About the number of oriented Hamiltonian paths and cycles in tournaments
aa r X i v : . [ m a t h . C O ] J a n ABOUT THE NUMBER OF ORIENTED HAMILTONIAN PATHS ANDCYCLES IN TOURNAMENTS
AMINE EL SAHILI AND ZEINA GHAZO HANNA Abstract.
We prove that a tournament T and its complement T contain the samenumber of oriented Hamiltonian paths (resp. cycles) of any given type, as a generalizationof Rosenfeld’s result proved for antidirected paths. Introduction
An oriented Hamiltonian path in a tournament is an oriented path containing all itsvertices, and if this path is directed, then it is said to be a directed Hamiltonian path. Thedefinitions are similar for cycles. Counting Hamiltonian paths and cycles in a tournamentis a well known problem. Given a certain type of oriented Hamiltonian paths (resp.cycles), one may ask how many such paths (resp. cycles) can be found in a tournament.No exact value of these numbers was given. What was done in this area is boundingthe number of only the directed Hamiltonian paths (resp. cycles) in tournaments, andworking on characterizing the tournaments having this minimum or maximum number.The oldest result through this investigation was given by Szele [7], more than seventyyears ago, who gave lower and upper bounds for the maximum number P ( n ) of directedHamiltonian paths in a tournament on n vertices, n !2 n − ≤ P ( n ) ≤ c n !2 n , where c is apositive constant independent of n . Then, the upper bound of P ( n ) was improved byAlon [1]: P ( n ) ≤ c .n n !2 n − , where c > is independent of n . For the minimum numberof directed Hamiltonian paths in a tournament, we can easily verify that it is equal to , and this value corresponds to the transitive tournament. But in the case of strongtournaments, this number increases a lot, as for the nearly-transitive tournament of order n , where the number of directed Hamiltonian paths is equal to n − + 1 . So in 1972, Moon[4] gave upper and lower bounds for the minimum number h P ( n ) of directed Hamiltonianpaths in strong tournaments of order n , and in 2006, after finding a characterization ofstrong tournaments, Busch [2] improved this result by proving that the exact value ofthis minimum number is exactly equal to the upper bound given by Moon. Later on,Moon and Yang [5], constructed some tournaments, called the "special chains", linking Date : Thursday 16 th May, 2019. AND ZEINA GHAZO HANNA between many nearly-transitive tournaments, and proved that they contain the minimumnumber of directed Hamiltonian paths, and that they are the only tournaments verifyingthis minimum. Concerning the maximum and minimum number of directed Hamiltoniancycles (also called Hamiltonian circuits) in tournaments, Thomassen [8] was able, in 1980,giving an extension of Moon’s result previously mentioned, to find the minimum numberof these cycles in a 2-connected tournament. On the other hand, it can be proven, usingthe probabilistic methods, that the maximum number of directed Hamiltonian cycles ina tournament of order n is greater than ( n − n . However, Moon observed that it seemsdifficult to give explicit tournaments with such a large number of directed Hamiltoniancycles.An antidirected path is an oriented path whose arcs have successively opposite direc-tions. Rosenfeld [6] proved in 1974 that the number of antidirected Hamiltonian pathsstarting with a forward arc is equal to the number of antidirected Hamiltonian pathsstarting with a backward arc, in any tournament, which can be stated as: the num-ber of antidirected Hamiltonian paths in any tournament T is equal to the number ofantidirected Hamiltonian paths in the complement of T , denoted by T .In this paper, we generalize Rosenfeld’s result for any type of oriented Hamiltonianpaths, and also for cycles: We prove that a tournament T and its complement T containthe same number of oriented Hamiltonian paths (resp. cycles) of any given type. Thenwe establish this fact for any digraph H whose maximal degree is less than or equal to 2.2. Basic definitions and preliminary results
We will follow in this paper the same definitions given in [3].Let α = ( α , α , . . . , α s ); s ≥ , α i ∈ Z , α i · α i +1 < ∀ i = 1 , . . . , s − .An oriented path P is said to be of type P ( α , α , . . . , α s ) if P is formed by s blocks(i.e. maximal directed subpaths) I , I , . . . , I s such that length ( I i ) = | I i | = | α i | andwith x i , y i being the ends of the block I i , I i ∩ I i +1 = { y i } = { x i +1 } , the followingcondition is verified: ∀ i = 1 , . . . , s, α i > ⇐⇒ I i is directed from x i to y i . Wenote end ( I i ) = { x i , y i } , and we write P = I I . . . I s . For u, v ∈ I i , I i [ u, v ] denotesthe subpath of I i of ends u and v . This notation can be extended by allowing α i to be , by considering P ( α , ..., α i , , α i +2 , ..., α s ) = P ( α , ..., α i + α i +2 , ..., α s ) (remarkthat in this case, α i and α i +2 have the same sign), P (0 , α , ..., α s ) = P ( α , ..., α s ) , and P ( α , ..., α s − ,
0) = P ( α , ..., α s − ) , and we say that P ( α ) is a standard type of a path P if α contains no zero components. In this paper, we will always consider standard typesof paths unless a non-standard type appears in calculations. RIENTED HAMILTONIAN PATHS AND CYCLES 3
Note that a path P = v v . . . v r that is of type P ( α , . . . , α s ) with respect to thisenumeration is also of type P ( − α s , . . . , − α ) with respect to the other enumeration v r v r − . . . v denoting it, so we remark that any path has at most two types. Moreover,two paths P and P ′ in a tournament T are said to be equal if they have the same set ofarcs, i.e. E ( P ) = E ( P ′ ) .For α = ( α , . . . , α s ) in Z s , we denote by − α the tuple ( − α , . . . , − α s ) and by α thetuple ( α s , α s − . . . , α ) . Let T be a tournament, then P T ( α , . . . , α s ) is defined to bethe set of oriented paths in T of type P ( α , . . . , α s ) and f T ( α , α , . . . , α s ) denotes thecardinality of this set. It can be easily verified that: P T ( α ) = P T ( β ) ⇐⇒ α = β or α = − β. Let α = ( α , . . . , α s ); α i ∈ Z , α i · α i +1 < ∀ i = 1 , . . . , s − , α s · α < .An oriented cycle C is said to be of type C ( α , . . . , α s ) if C is formed by s blocks I , I , . . . , I s , with end ( I i ) = { x i , y i } , | I i | = | α i | and I i ∩ I i +1 = { y i } = { x i +1 } , ≤ i ≤ s − and I s ∩ I = { y s } = { x } , such that ∀ i = 1 , . . . , s, α i > ⇐⇒ I i is directed from x i to y i . We write C = I I ...I s . Note that for cycles, if s = 1 ( s = 1 is the case of Hamiltonian circuits), then s must be even. As for paths, we may also allow α i to be for cycles, by considering C ( α , . . . , α i − , , α i +1 , . . . , α s ) = C ( α , . . . , α i − + α i +1 , . . . , α s ) , C (0 , α , . . . , α s ) = C ( α + α s , α , . . . , α s − ) and C ( α , . . . , α s − ,
0) = C ( α + α s − , α , . . . , α s − ) . We say that C ( α ) is a standard type of a cycle C if α contains no zerocomponents. In this paper, we will also always consider standard types of cycles unless anon-standard type appears in calculations.If T be a tournament, then C T ( α , . . . , α s ) is defined to be the set of oriented cycles of T of type C ( α , . . . , α s ) and g T ( α , . . . , α s ) denotes the cardinality of this set. We mayalso verify that: C T ( α ) = C T ( β ) ⇐⇒ β = ( α i , α i +1 , . . . , α s , α , . . . , α i − ) or β = ( − α i , − α i − , . . . , − α , − α s , . . . , − α i +1 ) , for some ≤ i ≤ s .A tuple α is said to be symmetric if α = − α .An oriented path P (resp. cycle C ) is said to be symmetric if there exists a tuple α that is symmetric, such that P (resp. C ) is of type P ( α ) (resp. C ( α ) ). Otherwise, thepath P (resp. cycle C ) is not symmetric. AMINE EL SAHILI AND ZEINA GHAZO HANNA Let T be a tournament on n vertices. An oriented cycle C in T is said to be gener-ated by an oriented path P = x x . . . x n if C = P ∪ h{ x , x n }i . We write C = C P . Formore simplicity, we write uv instead of < { u, v } > .The relation R defined on the set of oriented paths in T by: P R P ′ ⇐⇒ C P = C P ′ is an equivalence relation, and so is R α , the restriction of R on the set P T ( α ) .Let P = v v . . . v n and P ′ be two oriented paths in a tournament T of order n , it canbe easily remarked that P R P ′ if and only if P = P ′ or P ′ = v i v i +1 . . . v n v v . . . v i − forsome ≤ i ≤ n . Remark 1.
