aa r X i v : . [ m a t h . C O ] D ec Matrix patterns with bounded saturationfunction
Benjamin Aram Berendsohn ∗ January 1, 2021
A 0-1 matrix M contains a 0-1 matrix pattern P if we can obtain P from M by deleting rows and/or columns and turning arbitrary 1-entries into 0s.The saturation function sat( P, n ) for a 0-1 matrix pattern P indicates theminimum number of 1s in a n × n P , butchanging any 0-entry into a 1-entry creates an occurrence of P . Fulek andKeszegh recently showed that the saturation function is either bounded orin Θ( n ). Building on their results, we find a large class of patterns withbounded saturation function, including both infinitely many permutationmatrices and infinitely many non-permutation matrices.
1. Introduction
In this paper, all matrices are 0-1 matrices. For a cleaner presentation, we write matriceswith dots ( • ) instead of 1s and spaces instead of 0s, for example: (cid:16) (cid:17) = (cid:16) • •• (cid:17) In line with this notation, we call a row or column empty if it only contains 0s. Further-more, we refer to changing an entry from 0 to 1 as adding a 1-entry, and to the reverseas removing a 1-entry.We index matrices as follows: The entry ( i, j ) is in the i -th row (from top to bottom)and the j -th column (from left to right). For example, the above matrix has 1-entries(1 , ,
3) and (3 , pattern is a matrix that is not all-zero. A matrix M contains a pattern P if we canobtain P from M by deleting rows and/or columns, and turning arbitrary 1-entries into0s. If M does not contain P , we say M avoids P . Matrix pattern avoidance can be seenas a generalization of two other areas in extremal combinatorics: Pattern avoidance in ∗ Institut f¨ur Informatik, Freie Universit¨at Berlin, [email protected] . Work supported by DFGgrant KO 6140/1-1. M and P are permutation matrices; and forbidden subgraphs in bipartite graphs,which corresponds to avoiding a pattern P and all other patterns obtained from P bypermutation of rows and/or columns. A classical question in extremal graph theory is to determine the minimum number ofedges in a n -vertex graph avoiding a fixed pattern graph H . The corresponding problemin forbidden submatrix theory is determining the maximum weight (number of 1s) ofa m × n matrix avoiding the pattern P , denoted by ex( P, m, n ). We call ex(
P, n ) =ex(
P, n, n ) the extremal function of the pattern P . The study of the extremal func-tion originates in its applications to (computational) geometry [Mit87, F¨ur90, BG91].A systematic study initiated by F¨uredi and Hajnal [FH92] has produced numerous re-sults [Kla00, Kla01, MT04, Tar05, Kes09, Ful09, Gen09, Pet11b, Pet11a] and furtherapplications in the analysis of algorithms have been discovered [Pet10, CGK + P, n ) is at least linear and at most quadratic. Largeclasses of patterns with linear and quasi-linear extremal functions have been identified[Kes09, Pet11b]. On the other hand, there are patterns with nearly quadratic extremalfunctions [ARSz99].A natural counterpart to the extremal problem is the saturation problem . A matrixis saturated for a pattern P if it avoids P and is maximal in this respect, i.e., turningany 0-entry of M into a 1 creates an occurrence of P . Clearly, ex( P, m, n ) can also bedefined as the maximum weight of a m × n matrix that is saturated for P . The functionsat( P, m, n ) indicates the minimum weight of a m × n matrix that is saturating for P .We focus on square matrices and the saturation function sat( P, n ) = sat(
