Pattern Avoidance in Task-Precedence Posets
aa r X i v : . [ m a t h . C O ] J un Discrete Mathematics and Theoretical Computer Science
DMTCS vol. :2, 2016, Pattern Avoidance in Task-Precedence Posets
Mitchell Paukner ∗ Lucy Pepin † Manda Riehl ‡ Jarred Wieser § Department of MathematicsUniversity of Wisconsin – Eau ClaireEau Claire, WI, USA received 3 rd Nov. 2015 , accepted 15 th Jan. 2016 . We have extended classical pattern avoidance to a new structure: multiple task-precedence posets whose Hasse diagramshave three levels, which we will call diamonds. The vertices of each diamond are assigned labels which are compatiblewith the poset. A corresponding permutation is formed by reading these labels by increasing levels, and then from leftto right. We used Sage to form enumerative conjectures for the associated permutations avoiding collections of patternsof length three, which we then proved. We have discovered a bijection between diamonds avoiding 132 and certaingeneralized Dyck paths. We have also found the generating function for descents, and therefore the number of avoiders,in these permutations for the majority of collections of patterns of length three. An interesting application of this work(and the motivating example) can be found when task-precedence posets represent warehouse package fulfillment byrobots, in which case avoidance of both 231 and 321 ensures we never stack two heavier packages on top of a lighterpackage.
Keywords: permutation pattern, poset
In this paper, we continue a rich tradition of extending the notion of classical pattern avoidance in permuta-tions to other structures. Given permutations π = π π · · · π n and ρ = ρ ρ · · · ρ m we say that π contains ρ as a pattern if there exist ≤ i < i < · · · < i m ≤ n such that π i a < π i b if and only if ρ a < ρ b . In thiscase we say that π i π i · · · π i m is order-isomorphic to ρ and that π i π i · · · π i m reduces to ρ . If π does notcontain ρ , then π is said to avoid ρ . The classical definition of pattern avoidance in permutations has shownitself to be worthwhile in many fields including algebraic geometry [17] and theoretical computer science[9]. Analogues of pattern avoidance have been developed for a variety of combinatorial objects includingDyck paths [1], tableaux [11], set partitions [15], trees [14], posets [8], and many more. We use a definitionof pattern avoidance that is similar to that used in the study of heaps [10], but distinct from that used inprevious studies of trees. Unlike the question studied by Hopkins and Weiler [8] which identified classes of ∗ Student Blugold Commitment Differential Tuition funds through the University of Wisconsin-Eau Claire Summer Research Expe-riences for Undergraduates † Student Blugold Commitment Differential Tuition funds through the University of Wisconsin-Eau Claire Summer Research Expe-riences for Undergraduates ‡ University of Wisconsin - Eau Claire Office of Research and Sponsored Programs § University of Wisconsin - Eau Claire Ronald E. McNair Postbaccalaureate Achievement ProgramISSN 1365–8050 c (cid:13)
Mitchell Paukner, Lucy Pepin, Manda Riehl, Jarred Wieser
15 7 26 34 10 89
Fig. 1.1:
An element of D , (321) . posets for which certain properties are preserved, we extend the enumerative question of pattern avoidanceto a particular class of posets.A task-precedence poset is a poset which represents the order relations between several tasks to be com-pleted. We are particularly interested in considering d identical task-precedence posets, and here we focusour attention on those sets of tasks that require one task be completed before any others, and one final taskafter any others, with no restrictions on the rest of the tasks in the list. When considering a list of tasks,the Hasse diagram of this poset is a diamond, and as such we will refer to a task-precedence poset of thistype with v tasks as a diamond with v vertices (each with v − vertices in the middle level). We thenassign unique labels from { , , . . . , vd } to each vertex such that the labels obey the order relations of eachdiamond. We then refer to the set of all such labelled collections of diamonds as D v,d . Given an element D of D v,d we associate a permutation π D by recording the vertex labels as they areencountered reading the labels on each diamond consecutively, left to right by levels, beginning with theleast element. For example, if D is as pictured in Figure 1.1, then π D = 156273498(10) . We say that D contains (respectively avoids) ρ as a pattern if π D contains (respectively avoids) ρ as a classical pattern, usingthe definition above. We will abuse notation and sometimes refer to an element of D v,d and it’s associatedpermutation interchangeably. Let D v,d ( P ) be the elements of D v,d that avoid all patterns in list P. WhileFigure 1.1 contains , , , , , it is a member of D , (321) . Two patterns on diamonds, α and β , are said to be d -Wilf-equivalent if they have the same enumeration, that is, if |D v,d ( α ) | = |D v,d ( β ) | forall v and d . If so, we write α ∼ W β .Our motivation comes from a real-life application, namely a fleet of robots all completing the same se-quence of tasks in a warehouse for package fulfillment. In 2011, instead of having human employees walkthe warehouse floor retrieving items one after another to