Let P = v v . . . v n be an oriented path in a tournament T , of some type P ( α ) = P ( α , . . . , α s ) , and let C = C P be the cycle generated by P in T which is of sometype C ( β ) . We will see what different values β could take: • Case 1: s is even. Then if α > (which means α s < ), we have β = ( α +1 , α , . . . , α s ) or β = ( α , . . . , α s − , α s − whether ( v n , v ) or ( v , v n ) ∈ E ( T ) respectively, while if α < (i.e. α s > ), then β = ( α − , α , . . . , α s ) or β = ( α , . . . , α s − , α s + 1) whether ( v , v n ) or ( v n , v ) ∈ E ( T ) respectively. • Case 2: s is odd. Then if α > (which means α s > also), we have β =( − , α , . . . , α s ) or β = ( α s + 1 + α , α , . . . , α s − ) whether ( v , v n ) or ( v n , v ) ∈ E ( T ) respectively, while if α < (and so is α s ), then β = (1 , α , . . . , α s ) or β = ( α s − α , α , . . . , α s − ) whether ( v n , v ) or ( v , v n ) ∈ E ( T ) respectively.So we remark that every oriented path P in a tournament T may generate 2 types ofcycles, that we will denote by C ( β ) and C ( β ′ ) in the latter sections. Remark 2.
If a path P has the type P ( α ) = P ( α , . . . , α s ) where α is symmetric, thenthe cycle C P generated by P cannot be symmetric.In fact, if P has the type P ( α ) = P ( α , . . . , α s ) and α is symmetric, thus α = − α s , so α and α s have opposite signs, which means that s should be even. Thus by the previousremark, C P has one of these types: C ( α +1 , . . . , α s ) or C ( α − , . . . , α s ) or C ( α , . . . , α s − or C ( α , . . . , α s + 1) . But in all these cases, and due to the fact that α is symmetric, C P cannot be written as a succession of blocks having the type C ( β ) where β is symmetric,thus the cycle C P cannot be symmetric. Let α = ( α , . . . , α s ) ∈ Z s .An integer ≤ r ≤ s is said to be a period of α if [ i ≡ j ( mod r ) ⇒ α i s = α j s ] where i s is the unique integer in { , , . . . , s } such that i ≡ i s ( mod s ) .Let r ( α ) = min { r ; r is a period of α } . RIENTED HAMILTONIAN PATHS AND CYCLES 5
It can be shown that r is a period of α ⇐⇒ r ( α ) divides r , and consequently r ( α ) divides s , since s is a trivial period of α .Let t ( α ) = sr ( α ) . Proposition 3.
Let α = ( α , . . . , α s ) ∈ Z s and α ′ = α = ( α s , . . . , α ) , then r ( α ) = r ( α ) and t ( α ) = t ( α ) .Proof. Let α ′ = ( α ′ , α ′ , . . . , α ′ s ) = ( α s , . . . , α ) = α , and r ( α ′ ) = r ′ , r ( α ) = r .Let l, p be two integers such that p ≡ l ( mod r ) . We would like to prove that α ′ p s = α ′ l s .We have α ′ p s = α k ; k = s − p s + 1 , i.e. k ≡ − p + 1 ( mod s ) thus k ≡ − p + 1 ( mod r ) .Also, α ′ l s = α j ; j = s − l s + 1 , i.e. j ≡ − l + 1 ( mod s ) thus j ≡ − l + 1 ( mod r ) . Now p ≡ l ( mod r ) ⇒ − p + 1 ≡ − l + 1 ( mod r ) ⇒ k ≡ j ( mod r ) ⇒ α k = α j ⇒ α ′ p s = α ′ l s .Thus, r is a period of α ′ and r ′ ≤ r . Similarly, we can prove that r ≤ r ′ . Hence, r = r ′ .Consequently, t ( α ) = t ( α ′ ) . (cid:3) We may also remark that t ( α ) = t ( − α ) , and we can prove that ∀ ≤ i ≤ s , we have t ( α ) = t ( α i , α i +1 , . . . , α s , α , . . . , α i − ) . As a result, ∀ ≤ i ≤ s we have: t ( α ) = t ( − α i , − α i +1 , . . . , − α s , − α , . . . , − α i − )= t ( α i , α i − , . . . , α , α s , . . . , α i +1 )= t ( − α i , − α i − , . . . , − α , − α s , . . . , − α i +1 ) . Let C = I I . . . I s be an oriented cycle of type C ( β ) = C ( β , . . . , β s ) and let r = r ( β ) .For ≤ i < j ≤ s , I i and I j are said to be similar if j ≡ i ( mod r ) . This is equivalent tosay that j − i is a period of β . For every ≤ i ≤ s , there are t ( β ) − blocks similar to I i . It follows that if I i and I j are similar, then β i = β j , and for any non-negative integer k , I [ i + k ] s and I [ j + k ] s are similar.If C = I I . . . I s is an oriented cycle of type C ( β ) = C ( β , . . . , β s ) , two vertices u, v ∈ C are said to be clones (with respect to C ) if: • u and v belong to similar blocks, say I i and I j . • l ( I i [ x i , u ]) = l ( I j [ x j , v ]) .It obviously follows that l ( I i [ u, y i ]) = l ( I j [ v, y j ]) .If C ∈ C T ( β ) , then each vertex of C has t ( β ) − clones.3. Oriented Hamiltonian paths
Recall that Rosenfeld [6] proved in 1974 that the number of antidirected Hamiltonianpaths starting with a forward arc is equal to the number of antidirected Hamiltonianpaths starting with a backward arc, in any tournament. In this section, we will generalize
AMINE EL SAHILI AND ZEINA GHAZO HANNA Rosenfeld’s result, showing that in a tournament, the number of Hamiltonian paths of acertain type P ( α ) is equal to the number of Hamiltonian paths of type P ( − α ) : Theorem 4.
Let α = ( α , . . . , α s ) ∈ Z s ; α i · α i +1 < , α ≥ , and let T be a tournamentof order n ; n = s P i =1 | α i | +1 . We have: f T ( α ) = f T ( − α ) . In order to prove this result, it is more adequate to work on enumerations of orientedpaths. To this purpose, we define the following:Let α = ( α , . . . , α s ); α i ∈ Z , α i · α i +1 < ∀ i = 1 , . . . , s − , and let T be a tour-nament on n ≥ s P i =1 | α i | +1 vertices. Definition 1.
An enumeration E = v v . . . v r of some vertices of T is said to be oftype E ( α , . . . , α s ) if the path P = v v . . . v r is of type P ( α , . . . , α s ) with respect to thisenumeration. Definition 2.
Considering the tournament T of order n , E T ( α , . . . , α s ) is defined tobe the set of enumerations of any m = s P i =1 | α i | +1 vertices of T , m ≤ n , of type E ( α , . . . , α s ) . We denote by e T ( α , α , . . . , α s ) the cardinality of this set. Remark that, unlike the case of paths, where every path has two types, if two enu-merations E and E ′ have different types, then E = E ′ . In fact, we can easily prove thefollowing property: Proposition 5.
Let T be a tournament of order n , and α = ( α , α , . . . , α s ) , β =( β , β , . . . , β s ′ ) , s P i =1 | α i | ≤ n , and s ′ P i =1 | β i | ≤ n , we have: E T ( α ) = E T ( β ) ⇐⇒ α = β. Proposition 6.
