1 Introduction
Wilf equivalence, and the closely related notion of Wilf order, are undoubtedly among the key notions in the study of combinatorics of permutations. Despite considerable amount of research, many key questions related to these concepts remain open.
One of the most successful approaches in the study of Wilf equivalence is based on the concept of pattern avoidance in fillings of specific shapes. Of particular interest are the socalled transversal fillings, which contain exactly one 1cell in each row and each column, and can be viewed as generalizations of permutation matrices. Transversal fillings of Ferrers shapes have played a crucial part in establishing most of the known results on Wilf equivalence [1, 16], as well as in the study of many related combinatorial objects, such as involutions [2], words [7], matchings [9, 12], graphs [5, 6] or set partitions [4, 11].
Most of the research into patternavoiding fillings focuses on the avoidance of diagonal patterns. We let and denote the decreasing and increasing diagonal pattern of size , respectively. Backelin, West and Xin [1] have shown that the number of transversal fillings of a Ferrers shape that avoid is the same as the number of such fillings that avoid . This was later generalized to nontransversal fillings [13], and to more general shapes, such as stack polyominoes [8], moon polyominoes [15], or almostmoon polyominoes [14].
In this paper, we consider yet another family of shapes, namely the socalled skew shapes, which can be seen as a difference of two Ferrers shapes, or equivalently, as a shape whose boundary is a union of two nondecreasing lattice paths with common endpoints. Our interest in skew shapes is motivated by the following conjecture.
Conjecture 1.1 (Skew shape conjecture).
For any skew shape and any number , the number of transversal fillings of that avoid is greater than or equal to the number of such fillings that avoid .
Verifying the skew shape conjecture for any particular value of would imply that for any two permutations and , the pattern is more Wilfrestrictive than , where denotes the skewsum of permutations (see, e.g., Vatter’s survey [17] for permutationrelated terminology). However, the conjecture has so far only been verified for [3, 10].
In this paper, we prove two new results on diagonal avoiding fillings of skew shapes, both of which indirectly support the skew shape conjecture. As our first main result, we will show that the conjecture holds for skew shapes that avoid the shape from Figure 1 as subshape. In fact, we will show that for avoiding skew shapes the conjecture holds with equality, via a bijection that extends to general integer fillings. Since a Ferrers shape is a special case of a avoiding skew shape, this result can be seen as another generalization of the result of Backelin et al. [1] on transversals of Ferrers shapes and its generalization by Krattenthaler [13] to nontransversal fillings.
As our second main result, we show that for any skew shape , the number of 01fillings of avoiding the pattern is the same as the number of 01fillings of that avoid both and the pattern from Figure 1. This has been previously known for transversal fillings, where the proof follows from the fact that each skew shape that admit at least one transversal filling admits exactly one avoiding filling and exactly one avoiding filling. To extend the result to general, i.e. nontransversal, fillings, we devise a new bijective argument.
2 Preliminaries
Shapes.
A cell is a unit square whose vertices are points of the integer grid, and a shape is the union of finitely many cells. The coordinates of a cell are the Cartesian coordinates of its topright corner. If two shapes differ only by a translation in the plane, we shall treat them as identical. We will henceforth assume that in each nonempty shape we consider, the leftmost cell has horizontal coordinate equal to 1, and the bottommost cell has vertical coordinate equal to 1, unless explicitly stated otherwise. We use the term cell for the cell in column and row . The height of a shape is the vertical coordinate of its topmost cell, and its width is the horizontal coordinate of its rightmost cell. We say that a cell is strictly to the left of a cell if , and we say that it is weakly to the left if .
Two cells are adjacent if their boundaries share an edge. For a shape , the transitive closure of the adjacency relation on the cells of yields an equivalence, whose blocks are the connected components of . If has a single component, we say that it is connected. Connected shapes are known as polyominoes.