complete an order, Amazon began utilizing Kivarobots in their package fulfillment warehouses [12]. Each robot executes pieces of the larger task. Weassign robots to diamonds ordered by the weight of the object they will deliver, heaviest object first, so thatthe tasks to retrieve the first, heaviest object are represented in diamond , and the lightest object by the finaldiamond. First the robot drives to the appropriate inventory rack and mounts the rack on its back. Then itcan either drive through the warehouse highways to its picker (the human employee who will retrieve theitem off the rack without leaving their station), or it can rotate itself so that the appropriate side of the rackis facing the picker. Both of these need to be completed before the final step: having the item picked off therack by the human employee in order to place it in its shipping box. In this way, completing one order of d items from Amazon.com is exactly the task-precedence poset represented by d diamonds with verticeseach.We now give an example of this process, referring throughout to Figure 1.2. A customer has made an orderfor objects, o , o , o , with weights w ( o ) > w ( o ) > w ( o ) . Thus the leftmost diamond will representthe tasks completed by the robot retrieving object , the center diamond for retrieving object , and the attern Avoidance in Task-Precedence Posets . The labels represent the order in which each task of the totaltasks is executed. Each robot operates autonomously and independently, and each faces its own challenges.For example one of the objects may be at the back of the warehouse, there may be significant traffic alongsome of the paths the robots travel through the warehouse, or the robot assigned to retrieve an object maystill be executing its previous assignment. Thus the labels on the least elements of each diamond can varysignificantly, and there can be a large difference in the labelling of the least element of a particular diamondand its greatest element. In Figure 1.2, the first task completed is that the robot for object arrives and picksup the rack containing object . Next, the robot retrieving object arrives at the rack containing object .Next, the robot carrying object rotates its rack on its back to have the correct orientation to the picker.This continues, and based on the labelling of the elements, we see that object (the lightest) is placed in itsshipping box first (in step ), then object (in step ), and then object (in step ). So our human pickerhas placed two heavier objects on a lighter object (unless they rearrange the objects after packing). Then aa sufficient (though not necessary) condition to ensure that two heavier objects do not arrive after a lighterobject is that the associated permutation avoid and .210 11 5 46 12 8 17 9 3 Fig. 1.2:
An example of a robot task-precedence poset whose associated permutation does not avoid and . One could consider other applications that arise from task-precedence problems, but our motivating ex-ample can be generalized most appropriately by changing tasks per autonomous robot to v tasks.The generating f unction f or descents ( gf d ) for D v,d ( P ) is f Pv,d ( x, y ) = P D ∈D v,d ( P ) x des y d , and f Pv ( x, y ) = P ∞ d =1 f Pv,d ( x, y ) . For example, D , (213) is the set of diamonds with associated permuta-tions , , , , and . So, f , ( x, y ) = y (1 + 4 x ) .Throughout this paper, the main question we answer is “How many elements are in D v,d ( P ) ?” for anycollection P of patterns of length . In general we fix v ≥ and a set of patterns P and then determine a for-mula for the sequence {|D v,d ( P ) |} d ≥ , with key results for v = 4 shown in Table 1. The third column of thetable gives entries from the Online Encyclopedia of Integer Sequences [13]. Our results for pattern-avoidingdiamonds have connections with many other combinatorial objects, as evidenced by the low reference num-bers. Sequences A260331, A260332 and A260579, however, are new results particular to this study of taskprecedence posets.Our task, which answers our primary question, is to find f Pv,d ( x, y ) . Then when we substitute x = 1 andtake the coefficient of y d , we obtain |D v,d ( P ) | .In Section 2 we consider collections of diamonds that avoid a single pattern of length 3. In Section 3we consider collections of diamonds that avoid a pair of patterns of length 3, and in Section 4 we considercollections of diamonds avoiding three or more patterns of length 3. Finally in Section 5, we list some openproblems relating to this work. Mitchell Paukner, Lucy Pepin, Manda Riehl, Jarred Wieser
Patterns P {|D ,d ( P ) |} d ≥ OEIS Result ∅ , , , , . . . A260331 Theorem 1
123 0 , , , , , , , , , . . . A000007 Theorem 2
132 1 , , , , , , , , . . . A002294 Theorem 3 , , , , , , . . . A260332 Theorem 4 , , , . . . A260579 OPEN ,
213 1 , , , , , , , , , . . . A000079 Theorem 5 ,
312 1 , , , , , , , , , . . . A000079 Theorem 6 , ,
321 1 , , , , , , , , , . . . A001844 Theorem 7 , ,
312 2 , , , , , , . . . A004171 Theorem 8 ,
321 2 , , , , , , , . . . A109808 Theorem 9 , , ,
321 1 , , , , , , , , , . . . A000027 Theorem 10 , ,
321 2 , , , , , , , , . . . A081294 Theorem 11
Tab. 1:
Enumeration of pattern-avoiding diamonds when v = 4 Before we count pattern-avoiding diamonds, it is useful to enumerate all diamonds.
