Consider a soccer competition among various teams playing against each other in pairs (matches) according to a previously determined schedule. At some stage of the competition one may ask whether a particular team still has a (theoretical) chance to win the competition. The complexity of this question depends on the way scores are allocated according to the outcome of a match. For example, the problem is polynomially solvable for the ancient FIFA rules (2:0 resp. 1:1) but becomes NP-hard if the new rules (3:0 resp. 1:1) are applied. We determine the complexity of the above problem for all possible score allocation rules. \u

Kern, W.

Paulusma, D.

English

Consider a soccer competition among various teams playing against each other in pairs (matches) according to a previously determined schedule. At some stage of the competition one may ask whether a particular team still has a (theoretical) chance to win the competition. The complexity of this question depends on the way scores are allocated according to the outcome of a match. For example, the problem is polynomially solvable for the ancient FIFA rules (2:0 resp. 1:1) but becomes NP-hard if the new rules (3:0 resp. 1:1) are applied. We determine the complexity of the above problem for all possible score allocation rules

Kern, Walter

Paulusma, Daniël

NARCIS 

The new FIFA rules are hard: Complexity aspects of sports competitions

University of Twente Research Information

Faculty of Mathematical SciencesUniversity of TwenteUniversity for Technical and Social SciencesP.O. Box 2177500 AE EnschedeThe NetherlandsPhone: +31-53-4893400Fax: +31-53-4893114Email: memo@math.utwente.nlMemorandum No. 1505The new FIFA rules are hard:Complexity aspects of sports competitionsW. Kern and D. PaulusmaNovember 1999ISSN 0169-2690THE NEW FIFA RULES ARE HARD: COMPLEXITY ASPECTS OFSPORTS COMPETITIONSWALTER KERN AND DANIËL PAULUSMAABSTRACT. Consider a soccer competition among various teams playing againsteach other in pairs (matches) according to a previously determined schedule. Atsome stage of the competition one may ask whether a particular team still has a(theoretical) chance to win the competition. The complexity of this question de-pends on the way scores are allocated according to the outcome of a match. Forexample, the problem is polynomially solvable for the ancient FIFA rules (2:0resp. 1:1) but becomesN P-hard if the new rules (3:0 resp. 1:1) are applied. Wedetermine the complexity of the above problem for all possible score allocationrules.1. INTRODUCTIONConsider a sports competition like a national soccer league in which all participat-ing teams play against each other in pairs (matches) according to a prefixed sched-ule. Initially all teams have total score zero. When a team participates in a match,its total score is increased byÞ ∈R if it loses the match, byþ ∈R if the match endsin a draw, and by ∈R if it wins the match. We always assume thatÞ ≤ þ ≤  andcall the triple (Þ; þ; ) the rule (score allocation rule) of the competition. In caseof a soccer competition, the former FIFA rule was.Þ; þ; / = .0;1;2/, but thishas been changed into the new rule.Þ; þ; / = .0;1;3/. Other sports like chessor draughts still use the rule.Þ; þ; / = .0;1;2/, while stratego, also a strategicboard game, has as score allocation rule.Þ; þ; /= .0;1;6/.At a given stage of the competition one may ask whether a particular teamT0 stillhas a (theoretical) chance of “winning” the competition,i.e., ending up with thehighest final total score. To analyze this, we may w.l.o.g. assume thatT0 wins allremaining matches, resulting in a final total scores0 for T0 and a current total scoresi for all other teamsTi 6= T0. The question is now whether the teamsTi 6= T0 canfinish the remaining matches in such a way that eachTi ollects at mostci := s0− siadditional score points.This can be modeled by a multigraphG = .V; E/ whose vertices correspond toteamsTi 6= T0 and edges are in 1-1 correspondence with remaining matches. Eachnodei ∈ V has a capacityci ∈ R. We represent the outcome of a matche= .i; j/by directing the edge from the winner to the loser (and leaving the edge undirectedDate: September 23, 1999.1991Mathematics Subject Classification.03D15, 90C27.Key words and phrases. N P-complete, network flow.12 WALTER KERN AND DANIËL PAULUSMAin case of a draw). Our sports competition problem (“SC”) can now be formulatedas follows:SC(Þ; þ; )Given a multigraphG = .V; E/ and node capacitiesc ∈ RV can G be partiallyoriented such that for each nodei ∈ V:ÞŽ−.i /+ þŽ0.i /+ Ž+.i / ≤ ci ? .1:1/Here, as usual,Ž+ andŽ− denote the outdegree and indegree of a node, whereasŽ0 denotes the number of incident unoriented edges. A partial orientation ofGsatisfying the capacity constraints (1.1) is called asolutionof the instance (G; c).A simplified version of this (disallowing draws) was presented in Cooket al.[1998].In this case the problem reduces to a flow problem, cf. Cooket al. [1998] orsection 2 below. As we shall see, however, the question becomes more interest-ing if draws may occur. Our main result implies that in this case the problemis polynomially solvable ifÞ +  = 2þ (assumingP 6= N P). This means thatfor games like draughts and chess the problem is polynomially solvable. How-ever, for soccer competitions, by changing the score allocation rule into the rule.Þ; þ; / = .0;1;3/, the problem has becomeN P-complete. Also for strategocompetitions the problem isN P-complete.We end our introduction with the following simple observation. Given an instance(G; c) of SC(Þ; þ; ), we can derive an equivalent instance (G; c′) of SC(0; þ−Þ;  − Þ) by settingc′i := ci − ÞŽ.i /. (Here,Ž refers to the degree inG.) So withrespect to computational complexity of SC(Þ; þ; ) we may always assume that(Þ; þ; ) is normalized, i.e., Þ = 0≤ þ ≤ .2. COMPLEXITY RESULTSOur main result completely determines the computational complexity of the sportscompetition problem:Theorem 2.1. SC(Þ; þ; ) is polynomially solvable in each of the following threecases:(i) Þ = þ(ii) þ = (iii) Þ+  = 2þIn all other cases, the problem is N P-complete.Proof: First recall that we may assume (Þ; þ; ) is normalized, soÞ= 0. (Note thatnormalization does not affect (i)-(iii).)Case (i)is then trivial. Indeed, an instance(G; c) has a solution if and only ifc≥ 0. (Leave all edges unoriented.)In all other cases we haveþ > 0. By scaling, we may assume thatþ = 1. (Divideþ;  as well asc by þ.)Case (ii) þ =  = 1.Consider an instance given byG= .V; E/ andc ∈ RV. Construct a directed bi-partite graph with node setsV and E and arcs linking eachi ∈ V to all edges inCOMPLEXITY ASPECTS OF SPORTS COMPETITIONS 3E incident with i in G. Then add an additional sources and sinkt as indicated inFigure 2.1:.........(i,j)s tjiVEFigure 2.1The arcs froms to V all get lower capacity 0 and upper capacitybcic .i ∈ V/. Thearcs fromV to E get lower capacity 0 and upper capacity 1. The arcs fromE to tget lower capacity and upper capacity 1. The resulting network has a feasibles− tflow x ∈ R|V|+3|E| if and only if our instance (G; c) has a solution. Indeed, as allcapacities are integral, a feasible flow may also be assumed to be integral. Givenan integral feasible flow we can interpret an arc (i; .i; j/) from V to E which carries1 unit of flow asi winning the matche= .i; j/ and conversely (cf. also Cooketal [1998]).Case (iii) þ = 1;  = 2 (ancient FIFA rule).This can be solved similarly. In the network of Figure 2.1 we simply redefine theupper capacities of all arcs fromV to E to be 2. The lower and upper capacitiesof arcs fromE to t are also set to 2. Again, feasible integral flows are in 1-1correspondence with solutions of our instance (G; c). Each nodee ∈ E in ournetwork has two incoming arcs which carry a total flow of 2 units, distributed as2 : 0 or 1 : 1, corresponding to a win/loss match or a draw.Case (iv)þ = 1;  > 2.We proveN P-completeness by reduction from 3DM (cf. Garey and Johnson [1979]).Suppose|X| = |Y| = |W| = q andR⊆ X× Y×W is given. We are to determinewhetherR contains a matchingR′ ⊆ R, i.e., a set of triples covering each elementof X ∪ Y ∪W exactly once. Assume w.l.o.g. that each elementz ∈ X ∪ Y ∪W4 WALTER KERN AND DANIËL PAULUSMAactually occurs in some tripler ∈ R. We write