Abstract
An -path cover of a simple graph is a set of vertex disjoint paths of , each with vertices, that span . With every we associate a weight, , and define the weight of to be . The -path cover polynomial of is then defined as where the sum is taken over all -path covers of . This polynomial is a specialization of the path-cover polynomial of Farrell. We consider the -path cover polynomial of a weighted path and find the -term recurrence that it satisfies. The matrix form of this recurrence yields a formula equating the trace of the recurrence matrix with the -path cover polynomial of a suitably weighted cycle . A directed graph, , the edge-weighted -trellis, is introduced and so a third way to generate the solutions to the above -term recurrence is presented. We also give a model for general-term linear recurrences and time-dependent Markov chains.
1. Introduction, -Path Cover Polynomial, and Notation
Let be a graph with no loops or multiple edges, with vertex set .
First we review some basic concepts to establish notation.
A path in is a sequence of distinct vertices where each pair for is an edge. The length of a path is the number of vertices in it. Thus a path of length is a vertex and a path of length is an edge, and has length . Path begins at vertex , its first vertex, and ends at vertex , its last vertex. The path and its reverse are considered to be the same path. The set of vertices in is . Two paths and in are disjoint if . The empty path has vertices. Finally, recall that a subgraph of spans if it has the same vertex set as .
Now we introduce the central concept of this paper.
An -path has ; that is, it is a path of length at most for some fixed with .
An -path cover of is a set of pairwise disjoint -paths of that span . Thus each satisfies , and every vertex of lies in exactly one -path; that is, is a partition of .
With every -path we associate a weight, , and then the weight of is .
Definition 1. The -path cover polynomial of , , is the sum of the weights of all -path covers of ; that is, where is an -path cover of .
The path-cover polynomial (or path polynomial) of a graph is a specialization of the -cover polynomial of Farrell [1] where is restricted to be a path; see Farrell [2]. Thus our -path cover polynomial is a further specialization to paths of length . See also Chow [3] and D’Antona and Munarini [4].
It seems that this research is the first direct consideration of the -path cover polynomial of a graph. See McSorley et al. [5] for specialization to the case , where all classical orthogonal polynomials are generated as -path cover polynomials of suitably weighted paths. For related work see the theory of weighted linear species developed in Joyal [6] and Bergeron et al. [7]. In particular, Munarini [8] uses the -filtered linear partitions of a linearly ordered set to achieve some similar results; see especially our Sections 7 and 8.
In Section 2 we introduce a weighted path and find the -term recurrence that its -path polynomial satisfies. In Section 3 the matrix form of this recurrence is presented and yields a trace formula that, in Section 4, gives the -path cover polynomial of a suitably weighted cycle . Section 5 interprets our results in terms of a model for time-dependent Markov chains. In Section 6 a directed graph, , the edge-weighted -trellis, is introduced and so a third way to generate the solutions to the above recurrence and trace is found. In Section 7 we model general constant coefficient linear recurrences, and we derive various relevant formulas with both algebraic and combinatorial proofs. Finally, in Section 8, we obtain a relevant new integer sequence and relate this sequence to known sequences in the literature.
Notation. We write , instead of , for the -cover polynomial of the path ; similarly we write instead of , and so forth.
Vertices in (Section 2) and in subpaths of will be labelled ; vertices in (Section 4) will be labelled ; and vertices in (Section 6) will be labelled .
For we use indeterminate as the weight of a path of length in . Throughout the paper is fixed. In all the examples we set , and many examples have .
2. Weighted Path
For and the path has vertices . The first vertices are weighted with weight and the remaining vertices are weighted, one by one, with the indeterminates from the set . Thus all vertices, that is, all paths of length , in are weighted. For a path of length in is weighted with if its last vertex has weight and with if its last vertex has weight . The path is the empty path with no vertices.
Definition 2. For let be the -path cover polynomial of the weighted .
Starting conditions are for .
As mentioned in Section 1, throughout this paper the path is a subpath of the weighted .
We now derive our main -term recurrence.