Theorem 1. |D v,d ( ∅ ) | = ( vd )! v d ( v − d Proof:
Let v ≥ and d ≥ , first we choose v labels for each diamond, and then there are ( v − ways toarrange the internal vertex labels of any given diamond. We obtain (cid:18) vdv, . . . , v (cid:19) ( v − d = ( vd )!( v !) d (( v − d = ( vd )! v d ( v − d . Theorem 2. |D v,d (123) | = 0 . Proof:
It is impossible to avoid while having a diamond since the pattern is inherent in all valid diamondlabellings.
The complement of a permutation π of length n , denoted by π c , is obtained by replacing each letter j bythe letter n − j + 1 . The reverse of π = π π . . . π n , denoted by π r , is π n π n − . . . π . We let π rc be the reverse-complement of π and D v,d ( p ) rc be { π rcD | D ∈ D v,d ( p ) } . attern Avoidance in Task-Precedence Posets Proposition 1.
The reverse-complement of a task precedence poset remains a legal poset and D v,d ( p ) rc = D v,d ( p rc ) . In addition, for k ≥ , D v,d ( p , p , . . . , p k ) rc = D v,d ( p rc , p rc , . . . , p rck ) . Proof:
This is clear from the definitions and from how D v,d ( p ) rc is created from p .Thus we immediately see that a) ∼ W , b) ∼ W , c) , ∼ W , , d) , ∼ W , , and e) , ∼ W , .Given a permutation π in S n , lis ( π ) is the length of a longest increasing subsequence in π . For example, inthe permutation a longest increasing subsequence is and lis ( ) = 6 .Given a permutation π in S n , rlmax ( π ) is the number of right-left maxima in π . For example, in the permu-tation a maximum is reached when reading right-to-left twice and rlmax ( ) =2 . Let Dyck v,d be the set of all paths from (0 , to ( d, vd ) using only (0 , and (1 , steps (East andNorth steps) which stay weakly under y = vx . Given any p ∈ Dyck v,d , touchpoints ( p ) is the number oftimes p touches the line y = vx , excluding the point ( v, vd ) . In Figure 2.1, the Dyck path touches the line y = 4 x three times and touchpoints ( p ) = 3 . Given any p ∈ Dyck v,d , corners ( p ) is the number of Northsteps that are followed by one or more East steps in p . In Figure 2.1, there are three places where the Dyckpath has one or more North steps followed by one or more East steps and corners ( p ) = 3 . Given any p ∈ Dyck v,d , height ( p ) is the greatest vertical distance from any point on p to the line y = vx . In figure 2.1,the longest distance from a corner in the Dyck path to the line y = 4 x is seven (from (3 , to (3 , ) and height ( p ) = 7 . Lemma 1.
Any element of D v,d (132) has the elements on each diamond labelled in increasing order. Otherwise the label of the first element of the diamond together with the first descent would form a pattern.