A shape is convex if for any two of its cells in the same row or column it also contains every cell in between. Two columns of a shape are comparable if the set of row coordinates of one column is contained in the set of row coordinates of the other; comparable rows are defined analogously. We say that a shape is intersectionfree if any two of its columns are comparable. Note that this is equivalent to having any two of its rows comparable. Convex, intersectionfree shapes are known as moon polyominoes (see Figure 2).
A shape is topjustified if the top ends of all of its columns are in the same row and it is leftjustified if the left ends of all of its rows are in the same column. Bottomjustified and rightjustified shapes are defined analogously. A shape is topleft justified if it is both topjustified and leftjustified, and similarly for other combinations of horizontal and vertical directions.
A northwest Ferrers shape, or NW Ferrers shape for short, is a topleft justified moon polyomino, and similarly, a SE Ferrers shape is a bottomright justified moon polyomino; see Figure 3.
A skew shape is a shape that can be obtained by taking two NW Ferrers shapes and with a common topleft cell and putting , i.e., is the union of those cells of that do not belong to .
Fillings of shapes.
A filling of a shape is a function that assigns a nonnegative integer to every cell of . We write for the value assigned by to a cell . A binary filling (also known as 01filling) is a filling that only uses the values 0 and 1. A cell that has been assigned the value 0 (or 1) by a binary filling is referred to as a 0cell (or 1cell, respectively) of . If a row or column only contains 0cells in a filling , we say that it is empty in .
A binary filling of a shape is sparse if every row and every column of has at most one 1cell. A binary filling is a transversal filling (or just a transversal) if every row and every column of has exactly one 1cell.
In figures, we will often represent 0cells by empty boxes and 1cells by boxes with crosses; see Figure 5 for examples.
Containment of shapes and fillings.
Let be a shape of height and width , and let be a shape of height and width . We say that contains , or that is a subshape of , if there is a sequence of row indices and column indices , such that for every and , the cell is contained in if and only if the cell is in . The rows and columns together form an occurrence of in . If does not contain , we say that avoids .
With and as above, if is a filling of and a filling of , we say that contains , or that is a subfilling of , if there is an occurrence of in in rows and columns with the additional property that for every cell of , we have . We again say that the rows and columns form an occurrence of in . In the case when is binary, the condition simply says that each 1cell of must get mapped to a nonzero cell of by the occurrence.
Note that any subshape of a moon polyomino is again a moon polyomino, any subshape of a skew shape is a skew shape, and any subshape of a Ferrers shape is a Ferrers shape.
Let be the square shape of size , i.e, the shape consisting of the cells with and . The transversal fillings of correspond in a natural way to permutations of order : let be a permutation of order , represented by a sequence of numbers in which every value from 1 to appears exactly once. We represent by a transversal filling of whose 1cells are the cells for . Such a transversal filling is usually called the permutation diagram of . In this paper, we shall make no explicit distinction between a permutation and its diagram, and if there is no risk of ambiguity, we will use the term permutation to refer both to the sequence of integers and to the diagram. Note that the containment relation defined above for binary fillings is a generalization of the classical Wilf containment of permutations.
In this paper, we shall be mostly working with two types of forbidden patterns, corresponding to increasing and decreasing permutations, respectively. For any let denote the diagram of the increasing permutation ; in other words, is the filling of the square shape with 1cells for . Symmetrically, let be the diagram of the decreasing permutation . An occurrence of in a filling is also referred to as a NEchain of length in , while an occurrence of is a SEchain of length . See Figure 6 for examples.
Results.
Our first main result deals with transversal fillings of skew shapes of a special type. Let be the skew shape in Figure 1 ( stands for ‘dented shape’). We say that a skew shape is free if avoids .
Let be the number of transversal fillings of the shape that avoid the filling . Backelin et al. [1] have shown that for any and any Ferrers shape , we have the identity . This was later generalized by Krattenthaler [13] to general fillings of Ferrers shapes. Our first main result generalizes this identity to a broader class of shapes, namely to free skew shapes.
Theorem 2.1.
For any free skew shape , there is a bijection transforming a filling of to a filling of with these properties.

