Avoiding patterns in irreducible permutations

We explore the classical pattern avoidance question in the case of irreducible permutations, i.e. , those in which there is no index i such that σ ( i + 1) − σ ( i ) = 1 . The problem is addressed completely in the case of avoiding one or two patterns of length three, and several well known sequences are encountered in the process, such as Catalan, Motzkin, Fibonacci, Tribonacci, Padovan and Binary numbers. Also, we present constructive bijections between the set of Motzkin paths of length n − 1 and the sets of irreducible permutations of length n (respectively ﬁxed point free irreducible involutions of length 2 n ) avoiding a pattern α for α ∈ { 132 , 213 , 321 } . This induces two new bijections between the set of Dyck paths and some restricted sets of permutations.

Let q = q 1 • • • q k , k ≥ 1, be a sequence of pairwise different positive integers.The reduction red(q) of q is the permutation in S k obtained from q by replacing the i-th smallest number of q with i for 1 ≤ i ≤ k.For instance, if q = 53841 then we have red(q) = 42531.A permutation σ ∈ S n avoids the pattern π ∈ S k , k ≥ 1, if and only if there does not exist a sequence of indices [15,20]), i.e. such that π =red(σ i1 σ i2 • • • σ i k ).We denote by S n (π) the set of permutations in S n avoiding the pattern π.For example, if π = 123 then 52143 ∈ S 5 (π) while 21534 / ∈ S 5 (π).In the case where σ does not avoid π (or equivalently contains π), π is said to be involved in σ which is denoted π σ.A set F of permutations is called a permutation class if it is closed downwards under this involvement relation.Many classical sequences in combinatorics appear as the cardinality of pattern-avoiding permutation classes.A large number of these results were firstly obtained by West and Knuth [8,15,20,21,22] (see the surveys of Kitaev and Mansour [7,10]).Also Bóna [3] has written a book that is dedicated to the notion of pattern avoiding permutations.Some generalizations of pattern avoidance can be viewed in [2,12].
A succession in a permutation σ ∈ S n is a pair (σ i , σ i+1 ), 1 ≤ i ≤ n − 1, with σ i+1 − σ i = 1.For instance, the pair (3,4) is the only one succession in σ = 53421.Using the terminology of Atkinson and Stitt (see [1]), a permutation with no successions will be also called irreducible.They show how the notion of irreducibility plays, in some sense, a dual role compared to indecomposability (a permutation σ = σ 1 • • • σ n is indecomposable if there does not exist p ≤ n − 1 such that σ 1 • • • σ p is a permutation of [p]).Indeed, irreducible permutations are crucial for the construction of a basis of some wreath products C I where C is a class of permutations and I is the class of identity permutations (see [1]).On the other hand, an irreducible permutation can be also viewed as a permutation avoiding the bi-vincular pattern (12, {1}, {1}) (see [11]).
Let S irr n be the set of irreducible permutations of length n.For example, we have S irr 3 = {132, 213, 321}.The cardinality of the sets S irr n is known (see [14,18,19]) and given by More generally, permutations with a given number of successions also appeared in combinatorial theory literature (see [6,13]).
In this paper, we explore the classical pattern avoidance question in the case of irreducible permutations.The problem is addressed completely in the case of avoiding one or two patterns of length three, and several well-known sequences are encountered, such as the Catalan, Motzkin, Fibonacci, Binary, Tribonacci and Padovan numbers.In Section 2, we present general enumerative results for sets of permutations that are expanded (i.e., closed under inflation) and closed under deflation.In particular, we give bivariate generating functions according to the length of permutations and the number of successions.Generalizing the two concepts of inflation and deflation for sets of fixed point free involutions, we obtain similar results for involutions.In Section 3, we give exhaustive enumerative results for sets of irreducible permutations avoiding one pattern of length three.For the sets counted by the Motzkin numbers, we exhibit bijections between them and the set of Motzkin paths.In Section 4, we focus on sets of irreducible permutations avoiding two patterns of length three.In Sections 5 and 6, we study irreducible (fixed point free) involutions avoiding one pattern of length three.Exhaustive enumerative results are obtained, and we construct bijections between fixed point free irreducible involutions of length 2n avoiding α ∈ {132, 213, 321} and Motzkin paths of length n − 1.