Theorem 3. X σ ∈D v,d (132) w rlmax ( σ ) x des ( σ ) y d z lis ( σ ) = X p ∈ Dyck v,d w touchpoints ( p ) x corners ( p ) y d z height ( p ) . Proof:
We define a map φ from Dyck v,d to D v,d (132) . To find φ ( p ) , first write out the heights of the East steps.For each height, include a subscript j that indicates how many East steps are at that height. Reverse thissequence and add to every item in the list, leaving the subscripts unchanged. Each of the elements of thislist becomes the first label of a diamond, and then place vj labels in increasing order using the smallestelements that have not already been used as labels.As an example, refer to Figure 2.1. The heights of the East steps are , , , . When this sequenceis reversed and is added to each term, the resulting sequence is , , , . Thus the permutationassociated with this Dyck path is
13 14 15 16 6 7 8 9 5 10 11 12 1 2 3 4 .The importance of the subscripts j are evident from the image of Figure 2.3 under φ . The heights of theEast steps are , , , , and the resulting sequence is , , . Thus the permutation associated to imageis
11 12 13 14 4 5 6 7 8 9 10 15 1 2 3 16 . This map is certainly reversible, with the first label on each diamond forming a list, unless there is anincrease between diamonds, in which case the first label is repeated. Then the list is reversed and issubtracted from each element, giving us the heights of the East steps in the Dyck path.This bijection is particularly natural when you examine common statistics on both paths and permutations.Following touchpoints, corners, and height through the bijection, we find they correspond exactly to right-left maximum, descents, and longest increasing sequence on the permutation. Mitchell Paukner, Lucy Pepin, Manda Riehl, Jarred Wieser (0,0) (4,16)(1,4) (2,5)(3,12)
Fig. 2.1:
A Dyck path from (0,0) to (4,16)
Fig. 2.2:
Diamonds labelled according to the image of Figure 2.1 under the bijection (0,0) (4,16)(1,3) (3,3)(3,10)
Fig. 2.3:
A second Dyck path from (0 , to (4 , . attern Avoidance in Task-Precedence Posets Fig. 2.4:
An unlabelled member of D dv,j for d/geq , v/geq , and j/geq . Corollary 1. |D v,d (132) | = |D v,d (213) | = | Dyck v,d | = ( d ( v +1) d ) ( vd +1) . [5] Proof:
These equalities hold by the bijection in Theorem 3 and trivial Wilf equivalence from Proposition1.
Consider D in D v,d (231) , and suppose label vd occurs in position k . Then for all i < k and for all j >k , a i < a j . Consequently, if label vd is in position k , then labels (1 , . . . , k − appear in positions (1 , . . . , k − . We define D dv,j to be the collection of labelled diamonds for d − full diamonds with v vertices each followed by an incomplete diamond with j vertices for j = 1 , . . . , v − . Likewise D dv,j ( p ) are those diamonds that avoid pattern p . Note, when j = 1 there exist no order relations in the final partialdiamond. An example is shown in Figure 2.4. α dv,j ( x ) (or sometimes simply α dv,j for brevity) is the generating function for descents in D dv,j (231) . Inother words, α dv,j ( x ) = X D ∈D dv,j (231) x des ( π D ) . For example, D , (231) contains the diamonds with the following associated permutations: , , , , , , , , , . Counting descents in theseten permutations gives the generating function for descents α , ( x ) = 1 + 4 x + 4 x + x . Theorem 4. f (231) v,d ( x,
1) = α dv,v ( x ) = α dv, ( v − + x d − X i =1 α iv, ( v − α d − iv,v where α v,j = , if j = 1 C j − , if j = 2 , . . . , v − C v − , if j = v . and C i is the i th Catalan number.
Proof:
We proceed by partitioning elements of D dv,j (231) by where the largest label occurs. Let m = v ( d −
1) + j be the largest label in ( d − diamonds with v vertices followed by an incomplete diamondwith j vertices.Now, assume j = 1 . The m label can appear on the final element or on the greatest element of any of thefull diamonds. When m occurs on the final least element there are ( d − diamonds with v vertices thatprecede m, so we then have α d − v,v as the generating function for descents (gfd) for the vertices before m that Mitchell Paukner, Lucy Pepin, Manda Riehl, Jarred Wieser m | {z } | {z } α d − v,v α v, Fig. 2.5: α dv, when m appears on the greatest element of the last full diamond. m | {z } | {z } α dv, ( j − g ) α v,g Fig. 2.6: α dv,j when m appears on the final (partial) diamond. will avoid . When m appears on the greatest element of the i th complete diamond, /leqi/leqd − , wehave α iv,v − as the gfd for the vertices before m , and α d − iv, as the gfd for the vertices following m . Becausewe have created a descent from m to the least element of the next diamond or partial diamond, we