1. Introduction
A multiset is an unordered collection of elements which may be repeated. A multiset of height is a multiset whose elements are positive integers and whose largest element is . A finite multiset may be represented by , where is the multiplicity of in , i.e., the number of copies of in . We call a multiset of height reduced if each member of appears at least once in , or equivalently, each multiplicity is at least 1. We will assume, unless otherwise noted, that the multisets we work with are reduced. The size of a multiset , denoted , is the sum of the multiplicities of its elements.
A multipermutation of a multiset is an arrangement of the elements of into a sequence. We identify such a multipermutation with a word that has exactly occurrences of each symbol . The height of , denoted , is the height of the underlying multiset, i.e., the maximum of .
For two multipermutations and , we say that contains , if has a subsequence whose elements have the same relative order as , i.e., if and only if and if and only if for every . If does not contain , it avoids . In this context, is usually referred to as a pattern.
For a word of height , its reversal is the word , and its complement is the word . Note that the reversal represents the same multiset as , while the complement may represent a different one.
For a multiset and a multipermutation , we let denote the set of all the multipermutations of that avoid , and we let be the cardinality of . Two multipermutations and are equivalent, denoted by , if for every multiset , equals .
The notion of equivalence can be straightforwardly extended to sets of patterns. For instance, suppose that is a set of multipermutations. We let be the set of multipermutations of that avoid all the patterns contained in , and we let be its cardinality. We again call two sets and of multipermutations equivalent, denoted , if for every multiset . To avoid clutter, we often omit nested braces and write, e.g., instead of .
We may easily observe that each multipermutation is equivalent to its reversal , and that for every pair and of equivalent patterns, we also have and . Moreover, these symmetry relations can be generalized in an obvious manner to equivalences involving sets of patterns.
The notion of equivalence has been previously studied by Jelínek and Mansour [6], who called it ‘strong equivalence’. They focused on the classification of this equivalence for single patterns of fixed size, and they characterized the equivalence classes of patterns of size at most six. From their results, we will use here the following fact [6, Lemma 2.4].
Fact 1.1 (Jelínek and Mansour [6]).
For any , all the patterns that consist of a single symbol ‘1’, a single symbol ‘3’ and symbols ‘2’ are equivalent.
The main purpose of this paper is to classify the
equivalence for sets of patterns containing a pattern of size three and a pattern of size at most four. This extends earlier results concerning avoidance by multisets of a single permutation [8] or word [4] pattern of length three. With the help of computer enumeration, we identified the plausible equivalences and have managed to verify all of these equivalence. This also yields all of the nonsingleton (3,3) and (3,4) Wilfequivalence classes for compositions, up to at most two sporadic cases. Many of our results are based on arguments that generalize to larger patterns. However, to keep the presentation simple, we mostly state our theorems and proofs for the special case of patterns of size up to four, which is our main focus. We point out the possible generalizations separately as remarks.2. Avoidance results for multisets
We may represent words of height and length as binary matrices with rows and columns and exactly one cell in each column. We assume that the rows of a matrix are numbered bottomtotop, and the columns are numbered lefttoright. For a multipermutation of height , let be the matrix with a 1cell in row and column if and only if the th letter of is equal to . For example,
Conversely, if is a matrix with exactly one 1cell in each column and at least one 1cell in each row, then there is a unique reduced multipermutation such that . If there is no risk of confusion, we will identify a multipermutation with its corresponding matrix , and we will say, for instance, that two matrices and are equivalent, if they represent two equivalent multipermutations.
The Ferrers diagram (or Ferrers shape) is an array of boxes (called cells) arranged into downjustified columns, which have nonincreasing length. A filling of a Ferrers diagram is an assignment of zeros and ones into its cells. A filling is columnsparse if every column has at most one 1cell. A filling is sparse if every row and every column has at most one 1cell. A transversal filling, or a transversal, is a filling in which every row and every column has exactly one 1cell. In this paper, we only deal with columnsparse fillings and their various restrictions. We treat binary matrices, i.e. matrices containing only values 0 and 1, as fillings of rectangular Ferrers diagrams.
The deletion of the th column in a Ferrers diagram is the operation that removes from all the boxes in the th column, and then shifts the boxes in columns one unit to the left in order to fill the created gap. Note that the deletion transforms into a smaller Ferrers diagram. Deletion of a row is defined analogously.