Theorem 3. For a fixed and any ,
Proof. The last vertex of lies in every -path cover of . Suppose, in such an -path cover, it is present as the last vertex in an -path of length . Then this -path has weight and begins at . The sum of the weights of all such -path covers is therefore where is a subpath of . Now summing over gives the result. The initial conditions for ensure that this equation holds when .
Example 4. For the weighted path is
(4)
The weights of paths of lengths and (vertices and edges) are shown above the path. Vertex labels and weights of paths of length are shown below the path. Let
All 3-path covers of and their weights are shown below:
(6)
So .
Example 5. Theorem 3 with gives the -term recurrence for a fixed , Then the starting conditions give We check from Example 4.
Definition 6. For we define as above for , except that we have vertices instead of vertices of weight at the beginning of the path. Thus has vertices and is formed from by truncating from the right. All -paths in are weighted as in . We let be the -path cover polynomial of the weighted . We note that .
Example 7. For and ,
see Example 5.
For a fixed with we define the starting conditions,
We then have the following recurrence; the proof is similar to the proof of Theorem 3, and setting recovers Theorem 3.
Theorem 8. For a fixed with and any ,
We now work with the fundamental solutions to recurrence (2).
For let denote the th fundamental solution to (2). Thus the obey the recurrence with starting conditions where .
We have
Our next result expresses as the difference of two -path cover polynomials. Consistent with (10) we set for every .
Lemma 9. For and ,
Proof. By induction on , first consider . Now when and otherwise. Each satisfies (12), so . Now consider the path shown below:
(16)
The first vertex lies in every -path cover of so, similar to the proof of Theorem 3, we have
Thus, from above, ; that is, (15) is true for .
Now we have
using (12) again at the first line, the induction hypothesis at the second line and Theorem 8 at the last line. Hence the induction goes through and (15) is true for all .
Example 10. Using (12) and the starting conditions following (12) for and the fundamental solutions to recurrence (2) are
We check (14) using Example 5, We also check Lemma 9 using and Example 7,
By iteration of such formulas, we have Corollary 11, where (ii) is a specialization of (i) with .
Corollary 11. (i) For ,
(ii) the first fundamental solution to recurrence (2) is given by
Corollary 12 is a useful technical result.
Corollary 12. For and ,
Proof. For from Corollary 11(ii) we have . Now in the weighted path let vertex be covered by a path of length where . Then begins at vertex and ends at vertex , which has weight ; so . Now in every -path cover of vertex must be covered by such a path , so , which is the above formula for .
For any the path is a subpath of . In fact the weighted paths and (except for vertex labels) are identical, so . From Lemma 9 we have
So let vertex be covered by a path of length . Then begins at vertex and ends at vertex , which has weight . Hence . Furthermore, because ends at if , then ; hence , a contradiction; so .
Now, similar to the above, the sum of the terms of that contain is . Finally, summing over the lengths of all possible paths , namely, summing over with , gives the result.
This completes study of the weighted path .
3. Matrix Formulation and Trace
We set up our -term recurrence (2) in matrix form.
Let be the identity matrix, and for let be the matrix Let denote transpose, and let be the vector . Then recurrence (2) can be written as where . By iterating this equation we have , where
With tr denoting trace, we have the following.
Lemma 13. For ,
We now apply these results to the weighted cycle .
4. Weighted Cycle and Trace
We introduce the weighted cycle for shown in Figure 1. It has vertices labelled and edges.
It is weighted as follows: for , let be a path of length that traverses clockwise and ends at vertex . We define .
Thus the weighted cycle is an isolated vertex with weight ; and the weighted cycle has vertices with and and edges: edge with and edge with .
In Figure 1 only the weights of paths of lengths and are shown.
Lemma 14. For the following -path cover polynomials, the first which comes from and the second from , are equal:
Proof. Except for vertex labels, the weighted paths in and in are identical. Hence the result is obtained.
Definition 15. For let be the -path cover polynomial of the weighted .
In the following, when necessary, we reduce subscripts on , , and the second subscript on , all modulo . We write , , , and so forth.
Theorem 16 is the main result of this section. Recall the matrix from (28).