For any , avoids if and only if avoids .

There is a permutation of rowindices of and a permutation of columnindices of such that for any , the entries in the th row of have the same sum as the entries in the th row of , and the entries in the th column of have the same sum as the entries in the th row of .
The proof of this result is presented in Section 3. A direct consequence of the theorem is the following identity for transversals.
Corollary 2.2.
For any free skew shape and any , we have .
Theorem 2.1 cannot in general be extended to nonfree skew shapes. In fact, for the shape itself, we easily verify that and . However, computational evidence suggests the following conjecture.
Conjecture 2.3.
[Skew shape conjecture] For any skew shape and any , we have .
The conjecture is trivial for and is known to be true for . Indeed, for the case , a more precise result is known, proven by Jelínek [10, Lemmas 29 and 30] and independently by Burstein and Pantone [3, Lemmas 1.4 and 1.5].
Theorem 2.4 ([10, 3]).
For any skew shape that admits at least one transversal, there is exactly one transversal of that avoids , and exactly one transversal of that simultaneously avoids both and the filling from Figure 1.
Note that this theorem implies not only the case of the skew shape conjecture, but also the case of Corollary 2.2.
As our second main result, we will extend Theorem 2.4 to an identity for general (i.e., not necessarily transversal) binary fillings.
Theorem 2.5.
For any skew shape , the number of binary fillings of that avoid is the same as the number of binary fillings of that avoid both and . Moreover, this identity is witnessed by a bijection that preserves the number of 1cells in every row.
3 Proof of Theorem 2.1
3.1 The structure of free skew shapes
An important feature of free skew shapes is that they admit a natural decomposition into a concatenation of Ferrers diagrams. Before describing this decomposition properly, we need some preparation.
Let be a shape, and let be a cell of . We let denote the subshape of formed by those cells of that satisfy and . We also use the analogous notations , , with obvious meanings.
Lemma 3.1.
A skew shape is free if and only if for every cell of at least one of the two subshapes and is a rectangle.
Proof.
Suppose that a skew shape has an occurrence of in columns and rows . Then is not a rectangle, since it contains the two cells and but does not contain , which is not a cell of . Symmetrically, is not a rectangle either, and the righthand side of the equivalence in the statement of the lemma fails for and .
Conversely, suppose that contains a cell such that neither nor is a rectangle; see Figure 7. Fix row indices and so that is the bottommost cell of in column , and is the topmost cell of in column . Note that : if we had, e.g., , then would consist of a single row, and therefore it would be a rectangle. Fix column indices and so that and are the leftmost and rightmost cell of in row . Again, we see that .
Note that neither nor is a cell of : if, e.g., contained the cell , then would be a rectangle. On the other hand, since is a skew shape, it must contain both and . It follows that the columns and rows induce an occurrence of in . ∎
Let be a connected skew shape, and let a vertical line leading between two adjacent columns of divide it into two skew shapes and , where the leftmost column of is directly to the right of the rightmost column of , and the bottommost row of is at least as high as the bottommost cell of the rightmost column of . Then we say that is a vertical concatenation of and and we write . Similarly, we define horizontal concatenation and write .
We say that a skew shape is Ferrersdecomposable if it can be written as
with the following properties (see Figure 8):

All are Ferrers shapes, and all are Ferrers shapes.

All and are nonempty with the possible exception of .

For , if the dividing line between and is between columns and , then the topmost cell of in column is below the topmost cell of in column . Similarly, if the dividing line between and is between rows and , then the rightmost cell of in row is to the left of the rightmost cell of in row .