Preliminaries
A set F of permutations is said to be closed under inflation (also called expanded, [1]) if whenever a permutation σ = β i γ ∈ F so is the permutation obtained by replacing i with i (i + 1), after increasing by one each other's values greater than i in σ.A set F of permutations is said to be closed under deflation if whenever a permutation σ = β i (i + 1) γ so is the permutation obtained by replacing i (i + 1) with i, after decreasing by one each other's values greater than (i + 1) in σ.In the following, these two last elementary transformations will be respectively called inflation and deflation of σ.Obviously, a set F of irreducible permutations is closed under deflation since its elements do not contain any successions.A set closed under pattern involvement is closed under deflation.Moreover, a set closed under pattern involvement is expanded if and only if every basis element is irreducible (see Lemma 5,[1]).
In this section we provide general lemmas about generating functions for sets of permutations and fixed point free involutions which are expanded and closed under deflation.
Lemma 1 Let F be a set of permutations which is expanded and closed under deflation; let G be the set of its irreducible permutations.We denote by g(x) the generating function of G, and f (x, y) the bivariate generating function of F where the coefficient of x n y k is the number of permutations of length n with exactly k successions.Then we have In particular, the generating function of F is g (x/(1 − x)). Proof: be the set of permutations of length n in F containing exactly k successions.Obviously, we have Let σ be a permutation in F k n , i.e, σ contains exactly k successions.We define the succession set Since F is expanded and closed under deflation, a permutation π ∈ F irr n−|Iσ| can be uniquely obtained from a permutation σ ∈ F k n by deleting all entries σ i , i ∈ J σ and reducing the result to a permutation of F irr n−|Iσ| .This construction produces a simple bijection from F irr n−|Iσ| to the set F Iσ n of permutations in F n having the succession set equal to I σ .Since inflations of terms in an irreducible permutation correspond exactly to the terms in the geometric series expansion x 1−yx = x + yx 2 + y 2 x 3 + . .., the above bijection implies Now we give a similar Lemma for fixed point free involutions.In this case, we slightly modify the two concepts of inflation and deflation.Let F be a set of involutions with no fixed points.We say that F is expanded (closed under inflation) if whenever an involution σ ∈ F and (i, j), 1 ≤ i < j ≤ n, such that σ i = j, so is the permutation σ obtained from σ by replacing σ i with (σ i + 1)(σ i + 2), σ j with σ j (σ j + 1), and increasing by inc(k) =card{ ∈ {i + 1, j + 1}, ≤ k} each other's values k.For instance, if σ = 3 4 1 2 7 8 5 6, i = 2, j = 4, then σ = 4 5 6 1 2 3 9 (10) 7 8.On the other hand, we say that F is closed under deflation if whenever an involution σ ∈ F and (i, j), 1 ≤ i < j ≤ n − 1, such that σ i = j, σ i+1 = j + 1, so is the permutation σ obtained from σ by deleting (i + 1) and (j + 1) and reducing the result to a permutation.Obviously the set of all fixed point free involutions is expanded and closed under deflation.
Lemma 2 Let F be a set of fixed point free involutions which is expanded and closed under deflation; let G be the set of its irreducible involutions.We denote by g(x) the generating function of G, and f (x, y) the bivariate generating function of F where the coefficient of x n y k is the number of involutions of length n with exactly k successions.Then we have In particular, the generating function of Proof: This proof is a simple counterpart of the previous one.It suffices to remark that an involution with no fixed points is necessarily of even length.2 Lemma 1 ensures that if we know the generating function of a set G of irreducible permutations, then we can easily obtain the bivariate generating function (according to the length and the number of successions) of the wreath product G I where I is the class of identity permutations (the wreath product G I is the closure under inflation of G).An immediate consequence is: if G is a set of irreducible permutations such that the wreath product G I is enumerated by the Catalan numbers ( [16], A00018) then G is enumerated by the Motzkin numbers ( [16], A001006).Therefore, irreducible permutations of S n (α), n ≥ 1, α ∈ {321, 213, 132}, are enumerated by the Motzkin numbers (see Theorem 2).
3 Avoiding a pattern of length 3 In this section, we study the avoidance of one pattern of length 3 for irreducible permutations.The first part uses arguments with generating functions while the second part presents bijective proofs for the sets of permutations counted by Motzkin numbers.All enumerative results of this section are listed in Table 1.
