The Massey's method for sport rating: a network science perspective

0.17 MB·

by  Massimo Franceschet

The Massey’s method for sport rating:
a network science perspective
EnricoBozzo
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1 DepartmentMathematics,ComputerScience,andPhysics
0 UniversityofUdine
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[email protected]

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a MassimoFranceschet
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DepartmentMathematics,ComputerScience,andPhysics
2
UniversityofUdine
1
[email protected]

]
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S January13,2017
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Abstract

1 WerevisittheMassey’smethodforratingandrankinginsportsandcontextu-
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alizeitasageneralcentralitymeasureinnetworkscience.
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3 1 Introduction
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. Rating and ranking in sport have a flourishing tradition. Each sport competition has
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its own official rating, from which a ranking of players and teams can be compiled.
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The challenge of many sports’ fans and bettors is to beat the official rating method:
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1 todevelopanalternativeratingalgorithmthatisbetterthantheofficialoneinthetask
: ofpredictingfutureresults. Asaconsequence, manysportratingmethodshavebeen
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i developed.AmyN.LangvilleandCarlD.Meyerevenwrotea(compelling)bookabout
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(general)ratingandrankingmethodsentitledWho’s#1? [11].
r In 1997, Kenneth Massey, then an undergraduate, created a method for ranking
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college football teams. He wrote about this method, which uses the mathematical
theoryofleastsquares,ashishonorsthesis[12]. Informally,atanygivendaykofthe
season,Massey’smethodratesateamiaccordingtothefollowingtwofactors: (a)the
differencebetweenpointsforandpointsagainsti,orpointspreadofi,uptodayk,and
(b)theratingsoftheteamsthatimatcheduptodayk. Hence,highlyratedteamshave
a large point differential and matched strong teams so far. Below in the ranking are
teamsthatdidwellbuthadaneasyscheduleaswellasteamsthatdidnotsowellbut
hadatoughschedule.
In this paper, we review the original Massey’s method and explicitly provide the
algebraic background of the method. Moreover, we interpret Massey’s technique in
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thecontextofnetworkscience. Thepaperisorganizedasfollows. Section2reviews
theoriginalMassey’smethod. Inparticular,Section2.1providesafullexampleofthe
method,Section2.2embedsthemethodinthecontextofnetworkscience,andSection
2.3givesthealgebraicbackgroundoftheMassey’smethodforthecuriousreader. We
review related methods for sport rating in Section 3. Finally, we discuss alternative
(outsidethesportcontext)usesofMassey’smethodsinSection4.
2 The Massey’s method for sports ranking
The main idea of Massey’s method, as proposed in [12], is enclosed in the following
equation:
r −r =y
i j k
wherer andr aretheratingsofteamsiand jandy istheabsolutemarginofvictory
i j k
forgamekbetweenteamsiand j. Iftherearenteamswhoplayedmgames,wehave
alinearsystem:
Xr=y (1)
where X is a m×n matrix such the k-th row of X contains all 0s with the exception
of a 1 in location i and a −1 in location j, meaning that team i beat team j in match
k (ifmatchk endswithadraw, eitherior j locationcanbeassigned1, andtheother
−1). Observethat, ifedenotesthevectorofall1’s, thenXe=0. LetM=XTX and
p=XTy. Noticethat
(cid:26)
thenegationofthe#ofmatchesbetweeniand j ifi(cid:54)= j,
M =
i,j #ofgamesplayedbyi ifi= j.
and p isthesignedsumofpointspreadsofeverygameplayedbyi. Clearlytheentries
i
ofpsumto0,infacteTp=eTXTy=(Xe)Ty=0.TheMassey’smethodisthendefined
bythefollowinglinearsystem:
Mr=p (2)
whichcorrespondstotheleastsquaressolutionofsystem(1).
