RSK tableaux and box-ball systems

A box-ball system is a discrete dynamical system whose dynamics come from the balls jumping according to certain rules. A permutation on n objects gives a box-ball system state by assigning its one-line notation to n consecutive boxes. After a finite number of steps, a box-ball system will reach a steady state. From any steady state, we can construct a tableau called the soliton decomposition of the box-ball system. We prove that if the soliton decomposition of a permutation w is a standard tableau or if its shape coincides with the Robinson-Schensted (RS) partition of w, then the soliton decomposition of w and the RS insertion tableau of w are equal. We also use row reading words, Knuth moves, RS recording tableaux, and a localized version of Greene's theorem (proven recently by Lewis, Lyu, Pylyavskyy, and Sen) to study various properties of a box-ball system.


Introduction
A box-ball system (BBS) is a collection of discrete time states.At each state, we have an injective map from n balls (labeled by the integers from 1 to n) to boxes (labeled by the natural numbers); each box can fit at most one ball.The dynamics come from the balls jumping according to certain rules.Let S n denote the set of permutations on [n] = {1, 2, . . ., n}.A permutation w in S n gives a box-ball system state by assigning the one-line notation of the permutation to n consecutive boxes.Given a BBS state at time t, we compute the BBS state at time t + 1 by applying one BBS move, which is the process of moving each integer to the nearest empty box to its right, beginning with the smallest.See Figure 1.1.This version of the box-ball system was introduced in [Tak93] and is an extension of the box-ball system first invented by Takahashi and Satsuma in [TS90].A soliton is a maximal consecutive increasing sequence of balls which is preserved by all subsequent BBS moves.After a finite number of BBS moves, a box-ball system containing a configuration w will reach a steady state, decomposing into solitons whose sizes are weakly decreasing from right to left, that is, forming an integer partition shape.From such a state, we can construct the soliton decomposition of the box-ball system, denoted SD, by stacking solitons so that the rightmost soliton is placed on the first row, the soliton to its left is placed on the second row, and so on.We obtain a tableau where each row is increasing but which may or may not be standard.The soliton decomposition of a permutation w is the soliton decomposition of the box-ball system containing w.
Figure 1.2 shows the state of the box-ball system containing w = 452361 from t = 0 to t = 4.Note that steady state is first reached at t = 3.The soliton decomposition of w = 452361 is the tableau SD(w) = .
In this example, the soliton decomposition is a standard tableau, but most permutations have soliton decompositions which are not standard.The tableau SD(w) has shape (3, 2, 1).We will refer to the shape of the soliton decomposition as the BBS soliton partition.The well-known Robinson-Schensted (RS) insertion algorithm is a bijection w → (P(w), Q(w)) from S n onto pairs of standard size-n tableaux of the same shape [Sch61].The tableau P(w) is called the insertion tableau of w, and the tableau Q(w) is called the recording tableau of w.The shape of these tableaux is called the RS partition of w.
The row reading word of a tableau is the permutation formed by concatenating the rows of the tableau from bottom to top, left to right.
The tableau P(w) has row reading word r = 425136.The insertion tableau of r is the tableau P(w).For more information, see for example the textbook [Sag20, Section 7.5].
The carrier algorithm given in [Fuk04] (which we review in Section 3) is a way to transform a box-ball configuration at time t into the configuration at time t+1.At each step in the algorithm, we insert and bump numbers in and out of a carrier filled with a weakly increasing sequence, following a rule which should remind the reader of the RS insertion algorithm.
Our goal is to study the connection between the soliton decompositions and RS tableaux of permutations.We now describe our main results.

Insertion tableaux and soliton decompositions
For the permutation w = 452361 used in the above example, we have SD(w) = P(w).However, in general the soliton decomposition and the RS insertion tableau of a permutation do not coincide.Surprisingly, having a standard soliton decomposition tableau or having a BBS soliton partition which equals the RS partition is enough to guarantee that the soliton decomposition and the RS insertion tableau coincide.
Theorem A (Theorem 4.2).Suppose w is a permutation.Then the following are equivalent: 1. SD(w) = P(w).
3. The shape of SD(w) equals the shape of P(w).
The key ingredients of our proof are Greene's theorem (Theorem 2.2) and a result of Fukuda which says that the RS insertion tableau is an invariant of a box-ball system (Theorem 3.3).The proof that part (3) implies part (2) was suggested to us by Darij Grinberg.

Tableau reading words
We study the connection between steady-state configurations and row reading words.

Proposition B (Proposition 5.1). A permutation r is in steady state if and only if r is the row reading word of a standard tableau.
Next, we represent a box-ball system state as an array containing integers from 1 to n called the configuration array.This array has increasing rows but not necessarily increasing columns; it also may not have a valid skew shape and it may be disconnected.Proposition B turns out to be a special case of the following.

Proposition C (Proposition 5.2). A BBS configuration w is in steady state if and only if the configuration array of w is a standard skew tableau whose rows are weakly decreasing in length.
We will prove Proposition C in Section 5 using the carrier algorithm.Note that Proposition C is a corollary of a characterization for steady state given by Lewis, Lyu, Pylyavskyy, and Sen in [LLPS19, Proof of Lemma 2.1 and 2.3].