Theorem 1 The generating function for the sets S irr n (123), n ≥ 0, of irreducible permutations of length n avoiding the pattern 123 is given by (see [16], A114487): Proof: Let σ be a permutation in S n (123)\S irr n (123).It necessarily contains a succession (k, k + 1), 1 ≤ k ≤ n − 1; let us take the leftmost succession.Then σ can be written σ = uk(k + 1)v where 1 ≤ k ≤ n − 1, such that u and v are two subsequences of [n].Since σ avoids 123, the set of values in u necessarily equals to the interval [k + 2, n].So v belongs to S k−1 (123), u avoids the pattern 123 and does not contain any successions.Let f (x) be the generating function for the sets S irr n (123), n ≥ 0, the above decomposition of σ implies is the generating function for the sets S n (123), i.e., the generating function for the well-known Catalan numbers (see [16], A00018).We deduce: Theorem 2 For α ∈ {321, 213, 132}, the sets S irr n (α), n ≥ 1, are enumerated by the Motzkin numbers (see [16], A001006).
S n (α) is expanded and closed under deflation.Let g(x) be the generating function of S irr (α); Lemma 1 of Section 2 proves that c(x) = g(x/(1 − x)) and thus: More precisely, the set S irr n (α), n ≥ 1, is enumerated by the (n − 1)-th term of the Motzkin sequence ( [16], A001006). 2 In fact, Lemma 1 yields the more general result: Theorem 3 For α ∈ {321, 213, 132}, let S(α) be the set of permutations avoiding α, and f (x, y) its bivariate generating function, where the coefficient of x n y k is the number of permutations of length n with k successions in S(α).Then we have where c(x) is the generating function for the Catalan numbers.

Bijective proofs
A Motzkin path of length n is a lattice path starting at (0, 0), ending at (n, 0), and never going below the x-axis, consisting of up steps U = (1, 1), horizontal steps H = (1, 0), and down steps D = (1, −1).A Dyck path of length 2n is a Motzkin path of the same length that does not contain any horizontal steps.Dyck paths of length 2n are enumerated by the well-known Catalan numbers ( [16], A00018) and Motzkin paths by the Motzkin numbers ([16], A001006).We refer to Donaghey, Shapiro [5] and Stanley [17] for several combinatorial classes enumerated by the Motzkin and Catalan numbers.
• Bijective proof for S irr n (132).Here we construct a bijection between S irr n (132) and the set of Motzkin paths of length n − 1.We define the map f that transforms a permutation σ = σ 1 σ 2 • • • σ n ∈ S irr n (132) into the Motzkin path M of length n − 1 obtained by the following process: For each i from 1 to n − 1, (a) if σ i < σ i+1 , then the i-th step of M is a down step D; (b) if σ i > σ i+1 and there exists j > i + 1 such that σ j = σ i + 1, then the i-th step of M is an up step U ; (c) otherwise, the i-th step of M is an horizontal step H.
For instance, the permutations 21, 321, 213, 4213, 3214, 3241 and 4321 are respectively transformed by f into the Motzkin paths H, HH, U D, HU D, U HD, U DH, HHH (see Figure 1 for an example with n = 14).
Let us prove that f is a one-to-one correspondence between S irr n (132) and the set of Motzkin paths of length (n − 1).Let σ be a permutation of S irr n (132).Let us take i, 1 ≤ i ≤ n − 1, such that (b) is verified.Then, there exists j > i + 1 such that σ j = σ i + 1.Since σ avoids 132, σ j−1 < σ j and the index j − 1 > i verifies the case (a).Conversely, if i, 1 ≤ i ≤ n − 1, satisfies (a) then σ i < σ i+1 .Since σ is irreducible, we have σ i+1 = σ i + 1.Thus, there exists j = i and j = i + 1 such that σ j = σ i+1 − 1.As σ avoids 132, we necessarily have j < i and j verifies (b).Hence, there is a one-to-one correspondence between up steps and down steps such that each up step is associated with a down step on its right.This is precisely the characterization of the Motzkin paths.
Conversely, let M be a Motzkin path of length (n − 1) and let us prove that there exists a permutation σ ∈ S irr n (132) such that f (σ) = M .We proceed by induction on n in order to construct the permutation σ ∈ S irr n (132).We distinguish two cases: )) is recursively obtained from the Motzkin path m (resp.m).Since m does not belong to (1), its associated permutation necessarily have its greatest element on the last position.We conclude with a simple induction.