We observe how the Massey’s team ratings are in fact interdependent. Indeed,
Massey’smatrixMcanbedecomposedas
M=D−A,
whereDisadiagonalmatrixwithD equaltothenumberofgamesplayedbyteam
i,i
i,andAisamatrixwithA equaltothenumberofmatchesplayedbyteamiagainst
i,j
team j. Hence,linearsystem(2)isequivalentto
Dr−Ar=p, (3)
2
or,equivalently
r=D−1(Ar+p)=D−1Ar+D−1p. (4)
Thatis,foranyteami
1 p
r = ∑A r + i .
i i,j j
D D
i,i j i,i
(1) (2)
Thismeansthattheratingr ofteamiisthesumr +r oftwomeaningfulcompo-
i i i
nents:
1. themeanratingofteamsthatihasmatched
1
r(1)= ∑A r ;
i D i,j j
i,i j
2. themeanpointspreadofteami
p
(2) i
r = .
i D
i,i
How do we solve the Massey’s system Mr= p? Unfortunately, M =D−A is a
singular matrix, hence it is not possible to solve this system computing the inverse
of M. Notice that the symmetric matrix A can be interpreted as the adjacency matrix
of an undirected weighted graph that we denote with G . It is well known that A is
A
irreducible if and only if G is connected. Assuming that matrix A is irreducible, or
A
equivalentlythatthegraphG isconnected, asolutionofthesystemcanbeobtained
A
asfollows: letMˆ bethematrixobtainedbyreplacinganyrow,saythelast,ofMwitha
rowofall1’s,andlet pˆbethevectorobtainedbyreplacingthelastelementof pwith
a0. Then,solvethesystem:
Mˆr=pˆ (5)
Notice that the added constraint forces the ranking vector r to sum to 0. See Section
2.3forthemathematicaldetails.
WhatabouttheconnectivityhypothesisofthegraphofmatchesG ? Withoutsuch
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assumption, the ranking cannot be computed. Indeed, this is not a strong constraint.
Assume, realistically, a competition in which there are n teams matching. At each
seasondayeachteammatchesanotherteamnotmatchedbefore. Hence,forallteams
to be matched at any day, n must be even and we have n/2 matches each day for
a maximum of n−1 days before teams match twice. This is known as round-robin
competition. AtseasondaykthegraphG isobtainedasthedisjointunionofkperfect
A
matchings, hence, in particular, it is regular with degree k. Moreover, we have the
followingresult:
Lemma1. Givenn>2teams,withneven,competinginaround-robincompetition,
then:
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1. theminimumnumberofdaysforthegraphofmatchestobeconnectedis2;
2. themaximumnumberofdaysforthegraphofmatchestobeconnectedisn/2.
Proof. Weproveitem1. Afteroneseasondaythegraphofmatchesisnotconnected
beingjustaperfectmatchingoftheteams. Aftertwodaythegraphcanbecomeacycle
andhencecanbeconnected.
Wenowproveitem2. Firstofall,letusobservethataftern/2−1seasondaysthe
graphcanbenotconnected,buttheonlypossibilityisthatitisformedbytwocomplete
graphs with nodes inV andV having each n/2 nodes. At day n/2, all matches are
1 2
betweenpairsofteamsoneofwhichisinV andtheotherinV ,resultinginaconnected
1 2
graph.
Typically,theactualnumberofdaysforthematchgraphtobeconnectediscloseto
theminimum. Forinstance,weexperimentedthatinthelast11editionsoftheItalian
soccerleague(SerieA),after2or3daysthematchgraphisconnected,withameanof
2.6.
Masseyalsoproposedanoffensiverating(o)andadefensiverating(d)character-
izingtheoffensiveanddefensivestrengthsofteams. Masseyassumesthatr=o+d,
thatis,theoverallstrengthofateamisthesumofitsoffensiveanddefensivepowers.
Let’sdecomposethepointspreadvector p= f−a,where f holdsthetotalnumberof
pointsscoredbyeachteamandaholdsthetotalnumberofpointsscoredagainsteach
team. Then,theequationsdefiningoandd are:
(cid:40)
Do−Ad= f
Ao−Dd=a.
