A PROOF OF SOME SCHOTZENBERGER-TYPE RESULTS FOR EULERIAN PATHS AND CIRCUITS ON DIGRAPHS

This paper shows that the number of even Eulerian paths equals the number of odd Eulerian paths when the number of arcs is at least twice the number of vertices of a digraph.


INTRODUCTION AND CONVENTIONS.
This paper shows that in a digraph of order n with m arcs that satisfies rn _> 2n, the num- ber of even Eulerian paths equals the number of odd Eulerian paths.This result generalizes Schiitzenberger's theorem (see [1] and [2]), which says that in a digraph of order n with rn arcs that satisfies m _> 2n + 1, the number of even Eulerian circuits equals the number of odd Eule- rian circuits.The proof is also perhaps more intuitive, not depending on too much terminology or disparate graph theory results.
In this paper, a digraph is defined by a sequence of arcs, where an arc is indicated by an ordered pair such as (a, b), and multiple arcs and loops are allowed.Those letters which appear in describing the arcs in the sequence defining the graph are therefore considered as vertices or points.Note that with this definition, there can be no isolated vertices.If D is a sequence of arcs, its length is denoted by ]DI.If VD denotes the set of vertices which appear in D, then ]VD] denotes its cardinality.
Two digraphs D and D are considered to be the same, under appropriate relabeling of vertices, when D2, as a sequence of arcs, is a permutation of the sequence of arcs which define D. We Unlike the usual way, we define a path of D as a subsequence of a permutation of the sequence describing the digraph D. If a path (a,b).(a2, b2) (a,,bn) is a permutation of D and has the additional property that bl a2, b2 a3 bn-1 an, then it is an Eulerian path of D. The added condition b, al identifies an Eulerian circuit of D. A path (a, b).(c, d) (e, f) is said to start from a and end at f; the vertex a is the starting point and the vertex f is the end point.If a and /are paths of D, then the sequence obtained by putting /after a is written provided that the resulting sequence is also a path of D.

B. CHWE
Next, we denote by Ax D the set of all Eulerian paths of tire digraph D which start froxl le point x.If the starting point is fixed but unimportant we write A D. Also, vhen Ax D. f we consider as a digraph, then A c A D.
If and /3 are two paths and if there are two paths " and3, where either or both can be empty, such that 6 3, then o is a part of/3 and ve se the notation o _ J to indic'ate this fact.Hence, if an arc (a, b) is a term in a path/, we write (a, b) _ 3. By/3 is mealt tile subsequence of/3 obtained by eliminating all terms of o one by one.Thus, if arcs in/3 or ct'tr more than once the procedure Inay leave copies left over and the result may not be uniqtle.notations od(a), id(a), and d(o) are defined as follows: od(a) i. the nulnber of terms in D wl,'h start from a, id(a) is the number of terms in D which end at o, and d(a) od(a) + ,d(o) is clll,d the degree of a.An even (odd) point is a point with even (odd) degree.if c is a multiple of an even number of transpositions, and r(c) -1 if c is a multiple of an odd number of transpositions.Let g(c) e(c).r(c).
What we are looking for is a natural, intuitive proof of the formula: Z g(c) 0 ZaeA,/ g(c) for any x VD, if IVD[ n and IDI _> 2n for all integers n _> 1.
For convenience of notation, we define g(A D) ]-aeA.D g(c) and define g(A D) similarly.
For all positive integers n, let Bn be the family of digraphs D such that For all positive integers n, let .4,be the family of digraphs D sch that For some q 6_ VD, zd(q) / od(q) or 2.
The proposition Sn is the following: "For j 1,...,n, if D 6_ mj U Bj, then g(A D) 0." In Lemma 3.1 we demonstrate that $1, 6'2, and $3 are true.In order to prove Sn in general we proceed by induction on n, assuming Sn-.In computing g(A D), we seek to identify A D with a union of suitable A D:, where lYD:] < ]YD, I.This is accomplished by identification of a length 3 path with an arc or by deleting some paths in order to maintain the relation that g(A D) 0 if 9(h O:) 0 for all i.