Notice that paths of type (1) are mapped to permutations whose last element is not n, and paths of type (2) are mapped to permutations that end with n.Finally, the map f is a one-to-one correspondence between S irr n (132) and the set of Motzkin paths of length (n − 1). 2 • Bijective proof for S irr n (321).Here we construct a bijection between S irr n (321) and the set of Motzkin paths of length n − 1.Let σ = σ 1 σ 2 • • • σ n be a permutation in S irr n (321), and for i ∈ [n] we denote by s i the maximum of the set {σ 1 , σ 2 , . . ., σ i }.We define the map g that transforms a permutation σ ∈ S irr n (321) into the Motzkin path M of length n − 1 obtained by the following process: For each i from 1 to n − 1, (a) if s i + 1 < σ i+1 , then the i-th step of M is an up step U ; (b) if there exists j < i such that σ i+1 = s j + 1 and s j + 1 < σ j+1 , then the i-th step of M is a down step D; (c) otherwise, the i-th step of M is an horizontal step H.
For instance, the permutations 21, 132, 213, 1324, 3142, 2143 and 2413 are respectively transformed by g into the Motzkin paths H, U D, HH, U DH, HU D, HHH, U HD (see Figure 1 for an example of length 14).
Let us prove that g is a one-to-one correspondence between S irr n (321) and the set of Motzkin paths of length (n − 1).Let σ be a permutation of S irr n (321).Let us take i, . So there is j < i such that σ i+1 = s j + 1 and s j + 1 < σ j+1 and the index j verifies (a).Thus, there is a one-to-one correspondence between up steps and down steps such that each up step is associated with a down step on its right.This is precisely the characterization of the Motzkin paths.
Conversely, let M be a Motzkin path of length (n − 1) and let us prove that there exists a permutation σ ∈ S irr n (321) such that g(σ) = M .We proceed by induction on n in order to construct the permutation σ ∈ S irr n (321).We distinguish four cases: (1)  -If M = U k Dm, then we have M = g(σ) where σ is recursively obtained from the permutation π such that g(π) = U k−1 Hm by adding 1 on the left of π and after increasing by one all values of π.Notice that π cannot begin with 1 since there would be a down step just after U k−1 .Thus σ does not contain any successions and avoids 321.
-If M = U k H m where m does not begin with H.If = 1, then we have M = g(σ) where σ is recursively obtained from the permutation π such that g(π) = U k m by inserting 1 between the (k + 1) and (k + 2)-th positions of π and after increasing by one all values of π greater or equal than 1.Since g(π) = U k m and m does not start with H, we have π k+2 = 1 and the previous insertion of 1 does not create any succession and any pattern 321.
If M = U k H m where m does not begin with H and with = 1, then we have M = g(σ) where σ is recursively obtained from the permutation π such that g(π) = U k H −1 m by inserting π k+ −1 + 1 between the (k + ) and (k + + 1)-th positions of π and after increasing by one all values of π greater or equal than π k+ −1 + 1.
Since the first step of m is not H, we necessarily have either π k+ +1 > s k+ + 1 or π k+ +1 < s k+ which implies that the insertion of π k+ −1 + 1 does not create a succession at position (k + + 1) in σ.Also, the insertion of π k+ −1 + 1 between the (k + ) and (k + + 1)-th positions cannot create a succession at position (k + ).So, the permutation σ does not contain any succession.
-If M = H k U m, then we have M = g(σ) where σ = πuπ such that g(πu) = H k and g(red(uπ )) = U m.
-If M = H k , then we have M = g(σ) where σ is recursively obtained from the permutation π such that g(π) = H k−1 by inserting π k−1 + 1 in the last position and after increasing by one all values of π greater or equal than π k−1 + 1.The bijection f (resp.g) from S irr n (132) (resp.S irr n (321)) to the set of Motzkin paths of length (n − 1) induces a new constructive bijection between Dyck paths and some restricted irreducible permutations: Corollary 1 Let P 2n+1 be the set of permutations σ ∈ S irr 2n+1 (132) such that for all i, 1 ≤ i ≤ 2n − 1, with σ i > σ i+1 there exists j > i + 1 with σ j = σ i + 1.Then the map f induces a constructive bijection between P 2n+1 and the set of Dyck Paths of length 2n.

Avoiding two patterns of length 3
In this section, we explore the avoidance of two patterns of length 3 for irreducible permutations.All enumerative results of this section are listed in Table 2.