P, n, n ).The saturation problem for matrix patterns was first considered by Brualdi and Cao[BC20] as a counterpart of saturation problems in graph theory. Fulek and Keszegh[FK20] started a systematic study. They proved that, perhaps surprisingly, every pattern P satisfies sat( P, n ) ∈ O (1) or sat( P, n ) ∈ Θ( n ). This is in stark contrast to the extremalproblem. Further, they present large classes of patterns with linear saturation functions,and a single non-trivial pattern with bounded saturation function.Most interesting for our purposes is a class of patterns we call once-separable : Apattern is once-separable if it has the form (cid:18) A B (cid:19) or (cid:18) AB (cid:19) for two patterns A and B , where denotes an all-0 matrix of arbitrary dimensions. Theorem 1.1 ([FK20, Theorem 1.7]) . If P is once-separable, then sat( P, n ) ∈ Θ( n ) . In this paper, for the sake of simplicity, we only consider patterns with no empty rowsor columns. However, we note that the saturation function, unlike the extremal function,may change considerably by the addition of an empty row or column. In particular, Fulekand Keszegh proved that if the first or last row or column of a pattern P is empty, thensat( P, n ) ∈ Θ( n ). For this, we interpret the M and P as adjacency matrices of bipartite graphs. = (cid:18) •• • •• (cid:19) , Q = (cid:18) •• • •• (cid:19) , Q = (cid:18) •• •• • (cid:19) , Q = (cid:18) •• •• • (cid:19) Figure 1: 5 × P ′ can be obtained from P by rotation, inversion , or reflection , thensat( P, n ) = sat( P ′ , n ). Permutation matrix patterns.
In this paper, we give special attention to permutationmatrix patterns. A permutation matrix is a square matrix where every row and everycolumn contains exactly one 1-entry. Theorem 1.1 already covers the once-separablepermutation matrices.We call the 1-entries in the first or last row or column the outer Q and Q , where Q = (cid:18) •• •• (cid:19) , Q = (cid:18) •• •• (cid:19) . In particular, all 3 × Q is the only4 × × Q has bounded saturationfunction, and ask whether the same is true for Q .Call a permutation matrix Q -like ( Q -like ) if the outer 1-entries form Q (respec-tively, Q ). We prove that all Q -like permutation matrices have bounded saturationfunction. Theorem 1.2.
Let P be a Q -like k × k permutation matrix. Then sat( P, n ) ∈ O (1) . This covers the pattern Q (thus answering the question of Fulek and Keszegh) andthe patterns Q , Q , and Q in Figure 1. For permutation matrices of size at most 6,we obtain a full characterization of the saturation functions with the following theorem. Theorem 1.3.
Let P be a not-once-separable k × k permutation matrix with k ≤ .Then sat( P, n ) ∈ O (1) . Swapping the role of rows and columns. Reversing all rows or all columns. ther patterns. We call a pattern non-trivial if it has two rows that only have a 1 inthe leftmost (respectively rightmost) position, and two columns which only have a 1 inthe topmost (respectively bottommost) position. Otherwise, we call the pattern trivial .Fulek and Keszegh show that each trivial pattern has linear saturation function [FK20,Theorem 1.11]. Note that every permutation matrix is non-trivial. (cid:18) •• •• • (cid:19) , (cid:18) •• •• • (cid:19) . Figure 2: A non-trivial pattern (left), and a trivial pattern (right).Our techniques easily generalize to a more general class of non-trivial patterns (in fact,we only prove them in the general form). We restrict ourselves to the patterns withoutempty rows or columns where the first and last row and column each contain only asingle 1-entry. Since the case of once-separable patterns is already solved, this againleaves us with patterns where the outer 1-entries form either Q or Q (up to reflection).We extend our previous definitions as follows: An arbitrary pattern is called Q -like ( Q -like ) if it has no empty rows and columns, and exactly four outer entries that forman occurrence of Q (respectively, Q ). We prove a generalization of Theorem 1.2. Theorem 1.4.
Let P be a non-trivial Q -like k × k pattern. Then sat( P, n ) ∈ O (1) . We prove Theorem 1.4 (which implies Theorem 1.2) and Theorem 1.3 in Section 3.All our results are based on the construction of a witness , a concept introduced by Fulekand Keszegh. In Section 2, we formalize and develop this notion, based on the proof byFulek and Keszegh that Q has bounded saturation function.