For instance, as a trivial example, let (a, v), (v, w), (w, b) 6_ D and (a,v). (v,w). (w,b) C_ for all a 6_ AD.Let, when x # v and x # w, D' (D -(a,v) (v,w) (w,b)). (a,b), and let F be a map of A D' to A D defined by V(c) =/3(a, b)7 where a B(a, v). (v, w). (w, b)7.Then V is a one-to-one map and when a,/3 6_ A D,/3 is an even permutation of c if and only if F(B) is an even permutation of F(a).Thus 9(A D') 0 implies 9(A D) 0.
In particular we assume, based on this observation, that there are no vertices v and w of VD for which there is a path of the type just described, when we assume S,_l.
In general, if there are digraphs D, D2 D and such injeetive maps Fi from A Di to A D such that [,JF,(A D,) forms a partition of A D and each 9(A Di) O, then we can conclude g(A D) 0.
As another example of an identification of the type described above, if D contains (a, b) and (b, c) and if every c 6_ A D starts or ends with (a, b). (b, c), then take D' D (a, b) (b, c).In this ease it is also clear that if 9(A D') 0, then 9(A D) 0.
In all our arguments, it is the case that if a statement is true for a digraph D, then it is also true for the digraph obtained by reversing all ares (a, b) of D to ares (b, a).
If a digraph D contains an are (a, b) with multiplicity at least two, then trivially 9(A D) 0 since by a simple transposition of one (a, b) with another leaves D unchanged while 9(A D) changes sign.Therefore we assume no multiple arcs.
If d(q) 2, and q is the starting point, then AxD {(q,h)c(t,q) c 6_ A Do} where Do D-(q,h) -(t,q) and it is clear that 9(A D) =kg(AxDo).
If d(q) 1, then AD {(q,h) c 6_ A D0}, where Do D-(q,h), and it is also clear that g(h D) :hg(A Do).
Also it is clear that VDi q and IDol _> Inl-2, Iunil <_ IVD 1.Note that d(h) in D, becomes smaller by 2 except when h is an end point, where it becomes smaller by 1.We need this fact for the following Lemma 1.1.
B. CHWE (B).If D 6-A,+I and there is an arc (q,b) or (h,q) in D such that d(q) 2 and d(h) 3 ()r 4, then g(Ax D) 0 if xq and x # b.
PROOF.(A).The digraphs D, in Proposition belong to A,-I U B,-I. Hence g(A D,) 0 because of Sn-1, and g(A D) 0.
(B).In Proposition 1, whcn (q,h), (h,b,) C_ D, if d(h) 3 or 4 in D, then d(h) in D, or 2, except the case in which q or h is an end point.Thus D, 6-A, and g(D,) 0. Therefore g(Ax D) (-1),g(A D,) 0 if x : q and x # h.When (h,q) C_ D, we consider (b,,h) C_ n instead of (h, b,); then the argument is sinilar to the case of (q, h) C_ D. I-I LEMMA 1.2.When we assume S,-l, if a digraph D 6-B, has an arc (t, h) such that an(! h have degree 3 or 4, then g(A D) 0.
PROOF.If h, (t,t) C_ D, or (h,h) C_ D, then the statement reduces to a case of S,,_.So we assume that h, (t,t) D, and (h,h) D. According to our assumption, situations illustrated in the following drawings may occur: h h In those diagrams we can interchange h and t, and also we can place the arrows in any way so that AD#.
Assuming A D # , () and () or 4, we construct a digraph D 6-An+1 from D as follows.