Theorem 16. For ,
Proof. Consider the weighted . Vertex lies in every -path cover of . Suppose, in such an -path cover, it is covered by a path of length that begins at and ends at , for some . Now ; that is, . The sum of the weights of all such paths is then But , so letting at the second line, and using subscript reduction modulo and Lemma 14 at the third line, then Corollary 12 at the fourth line, and Lemma 13 at the last line.
Example 17. For and consider the weighted in Figure 2.
The -paths are weighted as follows: By considering all -path covers, the -path cover polynomial of the weighted is
Similar to Example 10, the recurrence (12) and the starting conditions following (12) give
Together with the following matrices we may check the results from Lemma 13 and Theorem 16,
5. Markov Chain Interpretation
In this section we consider an interesting special case, where in the matrix formulation of the recurrence we have stochastic matrices. A matrix of Form (26) can be considered a transition matrix for a Markov chain with states under the conditions Because the probabilities vary with , these are the transition matrices for a nonhomogeneous Markov chain. Note also that, as transition matrices are multiplied from left to right, the process is effectively time reversed. In fact, This process is often referred to as a ladder process. From any state , with , the process jumps with certainty to , then to , and so forth, up the ladder, till it reaches state . At that point it jumps randomly back down the ladder to one of the intermediates states , , and the procedure repeats. Because all of the matrices are stochastic, the row sums of matrices such as , see (28), will all equal 1. Recall from Section 3 that Thus, we have the following.
Proposition 18. In the stochastic case, all of the path polynomials evaluate to 1.
5.1. Homogeneous Case
In the case of constant coefficients (see (26)), sending , for all , we drop the dependence on and write with . Now is the -step transition matrix. It is easy to see that a row vector (on the left) fixed by is Furthermore, under the assumption , for all , it is immediate that the chain is irreducible and aperiodic, hence ergodic. That is, exists and has equal rows, and each row proportional to the left-invariant vector indicated above normalized to row sum 1.
Example 19. Take the uniform case , . Then we have the fixed vector and the limits
Thus, for large , if we randomly choose an -path cover of then the probability that it belongs to the -fundamental solution is . In particular, the first fundamental solution satisfies
So the -path cover polynomial model provides a combinatorial model for nonhomogeneous Markov chains. A closely related model, the trellis, is discussed in detail below.
6. Edge-Weighted -Trellis
In this section we deal with the edge-weighted -trellis, , shown in Figure 3, and give another method of generating and .
The vertices of are labelled . All edges in are directed, with arrows as shown. All circuits in are directed and are traversed in the direction of the arrows. We use to denote a directed circuit in , which we simply call a circuit. A circuit is based at vertex if it begins and ends at vertex . A circuit may pass through the same vertex more than once. The length of a circuit is the number of edges in it.
The weights on the edges of are taken from where , as shown. The weight of circuit , , is the product of the weights of all the edges in . If the edge with weight is traversed as the th edge in , then is a factor in ; thus the meaning of here is different from that in Sections 2 and 4. We allow empty circuits with length .
Definition 20. Let and, for , let be the sum of the weights of all circuits in that are based at vertex with length .
Notation. We use standard multiset notation: , and means no occurrences of .
Theorem 21. For ,
Proof. By strong induction on . Now ; hence (48) is true for . We now assume that for all . Consider any term in is the weight of some circuit in based at vertex with length . Clearly ends with a -cycle based at vertex , for some with . Thus the last edge of is , with weight , and the previous edges are , each of weight . Hence . Thus using the strong induction hypothesis and then Theorem 8. So the induction goes through and (48) is true for all .
Let be the expression obtained when every indeterminate in is replaced by ; similarly for other expressions.
Recall that is a subpath of for ; for the path is the empty path , and .
Corollary 22. For and ,
Proof. For we have . For then is a subpath of so, for every , we have . Now, from Theorem 21, , so , as required.
We now connect and the fundamental solutions of the -term recurrence (2).
Theorem 23. For ,
Proof. Consider a circuit in based at vertex with edges. Then, for some , the first edges in this circuit are , followed by edge ending at vertex . These edges contribute to . Now, starting at vertex , the last edges traversed in are , contributing to . Hence .