2. Witnesses
Let P be a matrix pattern without empty rows or columns. An explicit witness (calledsimply witness by Fulek and Keszegh [FK20]) for P is a matrix M that is saturatedfor P and contains at least one empty row and at least one empty column. Clearly, ifsat( P, n ) ∈ O (1), then P has an explicit witness. Fulek and Keszegh note that the reverseis also true: We can replace an empty row (column) by an arbitrary number of emptyrows (columns), and the resulting arbitrarily large matrix will still be saturating for P . As such, an m × n explicit witness for P of weight w implies that sat( P, m, n ) ≤ w foreach m ≥ m and n ≥ n .We call a row (column) of a matrix M expandable w.r.t. P if the row (column) is emptyand adding a single 1-entry anywhere in that row (column) creates a new occurrence of P in M . A explicit witness for P is thus a saturated matrix with at least one expandablerow and an expandable column w.r.t. P . We define a witness for P (used implicitly by Note that it is critical here that P has no empty rows or columns. Otherwise, increasing the numberof empty rows or columns in M might create an occurrence of P . P and has at least one expandable rowand at least one expandable column w.r.t. P . Clearly, an explicit witness is a witness.The following lemma shows that finding a (general) witness is sufficient to show thatsat( P, n ) ∈ O (1). Lemma 2.1.
If a pattern P without empty rows or columns has a m × n witness, then P has a m × n explicit witness.Proof. Let M be a m × n witness for P . If M is saturating for P , then we are done.Otherwise, there must be a 0-entry ( i, j ) in M that can be changed to 1 without creatingan occurrence P . Note that ( i, j ) cannot be contained in an expandable row or columnof M , so the resulting matrix is still a witness. Thus, we obtain an explicit witness afterrepeating this step at most m · n times. Fulek and Keszegh also considered the asymptotic behavior of the functions sat(
P, m , n )and sat( P, m, n ), where m and n are fixed. The dichotomy of sat( P, n ) also holds inthis setting:
Theorem 2.2 ([FK20, Parts of Theorem 1.3]) . For every pattern P , and constants m , n ,(i) either sat( P, m , n ) ∈ O (1) or sat( P, m , n ) ∈ Θ( n ) ;(ii) either sat( P, m, n ) ∈ O (1) or sat( P, m, n ) ∈ Θ( m ) . We can adapt the notion of witnesses in order to classify sat(
P, m , n ) and sat( P, m, n ).Let P be a matrix pattern without empty rows or columns. A horizontal (vertical) wit-ness for P is a matrix M that avoids P and contains an expandable column (row). Clearly, P has a horizontal witness with m rows if and only if sat( P, m , n ) is bounded;and P has a vertical witness with n columns if and only if sat( P, m, n ) is bounded.Further note that M is a witness for P if and only if M is horizontal witness and avertical witness.Observe that rotation and inversion of P may affect the functions sat( P, m, n ) orsat( P, m , n ), but reflection does not. Lemma 2.3.
Let P be a matrix pattern without empty rows or columns, and only oneentry in the last row (column). Let W be a horizontal (vertical) witness for P . Then,appending an empty row (column) to W again yields a horizontal (vertical) witness.Proof. We prove the lemma for horizontal witnesses, and appending a row. The othercase follows by symmetry. Let W be a m × n horizontal witness for P , where the j -thcolumn of W is expandable. Let W ′ be a matrix obtained by appending a row. Clearly, W ′ still does not contain P . Moreover, adding an entry in W ′ at ( i, j ) for any i = n + 1 A horizontal witness can be expanded horizontally, a vertical witness can be expanded vertically. P . It remains to show that adding an entry at ( n + 1 , j )creates an occurrence of P .We know that adding an entry at ( n , j ) in W ′ creates an occurrence of P . Let I theset of positions of 1-entries in W ( P ) that form the occurrence of P . Since P has onlyone entry in the last row, all positions ( i ′ , j ′ ) ∈ I \ { ( n , j ) } satisfy i ′ < n + 1. Thus,adding a 1-entry at ( n + 1 , j ) instead of ( n , j ) creates an ocurrence of P at positions I \ { ( n , j ) } ∪ { ( n + 1 , j ) } , which implies that W ′ is a horizontal witness.We now prove the following handy lemma, that allows us restrict our attention to theclassification of sat( P, m , n ) and sat( P, m, n ). It essentially is a generalization of thetechnique used by Fulek and Keszegh to prove that sat( Q , n ) ∈ O (1). Lemma 2.4.