A filling of a Ferrers diagram contains a binary matrix if can be transformed into via a sequence of deletions of rows and columns, possibly followed by changing some 1cells of into 0cells. If does not contain , we say that avoids . Notice that a multipermutation contains a multipermutation if and only if the matrix , understood as a Ferrers diagram, contains in the sense defined above.
We say that two binary matrices and are strongly Ferrersequivalent, denoted , if for every Ferrers shape , there is a bijection between avoiding and avoiding columnsparse fillings of that preserves the number of 1cells in each row and column. We also say that and are Ferrersequivalent for transversals, or FTequivalent for short, if for every Ferrers diagram , the number of its avoiding transversals is equal to the number of its avoiding transversals. This relation is denoted by . For a pair of multipermutations and , we will often write or for and , respectively. As with equivalence, we will also extend strong Ferrersequivalence and FTequivalence from individual patterns to sets of patterns.
Clearly, if two patterns (or sets of patterns) and are strongly Ferrersequivalent, then they are also FTequivalent. Moreover, as the next simple lemma shows, FTequivalence can always be extended from transversals to general sparse fillings, provided the two patterns have no zero rows or columns.
Lemma 2.1.
Suppose that and are FTequivalent matrices, and that every row and column of and of contains at least one 1cell. Then for every Ferrers diagram , there is a bijection between avoiding and avoiding sparse fillings of , which has the additional property of preserving the number of 1cells in each row and column of .
Proof.
Let be a Ferrers shape, and let be an avoiding sparse filling of . We delete all the rows and columns of that have no 1cell in . This transforms the filling of the diagram into an avoiding transversal of a Ferrers diagram . Since and are FTequivalent, there is a bijection that maps avoiding transversals of into avoiding transversals of . We define . We then reinsert the rows and columns we deleted in the first step into , filling the newly inserted boxes by zeros. This transforms into a sparse filling of the original diagram . Since has a 1cell in every row and column, the insertion of an allzero row or column into cannot create an occurrence of . Thus, is an avoiding sparse filling of , and we may easily observe that the transformation is the required bijection. ∎
Recall that a multipermutation contains a multipermutation if and only if the matrix contains . Thus, for a multiset of size and a pattern , there is a bijective correspondence between the set of avoiding multipermutations of and the set of all the avoiding matrices of shape having exactly one 1cell in each column and exactly 1cells in row , for each . It follows that for two multipermutations and , implies .
If is a word and an integer, we denote by the word obtained by increasing each letter of by . Recall that the height of a word is the maximum value appearing in . For two words and , we let their direct sum be the concatenation of and . For instance, for and , we have . For a set of patterns and a pattern , we write as a shorthand for the set .
An important feature of the various flavors of Ferrersequivalence is that they are closed with respect to direct sums. This follows from a standard argument appearing, among others, in the works of Backelin, West and Xin [1, Proposition 2.3] or of Stankova and West [10, Proposition 1] in the context of permutations, and later in the works of Jelínek and Mansour [6, Lemma 2.1] and [5, Lemma 14] in the more general setting of words. We omit repeating the argument here, and merely state the required result as a fact.
Fact 2.2 ([1, 5, 6, 10]).
Let and be two sets of multipermutations, and let be another multipermutation. If and are strongly Ferrersequivalent, then and are also strongly Ferrersequivalent. Likewise, if and are FTequivalent, then and are also FTequivalent.
2.1. Results based on Ferrersequivalence arguments.
We now state several known results on various forms of Ferrersequivalence which will be useful for our purposes. The first such result is the strong Ferrersequivalence, for any , of the increasing pattern and the decreasing pattern . This equivalence has been established by Backelin et al. [1] for transversal fillings, and Krattenthaler [7] then obtained more general results which imply the strong Ferrersequivalence of the two patterns.
Fact 2.3 (Krattenthaler [7]).
For any , the pattern is strongly Ferrersequivalent to .
Another family of strongly Ferrersequivalent patterns has been found by Jelínek and Mansour [5, Lemma 39].
Fact 2.4 (Jelínek and Mansour [5]).
For any , the pattern is strongly Ferrersequivalent to .
The next result, due to Stankova and West [10], is specific to FTequivalence.
Fact 2.5 (Stankova and West [10]).
The patterns and are FTequivalent.
In the statement of Fact 2.5, FTequivalence cannot be directly replaced with strong Ferrersequivalence, as was pointed out by Guo et al. [3]. However, Guo et al. [3] have found a different way of generalizing Fact 2.5 to a strong Ferrersequivalence result, which we now state.
Fact 2.6 (Guo et al. [3]).
We have the following strong Ferrersequivalences for sets of patterns:

and

.
There is another, simpler way to translate an arbitrary FTequivalence result into a strong Ferrersequivalence, which involves the pattern . Clearly, a multipermutation avoids if and only if each of its elements has multiplicity 1, i.e., is actually a permutation. Similarly, a columnsparse filling of a Ferrers diagram avoids if and only if each row has at most one 1cell, that is, the filling is sparse.
A pair of patterns , is said to be Wilfequivalent, denoted , if for every , the number of permutations of that avoid is the same as the number of those that avoid . Intuitively speaking, strong Ferrersequivalence refines equivalence in the same way as FTequivalence refines Wilfequivalence. This intuition is made more rigorous by the next easy observation, which can be easily deduced from the definitions and from Lemma 2.1. We omit its proof.
Observation 2.7.
For any two multipermutations and , if then , and if then .
A similar observation states that equivalence, as well as strong Ferrersequivalence, is preserved when we add a pattern for any .
Observation 2.8.
For any two patterns and and for any , implies , and implies .
By combining the previous facts and observations, we obtain the following equivalences among pairs involving a pattern of size 3 and a pattern of size 4.
Proposition 2.9.
The following equivalences hold:

[(a)]

,

,

,

,

,

,

,

.
Proof.
Parts (a) to (e) all use Obs. 2.8 to conclude the equivalence in conjunction with Fact 2.4 for (a), Fact 1.1 for (b) and Facts 2.2 and 2.3 for (c)(e). Part (f) and the equivalence make use of Fact 2.3 and Obs. 2.8 first and then Fact 2.2. The equivalence follows from the strong Ferrersequivalence of and , together with Obs. 2.8 and Fact 2.2. Finally, part (h) follows from combining Fact 2.5, Obs. 2.7 and Fact 2.2 in that order. ∎
2.2. Sorting minimal/maximal letter technique.
In this subsection, we prove some equivalences by defining bijections which reorder the relevant patternavoiding multiset permutations, expressed as words.
Theorem 2.10.
The following pair of patterns are equivalent:

,

.
Proof.
(1) Fix a multiset , and let and denote the sets and , respectively. Given , let denote the set of letters within which occur twice. Let be obtained from by replacing each occurrence with for , leaving all other letters unchanged in their positions. Note that these other letters must occur once and therefore cannot affect the avoidance of any of the patterns we consider. It is then seen that the mapping is a bijection between and , as desired.
(2) Equivalently, we show . Let us write and . With as above, we will describe a bijection between and . We proceed by induction on the number . Clearly, if , the required bijection is the identity mapping, since there is only one multipermutation of , and it avoids all patterns with two or more symbols.
Define now the multiset obtained by removing all copies of from . By induction, there is a bijection between and . Let be a multipermutation from . We want to insert copies of into to create a avoiding multipermutation of . If , then we may insert the symbol in any position of while preserving avoidance. If , then we observe that one of the copies of must be the leftmost symbol of in order to preserve avoidance, while the other can be placed arbitrarily. If , then all the symbols in must appear consecutively at the leftmost positions.
Similarly, when extending a multipermutation to a multipermutation , we proceed as follows: if , the symbol can be placed arbitrarily, if , the only restriction is that the two symbols must appear consecutively, and if , then the symbols must form the leftmost symbols of .
The above description shows that in both the avoiding and the avoiding multipermutations of , the position of all the symbols is uniquely determined by the position of the rightmost copy of . This yields a straightforward bijection between and , defined as follows: fix a , remove from all the occurrences of to obtain a , define , and finally, let be the unique member of in which the rightmost occurrence of appears at the same position as the rightmost occurrence of in . We easily see that this provides the required bijection. ∎
Remark: Extending the bijections described above shows more generally
where denotes any permutation of the multiset and is the complement of .
2.3. Equivalences by analysis of active sites.
In this subsection, we establish several equivalences by considering active sites within multiset permutations and the associated generating trees for the patterns in question. An active site of a parent multipermutation is in general a position in which we may insert one or more copies of a letter in producing its offspring without introducing a given set of patterns. In some instances, it will be convenient to modify this definition somewhat to accommodate the patterns in question. Throughout, we consider permutations of multisets of , though at times it will be more convenient notationally to insert either successively smaller or larger letters into a parent permutation in producing its offspring. For examples of the generating tree method applied to the avoidance problem on ordinary permutations, see, e.g., [11, 12].
We first establish the equivalence of and via an active site analysis where we successively insert smaller and smaller letters into a parent permutation.
Theorem 2.11.
The sets of patterns and are equivalent, that is
Proof.
Let , where and are fixed. We first enumerate members . To do so, we consider the various partial permutations , where for . We form the permutations by inserting copies of appropriately into the . By an active site, within a permutation of the stated form where , we mean a position where one may insert a single copy of the letter without introducing an occurrence of 2212 (where one may assume in the case ). It is understood that if , then all other letters are to be added at the very end of in order to avoid 112.
Let if and if for . We now show by induction on that each has exactly active sites. The case is apparent since there are (active) sites in the composition corresponding to the very first and very last positions of for all exponents and also to the position directly after the first if . Now assume that the hypothesis is true in the case for some and we show it holds in the case. If , then a single can be inserted into any one of the sites of some without introducing either pattern, and it is seen that regardless of where is inserted, the number of sites increases by one (essentially, one of the present sites is split into two). Also, replacing with in raises its value by one since , which accounts for the additional site. If , then there are two new sites introduced by the insertion of the letters , i.e., one directly following the leftmost added and another at the very end following the last . Since we have in this case, the induction is complete.
As all have the same number of active sites for each (with this number depending only on ), we have that the number of possible is given by the product of with the number of for each . Thus, the number of possible of the stated form is given by . A similar argument whose main details we describe briefly shows that there are the same number of . Let for and consider forming the from the various by inserting copies of appropriately. Define active site analogously except that now we insert all letters at the site as a single run (so as to avoid 121). Reasoning by induction as before, one can show for each that there are sites in all , which implies the same product formula as above for the number of possible . ∎