z ∈ r to indicate thatz occurs inr ∈ R.GivenR⊆ X×Y×W, we construct a graphG= .V; E/ as follows. We first makeone copy of each elementz∈ X ∪ Y∪W for each occurrence ofz in R, i.e., wedefineX̄ := {.x; r / | x ∈ X; r ∈ R; x ∈ r}Ȳ := {.y; r / | y ∈ Y; r ∈ R; y∈ r}W̄ := {.w; r / | w ∈W; r ∈ R; w ∈ r}:Construct a graphG = .V; E/ with node setV = X ∪ Y∪W∪ X̄ ∪ Ȳ∪ W̄∪ Rand edges as defined by the incidence relations in a straightforward way,i.e.,E = {.x; .x; r // | .x; r / ∈ X̄}∪ {.y; .y; r // | .y; r / ∈ Ȳ}∪ {.w; .w; r // | .w; r / ∈ W̄}∪ {.r; .x; r // | .x; r / ∈ X̄}∪ {.r; .y; r // | .y; r / ∈ Ȳ}∪ {.r; .w; r // | .w; r / ∈ W̄} (cf. Figure 2.2 below):...R...r...  W(w,r)wW......X(x,r)xX(y,r)...YYy...Figure 2.2COMPLEXITY ASPECTS OF SPORTS COMPETITIONS 5Next define node capacitiesc ∈ RV as follows:c ≡ 1 on X∪Yc ≡ 1+  on X̄∪ Ȳc ≡ max{;3} on Rc ≡ 1 onW̄c ≡ .Ž− 1/+ 1 onW:(Again,Ž refers to the degree function ofG.)We claim that this instance (G; c) has a solution if and only ifR contains a match-ing.“⇐” SupposeR′ ⊆ R is a matching. Define a corresponding partial orientationof G as follows. For eachw ∈ W choose the uniquer ′ ∈ R′ with .w; r ′/ ∈ W̄.We leave the edge.w; .w; r ′// unoriented and orient all other edges fromw to W̄.This way the capacity constraints ofw are met. For eachr ′ = .x; y; w/ ∈ R′ weorient the edge.r ′; .w; r ′// from r ′ towards.w; r ′/ and the edges.r ′; .x; r ′// and.r ′; .y; r ′// from X̄ respectivelyȲ towardsr ′. All edges incident withr ∈ R\R′remain unoriented. This way we ensure that the capacity constraints onW̄ and Rare respected. Finally, orient all edges betweenX̄ andX from X̄ towardsX exceptthose that correspond to an element inR′ (these remain unoriented). This way thecapacity constraints forX and X̄ are met. We orient edges betweenȲ andY in thesame way. This partial orientation gives a solution of the instance (G; c).”⇒” Conversely, suppose we are given a partial orientation ofG respecting thecapacity constraints. The latter imply that forx ∈ X we haveŽ−.x/ ≥ Ž.x/− 1 andŽ+.x/ = 0. We may assume w.l.o.g. that actuallyŽ−.x/ = Ž.x/− 1. (Otherwise,i.e., if Ž−.x/= Ž.x/, pick an arbitrary edge incident withx and make it unoriented.The modified orientation will still respect all capacity constraints.) A similar argu-ment holds for elementsy∈ Y. Nodes inX̄ have degree 2. In view of their capacitybound 1+ , we may assume w.l.o.g. that each.x; r / ∈ X̄ hasŽ0 = 1 andŽ+ = 1.(Otherwise, again modify the solution without violating the capacity constraints.)As eachx ∈ X hasŽ−.x/ = Ž.x/− 1 andŽ0.x/ = 1, we conclude that• There are exactly|X| arcs directed fromX̄ to R. Moreover, if..x; r /; r / is di-rected towardsr and..x′; r ′/; r ′/ is directed towardsr ′, thenx 6= x′.The same holds for the directed arcs fromȲ to R.Arguing similarly for nodes inW, we find that eachw ∈ W has w.l.o.g.Ž+.w/ =Ž.w/− 1 andŽ0.w/ = 1. (Otherwise modify the orientation such thatw actuallyuses its full capacity.) Because nodes inW̄ have degree 2 and capacity bound 1,this implies that• There are exactly|W| arcs directed fromR towardsW̄. Moreover, if.r; .w; r // isdirected fromr towards.w; r / and.r ′; .w′; r ′// is directed fromr ′ towards.w′; r ′/,thenw 6= w′.6 WALTER KERN AND DANIËL PAULUSMAFinally, the capacity constraints onR imply that a noder ∈ Rcan haveŽ+ ≥ 1 onlyif Ž− ≥ 2. From this and the above observations, it is straightforward to check thatR′ = {r ∈ R | Ž+.r / = 1}actually is a matching.Case (v) þ = 1<  < 2.Again, we proveN P-completeness by reduction from 3DM. In the graphG ofFigure 2.2 we redefine the node capacitiesc ∈ RV as follows:c ≡ .Ž− 1/+ 1 on X∪Yc ≡ 1 on X̄∪ Ȳc ≡ max{2;3} on Rc ≡ 1+  on W̄c ≡ 1 onW:Analogously to Case (iv) one can prove that the instance (G; c) has a solution ifand only if R contains a matching.♦3. REMARKSAs noted already, our results imply that sport competition problems with the newFIFA rules (Þ = 0; þ = 1;  = 3) are hard. The reason for this is that the networkmodel we used for solving cases (ii) and (iii) of our main theorem does not applyfor this case. Indeed, if we increase the upper capacities to 3 on all arcs fromV tE and fromE to t in the network of Figure 2.1, then a feasible flow does no longernecessarily represent a solution of our instance.