Thus
putting at the second line, then using Corollary 22 with and at the third line, and finally using Corollary 12 at the last line.
Example 24. Consider , the edge-weighted -trellis; see Figure 4.(a) = sum of weights of circuits of based at with length + , as in Example 10.(b) sum of weights of circuits based at with length 6. We observe that the first edge in such a circuit is edge of weight ; hence is a factor of every term in , consistent with Example 10 again.
Finally, we bring the results from Lemma 13 and Theorems 16 and 23 together in Theorem 25.
Theorem 25. For ,
Example 26. Again, from , we have : which are consistent with the above definitions and results and with Example 17.
7. Homogeneous Case,
In this section, we consider the case of constant coefficients, that is, where the indeterminates are independent of .
Notation. We use to modify a path or expression or matrix in which weights or indeterminates are replaced with .
First we review some known properties of -path polynomials using standard techniques. Then we show how our model recovers these results combinatorially.
7.1. Constant Coefficient Recurrences
This subsection mainly establishes notation and recalls basic results of interest.
Consider the recurrence
We begin with the first fundamental solution. The following is standard and readily derived via geometric series and multinomial expansion.
Proposition 27. One has the generating function and formula giving the (first) fundamental solution, , to the recurrence, that is, with initial values , , .
The matrix takes the form, confor Section 5.1, so that . Define the st fundamental solution to recurrence (55) to be the one with initial conditions and denote this fundamental solution by , with . Then the entries in the bottom row of are exactly the values In general,
The fundamental solutions for can be expressed in terms of the first fundamental solution as follows.
Proposition 28. The st fundamental solution to the recurrence (55) is given by where denotes the first fundamental solution.
Proof. We will illustrate for that shows how the general case works. We have For , we obtain 0 for nonpositive , except for , as required. Similarly, for , for nonpositive we obtain 1 precisely for ; otherwise we get 0. Note that the subtractions are necessary to cancel off terms when . Since the coefficients are independent of , these are indeed solutions to the recurrence. Thus the result is obtained.
Now for the trace, we have the following.
Proposition 29. The trace of is given by
Proof. From (60), we have, using Proposition 28,
Remark 30. These are a variation on Newton’s Identities relating power sum symmetric functions and elementary symmetric functions. Here, the homogeneous symmetric functions, , play a role as well.
7.2. Combinatorial Proofs
We now show how these formulas may be derived combinatorially by our model with the specialization . The weighted path looks like
(65)
Notation. Consistent with the above, we use or to represent expressions in which we have replaced with . Thus , for ; see Definition 6 of weighted path .
7.2.1. First Fundamental Solution
Proposition 27 is readily seen from the weighting of path . For the first fundamental solution, there are no vertices with weight , and no edges weighted . The first vertex has weight , and so on. In an -path cover the exponent is the number of paths of length , for each , and the multinomial coefficient counts the number of -path covers obtained from any fixed set of -paths. So this model gives a visual interpretation to the analytic formula.
7.2.2. Higher Fundamental Solutions
Start with the following.
Lemma 31. For a fixed with and any ,
Proof. For , consider the weighted path . The first vertex must lie in every -path cover of this path, say on a path of length for , starting at . If then , and the sum of all such -path covers is thus . If then finishes at vertex where , so . And if then finishes at vertex where and so , and the sum of all such -path covers is . Hence , and so the result is obtained.
Now for a combinatorial proof of Proposition 28.
Theorem 32. For the fundamental solutions to the recurrence for the homogeneous path polynomials, one has
Proof. By our definitions and Corollary 11(ii) we have . And, from Lemmas 9 and 31, we have
Now
where, at the second line, we note that in every -path cover of the weighted path vertex must lie on a path of length and weight where , and at the last line we use (68). This gives the result.
7.2.3. Trace Formula
We now give a combinatorial derivation of the trace formula in Proposition 29.
First let be the sum of the weights of all circuits of length in and the -trellis with edge-weights replaced by ; that is, ; see Section 6.