Let T be a tournament of order n .The sets E T ( α ) , α = ( α , α , . . . , α s ) ∈ Z s , α i .α i +1 < , s ≥ , with s P i =1 | α i | = n − , forma partition of the set E T of all the enumerations on n vertices of T .Proof. This proposition follows immediately from the fact that the binary relation R E defined on the set E T by E R E E ⇔ E and E belong to the same set E T ( α ) , is an equiv-alence relation on E T whose equivalences classes are the sets E T ( α ) , α = ( α , α , . . . , α s ) ∈ Z s , α i .α i +1 < , s ≥ , and s P i =1 | α i | = n − . (cid:3) RIENTED HAMILTONIAN PATHS AND CYCLES 7
Let α = ( α , . . . , α s ) ∈ Z s ; α i · α i +1 < , α ≥ , such that s P i =1 | α i | = n − . Proposition 7. If α is symmetric, then we have | E T ( α ) | = 2 . | P T ( α ) | , while | E T ( α ) | = |P T ( α ) | otherwise. The above proposition can be easily verified. In fact, this result follows from the ob-servation that the automorphism group of an oriented path has either order one or two.We may now give the
Proof of Theorem 4 : Proof.
First of all remark that if α is symmetric, so is − α and vice versa. Thus to provethat f T ( α ) = f T ( − α ) , and using Proposition 7, it is enough to prove that e T ( α ) = e T ( − α ) .The proof will be done by induction on s .If s = 1 , α = ( α ) = ( n − and − α = ( − α ) = (1 − n ) . Since every directed Hamil-tonian path P = v v . . . v n in T corresponds to two enumerations E = v v . . . v n and E ′ = v n v n − . . . v of types E ( α ) = E ( n − and E ( − α ) = E (1 − n ) respectively, and viceversa, thus | E T ( α ) | = | E T ( − α ) | = | P T ( α ) | and we have e T ( α ) = e T ( − α ) .Suppose that the result is true when α has s components, i.e. if α = ( α , . . . , α s ) ∈ Z s ; α i · α i +1 < , α ≥ , and T is a tournament of order n = s P i =1 | α i | +1 , we have: e T ( α , . . . , α s ) = e T ( − α , . . . , − α s ) , and let’s prove the result for s + 1 components.Let α = ( α , . . . , α s , α s +1 ) ∈ Z s +1 ; α i · α i +1 < , α ≥ , and T be a tournament of order n = s +1 P i =1 | α i | +1 .We argue now by induction on α . If α = 0 , then by induction on s , we have e T (0 , α , . . . , α s +1 ) = e T ( α , . . . , α s +1 ) = e T ( − α , . . . , − α s +1 ) = e T (0 , − α , . . . , − α s +1 ) . So suppose that α > , and that the result is true when the first component is equal to α − , and let’s prove it when the first component is equal to α .Let X ⊆ V ( T ) such that | X | = α . Set T ′ = T − X , and let’s define the followingsets: A X = E h X i ( α − × E T ′ ( α , . . . , α s +1 ) , A ′ X = { ( E, E ′ ) ∈ A X ; E = v . . . v α , E ′ = v α +1 . . . v n , and ( v α , v α +1 ) ∈ E ( T ) } , A ′′ X = { ( E, E ′ ) ∈ A X ; E = v . . . v α , E ′ = v α +1 . . . v n , and ( v α +1 , v α ) ∈ E ( T ) } .Obviously we have: A ′ X ∩ A ′′ X = ∅ , and A X = A ′ X ∪ A ′′ X , thus | A X | = | A ′ X | + | A ′′ X | . AMINE EL SAHILI AND ZEINA GHAZO HANNA Consider the two sets E X ( α , . . . , α s +1 ) = { E = v . . . v α v α +1 . . . v n ∈ E T ( α , . . . , α s +1 ); { v , . . . , v α } = X } , E X ( α − , α − , α . . . , α s +1 ) = { E = v . . . v α v α +1 . . . v n ∈ E T ( α − , α − , α , . . . , α s +1 ); { v , . . . , v α } = X } . We have | A ′ X | = | E X ( α , . . . , α s +1 ) | and | A ′′ X | = | E X ( α − , α − , α , . . . , α s +1 ) | . In fact, consider the two correspondences f : E X ( α , . . . , α s +1 ) −→ A ′ X such that ∀ E = v . . . v α v α +1 . . . v n ∈ E X ( α , . . . , α s +1 ) , f ( E ) = ( v . . . v α , v α +1 . . . v n ) , and f ′ : E X ( α − , α − , α , . . . , α s +1 ) −→ A ′′ X such that ∀ E = v . . . v α v α +1 . . . v n ∈ E X ( α − , α − , α , . . . , α s +1 ) , f ′ ( E ) = ( v . . . v α , v α +1 . . . v n ) . We can verify that both are bijectivemappings.Hence, | A X | = | E X ( α , . . . , α s +1 ) | + | E X ( α − , α − , α , . . . , α s +1 ) | . Now let’s consider − α = ( − α , . . . , − α s ) , and let X ⊆ V ( T ) such that | X | = α . Set T ′ = T − X , and let’s also define the following sets: B X = E h X i ( − α + 1) × E T ′ ( − α , . . . , − α s +1 ) , B ′ X = { ( E, E ′ ) ∈ B X ; E = v . . . v α , E ′ = v α +1 . . . v n , and ( v α +1 , v α ) ∈ E ( T ) } , B ′′ X = { ( E, E ′ ) ∈ B X ; E = v . . . v α , E ′ = v α +1 . . . v n , and ( v α , v α +1 ) ∈ E ( T ) } .We also have: B ′ X ∩ B ′′ X = ∅ , and B X = B ′ X ∪ B ′′ X , thus | B X | = | B ′ X | + | B ′′ X | .Consider the two sets E X ( − α , . . . , − α s +1 ) = { E = v . . . v α v α +1 . . . v n ∈ E T ( − α , . . . , − α s +1 ); { v , . . . , v α } = X } , E X ( − α + 1 , − α + 1 , − α , . . . , − α s +1 ) = { E = v . . . v α v α +1 . . . v n ∈ E T ( − α + 1 , − α + 1 , − α , . . . , − α s +1 ); { v , . . . , v α } = X } . Similarly as before, we can prove that | B ′ X | = | E X ( − α , . . . , − α s +1 ) | and | B ′′ X | = | E X ( − α + 1 , − α + 1 , − α , . . . , − α s +1 ) | , Hence | B X | = | E X ( − α , . . . , − α s +1 ) | + | E X ( − α + 1 , − α + 1 , − α , . . . , − α s +1 ) | . RIENTED HAMILTONIAN PATHS AND CYCLES 9
On the other hand, we have that | A X | = e X ( α − .e T ′ ( α , . . . , α s +1 ) , | B X | = e X ( − α + 1) .e T ′ ( − α , . . . , − α s +1 ) , but since we have here less than s + 1 blocks, thus by induction, e X ( α −
1) = e X ( − α + 1) and e T ′ ( α , . . . , α s +1 ) = e T ′ ( − α , . . . , − α s +1 ) , thus | A X | = | B X | . Also, ∀ β = ( α , . . . , α s +1 ) or ( α − , α − , α , . . . , α s +1 ) or ( − α , . . . , − α s +1 ) or ( − α +1 , − α + 1 , − α , . . . , − α s +1 ) , we have E T ( β ) = ∪ X ⊆ V ( T ); | X | = α E X ( β ) . (The union is disjointsince if X = X ′ , the enumerations differ).Since | A X | = | B X | , then | E X ( α , . . . , α s +1 ) | + | E X ( α − , α − , α , . . . , α s +1 ) | = | E X ( − α , . . . , − α s +1 ) | + | E X ( − α + 1 , − α + 1 , − α , . . . , − α s +1 ) | .Doing the summation over all the sets X ⊆ V ( T ) , | X | = α , we have: X X ⊆ V ( T ); | X | = α | E X ( α , . . . , α s +1 ) | + X X ⊆ V ( T ); | X | = α | E X ( α − , α − , α , . . . , α s +1 ) | = X X ⊆ V ( T ); | X | = α | E X ( − α , . . . , − α s +1 ) | + X X ⊆ V ( T ); | X | = α | E X ( − α +1 , − α +1 , − α , . . . , − α s +1 ) | , thus, | E T ( α , . . . , α s +1 ) | + | E T ( α − , α − , α , . . . , α s +1 ) | = | E T ( − α , . . . , − α s +1 ) | + | E T ( − α + 1 , − α + 1 , − α , . . . , − α s +1 ) | , which implies that e T ( α , . . . , α s +1 ) + e T ( α − , α − , α , . . . , α s +1 )= e T ( − α , . . . , − α s +1 ) + e T ( − α + 1 , − α + 1 , − α , . . . , − α s +1 ) . But by induction, since α − < α , we have that e T ( α − , α − , α , . . . , α s +1 ) = e T ( − α + 1 , − α + 1 , − α , . . . , − α s +1 ) . So we finally get e T ( α , . . . , α s +1 ) = e T ( − α , . . . , − α s +1 ) , which concludes the proof. (cid:3) Oriented cycles and generating paths
Let T be a tournament. In this section we find a relation between f T ( α ) , g T ( β ) and g T ( β ′ ) , where P ( α ) is some type of oriented Hamiltonian paths in T , and C ( β ) and C ( β ′ ) AND ZEINA GHAZO HANNA are the two types of cycles that can be generated by a path of type P ( α ) in the tourna-ment T , (see Remark 1), and this result will be of great use in the next section.We first start by proving the following theorem: Theorem 8.