(A total flow of 2 entering a nodee= .i; j/ ∈ E distributed as 2 : 0 on the two entering arcs does not correspond toa win /loss or a draw.) If we ”repair” this by introducing a ”capacity gap” ]1;3[ onall arcs fromV to E we get a flow problem with capacity gaps which again nicelydescribes our sports competition problem. So as a consequence of our result, thefollowing class of problems is alsoN P-complete (this might be known, but wecould not find it in the literature):FLOWS WITH CAPACITY GAPS (“FCG”)Instance:A digraph D = .V; A/ with sources and sinkt and for each arca ∈ A two dis-joint capacity intervalsI1.a/= [c1.a/; c2.a/] and I2.a/= [c3.a/; c4.a/] .ci.a/ ∈Z; i = 1; : : : ;4/.Question:Does a (w.l.o.g. integral)s− t flow x ∈ ZA exist with x.a/ ∈ I1.a/∪ I2.a/ .a ∈A/?Corollary 3.1. FCG is N P-complete.COMPLEXITY ASPECTS OF SPORTS COMPETITIONS 7Finally, as to sports competitions, we would like to remark that also other questionscan be treated in the same way. For example ”Is there a chance thatT0 ends upwith the lowest final score?” turns out to be of exactly the same complexity as SC:Assume thatT0 has a current total scores0 and loses all remaining matches. Thisresults in a current total scoresi for all other teamsTi 6= T0. The first question isnow whether the teamsTi 6= T0 can finish the remaining matches in such a way thateachTi collectsat least ci := s0− si additional score points. Again we model thisby a multigraphG= .V; E/whose vertices correspond to teamsTi 6= T0 and edgesare in 1-1 correspondence with remaining matches. Each nodei ∈ V has a (lower)capacityci ∈ R. Our “reverse” sports competition problem (“RSC”) can now beformulated as follows:RSC(Þ; þ; ):Given a multigraphG = .V; E/ and node capacitiesc ∈ RV can G be partiallyoriented such that for each nodei ∈ V:ÞŽ−.i /+ þŽ0.i /+ Ž+.i / ≥ ci ? .3:1/It is easy to see that fori ∈ V, (3.1) is equivalent to. − þ/Ž0.i /+ . − Þ/Ž−.i / ≤ Ž.i /− ci :Hence an instance.G; c/ of RSC(Þ; þ; ) corresponds to an instance.G; Ž− c/of SC(0; − þ; − Þ) and the corollary below immediately follows from Theorem1.1.Corollary 3.2. RSC(Þ; þ; ) is polynomially solvable in each of the following threecases:(i) Þ = þ(ii) þ = (iii) Þ+  = 2þIn all other cases, the problem is N P-complete.Questions such as ”Is there a chance thatT0 ends up being one of the three teamsthat have the three lowest final scores?” can also be treated in a similar way. Again,assume thatT0 has a current total scores0 and loses all remaining matches. Further-more choose two teamsTi; Tj 6= T0 and letTi andTj lose their remaining matchesagainst teamsTk .k 6= 0; i; j/. (Choose, if necessary, an arbitrary outcome for thematches betweenTi andTj.) These outcomes result in final total scoress0, si andsj , and current total scoressk for all other teamsTk .k 6= 0; i; j/. If it is possiblethat the teamsTk .k 6= 0; i; j/ can finish the remaining matches in such a way thateachTk collects at leastck := s0− sk additional score points, thenT0 can indeedend up being one of the three lowest teams. If this is not possible for any pairTi ; Tj,thenT0 can never end up being one of the three lowest teams. So one has to solveat most12|V|.|V|− 1/ problem instances in RSC(Þ; þ; ). Hence also this questionis of the same complexity as SC.8 WALTER KERN AND DANIËL PAULUSMAREFERENCES[1] W.J. Cook, W.H. Cunningham, W.R. Pulleyblank, and A. Schrijver [1998]:Combinatorial Op-timization. Wiley, New York.[2] M. Garey and D. Johnson [1979]:Computers and Intractability. A Guide to the Theory of NP-Completeness. Freeman, New York.FACULTY OF MATHEMATICAL SCIENCES, UNIVERSITY OF TWENTE, P.O.BOX 217, 7500 AEENSCHEDE, THE NETHERLANDSE-mail address: {kern,paulusma }@math.utwente.nl

[1979]: Computers and Intractability. A Guide to the Theory of NPCompleteness.

Combinatorial Optimization.

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The new FIFA rules are hard: Complexity aspects of sports competitions

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