Thatisforeachteami:
(cid:40)oi= D1i,i(∑jAi,jdj+fi)
di= D1i,i(∑jAi,joj−ai)
Thismeansthattheoffensiveratingofteamimultipliedbythenumberofgamesplayed
byiisequaltothedefensiveratingsofopponentsofiplusthenumberofpointsscored
by i. On the other hand, the defensive rating of team i multiplied by the number of
gamesplayedbyiisequaltotheoffensiveratingsofopponentsofiminusthenumber
ofpointsscoredagainsti.
Sinceweknowthatr=o+d,wehavethat,knowingr,thedefensiveratingd can
becomputedas:
(D+A)d=Dr−f (6)
and,knowingd andr,theoffensiveratingo=r−d.
Howdowesolvesystem(6)? Infact,itmighthappenthatmatrixD+Aissingular,
hence it cannot be inverted. Assuming that the graph G is connected, it holds that
A
D+AissingularpreciselywhenG isbipartite. LetN=D+Aandq=Dr−f. IfG
A A
isconnectedandbipartitewithnodesetsU andV,wecanapplytothesystemNd=q
aperturbationtricksimilartotheoneexploitedfortheMassey’ssystemMr= p. Let
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Nˆ bethematrixobtainedbyreplacinganyrow,saythelast,ofN withvectorvdefined
as v =1 if i∈U and v =−1 if i∈V and let qˆ be the vector obtained by replacing
i i
the last element of q with a 0. Then, a solution of system (6) is obtained solving the
followingperturbedsystem:
Nˆd=qˆ (7)
Again, seeSection2.3forthemathematicaldetails. Assumingaround-robincompe-
tition, the hypothesis that the match graph G is non-bipartite is not a strong one, as
A
provedinthefollowingresult.
Lemma2. Givenn>2teams,withneven,competinginaround-robincompetition,
then:
1. theminimumnumberofdaysforthegraphofmatchestobenon-bipartiteis3;
2. the maximum number of days for the graph of matches to be non-bipartite is
n/2+1.
Proof. Weproveitem1. Aftertwodaysthegraphofthematchescannothavecycles
oflength3. After3daysitispossibletohavecyclesoflength3sothatthegraphcan
benon-bipartite.
Wenowproveitem2. Aftern/2daysthegraphofthematchescanbebipartitebut
theonlypossibilityisthatitisacompletebipartitegraphwithnodesetsV andV such
1 2
that|V |=|V |=n/2. Atdayn/2+1allmatchesarebetweenpairsofnodeseitherin
1 2
V orinV . Hence, somecyclesoflength3areformedandthereforethegraphisno
1 2
morebipartite.
Typically,theactualnumberofdaysforthematchgraphtobenon-bipartiteisclose
totheminimum.Forinstance,weexperimentedthatinthelast11editionsoftheItalian
soccer league (Serie A), within 5 days the match graph is non-bipartite, with a mean
of3.7. Hence,weexpectthattheMassey’smethodisfullyapplicableafterfewseason
daysinthecompetition.
2.1 ExampleofMassey’smethod
The table below shows the results of 4 matches (numbered 1, 2, 3, 4) involving 4
fictitiousteams(labelledA,B,C,D):
match team1 team2 score1 score2
1 A C 2 0
2 A D 3 0
3 B C 1 1
4 B D 2 1
Thematch-teammatrixX isgivenbelow:
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A B C D
1 1 0 -1 0
2 1 0 0 -1
3 0 1 -1 0
4 0 1 0 -1
Thematchspreadvectoryis:
team y
1 2
2 3
3 0
4 1
TheMassey’smatrixM=XTX is:
A B C D
A 2 0 -1 -1
B 0 2 -1 -1
C -1 -1 2 0
D -1 -1 0 2
andtheteamspreadvector p=XTyis:
team p
A 5
B 1
C -2
D -4
TheresultingMassey’sratingisthefollowing:
team r
A 1.75
B -0.25
C -0.25
D -1.25
NoticethatteamsBandChavethesamerating(andhenceranking),despitethepoint
spreadofB(1)ishigherthanthepointspreadofC(-2). ThisisbecauseBagainstC
endedinadraw,Bwonthesecondmatch,butagainsttheweakestteam(D),whileC
lost the second match, but against the strongest team (A). Indeed, the rating r can be
(1) (2)
decomposedinthepartialratingsr andr thatfollows:
i i
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(1) (2)
team r r r
i i
A 1.75 -0.75 2.5
B -0.25 -0.75 0.5
C -0.25 0.75 -1.0
D -1.25 0.75 -2.0
Hence,BhasabetterpointspreadbutaneasierschedulewithrespecttoC. Summing
thetworatingcomponenteswegetthesameratingforteamsBandC.
