 105120136153171190
2
N= 0 0 0 1 1 2 4 4 5 7 10 10 11 13 16 20 20 21 23 26 30
3
N= 0 0 0 0 1 1 1 2 2 3 5 5 5 6 6 7 9 9 10 12 15
4N= 0 0 0 0 0 1 1 1 1 2 2 2 3 3 4 6 6 6 6 7 7
5
N= 0 0 0 0 0 0 1 1 1 1 1 2 2 2 2 3 3 3 4 4 5
1N= 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2N= 0 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6
3
N= 0 1 2 3 3 4 5 5 6 6 6 7 8 8 9 9 9 10 10 10 10
4N= 0 1 2 3 4 4 5 6 7 7 8 9 9 10 10 10 11 12 13 13 14
5
N= 0 1 2 3 4 5 5 6 7 8 9 9 10 11 12 12 13 14 14 15 15
0 22
14
7
0 7
5
8
5 5N N
8
5
+ 7
4
9
5
0 2/3
1/2
1/4
0
1/4 1/2 (x )
3/4 1
Fig. 27. ApproximatingaKruskal funtionwith the Takagifuntion. (The
smoothurveintheleft-handgraphisthelowerbound
5
N N ofexerise80.)
omittingparentheses andommasforonveniene. Theomplement ofavetor
inT(m
1
;:::;m
n )is
x = (m
1
1 x
1
;:::;m
n
1 x
n
): (70)
Notiethatxyholds ifandonlyif xy. Thereforewehave
rank(x)+rank(x) = T 1; whereT =m
1 :::m
n
, (71)
ifrank(x)denotesthenumberofvetorsthat preedexinrossorder.
Wewillnditonvenienttoallthevetors\points"andtonamethepoints
e
0 ,e
1 ,:::, e
T 1
ininreasingrossorder. Thus wehavee
7
=002 in(69),and
e
r
=e
T 1 r
ingeneral. Notiethat
e
1
=100:::00; e
2
=010:::00; :::; e
n
=000:::01; (72)
thesearetheso-alledunitvetors. Theset
S
N
= fe
0
;e
1
;:::;e
N 1
g (73)
onsisting of thesmallest N points is alleda standard set, and inthe speial
aseN =n+1wewrite
E =fe
0
;e
1
;:::;e
n
g=f000:::00;100:::00;010:::00;:::;000:::01g: (74)
Anyset of pointsX hasa spread X +
, a ore X Æ
, and a dual X
, dened
bytherules
X +
= fx2S
T
jx2X orx e
1
2X or orx e
n
2Xg; (75)
X Æ
= fx2S
T
jx2X andx+e
1
2X and andx+e
n
2Xg; (76)
X
= fx2S
T
jx2=Xg: (77)
Wean alsodenethespreadofX algebraially,writing
X +
= X+E; (78)
whereX+Y denotesfx+yjx2X andy2Y g. Clearly
X +
Y ifandonlyif X Y Æ
: (79)
Thesenotionsanbeillustratedinthetwo-dimensionalasem
1
=4,m
2
=6,by
themore-or-lessrandomtoroidalarrangementX =f00;12;13;14;15;21;22;25g
forwhihwehave,pitorially,
Æ
+ +
+ +
+ + +
+
Æ
Æ Æ
Æ Æ
Æ Æ Æ
+ + + + + +
+
; (80)
X X
Æ
andX +
X
X Æ
andX +
hereX inthersttwodiagramsonsistsofpointsmarked orÆ,X Æ
omprises
justthe Æs,and X +
onsistsof +spluss plusÆs. Notiethat ifwerotate the
diagramforX Æ
andX +
by180 Æ
,weobtainthediagramforX Æ
andX +
,but
with(;Æ;+; )respetivelyhanged to(+; ;;Æ); andinfattheidentities
X Æ
= X +
; X
+
= X Æ
(81)
holdingeneral(seeexerise86).
NowwearereadytostatethetheoremofWangandWang:
TheoremW. LetXbeanysetofN pointsinthedisretetorusT(m
1
;:::;m
n ),
wherem
1
m
n
. ThenjX +
jjS +
N
jand jX Æ
jjS Æ
N j.
Inotherwords,thestandardsetsS
N
havethesmallestspread andlargestore,
amongallN-pointsets. Wewillprovethisresultbyfollowingageneralapproah
rstusedbyF.W.J.Whipple toproveTheoremM[Pro.London Math. So.
(2)28(1928),431{437℄. Therststepisto provethat thespread andtheore
ofstandardsetsarestandard:
Lemma S. Therearefuntionsand suh thatS
N
=S
N andS
Æ
N
=S
N .
Proof. WemayassumethatN >0. Letrbemaximumwithe
r 2S
+
N
,andlet
N = r+1; we must provethat e
q 2 S
+
N
for 0 q < r. Suppose e
q
= x=
(x
1
;:::;x
n )and e
r
=y =(y
1
;:::;y
n
), and letk bethe largest subsript with
x
k
>0. Siney 2S +
N
,thereisasubsriptj suhthaty e
j 2S
N
. ItsuÆesto
provethat x e
k
y e
j
,andexerise 88doesthis.
The seond part follows from (81), with N = T (T N), beause
S
N
=S
T N .
Theorem Wis obviouslytruewhenn=1,so weassumebyindutionthat
ithasbeenprovedinn 1dimensions. Thenextstepisto ompress thegiven
setX inthekthoordinateposition,bypartitioningitintodisjointsets
X
k
(a) = fx2X jx
k
=ag (82)
for0a<m
k
and replaingeahX
k (a)by
X 0
k
(a) = f(s
1
;:::;s
k 1
;a;s
k
;:::;s
n 1 )j(s
1
;:::;s
n 1 )2S
jX
k (a)j
g; (83)
asetwiththesamenumberofelements. ThesetsSusedin(83)arestandardin
the(n 1)-dimensionaltorus T(m
1
;:::;m
k 1
;m
k+1
;:::;m
n
). Notie that we
have (x
1
;:::;x
k 1
;a;x
k+1
;:::;x
n ) (y
1
;:::;y
k 1
;a;y
k+1
;:::;y
n
) if and only
if(x
1
;:::;x
k 1
;x
k+1
;:::;x
n ) (y
1
;:::;y
k 1
;y
k+1
;:::;y
n
); therefore X 0
k (a)=
X
k
(a)ifandonlyifthe(n 1)-dimensionalpoints(x
1
;:::;x
k 1
;x
k+1
;:::;x
n )
with(x
1
;:::;x
k 1
;a;x
k+1
;:::;x
n
)2X areassmallaspossiblewhenprojeted
onto the(n 1)-dimensionaltorus. Welet
C
k
X = X 0
k (0)[X
0
k
(1)[[X 0
k (m
k
1) (84)
bethe ompression of X inposition k. Exerise 90 proves the basi fat that
ompressiondoesnotinreasethesizeofthespread:
jX +
j j(C
k X)
+
j; for1kn: (85)
Furthermore,ifompressionhangesX,itreplaessomeoftheelementsbyother
elementsoflowerrank. ThereforeweneedtoproveTheoremWonlyforsetsX
thataretotallyompressed,havingX =C
k
X forallk.
Consider, for example, the ase n = 2. A totally ompressed set in two
dimensionshasallpointsmovedtotheleftoftheirrowsandthebottomoftheir
olumns,as intheeleven-pointsets
+ + + + + +
or
+ + + + +
or
+ + + + +
or
+ + + + +
or
+ + + +
;
the rightmost of these is standard, and has the smallest spread. Exerise 91
ompletestheproofofTheoremWintwo dimensions.