For , is to the left of the vertical line separating from , and is below the horizontal line separating from .
We call this representation of the Ferrers decomposition of .
Lemma 3.2.
A connected skew shape is free if and only if it is Ferrersdecomposable.
Proof.
In a Ferrersdecomposable skew shape , every cell is in either a Ferrers subshape or a Ferrers subshape and thus satisfies the condition of Lemma 3.1. Therefore is free.
To prove the converse, let be a connected free skew shape; see Figure 9. We proceed by induction over the number of cells of , noting that the statement is trivial if has a single cell. Let be the topmost cell in the first column of . If is the topmost row of , then is a NW Ferrers shape, and it admits the trivial Ferrers decomposition .
Suppose that is not the topmost row of . Since is connected, it contains a cell in row , and let be the leftmost such cell. We define to be the subshape of formed by the cells in the first columns. Clearly, is a nonempty NW Ferrers shape. Since is connected, the cell is in . Applying Lemma 3.1 to this cell, we conclude that is a rectangle – note that cannot be a rectangle, since contains the cells and but not .
Let be the bottomright cell of the rectangle . If is the rightmost column of , then is a SE Ferrers shape , and we have a Ferrers decomposition . Suppose that is not the rightmost column, and let be the bottommost cell of in column . Note that , otherwise would not be a rectangle. We then let be the subshape of formed by the cells in columns and rows . is then a SE Ferrers shape, and can be written as , where is a connected free skew shape.
By induction, admits a Ferrers decomposition . We claim that is a Ferrersdecomposition of . Clearly, it satisfies conditions (a), (b) and (c) in the definition of Ferrersdecomposition. To verify condition (d), we only need to argue that the dividing line between and is to the right of . To see this, we note that by Lemma 3.1, is a rectangle, and in particular, the topmost cells in the columns intersecting this rectangle are all in the same row. Thus, by condition (c) of Ferrersdecomposition, this rectangle cannot be separated by a vertical dividing line, and in particular the line separating from is to the right of . ∎
3.2 Rubey’s bijection
Rubey [15] has proved several general bijective results about fillings of moon polyominoes. Here we formulate a part of his results which will be useful in our argument.
Given a finite sequence and a permutation of length , we let be the sequence . In addition, given a moon polyomino of width and a permutation of length , we denote by the shape created by permuting the columns of according to . A maximal rectangle in a moon polyomino is a rectangle which is not a proper subset of any other rectangle contained in . Note that a moon polyomino contains a maximal rectangle of width if and only if it has a row with exactly cells, and this maximal rectangle is then unique. The analogous property holds for columns as well. In fact, if and are moon polyominoes such that is obtained from by permuting its columns, then there is a sizepreserving correspondence between the maximal rectangles of and the maximal rectangles of ; see Rubey [15].
Definition 3.3.
Let be a moon polyomino of height and width , let and be sequences of integers of lengths and respectively, and let be a mapping which assigns to every maximal rectangle in a nonnegative integer . Then is the set of all integer fillings of such that

the sum of the entries in the th row of is equal to ,

the sum of the entries in the th column of is equal to , and

for every maximal rectangle of , the length of the longest chain in the filling of is equal to .
The following theorem is a special case of a result of Rubey [15, Theorem 5.3].
Theorem 3.4 (Rubey [15]).
Let be a moon polyomino of height and width , and let be a permutation of its columns such that is again a moon polyomino. Let be a function assigning nonnegative integers to maximal rectangles of , let and
be nonnegative integer vectors of length
and , respectively. Then the two sets and have the same size, and there is an explicit bijection between them.3.3 A bijection on Ferrers fillings
Using Rubey’s bijection as our main tool, we will now state and prove a technical result on fillings of Ferrers shapes satisfying certain restrictions.
Let be a NW Ferrers shape, let and be integers. Suppose that the leftmost columns of all have the same length, and the topmost rows of have the same length. We will call the leftmost columns of special columns and the topmost rows special rows of . For any , let denote the rectangle formed by the leftmost special columns of (i.e., columns ), and let be the rectangle formed by the rightmost special columns of (i.e., columns ). Similarly, for , is the rectangle formed by the topmost special rows of , and the rectangle formed by the bottommost special rows of . Additionally, for , let and denote the th special column from the left and the th special column from the right, respectively (i.e., and ), and for , let and be the th special row from the top and the th special row from the bottom, respectively (i.e., and , where is the height of ).
Lemma 3.5.
With , and as above, there is a bijection on the set of fillings of , such that for any filling of and its image , the following holds.

The longest SEchain in has the same length as the longest NEchain in .

For every , the longest SEchain of contained in has the same length as the longest NEchain of contained in .

For every , the longest SEchain of contained in has the same length as the longest NEchain of contained in .

The entries of in column have the same sum as the entries of in column , and the entries of in row have the same sum as the entries of in row .