Recording tableaux and time to steady state
We also study the relationship between the RS recording tableau of a permutation and the behavior of its box-ball system.The number of BBS moves required for a permutation w to reach steady state is called the steady-state time of w.For example, as illustrated in Figure 1.2, the steady-state time of the permutation 452361 is 3.
Theorem D (Theorem 6.7).If n 5, let This particular recording tableau is special; we conjecture that all other permutations in S n have steady-state time smaller than n − 3.

Conjecture 1.1.
A permutation in S n whose recording tableau is not equal to Q has steady-state time smaller than n − 3.
Furthermore, we conjecture that Theorem D is a special case of the following general phenomenon.

Types of Knuth moves
The RS insertion tableau is preserved under any Knuth move [Knu70].In contrast, the soliton decomposition is only preserved under certain types of Knuth moves.In addition, We say that v and w differ by a Knuth relation of both kinds (K B ) if they differ by a Knuth relation of the first kind (K 1 ) and of the second kind (K 2 ), that is, v = v 1 . . .y 1 xzy 2 . . .v n and w = v 1 . . .y 1 zxy 2 . . .v n or vice versa where x < y 1 < z and x < y 2 < z.
Note that, when we apply a K 1 move (respectively, a K 2 move), the move may or may not be a K B move.If we apply a K B move, then it is both a K 1 move and a K 2 move.
A proper K 1 move is a K 1 move which is not K B , and a proper When performing a Knuth move, if we replace an "xz" pattern with a "zx" pattern, we denote this with a superscript "+."Otherwise, if we replace a "zx" pattern with an "xz" pattern, we denote this with a superscript "−."For example, if x < y 1 < z and x < y 2 < z, the move y 1 xzy 2 → y 1 zxy 2 is denoted K + B .We say that v and w are Knuth equivalent if they differ by a finite sequence of Knuth relations.
Using the localized version of Greene's Theorem given in Section 2.2, we prove a partial characterization of the BBS soliton partition in terms of types of Knuth moves.
Theorem E (Theorem 7.1).If v and w are related by a sequence of Knuth moves containing an odd number of K B moves, then SD(v) = SD(w).If v and w are related by a sequence of non-K B Knuth moves, then sh SD(v) = sh SD(w).
We also use non-K B Knuth moves to give a family of permutations which have steady-state time 1.
Theorem F (Theorem 7.4).Let r be the row reading word of a standard tableau.If w is a permutation which is related to r by one proper K 1 move or one proper K 2 move, then the steady-state time of w is 1.
The paper is organized as follows.In the next two sections, we review materials in the literature that we will use to prove our results.First, we review Greene's theorem in Section 2.1 and Lewis, Lyu, Pylyavskyy, and Sen's localized Greene's theorem in Section 2.2.Next, we review Fukuda's carrier algorithm and its connection to the RS insertion tableaux in Section 3. In Section 4, we prove Theorem A. In Section 5, we define the configuration array and use the carrier algorithm to prove Proposition C. Section 6 is devoted to the proof of Theorem D. We prove the two results involving types of Knuth moves (Theorem E and Theorem F) in Section 7.

Greene's theorem and a localized version of Greene's theorem
In the 1970s, Greene showed that the RS partition of a permutation and its conjugate record the numbers of disjoint unions of increasing and decreasing sequences of the permutation, which we explain in Section 2.1.Lewis, Lyu, Pylyavskyy, and Sen recently showed that the BBS soliton partition of a permutation and its conjugate record a localized version of Greene's theorem statistics.They studied an alternate version of the box-ball system, so in Section 2.2 we reframe their result to match our box-ball convention.

Greene's theorem and RS partition
In this section, we review Greene's theorem [Gre74, Theorem 3.1], which states that the RS partition of a permutation and its conjugate record the numbers of disjoint unions of increasing and decreasing sequences of the permutation.For more details, see for example Chapter 3 of the textbook [Sag01].Definition 2.1 (longest k-increasing and k-decreasing subsequences).A subsequence σ of w is called k-increasing if, as a set, it can be written as a disjoint union where each σ i is an increasing subsequence of w.If each σ i is a decreasing subsequence of w, we say that σ is k-decreasing.Let i k (w) denote the length of a longest k-increasing subsequence of w and d k (w) denote the length of a longest k-decreasing subsequence of w.Theorem 2.2 ([Gre74, Theorem 3.1]).Suppose w ∈ S n .Let λ = (λ 1 , λ 2 , λ 3 , . . . ) denote the RS partition of w, that is, let λ = sh P(w).Let µ = (µ 1 , µ 2 , µ 3 , . . . ) denote the conjugate of λ.Then, for any k, The longest 2-increasing subsequence is given by 562374 = 567 ⊔ 234.
A longest 3-decreasing subsequence (among others) is given by 5623714 = 52 ⊔ 631 ⊔ 74. Thus, By Theorem 2.2, the RS partition is equal to λ . We can verify this by computing the RS tableaux .