Moreover, the rating r can be decomposed in the offensive and defensive ratings
thatfollows:
team r o d
A 1.75 1.875 -0.125
B -0.25 0.875 -1.125
C -0.25 -0.125 -0.125
D -1.25 -0.125 -1.125
wherethepointsforandagainstvectorsareasfollows:
team p f a
A 5 5 0
B 1 3 2
C -2 1 3
D -4 1 5
Notice that team B has a better offensive rating thanC, but a worse defensive rating,
andthesumofoffensiveanddefensivepowersisthesameforbothteams. Also,Aand
Chavethesamedefensiverating,althoughAhas0pointsagainstandChas3points
again;again,thisbecausethescheduleofCisharderthanthescheduleofA.
2.2 Massey’smethodandnetworkscience
WeobservethatEquation(4)isclosetotheequationdefiningKatz’scentrality[7]:
r=αAr+β (8)
whereα isagivenconstantandβ isagiven(exogenous)vector. TheKatz’sequation
can be solved by inverting matrix I−αA as soon as α does not coincide with the
reciprocal of an eigenvalue of A [13]. In particular, making the assumption that all
teamsplayedthesamenumberkofmatches,thenEquation(3)becomes
1
r= (Ar+p), (9)
k
which corresponds to Katz’s centrality with parameters α =1/k and β = p/k. Nev-
ertheless, notice that in this setting k is an eigenvalue of A. It follows that Massey’s
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equationisaninstanceofKatz’scentrality,butcannotbecomputedbyinvertingmatrix
I−αA.
Furthermore,inthefollowing,weproposeanelectricalinterpretationoftheMassey’s
rating. If we view the symmetric matrix A as the adjacency matrix of an undirected
graphG ,thenMistheLaplacianmatrixofthegraphG . Theratingvectorrdefined
A A
insystem(2)isthenequivalenttothepotentialvectoroveraresistornetworkdefined
byAwithsupplyvector p[5].
Following this metaphor, the resistor network has a resistor between nodes i and
j as soon as teams i and j has matched with conductance (reciprocal of resistance)
equal to the number A of matches between i and j. Sources, that are nodes i with
i,j
positive supply p through which current enters the network, are teams with positive
i
point spread, while targets, that are nodes i with negative supply p through which
i
current leaves the network, are teams with negative point spread. Notice that current
enteringandleavingthenetworkmustbeequal,indeedthepointspreadvector psums
to0. Thepotentialr fornodeicorrespondstotheratingofteami: teamswithlarge
i
potentialareteamshighintheranking. Moreover,thecurrentflowthroughedge(i,j)
is,byOhm’slaw,thequantityA (r −r ),whichcorrespondstotheratingdifference
i,j i j
betweenteamsiand j(whichisalsoanestimateofthepointspreadinamatchbetween
iand j)multipliedbythenumberoftimestheymatched: currentflowsmoreintensely
betweenteamsofdifferentstrengths(asmeasuredbyMassey’smethod)thatmatched
manytimes.