must alsomultiply by x to account for this extra descent.Hence α dv, ( x ) = α d − v,v + x d − X i =1 α iv, ( v − α d − iv, . Now, assume we have ( d − diamonds followed by an incomplete diamond with j vertices where j =2 , . . . , v − . The m th element can appear on any of the interior vertices but not on the least element of theincomplete diamond, or m can appear on the greatest element of any complete diamond. When m appearson any of the interior vertices of the final diamond we need to count the descents before m, after m, andfrom m itself. The descents that occur before m can be counted by α dv,j − g where g is the number of interiorvertices following m including m. The descents following m are counted by α v,g because the same numberof descents can occur in the remaining interior vertices as when we have a single incomplete diamond. Wethen count the descent that results from m by multiplying our gfd by x, but we do not get a descent from m when it appears on the final interior vertex. We then sum over all possible values of g to give us the gfdwhen m appears on the interior vertices of the final diamond which gives us α dv,j − α v, + x j − X g =2 α dv,j − g α v,g . Also, m can appear on the greatest element of any of the full diamonds. When m appears on the greatestelement of the i th complete diamond the gfd for vertices that appear before m is α iv, ( v − and α d − iv,j for thevertices following m. We count the descent from m by multiplying our gfd by x. The total gfd when m appears on the greatest element of the i th diamond is then α iv, ( v − α d − iv,j x. attern Avoidance in Task-Precedence Posets α d , x + 4 x + x x + 54 x + 95 x +74 x + 25 x + 3 x α d , x + 7 x + 2 x x + 72 x + 149 x +138 x + 53 x + 7 x α d , x x +15 x +10 x +2 x x + 106 x + 281 x +362 x + 225 x + 65 x + 7 x α d , x + x x + 31 x + 36 x +15 x + 2 x x + 161 x + 544 x +938 x + 840 x + 383 x +84 x + 7 x α d , x + x x + 37 x + 47 x +21 x + 3 x x + 188 x + 677 x +1246 x + 1193 x + 579 x +135 x + 12 x Tab. 2:
The recursive steps necessary to find the generating function for descents in D , (231) . Thus α dv,j ( x ) = α dv,j − α v, + x j − X g =2 α dv,j − g α v,g + x d − X i =1 α iv, ( v − α d − iv,j for j = 2 , . . . , v − . Lastly, we look at when we have d complete diamonds. The m th element can appear on any of the greatestelements. When m appears on the greatest element of the last diamond, the gfd is α dv, ( v − which countsdescents before m. When m appears on the greatest element of the i th complete diamond ( ≤ i ≤ d − ), the gfd for verticesthat appear before m is α iv, ( v − and α d − iv,v for vertices following m. We count the descent from m to thefollowing least element by multiplying the gfd by x. Hence α dv,v ( x ) = f (231) v,d ( x,
1) = α dv, ( v − + x d − X i =1 α iv, ( v − α d − iv,v . We can use this result to recursively obtain f v,d ( x, for any v and d . Corollary 2. f v,d (1 , y ) (cid:12)(cid:12)(cid:12) y d = α dv,v (1) = |D v,d (231) | Tables 2 and 3 are an example of the steps of such a computation for α , ( x ) and D , (231) . Corollary 3. |D v,d (231) | = |D v,d (312) | . Proof:
By Proposition 1, is d -Wilf-equivalent to . We were unable to find a closed formula for the pattern . In Table 4, we present the first few termsof the sequence and the first few generating functions for descents, which we found using Sage. We are0
Mitchell Paukner, Lucy Pepin, Manda Riehl, Jarred Wieser v=5d 1 2 3 α d , α d , α d , α d , α d , Tab. 3:
The total number of permutations for D ,d that avoid the pattern for d = 1 , , . v=4 d |D v,d (321) | f v,d ( x ) x x + 29 x + 5 x x + 2747 x + 1765 x + 430 x + 42 x Tab. 4: |D v,d (321) | and f v,d ( x ) for d = 1 , , . confident that a technique recently used by Bevan, et.al. [2] would be successful in this case too. Theirtechnique involved refining a bivariate generating function via a statistic called last inversion foot, usinga result of Bousquet-M´elou, and finding a functional equation, to eventually give a growth rate for thesequence. This suspicion was confirmed by Bevan, and in fact the sequence begins: 2, 106, 5976, 387564,27247446, 2020632046, 155622020610, 12327937844924, 998103225615208, 82224228576059340 [3].However the authors were unable, in the time available for this project, to learn all the tools necessary toenact the technique and so the problem remains officially open. Next, we study pairs of patterns of length . While there are such pairs of patterns, we focus on the pairs of patterns σ, ρ where |D v,d ( σ, ρ ) | is non-trivial. , Lemma 2.
In order to avoid and , the labels on each diamond must be increasing and consecutive.