Localized Greene's theorem and BBS soliton partition
In [LLPS19, Lemma 2.1] and the blog post [Lew19], Lewis, Lyu, Pylyavskyy, and Sen presented a localized version of Greene's theorem.They studied an alternate version of the box-ball system, and in this section we reframe their result to match our box-ball convention.
Definition 2.4 (A localized version of longest k-increasing subsequences).If u is a sequence, let i(u) denote the length of a longest increasing subsequence of u.
For w ∈ S n and k 1, we define where the maximum is taken over ways of writing w as a concatenation That is, we consider all ways to break w into k consecutive subsequences, sum the i(u j ) values for each way, and let I k (w) be the maximum sum.
If u is a sequence of ℓ elements, an integer m Definition 2.5 (A localized version of longest k-decreasing subsequences).Let For w ∈ S n and k 1, we define where the maximum is taken over ways to write w as the union of disjoint subsequences u j of w.Notice that we only require u 1 , . . ., u k to be disjoint, not consecutive, in contrast to the procedure for calculating I k (w).
The following lemma is a corollary of [LLPS19, Lemma 2.1].
Note that, in this example, SD(w) = P(w), demonstrating Theorem A. Also, in this example, sh SD(w) = (3, 2, 2) is smaller than sh P(w) = (3, 3, 1) in the dominance partial order.Corollary 2.8.If w ∈ S n , then the BBS soliton partition of w is smaller or equal to the RS partition of w in the dominance partial order.

Fukuda's carrier algorithm
In this section, we review the carrier algorithm and the fact that the RS insertion tableau is an invariant of a box-ball system (BBS).

Carrier algorithm
The carrier algorithm is a way to describe a BBS move as a sequence of local operations of inserting and bumping numbers in and out of a carrier filled with a weakly increasing string.A version of the carrier algorithm was first introduced in [TM97], and the version of the carrier algorithm we use in this paper comes from [Fuk04, Section 3.3].Given a BBS state at time t, the carrier algorithm is used to calculate the state at time t + 1.We describe the process in Algorithm 1.Note that, after each insertion and ejection step, the sequence in the carrier is weakly increasing.
1: begin carrier algorithm 2: Set e := n + 1, so that e is considered to be larger than any ball

The RS insertion tableau is an invariant of a box-ball system
Remark 3.2 ([Fuk04, Remark 4]).The carrier algorithm can be viewed as a sequence of Knuth moves.Consider the insertion of p into the carrier.Note that, since our carrier can carry n elements, if p = e, then the carrier must contain a number (possibly e) greater than p.If p = e, then no number in the carrier is greater than p.First, suppose p = e, and let C p denote the smallest element in the carrier which is greater than p.
(i) If C p is the smallest element in the carrier, then the insertion process is equivalent to applying a sequence of K − 1 moves (ii) If C p is the largest element in the carrier, then the insertion process is equivalent to applying a sequence of K + 2 moves (iii) If C p is neither the smallest nor the largest element in the carrier, then the insertion process is equivalent to applying a sequence of K − 1 moves followed by a sequence of K + 2 moves Next, suppose p = e.Then p is greater than or equal to every element in the carrier, and the insertion process is equivalent to applying the trivial transformation  Proof.Let r be the row reading word of SD(w).By definition of the soliton decomposition tableau, we know that r is the order in which the balls of w are configured once we reach a steady state.Therefore, r is a state in the box-ball system containing w. Theorem 3.3 tells us that the RS insertion tableau is preserved under a sequence of box-ball moves, so P(w) = P(r).
In Example 3.4, the soliton decomposition coincides with the RS insertion tableau of the box-ball system, but in Example 3.6 these two tableaux do not coincide.In the next section we discuss when SD(w) = P(w).

When the soliton decomposition and the RS insertion tableau coincide
In this section, we will prove Theorem 4.2.One direction of our proof uses the following lemma, which was communicated to us by Darij Grinberg.Lemma 4.1.Suppose S is a row-strict tableau, that is, every row is increasing (with no restrictions on the columns).Let r be the row reading word of S. If sh S = sh P(r), then S is standard, that is, every column of S is increasing.
Proof.Suppose S is not standard.Then S has two adjacent entries in a column which are out of order.Indexing our rows from top to bottom and our columns from left to right, this means there is a column (say, column c) for which the entry in some row k is bigger than the entry immediately below it.Let y be the entry in the k-th row, c-th column of S, and let x be the entry immediately below it (in the k + 1-th row, c-th column of S).
Since r is the row reading word of S and since each row of S is increasing, we can construct a list of k disjoint increasing subsequences of r: The first k − 1 increasing subsequences of r are the first k − 1 rows of S. The k-th increasing subsequence starts in row k + 1, column 1 of S, moving along the same row until we get to column c (with entry x), then going up to row k above (which has entry y), then continuing to the end of row k.
The length of the k-th increasing subsequence is larger (by 1) than the length of the k-th row of S. So the total number of letters in our list of k disjoint increasing subsequences of r is larger by 1 than the total length of the first k rows of S. Thus, Greene's theorem (Theorem 2.2) says that the total length of the first k rows of the RS insertion tableau P(r) of r is larger (at least by 1) than the total length of the first k rows of S. Therefore, the shape of S is not equal to the shape of P(r).
The following theorem gives a characterization of permutations whose soliton decompositions are equal to their RS insertion tableaux.Theorem 4.2.Let w be a permutation.Then the following are equivalent: 1. SD(w) = P(w).
3. The shape of SD(w) equals the shape of P(w).
Let r be the row reading word of SD(w).By Corollary 3.5, we have First, we show that (2) implies (1).Suppose that SD(w) is a standard tableau T .Since r is the row reading word of T , we have P(r) = T by (1.1).Combining this equality with (4.1), we get P(w) = P(r) = T = SD(w).
Next, we show that (3) implies (2).Let S denote SD(w), and note that SD(w) is a rowstrict tableau by construction.Suppose sh S = sh P(w).Since P(w) = P(r) by (4.1), we have sh P(w) = sh P(r), so sh S = sh P(w) = sh P(r).Because S is a row-strict tableau, the permutation r is the row reading word of S, and sh S = sh P(r), Lemma 4.1 tells us that S is standard.
Corollary 4.3.Let w be a permutation.Then the following five statements are equivalent: 1. SD(w) = P(w).