Let P be a not-once-separable pattern without empty rows or columns,and with only one 1-entry in the last row and one 1-entry in the last column. Then sat( P, n ) ∈ O (1) if and only if there exist constants m , n such that sat( P, m , n ) ∈ O (1) and sat( P, m, n ) ∈ O (1) .Proof. Suppose that sat(
P, n ) ∈ O (1). Then P has a m × n witness M , and thussat( P, m , n ) is at most the weight of M , for every n ≥ n . Similarly, sat( P, m, n ) ∈O (1).Now suppose that sat( P, m , n ) ∈ O (1) and sat( P, m, n ) ∈ O (1). Then, for some m , n , there exists a m × n horizontal witness W H and a m × n vertical witness W V . Consider the following ( m + m ) × ( n + n ) matrix, where m × n denotes the all-0 m × n matrix: W = (cid:18) m × n W H W V m × n (cid:19) We first show that W does not contain P . Suppose it does. Since P is containedneither in W H nor in W V , an occurrence of P in W must contain 1-entries in both thebottom left and top right quadrant. But then P must be once-separable, a contradiction.By Lemma 2.3, W ′ V = ( W V , m × n ) is a vertical witness, and W ′ H = (cid:0) W H m × n (cid:1) is ahorizontal witness. The expandable row in W ′ V and the expandable column in W ′ H areboth also present in W . This implies that W is a witness for P , so sat( P, n ) ∈ O (1).Figure 3 shows an example of a witness for Q , constructed with Lemma 2.4, us-ing vertical/horizontal witnesses presented later in Section 3, and an explicit witnessconstructed using Lemma 2.1.For certain classes of patterns closed under rotation or inversion, we can further restrictour attention to only vertical witnesses. Lemma 2.5.
Let P be a class of not-once-separable patterns without empty rows orcolumns, and with only one 1-entry in the last row and one 1-entry in the last column.If P is closed under rotation or inversion and each pattern in P has a vertical witness,then sat( P, n ) ∈ O (1) for each P ∈ P . · •• ·· •• ·· •• ·• ·• • ·· · · · · · · · · · ·• • ·• · • • • • • • • · • •• · • •• • • · • •• • • · •• • · •• • • • • · • •• • • • ·• • • • • ·· · · · · · · · · · ·• • • • • ·• • • • · • • Figure 3: A witness (left) and an explicit witness (right) for the pattern Q . The smalldots indicate the expandable row/column. (cid:18) •• •• (cid:19) → (cid:18) •• ••• •• (cid:19) → (cid:18) •• •· · · · · ·• •• (cid:19) Figure 4: Construction of W ( Q ) from Q . The small dots indicate the expandable row. Proof.
By Lemma 2.4, it suffices to show that each pattern in P has a horizontal witness.Let P ∈ P and let P ′ ∈ P be obtained by rotating P by 90 degrees clockwise (respec-tively, by inverting P ). Let W ′ be a vertical witness for P ′ , and let W be obtained byrotating W ′ by 90 degrees counterclockwise (respectively, by inverting W ′ ). Clearly, W is a horizontal witness for P . Lemma 2.4 concludes the proof.