Theorem 2.12.
We have .
Proof.
We show that , where is as in the preceding proof. Let denote an arbitrary member of , where is as before. By an offspring of , we mean some that can be obtained from by inserting copies of appropriately. Define an active site of to be a position in which a (single) may be inserted without introducing 2122. Note that an offspring of is produced when a single is added at an active site and all other are added at the end.
Suppose and that has exactly (active) sites. We consider the nature of the offspring of based on cases for the exponent . If or , then one may verify that each of the offspring of has or sites, respectively. If , first note in this case that every site of to the right of the leftmost is lost, as all offspring in this case must end in at least two letters . Allowing the leftmost to occur in each of the possible positions, it is seen that there is exactly one offspring of that has sites for each .
Now consider forming for from . Define offspring and (active) site analogously as before. Suppose that has sites and we describe its offspring. If or , then it is seen again that each offspring of has or sites, respectively. If , then inserting the run into a site effectively nullifies all sites of occurring to the left of the run. Thus, the number of sites in the offspring of ranges from to in this case. Comparing the offspring of the various and , one can show by induction on (upon considering cases based on the exponent ) that the number of members of and having exactly sites is the same as the corresponding number of members of for all . Allowing to vary over all possible values then implies the desired result. ∎
Theorem 2.13.
We have .
Proof.
Let , , for and offspring be defined as in the proof of Theorem 2.11, but now in conjunction with the pattern sets and , respectively. By an active site in , we mean a position in which one can insert a single letter such that no occurrence of arises when another copy of is appended to the end of the resulting multipermutation. Let denote the number of (active) sites of a multipermutation and we will make use of this notation in subsequent proofs. Note that inserting a single (i.e., when ) into any position of introduces neither 112 nor 2121 and changes the act statistic value (always increasing it by one) if and only if the is inserted into a present site of . On the other hand, if , then inserting into a site of not the last (and placing copies of at the end) nullifies all sites of to the right of , with effectively preserved; moreover, the site at the very end of is in essence replaced by a site at the very end of .
Suppose now where . By the previous observations, if , then has offspring with sites and with sites, where . If , then it is seen that the set of act values in the offspring of comprise the interval .
Now define an active site in to be a position of in which one can insert a run of of length two or more without introducing , with the corresponding statistic again denoted by act. Upon considering cases based on whether or , one can show that the set of act values of the offspring of where is the same as those of the offspring of above. By induction on (the case trivial), the corresponding act statistics on and are identically distributed for all . Taking in particular implies the desired equivalence of patterns. ∎
Theorem 2.14.
We have .
Proof.
We again make use of the same notation. For the first pattern set, let us define an active site to be a position of in which a single may be inserted without introducing , where any remaining must be added at the end of . Suppose , where . If , then each offspring of is seen to have sites. On the other hand, if , then inserting the leftmost into any site other than the last destroys all sites of to the right of . Note that itself is split into two sites, with a new site reemerging at the end of corresponding to the final added . Thus as varies, one gets offspring whose act values comprise the interval . If all of the letters are added at the very end of , then every site of is preserved with each position directly following a active as well in this case, which implies that the offspring will have sites altogether.
Now suppose with or with , where active sites are defined analogously. One can show by comparable reasoning as before that if , then the offspring of both and all have act values of , whereas if , then the values comprise the set . By induction on (the case trivial), it is seen that the various act statistics defined on the sets consisting of the possible , or are identically distributed for , which in particular implies the desired equivalences.
It is also possible to establish the second equivalence via a bijection. It is instructive to describe such a bijection since it will be seen to preserve further statistics within the framework of multiset equivalence. It suffices to define a bijection between the set of permutations of that avoid and those that avoid , where and are fixed. Suppose , expressed as a word, is a permutation belonging to the former set, where . We first decompose as , where contains no . Define by , where denotes the reversal of . If , then set . Otherwise, let denote the largest letter occurring in and suppose , where contains no .