Let P ∈ P T ( α ) be an oriented Hamiltonian path in a tournament T and let C P be the cycle generated by P in T , of type C ( β ) , such that C P has at least 2 blocks (i.e. C P is not a circuit). Then if C P is non-symmetric, we have | P | = t ( β ) , while if C P issymmetric, then | P | = 2 .t ( β ) , where P is the equivalence class of P with respect to R α . In [3], one actually proved that if C P is non-symmetric, then | P | = t ( β ) . In thefollowing, we will present arguments useful for both the symmetric and the non-symmetrictypes. Remark 9. If P ∈ P T ( α ) is an oriented Hamiltonian path in a tournament T of order n , and C P the cycle generated by P in T , such that C P is a Hamiltonian circuit, then | P | = n , where P is the equivalence class of P with respect to R α . In fact, since C P is a circuit, then P must be a directed path, also every Hamiltoniancircuit is generated by exactly n directed Hamiltonian paths, starting each from a vertexof C P .Note that if C P is a circuit, say of type C ( β ) , ( β in this case is a singleton, that is hasone component), then t ( β ) = 1 .In order to prove Theorem 8, we first give the three following lemmas: Lemma 10.
Let T be a tournament of order n , and let C = v v . . . v n be a Hamil-tonian cycle in T . Then C is symmetric if and only if ∀ ≤ i ≤ n , and for ev-ery Hamiltonian path P = v i v i +1 . . . v n v . . . v i − , there exists ≤ i ′ ≤ n such that P = v i v i +1 . . . v n v . . . v i − and P ′ = v i ′ v i ′ − . . . v v n v n − . . . v i ′ +1 have the same type withrespect to these enumerations.Proof. For the necessary condition, since C is symmetric, we can suppose without loss ofgenerality that C = v v . . . v n is of type C ( β , β , . . . , β s ) = I I . . . I s with respect to thisenumeration, where β is symmetric. We have | I j | = | β j | , and let end ( I j ) = { x j , y j } , ∀ ≤ j ≤ s .Suppose that v i ∈ I j , for some ≤ j ≤ s , and suppose without loss of generality that β j > , (the case β j < is similar).Let i ′ = n − ( i − , (assuming that if i = 1 , i ′ = n + 1 simply denotes i ′ = 1 ), so P ′ = v n − ( i − v n − ( i − − . . . v v n . . . v n − ( i − . We will show that this value of i ′ satisfies RIENTED HAMILTONIAN PATHS AND CYCLES 11 the necessary condition.In fact, let x = l ( I j [ x j , v i ]) , then the path P = v i v i +1 . . . v n v . . . v i − is of type P ( β j − x, β j +1 , . . . , β s , β , . . . , β j − , x − with respect to this enumeration.Since β is symmetric, then ( β , β , . . . , β j ) = ( − β s , − β s − , . . . , − β s − ( j − ) , so | β i | = | β s − i +1 |∀ ≤ i ≤ j , and since l ( C [ v v . . . v i ]) = l ( C (cid:2) v v n v n − . . . v n − ( i − (cid:3) ) , we deduce that v n − ( i − ∈ I s − ( j − and l ( I j [ x j , v i ]) = l ( I s − ( j − (cid:2) y s − ( j − , v n − ( i − (cid:3) ) = x .As a result, the path P ′ = v n − ( i − v n − ( i − − . . . v v n . . . v n − ( i − is of type P ( − β s − ( j − − x, − β s − ( j − − , . . . , − β , − β s , . . . , − β s − ( j − , x − with respect to this enumeration.But ( β , β , . . . , β s ) = ( − β s , − β s − , . . . , − β ) (since β is symmetric), so we get that P ′ = v n − ( i − v n − ( i − − . . . v v n . . . v n − ( i − is of type P ( β j − x, β j +1 , . . . , β s , β , . . . , β j − , x − with respect to this enumeration.For the sufficient condition, suppose to the contrary that C is non-symmetric but ∀ ≤ i ≤ n , and for every Hamiltonian path P = v i v i +1 . . . v n v . . . v i − , there exists some i ′ , ≤ i ′ ≤ n , such that P = v i v i +1 . . . v n v . . . v i − and P ′ = v i ′ v i ′ − . . . v v n v n − . . . v i ′ have the same type with respect to these enumerations.Suppose without loss of generality that v i ∈ I , and that β > . (The case β < is simi-larly treated). The path P = v i v i +1 . . . v n v . . . v i − is of type P ( β − x, β , . . . , β s , x − with respect to this enumeration, for some ≤ x ≤ β . Thus the path P ′ = v i ′ v i ′ − . . . v v n . . . v i ′ +1 has the type P ( β − x, β , . . . , β s , x − with respect to this enumer-ation. But the vertex v i ′ belongs to C , thus v i ′ belongs to a block I j of C of length | β j | ,then P ′ = v i ′ v i ′ − . . . v v n . . . v i ′ +1 is of type P ( − β j − y, − β j − , . . . , − β , − β s , . . . , − β j +1 , y − with respect to this enumeration, where − β j > in this case (since we should have β − x = − β j − y and β − x > ), and ≤ y ≤ − β j . We get ( β − x, β , . . . , β s , x −
1) =( − β j − y, − β j − , . . . , − β , − β s , . . . , − β j +1 , y − , thus x − y − so x = y .As a result, we have: ( β − x, β , . . . , β j , β j +1 , . . . , β s , x − ( ∗ ) = ( − β j − x, − β j − , . . . , − β , − β s , . . . , − β j +1 , x − . Property ( ∗ ) implies that β − x = − β j − x and ∀ ≤ p ≤ j , β p = − β j − ( p − , thus ∀ ≤ p ≤ j , β p = − β j − ( p − , that is β ′ = ( β , β , . . . , β j ) = ( − β j , − β j − , . . . , − β ) , hence β ′ is symmetric, and we can write β ′ as ( β , β , . . . , β j , − β j , . . . , − β , − β ) .Also, property ( ∗ ) implies that ∀ ≤ p ′ ≤ s − j , β j + p ′ = − β s − ( p ′ − , that is β ′′ =( β j +1 , . . . , β s − , β s ) = ( − β s , − β s − , . . . , − β j +1 ) , which means that β ′′ is symmetric, andwe can write β ′′ as ( − β s , − β s − , . . . , − β s − j , β s − j , . . . , β s − , β s ) . AND ZEINA GHAZO HANNA So finally we have: ( β , β , . . . , β j , β j +1 , . . . , β s ) = ( β , β , . . . , β j , − β j , . . . , − β , − β , − β s , − β s − ,. . . , − β s − j , β s − j , . . . , β s − , β s ) , which is the type of the cycle C .Now, if we consider β ∗ = ( − β j , . . . , − β , − β , − β s , − β s − , . . . , − β s − j , β s − j , . . . , β s − , β s , β , β , . . . , β j ) ,β ∗ is symmetric, and is also a type of the cycle C , thus C is symmetric, which leads to acontradiction since C is non-symmetric. (cid:3) Now, let T be a tournament of order n , and let C be a Hamiltonian cycle in T ,such that C = v v . . . v n is of type C ( β ) with respect to this enumeration, where β is symmetric. We now know by the proof of the necessary condition of Lemma10 that ∀ ≤ i ≤ n , the Hamiltonian paths P = v i v i +1 . . . v n v . . . v i − and P ′ = v n − ( i − v n − ( i − − . . . v v n . . . v n − ( i − have the same type with respect to these enumer-ations.So let A be the set of all the paths in T that generate the cycle C and that havethe form v i v i +1 . . . v n v . . . v i − and are of a certain type P ( α ) with respect to this enu-meration, and let B be the set of all the paths that generate C and that have the form v n − ( i − v n − ( i − − . . . v v n . . . v n − ( i − and that also have the type P ( α ) with respect tothis enumeration. Lemma 11.