Finally,itisinterestingtoanalysewhathappenstoMassey’ssystemattheendof
the season, assuming that all n teams matched all other teams once. In this case, the
opponentsratingcomponent
r
(1) i
r =− ,
i n−1
wherewehaveusedthefactthat∑iri=0,andthepointspreadcomponent
p
(2) i
r = ,
i n−1
hence
r p
(1) (2) i i
r =r +r =− + ,
i i i n−1 n−1
andthus
p
i
r = .
i
n
Thesameresultisobtainednoticingthat
Mp=(D−A)p=(n−1)p−Ap=np−(p+Ap)=np
where we have used the fact that p sums to 0. Hence p is an eigenvector of M with
eigenvalue n and hence, once again, r = p/n is a solution of the Massey’s system.
Hence,thefinalratingofateamissimplythemeanpointspreadoftheteam.
It is possible to be a bit more precise about this property of Massey’s method by
exploitingthepropertiesofthesetofeigenvalues,orspectrum,oftheLaplacianmatrix
M =D−A. The spectrum reflects various aspects of the structure of the graph G
A
associated with A, in particular those related to connectedness. It is well known that
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theLaplacianissingularandpositivesemidefinite(recallthatM=XTX andXe=0)
sothatitseigenvaluesarenonnegativeandcanbeorderedasfollows:
λ =0≤λ ≤λ ≤...≤λ .
1 2 3 n
It can be shown that λ ≤n, see for example [2]. The multiplicity of λ =0 as an
n 1
eigenvalueoftheLaplaciancanbeshowntobeequaltothenumberoftheconnected
components of the graph, see again [2]. If the graph of the matches is connected or,
equivalently, M is irreducible, as we assume in the following, λ (cid:54)=0 is known as
2
algebraicconnectivityofthegraphandisanindicatoroftheefforttobeemployedin
ordertodisconnectthegraph.
WecanwritethespectraldecompositionofMasM=UDUT whereUisorthogonal
√
anditsfirstcolumnisequaltoe/ n,andD=diag(0,λ ,...,λ ). FromMr= pwe
2 n
obtainr=UD+UTpwhereD+=diag(0, 1 ,..., 1 ). Now
λ2 λn
p p (cid:104) I(cid:105)
r− =UD+UTp− =U D+− UTp,
n n n
where I is the identity matrix. Observe that the first component of the vectorUTp is
equaltozerosothat
p (cid:104) I(cid:105) (cid:104) I˜(cid:105)
r− =U D+− UTp=U D+− UTp,
n n n
whereI˜=diag(0,1,...,1). Ifwedenotewith(cid:107)·(cid:107)theEuclideannormweobtain
(cid:107)r− p(cid:107)=(cid:107)U(cid:2)D+− I˜(cid:3)UTp(cid:107)≤(cid:107)p(cid:107) max (cid:12)(cid:12) 1 −1(cid:12)(cid:12)≤(cid:107)p(cid:107)n−λ2,
n n k=2,...,n(cid:12)λk n(cid:12) nλ2
whereweusedthefactthattheEuclideannormofanorthogonalmatrixisequaltoone.
Hence, asthealgebraicconnectivityλ , aswellastheothereigenvalues, approachn,
2
that is, as more and more matches are played, the vector r approaches p/n and the
equalityisreachedwhenthegraphofthematchesbecomescomplete.