3. A simple witness construction
We present a construction that yields vertical witnesses for certain non-trivial matriceswith only one 1-entry in the first and last column. Figure 4 shows an example of theconstruction. The idea is simple: Make two copies P and P of P , and arrange them ina way that the rightmost 1-entry of P coincides with the leftmost 1-entry of P (increasethe matrix size as necessary, without creating empty rows or columns). Then, deletethe column where P and P intersect. Note that this creates an empty row, whichformerly contained the intersection of P and P . Adding a 1-entry in that row createsan occurrence of P by either completing P or P , so that row is expandable. If now theconstructed matrix also avoids P (which is not necessarily the case), then it is a verticalwitness for P . We now proceed with the formal definition and proof.Let P = ( p i,j ) i,j be a k × k pattern with exactly one entry in the first column andexactly one entry in the last column. Let s and t be the rows of the leftmost andrightmost 1-entry in P , i.e., p s, = 1 and p t,k = 1. Without loss of generality, assumethat s < t . We define the ( k + t − s ) × (2 k −
2) matrix W ( P ) = ( w i,j ) i,j as follows: w i,j = p i,j , if j < k, i ≤ kp i − ( t − s ) ,j − ( k − , if j ≥ k, i ≥ ( t − s ) + 10 , otherwise.7 emma 3.1. Let P be a non-trivial pattern without empty rows and columns and withexactly one entry in the first and last column. If W ( P ) avoids P , then W ( P ) is a verticalwitness for P .Proof. Since P is non-trivial, the s -th and t -th rows of P each only contain one entry,so the t -th row of W ( P ) is empty. It remains to show that adding a 1-entry in the t -throw of W ( P ) creates a new occurrence of P .Let M be the matrix obtained by adding a 1-entry ( t, u ) in W ( P ). If u ≤ k −
1, weremove the first t − s rows and all columns other than the u -th and the last k −
1. Theresult is P with an additional 1-entry in the first column (which was the u -th column in M ). If u > k −
1, we remove the last t − s rows and all columns except the u -th and thefirst k −
1. The result is P with an additional 1-entry in the last column. Q -like patterns Lemma 3.2.
Let P be a non-trivial Q -like pattern. Then W ( P ) avoids P .Proof. Suppose W ( P ) contains an occurrence of P , say at positions I . Consider thebottommost and topmost positions ( i B , j B ) , ( i T , j T ) ∈ I . Since P is Q -like, we have j B < j T . Moreover, i T − i B ≥ k − P has k − j B ≤ k −
1. Then, by construction, i B ≤ k , which impliesthat I is completely contained in the first k rows, including the empty t -th row. However,an occurrence of P must have 1-entries in k distinct rows, a contradiction.Second, consider the case that j B > k −
1. Then j T > k . By construction, this impliesthat i T ≥ t − s + 1. Since W ( P ) has t − s + k rows in total, I is contained in the last k rows, including the empty t -th row. This is again a contradiction.The class of non-trivial Q -like patterns is closed under rotation, so Lemma 2.5 andLemma 3.1 imply Theorem 1.4. Theorem 1.4.
Let P be a non-trivial Q -like k × k pattern. Then sat( P, n ) ∈ O (1) . Q -like permutation matrix patterns One can manually check that W ( P ) avoids P even for many Q -like patterns, such as Q . We refine Lemma 3.2 to cover more patterns, including Q and all but four ofthe not-once-separable 6 × W ( P ) yields a witness. For the lastpattern, we construct a witness by modifying W ( P ) slightly. This shows that every Q -like permutation matrix of size at most 6 has a vertical witness. Since these patternsare closed under inversion, they all have bounded saturation function. Together withTheorem 1.4, we obtain: Theorem 1.3.
Let P be a not-once-separable k × k permutation matrix with k ≤ .Then sat( P, n ) ∈ O (1) . I = { ( i B , j B ) , ( i T , j T ) } be an occurrence of ( •• ) in some matrix, i.e., two 1-entrieswith i B > i T and j B < j T . We define the height if I as i T − i B + 1, the number of rowscontaining an entry in I or between the two entries of I .We first consider Q -like patterns that contain an occurrence of ( •• ) with height k −
1, which, among others, covers all but four permutation matrices of size at most6. Observe that permutation matrices of this type can be though of as almost Q -like:Removing the top (or bottom) row and then the new empty column creates a Q -likepermutation matrix. We first prove some facts about occurrences of ( •• ) in W ( P ). Lemma 3.3.