We have
A ∩ B = ∅ , and |A| = |B| .Proof. Suppose to the contrary that ∃ P ∈ A ∩ B . Since P ∈ A , then ∃ ≤ i ≤ n such that P = v i v i +1 . . . v n v . . . v i − v i − and is of type P ( α ) = P ( α , . . . , α s ) withrespect to this enumeration. Since P ∈ B also, then P = v i − v i − . . . v v n . . . v i +1 v i isof type P ( α ) = P ( α , . . . , α s ) with respect to this enumeration, which means that P = v i v i +1 . . . v n v . . . v i − v i − is of type P ( − α ) = P ( − α s , . . . , − α ) . Thus, ( α , . . . , α s ) =( − α s , . . . , − α ) which means that α is symmetric. But, since C = C P , and since P hasthe type P ( α ) where α is symmetric, then the cycle C cannot be symmetric by Remark2, thus β cannot be symmetric, which leads to a contradiction since β is symmetric.For the second part, consider the correspondence f : A −→ B , such that for every P = v i v i +1 . . . v n v . . . v i − of type P ( α ) with respect to this enumeration in A , corresponds thepath P ′ = v n − ( i − v n − ( i − − . . . v v n . . . v n − ( i − , which belongs to B since it is of type P ( α ) with respect to this enumeration, by Lemma 10. The correspondence f is triviallya bijective mapping, so |A| = |B| which concludes the proof. (cid:3) RIENTED HAMILTONIAN PATHS AND CYCLES 13
The last lemma is a result proven implicitely in [3]:
Lemma 12. [3]
Let P = v v . . . v n and P ′ = v i v i +1 . . . v n v . . . v i − be two distinct orientedHamiltonian paths in a tournament T of order n , that generate a cycle C = C P = C P ′ in T . Then P and P ′ have the same type with respect to these enumerations if and only if v and v i are clones. We may now give the
Proof of Theorem 8 : Proof.
The set P contains P as well as the paths P ′ in T that have the same type as thetype of P and such that C P ′ = C P . Let C P = v v . . . v n such that C P is of type C ( β ) ,with respect to this enumeration and P = v i v i +1 . . . v n v . . . v i − for some ≤ i ≤ n . If weconsider all the paths P ′ having the form v j v j +1 . . . v n v . . . v j − , and that have the sametype as P with respect to these enumerations, and such that C P = C P ′ , then by Lemma12, the number of such paths is exactly the number of clones that an end of P could have,that is t ( β ) − .Let A be the set of paths that generate C and have the same type as the type of P ,following the order v v . . . v n of the vertices, and B be the set of paths that generate C and have the same type as the type of P , following the order v v n . . . v of the vertices.We have | A | = t ( β ) . Now let’s count the number of paths in B .If the cycle C is non-symmetric, then by Lemma 10, ∀ ≤ i ≤ n , the path P ′ = v i v i − . . . v v n . . . v i +1 cannot have the same type of P with respect to this enumeration,thus the set B is empty. As a result, | P | = | A | = t ( β ) .If the cycle C is symmetric, (we may suppose without loss of generality that the cycle C = v v . . . v n is of type C ( β ) with respect to this enumeration, where β is symmetric)then by Lemma 10, the set B is non-empty, and by Lemma 11 we have that | A | = | B | andthat the sets A and B are disjoint, thus we deduce that | P | = | A | + | B | = t ( β ) + t ( β ) =2 .t ( β ) . This concludes our proof. (cid:3) Note that all of the above results of this section are true for any oriented paths and cy-cles that are not necessarily Hamiltonian, since any path or cycle defines a set of vertices,and hence a subtournament in which the path and the cycle are Hamiltonian.The following lemma, proved in [3], is of practical use in the next theorem:
Lemma 13. [3]
Let α , α , . . . , α s , β , . . . , β s ∈ Z .If ( α , α , . . . , α s ) = ( β i , β i +1 , . . . , β s , β , . . . , β i − ) , then for any integer k ≥ , we have α k s = β [ k + i − s . AND ZEINA GHAZO HANNA Remark 14.
We saw in the second case of Remark 1 that if α = ( α , . . . , α s ) ∈ Z s , α i · α i +1 < , s is odd, T is a tournament of order n = s P i =1 | α i | +1 , and P be aHamiltonian path of type P ( α ) in T , then the two types of cycles that can be generated by P have either s − or s + 1 blocks. So if we call C ( β ) and C ( β ′ ) these two types, thenobviously, the sets C T ( β ) and C T ( β ′ ) are different. Remark also that when s is odd, α isalways non-symmetric, because we can’t have α = − α s since α and α s have the samesign. However, when s is even, it’s a different story. In fact, we have the following result: Theorem 15.