For any nonspecial row or column, the sum of its entries in is the same as the sum of its entries in .
Proof.
Let be a filling of . Refer to Figure 10 for the illustration of the steps of the construction. As the first step, we add new cells to , filled with zeros, as follows: for every , we add new cells to the bottom of column , and for every , we add new cells to the right of row . We refer to these newly added cells as dummy cells. We let denote the Ferrers shape obtained from by the addition of the dummy cells, and the corresponding filling.
The dummy cells will remain filled with zeros throughout all the steps of the construction and will not contribute to any chain; their purpose is to ensure that for every and , the shape has a maximal rectangle containing the cells of and no other cell of , and similarly for .
As the next step, we reverse the order of the columns of and , while keeping the order of the rows preserved. Thus, we replace the filling of with its ‘mirror image’ of the shape , changing all SEchains into NEchains.
Let be the permutation of the columns of which leaves the special columns of fixed, and places the nonspecial columns to the right of the special ones, in the same order as in . This permutation of columns yields a moon polyomino that only differs from by having the order of the special columns reversed. We call this polyomino . We now apply to the bijection from Theorem 3.4, with as above, and let be the resulting filling of .
Next, we permute the rows of , by reversing the order of the special rows. Let be the resulting moon polyomino. The corresponding filling is obtained by again invoking Theorem 3.4, or more precisely its symmetric version for row permutations.
Finally, we remove from and the dummy cells, to obtain a filling of the original Ferrers shape . We define by . It follows from Theorem 3.4 that has all the properties stated in the lemma. ∎
3.4 Proof of Theorem 2.1
We now have all the ingredients to prove the main result of this section.
Proof of Theorem 2.1.
Clearly, it suffices to prove the theorem for connected skew shapes, since for a disconnected skew shape, we can treat each component separately. Let be a free connected skew shape. Fix its Ferrersdecomposition
For two consecutive Ferrers shapes and from the decomposition, say that a row is special if contains a cell from both and in this row. In particular, and have the same number of special rows. Clearly, all the special rows have the same length in as well as in . Similarly, for consecutive Ferrers shapes and , define a special column to be any column containing cells from both these shapes.
Fix now a filling of . We transform by applying the bijection separately to the restriction of in every Ferrers shape of the decomposition. More precisely, in we apply directly, while in we first rotate the filling by degrees, apply , and then rotate back. Let be the resulting filling of , obtained by combining the transformed fillings of the individual Ferrers shapes in the decomposition.
We claim that has all the properties stated in Theorem 2.1. Suppose first that contains for some , and let us show that contains . If the occurrence of in is confined to a single shape or , then contains in the same shape by the properties of .
Suppose now that has an occurrence of not confined to a single Ferrers shape of the decomposition. The occurrence must then be confined to the union of two consecutive shapes of the decomposition, since a pair of shapes further apart does not share any common row or column. Suppose, without loss of generality, that has an occurrence of inside , with 1cells from the occurrence being in and 1cells from the occurrence being in . Moreover, let be the number of special rows in (and therefore also in ). There are then some values with , such that contains a SEchain of length in its topmost special rows while has a SEchain of length in its bottommost special rows (which will become topmost once is rotated before the application of ).
Consequently, in , the shape has a NEchain of length in its bottommost special rows, and has a NEchain of length in its topmost special rows. The two chains together form a NEchain of size , i.e., an occurrence of in . The same reasoning shows that an occurrence of in implies an occurrence of in .
Let us now consider the row sums of and . Let be the set of row indices of the special rows shared by and . The sets are pairwise disjoint. If a row does not belong to any , then it intersects only one shape from the decomposition and is not special. Such a row then has the same sum in as in , by the properties of . On the other hand, for the rows in a set , the properties of ensure that the order of their row sums is reversed by the mapping , that is, the sum of the topmost row of in is the same as the sum of the bottommost row of in , and so on. Thus, the row sums of are obtained from the row sums of by a permutation that reverses the order of each and leaves the remaining elements fixed. An analogous property holds for column sums. ∎
4 Binary fillings avoiding a chain of length 2
In this section, we prove Theorem 2.5.
Recall the skew shape and its filling from Figure 1.
Given a skew shape with a total of cells, we assign labels to every cell starting from the lower left corner, iterating over rows from the bottom to the top of and labeling cells in a row from left to right, as indicated in Figure 11.
We associate with every cell a threepart piecewise linear curve consisting of the ray going from the topright corner of to the left, the right border of and the ray going from the bottomright corner of to the right. The curve of the cell of a skew shape clearly divides it into two parts, with cells below and above the curve, as illustrated in Figure 11.
Let be a filling. Let be a skew shape with cells numbered as above, and let be one of its cells. We say that an occurrence of in a filling of is low, if the topright cell of the occurrence is one of the cells . Intuitively, this means that the entire occurrence is below the line . If an occurrence is not low, then it is high. Note that an occurrence may be high even when all its 1cells are below .
For a skew shape with cells numbered as above, let be the set of all the binary fillings of that contain neither an high occurrence of nor an low occurrence of or of . See Figure 12 for an example.
Note that is precisely the set of all avoiding binary fillings of , while is the set of avoiding binary fillings of . Thus, Theorem 2.5 will follow directly from the next lemma.
Lemma 4.1.
Let be a skew shape with cells and let . Then there is a bijection between and .
Proof.
First, consider the case when the cells and are not in the same row, i.e. is the first cell of a new row and is the last cell of the previous row. In this case we will show that the sets and are identical.
To see this, note that since the three relevant patterns , and all have two cells in the topmost row, they can never have an occurrence whose topright cell is mapped to . Thus, in a filling of , an occurrence of any of these patterns is low if and only if it is low, and hence .
Suppose now that and are adjacent cells in a single row. We will partition each of the sets and into two disjoint parts and construct bijections between corresponding parts. Let consists of all the cells in the same row as and strictly to the left of it, let consist of all the cells in the column of and strictly below it, and let be the rectangle consisting of all cells that are in a column intersecting and a row intersecting ; see Figure 13. Note that , , and form a rectangle which is maximal among all the rectangles contained in with as topright corner. Therefore, in any occurrence of with the upper 1cell in , the lower 1cell is inside the rectangle .
We divide into two disjoint sets as follows:

contains the fillings in which there is either no 1 in or no 1 inside ,

contains the fillings with a 1 in and at least one 1 inside .
We divide into two disjoint sets as follows:

contains the fillings in which there is either no 1 inside or no 1 inside ,

contains the fillings in which both and are nonempty.
First of all we show that the sets and contain the same fillings. Choose a filling . Clearly, since has no high occurrence of , it has no high occurrence either. The only possibility for to have an low occurrence of or is that is actually the topright corner of the occurrence. The definition of guarantees that there is no such occurrence of . If contained an occurrence of with topright corner in , then it would contain an high occurrence of , contradicting . This also shows that belongs to .
Conversely, for , we see that has no low occurrence of or , since it has no such low occurrence. It also has no high occurrence of , because the topright cell of such an occurrence would have to coincide with , which is excluded by the definition of . Hence is in . And since has no low occurrence of , we may conclude that is in . This shows that .
For a filling we perform the following transformation into a filling :

Label all nonempty columns of from left to right as .

If is nonempty, then is empty and we move the 1 from the cell to the cell of above the column and finish.

If is empty, then replace the filling of by the filling of and for replace the filling of by the filling of . Replace the filling of by all zeros.

If there is a 1 in the cell of above the column , finish if or move the 1 from to the cell of above the column if .

Finally if there is no 1 in the cell of above the column , move the 1 from to this cell.
Observe that since has no occurrence of inside , all its 1cells in must be weakly to the left of , and for , all the 1cells in must be weakly above those of . Observe also that for any 1cell of inside , has a 1cell in in row weakly to the right of , and also has a 1cell in in column weakly above . Conversely, for any 1cell of in , has a 1cell in the same column and weakly below it, and for any 1cell of in , has a 1cell in the same row and weakly to the left of it.
Recall that is the rectangle . Note that in any cell not belonging to , the two fillings and coincide.
We now verify that the filling belongs to . If has an high copy of , then at least one 1cell of this copy must be in , because in the rest of , no 1cells were modified. The other 1cell must be strictly above or strictly to the right of , otherwise the occurrence would be low.
However, since for every 1cell of in , there is a 1cell of in the same column of and also a 1cell of in the same row of , it follows that to any high occurrence of in , we may find such an high occurrence in as well, contradicting .
If has an low occurrence of , we consider the following cases.