Proof:
By Lemma 1, the labels appear in increasing order on each diamond. Then any label “missing”from consecutive labelling would either create if it occurred before its surrounding labels, or a if itoccurred after. Therefore the labels on each diamond must be consecutive and increasing.
Theorem 5. f , v,d ( x, y ) = ∞ X d =1 X σ ∈D (132 , y d x des ( σ ) = 1 − yx − y (1 + x ) . Proof:
By Lemma 2, we know that the labels on each diamond are consecutive and increasing, so there is adiamond labelled , , . . . , v , another labelled v + 1 , . . . , v , etc. So the only thing we must ensure is thatthe entire collection of diamonds avoids and between the respective diamonds. In their foundational attern Avoidance in Task-Precedence Posets and [16], and the recursive nature oftheir proof can also be adapted to find our generating function for descents.The labels v ( d −
1) + 1 , . . . , vd must occur on either the first diamond, or the last. In the first case, theycreate a descent. In the second, they do not, giving a (1 + x ) term in the generating function. We continuerecursively and obtain: X D ∈D v,d (132 , y d x des ( σ ( D )) = 1 + ∞ X d =1 y d (1 + x ) d − = 1 + 11 + x (cid:18) − − y (1 + x ) (cid:19) = x x + 1(1 + x )(1 − y (1 + x ))= x (1 − y (1 + x )) + 1(1 + x )(1 − y (1 + x ))= 1 − yx − y (1 + x ) . Corollary 4. f , v,d (1 , y ) (cid:12)(cid:12)(cid:12) y d = |D v,d (132 , | = 2 d − . , and , Lemma 3.
In order to avoid and , the final diamond is labelled with either v ( d −
1) + 1 , v ( d −
1) +2 , . . . , vd or , v ( d −
1) + 2 , . . . , vd
Proof:
Since v ≥ , the label vd must appear on the final diamond in order to avoid . Likewise theinterior vertices on the final diamond must be in consecutive increasing order in order to avoid , so the v − final vertices are v ( d −
1) + 2 , . . . , vd . If the label on the first vertex of the last diamond were somenumber j other than or v ( d −
1) + 1 , then the first vertex of whichever diamond v ( d −
1) + 1 , along with v ( d −
1) + 1 , and j would form a . Theorem 6. f , v,d ( x, y ) = ∞ X d =1 X σ ∈D (132 , y d x des ( σ ) = 1 − yx − y (1 + x ) . Proof:
We proceed similarly to the proof of Theorem 5 with a recursive argument. By Lemma 2, the finaldiamond has only two possibilities, one of which forms a descent with the previous diamond, and one ofwhich doesn’t. Thus our descent generating function gains a (1 + x ) term for each additional diamond, andexactly as in Theorem 5, the result follows. Corollary 5. f , v,d (1 , y ) (cid:12)(cid:12)(cid:12) y d = |D v,d (132 , | = |D v,d (213 , | = 2 d − . Proof:
By Proposition 1, , is d -Wilf-equivalent to , . Mitchell Paukner, Lucy Pepin, Manda Riehl, Jarred Wieser , or , Lemma 4.
Any diamond avoiding and has at most one descent. Moreover, if there is a descent, itinvolves the label . Proof:
By examination of cases, any arrangement of two descents forms either a or a . If a descentdoes not involve the , then either the occurs before, causing a , or the occurs after, causing a . Theorem 7. f , v,d ( x, y ) = ∞ X d =1 X σ ∈D (132 , y d x des ( σ ) = 1 − y + y + vxy (1 − y ) . Proof:
By Lemma 4, we need only enumerate those diamonds with one descent where the descent involves the . Everything after the increases, as does everything before the . In fact, the permutations associated todiamonds that avoid and look like a portion of the identity permutation was deleted from the frontand inserted after position vi , for i = 1 , . . . , d − . When i = d − , there are v possibilities for how manynumbers appear consecutively with , including . When i = d − , there are v possibilities, etc. When i = 1 , there are ( d − v possibilities. Thus we have v d ( d − diamonds with one descent, and one withzero descents. Thus, X y d x des ( σ ) = ∞ X d =0 y d [1 + v d ( d − x ]= ∞ X d =0 y d + vx ∞ X d =0 y d d ( d − − y + vx y ) (cid:18) − y ) (cid:19) = 11 − y + vxy (1 − y ) = (1 − y ) (1 − y ) + vxy (1 − y ) = 1 − y + y + vxy (1 − y ) . Corollary 6. f , v,d (1 , y ) (cid:12)(cid:12)(cid:12) y d = |D v,d (132 , | = |D v,d (213 , | = 1 + v (cid:16) d ( d − (cid:17) Proof: D v,d (213 , and D v,d (132 , are d -Wilf Equivalent by Proposition 1. , Theorem 8. f , v,d ( x, y ) = ∞ X d =1 X σ ∈D (231 , y d x des ( σ ) = x + yx (1 + x ) v − + 1(1 + x )(1 − y (1 + x ) v − ) . attern Avoidance in Task-Precedence Posets Proof:
Let n = vd be the largest label on a diamond D ∈ D v,d (231 , . Avoiding the pattern meansthe must be at the beginning and avoiding the pattern implies everything after n must be decreasingwhich forces n to the end of the permutation. By a result of Simion and Schmidt on permutations, there are v − ways to arrange the middle-level vertices within each of the d diamonds in order to avoid both and creating between and v − descents [16]. There are also two ways to either swap or not swap the lastelement of each diamond with the first element of the next. This gives the following generating function. X D ∈D v,d (231 , y d x des ( σ ( D )) = 1 + ∞ X d =1 y d (1 + x ) ( v − d − = 1 + 1(1 + x ) ∞ X d =1 ( y (1 + x )) ( v − d = 1 − x ) + 1(1 + x )(1 − y (1 + x ) v − )= x + yx (1 + x ) v − + 1(1 + x )(1 − y (1 + x ) v − ) . Corollary 7. f , v,d (1 , y ) (cid:12)(cid:12)(cid:12) y d = |D v,d (231 , | = 2 d ( v − − , Lemma 5.
All labels that appear after n = vd must be consecutive and increasing, and if a n = n , then a n = n − . Let β dv,j be the generating function for descents in D v,d (231 , . Recall Figure 2.4 is an example of d − full diamonds with v vertices followed by an incomplete diamond with j vertices for j = 1 , . . . , v − . Theorem 9. f , v,d ( x,
1) = β dv,v = β dv, ( v − + x d − X i =1 β d − iv, ( v − where β v,j = , if j = 12 j − , if j = 2 , . . . , v − v − , if j = v is the generating function for descents for D v,d (231 , . Proof:
We approach the proof similarly to that of Theorem 4 and partition our diamonds by the position of thelargest element and proceed recursively. Because the proofs are very similar, we omit the details of thisproof for brevity. The only differences are that since we are now avoiding , we have no descents afterthe appearance of the largest label, and we have different initial conditions on one diamond.Table 5 is an example of using this recursive technique to find the generating function for descents in D , (231 , . Mitchell Paukner, Lucy Pepin, Manda Riehl, Jarred Wieser v=5d 1 2 3 β d , x + 3 x x +41 x +37 x +12 x β d , x + 6 x x +54 x +62 x +24 x β d , x x + 13 x + 3 x x + 80 x + 128 x +73 x + 12 x β d , x x + 25 x + 12 x x + 121 x + 248 x +184 x + 48 x β d , x x + 28 x + 12 x x + 134 x + 273 x +196 x + 48 x Tab. 5:
The recursive steps necessary to find the generating function for descents in D , (231 , . v=5d 1 2 3 β d , β d , β d , β d , β d , Tab. 6:
The total number of permutations for D ,d that avoid the patterns (231 , when d = 1 , , . Corollary 8. f , v,d (1 , y ) (cid:12)(cid:12)(cid:12) y d = β dv,v (1) = |D v,d (231 , | = |D v,d (312 , | . Proof:
By Proposition 1, , is d -Wilf-equivalent to , . There are only two nontrivial cases to examine when we avoid three patterns of length 3: , , and , , . , , Theorem 10. f , , v,d ( x, y ) = ∞ X d =1 X σ ∈D (132 , , y d x des ( σ ) = 1 − y + xy (1 − y ) . Proof:
Let n = vd be the largest label in d diamonds with v vertices. Avoiding the pattern forces alllabels before n to be larger than all labels after. Avoiding the pattern forces all labels before n to beincreasing. Avoiding the pattern forces all labels after n to be increasing. This indicates that all verticesthat appear before n will be the consecutive numbers prior to n and all vertices after n will be the remainingelements ordered consecutively. A label a i = n iff i = vs for some s = 1 , . . . , d , and there is only onearrangement for the rest of the elements. Therefore, there can only be, at most, one descent and it occursbetween diamonds. So, attern Avoidance in Task-Precedence Posets X y d x des ( σ ) = 1 + ∞ X d =1 y d (1 + ( d − x )= 11 − y + xy ∞ X d =1 ( d − y d − = 1 − y + xy (1 − y ) . Corollary 9. f , , v,d (1 , y ) | y d = |D v,d (132 , , | = d . We will proceed by examining what changes can be made to the identity permutation while still avoiding , , and . Lemma 6.