Let P be a non-trivial Q -like k × k pattern. Then each occurrence I of ( •• ) in W ( P ) has height at most k − .Proof. Suppose there is an occurrence I = { ( i B , j B ) , ( i T , j T ) } of ( •• ) in W ( P ) of heightat least k , i.e., j B < j T and i B − i T ≥ k −
1. We claim that I is completely contained inone of the two partial copies of P in W ( P ), i.e., either j B < j T ≤ k − k − < j B < j T .This implies that there is also a height- k occurrence of ( •• ) in P , which contradicts theassumption that P is Q -like. It remains to show our claim.Let s and t be the rows of the leftmost and rightmost 1-entry in P . Towards our claim,suppose on the contrary that j B ≤ k − < j T . Then i T ≥ ( t − s ) + 1 by construction,and thus i B ≥ k + ( t − s ) > k . But then ( i B , j B ) cannot be a 1-entry, a contradiction. Lemma 3.4.
Let P be a non-trivial Q -like k × k pattern. Then each occurrence I of ( •• ) in W ( P ) with height k − has the empty row between its two entries.Proof. Let s and t be the rows of the leftmost and rightmost 1-entry in P , so W ( P ) isa ( k + t − s ) × (2 k −
2) matrix where the t -th row is empty. Since P is Q -like, we have s ≥ t ≤ k −
1. Consider an occurrence I = { ( i B , j B ) , ( i T , j T ) } of ( •• ) in W ( P )where i B − i T = k −
2. We have i T = i B − k + 2 ≤ k + t − s − k + 2 = t − s + 2 ≤ t , and i B = k − i T ≥ k − ≥ t. Since the t -th row is empty, we also have i T = t = i B , and thus i T < t < i B . Proposition 3.5.
Let P be a non-trivial Q -like k × k pattern that contains an occurrence ( •• ) of height k − . Then W ( P ) avoids P .Proof. Suppose that P is contained in W ( P ). Then W ( P ) must contain an occurrence I of ( •• ), such that there are k − I . Thismeans that either I has height at least k , or I has height k − Q -like permutation matrices of size atmost 6. Figure 5 shows them along with vertical witnesses. Proposition 3.6.
For i ∈ { , , , } , the matrix W i is a vertical witness for Q i . = •• •• • • ! W = •• •· · · · · · · · · ·• •• •• • • Q = •• •• •• ! W = •• •· · · · · · · · · ·• •• •• •• Q = •• •• •• ! W = •• •• •· · · · · · · · · ·• •• •• Q = • •• •• • ! W = • ••• •· · · · · · · · · ·• •• • • Figure 5: Remaining Q -like 6 × Proof.