The occurrence of has a 1cell in and another 1cell in : this cannot occur, since in , the rightmost 1cell in is in column or , and there is no 1cell in to the left of this column.

The occurrence of has both 1cells in . However, the subfilling of induced by the nonempty columns is the same in as in , so contains an occurrence of in as well, which is impossible.

The occurrence of has one 1cell in and another outside . Since the transform preserves the set of nonempty rows and columns in , there must be an low occurrence of in , a contradiction.

The occurrence of uses the 1cell and a 1cell in . This is impossible, since is a 1cell in only if is empty.

The occurrence of has no 1cell in . Then it is also an low occurrence of in , which is impossible.
The above cases cover all possibilities, and show that avoids an low .
Suppose that has an low occurrence of . Let , and be the three 1cells of this occurrence, ordered left to right; see Figure 14. We note that in order to get an occurrence of , the column containing must not intersect the row containing inside . Since all the three 1cells are below the line , this implies that is strictly to the left of the leftmost column of . Moreover, at least one of and must be inside , otherwise the three 1cells would form a forbidden copy of in .
We claim that must be strictly below the bottom row of . If not, then would be strictly to the right of , since the column of does not intersect the row of inside . This implies that is in , and more precisely must be in , otherwise the copy of would not be low. But then has a 1cell in the same row as and weakly to the left, and , and form a forbidden occurrence of in . This shows that is indeed strictly below .
Define to be the largest rectangle inside whose bottomleft corner is . Note that has no 1cell in , since such a 1cell would form a forbidden occurrence of with . Note also that the column containing intersects the row containing inside , otherwise , and would not form a copy of .
The filling contains a 1cell in the same column as and weakly below it, and it contains a 1cell in the same row as and weakly to the left of it. Since neither nor are in , they form an occurrence of with , and this occurrence is low. In fact, this occurrence must be low as well, otherwise we would have and , in which case and would form a forbidden high occurrence of in . We conclude that has a forbidden occurrence of . This contradiction completes the proof that has no low occurrence of , and therefore that in in .
It follows directly from the definition of that for any , the filling has a 1cell both in and in , and hence .
Next, we describe a mapping that transforms a filling into a filling .

If there is exactly one 1cell in and the column of below this cell is nonempty, move this entry to and finish.

Otherwise there are either at least two 1cells in or the column below the single 1cell is empty. In both cases we can choose the rightmost empty column of such that there is a 1 in above it. We call this column . Label all nonempty columns of from left to right as .

For each copy the filling of to , copy the filling of to and replace the filling of by all zeros.

If finish. If and there is a 1 in above , this is the rightmost 1 in , move it to and finish.

Finally if the entry 1 in above is the rightmost 1cell in , move this 1 to .
We may again observe easily that in , either or is empty, and that and have the same nonempty rows and columns in . Moreover, for any 1cell of in , the preimage contains a 1cell in in the same column and weakly above , as well as a 1cell in in the same row and weakly to the right of .
We claim that the filling belongs to . Assume that has an high copy of . Its topright corner is not because or is empty in . Thus, exactly one of its 1cells is in , but since has the same nonempty rows and columns in as , this means there is an high copy of in , a contradiction.
Suppose contains an low copy if , and let and be its two 1cells, with left and below . If is in a column or row that intersects , we easily find a forbidden copy of in , using the fact that nonempty rows and columns of are preserved by . The only nontrivial case is when is strictly to the left and below . Then contains a 1cell in the same column as and weakly above it, as well as a 1cell in the same row as and weakly to the right of it. If does not form a copy of with or , it implies that , and form a forbidden copy of in , a contradiction.
Finally, suppose contains an low copy of , with 1cells , and ordered left to right. If is not strictly to the left and below , we may use the same reasoning we used in the analysis of the transform to show that contains a forbidden copy of . If is strictly to the left and below , we may find in a 1cell in the same column and weakly above , as well as a 1cell in the same row and weakly to the right of , and see that , and form a forbidden copy of .
It follows that is in , and since either or is empty in as we observed above, it follows that is in .
Overall, we have shown that maps
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