Let α = ( α , . . . , α s ) ∈ Z s , α i · α i +1 < , and let T be a tournament oforder n = s P i =1 | α i | +1 . We have: C T ( β ) = C T ( β ′ ) ⇐⇒ α is symmetric, where C ( β ) and C ( β ′ ) are the two types of Hamiltonian cycles in T that can be generatedby a Hamiltonian path of type P ( α ) in T .Proof. The case where s is odd being completely settled by Remark 13, we may assumethat s is even.By the first case of Remark 1, when α > , then β = ( β , β , . . . , β s ) = ( α +1 , α , . . . , α s ) and β ′ = ( β ′ , β ′ , . . . , β ′ s ) = ( α , . . . , α s − , α s − , while if α < , then β = ( β , β , . . . , β s ) =( α − , α , . . . , α s ) and β ′ = ( β ′ , β ′ , . . . , β ′ s ) = ( α , . . . , α s − , α s + 1) .We will treat the case where α > , and the other case is similar.For the sufficient condition, suppose that α is symmetric, thus α = ( α , . . . , α s ) =( − α s , . . . , − α ) , which implies that C T ( β ) = C T ( α + 1 , . . . , α s ) is equal to C T ( − α s +1 , . . . , − α ) . Moreover, this set is equal to the set C T ( β ′ ) = C T ( α , . . . , α s − , thus C T ( β ) = C T ( β ′ ) .For the necessary condition, suppose C T ( β ) = C T ( β ′ ) , i.e. C T ( α + 1 , α , . . . , α s ) = C T ( α , . . . , α s − , α s − . Thus, since α is different from α + 1 and − α − , then ( α , . . . , α s − is equal to one of these tuples:(1) ( α i , α i +1 , . . . , α s , α + 1 , α . . . , α i − ) for some ≤ i ≤ s (2) ( − α i , − α i − , . . . , − α , − α − , − α s , . . . , − α i +1 ) for some ≤ i ≤ s Suppose that the first case is true, i.e. ( α , . . . , α s −
1) = ( α i , α i +1 , . . . , α s , α +1 , α . . . , α i − ) =( β i , β i +1 , . . . , β s , β , β . . . , β i − ) for some ≤ i ≤ s .First observe that β [ i ] s = ( α [ i ] s if < i ≤ sα + 1 if i = 1 . RIENTED HAMILTONIAN PATHS AND CYCLES 15
We have: α = α s = β [1+ i − s (by Lemma 13) = α [1+ i − s (since otherwise we get α = α + 1 which is a contradiction) = β [1+2( i − s (also by Lemma 13) = α [1+2( i − s (also sothat we don’t get α = α + 1 , a contradiction). And so on, we may prove by inductionthat α = α [1+ k ( i − s ∀ k ∈ N ∗ , ∀ i ≥ . ( ∗ ) Now, observe that α = α i , α = α i +1 , . . . , α s − i +1 = α s and α s − i +2 = α + 1 .Moreover, we can write s − i +2 = 1+ k ′ ( i − λ.s = [1 + k ′ ( i − s , with k ′ = s − ∈ N ∗ and λ = 2 − i ∈ Z .It follows that α s − i +2 = α [1+ k ′ ( i − s = α by ( ∗ ). But α s − i +2 = α + 1 , thus we reach acontradiction. So the first case cannot occur.Now consider the second case. First suppose that i = s . We have α = − α i for some ≤ i ≤ s − , α = − α i − , α = − α i − , . . . , α i − = α i − (( i − − = − α and α i = − α − .Thus α = − α i = α + 1 and we reach a contradiction. So the second case is impossiblefor ≤ i ≤ s − . If i = s , we have α = − α s , α = − α s − , α = − α s − . . . , α s − = − α s − (( s − − = − α and α s − − α − which also means that α s = − α .Thus ( α , . . . , α s ) = ( − α s , . . . , − α ) and as a result α is symmetric. (cid:3) Let β = ( β , β , . . . , β s ) ∈ Z s ; s is even, and β i β i +1 < ∀ i = 1 , . . . , s − .Then ∀ ≤ i ≤ s , define β i ∗ as: β i ∗ ( β i − if β > β i + 1 if β < We are now ready to link between the number of Hamiltonian paths of some type P ( α ) in T and the number of Hamiltonian cycles of types C ( β ) and C ( β ′ ) (mentioned in thebeginning of this section): Theorem 16.
Let T be a tournament of order n , and ( β , . . . , β s ) ∈ Z s ; s P i =1 | β i | = n , s is even, and β i β i +1 < ∀ i = 1 , . . . , s − . Then:If ( β ∗ , β , . . . , β s ) is symmetric, we have: f T ( β ∗ , β , . . . , β s ) = g T ( β , β , . . . , β s ) .t ( β , β , . . . , β s ) . Otherwise, we have: f T ( β ∗ , β , . . . , β s ) = δ ( β , β , . . . , β s ) .g T ( β , β , . . . , β s ) .t ( β , β , . . . , β s )+ δ ( β ∗ , β , . . . , β s ∗ .g T ( β ∗ , β , . . . , β s ∗ .t ( β ∗ , β , . . . , β s ∗ AND ZEINA GHAZO HANNA where δ ( γ ) = if γ is non − symmetric and not a singleton if γ is symmetric nt ( γ ) if γ is a singleton Proof.
In order to prove this theorem, let us compute f T ( β ∗ , . . . , β s ) .Consider the set P T ( β ∗ , β , . . . , β s ) and let P = x . . . x n be an element of this set. C P is either of type C ( β , . . . , β s ) or of type C ( β ∗ , β , . . . , β s ∗ whether ( x n , x ) or ( x , x n ) ∈ E ( T ) .Let C T ( β ) = { C , . . . , C t } be the set of cycles of type C ( β , . . . , β s ) in T , and let C T ( β ′ ) = { C ′ , . . . , C ′ r } be the set of cycles of type C ( β ∗ , β , . . . , β s ∗ in T . We have two casesto consider:(a) If ( β ∗ , β , . . . , β s ) is symmetric, then by Theorem 15, C T ( β ) = C T ( β ′ ) . Thus weonly have to consider one of them, say C T ( β ) , to avoid counting the same cycletwice in the following step.Let C T ( β ) = { C , C , . . . , C t } . We have that for all C i ∈ C T ( β ) , there exists asubclass X i of P T ( β ∗ , β , . . . , β s ) with respect to R ( β ∗ ,β ,...,β s ) such that everypath in X i generates C i . Thus by Theorem 8, | X i | = | P | = t ( β ) for some P ∈ X i ,since if β ∗ , β , . . . , β s is symmetric, none of β or β ′ can be symmetric, nor asingleton. Hence, f T ( β ∗ , β , . . . , β s ) = t X i =1 | X i | = t X i =1 t ( β )= t.t ( β ) = | C T ( β ) | .t ( β )= g T ( β ) .t ( β ) . (b) If β ∗ , β , . . . , β s is non-symmetric, then by Theorem 15, C T ( β ) = C T ( β ′ ) , thus C T ( β ) ∩ C T ( β ′ ) = ∅ (because the sets of every type of Hamiltonian cycles form apartition of the set of all oriented Hamiltonian cycles in T ).For all C i ∈ C T ( β ) , there exists a subclass X i of P T ( β ∗ , β , . . . , β s ) with respectto R ( β ∗ ,β ,...,β s ) such that every path in X i generates C i . Thus by Theorem 8,and Remark 9, | X i | = | P | = t ( β ) or .t ( β ) or n for some P ∈ X i , whether β is non-symmetric and not a singleton, is symmetric, or is a singleton, so | X i | = | P | = δ ( β ) .t ( β ) .Similarly, ∀ C ′ j ∈ C T ( β ′ ) , ∃ a subclass X ′ j of P T ( β ∗ , β , . . . , β s ) with respectto R ( β ∗ ,β ,...,β s ) such that every path in X ′ j generates C ′ j . Thus | X ′ j | = | P ′ | = δ ( β ′ ) .t ( β ′ ) for some P ′ ∈ X ′ j . RIENTED HAMILTONIAN PATHS AND CYCLES 17
Hence, f T ( β ∗ , β , . . . , β s ) = t X i =1 | X i | + r X j =1 | X ′ j | = t X i =1 δ ( β ) .t ( β ) + r X j =1 δ ( β ′ ) .t ( β ′ )= t.δ ( β ) .t ( β ) + r.δ ( β ′ ) .t ( β ′ )= | C T ( β ) | .δ ( β ) .t ( β )+ | C T ( β ′ ) | .δ ( β ′ ) .t ( β ′ )= g T ( β ) .δ ( β ) .t ( β ) + g T ( β ′ ) .δ ( β ′ ) .t ( β ′ ) , and this concludes our proof. (cid:3) Moreover, for the case when the type of oriented paths in a tournament T is symmetric,we have the following property: Theorem 17.
Let α = ( α , . . . , α s ) ∈ Z s , α i · α i +1 < , α symmetric, and T a tournamentof order n = s P i =1 | α i | +1 . Let P be a Hamiltonian path in T of type P ( α ) and C P ∈ C T ( β ) .Then we have: t ( β ) = 1 . Proof.