For i ∈ { , , } , we simply have W i = W ( P i ). Thus, it suffices to show that W i avoids Q i . For i ∈ { , } , note that W i has height k + 1, and therefore only k non-emptyrows. As such, an occurrence of Q i in W i must map the topmost 1-entry of Q i to thetopmost 1-entry of W i (marked red in the figure). It is easy to see that then the 1-entryin the second column of Q i must be mapped to the second column of W i (marked blue).But then the 1-entries in the second and third column of Q cannot be correctly mappedto 1-entries in W i .Considering i = 8, observe that Q has two occurrences I , I of ( •• ) of height k − •• ) in W have height atmost 4, and out of the occurrences of height 4, only two, say I ′ and I ′ (marked red/blue),do not contain the empty row. Thus, an occurrence of Q in W must map I , I to I ′ , I ′ .However, I , I span overlapping rows, while I ′ , I ′ do not, a contradiction.Finally, consider i = 9. The matrix W is almost equal to W ( Q ); the only differenceis that the entry in the 6-th column is moved one row up. Note that this entry is thehighest 1-entry in the right partial copy of Q in W ( Q ). Since we only move the highestentry up, the right partial copy stays intact in some sense. In particular, adding a 1-entry in the left half of the t -th row will still complete an occurrence of Q . The same istrue for the right half of the t -th row, since the left partial copy is not changed. Thus,the t -th row of W is expandable.We still have to argue that W does not contain Q . Suppose otherwise. Observe that Q contains exactly one occurrence I of ( •• ) of height 4 (marked in red in the figure).All occurrences of ( •• ) in W of height at least 4 have the empty row in between theirtwo entries, so I must be mapped to the some occurrence I ′ of ( •• ) in W of heightlarger than 4. There are only two such occurrences (marked in red), both involving the10ntry in the sixth column of W . However, the top entry in I ′ is in the first row of W ,but the top entry of I is in the second row of Q , leaving no room for the top entry of Q . This means Q is not contained in W .Propositions 3.5 and 3.6 show that each not-once-separable Q -like permutation ma-trix of size at most 6 has a vertical witness. As discussed at the start of this section, thisimplies Theorem 1.3. For convenience, we list all not-once-separable Q -like permutationmatrices of size at most 6 in Appendix A.
4. Conclusion
Fulek and Keszegh [FK20] showed that the saturation function of once-separable patternsis linear. We extend their result by showing that all non-trivial Q -like patterns havebounded saturation function. In particular, this is another step towards the classificationof permutation matrices, leaving only the Q -like permutation matrices. We find manymore Q -like permutation matrices with bounded saturation function. This completesthe classification of permutation matrices of size at most 6, showing that a permutationmatrix of size at most 6 has linear saturation function if and only if it is once-separable.It seems possible that this is true for all permutation matrices. Open Question.
Is the saturation function bounded for each not-once-separable per-mutation matrix?
Our witness construction W ( P ) undoubtedly works for a larger class of matrices thanwe identified (cf. Proposition 3.6). However, we also provide an example of a not-once-separable Q -like permutation matrix ( Q ) for which our construction does not yielda vertical witness. It would be interesting to precisely identify the patterns where theconstruction works. Open Question.
Is there a simple characterization of patterns P where W ( P ) avoids P ? Our results also extend to certain non-permutation matrices, but we did not considermatrices with empty rows or columns or with more than one 1-entry in either of thefirst or last row or column. We note, however, that Lemma 2.5 still may be useful forpatterns that have only one 1-entry in the the last row and only one 1-entry in the lastcolumn, but multiple 1-entries in the first row and column.11 eferences [ARSz99] Noga Alon, Lajos R´onyai, and Tibor Szab´o. Norm-graphs: Variations and ap-plications.
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The following table lists all not-once-separable and Q -like permutation matrices of sizeat most 6 ×
6, up to reflection. For each matrix, we reference the proof that it hasbounded saturation function. Whenever Proposition 3.5 is used, the relevant occurrenceof ( •• ) is highlighted in red. (cid:18) •• •• • (cid:19) Proposition 3.5 ( Q ) •• • •• • ! Proposition 3.5 •• •• • • ! Proposition 3.6 ( Q ) •• •• •• ! Proposition 3.6 ( Q ) •• •• •• ! Proposition 3.6 ( Q ) •• •• • • ! Proposition 3.5 •• • •• • ! Proposition 3.5 •• ••• • ! Proposition 3.5 •• • •• • ! Proposition 3.5 •• ••• • ! Proposition 3.5 •• • •• • ! Proposition 3.5 •• ••• • ! Proposition 3.5 • •• •• • ! Proposition 3.6 ( Q ) • •• •• • ! Proposition 3.5 • •• ••• ! Proposition 3.5 • •• ••• ! Proposition 3.5 • •• •• • !!