SD(w) is a standard tableau.
3. The shape of SD(w) equals the shape of P(w).By Greene's theorem (Theorem 2.2), the shape of P(w) is (i 1 , i 2 − i 1 , i 3 − i 2 , . . . ) and the shape of the conjugate of

For all k 1, we have
Combining these facts, we conclude that sh SD(w) = sh P(w) if and only if

Reading words and steady states
We study the steady-state configurations of a box-ball system.The main result of this section (Proposition 5.2) is a corollary of [LLPS19, Proof of Lemma 2.1 and 2.3].

Reading words of standard tableaux
The permutations which reach their steady state at time 0 are precisely the row reading words of standard tableaux.Proposition 5.1.A permutation r has steady-state time 0 if and only if r is the row reading word of a standard tableau.
In particular, if r is the row reading word of a standard tableau T , then T = SD(r).In the next section, the standard tableau in Proposition 5.1 is generalized to standard skew tableaux whose rows are weakly decreasing in length.

Reading words of standard skew tableaux
A BBS state can be represented as a configuration array containing the integers from 1 to n as follows: scanning the boxes from right to left, each increasing run (maximal consecutive increasing string of balls) becomes a row in the array.A string of g empty boxes indicates that the next row below should be shifted g spaces to the left.Note that this array has increasing rows but not necessarily increasing columns; it may be disconnected and it may not have a valid skew shape.

Proposition 5.2. A BBS configuration is in steady state if and only if its configuration array is a standard (possibly disconnected) skew tableau whose rows are weakly decreasing in length.
We will give a proof in Section 5.In this box-ball system, all configurations at time t 1 are in steady state.
Example 5.4.The following is an example of a non-steady-state BBS configuration and its configuration array.Note that the configuration array is a standard skew tableau but its rows are not weakly decreasing in length.

Separation condition
A 'separation condition' for steady state is given in statement (43) in [LLPS19].In Lemmas 5.5 and 5.6, we reframe this characterization for steady state in terms of our version of the box-ball system.Proposition 5.2 follows directly from these two lemmas.

Lemma 5.5 (Separation condition).
Let a BBS configuration be in steady state.Suppose two adjacent solitons L (the left soliton with length ℓ) and R (the right soliton) are separated by g empty boxes, where g < ℓ.Then, for i = 1, 2, . . ., ℓ − g, the i-th smallest ball of the right soliton R is smaller than the (i + g)-th smallest ball of the left soliton L.
Proof.We apply one BBS move to the configuration via the carrier algorithm.Suppose Since we started with a steady-state configuration, the left soliton L must stay intact at the end of the carrier algorithm.So, for each i = 1, . . ., ℓ − g, as we insert R i , we must eject L g+i , and get e . . .e L 1 . .If we have a third soliton located to the right of R, we would be in the same situation as (5.1).We then repeat the same process for the rest of the solitons and arrive at the same conclusion.
Lemma 5.6 (Sufficient condition for steady state).Suppose a BBS configuration w satisfies the following.
1.The configuration array of w has rows of weakly decreasing length.
2. The configuration array of w is standard; that is, if two adjacent maximal consecutive increasing blocks L (the left block with length ℓ) and R (the right block) of w are separated by g empty boxes such that g < ℓ, then, for i = 1, 2, . . ., ℓ − g, the i-th ball of the right block R is smaller than the (i + g)-th ball of the left block L.
Then w is in steady state.
Proof.Suppose w is the configuration at time t.We apply the carrier algorithm to get the configuration at time t + 1. Next, we insert the g copies of e into the carrier and eject L 1 , . . ., L g .There are two cases: either (a) g ℓ or (b) g < ℓ.
(a) First, suppose that g ℓ.Then all of L 1 , . . ., L ℓ are ejected and the carrier is now empty: We proceed by inserting R 1 , • • • , R r into the carrier.We have ℓ r by assumption part (1) and R i < L g+i for i = 1, 2, . . ., l − g by assumption part (2).Therefore, as we insert R 1 , . . ., R ℓ−g , we must eject L g+1 , . . ., L ℓ , and we get In both cases, at time t+1 there are at least r−ℓ+g empty boxes to the right of L. Since ℓ r, we have g r − ℓ + g, so there are at least as many empty boxes to the right of L as at time t.Furthermore, the increasing run L stays together.
If we have a third increasing run S = S 1 . . .S s to the right of R (with a gap of g ′ empty boxes), we would be in the same situation as (5.2).After inserting the elements of S into the carrier, we would have At the end of the carrier algorithm, the increasing runs stay together, their order stays the same, and the gap of empty boxes between each pair of adjacent sequences is at least as large as at time t.The new configuration satisfies both part ( 1) and ( 2) of the assumption.By induction, subsequent carrier algorithm applications leave the order of the increasing runs unchanged, so these increasing runs are in fact solitons.
By the two lemmas above, we have Proposition 5.2: a box-ball configuration is in steady state if and only if (1) its configuration array has rows of weakly decreasing length and ( 2) each column of the configuration array is increasing.