We now introduce the following definition. Suppose is a ary word and . Then we will refer to a (maximal) string of consecutive letters in all of which belong to as an upper run and a string all of whose letters belong to as an lower run. We consider a lefttoright scan of the upper and lower runs of . Let denote the distinct lower runs in , where is possible. Then can be decomposed as if , with if , where are upper runs with (only) and possibly empty. Similarly, let if , with if , where are lower runs, are upper runs and are possibly empty.
We now define a multiset derived from as follows. If , then let be defined as
where it is seen that this definition may be extended to the case when , with if for all . If , then let be given by
which can be extended to the case when . That is, if is obtained from by replacing the runs with in that order, then is obtained from by repositioning the lower runs to the right of the last so that they now occur prior to those to the left of this , maintaining the order of the upper runs (as well as the order of the letters within all runs). Note that the upper runs of are the same as those in , with the relative order of lower and upper runs in a lefttoright scan also seen to be the same.
If is empty, then set . Note that by reordering the lower runs as described, we have eliminated any possible occurrences of in which the can correspond to the terminal . If is nonempty, then let be the largest letter of and write , where contains no . Then consider any lower runs within the section of . We arrange the lower runs of such that the reversals of those in occur (in reverse order) prior to the others, as we did with the ’s above in . If denotes the resulting word, then set if .
Otherwise, we continue in the manner described until for some , setting equal to the word that results after applying the procedure described above for a final time. By construction, it is seen that avoids 121, as does each word arising from an intermediate step of the algorithm. One may verify also that avoids 2213; note that it suffices to check that contains no 2213 in which the 3 corresponds to a (strict) rightleft maximum. To reverse , consider successively the rightleft maxima, starting with the last letter and working back to the rightmost , where we reverse each step of the algorithm described above starting with the last. Note that this may be done since the values of rightleft maxima are preserved by each step of the algorithm and hence by . ∎
Remark: From the preceding proof, we have in particular that the multiset equivalence of and respects the last letter and rightleft maxima statistics.
Theorem 2.15.
We have .
Proof.
Proceeding as in the prior proof and using the same notation, consider permutations of that avoid either , or . If and , then all offspring of have act value for each pattern set. If , then the act values of the offspring of are seen to comprise the interval in each case. Note that when avoiding , inserting copies of into a site of destroys all sites to the left of while splitting into sites. If avoiding , inserting the leftmost into a site of , not the last, is seen to destroy all sites to its left when . Similar reasoning applies to except that (nonterminal) sites to the right are destroyed. Since the pattern sets obey the same rules with regard to the number of sites in offspring, the result follows. ∎
Remark: Extending the previous proof shows more generally
One can provide a proof of the following result analogous to the previous ones by modifying appropriately the definition of a site. However, we find it more instructive to give a bijective argument which makes use of a certain encoding of the offspring and suggests how one might go about finding bijective proofs of other comparable results.
Theorem 2.16.
We have .
Proof.
Let be a multipermutation of the multiset . Given , let and be obtained from by considering only the relative positions of the parts . We construct an encoding for as follows. Let as a word where and denote the index such that with maximal, if it exists (i.e., corresponds to the rightmost ascent bottom of ), with otherwise (i.e., if is decreasing, perhaps weakly). We seek to form from for by making an appropriate insertion of the copies of .
Note that in forming from when that there are exactly positions to the left of the rightmost ascent bottom of in which to insert . Observe further that if a is inserted anywhere to the left of in , then necessarily , for otherwise would contain an occurrence of as the remaining copies of are forced to occur at the very end (in particular, to the right of the last ascent).
Let denote the smallest index such that , assuming such exists. Then implies is decreasing (i.e., where ). Then and avoiding implies contains only one ascent, with determining the position of that ascent. Note that is of the form , where corresponds to the position of the leftmost letter (i.e., corresponds to the ()th entry of where ).
Now assume . If it is the case that both and , then we must also specify, in addition to the value of , some element of the set , which gives the entry number of the only in . Note that when , we have if and only if the is inserted to the left of or within the rightmost ascent of , as the position of the rightmost ascent bottom is shifted to the right by one place in this case. Thus, the position number of must belong to . Let
denote the vector
consisting of the various values, where in addition an element of is specified parenthetically in the th component for each such that and . One can verify that the sequence satisfies the following succession rules for :
Comments
There are no comments yet.