Theorem 33. For any ,
Proof. We recall that the indeterminates in any term of are initially ordered according to the edges traversed in the corresponding circuit; see Example 26. Let be a typical ordered term in with all 1’s removed and with first indeterminate . We first show that term occurs times in .
When there are two successive indeterminates and in , then, in the corresponding circuit, the edges traversed are first of weight , followed by the edges each of weight , and then finishing with the edge of weight . Hence pair becomes when the indeterminates are considered as weights on edges in a circuit in .
Now, because the first indeterminate in is , any circuit corresponding to must be based at vertex for some . Hence will appear in as
for each in . There are such , so there are occurrences of term in .
Now consider an occurrence of in which , namely,
Let
Then the sequence of edges traversed in corresponding to begins at and ends at , and so is a circuit based at , with length . Thus . Conversely given any then is an occurrence of term starting with and ending with . Thus and .
Now we can partition the weighted circuits of of length by their first indeterminate (ignoring the edges of weight preceding this first indeterminate). That is, we can partition the terms of by their first indeterminate . So, using the above arguments, we have
Furthermore, ; the first equality is Theorem 21, the second is Corollary 11(ii), and the third is by definition of . So finally,
Example 34. See Examples 17 and 26. Here and : where, at line 2, we have rearranged the terms according to their first indeterminate , using Example 26, and combined like terms.
Remark 35. From Theorem 25 and our definitions of matrices and from Sections 3 and 5.1, respectively, we have the following equalities:
Thus, from Theorem 33,
8. Sequences,
In Section 7 we specialized by replacing weights with . In this section we specialize further by replacing all weights with . We denote this operation by . We then use these matrices to count -path covers of the path and cycle.
Recall matrix from (26), we define matrix : Similarly, let be the expression evaluated when all . So and . Thus counts the number of -path covers of the weighted or of an arbitrary -cycle. (cf., Corollary 11.1, Section 8, Farrell [2].)
Theorem 36. For , one has .
Proof. Let and let denote the cycle whose vertices are the elements of arranged clockwise in a circle. Now so any path cover of will be an -path cover. We show that the number of path covers of is .
Given a subset of with we define a path cover , of . Conversely, given a path cover of we take the first vertex from each path to form a subset of and then rearrange its elements to form a subset of with increasing elements. These two operations illustrate a bijection from the set of non-empty subsets of to the set of -path covers of . Hence .
From recurrence (2), Lemma 13, and Theorems 16 and 36, for we see that obeys the -anacci recurrence, with starting conditions for .
In the square array (see Table 1) is the entry, for . Column is determined by the above -anacci recurrence. We observe that the main diagonal entry is .
Consider the triangle, in bold, where is the entry for all and ; it counts the number of -path covers of a cycle with vertices. We have entered the sequence obtained from reading this triangle row-by-row to the Online Encyclopedia of Integer Sequences [9]; it is sequence A185722.
Each of the 10 columns of the square array (see Table 1) appears as a sequence in [9]; for example, the second column gives sequence A000204 and the third column gives A001644. Thus we have a new combinatorial interpretation for each of these sequences and a connection between them.
A closely related sequence is A126198 (replace “” by “” in its description). Let be the entry of the triangle corresponding to A126198, then counts the number of compositions of integer into parts of size ≤. Now consider vertices arranged in a path. A composition of into parts of size ≤ corresponds naturally to an -path cover of this path with vertices by identifying a part of size in the composition with a path of length in the corresponding -path cover. This correspondence can also be reversed. Thus in our terminology, is the number of -path covers of a path with vertices, and, from Corollary 11(ii) and our operation , we have . The main diagonal entry in this triangle is (as is well known, there are compositions of ) and column of this triangle is determined by the -anacci recurrence, for , with starting conditions for .
The entry in our triangle, , counts the number of -path covers of a cycle with vertices. We have starting conditions as opposed to above, for .
Furthermore, from above and the definition of matrix from (28), we have the entry of matrix . Thus both can be obtained from matrix . This gives a new derivation of , and so of sequence A126198.
Example 37. and . We have gives see Examples 17, 26, and 7.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of the paper.
Acknowledgment
The authors thank the referee for bringing to their attention various relevant papers.