A recording tableau giving n-3 steady-state time
In this section, we prove Theorem 6.7, which states that all permutations in S n with a certain recording tableau have box-ball steady-state time n−3.We conjecture that all other permutations in S n have steady-state time smaller than n − 3 (Conjecture 1.1).Theorem 6.7 turns out to be a special case of a general phenomenon, which is proven in a sequel to this paper [CFG + 22]: if two permutations have the same recording tableau, then they have the same BBS steady-state time (Conjecture 1.2).

A recording tableau giving n-3 steady-state time
Let S n ( Q) be the set of permutations w ∈ S n such that its recording tableau Q(w) is equal to Q.The rest of this section is devoted to proving Theorem 6.7, which states that every permutation in S n ( Q) has steady-state time n − 3.

Lemmas for Theorem 6.7
Lemma 6.4.Let n 5, and suppose w ∈ S n ( Q).Then w is not the union of two increasing subsequences.
Proof.The recording tableau of w is equal to Q, which has height 3. Therefore, the RS partition of w has three parts.By Greene's theorem (Theorem 2.2), w is not the union of two increasing subsequences.Lemma 6.5.Let n 5, and suppose w = w 1 w 2 . . .w n ∈ S n ( Q).Then w satisfies the following.
Proof.Since w ∈ S n ( Q), the recording tableau of w is equal to Q.We will use the inverse RS algorithm 1 to construct w.Let P = P(w) and Q = Q(w).Denote the entries in the top row of P by a 1 , a 2 , . . ., a q (where q = n − 3), the second row of P by b 1 and b 2 , and the entry in the third row of P by c 1 .Hence, the starting pair P and Q is After the first step in the inverse RS algorithm, we get the pair of tableaux .
We now pause to observe two facts that will be referenced at the end of this proof.First, note that P n−1 is standard by definition of the inverse RS algorithm.Thus, α 1 , α 2 , . . ., α q is increasing. (6.1) Second, we note that a x < β 2 , (6.2) as we now explain.Recall that w n = a x , so, using the original RS algorithm, we insert a x into P n−1 to get P .Since row 1 of P n−1 and row 1 of P have the same size, we know that a x bumps a number in row 1 of P n−1 to row 2. Let a i denote the smallest entry in row 1 of P n−1 which is greater than a x .
The RS algorithm replaces a i with a x and bumps a i to row 2. Since row 2 of P n−1 and row 2 of P have the same size, we know that a i bumps a number in row 2 of P n−1 .So a i must be smaller than β 2 .Since a x < a i , we have a x < β 2 .This concludes our explanation for (6.2).We also note that ) α 1 < β 1 , and (6.4) since P n−1 is standard.We will reference these inequalities at the end of this proof.
The new pair of tableaux is .
Note that 4 is the bottom right corner of Q 4 .Since α 2 < β 2 by (6.5), we know that α 2 is the largest element in row 1 of P which is smaller than β 2 .So w 4 = α 2 , and the last n − 3 letters of w are α 2 , α 3 , α 4 , . . ., α q , α q−1 , a x .The new pair of tableaux is Note that 3 is in the second row of Q 3 .We know from (6.3) that β 2 is larger than β 1 , so α 1 is the largest element in row 1 smaller than β 1 .Thus, w 3 = α 1 .So the last n − 2 letters of w are α 1 , α 2 , α 3 , α 4 , . . ., α q , α q−1 , a x .The new pair of tableaux is We then remove β 2 and β 1 from P 2 , in that order.Therefore, We now have all the necessary information to prove all parts of the lemma. Proof.

Proof of
Let j be the smallest number in {3, 4, . . ., n − 1} such that w n < w j .We claim that the box-ball configuration at time t = 1 is where x = w j , there are (n − 5) copies of e between w 2 and x, and To prove this claim, consider the following cases.Due to Lemma 6.6, these five cases cover all possibilities.First, suppose w n = 1.Lemma 6.5 tells us that w 3 is smaller than each w i except for w n = 1, so we must have w 3 = 2 and j = 3: where there are (n − 5) copies of e between w 2 and x = w 4 .In this case, w 3 = 1 is not bigger than w n = 2, but w 4 must be bigger than w n = 2 since w 4 / ∈ {1, 2}, so j = 4. Third, suppose w 3 = 1 and w 1 = 2 and w n = 3. Lemma 6.5 tells us that w 4 is smaller than each of the w i (except for w 3 = 1, w 1 = 2, and w n = 3), so w 4 must be 4: where there are (n − 5) copies of e between w 2 and x = w 4 .In this case, j = 4 since w 3 = 1 is not larger than w n = 3 but w 4 = 4 is.
Finally, suppose we have one of the last two cases, so w 3 = 1 and w 4 < w n : where there are (n − 5) e's between w 2 and x = w j .In this case, j 5 since w 4 is smaller than w n .This concludes the proof of our claim that the box-ball configuration at time t = 1 is as given in (6.In fact, at each BBS move, the increasing sequence w 1 , w 2 moves together two spaces to the right, the singleton x moves one space to the right, and the increasing sequence 1, y 1 , y 2 . . ., y n−4 moves n − 3 spaces to the right.So the number of e's between w We claim that x < w 2 , which we now prove.Recall that x = w j , where j is the smallest number in {3, 4, . . ., n − 1} such that w n < w j .If w 2 < w j , then w 1 < w 2 < w j < w j+1 < • • • < w n−1 and the remaining w i 's form two increasing subsequences of w whose union is w.This contradicts Lemma 6.4, so indeed x < w 2 .Since x < w 2 , we have either x < w 1 < w 2 or w 1 < x < w 2 .If x < w 1 < w 2 , then the configuration at t = n − 3 is Either way, the configuration array at t = n − 3 is a standard skew tableau whose rows have length n − 3, 2, and 1.By Proposition 5.2, the configuration at t = n − 3 is in steady state.
The configuration at t = n − 4 is not yet in steady-state, as the relative positions of w 1 , w 2 , and x in the configuration at t = n − 4 differ from the configuration at t = n − 3. Therefore, t = n − 3 is the minimum steady-state time of w.