If d(t) 3 and ((,,t) is not an arc of D, then 2 { (t,t) (t.l) } D2 ( -(t,t) -(t,d) -(, t) -(t.q) -(q,b)).(c, 1,) A3 { e A, #. (,t). (t,,). (c,d) and the results arc the same.Moreover, it is clear that g(A D) g(A D).Since D2 and D3 do not contain vertices and q, D2, D3 A,_t U B,_. Th,s g(A D2) 0 and 9(A D3) 0. Together with g(A, ) o, we get g(AD) =0, so we get g( A D, =0for:#h.In theceh=z, wecanswitcht and h.Thus g(A, D) 0 for all z VD, or #(A D) 0. PROPOSITION 2. Supposc D B and A D # O. Also wc sume that n 4, and that D does not contain any arc of multiplicity more than one, nor any vertex v such that d(v) 4 and arc (,,, v) D. Then D contains an arc (t, h) such that # h, one of and h h degree 4 and the other h degree 4 or 3.
PROOF.The fact that D B implies that vVD d(v) 4n. and d(v) 4 for all even points v of D. Since A D , the number of odd points is 0 or 2. If it is 0, then d(v) 4 for all thus our sertion is true.In thc ce of 2, say a and b are odd points, and d(a) + d(b) 6 or 8.If d(a) + d(b) 6, then there must be c VD such that d(c) 6 and all other vertices have degree 4, while n 4 implies there is a vertex d beside a, b, and c.If d incidents only to c, then the arc (c, d) or (d, c) h the multiplicity more than 1.So d must incident to a or b or a point of degree 4. Thus the sertion is true, because d(d) 4 and d(a) d(b) 3.
In the ce d(a) + d(b) 8, then say d(a) 3 and d(b) 5. Since n 4, there are at let two more vertices, say c and d, such that d(c) d(d) 4. If any vertices of degree 4 do not incident to each other and also do not incident to a then all vertices of degree 4 must incident to b, thus d(b) 8.This contradicts with d(b) 5. Thus some vertex of degree 4 h to be incident to the vertex a or a vertex of degree 4. Thus our sertion is true.LEMMA 2.1.If D B, n 4, and we sume S_, then g(A D) 0.
PROOF.From Proposition 2, D h an arc (t, h), # h, such that one of and h h degree 4 and the other h degree 3 or 4. Then by Lemma 1.2, g(A D) 0. PROPOSITION 3. If g(A D') 0 for all D' B, then for all D B, g(A D) 0 when we sume S_ .
PROOF.Let D B, a VD, and a D. Considering a digraph to be a sequence of arcs, it is true that A= a Aa D. Now let VD] n, D]-2n r and a A= D. Define p(a) to be the sequence of the first r terms of a, while q(a) to be the subsequence of a consisting of the next 2n terms.Thus a p(a)q(a).Let Ap() { A= D p() p(a)} {p(a)f f Aq(a)}, where i th d point o p(), d () A o om j < o () U . () om j < , th (A ()) 0 b o S_. f () A thn (A ()) 0 om Lemmn 1.1.A.If q(a) B then g(A q(a)) 0 from our hypothesis, g(A D') 0 if D' B. Since A D {p(a) A q(a)} where a runs all elements of A D, and g({p(a) A q(a)}) g(A q(a)) 0, hence g(A D) 0. LEMMA 3.1.S, Se, and $3 are true.
PROOF.S is clearly true.If D B, then D is one of the following (with the orientation arbitrary)" write a sequence such as ((a,b),(c,d),(d,y)) D in the form D (a,b).(c,d).(d,f).For example, as D1 (a,b). (b, c). (a,b) or D2 (a,b). (a,b). (b, c) with D1 D2 according to our definition.Also, IO1=3 and IVDI 3.
For example, if D is the digraph D (a, o).(a, d). (c, a). (b, a). (b, a) then od(a) 2, ,d(a) od(b) 2, id(b) O, od(c) 1, ,d(c) O, od(d) O, id(d) 1, and A D ).The diagram for D can be constructed as: b a d Let a be a permutation of D, and define the following functions.Let e() if is an Eulerian path or circuit, and let e(c) 0 if c is not.Let r() be the sign of the permutation o; r(c) + For example, if D (a, b).(a, c).(b, a).(a, a), D {(a,a)-(a,b)-(b,a). (a,c), (a,b). (b,a). (a,a). (o,c)}, AD AD b, g((a,o) (a,b) (b,a)-