Suppose that α > . Since α is symmetric, then s is even, and by Theorem 15,we can assume that C P ∈ C T ( α + 1 , . . . , α s ) = C T ( β ) . If α < , then also by Theorem15, we can assume that C P ∈ C T ( α − , . . . , α s ) , but we will treat the case α > , andthe other case is similar.Since α is symmetric then α = ( α , α , . . . , α l , − α l , . . . , − α , − α ) where l = s , and β = ( α + 1 , α , . . . , α l , − α l , . . . , − α , − α ) .Set r ′ = r ( β ) , we have t ( β ) = sr ′ . Suppose to the contrary that t ( β ) > , We have 2 cases:(a) t ( β ) is even. Set t ( β ) = 2 k , thus β is divided into k tuples ( β , . . . , β r ′ ) .Set a be the first component of the first tuple, we have a = α + 1 . Set b be thelast component of the last tuple ( k th tuple), we have b = − α .Since r ′ is a period, then the first component a ′ of the ( k + 1) th tuple is equal to a , and the last component b ′ of the k th tuple is equal to b .But since α is symmetric, a ′ = − b ′ because a ′ = α l and b ′ = − α l . Thus a = − b which implies that α + 1 = − ( − α ) = α and this leads to a contradiction. So t ( β ) cannot be even. AND ZEINA GHAZO HANNA (b) t ( β ) is odd. Set t ( β ) = 2 k + 1 , k ≥ , thus β is divided into k + 1 tuples ( β , . . . , β r ′ ) , by noting that since α is symmetric, the ( k + 1) th tuple should beof the form ( β , . . . , β r ′ , − β r ′ , . . . , − β ) where β r ′ = α l , thus it is symmetric.(Obviously all the other k tuples have this form since they are all equal).Set a be the first component of the first tuple, we have a = α + 1 . Set b be thelast component of the last tuple ( (2 k + 1) th tuple), we have b = − α .Since r ′ is a period, then the first component a ′ of the ( k + 1) th tuple is equal to a , and the last component b ′ of the ( k + 1) th tuple is equal to b .But since the ( k + 1) th tuple is symmetric, a ′ = − b ′ . Thus a = − b which impliesthat α + 1 = − ( − α ) = α and this leads to a contradiction. So T ( β ) cannot bean odd integer strictly greater than 1.Thus we conclude that t ( β ) = 1 . (cid:3) And finally, with the same hypothesis of Theorem 16, we can deduce the following:
Corollary 18. If ( β ∗ , β , . . . , β s ) is symmetric, Then: f T ( β ∗ , β , . . . , β s ) = g T ( β , β , . . . , β s ) Proof.
The result follows immediately from Theorem 16 and Theorem 17. (cid:3) Oriented Hamiltonian cycles
Based on Theorem 16, linking between the number of oriented Hamiltonian paths ofsome type, and the number of oriented Hamiltonian cycles that can be generated by thesepaths in a tournament, we are now able to establish the main result of Section 2, Theorem4, for oriented cycles:
Theorem 19.
Let α = ( α , . . . , α s ) ∈ Z s ; α i · α i +1 < , α ≥ , s is even if s = 1 , andlet T be a tournament of order n ; n = s P i =1 | α i | . We have: g T ( α ) = g T ( − α ) . Proof.
The proof will be done by induction on s .If s = 1 , α = ( α ) = ( n ) and − α = ( − α ) = ( − n ) and we have g T ( n ) = g T ( − n ) .Suppose that the result is true for s − blocks, s > . That is, if α = ( α , . . . , α s − ) ∈ Z s − ; α i · α i +1 < , α ≥ , s − is even, and T is a tournament of order n ; n = s − P i =1 | α i | , we RIENTED HAMILTONIAN PATHS AND CYCLES 19 have: g T ( α ) = g T ( − α ) . Let’s prove the result for s blocks. Let α = ( α , . . . , α s ) ∈ Z s ; α i · α i +1 < , α ≥ , and let T be a tournament of order n ; n = s P i =1 | α i | . We argue byinduction on α .If α = 0 , then by induction, g T ( α ) = g T (0 , α , . . . , α s ) = g T ( α + α s , α , . . . , α s − ) = g T ( − α − α s , − α , . . . , − α s − ) = g T (0 , − α , − α , . . . , − α s − , − α s ) = g T ( − α ) .So suppose that α > and the result is true when the first block is of length α − , andlet’s prove it when the first block is of length α .We will consider two cases:(a) The tuple ( α − , α , . . . , α s ) is non-symmetric.We have α − ≥ , thus by Theorem 16, f T ( α − , α , . . . , α s ) = δ ( α , . . . , α s ) .t ( α , . . . , α s ) .g T ( α , . . . , α s )+ δ ( α − , α , . . . , α s − , α s − .t ( α − , α , . . . , α s − , α s − .g T ( α − , α , . . . , α s − , α s − δ ( β ) .t ( β ) .g T ( β ) + δ ( β ′ ) .t ( β ′ ) .g T ( β ′ ) where δ ( γ ) = if γ is non − symmetric and not a singleton if γ is symmetric nt ( γ ) if γ is a singleton Now consider the tuple ( − α + 1 , − α , . . . , − α s ) which is also non-symmetric.We have − α + 1 ≤ , thus by Theorem 16, f T ( − α + 1 , − α , . . . , − α s ) = δ ( − α , . . . , − α s ) .t ( − α , . . . , − α s ) .g T ( − α , . . . , − α s )+ δ ( − α + 1 , − α , . . . , − α s + 1) .t ( − α + 1 , − α , . . . , − α s + 1) .g T ( − α + 1 , − α , . . . , − α s + 1)= δ ( β ) .t ( β ) .g T ( β ) + δ ( β ′ ) .t ( β ′ ) .g T ( β ′ ) where δ ( γ ) = if γ is non − symmetric and not a singleton if γ is symmetric nt ( γ ) if γ is a singleton Since ( − α + 1 , − α , . . . , − α s ) = − ( α − , α , . . . , α s ) , then by Theorem 4 wehave f T ( α − , α , . . . , α s ) = f T ( − α + 1 , − α , . . . , − α s ) . As a result, δ ( β ) .g T ( β ) .t ( β ) + δ ( β ′ ) .g T ( β ′ ) .t ( β ′ ) = δ ( β ) .g T ( β ) .t ( β ) + δ ( β ′ ) .g T ( β ′ ) .t ( β ′ ) . AND ZEINA GHAZO HANNA But, since β = − β and β ′ = − β ′ thus if β is non-symmetric and not a singleton(resp. is a singleton, or is symmetric), so is β , and similarly for β ′ and β ′ , so δ ( β ) = δ ( β ) , and δ ( β ′ ) = δ ( β ′ ) , and also by Proposition 3 we have t ( β ) = t ( β ) and t ( β ′ ) = t ( β ′ ) .Moreover, since α − < α , then by induction g T ( β ′ ) = g T ( β ′ ) , hence we have g T ( β ) = g T ( β ) , and the result follows.(b) The tuple ( α − , α , . . . , α s ) is symmetric.We have α − ≥ , thus by Corollary 18, f T ( α − , α , . . . , α s ) = g T ( α , α , . . . , α s ) . Now consider the tuple ( − α + 1 , − α , . . . , − α s ) which is also symmetric. We have − α + 1 ≤ , thus by Corollary 18, f T ( − α + 1 , − α , . . . , − α s ) = g T ( − α , − α , . . . , − α s ) . Since by Theorem 4 we have f T ( α − , α , . . . , α s ) = f T ( − α + 1 , − α , . . . , − α s ) , we get g T ( α , . . . , α s ) = g T ( − α , . . . , − α s ) . (cid:3) Digraphs of maximal degree ∆ ≤ After establishing Theorem 4 and Theorem 19, proving that a tournament and its com-plement contain the same number of oriented Hamiltonian paths and cycles of any giventype, we may generalize this fact to any digraph of maximal degree 2: If H is a digraphwith maximal degree ∆( G ( H )) ≤ , then f T ( H ) = f T ( H ) , where f T ( H ) is the number ofcopies of the digraph H in a tournament T .For this purpose, we first need to prove several lemmas: Lemma 20.
Let α = ( α , . . . , α s ) ∈ Z s ; α i · α i +1 < , α ≥ , and let T be a tournamentof order n ; n ≥ s P i =1 | α i | +1 . We have: f T ( α ) = f T ( − α ) . RIENTED HAMILTONIAN PATHS AND CYCLES 21
Proof.