Knuth moves
We study how types of Knuth moves (Definition 1.4) play a role in a box-ball system.In Section 7.1, we prove that a non-K B Knuth move preserves the shape of a soliton decomposition and that a K B move changes it (Theorem 7.1).In Section 7.2, we prove that every permutation which is one non-K B Knuth move from a row reading word has steady-state time 1 (Theorem 7.4).

Soliton decompositions are preserved by certain Knuth moves
Using the localized version of Greene's Theorem given in Section 2.2, we prove a partial characterization of the shape of SD in terms of types of Knuth moves.Proof.To prove part (1), we observe that a K + B move decreases the number of descents by 1, and a K − B move increases the number of descents by 1.Since the height the partition sh SD(w) is equal to D 1 (w) = 1 + |{descents of w}| by Lemma 2.6, it follows that applying an odd number of K B moves to w changes sh SD(w).
To prove part (2), suppose x, y ∈ S n are related by a proper K 1 or proper K 2 move.Due to Lemma 2.6, it suffices to prove that D k (x) = D k (y) for all k.This breaks down into two main cases: case (i), where y = K + 1 (x), and case (ii), where y = K + 2 (x).These further divide into the following subcases, where a < b < c in all cases: The proofs are similar for each case.We include a partial proof of case (ia).Suppose  In Figure 7.1, the permutation 362154 is one K − B move from r, and its steady-state time is t = 2. Another permutation, 326514, is also one K − B move from r, and its steady-state time is t = 1.
In Figure 7.2, we can perform a K − B move and also a proper K + 1 move on r (see Lemma 7.6).The permutation 635214 is one proper K + 1 move from r, and its steady-state time is t = 1, illustrating Lemma 7.9.r = 362514, t = 0   where The tableau P must be of the form given in Figure 7. 3, where the entry y is in its own row, and the row immediately above y starts with entries x, z.Proof.First, we prove part (1) of the lemma.Suppose we perform a K − 1 move yzx → yxz (where x < y < z) on r.Since r is the row reading word of P , the tableau P must contain a subtableau x b y z or x . . .b . . .y z .

If one performs a
Since the rows and columns of P are increasing, we must have x < b < z.Thus, r must contain a consecutive subsequence yzxb ′ where x < b ′ b < z, so the K − 1 move yzx → yxz is K − B .Now suppose we perform a K + 1 move yxz → yzx on r.First, we prove part (2).Since x < y < z and P is standard, the entry y must be the only element in its row in P , that is, the rows of P containing x, y, z are of the form x z . . .y If r 1 = y, then we are done.Suppose r 1 = y, and write r = r 1 r 2 . . .r ℓ yxz . . .r n .Since the rows of P are weakly decreasing in length, the rows of P below y are of size 1.Since P is standard, we have r 1 > r 2 > • • • > r ℓ > y.So r is of the form given in (7.1) and P is of the form given in Figure 7.3.
Finally, to prove part (3) of the lemma, we prove that this K + 1 move is not a K B move.If r n = z, then we know this K + 1 move is not K B .Suppose r n = z, so r 1 . . .yxzb . . .r n for some b.Since r is the row reading word of P , either the entry b is immediately above x in P or the entry b is immediately to the right of z in P :  We apply the carrier algorithm to w.First, we insert r 1 , r 2 , . . ., r ℓ , y into the carrier.Since these are decreasing, we eject e, r 1 , r 2 , . . ., r ℓ from the carrier in consecutive order: The configuration array at t = 1 is the skew tableau created by taking P and shifting some of the rows to the right.Since P is standard tableau with partition shape to begin with, the configuration array is a standard skew tableau with weakly increasing rows.By Proposition 5.2, the configuration at t = 1 is in steady state.
Lemma 7.10.Suppose r = r 1 r 2 . . .r n ∈ S n is the row reading word of a standard tableau P .Let w be a permutation which differs from r by one proper K 2 move.Then w first reaches its steady state at t = 1.
As in the proof of Lemma 7.9, we apply the carrier algorithm to w.We insert the decreasing sequence r 1 , r 2 , . . ., r ℓ , x into the carrier and eject e, r 1 , r 2 , . . ., r ℓ , in that order.As we insert z and y, we eject e and z, in that order: The configuration array at t = 1 is the skew tableau created by taking P and shifting some of the rows to the right.Since P is standard tableau with partition shape to begin with, the configuration array is a standard skew tableau with weakly increasing rows.By Proposition 5.2, the configuration at t = 1 is in steady state.