For labels a i , a j , a k if a i , a j < a k , then i < k or j < k in order to avoid the patterns and . A swap is when two consecutive labels from the identity permutation switch positions in the permutation.Since any permutation can be created from the identity using swaps, restricting our changes to swaps willnot exclude any possibilities. Lemma 7.
All swaps must be disjoint in order to avoid . Proof:
We simply examine the cases when two swaps overlap in some way, either with two swaps executedon elements, or two overlapping swaps on elements. Theorem 11.
The generating function for descents in D v,d (231 , , is f , , v,d ( x ) = (1 + x ) d − d ⌊ v − ⌋ X k =0 (cid:18) v − − kk (cid:19) x k . Proof:
Every final element of a diamond can either remain unchanged or be swapped with the least elementof the next diamond. This then gives the generating function (1 + x ) d − for each possible swap. Let k represent the nonconsecutive positions from which to choose a swap among the interior vertices. Note thatin a diamond there are v − positions to swap since there are v − interior vertices. By Lemma 6 and Lemma7 any consecutive interior vertices can only be swapped disjointly. Since the swaps must be nonconsecutive, k must be chosen from v − − ( k − . This gives (cid:0) v − − kk (cid:1) . We then sum over all k in order to generateall possible descents for a single diamond. Since we have d diamonds in which to execute these swaps, weraise to the d th power. The gfd for D v,d (231 , , is then (1 + x ) d − (cid:0) Σ ⌊ v − ⌋ k =0 (cid:0) v − − kk (cid:1) x k (cid:1) d . Corollary 10. f , , v,d (1 , y ) (cid:12)(cid:12)(cid:12) y d = |D v,d (231 , , | = 2 · d − . Corollary 11. {|D v, (231 , , |} v ≥ is the Fibonacci numbers. Mitchell Paukner, Lucy Pepin, Manda Riehl, Jarred Wieser
Proof:
The base cases are |D , (231 , , | = 1 which is the identity, and |D , (231 , , | = 2 , which is the identity permutation and the permutation with the interior vertices swapped. Let there bea single diamond with v vertices, where are v − interior vertices that can be swapped which givesthe the D v, (231 , , permutations that avoid the three patterns. Now consider a single diamondwith v + 1 vertices. The final interior vertex will either be the v element when there is no descent inthe last two interior vertices, or the v − element when there is a descent between the final two inte-rior vertices. When v is the final interior vertex, there are v − vertices that can be re-arranged. Thusthere are the D v, (231 , , permutations. When there is a descent in the final two interior vertices,there are v − interior vertices that can be re-arranged, thus there are the D v − , (231 , , per-mutations. Hence |D v +1 , (231 , , | = |D v, (231 , , | + |D v − , (231 , , | . Therefore, |D v, (231 , , | follows the Fibonacci numbers. Theorem 12.
Let S be a set of at least distinct permutations of length .Then |D v,d ( S ) | = ( , if ∈ S , if S .
Proof:
Let n be the largest label in any permutation. Due to the structure of the diamonds, any set ofpermutations involving cannot be avoided. For any other collection of or more patterns, the result iseasily seen using the lemmas for avoiding a single pattern earlier in the paper. This investigation leaves several directions open for future study. We did not touch on patterns of length , they all remain open. We are confident the techniques of Bevan et.al. [2] will give the growth rate andminimal polynomial for diamonds avoiding , but in addition it is likely that these techniques wouldalso work for some patterns of length . Although the minimal polynomials are unlikely to generalize,the transition operators in particular cases could potentially even generalize to length k for the decreasingpattern k k − . . . . There are also a wide variety of other poset classes that could be approached inthis manner other than diamonds. We generalized our diamonds by adding additional elements and orderrelations between the least and greatest elements, but one could also imagine creating a diamond-type posetwith more than levels as another generalization. Acknowledgements
The authors would like to thank both referees for their careful readings of the paper with many constructivesuggestions that not only improved the paper but also gave the authors excellent examples of the mathemat-ical publishing process and language and notation standards. The authors would also like to thank DavidBevan for confirming their suspicion that the techniques used in previous papers would also be successfulhere for avoiders and for calculating the first terms of the sequence.
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