Proof: The result is well known for α = 123.Now, let us take α = 213 (the case α = 132 will be obtained with a simple Wilf equivalence).Let σ be a permutation of S irr n (321, 213).Writing σ = βnγ with β and γ are two subsequences of [n − 1], the avoidance of 213 implies that Since σ is irreducible and avoids 213, γ and β contain at most one element which implies that there does not exist any permutations σ in S irr n (321, 213) whenever n ≥ 4. 2 Theorem 6 For n ≥ 1, the set S irr n (213, 132) is reduced to the unique permutation n(n − 1) • • • 321.
Tab. 2: Wilf-equivalence classes for two pattern subsets of S3 in irreducible permutations.

Fixed point free involutions
In this section, we study the avoidance of one pattern of length 3 for fixed point free irreducible involutions, i.e., for involutions with no fixed points and no successions.All enumerative results of this section are listed in Table 3.Let DI (resp.DI irr ) be the set of fixed point free (resp.fixed point free irreducible) involutions.These sets restricted to their length n elements will be respectively denoted DI n and DI irr n .
Proof: Obviously, there does not exist any fixed point free involutions of odd length; thus a 2n+1 = 0 for n ≥ 0. Now, let σ be an irreducible involution of length 2n, n ≥ 2, with no fixed points.Let k be the index in [2n − 1] such that σ k = 2n (and also σ 2n = k).
Let π be the involution of length (2n − 2) obtained from σ by deleting k and 2n, and reducing the result to an involution of length (2n−2).Two cases occur: (a) π is irreducible; and (b) π has the two successions (π k−1 , π k ) and (k − 1, k).Conversely, each irreducible involution of length 2n can be obtained from an involution π belonging to one of the two previous cases (a) and (b).
Thus, the generating function for the permutations σ of type (a) is (2n − 2)a 2n−2 .Now let us consider involutions of type (b).Let A 2n−2 be the set of involutions of length (2n − 2) with exactly two successions and with no fixed points.An involution in A 2n−2 can be extended into an irreducible involution of length 2n of type (b) in two possible ways.So, let π be an involution in A 2n−2 such that (π k , π k+1 ) and (k, k + 1) are the two successions.From π, we construct the pair where π is obtained from π by deleting the two entries π k+1 and (k + 1), and reducing the result to a permutation of length (2n − 4).Since we obtain the same permutation in π ∈ DI More generally, the proofs of Lemmas 1 and 2 imply that there are n−1 k •a 2n fixed point free involutions of length 2n with exactly 2k successions, where a 2n is the sequence defined in Theorem 11.
Proof: Since the set DI(α) is expanded and closed under deflation, we apply Lemma 2 of Section 2. As the generating function for is the generating function for the Catalan numbers, the generating function for DI irr n (α) is

2
A simple application of Lemma 2 yields the bivariate generating function f (x, y) = c x 2 1+(1−y)x 2 of the set DI(α) of fixed point free involutions avoiding the pattern α for α ∈ {321, 213, 132}, where the coefficient of x n y k gives the number of elements in DI n (α) with exactly k successions.2 Theorem 14 The generating function for the sets DI irr n (123), n ≥ 0, of irreducible involutions of length n without fixed points and avoiding the pattern 123 is Proof: Let σ be an involution of length 2n without fixed points containing at least one succession.We suppose that (σ k , σ k+1 ) is the leftmost succession.Since σ avoids 123, it is straightforward to see that be the generating function for the set of involutions without fixed points and avoiding 123 (see for instance [4]).Let f (x) be the generating function of irreducible involutions without fixed points and avoiding 123.We have where h(x) is the generating function for irreducible permutations avoiding the pattern 123 (see Theorem 1).Finally, a simple calculation gives the desired results.