Conjecture 1 . 2 .
If two permutations v and w are such that Q(v) = Q(w), then v and w have the same steady-state time.Conjecture 1.2 is proven in a sequel to this paper [CFG + 22].Remark 1.3.Conjecture 1.2 would simplify the proof of Theorem D; it would simply require demonstrating that one single permutation whose recording tableau is Q has steady-state time n − 3.

Definition 1 . 4 (
Knuth Moves).Suppose v, w ∈ S n and x < y < z.1.We say that v and w differ by a Knuth relation of the first kind(K 1 ) if v = v 1 . ..yxz . . .v n and w = v 1 . . .yzx . . .v n or vice versa.2. We say that v and w differ by a Knuth relation of the second kind (K 2 ) if v = v 1 . . .xzy . . .v n and w = v 1 . . .zxy . . .v n or vice versa.

3 : 4 : 5 : 6 :◮◮◮ 3 . 1 .
Set B := the configuration of the BBS at time t, where each empty box is replaced with an e and the first (leftmost) element of B is the integer in the first (leftmost) nonempty box in the configuration and the last (rightmost) element of B is the integer in the last (rightmost) nonempty box of the configuration Let ℓ denote the number of elements (including the e's) of B Fill the "carrier" C -depicted -with n copies of e Write B to the right of C 7: begin Process 1: insertion process 8: for all i in {1, 2, . . ., ℓ} do 9: Set p to be the i th leftmost element of B 10: begin element ejection process 11: if an element in C is larger than p then 12: Set s := smallest element in C larger than p.If s = e, pick the leftmost e 13: Eject s from C and put it immediately to the left of C s := the leftmost (smallest) element in C 17: Eject s from C and put it immediately to the left of C 18: Note: There are now n − 1 elements in C 19: Shift each element of C to the left by one 20: Place p in the rightmost location in C 21: Note: There are now n elements in C Note: The current elements to the left of C correspond to the t + 1 state of the BBS 33: end carrier algorithm Example We compute the configuration at time t = 3 of the box-ball system from Figure 1.2 by applying the carrier algorithm to the configuration at time t = 2. Following Algorithm 1, we set B := 452ee136.The carrier algorithm then proceeds as follows.The elements ee425eee136 to the left of C correspond to the configuration at time t = 3 given in Figure1.2.

5 .
For all k 1, we have D k (w) = d k (w).The symbols I k and D k are the statistics from localized Greene's theorem (Section 2.2) and i k and d k are the statistics from Greene's theorem (Section 2.1).Proof.For short, we write i k := i k (w), I k := I k (w), d k := d k (w), and D k := D k (w).By localized Greene's theorem (Lemma 2.6), the shape of SD(w) is (I 1 , I 2 − I 1 , I 3 − I 2 , . . . ) and the shape of the conjugate of SD(w) is (D 1 , D 2 − D 1 , D 3 − D 2 , . . .).

3 . 5 . 3 .e e e e e e e e 5 6 e 2 7 e e e 1
Example Let w = 5623714, the example we use in Section 2. The following are the box-ball system states from time t = 0 to t = 4 and their configuration arrays.e e e 5 6 e 2 7 e e 1 3 4 e . . .
So we must have R i < L g+i for i = 1, 2, . . ., ℓ − g, as needed.After we insert the rest of the elements of R into the carrier, we have e . . .e L 1 . . .L ℓ r−ℓ+g copies ee . . .e R 1 • • • R r ee • • • e . . .

3 4 n
Since P is standard, we know that b 1 < c 1 .The other entry b 2 in the second row is larger thanb 1 .If b 2 < c 1 , let b y equal b 2 .Otherwise, let b y be b 1 .In other words, we let b y denote the largest element in the second row which is smaller than c 1 .Similarly, let a x denote the largest element in the first row which is smaller than b y .The first step of the inverse RS algorithm tells us that w n = a x .

w 5 3 .e w 1 2 w 2 n− 5 4 x 1 w
. . .w n−1 w n Using the same reasoning as in the previous two cases, applying one box-ball move to w results in the configuration e copies e e e . . .e w n w 5 w 6 . . .w n−1
2 and x decreases by 1 after each BBS move.The configuration at t = n − 4 is . . .e e e w 1 w 2 x e e e . . .e e e 1 y 1 y 2 . . .y n−4 increasing block .

Theorem 7. 1 .
Suppose v and w are two permutations in the same Knuth equivalence class.1.If v and w are related by a sequence of Knuth moves containing an odd number of K B moves, then SD(v) = SD(w).

2 .
If v and w are related by a sequence of non-K B Knuth moves, then sh SD(v) = sh SD(w).