Let m = s P i =1 | α i | +1 . Every oriented path in T of type P ( α ) is a Hamiltonianpath of type P ( α ) contained in a subtournament T ′ of T of order m . By Theorem 4, f T ′ ( α ) = f T ′ ( − α ) . Moreover, if we consider another subtournament T ′′ of T , of order m , T ′′ = T ′ , then P T ′ ( α ) ∩ P T ′′ ( α ) = ∅ , because every Hamiltonian path in T ′ differs with aleast one vertex from every Hamiltonian path in T ′′ .So let V ( T ) = S X ⊆ V ( T ) , | X | = m X , we have: f T ( α ) = X X ⊆ V ( T ) , | X | = m f h X i ( α ) = X X ⊆ V ( T ) , | X | = m f h X i ( − α ) = f T ( − α ) , and we get our result. (cid:3) Similarly, we may prove the same result for cycles in tournaments:
Lemma 21.
Let α = ( α , . . . , α s ) ∈ Z s ; α i · α i +1 < , α ≥ , and let T be a tournamentof order n ; n ≥ s P i =1 | α i | . We have: g T ( α ) = g T ( − α ) . Lemma 22.
Let T be a tournament, and let H be a digraph with ∆( G ( H )) ≤ and suchthat its connected components are mutually isomorphic. Then the number of copies of H in T and that in its complement T are the same.Proof. Since H is a digraph with ∆( G ( H )) ≤ and such that its connected componentsare isomorphic, then H = H ∪ H ∪ · · · ∪ H r where the digraphs H i , ≤ i ≤ r , are itsconnected components, with | H i | = m ∀ ≤ i ≤ r , and such that they are either all pathsof the same type, say P ( α ) , or all cycles of the same type, C ( β ) . If T contains a copyof H , then since the digraphs H i , ≤ i ≤ r , are disjoint, the copy of every digraph H i is a spanning subdigraph of a subtournament T i of T , such that the subtournaments T i , ≤ i ≤ r , are also disjoint, with | V ( T i ) | = m ∀ ≤ i ≤ r . Note that rm ≤ n .Let’s consider r disjoint subtournaments of T , T i , ≤ i ≤ r , all of order m , and supposethat T contains a copy of H such that ∀ ≤ i ≤ r , H i has a copy in T i . As f T i ( H i ) denotes the number of copies of H i in the subtournament T i , then the number of copiesof H in T , such that the copy of H i is a spanning subdigraph of T i , is: r Y i =1 f T i ( H i ) . Now if we consider any permutation σ of the subtournaments T i , and since all the digraphs H i are isomorphic, then ∀ ≤ i ≤ r , if H i has a copy in T i , then H i also has a copy in T σ ( i ) . But, also since all the digraphs H i are isomorphic, then the copies of H obtained in AND ZEINA GHAZO HANNA T such that the copy of each H i is a spanning subdigraph of T i are the same as the onesobtained in T such that the copy of each H i is a spanning subdigraph of T σ ( i ) .Let’s compute now f T ( H ) , the total number of copies of H in T .Let L = { ( T , T , . . . , T r ); T i subtournament of T ∀ ≤ i ≤ r , T i ∩ T j = ∅ ∀ ≤ i, j ≤ r, | V ( T i ) | = m } . We have: f T ( H ) = X ( T ,T ,...,T r ) ∈L Q ri =1 f T i ( H i ) r ! . However, by Lemma 20 and Lemma 21, we have that ∀ ≤ i ≤ r , f T i ( H i ) = f T i ( H i ) . So let L ′ = { ( T , T , . . . , T r ); T i subtournament of T ∀ ≤ i ≤ r , T i ∩ T j = ∅ ∀ ≤ i, j ≤ r, | V ( T i ) | = m } , we get: f T ( H ) = X ( T ,T ,...,T r ) ∈L Q ri =1 f T i ( H i ) r ! = X ( T ,T ,...,T r ) ∈L ′ Q ri =1 f T i ( H i ) r ! = f T ( H ) , and the result follows. (cid:3) We may now prove the main result of this section:
Theorem 23.
Let T be a tournament and let H be a digraph with ∆( G ( H )) ≤ . Thenthe number of copies of H in T and its complement T is the same.Proof. Since ∆( G ( H )) ≤ , then H is a disjoint union of directed paths and cycles. Write H as H = S ti =1 H i , where each H i is a subdigraph of H whose all connected componentsare isomorphic, and which is maximal with this property. The connected components ofeach H i are either all paths of the same type or cycles of the same type. Note that thedigraphs H i , ≤ i ≤ t , are disjoint, and non-isomorphic.If T contains a copy of H , then since the digraphs H i , ≤ i ≤ t , are disjoint, the copyof every digraph H i is a spanning subdigraph of a subtournament T i of T , and such thatthe subtournaments T i , ≤ i ≤ t , are also disjoint, with | V ( T i ) | = | V ( H i ) | ∀ ≤ i ≤ t .As we did in the previous lemma, let’s consider t disjoint subtournaments of T , T i , ≤ i ≤ t , and such that | V ( T i ) | = | V ( H i ) | , and suppose that T contains a copy of H such that ∀ ≤ i ≤ t , H i has a copy in T i . The number of copies of H in T , such thatthe copy of H i is a spanning subdigraph of T i , is: t Y i =1 f T i ( H i ) . RIENTED HAMILTONIAN PATHS AND CYCLES 23
However, if we consider any permutation σ of the subtournaments T i , and since all thedigraphs H i are non-isomorphic, then if T contains a copy of H such that ∀ ≤ i ≤ t , H i has a copy in T σ ( i ) , the copies of H obtained in T such that the copy of each H i is aspanning subdigraph of T i are all different from those obtained in T such that the copyof each H i is a spanning subdigraph of T σ ( i ) .So let’s compute now the total number of copies of H in T , f T ( H ) .Let L = { ( T , T , . . . , T t ); T i subtournament of T ∀ ≤ i ≤ t , T i ∩ T j = ∅ ∀ ≤ i, j ≤ t, | V ( T i ) | = | V ( H i ) |} . We have: f T ( H ) = X ( T ,T ,...,T t ) ∈L t Y i =1 f T i ( H i ) . However, by Lemma 22, since the connected components of each digraph H i are iso-morphic, we have that ∀ ≤ i ≤ t , f T i ( H i ) = f T i ( H i ) .So let L ′ = { ( T , T , . . . , T t ); T i subtournament of T ∀ ≤ i ≤ t , T i ∩ T j = ∅ ∀ ≤ i, j ≤ t, | V ( T i ) | = | V ( H i ) |} , we get: f T ( H ) = X ( T ,T ,...,T t ) ∈L t Y i =1 f T i ( H i ) = X ( T ,T ,...,T t ) ∈L ′ t Y i =1 f T i ( H i ) = f T ( H ) , hence: f T ( H ) = f T ( H ) , and this concludes the proof. (cid:3) Remark 24.
Let T be a tournament on n + 1 vertices, formed by a directed n -cycle C = v v . . . v n , with its internal edges, where these edges may have any orientations, anda vertex v of in-degree equal to zero (a source), adjacent to the n vertices of the cycle( d + T ( v ) = n ). Then the complement T of this tournament is formed by a directed n -cycle, C ′ = v v n v n − . . . v , and its internal edges which have opposite orientations of those of T h C i , and a vertex v of out-degree equal to zero (a sink) adjacent to all the vertices of C ′ . Also note that since C and C ′ are directed cycles, then ∀ x ∈ C , d + T ( x ) ≤ n − and ∀ y ∈ C ′ , d + T ( y ) ≤ n − .Thus if we consider a digraph H on n + 1 vertices, formed by a vertex y and n out-neighbors of y , which is a digraph of maximal degree ∆( G ( H )) = n , the number of copiesof H in T is equal to one, while there are no such copies in T . AND ZEINA GHAZO HANNA Based on the remark above, we finally ask the following:Let f T ( H ) denote the number of copies of a digraph H in a tournament T . Problem 25.
Can we characterize the set H of all digraphs H such that f T ( H ) = f T ( H ) for any tournament T ? Acknowledgments.
We would like to thank the Lebanese University for the PhD grant,and Campus France for the Eiffel excellence scholarship (Eiffel 2018).
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