Let us prove that h is a one-to-one correspondence between DI irr 2n (321) and the set of Motzkin paths of length n − 1.Let σ be an involution in DI irr 2n (321).Using Remark 2, σ is associated with a Dyck path where left-to-right maxima (resp.right-to-left minima) correspond to up steps (resp.down steps).So, there is a one-to-one correspondence between up steps and down steps such that the image of an up step is a down step lying on its right.This induces a one-to-one correspondence c between the set of all occurrences of U U and the set of all occurrences of DD such that the image by c of a UU-occurrence lies on its right.This implies that there is a one-to-one correspondence c between the sets U p = {i, 1 ≤ i ≤ n − 1, (a) is verified} and Down = {i, 1 ≤ i ≤ n − 1, (b) is verified} such that c(i) is greater than i (we have c(i) = i since the configuration (2) of Figure 3 does not occur).As the function h associates an up step when i ∈ U p, a down step when i ∈ Down and an horizontal step otherwise, h(σ) is a Motzkin path of length (n − 1).Moreover this construction ensures that the images by h of two different involutions necessarily yields two different Motzkin paths.
Conversely, from any Motzkin path M of length n − 1, we construct an involution of length 2n without fixed points and that avoids 321, by the following process.More precisely, we will construct a sequence of red and green points where red (resp.green) points correspond to left-to-right maxima (resp.right-to-left minima) of the involution of length 2n.This sequence will characterize the desired involution.
We start the process with a red point followed by a green point.Through the Motzkin path M from left to right: -if we meet an up step U , then we add one red point just after the last red point, and we add one red and one green points (in this order) on the right; -if we meet a down step D, then we add one green point on the right; -if we meet horizontal step H, then we add one red and one green points (in this order) on the right.
For instance, this process applying to the Motzkin path M = U HD provides the following steps: Step 0: Step 1: Step 2: Step 3: At each step of this process, we add only one green point on the right.Moreover, the number of red points is, at each step, at most the number of green points; at the end of the process, there is equality.The configuration (2) of Figure 3 does not occur.Therefore, the obtained matching corresponds to that of a fixed point free irreducible involution that avoids 321 and such that its image by h is exactly the Motzkin path M .We conclude that h is a one-to-one correspondence between DI irr 2n (321) and the set of Motzkin paths of length n − 1. 2 Notice that the above construction appears as a generalization of the bijection of P. Manara and C. Perelli Cippo, [9], which transforms a restricted set of Motzkin paths into the set of simple involutions avoiding the pattern 321.

Pattern avoiding involutions
In this section, we present enumerative results for sets of irreducible involutions avoiding one pattern of length three (see Table 4).
Proof: A length n irreducible involution σ avoiding 231 can be written βnγk, 1 ≤ k ≤ n, where β ∈ I irr k−1 (α) such that the last value of β is different from n − 1 whenever k = n, and where nγk = n(n − 1) • • • (k + 1)k.Let g(x) (resp.h(x)) be the generating function for the sets I irr n (α), n ≥ 0, (resp.for the sets of irreducible involutions of length n avoiding α such that the last value is different from n).
(1) M = mHm where m and m are two Motzkin paths and such that m does not contain any horizontal steps on the x-axis, and (2) M = mU m D where m and m are two Motzkin paths such that m does not belong to the case (1).-(1) If M = mHm , then we have M = f (σ) where σ = βnγ such that γ (resp.red(βn)) is recursively obtained from the Motzkin path m (resp.m).Notice that the position of n in σ creates the horizontal step H between m and m .-(2) If M = mU m D, then we have M = f (σ) where σ = β(n − 1)γn such that γ (resp.red(β(n − 1) For example, if M = mU m D with m = U HD and m = U DH (n = 9), then β8 = 7658, red(β8) = 3214, γ = 3241 and σ = 765832419.If M = mHm with m = U HD and m = U DH (n = 8), then β8 = 7658, red(β8) = 3214, γ = 3241 and σ = 76583241.
and (4) M = HH . . .H, with k ≥ 1, ≥ 0, and where m is a suffix of the Motzkin path M .

1
where a 1 = 1 and a 2 = 1 which corresponds to the Sloane's sequence A005251.The case α = 213 is obtained by symmetry.

Fig. 2 : 1 {} a 2n+1 = 0 a
Fig. 2: Proof of Theorem 14.The special structure of a fixed point free involution avoiding 123 and with at least one succession.
γ is empty.Since γ avoids 231, β does not contain any values greater than γ k .Thus we have either γ is empty or (18)9)(10)(12)(20)(15)(18).(b) if A i is verified but not B i , then the i-th step of M is a down step D; (c) if A i and B i are verified, then the i-th step of M is an horizontal step H.For instance, involutions 2143, 351624, 214365, 35172846, 35162487, 21573846 and 21436587 are respectively transformed by h into H, U D, HH, U HD, U DH, HU D and HHH (see Figure