Theorem 7. 1 2 .Example 7 . 3 .
allow us to use Knuth moves to find a subset of permutations whose soliton decomposition and RS insertion tableau coincide.Corollary 7.2 (Corollary of Theorem 4.2 and Theorem 7.1).Let w ∈ S n , let T = P(w), and let r be the row reading word of T .1.If w is related to r by a sequence of Knuth moves containing an odd number of K B moves, then SD(w) = P(w) = T .If w is related to r by a sequence of non-K B moves, then SD(w) = P(w) = T .The permutations 362514 and 632514 are the reading words of the tableaux

Figure 7 . 1 (7. 2 .. 4 . 1 .Example 7 . 5 .
Figure 7.1 (respectively, Figure 7.2) shows all permutations in the Knuth equivalence class of r = 362514 (respectively, r = 632514).The corresponding soliton decomposition is drawn next to each permutation.An edge with label K 1 (respectively, K 2 ) indicates that the Knuth move is a proper K 1 (respectively, K 2 ) move.An edge with label K B indicates that the Knuth move is both K 1 and K 2 .The permutations are arranged such that they form a subdiagram of the Hasse diagram of the right weak order 2 on the symmetric group S 6 .

Figure 7 . 2 :
Figure 7.2: The Knuth equivalence class of r = 632514, with their soliton decompositions and steady-state times.

7. 2 . 1
Proof of Theorem 7.4 Theorem 7.4 follows from the following four lemmas.Lemma 7.6.Let r = r 1 r 2 . . .r n be the row reading word of a standard tableau P .1.If one performs a K − 1 move on r, the move is K B . 2. Suppose we are able to perform a K + 1 move yxz → yzx (where x < y < z) on r.If r 1 = y, we must have r = r 1 . . .r ℓ y x decreasing z . . .r n−1 r n (7.1)

r 1 Figure 7 . 3 :
Figure 7.3: General form of a standard tableau P whose row reading word can undergo a K + 1

y 2 .
Since P is standard, either b < x or z < b.Either way, this K + 1 move is not K B .Lemma 7.7.Let r = r 1 r 2 . . .r n be the row reading word of a standard tableau P .1.It is impossible to perform a K + 2 move on r.Suppose we are able to perform a K − 2 move zxy → xzy (where x < y < z) which is not a K B move on r.If r 1 = z, we haver = r 1 . . .r ℓ z x decreasing y . . .r n−1 r n (7.2)wherer 1 > r 2 > • • • > r ℓ > z.The tableau P must be of the form given in Figure7.4,where the entry z is in its own row, and the row immediately above z starts with entries x, y. . . .• • • a 1 a 2 a 3 . . .x y b 1 b 2 b 3 • • • b m z (possibly with no b i 's) r ℓ . . .

r 2 r 1 .Figure 7 . 4 :
Figure 7.4: General form of a standard tableau P whose row reading word can undergo a K − 2

e e • • • e carrier r 1
r 2 r 3 . . .r ℓ−1 r ℓ y decreasing z . . .r n e r 1 e e • • • e r 2 r 3 . . .r ℓ−1 r ℓ y decreasing z x . . .r n e r 1 r 2 e e • • • e r 3 . . .r ℓ−1 r ℓ y decreasing z x . . .r n . . .e r 1 r 2 . . .r ℓ y e e • • • e z x . . .r n Next, we insert z into the carrier.Since the only non-e entry in the carrier, y, is smaller than z, we eject an e: e e r 1 r 2 . . .r ℓ e y z e e • • • e x r ℓ+4 . . .r n Next, we insert x into the carrier.Since x < y < z, we eject y and get e e r 1 r 2 . . .r ℓ e y x z e e • • • e r ℓ+4 . . .r n Note that the string x z r ℓ+4 . . .r n−1 r n is equal to the consecutive subsequence r ℓ+2 . . .r n−1 r n of r.This string is the row reading word of the subtableau (possibly with no b i 's) . . .• • • a 1 a 2 a 3 • • • x z b 1 b 2 • • • of P , where P is given in Figure 7.3.Since this subtableau has the shape of a partition and has increasing rows and columns, completing the carrier algorithm yields the configuration at time t = 1: e e r 1 r 2 . . .r ℓ e y 0 or more copies e e . . .e x z b 1 b 2 b 3 . . .a 1 a 2 . . . . . .r n−1 r n e e • • • e .
The RS insertion tableau is a conserved quantity under the time evolution of the BBS, i.e., the RS insertion tableau is preserved under each BBS move.More precisely, let B t be the state of a box-ball system at time t.Let B ′ t be the permutation created from B t by removing all e's.Then P(B ′ t ) is identical for all t.
First, we simply insert L 1 , . . ., L ℓ into the carrier.Since L is increasing, each time we insert a ball of L, we eject a copy of e.We get Next, we insert the g copies of e into the carrier and eject L 1 , . . ., L r are the two leftmost solitons.Our initial setup with n copies of e in the carrier is ee • • • e carrier L 1 . . .L ℓ g copies e . . .e R 1 . . .R r . . .g : e . . .e L 1 . . .L g first g balls ℓ − g balls r are the two leftmost increasing runs (maximal consecutive increasing blocks of balls).First, we insert each of L 1 , . . ., L ℓ into the carrier and eject an e each time.We get [OEI,le 6.2.For n = 5, the five permutations of S n ( Q) are the following.Note that one of our running examples, 452361, is in S 6 ( Q).As illustrated in Figure1.2, its steady-state time is 3 = 6 − 3. Remark 6.3.It follows from Definition 6.1 that the RS algorithm induces a bijection from S n ( Q) onto the set of standard tableaux of shape (n−3, 2, 1), so S n ( Q) is counted by the sequence[OEI,  A077415].