2.3 Theunderlyinglinearalgebra
Let n≥2. An n×n matrix A is defined to be strictly diagonally dominant if |A |>
i,i
∑j(cid:54)=i|Ai,j|fori=1,...,n. ByusingGers˘gorintheoremitiseasytoprovethatifAis
strictly diagonally dominant then it is nonsingular, see [6]. A matrix A is defined to
bediagonallydominantif|Ai,i|≥∑j(cid:54)=i|Ai,j|fori=1,...,n. AmatrixAisdefinedto
be irreducibly diagonally dominant if it is irreducible, it is diagonally dominant, and
foratleastonevalueofithestrictinequality|Ai,i|>∑j(cid:54)=i|Ai,j|holds. Anirreducibly
diagonallydominantmatrixisnonsingular, seeagain[6]. Withanonstandardtermi-
nology,wedefinethematrixAtobeexactlydiagonallydominantif|Ai,i|=∑j(cid:54)=i|Ai,j|
fori=1,...,n. WeintroducethisconceptsincetheLaplacianandthesignlessLapla-
cianofagraphareexactlydiagonallydominant. ThesignlessLaplacianofagraphis
definedasthematrixwhoseentriesaretheabsolutevaluesoftheentriesoftheLapla-
cian [3]. For example, the matrix M =D−A of system (2) is the Laplacian of the
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graphG , andthematrixN =D+Aofsystem(6)isitssignlessLaplacian. Thefol-
A
lowinglemmapointsoutaninterestingpropertyofasymmetricirreducibleandexactly
diagonallydominantmatrix.
Lemma3. Letn≥2,Abeasymmetricirreducibleexactlydiagonallydominantn×n
matrix and k∈{1,2,...,n}. The (n−1)×(n−1) matrix A(k) obtained by deleting
fromAthek-throwandthek-thcolumnisnonsingular.
Proof. First of all notice that A cannot have diagonal elements equal to zero, since,
beingexactlydiagonallydominant,thiswouldimplythatAhasarowcompletelyequal
tozero,andthisconflictswiththeassumptionthatAisirreducible.
The matrix A(k) is the adjacency matrix of the subgraph G obtained by the
A(k)
elimination of node k and of the edges incident in that node from G . The matrix
A
A(k) can be transformed by a simultaneous permutation of rows and columns into a
block diagonal matrix where each diagonal block is relative to one of the connected
componentsofthesubgraph.Wewanttoshowthateachoftheseblocksisnonsingular.
If the block has dimension 1, then it contains one of the diagonal elements of A,
henceitsonlyentryisdifferentfromzero. Ifthedimensionoftheblockisbiggerthan
1,thentheblockisirreduciblebecauseitsgraphisconnectedbyconstruction. More-
over,sinceAisexactlydiagonallydominant,eachoftheblocksofA(k)isdiagonally
dominantandthestrictinequalityholdsforoneofthevaluesoftheindexineachblock
becauseatleastoneofthedeletededgeswasincidentinoneofthenodesofthecon-
nected component. Hence the block is irreducibly diagonally dominant so that it is
nonsingular.
Finally, the matrix A(k) is nonsingular since the diagonal blocks are nonsingular.
From Lemma 3 we deduce that a symmetric irreducible and exactly diagonally
dominantn×nmatrixAhasrankatleastn−1andmorespecifically,ifwedeletean
arbitrary row from A, the remaining rows are linearly independent. Hence, either A
is nonsingular, or its nullspace has dimension one. For example the Laplacian of a
connectedgraphisalwayssingular,whileitssignlessLaplacianissingularifandonly
ifthegraphisbipartite[3]. IfAissingularandv(cid:54)=0issuchthatAv=0thenvcannot
haveentriesequaltozerosinceallthesubsetsofn−1rows,orequivalentlycolumns,
of A are linearly independent. For example for the Laplacian of a connected graph
v=e while for its signless Laplacian, in the case where the graph is bipartite, v has
componentsequalto1or−1incorrespondencewiththetwopartsofthegraph.
Theorem1. LetAbeasymmetricirreducibleandexactlydiagonallydominantmatrix
andletv(cid:54)=0besuchthatAv=0. Thefollowingassertionsholdtrue.
1. ThelinearsystemAx= f issolvableifandonlyif f isavectorsuchthatvTf =0.
2. Bysubstitutingthek-throwofAwithvT weobtainanonsingularmatrix.
3. LetAˆ thematrixobtainedfromAbysubstitutingthek-throwofAwithvT and
let fˆthevectorobtainedfrom f bysubstitutingthek-thentryof f withzero. The
solutionx∗ofthenonsigularsystemAˆx= fˆissuchthatAx∗= f.
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